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[ "Request for Quotation\n\nYou can get the price list and a A&C representative will contact you within one business day.(", null, "[email protected])\n\nmotor power calculation for conveyor xls Description\n\n•", null, "Conveyor Horsepower Calculator - Superior …\n\nConveyor Horsepower Calculator Sara Hoidahl 2017-08-28T21:17:11+00:00. Conveyor Length (center-to-center) Belt Width. Vertical Lift . Belt Capacity. Calculated Minimum HP. 0.0 HP. Minimum HP + 10%. 0.0 HP. Backstop. Not needed. This required horsepower calculator is provided for reference only. It provides a reasonable estimation of required horsepower given user requirements. Superior ...\n\n•", null, "Motor Torque Calculations - NEPSI - Northeast …\n\nMOTOR TORQUE. The following calculators compute the various torque aspects of motors. These equations are for estimation only, friction, windage, and other factors are not taken into consideration. Calculator-1. Known variables: Horse Power and Speed in RPM Torque is the action of a force producing or tending to produce rotation. Torque = force x distance Torque Input Horse Power, hp : …\n\n•", null, "Belt Conveyors for Bulk Materials Practical Calculations\n\nBELT CONVEYORS - BASIC CALCULATIONS: 1. Mass of the Load per Unit Length: Load per unit length. Given the production capacity Qt = tph, the weight of the load per unit length (kg/m) – (lbs per ft) is calculated by: Wm = 2000. Qt or Wm = 33.333.Qt = (lb/ft) 60 x v v Q = 0.278.Qt or Q = Qt = (Kg/m) v 3.600 x v 2. Belt Tensions: In order to find the maximum tension is necessary to calculate the ...\n\n•", null, "Conveyor Power and Torque Calculator - EICAC\n\nCONVEYOR POWER CALCULATOR. Use this calculator to calculate the force, torque and power required from a conveyor to move a load at an angle. If your conveyor is horizontal enter an angle of 0. Enter your values for the Mass, Diameter, Beltspeed, Friction and Angle; select your units as required. MASS TO MOVE (M) DIAMETER OF DRIVE DRUM (D): BELTSPEED (S): COEFFICIENT OF …\n\n•", null, "Calculations for Screw conveyors - Bechtel\n\nCalculations for screw conveyors Power in Kw (P) Q x L x K 3600 x 102 P = power in Kw Q = capacity in 1000 kg per hour L = conveyor screw length (m) K = friction coeffi cient P = v = speed in m per sec v = estring 395 T +49 (0)212 64 50 94-0 [email protected] Wuppertal F +49 (0)212 64 50 94-10 K 102 Calculations for screw conveyors Capacity in m2 per hour (Q) Q = 47 ...\n\n•", null, "(DOC) erhitungan Daya Motor Conveyor …\n\nerhitungan Daya Motor Conveyor (Calculation of Conveyor Power Equipment\n\n•", null, "Conveyors - Load & Power Consumption\n\nLevel Ground Conveyors. Horsepower required for conveyors transporting material on level ground: 1 hp (English horse power) = 745.7 W = 0.746 kW; 1 ft (foot) = 0.3048 m = 12 in = 0.3333 yd; Lifting Conveyors. With lifting conveyors - add lifting power from the chart below to the level ground power from the chart above.\n\n•", null, "How to Calculate 3 Phase Motor Power …\n\nCalculate three-phase motor power consumption by multiplying amps by volts by the square root of three (W = AV(sqrt 3). For example, if the motor is drawing 30 amps at 250 volts, you have 30 x 250 x sqrt 3 (about 1.73) = 12,975 watts). Convert watts to kilowatts by dividing the number of watts by 1,000. Thus, a three-phase electric motor drawing 12,975 watts is consuming 12.975 kilowatts. For ...\n\n•", null, "Conveyor Belt Calculations - Bright Hub Engineering\n\nEvery calculation should contain a contingency factor to allow for occasional temporary overloads. It easy enough, given the low cost of low and fractional horsepower drives, to simply overpower your system. But your electrical controls contain a thermal overload which will trip the motor in the event of a jam or stall. This device not only protects the motor, it also protects from harm your ...\n\n•", null, "Electric Motor Calculator - Engineering ToolBox\n\nRLA - \"Running Load Amps\" - current drawn during normal operation of electric motor. FLA - \"Full Load Amps\" - amount of current drawn when full-load torque and horsepower is reached for the motor.FLA is usually determined in laboratory tests.Note! - in the calculator above FLA is RLA + 25%. 1 hp = 0.745 kW; Related Mobile Apps from The Engineering ToolBox ...\n\n•", null, "Motor Sizing Calculations\n\nCalculation for the Effective Load Torque ( Trms ) for Servo Motors and BX Series Brushless Motors. When the required torque for the motor varies over time, determine if the motor can be used by calculating the effective load torque. The effective load torque becomes particularly important for operating patterns such as fast-cycle operations ...\n\n•", null, "Electric Motor Calculator - Engineering ToolBox\n\nRLA - \"Running Load Amps\" - current drawn during normal operation of electric motor. FLA - \"Full Load Amps\" - amount of current drawn when full-load torque and horsepower is reached for the motor.FLA is usually determined in laboratory tests.Note! - in the calculator above FLA is RLA + 25%. 1 hp = 0.745 kW; Related Mobile Apps from The Engineering ToolBox ...\n\n•", null, "Conveyor Power and Torque Calculator - EICAC\n\nCONVEYOR POWER CALCULATOR. Use this calculator to calculate the force, torque and power required from a conveyor to move a load at an angle. If your conveyor is horizontal enter an angle of 0. Enter your values for the Mass, Diameter, Beltspeed, Friction and Angle; select your units as required. MASS TO MOVE (M) DIAMETER OF DRIVE DRUM (D): BELTSPEED (S): COEFFICIENT OF …\n\n•", null, "Conveyors - Load & Power Consumption\n\nLevel Ground Conveyors. Horsepower required for conveyors transporting material on level ground: 1 hp (English horse power) = 745.7 W = 0.746 kW; 1 ft (foot) = 0.3048 m = 12 in = 0.3333 yd; Lifting Conveyors. With lifting conveyors - add lifting power from the chart below to the level ground power from the chart above.\n\n•", null, "Screw Conveyor Interactive Calculators | …\n\nEng. Guide Index Download Guide PDF HORIZONTAL SCREW CONVEYOR CAPACITY & SPEED CALCULATION: Visit the online engineering guide for assistance with using this calculator. Click on symbol for more information. DESIGN CONDITIONS 1. Flowrate(m): lb/hr 2. Density: 3. Loading (K): % SPCL. FLIGHT […]\n\n•", null, "Calculating Conveyor Power for Bulk Handling | …\n\nOriginal Power Calculation Program (free downloadable Excel program, CEMA 4 version) Online Application Data Sheet (linked to our engineers) Application Data Sheet (downloadable pdf file you can send to us) We use a modified version of the Conveyor Equipment Manufacturers Association guidelines. The primary equation for Effective Tension, Te, is as follows: Te = LKt (Kx + KyWb + …\n\n•", null, "Motor Sizing Calculations\n\nCalculation for the Effective Load Torque ( Trms ) for Servo Motors and BX Series Brushless Motors. When the required torque for the motor varies over time, determine if the motor can be used by calculating the effective load torque. The effective load torque becomes particularly important for operating patterns such as fast-cycle operations ...\n\n•", null, "Power calculation for belt conveyor | Tecnitude\n\nPower calculation. We provide this calculation form to assist you with assessing the required power for your belt conveyor, depending on the weight carried. You can also use our product configurator to view your tailored conveyor. Feel free to contact us for any of your projects. Tecnitude's team is at your disposal. Name . Company . Email . Telephone . Country +33 (0)3 89 60 34 40. Rent ...\n\n•", null, "roller conveyor calculations? - Commercial …\n\n21.02.2005· roller conveyor calculations? bmw318s70 (Electrical) (OP) 1 Feb 05 23:15. hi. I am trying to buil a spreadsheet for calculation the required HP of roller conveyors. The setup of the conveyor is te following: Motor connected to a gearbox. Output sprocket linked to a roller sprocket. All rollers linked toghether. This conveyor should move a pallet-load of X lbs, at Y feet/minutes. The various ...\n\n•", null, "Understanding Conveyor Belt Calculations | …\n\nUnderstanding a basic conveyor belt calculation will ensure your conveyor design is accurate and is not putting too many demands on your system. We use cookies to personalize content and analyze traffic. We also share information about your use of our site with our social media, advertising, and analytics partners who may combine it with other information that you've provided or that we have ...\n\n•", null, "torque - Sizing a motor for a conveyor - Electrical ...\n\nCompanies that manufacture conveyor systems probably have software tools that calculate the motor size for them. But for someone who doesn't have this software, how does one go about determining what size motor they need to drive a 6000lbs load over a conveyor of length 20 feet. Let's assume 1800RPM since that's what most conveyor motors at our ...\n\n•", null, "How to Calculate 3 Phase Motor Power …\n\nCalculate three-phase motor power consumption by multiplying amps by volts by the square root of three (W = AV (sqrt 3). For example, if the motor is drawing 30 amps at 250 volts, you have 30 x 250 x sqrt 3 (about 1.73) = 12,975 watts). Convert watts to kilowatts by dividing the number of watts by 1,000." ]

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[ "Search\n\n# Average Directional Index", null, "The ADX or Average Directional Index is a technical indicator used to measure the overall strength of a trend.\n\nDeveloped by J. Welles Wilder, the Average Directional Index (ADX) helps traders measure how strongly price is trending and whether its momentum is increasing or falling.\n\nIt’s important to emphasize that while ADX measures the strength of a trend, it does NOT identify the trend’s direction.\n\nIt can be used to find out whether the market is ranging or starting a new trend.\n\nThe oscillator ranges between 0 and 100 with high readings indicating a strong trend and low readings indicating a weak trend. How to Calculate ADX:\n\nThe ADX is derived from two directional indicators, known as DI+ and DI-:\n\n• The positive directional indicator (+DI)\n\n• The negative directional indicator (-DI)\n\nThese two indicators are derived from the Directional Movement Index (DMI).\n\nADX is calculated by finding the difference between DI+ and DI-, as well as the sum of DI+ and DI-.\n\nThe difference is divided by the sum, and the resulting number multiplied by 100.\n\nThe result is known as the Directional Index or DX.\n\nA moving average is then taken of DX, typically over a fourteen-day period (although any number of periods can be used.)\n\nThis final moving average is the ADX." ]

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[ "[email protected] (309) 691-8741\n249\n\n# Topics in College Algebra and Trig\n\nCourse ID\n249\nLevel\n12\nSemester\n1, 2\nCredit\n1.000\nMethod\nRegular\n\nThis course will take a more detailed examination into topics discussed in 243 Algebra II and then introduces topics essential to the understanding of trigonometry. In Algebra, topics include linear equations, quadratic equations, polynomial equations, logarithms, exponentials, rational equations and conics. Trigonometry topics will include the trigonometric ratios and functions, right triangle trigonometry, use of the unit circle, graphing trigonometric functions, and solving trigonometric equations. A calculator is required. The math faculty recommends the TI 34 series calculator.\n\n#### Prerequisites\n\nCompletion of 243 Algebra II. An academic course review will be conducted to determine student’s academic placement. Course satisfies a requirement for graduation." ]

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[ "", null, "# Dikshant Bansal\n\nLet A = the number be abc Let B = the reverse cba Now write he value of each number (1) A = 100*a + 10*b + c and (2) B = 100*c + 10*b + a Now get the differnce, D, (3) D = 100*a + 10*b + c - 100*c - 10*b - a or (4) D = 100*(a - c) + (c - a) or (5) D = 100*(a - c) - (a - c) or (6) D = 99*(a - c) Answer: The greatest number that divides the difference of a 3 digit number and its reverse i\nAug 31, 2018 08:35" ]

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[ "", null, "", null, "", null, "Skip to content\n\n# MATH 105 College Mathematics/Applications\n\nCredits: 3\nPrerequisites:\n• Pass MATH 095 or ACT math score = 19 or an alternative placement mechanism as approved by the math department or instructor approval.\nDistance Learning Options:\n\nThis course is a survey of mathematical applications of functions. Topics that will be covered include: fundamental concepts of algebra, algebraic equations and inequalities; functions and graphs; zeros of polynomial functions; exponential and logarithmic functions; systems of equations and inequalities. The mathematics of personal finance will also be studied." ]

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[ "# Divisibility Rules Worksheet Grade 4", null, "### Image result for divisibility rules Divisibility rules", null, "### Label the number as 'divisible' or 'not divisible' based on the remainder, in this collection of divisibility test pdf worksheets for grade 4 and grade 5.\n\nDivisibility rules worksheet grade 4. Divisibility rules or divisibility tests have been mentioned to make the division procedure easier and quicker. 3030 is divisible by 5, because it ends in 0. Instruct students to finish the challenge independently, using their rules chart as a guide.\n\nThis free product introduces divisibility rules divisibility tests with a fun to use poem that your studen divisibility rules teaching mathematics math poems. Circulate the room as students work and provide support as needed. 25, 100, 365 9 the sum of the digits of the number is divisible by 9.\n\nB) circle the numbers that are divisible by 4. Direct students' attention to the divisibility challenge on the divisibility rules worksheet. The sum of the digits of 1,485 is 1+4+8+5 = 18 which is divisible by 9, so 1,485 is divisible by 9.\n\nFill in the digits to make this number divisible by 4 & 6. Divisibility rules add to my workbooks (18) embed in my website or blog add to google classroom add to microsoft teams 645 964,726 42,391 8,862 250,860 25,368 53,420 5,232 27,568 186 17 105,743 7,526 3,892 301,570 86,340 4,900 4,584 78 41,346\n\nDivisibility rules worksheet with answers to practice & learn 6th grade math problems on finding divisors is available online for free in printable & downloadable (pdf & image) format. Some of the worksheets for this concept are divisibility rules work, divisibility rules, divisibility rules practice directions use divisibility, divisibility rules workbook, divisibility rule, divisibility rule 1, math mammoth grade 5 a worktext, divisibility rules. Test if the numbers are divisible by 4, by dividing the last 2 digits of the number by 4.\n\nSome of the worksheets for this concept are divisibility work, divisibility and factors, divisibility rules workbook, divisibility rules practice directions use divisibility, divisibility rules, gr 7 divisibility, divisibility rules, divisibility rules a. If you answer incorrectly, your ship is shot by the pirate. Check divisibility (4) a) 3466 b) 1288 c) 39804 d) 64 684 21.", null, "### Working 4 the Classroom “Flipping” for a Year End Review", null, "### Divisibility Rules Worksheet 6th Grade Divisibility Test", null, "### Divisibility Rules Poster Large Printable Bulletin Board", null, "### FREE HandsOn Divisibiltiy Rules Worksheet For 2, 3, 6, 5", null, "### Divisibility Rules Reference Sheet & Poster Kraus Math", null, "### Number Theory, Divisibility Rules, Prime Factorization", null, "### Divisibility Rules Reference Sheet & Poster Kraus Math", null, "### Multiplication Strategies & Divisibility Rules Cheat", null, "### Random Posts", null, "" ]

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[ "# 54. International Winter Meeting on Nuclear Physics\n\nJan 25 – 29, 2016\nBormio, Italy\nEurope/Berlin timezone\n\n## Measurement of neutral mesons in pp and Pb-Pb collisions at midrapidity with the ALICE experiment at the LHC\n\nJan 29, 2016, 5:20 PM\n20m\nBormio, Italy\n\n#### Bormio, Italy\n\nShort Contribution Relativistic Heavy Ion Physics\n\n### Speaker\n\nMs Lucia Leardini (Physikalisches Institut (PI) Heidelberg)\n\n### Description\n\nNeutral mesons, such as $\\pi^{0}$ and $\\eta$, are probes for the study of the energy loss of partons traversing the hot and dense medium, the Quark-Gluon Plasma, that is formed in heavy-ions collisions. Moreover, they represent the largest background for the direct photon measurement and an accurate estimate is therefore necessary to determine the decay photon contribution. The ALICE experiment measures $\\pi^{0}$ and $\\eta$ mesons via the two-gamma decay channel. The photon detection can either be direct, using the electromagnetic calorimeters EMCal and PHOS, or by reconstructing the electron-positron pairs from photon conversions in the detector material (photon conversion method, PCM). With the PCM, photons are reconstructed using the ALICE Inner Tracking System (ITS) and the Time Projection Chamber (TPC). This method has full azimuthal coverage, which compensates for the small conversion probability. It provides a precise measurement at low transverse momentum. The calorimeters have reduced acceptance but trigger capabilities and provide the high $p_{\\mbox{\\tiny{T}}}$ measurement, where they also have good energy resolution. The errors from these measurements are independent thus their comparison is a good cross check and their combination gives a more precise result. In this presentation, we will show the $\\pi^{0}$ and $\\eta$ spectra in Pb--Pb collisions at $\\sqrt{s_{\\mbox{\\tiny NN}}}$ = 2.76~TeV and pp collisions at different center of mass energy obtained over a wide transverse momentum range.\n\n### Primary author\n\nMs Lucia Leardini (Physikalisches Institut (PI) Heidelberg)\n\n Slides" ]

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[ "# If statement with or condition\n\n4 views (last 30 days)\nluca on 15 Oct 2019\nAnswered: Fabio Freschi on 15 Oct 2019\nHi given the following code\nBDR= [175 175 175 175 175 175 175 175 175 175 175 175 175];\nSETTIMANA=[190 130 120 140 100 160 175 165 157 140 130 175 140 ];\nNEXTpro = [25 60 50 40 30 30 30 35 10 34 23 45 12];\nif (SETTIMANA(1) < BDR (1)) | (SETTIMANA(2)< BDR(2)) | (SETTIMANA(3)< BDR(3)) | (SETTIMANA(4)< BDR(4))| (SETTIMANA(5)< BDR(5))| (SETTIMANA(6)< BDR(6))| (SETTIMANA(7)< BDR(7))| (SETTIMANA(8)< BDR(8))| (SETTIMANA(9)< BDR(9)) | (SETTIMANA(10)< BDR (10)) | (SETTIMANA(11)< BDR(11))| (SETTIMANA(12)< BDR(12)) | (SETTIMANA(13)< BDR(13));\nY=NEXTpro\nelse\nY=NEXTpro\nSETT1 = SETTIMANA(1:numel(BDR)); % Equalise Vectors\nY = Y(1:numel(BDR)); % Equalise Vectors\nidxy = SETT1 <= BDR; % Logical Index Vecto\nY=BDR - SETT1;\nY=Y.*idxy\nend\nI cannot understand why the if condition cannot read the or operator in the right way.\nThe conditin is: if I have a value in SETTIMANA that exceed the value in BDR in the same column, then switch to condition else.\n##### 2 CommentsShowHide 1 older comment\nKALYAN ACHARJYA on 15 Oct 2019\nIts perfectly working as you mentioned\nIf any element of SETTIMANA< correcponding NEXTpro\n%...^...Note here\n% do\nelse\n%do\nend\nor\nIf all element of SETTIMANA< correcponding NEXTpro\n%...^...Note here\n% do\nelse\n%do\nend\nWhich one\nAlso you can use logical indexing directly without mentioning one by one of SETTIMANA and NEXTpro\n\nFabio Freschi on 15 Oct 2019\nif any(settimana < bdr)\n...\nelse\n...\nend\n\nR2019b\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# Linear interpolation", null, "Given the two red points, the blue line is the linear interpolant between the points, and the value y at x may be found by linear interpolation.\n\nIn mathematics, linear interpolation is a method of curve fitting using linear polynomials.\n\n## Linear interpolation between two known points", null, "In this geometric visualisation, the value at the green circle multiplied by the distance between the red and blue circles is equal to the sum of the value at the red circle multiplied by the distance between the green and blue circles, and the value at the blue circle multiplied by the distance between the green and red circles.\n\nIf the two known points are given by the coordinates", null, "$(x_0,y_0)$ and", null, "$(x_1,y_1)$, the linear interpolant is the straight line between these points. For a value x in the interval", null, "$(x_0, x_1)$, the value y along the straight line is given from the equation", null, "$\\frac{y - y_0}{x - x_0} = \\frac{y_1 - y_0}{x_1 - x_0}$\n\nwhich can be derived geometrically from the figure on the right. It is a special case of polynomial interpolation with n = 1.\n\nSolving this equation for y, which is the unknown value at x, gives", null, "$y = y_0 + (y_1-y_0)\\frac{x - x_0}{x_1-x_0}$\n\nwhich is the formula for linear interpolation in the interval", null, "$(x_0,x_1)$. Outside this interval, the formula is identical to linear extrapolation.\n\nThis formula can also be understood as a weighted average. The weights are inversely related to the distance from the end points to the unknown point; the closer point has more influence than the farther point. Thus, the weights are", null, "${\\textstyle \\frac{x-x_0}{x_1-x_0}}$ and", null, "${\\textstyle \\frac{x_1-x}{x_1-x_0}}$, which are normalized distances between the unknown point and each of the end points. Because these sum to 1,", null, "$y = y_0 * (1-\\frac{x - x_0}{x_1-x_0}) + y_1 * (1-\\frac{x_1 - x}{x_1-x_0}) = y_0 * (1-\\frac{x - x_0}{x_1-x_0}) + y_1 * (\\frac{x - x_0}{x_1-x_0})$\n\nwhich yields the formula for linear interpolation given above.\n\n## Interpolation of a data set", null, "Linear interpolation on a data set (red points) consists of pieces of linear interpolants (blue lines).\n\nLinear interpolation on a set of data points (x0, y0), (x1, y1), ..., (xn, yn) is defined as the concatenation of linear interpolants between each pair of data points. This results in a continuous curve, with a discontinuous derivative (in general), thus of differentiability class", null, "$C^0$.\n\n## Linear interpolation as approximation\n\nLinear interpolation is often used to approximate a value of some function f using two known values of that function at other points. The error of this approximation is defined as", null, "$R_T = f(x) - p(x) \\,\\!$\n\nwhere p denotes the linear interpolation polynomial defined above", null, "$p(x) = f(x_0) + \\frac{f(x_1)-f(x_0)}{x_1-x_0}(x-x_0). \\,\\!$\n\nIt can be proven using Rolle's theorem that if f has a continuous second derivative, the error is bounded by", null, "$|R_T| \\leq \\frac{(x_1-x_0)^2}{8} \\max_{x_0 \\leq x \\leq x_1} |f''(x)|. \\,\\!$\n\nAs you see, the approximation between two points on a given function gets worse with the second derivative of the function that is approximated. This is intuitively correct as well: the \"curvier\" the function is, the worse the approximations made with simple linear interpolation.\n\n## Applications\n\nLinear interpolation is often used to fill the gaps in a table. Suppose that one has a table listing the population of some country in 1970, 1980, 1990 and 2000, and that one wanted to estimate the population in 1994. Linear interpolation is an easy way to do this.\n\nThe basic operation of linear interpolation between two values is commonly used in computer graphics. In that field's jargon it is sometimes called a lerp. The term can be used as a verb or noun for the operation. e.g. \"Bresenham's algorithm lerps incrementally between the two endpoints of the line.\"\n\nLerp operations are built into the hardware of all modern computer graphics processors. They are often used as building blocks for more complex operations: for example, a bilinear interpolation can be accomplished in three lerps. Because this operation is cheap, it's also a good way to implement accurate lookup tables with quick lookup for smooth functions without having too many table entries.\n\n## Extensions\n\n### Accuracy\n\nIf a C0 function is insufficient, for example if the process that has produced the data points is known be smoother than C0, it is common to replace linear interpolation with spline interpolation, or even polynomial interpolation in some cases.\n\n### Multivariate\n\nLinear interpolation as described here is for data points in one spatial dimension. For two spatial dimensions, the extension of linear interpolation is called bilinear interpolation, and in three dimensions, trilinear interpolation. Notice, though, that these interpolants are no longer linear functions of the spatial coordinates, rather products of linear functions; this is illustrated by the clearly non-linear example of bilinear interpolation in the figure below. Other extensions of linear interpolation can be applied to other kinds of mesh such as triangular and tetrahedral meshes, including Bézier surfaces. These may be defined as indeed higher-dimensional piecewise linear function (see second figure below).", null, "Example of bilinear interpolation on the unit square with the z-values 0, 1, 1 and 0.5 as indicated. Interpolated values in between represented by colour.", null, "A piecewise linear function in two dimensions (top) and the convex polytopes on which it is linear (bottom).\n\n## History\n\nLinear interpolation has been used since antiquity for filling the gaps in tables, often with astronomical data. It is believed that it was used by Babylonian astronomers and mathematicians in Seleucid Mesopotamia (last three centuries BC), and by the Greek astronomer and mathematician, Hipparchus (2nd century BC). A description of linear interpolation can be found in the Almagest (2nd century AD) by Ptolemy.\n\n## Programming language support\n\nMany libraries and shading languages have a 'lerp' helper-function, returning an interpolation between two inputs (v0,v1) for a parameter (t) in the closed unit interval [0,1]:\n\n// Imprecise method which does not guarantee v = v1 when t = 1,\n// due to floating-point arithmetic error.\nfloat lerp(float v0, float v1, float t) {\nreturn v0 + t*(v1-v0);\n}\n\n// Precise method which guarantees v = v1 when t = 1.\nfloat lerp(float v0, float v1, float t) {\nreturn (1-t)*v0 + t*v1;\n}\n\n\nThis function is used for alpha blending (the parameter 't' is the 'alpha value'), and the formula may be extended to blend multiple components of a vector (such as spatial x,y,z axes, or r,g,b colour components) in parallel." ]

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[ "# td(0) Converges Provably Faster than the Residual Gradient Algorithm\n\nRalf Schoknecht and Artur Merke\n\nIn Reinforcement Learning (RL) there has been some experimental evidence that the residual gradient algorithm converges slower than the td(0) algorithm. In this paper, we use the concept of asymptotic convergence rate to prove that under certain conditions the synchronous off-policy td(0) algorithm converges faster than the synchronous off-policy residual gradient algorithm if the value function is represented in tabular form. This is the first theoretical result comparing the convergence behaviour of two RL algorithms. We also show that as soon as linear function approximation is involved no general statement concerning the superiority of one of the algorithms can be made." ]

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[ "Python 是一個高階的編程語言。基本上就是你可以在一個 runtime 下直接執行一些程式碼。 而且很多人也覺得 Python 的程式碼看上去很直覺，很像虛擬碼，因此初學者會很好上手。上一篇文章有提到了，Python 在資料科學、機器學習、深度學習方面有很多很棒的套件。如： Tensorflow、Keras、PyTorch 等... 但是在進入這些套件之前，讓我們先來了解一些 Python 的基礎。正如 C、Java 等課程，大家都會先從一些基本的 Data Type 開時。\n\n### 數字 Numbers\n\n# 這個是註解\nx = 2\ny = 3.2\nprint(x) # \"2\"\nprint(type(y)) # \"<class 'float'>\"\nprint(y ** x) # \"10.24\"\nx += 1\nprint(x) # \"3\"\n\n### 字串 Strings\n\nPython 的字串可以用 '\" 來表示字串。在 Python 裡面一個很特別的功能是，字串可以用 + 來做 concatenation。 一點值得注意的是，字串和數字是無法做 concatenation 的，因此要用 str() 先把數字轉成字串。字串有很多 function 可以在文字處理上應用，大家未來在做文字分析的時候可能會很常用到哦！\n\nh = 'hello'\nw = \"world\"\nhw = h + ' ' + w\nprint(hw) # \"hello world\"\nprint(\"OM\" + str(0)) # \"OM0\"\nhw1 = '%s %s %d' % (hello, world, 1) # sprintf style string formatting\nhw2 = '{} {} {}'.format(hello, world, '2') # format style string formatting\nprint(hw1) # \"hello world 1\"\nprint(hw2) # \"hello world 2\"\nstr = \"bird \"\nprint(len(w)) # 印出字串的長度 \"5\"\nprint(s.capitalize()) # 將第一個字轉為大寫 \"Bird \"\nprint(s.upper()) # 將整個字串轉為大寫 \"BIRD \", s.lower 則為轉小寫\nprint(s.replace('bi', '(ne)')) # 將字串中與第一組字串相似的組合換成第二組字串 \"(ne)rd \"\nprint(s.strip()) # 將字串前後的空格去掉 \"bird\"\nprint(s.split('i')) # 將字串遇到 'i' 的時候切段，並組成 list \"rd\"\n\n### 布林 Boolean\n\nt = True\nf = False\nprint(t and f) # Logical AND \"False\" (兩個都要 True 或 False 才會回傳 True)\nprint(t or f) # Logical OR \"True\" (其中一個 True 就會回傳 True)\nprint(not t) # Logical NOT \"False\" (not True = False, not False = True)\nprint(t != f) # Logical XOR \"True\" (兩者要不一樣才會回傳 True)" ]

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[ "# Implementing a PLC Calibration Routine to Ensure Accurate Instrument Readings\n\n## Why Calibrate?\n\nMany applications call for a means to take accurate analog measurements. Force, pressure, electrical current, lengths, positions and other analog values that are measured must be done so with a degree of accuracy. Factory and lab personnel use calibration procedures to ensure that equipment is reading accurately. If the readings are not as accurate as required, a calibration procedure should allow for the implementation of a correction factor. Personnel operating the equipment can perform their jobs more readily if the measurement system can easily be checked for accuracy and calibrated. Ideally, the process would be transparent to operators who do not need to understand the details of the operating system and only need to see and understand the measurement results reported by the equipment.\n\nAn engineer designing a measurement system will generally examine the voltage or current to be provided by the measuring transducer and the “counts” that should result after the analog input card converts the electrical signal to a number. The conversion factors can be set up to convert counts to real world units such as pressure or force. These theoretical or ideal factory published numbers are suitable for specifying equipment, but when the system is set up and running, the actual readings and numbers will differ from the theoretical. Wires will add resistances that cannot be calculated in advance and temperature changes will cause variations that must be accounted for. No analog system will run exactly per “book values” and thus a calibration routine must correct for the differences.\n\n## Calibration Method\n\nThe calibration method described here can be applied to any analog measurement such as pressure, amperage, weight, length or force. Let’s consider an example where the length of an object is being measured. We will assume that a device has been built that converts the length of the object to numerical readings via the analog input. The type of instrumentation used may be a laser measuring device, an LVDT, or any device that provides an analog signal that will vary with the length of the object being measured. We will assume the following:\n\n1. The object being measured ranges from 2 to 9 inches in length and the analog device is capable of measuring over this range.\n\n2. The analog device provides a voltage signal to the analog input card. The voltage is converted to a number or “counts” by the analog card. The “count” resides in real time in a register in the PLC and the ladder logic can perform math functions using this number. For example, let’s assume that an object is being measured 2 inches in length and the count is 100, and when an object 9 inches in length is measured the count is 3900.\n\nThe nice thing about the method described here is there is no need to be concerned with the amount of voltage provided by the transducer or received by the analog input card. By utilizing the counts and then converting the counts directly to length, the voltage becomes irrelevant as long as any change in counts is linear with any corresponding change in length. This provides the advantage of not needing to be concerned with calibration or setup of the transducer and its amplifier. Too often, time is spent setting the “zero” and “span” of amplifiers and signal conditioners where a person might determine: “This object measures 2.3 inches, therefore my voltage should be 1.2 volts”. Then they proceed to set the zero and span to achieve an exact voltage for the corresponding length. By using the calibration method described here, this process is no longer necessary and the exact voltage at any given length is of no concern. Time is saved because the calibration of the amplifier’s voltage is now rolled into the software calibration of the entire system. Additionally, if someone comes along later and adjusts the zero or span, the calibration procedure described below can be run and the PLC program will compensate for the changes.\n\n## Calibration Procedure\n\nCalibration routines are often set up to simply add in a correction factor; but this “one point” method leaves room for error over a range of readings. A car speedometer that is permanently stuck at 60 mph will appear accurate if checked when the vehicle is moving at 60 mph. A better calibration method uses two points; a “low point” and a “high point” and then calculates a line that goes through both points. We will use a 2-point calibration, which requires two calibration units of specific lengths. We will call these units “calibration masters”. In our example, they can be any length within the readable linear range. Let’s assume that one is 4.2 inches long and the other is 6.8 inches long.\n\nFour values will be captured during the calibration procedure:\n\na. The actual length of the “low” calibration master (4.2 inches)\n\nb. The analog reading (counts) when the low master is\nmeasured\n\nc. The length of the “high” cali- bration master (6.8 inches)\n\nd. The analog reading (counts) when the high master is\nmeasured\n\nFrom this point forward the four values, a through d, will be used to set up the math required to covert counts to inches. We will call the resulting value “e”.\n\nThe change in length for any corresponding change in counts is calculated as follows:\n\n(c-a)/(d-b) = e\n\nThe person performing the calibration enters the lengths of the high and low calibration masters (values a and c) via the operator interface and these values are stored in PLC registers. The corresponding analog counts for the high and low calibration masters (values b and d) are captured during the calibration routine and are also stored in PLC registers.\n\nWe will use the following data registers:\n\na = “low” calibration master = Register V3000 / V3001\n\nb = low master analog reading = Register V3002 / V3003\n\nc = “high” calibration master = Register V3004 / V3005\n\nd = high master analog reading = Register V3006 / V3007\n\nThe following ladder logic will calculate “e” and store this value in V3010/V3011. Data register V3014 is used to store temporary math results while the accumulator is in use performing other math. See Figure 1", null, "## Length Conversion\n\nAfter the calibration routine has been run and the values a through e are stored in the PLC, the formula for converting counts to length is:\n\n((counts – b) x e) + a\n\nThe following ladder logic will calculate the length from the counts and store the resulting length in V3012/V3013. The analog reading from the device being measured is stored in register V2000. See Figure 2", null, "Some of the advantages of this calibration procedure are:\n\nBecause the user chooses the calibration points, calibration masters can be at any length. Calibration masters no longer need to be made to a specified predetermined length.\n\nIf a specific range is the most critical, the calibration can be done at that specific range. In the example above, if the most critical readings are between 3.5 and 4.2 inches of length, the calibration masters can be approximately 3.5 and 4.2 inches. In this way, calculating the straight line over the critical range minimizes any non-linearity in the system and provides the most accurate readings in the critical range.\n\nBy displaying the actual calculated length and the analog counts on the operator interface, anyone who desires can follow along with the PLC and observe the math in action. This makes the system’s internal workings clear to anyone using the equipment.\n\nBy Gary Multer,\nMulteX Automation\n\nOriginally Published: Sept. 1, 2004" ]

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[ "We’d like to understand how you use our websites in order to improve them. Register your interest.\n\n# Earth magnetic field modeling from Oersted and Champ data\n\n## Abstract\n\nWe present a method to model geomagnetic field requiring only a restricted number of measurements on magnetic survey satellite orbits. These points are chosen in an optimal—or close to optimal—manner relying on recent developments in the problem of numerical integration over spheres. The method allows us to compute a series of models at short time intervals, namely 10 days in the present study. At each of these close dates several models are computed from independent sets of data; their redundancy in turn provides a control of results thanks to which the selection of data—for example, as a function of magnetic activity or latitude—may be reduced. We find that the internal low degree Gauss coefficients derived from Oersted and Champ data, respectively, differ from one another by 1 or 2 nT. We then take as a second example of the method application a brief study of the so-called external field. We compare the first-degree axisymmetric field with the Dst index.\n\n## 1. Introduction\n\nThe problem of computing a model of magnetic field that first fits ground observations, then satellite observations (since the years 1960) goes back to Gauss in the 1830s. The most recent models using Oersted and Champ data rely on a least squares technique providing the spherical harmonic coefficients, and some of these solve the Euler angles of the sensor attitude at the same time (Olsen et al., 2006). There is probably no need to abandon spherical harmonic expansion, which is so practical for all applications of the models. Nevertheless, even when keeping to this classical method, different options exist. We will describe and use one of these, the choice of which is guided by a few considerations.\n\nFirst, it is desirable to retain as much data as possible from high (>55°) latitude regions, despite the large disturbances which are present in those regions. Second, the key issue in spherical harmonic expansion modeling, i.e. the computation of Gauss coefficients, both internal and external, is the geographical distribution of data; integral orthogonality properties of the harmonics over the data points must be strongly adhered to, not loosely. Third, we call for a flexible algorithm, using the minimum but sufficient number of data points to compute the internal and external fields, at short time intervals, for night hours or day hours, different universal times, and different conditions of activity, etc.\n\nExcellent models have been computed by different teams according to their own methods: using data from Pogo, Magsat, Oersted and Champ satellites as well as from ground magnetic observatories (Sabaka et al., 2002) and Sabaka et al. (2004) have constructed comprehensive models covering the time interval (1960-2002). Maps of the internal field (core + crustal) and the external field (ionospheric and magnetospheric) are made available with a short time sampling. Other teams have produced models of the different ingredients of the geomagnetic field from satellite data covering the time interval (2000, 2005) (Maus et al., 2006; Olsen et al., 2006).\n\nWe present the general features of our method in the main text reserving more mathematical considerations for Appendix. Its efficiency will be demonstrated in the following sections.\n\n## 2. The Model\n\n### 2.1 Optimal arrays of points\n\nLet us consider a sphere S of unit radius. An optimal array—in the sense to be defined below—of N points on S is made of the points whose colatitude θ k and longitude ϕ k , k = 1,… N, are given by the formulae below", null, "A subroutine computes the longitudes ϕ k and colatitudes θ k of the optimal array, N being given (Appendix). In this paper, we will take N = 1000, 2000, 3000, 5000, 10000.\n\nLet the surface harmonic functions", null, "be ranked in the usual lexicographic order, and u j be the corresponding jth harmonic, internal or external (see Eq. (2)), in the corresponding series. The gradients u i , U j are orthogonal on the set of N points Q l , l = 1,… N. (see Appendix)", null, "((1))\n\nwith C i being the corresponding norm of u i .\n\n### 2.2 Array of points close to an optimal array\n\nLet us compute, as an example, a model based on night values of a given period of time. We first retain all measurement points P k satisfying this local time condition. Let E (Pk) = E (P k (r,θ,ϕ)) be this set of points. We pick up in E (P k ) the point P i which is the closest to the point Q i of a given optimal array, in the sense that the angle P i OQ i (O being the sphere center) is the smallest (Fig. 1). The number N of points P i , equal to the number of points Q i of the optimal array, is much smaller than the number of points P k . After this operation is made for each point Q i , we are left with an N array close to the optimal: E N (P i (r i ,θ i , ϕ i )). Now, let the geomagnetic potential to be computed be written in the usual form:", null, "((2))\n\nwith", null, ", and K[ext] being the degrees of the internal and external expansions.\n\nLet B⃗(P l ) = (X(P l ), Y(P l ), Z(P l )) be the vectorial measurement of B⃗ at P l ε E N (P). We want to compute the Gauss coefficients g and γ such that", null, "((3))\n\nis minimal.\n\nV contains K[tot] = K[int](K[int] + 2) + K[ext] (K[ext] + 2) unknown coefficients, and there are 3N equations (3). In this paper we will address the core field and, briefly, a component of the so-called external field, with K[int] = 165, and K[ext] = 2. For such values, N = 1000 already provides a largely overdetermined set of equations. The system (3) will be solved by the usual least squares technique, and the inversion of the resulting normal matrix by the Singular Value Decomposition algorithm. When computing this matrix, the following approximations of scalar products in Eq. (1) appear:", null, "with, for example,", null, "which allows for computing the following angles between the K[tot] column vectors of the matrix:", null, "Looking at the values of β ij is key to the method. Indeed, u j vectors are not expected to be strictly orthogonal on the {P i } array: their projections on the sphere (r = a) are not located exactly at points of the optimal array and, furthermore, radii of P i points may vary within 200 kilometers (see next paragraph).\n\nRemark We could correct B⃗(P i ) for the difference B⃗(P i )-B⃗(Q i ) using an a priori model. However, here we prefer to present a self-contained algorithm.\n\n### 2.3 A synthetic example\n\nFirst, we pick up a model {g j , γ j } from the literature (Langlais et al., 2003; K[int] = 16, K[ext] = 2 ) and compute the values X, Y, Z of the model field at M points which are “real” points in the sense that there are points of the orbits of Champ where real measurements were made (specifically the data points corresponding to 130 days centered at 2003.0). We select N points P l from the whole set, the one closest to an optimal array of 1000 or 3000 points {Q l }, as explained in Fig. 1, and compute back the Gauss coefficients {g j , γ j }. The differences between the initial Gauss coefficients and the recovered ones, g j and γ k , j = 1,… 288 k = 1,… 8, for N = 1000 and N = 3000, are of the order of 10-5 nT.\n\nLet us take the opportunity with this synthetic example— but with real orbit points—to be more specific about the orthogonality of u j vectors, taking the case N = 1000 (1000 points in array {Q l }). For K[int] = 16 and K[ext] = 2, we find that 99% of couples u i , u j vectors make angles", null, "with ε < 0.01 radian. The value of ε does not change much with the degree: from 0.005 for the low harmonic couples to 0.03 for the high harmonics. Simulations show that for such values of the departure ε from orthogonality, the coefficients g j and γ j can be recovered independently of one another with the required accuracy. Such will be the case for all the computations in the paper.\n\n### 2.4 The time sampling of the modeling\n\nA huge advantage of satellite data is their high density (108 per year!), which allows for a massive number of computations which could not be dreamed of in the pre-satellite times; this density makes it possible to split the data set into many subsets for various applications. In particular, it is possible to compute models at a succession of close time moments; in this paper, we will compute a model every 10 days, at times t k = t0 + k × 10 days. Nevertheless, those models are not genuine instantaneous models at t k . Indeed, to get a uniform enough distribution of local times when computing a model of the main field Eq. (2), it is necessary to consider data spanning a time interval τ around t k ;the duration τ is different for Champ and Oersted. So, our model at time t k is a model computed from data in a time interval of length τ centered on t k . But we keep g(t), h(t) (with a daily sampling) in the following temporal series, and not the decimated series with a sampling of τ days (τ = 130 in case of Champ; see below); a surprising amount of information on short term (τ) features is preserved in the process of computing the model from data in a window of length τ (e.g., Blanter et al., 2005).\n\n### 2.5 The choice of data amount\n\nWe have addressed the question of the number N of data points required to compute a model (an expansion) to degree d with an empirical point of view; indeed, not only the distribution of data points, but also the noise intervene in a broad sense, on the data. We compute a large number (100) of models {g i , h j } from suitable random simulations of data; we infer a mean value m and error bar σ on each coefficient. Each random simulation at N points is obtained by biasing the initial vector field (calculated from a 16-degree model) by a Gaussian random vector whose distribution parameters were estimated at each point using real data records. And we repeated those computations varying N from 700 to 6000 in 25 steps. For the large majority of coefficients, m is approximately stabilized at N = 2400 (Fig. 2). The value of σ decreases from 0.75 nT for N = 700 to 0.2–0.4 nT (depending on the coefficient) for N = 2400 and keeps decreasing slowly thereafter. From these results we retained N = 2400 in most of the computations of the present paper, for models of degree 16.\n\n### 2.6 Selecting the data as a function of magnetic activity We now switch to real data.\n\nIn most studies, measurements are selected according to the magnetic situation at the time of the measurement, as characterized by the value of the planetary Kp index. We made a series of experiments to evaluate the influence of magnetic activity on our modeling. We take advantage of redundant observations. Let us first retain all of the data, without any selection, versus magnetic activity. For each date t k (multiple of 10 days) we build a number of quasi-optimal arrays {P i } close to the same optimal array {Q i } (with, in general, N = 2400), from different disjoint sets of data; we then compute as many coefficient sets or models. Some of these appear to be perturbed by big magnetic storms; but it is always possible to find some which are not. The trend of the representative curves allows us to easily discard perturbed values due to the tightness of time sampling. Figure 3(a) has been chosen to illustrate the situation; it represents the evolution of the g 11 estimate. Crosses are for estimates computed from 130 days of Champ measurements without any selection in the function of magnetic activity. A segment of the curve, in the second half of 2002, is shifted 12 nT below the general trend. And a blank is observed in the first months of 2004. Both segments (S. Maus, personal communication) are characterized by a relatively low number of usable measurement points, which makes it harder to find points close to optimal data sets in the 130 days of data, especially if high magnetic activity is present during those intervals; in this situation, steps in (g, h) estimate may occur. In fact, due to the abundance of data, it is always possible—except in the case of long gaps—to find close to optimal subsets from which g, h estimates fill the gaps of the graphs of Fig. 3, and are on the general trend. Those conclusions hold for all low degree coefficients of the main field. We kept those gaps here for illustration. Nevertheless, it would be awkward not to take advantage of the large redundancy of data to avoid computations from data corresponding to high activity, e.g., a m > 20 nT (Mayaud, 1980). An extra verification is easily obtained by changing the threshold for a m . Coefficients computed with this condition are represented by dots in Fig. 3 together with those estimated without selection. Along the same lines, we keep the vectorial measurements in high latitudes (as Maus et al , 2006).\n\n## 3. Fitting Oersted and Champ Data\n\n### 3.1 The data\n\nWe use Champ data provided by the German team in the form of a list t, X(t), Y(t), Z(t), r⃗(t), r⃗(t) being the current point and τ being the time of the measurement on the orbit, counted in seconds from June 2001 to December 2004, and Oersted data provided by the Danish team in the same form, τ being then a multiple of 1.3 sec (see Stolle et al., 2006), running from March 1999 to June 2003. Note that these data are transformed data. To obtain geocentric components X, Y, Z, Euler angles of the sensor attitude have been determined by the teams in charge. We do not discuss this determination.\n\n### 3.2 Fitting the data\n\nA first opportunity to check the efficiency and accuracy of the algorithm is to look at how coefficients issued from Oersted and Champ fit together, paying special attention to the overlapping period (June 2001–June 2003). We treat the data from both satellites exactly the same way, except for the time span τ required to build a model from data reasonably uniformly distributed in local time; indeed, 130 days are required in the case of Champ, while 2 years are needed in the case of Oersted to get the same performance; actually we use a 90-day time span for Oersted to the cost of a less strict condition on the uniformity of local time distribution.\n\nResults are illustrated, for a few Gauss coefficients in Fig. 4. Let us look, for example, at h 11 (Fig. 4(b)); at each time t k there are eight estimates for Champ, five for Oersted. We will systematically use this control in all our computations. Clearly, h 11 values derived from Oersted and Champ agree within 2 nT most of the time, without any further averaging. Other examples are for h 35 and h 45 (mind the enlarged scale). the mean values from Oersted and Champ coincide within a nT. In the computations of this section, we retained data subsets built in such a way that local times are reasonably uniformly distributed. The agreement between Oersted- and Champ-derived coefficients is pretty good but not perfect. There is a physical limitation to this agreement. the ionospheric field is not sampled in exactly the same way by the two satellites. Resulting departures are small and quite variable with the coefficient. For example, for coefficient g 01 , a tiny drift of Oersted estimates with respect to Champ ones is observed Fig. 4(a) (the differences remaining smaller than 2 nT).\n\nWe also computed the dipole moment and the angle between the geographical axis and the dipole axis over the 1999–2004 time span (Fig. 5(a)). Except for some trouble at the beginning of Oersted life, the linear decreasing trend appears almost perfect.\n\n## 4. Internal and External Field\n\nThe satellite sees as a field of internal origin the sum of the main field generated by the dynamo, the lithospheric field, and the field generated by electric currents flowing in the ionosphere below the perigee of the satellite, especially according to the classical views, in the E layer at 110 km altitude (Ratcliffe, 1972). The satellites also encounter electric currents since they are flying in the upper part of the F layer. In particular, the fields associated with field-aligned electric currents in the polar regions can be large. But, as discussed in Section 2.6, despite their large magnitude, they do not severely affect the aimed modeling (the computation of internal low ( 13) degree Gauss coefficients from data covering a given time interval). So, we may reasonably assume that the essential part of the field which is external to the solid Earth but internal to the satellites orbits is generated by currents in the E layer, i.e., currents driven by the atmospheric dynamo and currents driven in high latitude regions by forces originating high in the ionosphere (the polar current system) (Ratcliffe, 1972; Encrenaz et al., 2004).\n\nTo obtain a model of the main field, we have to get rid of this ionospheric field. As is well known, this is not an easy task (Olsen, 1996; Thomson, 2000). It is nevertheless possible to estimate the magnitude of the ionospheric contribution to the internal Gauss coefficients. We will not develop this question at length in the present paper. A classical method is to select data on the basis of local time, e.g., all local times, only day times or only night times (e.g., 6:00–18:00 or 18:00–6:00 LT). Our results will be illustrated by a few graphs. Figure 6(a) shows estimates of g 01 , over a time span of 4 years, derived from Champ data, computed respectively from data at all local times, day times, and night times. The difference between night time estimates and the all times estimates is almost everywhere smaller than 3 nT (2 nT in the second half of 2004). Results for g 22 are shown in Fig. 6(b) (mind the scale). Generally, the amplitudes of the differences between the different estimates as well as their evolution in time depend on the considered coefficient in a way which is not straightforward to understand. For all of the coefficients, differences between different estimates do not exceed 2 or 3 nT. A last graph Fig. 7 shows two estimates of g 01 using all local times; for the first one, no selection is made in the function of activity; for the second one, only measurements corresponding to a m 20 nT are retained. A significant difference of 2 nT shows up, but the trend is the same.\n\n### 4.1 The internal field\n\nTo obtain the main (dynamo) field model, we first take the average of models of the internal field (sources within the sphere r = rperigee) computing on all the universal times (practically at 0:00 UT, 1:00 UT to 23:00 UT). That comes down to compute a model I from points uniformly distributed in longitude; to check it we choose points Q i (see Section 2.1) whose measurements correspond to local times drawn randomly in the interval 0:00 LT–24:00 LT and compute a model II derived from this set of points. Figure 8(c) shows that the two models are indeed identical. Of course, adopting such a model for the main field means that the 24-h averaged ionospheric field is supposed to be zero—when averaged over a full day. It is such a model that we compute at days t k = t0 + k × 10 days, from data in a 130-day time span τ centered on t k .\n\nWe compare our two models, relative to 2003.0, with the POMME-3 model (Maus et al., 2006) relative to the same epoch, in two ways. First, we compare the coefficients of our models and those of the POMME-3 model by computing their differences as well as the mean and standard deviation of these differences (Fig. 8(a) and 8(b)). Except for g 01 andg 03 , the models can be said to be very similar. Our computation of the main field supposes we eliminate the ionospheric field by averaging in longitude, i.e. that this latter field has no zonal component, or, for approximate symmetry reasons, no component g 02k+1 . Choosing only night values leads to slightly different estimates (Fig. 6(a)). We then draw the maps of the vertical components of models I and POMME-3 at the core-mantle boundary (CMB) more precisely on a sphere of radius 3480 km. We also map their differences (Fig. 9). We note that, despite the smallness of difference in coefficients illustrated by Fig. 8, a few small scale anomalies may reach a notable amplitude. The geometrical factor", null, "is indeed equal to 4500; the downward continuation to the CMB of a core field contaminated by crustal anomalies is known to require some precautions. As expected, the largest discrepancies are observed in high lattitudes, due to field aligned currents.\n\n### 4.2 The external (ring current) field\n\nWe keep here a rather formal point of view, without addressing the physical nature of the field. This is the way the question has been treated for decades—the external field was globally called the ring current field up to recent modelings of satellite data (Olsen et al., 2000). Only external field coefficients of the first degree were considered to be safely determined. So we compute, with the same sampling interval (every 10 days), from the same sets of data, coefficients (we note them γ, ν for simplicity, γ for the cosine and ν for sine term) of the external field, i.e., external to the sphere containing the satellites’ orbits (r > rapogee max). It is useful to recall again that we are not computing an instantaneous field, but, at day t k ,a field based on a data set extracted from 130 days of data centered at t k (for Champ). For each data set we check that the orthogonality conditions required in Section 2 are verified: the VM[ext] are orthogonal, with a high accuracy, to one another, and orthogonal to all the VM[int] (Section 3). There is no contamination of the external coefficients by the internal ones, i.e., no contamination of the external field by the internal field. The results depend on what is being looked for. For example, y° value is much larger when computed only from night (18:00-06:00 LT) values than when computed from day (06:00–18:00) values. This asymmetry was pointed out as early as in 1970 by Olson (1970).\n\nWe retain all the local times—which comes down to averaging in longitude—and focus on the axisymmetric coefficients γ 0 k . The graph of Fig. 10 represents the variation from April 1999 to December 2004 of γ 01 . Data from both satellites, Champ and Oersted, are used. Again, the fitofthe models derived from the two data sets is excellent, within 1 or 2 nT. γ 01 displays variations with time constants of a few months and amplitudes of some 15 nT. At the bottom of Fig. 10 the evolution of Dst index is presented, averaged over a running window of 90 days to make the two graphs comparable (such a comparison is not a new idea (e.g. Cain et al., 1967). Let us recall that Dst (Sugiura, 1964) is “the disturbance field which is axially symmetric with respect to the dipole axis, and which is regarded as a function of storm time”. The correlation between Dst and γ 01 is good up to the end of 2002; amplitudes of γ 01 variations are smaller. An interesting observation concerns the base level; it is zero by construction for Dst index, while γ 01 evolves between 15 and 40 nT.\n\nMaus and Lühr recently performed a study of the mag-netospheric field during magnetically quet times (Maus and Luhr, 2005) using Oersted and Champ data from the years 1999–2004. The field is decomposed into contributions from sources in the solar-magnetic frame, and those in the geocentric-solar-magnetospheric frame. Such a separation is probably necessary for a coherent study of the external field. We pointed out at the beginning of this section the limited scope of our study of the “external field“.\n\n## 5. Conclusion\n\nThe general objective of the analysis we presented here is to model different ingredients of the field altogether. The method reported here is the realization. This paper also has also a methodological character; we computed Gauss coefficients of the main (dynamo) field, and computed coefficients of the so-called ring current field. The characteristics of the analysis are the following. To compute a given model of the main field, for example, it is possible to use for each τ-interval centered on day τ a number of disjoint data sets, each including a rather small (1000, 2400) number of points. In this way, spurious values of coefficient estimates are made conspicuous; this provides a control of the model (g, h) which allows us to release data selection. An advantage of a close time-spacing of the model is to provide time series whose trends can be studied in the usual way.\n\n## References\n\n1. Blanter, E., M. Shnirman, and J. L. Le Mouël, Solar variability, evaluation of correlation properties, J. Atmos. Terr. Phys., 67, 521–534, 2005.\n\n2. Cain, J. C., S. J. Hendricks, R. A. Langel, and W. V. Hudson, A proposed model for the International Geomagnetic Reference Field-1965, J. Geomag. Geoelectr., 19, 335–355, 1967.\n\n3. Delsarte, P., J. Goethais, and J. Seidel, Spherical codes and designs, Geom. Dedicata, 6, 363–388, 1977.\n\n4. Encrenaz, T., J. Bibring, M. Blanc, M. Barucci, P. Zarka, and F. Roques, Le système solaire, CNRS editions, 2004.\n\n5. Habicht, W. and B. L. van der Waerden, Lagerung von Punkten auf der Kugel, Math. Ann., 123, 223–234, 1951.\n\n6. Hardin, D. P. and E. B. Saff, Discretizing manifolds via minimum energy points, Notices of AMS, 51(N10), 1186–1194, 2004.\n\n7. Hardin, R. and N. Sloane, McLaren/rss improved snub cube and other new spherical designs in three dimensions, /avmalgin. livejournal.com/1203330.html, Disc. Comp. Geom., 15, 429–442, 1996.\n\n8. Langlais, B., M. Mandea, and P. Ultre-Guerard, High-resolution magnetic field modeling: application to Magsat and Ørsted data, Phys. Earth Planet. Inter., 135,77–91, 2003.\n\n9. Maus, S. and H. Luhr, Signature of the quiet-time magnetospheric magnetic field and its electromagnetic induction in the rotating Earth, Geophys. J. Int., 162, 755–763, 2005.\n\n10. Maus, S., M. Rother, C. Stolle, W. Mai, S. Choi, H. Lühr, D. Cooke, and C. Roth, Third generation of the Potsdam Magnetic Model of the Earth (POMME), G3, 7, N7, 2006.\n\n11. Mayaud, P. N., Derivation, meaning and use of geomagnetic indices, Geophysical Monograph, 22, AGU, Washington D.C., 1980.\n\n12. Olson, W. P., Variations in the Earth’s surface magnetic field from the magnetopause current system, Planet. Space Sci., 18,1471–1484,1970.\n\n13. Olsen, N., A new tool for determining ionospheric currents from magnetic satellite data, Geophys. Res. Lett., 23, 3635–3638, 1996.\n\n14. Olsen, N., T. J. Sabaka, and L. Toffner-Clausen, Determination of the IGRF 2000 model, Earth Planets Space, 52, 1175–1182, 2000.\n\n15. Olsen, N., H. Lühr, T. J. Sabaka, M. Mandea, M. Rother, L. Toffner-Clausen, and S. Choi, CHAOS—a model of the Earths magnetic field derived from CHAMP, Oersted, and SAC-C magnetic satellite data, Geophys. J. Int., 166,67–75, 2006.\n\n16. Rakhmanov, E., E. Saff, and Y. Zhou, Minimal discrete energy on the sphere, Math. Res. Lett., 1, 647–662, 1994.\n\n17. Ratcliffe, J. A., An Introduction to the Ionosphere and Magnetosphere, Cabridge University Press, 1972.\n\n18. Sabaka, T. J., N. Olsen, and R. A. Langel, A comprehensive model of the quiet-time near-Earth magnetic field: Phase 3, Geophys. J. Int., 151,32–68, 2002.\n\n19. Sabaka, T. J., N. Olsen, and M. Purucker, Extending comprehensive models of the Earth’s magnetic field with Oersted and CHAMP data, Geophys. J. Int., 159, 521–547, 2004.\n\n20. Stolle, C., H. Lühr, M. Rother, and G. Balasis, Magnetic signatures of equatorial spread F as observed by the CHAMP satellite, J. Geophys. Res., 111, A02304, 2006.\n\n21. Sugiura, M., Hourly values of equatorial Dst for IGY, in Annals of the International Geophysical Year, 35, 945–948, Pergamon Press, Oxford, 1964.\n\n22. Thomson, A. W. P., Improving the modelling of the geomagnetic mainfield: Isolating the average ionospheric field in satellite data, Earth Planets Space, 52, 1199–1206, 2000.\n\n## Acknowledgments\n\nWe acknowledge fruitful discussions with S. Maus. We deeply thank the two referees, S. Macmillan and H. Utada, for the attention they paid to the first version of the paper and the resulting numerous and helpful comments and recommendations.\n\nAuthors\n\n## Appendix A. The Numerical Integration Over Spheres\n\n### Appendix A. The Numerical Integration Over Spheres\n\nThe question of distributing uniformly N points on a surface—we consider here the case of a sphere—is by no means a trivial one; it has been of interest to mathematicians since Antiquity, and is still the object of research (Hardin and Saff, 2004).\n\nWe want an algorithm which distributes a set of N points on a sphere in such a way that the distribution of these points converges to an uniform distribution when N gets large. But uniformity makes sense only for infinite sets. For N finite, we look for a configuration which is optimal with respect to some property, but which may be poor with respect to another property.\n\nFor example, we may want to approximate the integral of a function over the sphere S2 by an arithmetic sum (without weights) of values ƒ (⃗k) at some N well-chosen points ⃗k on S2, in other words we want the difference", null, "((A.1))\n\nto be small for a large class of functions. Such a configuration of points will be called optimal for the evaluation of the integral. The problem has not yet received a general precise solution, but explicit particular solutions have been found. For example Delsarte et al (1977) considered so-called {N, t} spherical designs: configurations of N points ⃗k such that for all polynomials Pm(⃗) (i.e., polynomials in three variables (x, y, z) = r⃗) of degree m t, the difference in Eq. (A.1) is equal to zero. By means of a computer search, Hardin and Sloane (1996) found spherical designs for all t 13 with a minimal number of points; for instance, they produced a (94, 13)-design. A variety of algorithms have also been proposed for explicitly constructing asymptotically uniform distributions of points on S2. The most recent approach is to look for configurations corresponding to the minimum potential energy of N repelling points (repelling force needs to be redefined). The problem with the minimum energy configurations algorithm is that it is long and cumbersome. That is why we consider in this paper much more straightforward “spiral sets” which are almost as good as the “polynomial adapted” or “energy sets” refered to above. They are also good with respect to the property of orthogonality of spherical harmonics; this essential property will be simply checked.\n\nLet us now describe the spiral set devised by Rakhmanov, Saff, and Zhou (Rakhmanov et al., 1994), used in the present study. In spherical coordinates (θ,ϕ), 0 θ π, 0 ϕ 2π 0 ≤ϕ ≤2π, we take the following coordinates of the N points as:", null, "((A.2))\n\nThe estimate of the maximum diameter of N nonoverlapping disks on the spherical surface (the so-called “best packing argument” (Habicht and van der Waerden, 1951)) suggests the constant in Eq. (A.2) to be chosen such that", null, "The following version of the latter construction with a good choice of the constant is easy to remember: we first generate N points (x, y) belonging to the unit square 0 x, y 1 and then use the cylindrical equal-area projection (i.e., θ = arcsin(2y - 1), ϕ = 2πx) onto the spherical surface. The generation of the initial sequence in the square is also simple: xk = {(k - 1)sfϕ} (here braces denote the non-integer part of the real value) and γk = (k - N - 1). The good value of the constant in Eq. (A.2) corresponds to", null, "(the reciprocal of the so-called “golden number”). This spiral set is illustrated in the main text for N = 1000 and N = 3000 (Fig. 1).\n\nLet the surface harmonic functions Pmn (cos θ) cos , Pmn (cos θ) sin be ranked in the usual lexicographic order, let uj be the jth surface harmonic in the corresponding series. Their gradients ui, uj are orthogonal to one another on the set of N points r⃗l, l = 1,… N:", null, "((A.3))\n\nThis ensures that computing Gauss coefficients g mn , h mn in the main text through non-weighted averages taken on the points r⃗l of the spiral set (Ql in the main text) is valid. To keep a practical point of view we checked that this orthogonality property is satisfied with a sufficient accuracy for our needs in the present study (see main text).\n\nAs we do not know of any other algorithm as simple to operate as this spiral set one, while producing better results for the problem at hand, we did not extend the analysis to other constructions.\n\n## Rights and permissions\n\nReprints and Permissions\n\nLe Mouël, J.L., Shebalin, P. & Khokhlov, A. Earth magnetic field modeling from Oersted and Champ data. Earth Planet Sp 62, 277–286 (2010). https://doi.org/10.5047/eps.2009.11.003\n\n• Revised:\n\n• Accepted:\n\n• Published:\n\n• Issue Date:\n\n### Key words\n\n• Geomagnetic field model\n• satellite\n• numerical integration over spheres", null, "" ]

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[ "", null, "Site Navigation", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Graphing Linear Functions\nThis lesson will make use of slopes and y-intercepts on a graph. If you need to review these topics, click here for slope (linear equations slope.doc) or click here for y-intercepts. (linear equations intercepts.doc)\n\nLet's start with a line that has a positive slope.\nSuppose we are given the function y = 2x + 5. Because this is in slope-intercept form of a line, we can see that the slope is 2 and the y-intercept is 5.\n\nThis is enough information to graph the function. Because we know the y-intercept is 5, we can start by placing a point at (0, 5) on the coordinate system.", null, "The slope of 2 gives us the additional information we need to complete the graph. Remember that slope is “rise over run” or “the change in y over the change in x”. In any case, we always need to think about slope as a fraction. Since we know the slope is 2, we should think of it in the fraction form of", null, ". This says that every time we change the y-value by 2, we must also change the x-value by 1. Most of the time this is done by moving up 2 (changing in the y-direction) and over 1 (changing in the x-direction) and then placing the new point. You can see on the graph below that the new point is at (1, 7).", null, "Once you know two points on a graph, you can connect those two points with a line and you then have the graph of the equation.", null, "To find other points on the line, simply choose a starting point, and then move right one space and up two spaces to satisfy the slope of", null, ". You should know that you can also move to the left one space and down two spaces and still find another point on the line. Why would this work? Think about this. When you move to the left one you are changing the x-value by -1. When you move down two you are changing the y-value by -2. This will give a slope of", null, "which is what our line indicates. Look at the graph below to verify moving left 1 and down 2 will give you the point (-1, 3) which is on the graph.", null, "Let’s now consider what should happen if we have a negative slope.\nLook at the function", null, ". The y-intercept for this graph will be at the point (0, -3). This will be our starting point for finding other points on the graph.\n\nSince the slope is", null, "there are several ways to approach this problem. We know that we will be moving in the y-direction two spaces and in the x-direction three spaces. We just need to figure out whether those moves are left, right, up or down.\n\nWe need to remember that when there is a negative fraction, as there is in this case, it can be written in one of three ways.", null, "• This indicates that if we move down 2, we should move right three because of", null, ".\n• It also says that if we move up 2, we should move left 3 because of", null, ".\nLet’s find two more points.\n• Using", null, ", we start from (0, -3) move right (x-direction) 3 and down (y-direction) 2. This will put a new point at (0+3, -3-2) = (3, -5)\n• Using", null, ", we again start from (0, -3) move left (x-direction) 3 and up (y-direction) 2. This will put a new point at (0-3, -3+2) = (-3, -1)\nThe graph of", null, "is shown below. You should now verify these three points on the graph.", null, "Examples\nGraph each line by plotting three points.", null, "", null, "What is your answer?", null, "", null, "", null, "What is your answer?", null, "", null, "", null, "What is your answer?", null, "S Taylor\n\nShow Related AlgebraLab Documents", null, "AlgebraLAB\nProject Manager\nCatharine H. Colwell\nApplication Programmers\nJeremy R. Blawn\nMark Acton" ]

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[ "# What is 95 percent in money? How much?\n\n95 percent is just 95*Money/100 of the total amount of money.\n\nMoney\nPercent: %\n\n 20 dollars = 100%. 95% = 19 dollars 100 dollars = 100%. 95% = 95 dollars 3500 dollars = 100%. 95% = 3325 dollars 3500 dollars = 95%. 100% = 3684.21 dollars\nHTML code:\nBB code:" ]

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[ "# 1.50x+8.45=26.95\n\n## Simple and best practice solution for 1.50x+8.45=26.95 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.\n\nIf it's not what You are looking for type in the equation solver your own equation and let us solve it.\n\n## Solution for 1.50x+8.45=26.95 equation:\n\n1.50x+8.45=26.95\nWe move all terms to the left:\n1.50x+8.45-(26.95)=0\nWe add all the numbers together, and all the variables\n1.50x-18.5=0\nWe move all terms containing x to the left, all other terms to the right\n1.50x=18.5\nx=18.5/1.50\nx=12+0.5/1.50\n\n`" ]

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[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## Compare 4/7 and 1/50\n\n4/7 is greater than 1/50\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 7 and 50 is 350\n2. For the 1st fraction, since 7 × 50 = 350,\n4/7 = 4 × 50/7 × 50 = 200/350\n3. Likewise, for the 2nd fraction, since 50 × 7 = 350,\n1/50 = 1 × 7/50 × 7 = 7/350\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 200/350 > 7/350 or 4/7 > 1/50\n\n#### Compare Fractions Calculator\n\nand\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS\n\n© everydaycalculation.com" ]

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[ "Explore BrainMass\n\n# Density of a Metal\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nA graduated cylinder weighs 46.3 g when empty and holds 35.1 ml water (d = 1.00 g/cm3). An unknown metal is submerged into the water and brings the total volume to 58.1 ml and the mass of the cylinder and contents to 332.3 g. What is the density of the metal?\n\nhttps://brainmass.com/chemistry/stoichiometry/density-metal-calculated-23695\n\n## SOLUTION This solution is FREE courtesy of BrainMass!\n\nPlease see the attached Word document for all calculations and detailed explanation of this exercise.\n\nA graduated cylinder weighs 46.3 g when empty and holds 35.1 ml water (d = 1.00 g/cm3). An unknown metal is submerged into the water and brings the total volume to 58.1 ml and the mass of the cylinder and contents to 332.3 g. What is the density of the metal?\n\nThe density of the metal is the mass of the metal divided by the volume of the metal (from the definition of density).\n\nFirst recall that one milliliter equals one cubic centimeter.\n\n1 ml = 1\n\nThe mass of the water in the cylinder is:\n\nThen it's simple to calculate the mass of the metal since:\n\ng\n\nSimilarly you calculate the volume of the metal:\n\nThen use the equation for density:\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!" ]

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[ "# AtCoder Regular Contest 092/AtCoder Beginner Contest 091 D問題 - Two Sequences\n\n## cLay（version 20191111-1）のコード\n\nC++に変換後のコードはこちら\n\nint N, A[2d5], B[2d5];\n\nint a0s, a0[2d5], a1s, a1[2d5];\nint b0s, b0[2d5], b1s, b1[2d5];\n\n{\nint res = 0, all;\nrd(N,A(N),B(N));\nrrep(bt,30){\nall = (1<<(bt+1)) - 1;\na0s = a1s = b0s = b1s = 0;\nrep(i,N){\nif(BIT_ith(A[i],bt)) a1[a1s++] = (A[i]&all) ^ BIT_ith(bt);\nelse a0[a0s++] = (A[i]&all);\nif(BIT_ith(B[i],bt)) b1[b1s++] = (B[i]&all) ^ BIT_ith(bt);\nelse b0[b0s++] = (B[i]&all);\n}\nsortA(a0s, a0); sortA(a1s, a1);\nsortA(b0s, b0); sortA(b1s, b1);\nif(counterSumIsLT(a0s, a0, b1s, b1, BIT_ith(bt)) % 2) res ^= BIT_ith(bt);\nif(counterSumIsLT(a1s, a1, b0s, b0, BIT_ith(bt)) % 2) res ^= BIT_ith(bt);\nif(((ll) a0s * b0s - counterSumIsLT(a0s, a0, b0s, b0, BIT_ith(bt))) % 2) res ^= BIT_ith(bt);\nif(((ll) a1s * b1s - counterSumIsLT(a1s, a1, b1s, b1, BIT_ith(bt))) % 2) res ^= BIT_ith(bt);\n}\nwt(res);\n}\n\n\nCurrent time: 2021年09月28日08時14分39秒" ]

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[ "Home » Modulo Division » mod 49 » 12/13 mod 49\n\n# 12/13 mod 49\n\nWelcome to 12/13 mod 49, our post which explains the mathematical operation 12/13 modulus 49.\n\nThis is also known as remainder of 12/13 divided by 49, and if you have been looking for 12/13 modulo 49, then you are right here, too.\n\nRead on to find the 12/13 mod 49 value as well as the math in a nutshell.\n\nReset\n\n## 12/13 Modulo 49\n\n12/13 modulus 49 stands for the Euclidean division discussed, defined and explained in full detail on our home page.\n\nThe result of this modulo operation is:\n\n12/13 mod 49 = 0.(923076)\n\n12/13 is the dividend, 49 is the divisor (modulo), 0 is the quotient explained below, and 0.(923076) is called the remainder.\n\nThe division rest of 12/13 by 49 equals 0.(923076), and the value of the quotient is 0.\n\nProof: 12/13 = (49×0) + 0.(923076).\n\nNote that there is no other quotient q than 0, and that there is no other remainder r than 0.(923076) which solves the equation 12/13 = (49×q) + r and 0 ≤ r < 49; r ∈ set of real numbers R; q ∈ set of whole numbers Z.\n\nNow that you understand what 12/13 mod 49 means, it’s time to zoom in on how this modulo operation is actually calculated.\n\nStep by step, easy and straight to the point.\n\nYou can find the math the next part of this post.\n\n## How is 12/13 mod 49 Calculated?\n\nTo obtain 12/13mod49 conduct these three steps:\n\n1. Integer division (result without fractional part) of dividend by modulus: 12/13 / 49 = 0\n2. Multiplication of the result right above (0) by the divisor (49): 0 × 49 = 0\n3. Subtraction of the result right above (0) from the dividend (12/13): 12/13 – 0 = 0.(923076).\n\nCalculation examples similar to the modulo division 12/13%49, but more detailed, can be found in our article in the header menu.\n\nOther operations belonging to the modulo 49 division category include, for example:\n\nNote that you can locate all of our calculations, including 12/13 modulus 49, quickly, by filling in the search box placed in the header and sidebar; the result page contains all relevant posts.\n\nAhead is the summary of our information.\n\n## Remainder of 12/13 Divided by 49\n\nYou have reached the final section of this post, and you should be able to answer questions like what is 12/13 mod 49?, compute is value, and name its part.\n\nHowever, if you are in doubt about something related to the quotient and remainder of 12/13 by 49, or if you like to leave feedback, then simply use the comment form at the bottom of this article.\n\nAlternatively, send us an email with a meaningful title such as 12/13 modulus 49 division.\n\nEither way you let us know your question, we will get back to you as soon as possible.\n\nIn conclusion:\n\nIf our 12/13 remainder 49 math has been of help to you, hit the sharing buttons and place a bookmark in your browser.\n\nWe recommend to you installing our absolutely free PWA app (see menu or sidebar).\n\nThanks for visiting our post about 12/13 modulo 49.\n\nSubmitting..." ]

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[ "In a RGB color space, hex #9fada4 is composed of 62.4% red, 67.8% green and 64.3% blue. Whereas in a CMYK color space, it is composed of 8.1% cyan, 0% magenta, 5.2% yellow and 32.2% black. It has a hue angle of 141.4 degrees, a saturation of 7.9% and a lightness of 65.1%. #9fada4 color hex could be obtained by blending #ffffff with #3f5b49. Closest websafe color is: #999999.\n\n• R 62\n• G 68\n• B 64\nRGB color chart\n• C 8\n• M 0\n• Y 5\n• K 32\nCMYK color chart\n\n#9fada4 color description : Dark grayish cyan - lime green.\n\nThe hexadecimal color #9fada4 has RGB values of R:159, G:173, B:164 and CMYK values of C:0.08, M:0, Y:0.05, K:0.32. Its decimal value is 10464676.\n\nHex triplet RGB Decimal 9fada4 `#9fada4` 159, 173, 164 `rgb(159,173,164)` 62.4, 67.8, 64.3 `rgb(62.4%,67.8%,64.3%)` 8, 0, 5, 32 141.4°, 7.9, 65.1 `hsl(141.4,7.9%,65.1%)` 141.4°, 8.1, 67.8 999999 `#999999`\nCIE-LAB 69.425, -6.647, 2.938 35.941, 39.938, 40.935 0.308, 0.342, 39.938 69.425, 7.267, 156.156 69.425, -7.339, 5.394 63.197, -9.078, 5.833 10011111, 10101101, 10100100\n\n``#9fada4` `rgb(159,173,164)``\n``#ad9fa8` `rgb(173,159,168)``\nComplementary Color\n``#a1ad9f` `rgb(161,173,159)``\n``#9fada4` `rgb(159,173,164)``\n``#9fadab` `rgb(159,173,171)``\nAnalogous Color\n``#ad9fa1` `rgb(173,159,161)``\n``#9fada4` `rgb(159,173,164)``\n``#ab9fad` `rgb(171,159,173)``\nSplit Complementary Color\n``#ada49f` `rgb(173,164,159)``\n``#9fada4` `rgb(159,173,164)``\n``#a49fad` `rgb(164,159,173)``\n``#a8ad9f` `rgb(168,173,159)``\n``#9fada4` `rgb(159,173,164)``\n``#a49fad` `rgb(164,159,173)``\n``#ad9fa8` `rgb(173,159,168)``\n• #768a7d\n``#768a7d` `rgb(118,138,125)``\n• #83968a\n``#83968a` `rgb(131,150,138)``\n• #91a197\n``#91a197` `rgb(145,161,151)``\n``#9fada4` `rgb(159,173,164)``\n``#adb9b1` `rgb(173,185,177)``\n• #bbc4be\n``#bbc4be` `rgb(187,196,190)``\n• #c8d0cb\n``#c8d0cb` `rgb(200,208,203)``\nMonochromatic Color\n\nBelow, you can see some colors close to #9fada4. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n``#9fada1` `rgb(159,173,161)``\n``#9fada2` `rgb(159,173,162)``\n``#9fada3` `rgb(159,173,163)``\n``#9fada4` `rgb(159,173,164)``\n``#9fada5` `rgb(159,173,165)``\n``#9fada6` `rgb(159,173,166)``\n``#9fada8` `rgb(159,173,168)``\nSimilar Colors\n\nThis text has a font color of #9fada4.\n\n``<span style=\"color:#9fada4;\">Text here</span>``\n\nThis paragraph has a background color of #9fada4.\n\n``<p style=\"background-color:#9fada4;\">Content here</p>``\n\nThis element has a border color of #9fada4.\n\n``<div style=\"border:1px solid #9fada4;\">Content here</div>``\nCSS codes\n``.text {color:#9fada4;}``\n``.background {background-color:#9fada4;}``\n``.border {border:1px solid #9fada4;}``\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #080a09 is the darkest color, while #fefefe is the lightest one.\n\n• #080a09\n``#080a09` `rgb(8,10,9)``\n• #111412\n``#111412` `rgb(17,20,18)``\n• #1a1f1c\n``#1a1f1c` `rgb(26,31,28)``\n• #232a26\n``#232a26` `rgb(35,42,38)``\n• #2d342f\n``#2d342f` `rgb(45,52,47)``\n• #363f39\n``#363f39` `rgb(54,63,57)``\n• #3f4942\n``#3f4942` `rgb(63,73,66)``\n• #48544c\n``#48544c` `rgb(72,84,76)``\n• #515e56\n``#515e56` `rgb(81,94,86)``\n• #5a695f\n``#5a695f` `rgb(90,105,95)``\n• #637469\n``#637469` `rgb(99,116,105)``\n• #6c7e72\n``#6c7e72` `rgb(108,126,114)``\n• #75897c\n``#75897c` `rgb(117,137,124)``\n• #7f9286\n``#7f9286` `rgb(127,146,134)``\n• #8a9b90\n``#8a9b90` `rgb(138,155,144)``\n• #94a49a\n``#94a49a` `rgb(148,164,154)``\n``#9fada4` `rgb(159,173,164)``\n• #aab6ae\n``#aab6ae` `rgb(170,182,174)``\n• #b4bfb8\n``#b4bfb8` `rgb(180,191,184)``\n• #bfc8c2\n``#bfc8c2` `rgb(191,200,194)``\n• #c9d1cc\n``#c9d1cc` `rgb(201,209,204)``\n``#d4dad6` `rgb(212,218,214)``\n• #dee3e0\n``#dee3e0` `rgb(222,227,224)``\n• #e9ecea\n``#e9ecea` `rgb(233,236,234)``\n• #f4f5f4\n``#f4f5f4` `rgb(244,245,244)``\n• #fefefe\n``#fefefe` `rgb(254,254,254)``\nTint Color Variation\n\nA tone is produced by adding gray to any pure hue. In this case, #a6a6a6 is the less saturated color, while #54f88e is the most saturated one.\n\n• #a6a6a6\n``#a6a6a6` `rgb(166,166,166)``\n``#9fada4` `rgb(159,173,164)``\n• #98b4a2\n``#98b4a2` `rgb(152,180,162)``\n• #91bba0\n``#91bba0` `rgb(145,187,160)``\n• #8ac29e\n``#8ac29e` `rgb(138,194,158)``\n• #84c89c\n``#84c89c` `rgb(132,200,156)``\n• #7dcf9a\n``#7dcf9a` `rgb(125,207,154)``\n• #76d698\n``#76d698` `rgb(118,214,152)``\n• #6fdd96\n``#6fdd96` `rgb(111,221,150)``\n• #68e494\n``#68e494` `rgb(104,228,148)``\n• #61eb92\n``#61eb92` `rgb(97,235,146)``\n• #5bf190\n``#5bf190` `rgb(91,241,144)``\n• #54f88e\n``#54f88e` `rgb(84,248,142)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #9fada4 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# 狄莫弗的初步结果\n\n$$b(m)\\sim 2.168\\frac{(1-\\frac{1}{N})^N}{\\sqrt{N-1}}\\tag{5}$$\n$$\\text{log}(\\frac{b(m)}{b(m+d)})\\sim (m+d-\\frac{1}{2})\\text{log}(m+d-1)\\\\ +(m-d+\\frac{1}{2})\\text{log}(m-d+1)-2m\\text{log}m+\\text{log}(\\frac{m+d}{m})\\tag{6}$$\n\n（5）的结论现在我们已经在书上见不到了，因为我们有更好的近似结论了，而且(5)对狄莫弗后续研究也没啥作用，\n\n## 总结\n\n### 说点什么", null, "Subscribe" ]

[ null, "https://secure.gravatar.com/avatar/", null ]

[ "# Constant Variable\n\n#### blastStu\n\n##### New Member\nHi folks,\nI was wondering if you could help me understand a problem I have with a regression. Basically, SPSS excludes one of my variables because it claims that the variable is constant, but I just don't believe it!\n\nI've got four variables that I'm concerned with, all are binary:\n\nisSpeaking: 0 = No, 1 = Yes\nisNodding: 0 = No, 1 = Yes\nTaskRole: 0 = Instructor, 1 = Learner\nRecipientRole: 0 = Secondary, 1 = Primary.\n\nEach data point is a time point per person. isSpeaking and RecipientRole are exclusive, by that I mean if isSpeaking is 1, then there will be no data for RecipientRole (as a person can't be a speaker and a recipient at the same time here). If isSpeaking is 0, then there should be a value for RecipientRole. So a person may be speaking, nodding, a learner and no entry for recipient role.\n\nWhen I perform a correlation analysis there is an interaction between isNodding and isSpeaking (and lots of other interactions):", null, "isSpeaking is constant w.r.t. RecipientRole as expected.\n\nHowever, when I try to predict isNodding (in a binary logistic regression), the isSpeaking variable is excluded all together (see point b):", null, "However, it's just not constant for the selected cases! It's constant w.r.t to RecipientRole in that whenever there is a value in the RecipientRole variable there is a 0 in isSpeaking.. but surely this just means there's no interaction between them?\n\nAm I missing something with how a regression works here?\n\nThanks,\n\nRegards\nStuart\n\n#### blastStu\n\n##### New Member\nOK. I have part of the answer now. I didn't realise that if there is a missing value for a variable then SPSS excludes that entire case. In my situation the isSpeaking and RecipientRoles variables are such that, if isSpeaking is 0, then RecipientRole will be 1 or 0. If isSpeaking is 1, then there will be no value for Recipient Role.\n\nAs this is the case, SPSS is excluding all cases where isSpeaking = 1, and hence isSpeaking is always a constant 0.\n\nIs there a way to force SPSS to analyse all the data, even when there's a missing value in a datapoint?\n\nThanks,\nS\n\n#### Mean Joe\n\n##### TS Contributor\nI'm not much help since I don't know SPSS.\n\nBut is there a command to tell SPSS to analyze subjects \"where isSpeaking=0\"? Or can you make a subset, where isSpeaking=0, and then analyze that subset? Or just drop the isSpeaking variable from your analysis?" ]

[ null, "http://www.eecs.qmul.ac.uk/~stuart/moz-screenshot-15.png", null, "http://www.eecs.qmul.ac.uk/~stuart/moz-screenshot-16.png", null ]

[ "# Number 112\n\nNumber 112 spell 🔊, write in words: one hundred and twelve . Ordinal number 112th is said 🔊 and write: one hundred and twelfth. The meaning of number 112 in Maths: Is Prime? Factorization and dividers. The square root and cube root of 112. The meaning in computer science, numerology, codes and images, writing and naming in other languages, other interesting facts.\n\n## Useful information about number 112\n\nThe decimal (Arabic) number 112 converted to a Roman number is CXII. Roman and decimal number conversions.\n The number 112 converted to a Mayan number is", null, "Decimal and Mayan number conversions.\n\n#### Weight conversion\n\n112 kilograms (kg) = 246.9 pounds (lb)\n112 pounds (lb) = 50.8 kilograms (kg)\n\n#### Length conversion\n\n112 kilometers (km) equals to 69.594 miles (mi).\n112 miles (mi) equals to 180.247 kilometers (km).\n112 meters (m) equals to 180.247 feet (ft).\n112 feet (ft) equals 34.138 meters (m).\n112 centimeters (cm) equals to 44.1 inches (in).\n112 inches (in) equals to 284.5 centimeters (cm).\n\n#### Temperature conversion\n\n112° Fahrenheit (°F) equals to 44.4° Celsius (°C)\n112° Celsius (°C) equals to 233.6° Fahrenheit (°F)\n\n#### Power conversion\n\n112 Horsepower (hp) equals to 82.36 kilowatts (kW)\n112 kilowatts (kW) equals to 152.30 horsepower (hp)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n112 seconds equals to 1 minute, 52 seconds\n112 minutes equals to 1 hour, 52 minutes\n\n### Codes and images of the number 112\n\nNumber 112 morse code: .---- .---- ..---\nSign language for number 112:", null, "", null, "", null, "Images of the number Image (1) of the number Image (2) of the number", null, "", null, "There is no copyright for these images. Number 112 infographic. More images, other sizes, codes and colors ...\n\n### Gregorian, Hebrew, Islamic, Persian and Buddhist year (calendar)\n\nGregorian year 112 is Buddhist year 655.\nBuddhist year 112 is Gregorian year 431 a. C.\nGregorian year 112 is Islamic year -526 or -525.\nIslamic year 112 is Gregorian year 730 or 731.\nGregorian year 112 is Persian year -511 or -510.\nPersian year 112 is Gregorian 733 or 734.\nGregorian year 112 is Hebrew year 3872 or 3873.\nHebrew year 112 is Gregorian year 3648 a. C.\nThe Buddhist calendar is used in Sri Lanka, Cambodia, Laos, Thailand, and Burma. The Persian calendar is official in Iran and Afghanistan.\n\n## Share in social networks", null, "## Mathematics of no. 112\n\n### Multiplications\n\n#### Multiplication table of 112\n\n112 multiplied by two equals 224 (112 x 2 = 224).\n112 multiplied by three equals 336 (112 x 3 = 336).\n112 multiplied by four equals 448 (112 x 4 = 448).\n112 multiplied by five equals 560 (112 x 5 = 560).\n112 multiplied by six equals 672 (112 x 6 = 672).\n112 multiplied by seven equals 784 (112 x 7 = 784).\n112 multiplied by eight equals 896 (112 x 8 = 896).\n112 multiplied by nine equals 1008 (112 x 9 = 1008).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 112\n\nHalf of 112 is 56 (112 / 2 = 56).\nOne third of 112 is 37,3333 (112 / 3 = 37,3333 = 37 1/3).\nOne quarter of 112 is 28 (112 / 4 = 28).\nOne fifth of 112 is 22,4 (112 / 5 = 22,4 = 22 2/5).\nOne sixth of 112 is 18,6667 (112 / 6 = 18,6667 = 18 2/3).\nOne seventh of 112 is 16 (112 / 7 = 16).\nOne eighth of 112 is 14 (112 / 8 = 14).\nOne ninth of 112 is 12,4444 (112 / 9 = 12,4444 = 12 4/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 112\n\n#### Is Prime?\n\nThe number 112 is not a prime number. The closest prime numbers are 109, 113.\n112th prime number in order is 613.\n\n#### Factorization and dividers\n\nFactorization 2 * 2 * 2 * 2 * 7 = 112\nDivisors of number 112 are 1 , 2 , 4 , 7 , 8 , 14 , 16 , 28 , 56 , 112\nTotal Divisors 10.\nSum of Divisors 248.\n\n#### Powers\n\nThe second power of 1122 is 12.544.\nThe third power of 1123 is 1.404.928.\n\n#### Roots\n\nThe square root √112 is 10,583005.\nThe cube root of 3112 is 4,820285.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 112 = loge 112 = 4,718499.\nThe logarithm to base 10 of No. log10 112 = 2,049218.\nThe Napierian logarithm of No. log1/e 112 = -4,718499.\n\n### Trigonometric functions\n\nThe cosine of 112 is 0,455969.\nThe sine of 112 is -0,889996.\nThe tangent of 112 is -1,951877.\n\n### Properties of the number 112\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 112 in Computer Science\n\nCode typeCode value\nAscii112 ascii is character p\nUnix timeUnix time 112 is equal to Thursday Jan. 1, 1970, 12:01:52 a.m. GMT\nIPv4, IPv6Number 112 internet address in dotted format v4 0.0.0.112, v6 ::70\n112 Decimal = 1110000 Binary\n112 Decimal = 11011 Ternary\n112 Decimal = 160 Octal\n112 Decimal = 70 Hexadecimal (0x70 hex)\n112 BASE64MTEy\n112 MD57f6ffaa6bb0b408017b62254211691b5\n112 SHA1601ca99d55f00a2e8e736676b606a4d31d374fdd\n112 SHA224a4067cf5e783771c4a958ecc76f0a6976bc666e899e4c8cb40c6f7b0\n112 SHA384a701b50984e65e1d718b8fb937889d9e7586656155120c3c4bb4f0b119220bf6c421957c9e89afba502227cafb78440d\nMore SHA codes related to the number 112 ...\n\nIf you know something interesting about the 112 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 112\n\n### The meaning of the number 1 (one), numerology 1\n\nCharacter frequency 1: 2\n\nNumber one (1) came to develop or balance creativity, independence, originality, self-reliance and confidence in the world. It reflects power, creative strength, quick mind, drive and ambition. It is the sign of individualistic and aggressive nature.\nMore about the meaning of the number 1 (one), numerology 1 ...\n\n### The meaning of the number 2 (two), numerology 2\n\nCharacter frequency 2: 1\n\nThe number two (2) needs above all to feel and to be. It represents the couple, duality, family, private and social life. He/she really enjoys home life and family gatherings. The number 2 denotes a sociable, hospitable, friendly, caring and affectionate person. It is the sign of empathy, cooperation, adaptability, consideration for others, supersensitivity towards the needs of others. The number 2 (two) is also the symbol of balance, togetherness and receptivity. He/she is a good partner, colleague or companion; he/she also plays a wonderful role as a referee or mediator. Number 2 person is modest, sincere, spiritually influenced and a good diplomat. It represents intuition and vulnerability.\nMore about the meaning of the number 2 (two), numerology 2 ...\n\n## Interesting facts about the number 112\n\n### Asteroids\n\n• (112) Iphigenia is asteroid number 112. It was discovered by C. H. F. Peters from Litchfield Observatory, Clinton on 9/19/1870.\n\n### Aircrafts and flights\n\n• 05/05/1972 the Alitalia plane, flight number 112, McDonnell Douglas DC-8-43 crashed in Cinisi, Sicily, Italy 108 passengers and 7 crew members died.\n\n### Areas, mountains and surfaces\n\n• The total area of Bolshoy Lyakhovsky Island is 1,991 square miles (5,157 square km). Country Russia (Sakha Republic). 112th largest island in the world.\n\n### Distances between cities\n\n• There is a 112 miles (180 km) direct distance between Āgra (India) and Delhi (India).\n• There is a 112 miles (180 km) direct distance between Al Jīzah (Egypt) and Alexandria (Egypt).\n• There is a 112 miles (179 km) direct distance between Allahābād (India) and Lucknow (India).\n• There is a 70 miles (112 km) direct distance between Anshan (China) and Fushun (China).\n• There is a 112 miles (179 km) direct distance between Bareilly (India) and Meerut (India).\n• There is a 70 miles (112 km) direct distance between Dombivli (India) and Pune (India).\n• There is a 70 miles (112 km) direct distance between Foshan (China) and Yunfu (China).\n• There is a 112 miles (179 km) direct distance between Gorakhpur (India) and Chandigarh (India).\n• There is a 112 miles (179 km) direct distance between Goyang-si (South Korea) and Pyongyang (North Korea).\n• There is a 112 miles (179 km) direct distance between Los Angeles (USA) and San Diego (USA).\n• There is a 112 miles (180 km) direct distance between Mendoza (Argentina) and Santiago (Chile).\n• There is a 112 miles (180 km) direct distance between Mysore (India) and Salem (India).\n• There is a 112 miles (179 km) direct distance between Teni (India) and Thiruvananthapuram (India).\n\n### Games\n\n• Pokémon Rhydon (Saidon, Sidon) is a Pokémon number # 112 of the National Pokédex. Rhydon is ground and rock type Pokémon in the first generation. It is a Monster and Field egg group Pokémon. The other Rhydon indexes are Johto index 207 , Hoenn index 170 , Sinnoh index 187 , Kalos Centro index 051 - Coastal Zone Pokédex.\n\n### In other fields\n\n• 112 is the emergency telephone number in the European Union. Calls to this number are free.\n\n### Mathematics\n\n• 112 is the side of the smallest square that can be tiled with distinct integer-sided squares.\n\n### Science\n\n• Copernicium is the chemical element in the periodic table that has the symbol Cn and atomic number 112.\n\n## Number 112 in other languages\n\nHow to say or write the number one hundred and twelve in Spanish, German, French and other languages.\n Spanish: 🔊 (número 112) ciento doce German: 🔊 (Anzahl 112) einhundertzwölf French: 🔊 (nombre 112) cent douze Portuguese: 🔊 (número 112) cento e doze Chinese: 🔊 (数 112) 一百一十二 Arabian: 🔊 (عدد 112) مائة و اثنا عشر Czech: 🔊 (číslo 112) sto dvanáct Korean: 🔊 (번호 112) 백십이 Danish: 🔊 (nummer 112) ethundrede og tolv Hebrew: (מספר 112) מאה ושנים עשרה Dutch: 🔊 (nummer 112) honderdtwaalf Japanese: 🔊 (数 112) 百十二 Indonesian: 🔊 (jumlah 112) seratus dua belas Italian: 🔊 (numero 112) centododici Norwegian: 🔊 (nummer 112) en hundre og tolv Polish: 🔊 (liczba 112) sto dwanaście Russian: 🔊 (номер 112) сто двенадцать Turkish: 🔊 (numara 112) yüzoniki Thai: 🔊 (จำนวน 112) หนึ่งร้อยสิบสอง Ukrainian: 🔊 (номер 112) сто дванадцять Vietnamese: 🔊 (con số 112) một trăm mười hai Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 112 or any natural number (positive integer) please write us here or on facebook." ]

[ null, "https://numero.wiki/s/numeros-mayas/numero-maya-112.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-2.png", null, "https://numero.wiki/img/a-112.jpg", null, "https://numero.wiki/img/b-112.jpg", null, "https://numero.wiki/s/share-desktop.png", null ]

[ "", null, "", null, "Multiplication And Division Word Problems 4Th Grade Pdf. Fastt math is proven effective for struggling students. If the nurses are equally assigned to the doctors, how many nurses are assigned to most.\n\nThese math word problems may require multiplication or division to solve. The following collection of free 4th grade maths word problems worksheets cover topics including addition, subtraction, multiplication, division, mixed operations, fractions, and decimals. Each nurse is assigned to assist one doctor.\n\n### We Have 100 Pictures About Multiplication Word Problem Worksheets 3Rd Grade Like Multiplication Word Problem Worksheets 3Rd Grade, Multiplication Word Problems 2 Interactive Worksheet And Also 428 Addition Worksheets For You To Print Right Now.\n\nUse multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Multiplying numbers in columns is a math skill which requires lots of practice. Kids use details from the word problems on this third grade math worksheet to construct and solve multiplication problems.\n\n### The Following Collection Of Free 4Th Grade Maths Word Problems Worksheets Cover Topics Including Addition, Subtraction, Multiplication, Division, Mixed Operations, Fractions, And Decimals.\n\nThese worksheets, sorted by grade level, cover a mix of skills from the curriculum. Math worksheet based on mixed operations division and Each pizza has 4 slices." ]

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[ "# Cost of Goods Sold\n\nCost of goods sold (COGS) denotes the carrying value of goods and raw materials sold to customers during specific period.\n\n### Calculator of Cost of Goods Sold\n\n Select Currency = Beginning inventory = Purchases = Ending inventory = COGS =\n\n### Formula of Cost of Goods Sold\n\nCost of goods sold = Beginning inventory + Purchases – Ending inventory\n\n### Example of Cost of Goods Sold\n\nBeginning inventory of a company was \\$16000 and the company purchased new inventory for the cost of \\$5000. Ending inventory of the company was \\$10000.Calculate the cost of goods sold in a year.\n\n#### Given:\n\nBeginning inventory = \\$16000\nPurchases = \\$5000\nEnding inventory = \\$10000\n\n### To Find Cost of goods sold\n\n#### Solution:\n\nCOGS = Beginning inventory + Purchases – Ending inventory\n= 16000 + 5000 – 10000\n= 21000 – 10000\n= \\$11000", null, "" ]

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[ "# scipy.interpolate.KroghInterpolator¶\n\nclass scipy.interpolate.KroghInterpolator(xi, yi, axis=0)[source]\n\nInterpolating polynomial for a set of points.\n\nThe polynomial passes through all the pairs (xi,yi). One may additionally specify a number of derivatives at each point xi; this is done by repeating the value xi and specifying the derivatives as successive yi values.\n\nAllows evaluation of the polynomial and all its derivatives. For reasons of numerical stability, this function does not compute the coefficients of the polynomial, although they can be obtained by evaluating all the derivatives.\n\nParameters: xi : array_like, length N Known x-coordinates. Must be sorted in increasing order. yi : array_like Known y-coordinates. When an xi occurs two or more times in a row, the corresponding yi’s represent derivative values. axis : int, optional Axis in the yi array corresponding to the x-coordinate values.\n\nNotes\n\nBe aware that the algorithms implemented here are not necessarily the most numerically stable known. Moreover, even in a world of exact computation, unless the x coordinates are chosen very carefully - Chebyshev zeros (e.g. cos(i*pi/n)) are a good choice - polynomial interpolation itself is a very ill-conditioned process due to the Runge phenomenon. In general, even with well-chosen x values, degrees higher than about thirty cause problems with numerical instability in this code.\n\nBased on [R44].\n\nReferences\n\n [R44] (1, 2) Krogh, “Efficient Algorithms for Polynomial Interpolation and Numerical Differentiation”, 1970.\n\nExamples\n\nTo produce a polynomial that is zero at 0 and 1 and has derivative 2 at 0, call\n\n>>> KroghInterpolator([0,0,1],[0,2,0])\n\n\nThis constructs the quadratic 2*X**2-2*X. The derivative condition is indicated by the repeated zero in the xi array; the corresponding yi values are 0, the function value, and 2, the derivative value.\n\nFor another example, given xi, yi, and a derivative ypi for each point, appropriate arrays can be constructed as:\n\n>>> xi_k, yi_k = np.repeat(xi, 2), np.ravel(np.dstack((yi,ypi)))\n>>> KroghInterpolator(xi_k, yi_k)\n\n\nTo produce a vector-valued polynomial, supply a higher-dimensional array for yi:\n\n>>> KroghInterpolator([0,1],[[2,3],[4,5]])\n\n\nThis constructs a linear polynomial giving (2,3) at 0 and (4,5) at 1.\n\nMethods\n\n __call__(x) Evaluate the interpolant :Parameters: x : array_like Points to evaluate the interpolant at. derivative(x[, der]) Evaluate one derivative of the polynomial at the point x :Parameters: x : array_like Point or points at which to evaluate the derivatives der : integer, optional Which derivative to extract. derivatives(x[, der]) Evaluate many derivatives of the polynomial at the point x Produce an array of all derivative values at the point x.\n\n#### Previous topic\n\nscipy.interpolate.BarycentricInterpolator.set_yi\n\n#### Next topic\n\nscipy.interpolate.KroghInterpolator.__call__" ]

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[ " mongodb中使用lookup联合查询及分组group排序sort_双黑个人网站\n\n# mongodb中使用lookup联合查询及分组group排序sort\n\n0 0条评论\n\npaper表\n\n``````const schema = baseSchema().add({\n/**\n* 会员ID\n*/\nuser: {\ntype: mongoose.Schema.Types.ObjectId,\nref: 'user',\nrequired: true\n},\n/**\n* 得分\n*/\nscore: { type: Number, required: true, index: true },\n/**\n* 所用时间，单位秒\n*/\ntime: { type: Number, required: true, index: true },\n/**\n* 答题时间\n*/\ncreatetime: { type: Number }\n})``````\n\nuser表\n\n(两张表实际项目中列肯定不止那么少，我只是找几个重点罗列)\n\n`````` Model.aggregate().lookup({\nfrom: 'users', localField: 'user',\nforeignField: '_id', as: 'users'\n})\n.sort('-score time createtime')\n.group({\n_id: '\\$user', score: { \\$first: '\\$score' }, time: { \\$first: '\\$time' },\ncreatetime: { \\$first: '\\$createtime' }, users: { \\$first: '\\$users' }\n})\n.project({\n_id: 0, score: 1, time: 1, createtime: 1,\nnickname: { \\$min: '\\$users.nickname' }\n})\n.sort('-score time createtime').limit(10)``````" ]

[ null ]

[ "# Show statistical significance (or not) between means\n\nSuppose I have two experiments A and B. Each experiment is composed of N elements that are clearly separated into two groups. In experiment B one of the groups has a higher mean than the same group in experiment A. This difference (shown in red in the image) is what I am trying to understand. I need to show that it is significant, or not, determined by any value. p-value? a test?", null, "To show statistical significance (that is, to show that the results of your experiments are not just due to chance), you first need to define your hypotheses (the null and the alternative hypothesis). In this case, the null hypothesis, $$H_0$$, could be \"There is no difference between the mean of the groups in experiments $$A$$ and $$B$$\". The alternative hypothesis, $$H_a$$, could be \"There is a difference\".\nYou now perform a statistical test. If the resulting p-value (returned by the statistical test) is less than a threshold value (often called the \"significance level\" and it is e.g. $$0.05$$), then you reject the null hypothesis and you say that this result (that is, the p-value less than the significance level) is statistical significant." ]

[ null, "https://i.stack.imgur.com/V8b1j.png", null ]

[ "# What is the Maximum Likelihood Estimate (MLE)?\n\nProbabilistic Models help us capture the inherant uncertainity in real life situations. Examples of probabilistic models are Logistic Regression, Naive Bayes Classifier and so on..  Typically we fit (find parameters) of such probabilistic models from the training data, and estimate the parameters. The learnt model can then be used on unseen data to make predictions.\n\nOne way to find the parameters of a probabilistic model (learn the model) is to use the MLE estimate or the maximum likelihood estimate.\n\nMaximum likelihood estimate is that value for the parameters that maximizes the likelihood of the data.\n\nWhat are some examples of the parameters of models we want to find? What exactly is the likelihood? How do we find parameters that maximize the likelihood? These are some questions answered by the video.", null, "##### What are some examples of parameters we might want to estimate?\n\nConsider the Gaussian distribution. The mathematical form of the pdf is shown below. The parameters of the Gaussian distribution are the mean and the variance (or the standard deviation). Given a set of points, the MLE estimate can be used to estimate the parameters of the Gaussian distribution.", null, "", null, "Consider the Bernoulli distribution. A good example to relate to the Bernoulli distribution is modeling the probability of heads (p) when we toss a coin. The probability of heads is p, the probability of tails is (1-p). By observing a bunch of coin tosses, one can use the maximum likelihood estimate to find the value of p.\n\n##### What is the likelihood of a probabilistic model?\n\nThe likelihood is the joined probability distribution of the observed data given the parameters.\n\nFor instance, if we consider the Bernoulli distribution for a coin toss with probability of heads as p. Suppose we toss the coin four times, and get H, T, T, H.\n\nThe likelihood of the observed data is the joint probability distribution of the observed data. Hence:", null, "##### How do we find the parameter that maximizes likelihood?\n\nThe MLE estimator is that value of the parameter which maximizes likelihood of the data. This is an optimization problem.", null, "For instance for the coin toss example, the MLE estimate would be to find that p such that  p (1-p) (1-p) p is maximized.", null, "" ]

[ null, "https://machinelearninginterview.com/wp-content/uploads/2021/03/MLE_Estimate_Thumbnail.png", null, "https://machinelearninginterview.com/wp-content/uploads/2021/03/normal_dist_formula-e1616395155211-300x172.jpg", null, "https://machinelearninginterview.com/wp-content/uploads/2021/03/normal-distribution-probability-300x108.jpg", null, "https://machinelearninginterview.com/wp-content/uploads/2021/03/bernoulli_distribution.png", null, "https://machinelearninginterview.com/wp-content/uploads/2021/03/picture_mle.jpeg", null, "https://machinelearninginterview.com/wp-content/uploads/2021/03/finding_MLE_coin_toss.png", null ]

[ "# Ben - a Light Following Breadboard Arduino Robot\n\n11,269\n\n29\n\n15\n\n## Introduction: Ben - a Light Following Breadboard Arduino Robot\n\nBen the Light Following Breadboard Arduino Robot is the second robot I have made to help teach robotics to high school students in a class I teach voluntarily. The first robot also has its own Instructable which can be found here: https://www.instructables.com/id/James-Your-first-Arduino-Robot/\nThe video shows the construction process however it is in fast motion and I will cover some of the more complicated things here more in depth.\n\nSo to make ben you will need the following components:\n·         A small sheet of acrylic\n·         An Arduino Nano\n·         Two Continuous Rotation Servos\n·         Two hobby wheels (I used model aeroplane wheels)\n·         A castor wheel\n·         9V battery\n·         4.8V rechargeable battery pack (or just 4 AA’s in a battery case)\n·         Two Light Dependant Resistors\n·         Two 10,000 ohm Resistors\n·         Double Sided Foam Tape\n·         A Power Switch (not necessary but would be handy)\n\nYou won’t need any tools for the construction however you could replace the Double Sided Foam Tape with Hot Glue in which case you would need a Hot Glue Gun.\n\nNow the first thing that may need further explanation is the use of the Light Dependant Resistors. Light Dependant Resistors (or LDR’s) are resistors whose value changes depending on the amount of ambient light, but how can we detect resistance with Arduino? Well you can’t really, however you can detect voltage levels using the analog pins, which can measure (in basic use) between 0-5V. Now you may be asking “Well how do we convert resistance values into voltage changes?”, it’s simple, we make a voltage divider. A voltage divider takes in a voltage and then outputs a fraction of that voltage proportional to the input voltage and the ratio of the two values of resistors used. The equation for which is:\nOutput Voltage = Input Voltage * ( R2 / (R1 + R2) )\nWhere R1 is the value of the first resistor and R2 is the value of the second.\n\nThe circuit schematic for which looks like this\nA diagram of this in our situation looks a little something like this\n\nNow this still begs the question “But what resistance values does the LDR have?”, good question. The less amount of ambient light the higher the resistance, more ambient light means a lower resistance. Now for the particular LDR’s I used their resistance range was from 200 – 10 kilo ohms, but this changes for different ones so make sure to look up where you bought them from and try to find a datasheet or something of the sort.\n\nNow in this case R1 is actually our LDR, so let’s bring back that equation and do some math-e-magic (mathematical electrical magic).\n\nNow first we need to convert those kilo ohm values into ohms:\n200 kilo-ohms = 200,000 ohms\n10 kilo-ohms = 10,000 ohms\nSo to find what the output voltage is when we are in pitch black we plug in the following numbers:\n5 * ( 10000 / (200000 + 10000) )\nThe input is 5V as that is what we are getting from the Arduino.\nThe above gives 0.24V (rounded off).\nNow we find what the output voltage is in peak brightness by using the following numbers:\n5 * ( 10000 / (10000 + 10000) )\nAnd this gives us 2.5V exactly.\n\nSo these are the voltage values that we are going to get into the Arduino’s analog pins, but these are not the values that will be seen in the program, “But why?” you may ask. The Arduino uses an Analog to Digital Chip which converts the analog voltage into usable digital data. Unlike the digital pins on the Arduino that can only read a HIGH or LOW state being 0 and 5V the analog pins can read from 0-5V and convert this into a number range of 0-1023.\n\nNow with some more math-e-magic we can actually calculate what values the Arduino will actually read. Because this will be a linear function we can use the following formula:\nY = mX + C\nWhere; Y = Digital Value\nWhere; m = slope, (rise / run), (digital value / analog value)\nWhere; C = Y intercept\nThe Y intercept is 0 so that gives us:\nY = mX\nm = 1023 / 5 = 204.6\nTherefore:\nDigital value = 204.6 * Analog value\n\nSo in pitch black the digital value will be:\n204.6 * 0.24\nWhich gives approximately 49.\n\nAnd in peak brightness it will be:\n204.6 * 2.5\nWhich gives approximately 511.\n\nNow with two of these set up on two analog pins we can create two integer variables to store their values two and do comparison operators to see which one has the lowest value, turning the robot in that direction.\n____________________________________________________________________________________________________\n\nNow that was probably the most complex thing about the whole robot build however there is just one more thing that I would like to mention and it’s to do with using servos with Arduino.\n\nThere are several tutorials and diagrams on the internet showing that you must connect the voltage in of the servo up to the 5V rail of the Arduino and the ground of the servo to the ground of the Arduino, this is dangerous! Servos can draw a lot of current, and in most cases that current draw will be more than the voltage regulator on the Arduino can supply, this will lead to bad things happening. The proper way to hook servos to your Arduino is to use an external power supply. In Bens cause I am running the continuous rotation servos of a 4.8V rechargeable Ni-Cd battery pack, this is ideal as the servos operate well from 4.8-6V, 6V being the peak charge voltage of the battery pack.\n\nNow you may be tempted to just hook up V+ of the battery to the V+ of the servos and the GND of the battery pack to the GND of servos and the signal pins to the Arduino, this won’t work either! You need to remember that electricity needs to flow from one ‘point’ back to its original point, not connecting the ground of the servos and battery pack to the Arduino’s ground won’t allow the electricity to flow from the signal pins.\n\nHere is a diagram showing the proper circuitry\n_____________________________________________________________________________________________________\n\nCombining the two diagrams shown earlier gives the complete circuitry required to make Ben.\nNow I won’t explain the code as it is heavily commented and should pretty much explain itself.\nCode", null, "Participated in the\nToy Contest", null, "Participated in the\nKit Contest", null, "Participated in the\nArduino Contest\n\n## Recommendations\n\nArduino: 1.6.7 (Windows 10), Board: \"Arduino Nano, ATmega168\"\n\nBuild options changed, rebuilding all\nMultiple libraries were found for \"Servo.h\"\nUsed: C:\\Users\\Shane\\Documents\\Arduino\\libraries\\Servo\nNot used: D:\\Shane\\Arduino\\libraries\\Servo\n\nSketch uses 2,552 bytes (17%) of program storage space. Maximum is 14,336 bytes.\nGlobal variables use 65 bytes (6%) of dynamic memory, leaving 959 bytes for local variables. Maximum is 1,024 bytes.\navrdude: ser_open(): can't open device \"\\\\.\\COM1\": The system cannot find the file specified." ]

[ null, "https://www.instructables.com/assets/img/pixel.png", null, "https://www.instructables.com/assets/img/pixel.png", null, "https://www.instructables.com/assets/img/pixel.png", null ]

[ "# wakefield: Generate Random Data Sets", null, "``````> library(wakefield)\n``````\n``````\nAttaching package: 'wakefield'\n\nThe following object is masked from 'package:raster':\n\narea\n``````\n\nバージョン: 0.2.0\n\n`age` Generate Random Vector of Ages\n`animal` Generate Random Vector of animals\n`animal_list` Animal List\n`answer` Generate Random Vector of Answers (Yes/No)\n`area` Generate Random Vector of Areas\n`as_integer` Convert a Factor Data Frame to Integer\n`car` Generate Random Vector of Cars\n`children` Generate Random Vector of Number of Children\n`coin` Generate Random Vector of Coin Flips\n`color` Generate Random Vector of Colors\n`date_stamp` Generate Random Vector of Dates\n`death` Generate Random Vector of Deaths Outcomes\n`dice` Generate Random Vector of Dice Throws\n`dna` Generate Random Vector of DNA Nucleobases\n`dob` Generate Random Vector of Birth Dates\n`dummy` Generate Random Dummy Coded Vector\n`education` Generate Random Vector of Educational Attainment Level\n`employment` Generate Random Vector of Employment Statuses\n`eye` Generate Random Vector of Eye Colors\n`grade` Generate Random Vector of Grades\n`grade_level` Generate Random Vector of Grade Levels\n`grady_augmented` Augmented List of Grady Ward's English Words and Mark Kantrowitz's Names List\n`group` Generate Random Vector of Control/Treatment Groups\n`hair` Generate Random Vector of Hair Colors\n`height` Generate Random Vector of Heights\n`hour` Generate a Random Sequence of H:M:S Times\n`id` Identification Numbers\n`income` Generate Random Gamma Vector of Incomes\n`internet_browser` Generate Random Vector of Internet Browsers\n`interval` Cut Numeric Into Factor\n`iq` Generate Random Vector of Intelligence Quotients (IQs)\n`language` Generate Random Vector of Languages\n`languages` Languages of the World\n`level` Generate Random Vector of Levels\n`likert` Generate Random Vector of Likert-Type Responses\n`lorem_ipsum` Generate Random Lorem Ipsum Strings\n`marital` Generate Random Vector of Marital Statuses\n`military` Generate Random Vector of Military Branches\n`minute` Generate a Random Sequence of Minutes in H:M:S Format\n`month` Generate Random Vector of Months\n`name` Generate Random Vector of Names\n`name_neutral` Gender Neutral Names\n`normal` Generate Random Normal Vector\n`peek` Data Frame Viewing\n`plot.tbl_df` Plots a tbl_df Object\n`political` Generate Random Vector of Political Parties\n`presidential_debates_2012` 2012 U.S. Presidential Debate Dialogue\n`print.available` Prints an available Object.\n`print.variable` Prints a variable Object\n`probs` Generate a Random Vector of Probabilities.\n`r_data` Pre-Selected Column Data Set\n`r_data_frame` Data Frame Production (From Variable Functions)\n`r_dummy` Title\n`r_insert` Insert Data Frames Into 'r_data_frame'\n`r_list` List Production (From Variable Functions)\n`r_na` Title\n`r_sample` Generate Random Vector\n`r_sample_binary` Generate Random Binary Vector\n`r_sample_factor` Generate Random Factor Vector\n`r_sample_integer` Generate Random Integer Vector\n`r_sample_logical` Generate Random Logical Vector\n`r_sample_ordered` Generate Random Ordered Factor Vector\n`r_sample_replace` Generate Random Vector (Without Replacement)\n`r_series` Data Frame Series (Repeated Measures)\n`race` Generate Random Vector of Races\n`relate` Create Related Numeric Columns\n`religion` Generate Random Vector of Religions\n`sat` Generate Random Vector of Scholastic Aptitude Test (SATs)\n`second` Generate a Random Sequence of Seconds in H:M:S Format\n`sentence` Generate Random Vector of Sentences\n`seriesname` Add Internal Name to Data Frame\n`sex` Generate Random Vector of Genders\n`sex_inclusive` Generate Random Vector of Non-Binary Genders\n`smokes` Generate Random Logical Smokes Vector\n`speed` Generate Random Vector of Speeds\n`state` Generate Random Vector of states\n`state_populations` State Populations (2010)\n`string` Generate Random Vector of Strings\n`table_heat` View Data Table Column Types as Heat Map\n`time_stamp` Generate a Random Sequence of Times in H:M:S Format\n`upper` Generate Random Letter Vector\n`valid` Generate Random Logical Vector\n`variables` Available Variable Functions\n`varname` Add Internal Name to Vector\n`wakefield` Generate Random Data Sets\n`year` Generate Random Vector of Years\n`zip_code` Generate Random Vector of Zip Codes\n\n## age\n\n``````> age(10)\n``````\n`````` 28 28 34 21 34 29 24 33 32 23\n``````" ]

[ null, "http://www.r-pkg.org/badges/version/wakefield", null ]

[ "# Chapter 11", null, "Chapter 11 The Cost of Capital Learning Objectives 1. Understand the different kinds of financing available to a company: debt financing, equity financing, and hybrid equity financing. 2. Understand the debt and equity components of the weighted average cost of capital (WACC) and explain the tax implications on debt financing and the adjustment to the WACC. 3. Calculate the weights of the components using book values or market values. 4. Explain how the WACC is used in capital budgeting models 5. Determine the beta of a project and its implications in capital budgeting problems. 6. Select optimal project combinations for a companys portfolio of acceptable potential projects.\n\n11-2 2013 Pearson Education, Inc. All rights reserved. 11.1 The Cost of Capital: A Starting Point 3 broad sources of financing available or raising capital: debt, common stock (equity), and preferred stock (hybrid equity). Each has its own risk and return profile and therefore its own rate of return required by investors to provide funds to the firm. Figure 11.1 Component sources of capital. 11-3 2013 Pearson Education, Inc. All rights reserved.\n\n11.1 The Cost of Capital: A Starting Point The weighted average cost of capital (WACC) is estimated by multiplying each component weight by the component cost and summing up the products. The WACC is essentially the minimum acceptable rate of return that the firm should earn on its investments of average risk, in order to be profitable. WACC discount rate for computing NPV IRR > WACC for acceptance of project. 11-4 2013 Pearson Education, Inc. All rights reserved. 11.1 The Cost of Capital: A Starting Point (continued) Example 1: Measuring weighted average cost of a mortgage\n\nJim wants to refinance his home by taking out a single mortgage and paying off all the other sub-prime and prime mortgages that he took on while the going was good. Listed below are the balances and rates owed on each of his outstanding home-equity loans and mortgages: Lender Balance First Cut-Throat Bank Rate \\$ 150,000 7.5% Second Considerate Bank \\$ 35,000 8.5% Third Pawn Mortgage Co.\n\n\\$ 15,000 9.5% Below what rate would it make sense for Jim to consolidate all these loans and refinance the whole amount? 11-5 2013 Pearson Education, Inc. All rights reserved. 11.1 The Cost of Capital: A Starting Point (continued) Example 1 Answer Jims weighted average cost of borrowing = Proportion of each loan * Rate (10,000/200,000)*.075+(35,000/200,000) *.085+(15,000/200,000)*.095 (.75*.075) + (.175*.085) (+.075*.095) = .07825 or 7.825%\n\nJims average cost of financing his home is 7.825%. Any rate below 7.825% would be beneficial. 11-6 2013 Pearson Education, Inc. All rights reserved. 11.2 Components of the Weighted Average Cost of Capital To determine a firms WACC we need to know how to calculate: 1. the relative weights and 2. costs of the debt, preferred stock, and common stock of a firm. 11-7 2013 Pearson Education, Inc. All rights reserved.\n\n11.2 (A) Debt Component The cost of debt (Rd) is the rate that firms have to pay when they borrow money from banks, finance companies, and other lenders. It is essentially measured by calculating the yield to maturity (YTM) on a firms outstanding bonds, as covered in Chapter 6. Although best solved for by using a financial calculator or spreadsheet, the YTM can also be figured out as follows: 11-8 2013 Pearson Education, Inc. All rights reserved. 11.2 (A) Debt Component (continued) YTM on outstanding bonds, indicates what investors require for lending the firm their\n\nmoney in current market conditions. However, new debt would also require payment of transactions costs to investment bankers reducing the net proceeds to the issuer and raising the cost of debt. We must adjust the market price by the amount of commissions that would have to be paid when issuing new debt, and then calculate the YTM. 11-9 2013 Pearson Education, Inc. All rights reserved. 11.2 (A) Debt Component (continued) Example 2: Calculating the cost of debt Kelloggs wants to raise an additional \\$3,000,000 of debt as part of the capital that would be needed to expand their operations into the Morning Foods sector.\n\nThey were informed by their investment banking consultant that they would have to pay a commission of 3.5% of the selling price on new issues. Their CFO is in the process of estimating the corporations cost of debt for inclusion into the WACC equation. The company currently has an 8%, AA-rated, non-callable bond issue outstanding, which pays interest semi-annually, will mature in 17 years, has a \\$1000 face value, and is currently trading at \\$1075. Calculate the appropriate cost of debt for the firm. 11-10 2013 Pearson Education, Inc. All rights reserved. 11.2 (A) Debt Component (continued) Example 2 Answer First determine the net proceeds on each bond\n\n= Selling price Commission =\\$1075-(.035*1075) = \\$1037.38 Using a financial calculator we enter: P/Y = C/Y = 2 Input 34 ? -1037.38 40 1000 Key N I/Y PV PMT FV Output 7.60% The appropriate cost of debt for Kelloggs is 7.6% 11-11 2013 Pearson Education, Inc. All rights reserved. 11.2 (B) Preferred Stock Component Preferred stock holders receive a constant dividend with no maturity point; The cost of preferred (Rp)can be estimated by dividing the annual dividend by the net\n\nproceeds (after floatation cost) per share of preferred stock: Rp = Dp/Net price 11-12 2013 Pearson Education, Inc. All rights reserved. 11.2 (B) Preferred Stock Component (continued) Example 3: Cost of Preferred Stock Kelloggs will also be issuing new preferred stock worth \\$1 million. They will pay a dividend of \\$4 per share which has a market price of \\$40. The floatation cost on preferred will amount to \\$2 per share. What is their cost of preferred stock? Answer Net price on preferred stock = \\$38; Dividend on preferred = \\$4\n\nCost of preferred = Rp = \\$4/\\$38 = 10.53% 11-13 2013 Pearson Education, Inc. All rights reserved. 11.2 (C) Equity Component The cost of equity (Re) is essentially the rate of return that investors are demanding or expecting to make on money invested in a companys common stock. The cost of equity can be estimated by using either the SML approach (covered in Chapter 8) or the Dividend Growth Model (covered in Chapter 7). 11-14 2013 Pearson Education, Inc. All rights reserved.\n\n11.2 (C) Equity Component (continued) The Security Market Line Approach: calculates the cost of equity as a function of the risk-free rate (rf) the market riskpremium [E(rm)-rf], and beta (i). That is, 11-15 2013 Pearson Education, Inc. All rights reserved. 11.2 (C) Equity Component (continued) Example 4: Calculating Cost of Equity with the SML equation Remember Kelloggs from the earlier 2 examples? Well, to reach their desired capital structure their CEO has decided to utilize all of their expected retained earnings in the coming\n\nquarter. Kelloggs beta is estimated at 0.65 by Value Line. The risk-free rate is currently 4%, and the expected return on the market is 15%. How much should the CEO put down as one estimate of the companys cost of equity? Answer Re = rf + [E(rm)-rf]i Re=4%+[15%-4%]0.65 Re= 4%+7.15% = 11.15% 11-16 2013 Pearson Education, Inc. All rights reserved. 11.2 (C) Equity Component (continued) The Dividend Growth Approach to Re: The Gordon Model, introduced in Chapter 7, is used to calculate the price of a constant growth stock. However, with some algebraic manipulation it can be\n\ntransformed into Equation 11.6, which calculates the cost of equity, as shown below: where Div0 = last paid dividend per share; Po = Current market price per share; and g = constant growth rate of dividend. 11-17 2013 Pearson Education, Inc. All rights reserved. 11.2 (C) Equity Component (continued) For newly issued common stock, the price must be adjusted for floatation cost (commission paid to investment banker) as shown in Equation 11.7 below: where F is the floatation cost in percent.\n\n11-18 2013 Pearson Education, Inc. All rights reserved. 11.2 (C) Equity Component (continued) Example 5: Applying the Dividend Growth Model to calculate Re Kelloggs common stock is trading at \\$45.57 and its dividends are expected to grow at a constant rate of 6%. The company paid a dividend last year of \\$2.27. If the company issues stock they will have to pay a floatation cost per share equal to 5% of selling price. Calculate Kelloggs cost of equity with and without floatation costs.\n\n11-19 2013 Pearson Education, Inc. All rights reserved. 11.2 (C) Equity Component (continued) Example 5 Answer Cost of equity without floatation cost: Re = (Div0*(1+g)/Po) + g (\\$2.27*(1.06)/\\$45.57)+.0611.28% Cost of equity with floatation cost: Re = [\\$2.27*(1.06)/(45.57*(1-.05)]+.06 11.56% 11-20 2013 Pearson Education, Inc. All rights reserved. 11.2 (C) Equity Component (continued)\n\nDepending on the availability of data, either of the two models, or both, can be used to estimate Re. With two values, the average can be used as the cost of equity. For example, in Kelloggs case we have (11.15%+11.28%)/211.22% (without floatation costs) or (11.15%+11.56%) /211.36%(with floatation costs) 11-21 2013 Pearson Education, Inc. All rights reserved. 11.2 (D) Retained Earnings Retained earnings does have a cost, i.e. the opportunity cost for the shareholders not being able to invest the money themselves. The cost of retained earnings can be calculated by\n\nusing either of the above two approaches, without including floatation cost. Also, since interest expenses are tax deductible, the cost of debt, must be adjusted for taxes, as shown below, prior to including it in the WACC calculation: After-tax cost of debt = Rd*(1-Tc) So if the YTM (with floatation cost) = 7.6%, and the companys marginal tax rate is 30%, the after-tax cost of debt7.6%*(1-3)5.32% 11-22 2013 Pearson Education, Inc. All rights reserved. 11.3 Weighting the Components: Book Value or Market Value? To calculate the WACC of a firm, each components cost is multiplied by its proportion in the capital mix and then\n\nsummed up. There are two ways to determine the proportion or weights of each capital component, using book value, or using market values. 11-23 2013 Pearson Education, Inc. All rights reserved. 11.3 (A) Book Value Book value weights can be determined by taking the balance sheet values for debt, preferred stock, and common stock, adding them up, and dividing each by the total. These weights, however, do not indicate the current proportion of each component. 11-24\n\n2013 Pearson Education, Inc. All rights reserved. 11.3 (B) Adjusted Weighted Average Cost of Capital Equation 11.9 can be used to combine all the weights and component costs into a single average cost which can be used as the firms discount or hurdle rate: 11-25 2013 Pearson Education, Inc. All rights reserved. 11.3 (B) Adjusted Weighted Average Cost of Capital (continued) Example 6: Calculating Adjusted WACC Using the market value weights and the component costs determined earlier, calculate Kelloggs adjusted WACC.\n\nCapital Component Weight After-tax Cost% Debt .38 7.6%*(1-.3) =5.32% Rd (1-Tc) Preferred Stock.14 10.53% Rp Common Stock .48 11.36%* Re *using average of SML and Div. Growth Model (with floatation cost) Answer WACC = .38*5.32% + .14*10.53%+.48*11.36% =2.02%+1.47%+5.45%=8.94% 11-26 2013 Pearson Education, Inc. All rights reserved. 11.3 (C) Market Value\n\nMarket value weights are determined by taking the current market prices of the firms outstanding securities and multiplying them by the number outstanding, to get the total value; and then dividing each by the total market value to get the proportion or weight of each If possible, market value weights should be used since they are a better representation of a companys current capital structure, which would be relevant for raising new capital. 11-27 2013 Pearson Education, Inc. All rights reserved. 11.3 (C) Market Value (continued) Example 7: Calculating capital component weights: Kelloggs CFO is in the process of determining the firms WACC and needs to figure out the weights of the various types of capital sources.\n\nAccordingly, he starts by collecting information from the balance sheet and the capital markets, and makes up the Table shown below: Component Debt Preferred Stock Common Stock Balance Sheet Value \\$ 150,000,000 \\$ 45,000,000 \\$ 180,000,000 Number outstanding 150,000 1,500,000 4,500,000\n\nWhat should he do next? 11-28 2013 Pearson Education, Inc. All rights reserved. Current Market Price \\$1,075 \\$40 \\$45.57 Market Value \\$161,250,000 \\$ 60,000,000 \\$205,065,000 11.3 (C) Market Value (continued) Example 7 Answer\n\n1) Calculate the total book value and total market value of the capital 2) Divide each components book value and market value by their respective totals. Total Book Value = \\$375,000,000; Total Market Value = \\$426,315,000 Book Value Weights: Market Value Weights: Debt = \\$150m/\\$375m=40%; Debt = \\$161.25m/\\$426.32m=38% P/ S=\\$45m/\\$375m=12%; P/S = \\$60m/\\$426.32=14% C/S = \\$180m/\\$375m=48%; C/S= \\$205.07m/\\$426.32m=48% (Rounded to nearest whole number) He should use the market value weights as they represent a more current picture of the firms capital structure. 11-29 2013 Pearson Education, Inc. All rights reserved.\n\n11.4 Using the Weighted Average Cost of Capital in a Budgeting Decision Once a firms WACC has been determined, it can be used either as the discount rate to calculate the NPV of the projects expected cash flow or as the hurdle rate which must be exceeded by the projects IRR. Table 11.1 presents the incremental cash flow of a \\$5 million project being considered by a firm whose WACC is 12%. Table 11.1 Incremental Cash Flow of a \\$5 Million Project 11-30 2013 Pearson Education, Inc. All rights reserved. 11.4 Using the Weighted Average Cost of Capital in a Budgeting Decision (continued)\n\nUsing a discount rate of 12%, the projects NPV would be determined as follows: Since the NPV > 0 this would be an acceptable project. Alternatively, the IRR could be determined using a financial calculator14.85% Again, since IRR>12%, this would be an acceptable project. 11-31 2013 Pearson Education, Inc. All rights reserved. 11.4 (A) Individual Weighted Average Cost of Capital for Individual Projects Using the WACC for evaluating projects assumes that the project is of average risk. If projects have varying risk levels, using the same discount rate could lead to incorrect\n\ndecisions. 11-32 2013 Pearson Education, Inc. All rights reserved. 11.4 (A) Individual Weighted Average Cost of Capital for Individual Projects Figure 11.3 Capital project decision model without considering risk. 4 projects, whose IRRs range from 8% to 11%, but the risk levels also go from lowmoderatehighvery high With a WACC of 9.5%, only Projects 3 and 4, with IRRs of 10% and 11% respectively would be accepted. However, Projects 1 and 2 could have been profitable lower risk\n\nprojects that are being rejected in favor of higher risk projects, merely because the risk levels have not been adequately adjusted for. 11-33 2013 Pearson Education, Inc. All rights reserved. 11.4 (A) Individual Weighted Average Cost of Capital for Individual Projects To adjust for risk, we would need to get individual project discount rates based on each projects beta. Using a risk-free rate of 3%; a market risk premium of 9%; a before-tax cost of 10%, a tax rate of 30%; equally-weighted debt and equity levels, and varying project betas we can compute each projects hurdle rate as follows: 11-34\n\n2013 Pearson Education, Inc. All rights reserved. 11.4 (A) Individual Weighted Average Cost of Capital for Individual Projects Under the risk-adjusted approach, Project 1 (IRR=8%>7.7%) and Project 2 (IRR=9%>8.6%) should be accepted, while Project 3 (IRR=10%<10.4%) and Project 4 (IRR=11%<13.1%) should be rejected. Figure 11.4 Capital project decision model with risk. 11-35 2013 Pearson Education, Inc. All rights reserved.\n\n11.5 Selecting Appropriate Betas for Projects It is important to adjust the discount rate used when evaluating projects of varying risk, based on their individual betas. However, since project betas are not easily available, it is more of an art than a science. There are two approaches generally used: 1. Pure play betas: i.e. matching the project with a company that has a similar single focus, and using that companys beta. 2. Subjective modification of the companys average beta: i.e. adjusting the beta up or down to reflect different levels of risk. 11-36 2013 Pearson Education, Inc. All rights reserved. 11.6 Constraints on Borrowing and\n\nSelecting Projects for the Portfolio Capital constraints prevent firms from funding all potentially profitable projects that come their way. Capital rationing -- select projects based on their costs and expected profitability, within capital constraints. Rank order projects (descending order) based on NPV or IRRs along with their costs choose the combination which has the highest combined return or NPV while using up as much of the limited capital budget. 11-37 2013 Pearson Education, Inc. All rights reserved. 11.6 Constraints on Borrowing and Selecting Projects for the Portfolio (continued)\n\nExample 8: Selecting Projects with Capital Constraints. The XYZ Companys managers are reviewing various projects that are being presented by unit managers for possible funding. They have an upper limit of \\$5,750,000 for this forthcoming year. The cost and NPV of each project has been estimated and is presented below. Which combination of projects would be best for them to invest in? 11-38 2013 Pearson Education, Inc. All rights reserved. 11.6 Constraints on Borrowing and Selecting Projects for the Portfolio (continued) Example 8 Answer\n\nProject Cost NPV 1 2,000,000 500,000 2 2,250,000 400,000 3 1,750,000 300,000 4 750,000 100,000 5 500,000 50,000 1) Form combinations of projects by adding the costs to sum up as close to the \\$5,750,000 limit as possible. Sum up the NPVs as well. Comb Total Cost NPV\n\n1,2,4,5 2m+2.25m+.75m+0.5m=\\$5.5m 1.5m 1,3,4,5 2m+1.75m+.75m+.5m=\\$5m 0.95m 2,3,4,5 2.25m+1.75m+.75m+.5m=\\$5.25m 0.85m 2) Select the combination which has the highest NPV i.e. Combination 1 including projects 1,2,4, and 5 with a total NPV of \\$1.5m. 11-39 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 1 Cost of debt for a firm: You have been assigned the\n\ntask of estimating the after-tax cost of debt for a firm as part of the process in determining the firms cost of capital. After doing some checking, you find out that the firms original 20-year 9.5% coupon bonds (paid semiannually), currently have 14 years until they mature and are selling at a price of \\$1,100 each. You are also told that the investment bankers charge a commission of \\$25 per bond when new bonds are sold. If these bonds are the only debt outstanding for the firm, what is the after-tax cost of debt for this firm if the marginal tax rate for the firm is 34 percent? 11-40 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 1 (Answer) Calculate the YTM on the currently outstanding bonds, after adjusting the price for the \\$25\n\ncommission. i.e. Net Proceeds = \\$1100-\\$25 \\$1075 Set P/Y=2 and C/Y = 2 Input 28 ? -1075 47.5 1000 Key N I/Y PVPMT FV Output 8.57% After-tax cost of debt = 8.57%(1-.34) 5.66% 11-41 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 2 Cost of Equity for a firm: R.K. Boats Inc. is in the process of making some major investments for growth and is interested in calculating their cost of equity so as to be able to correctly estimate their\n\nadjusted WACC. The firms common stock is currently trading for \\$43.25 and their annual dividend, which was paid last year, was \\$2.25, and should continue to grow at 6% per year. Moreover, the companys beta is 1.35, the risk-free rate is at 3%, and the market risk premium is 9%. Calculate a realistic estimate of RKBIs cost of equity. (Ignore floatation costs.) 11-42 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 2 (Answer) Using the SML Approach: Rf =3%; Rm-Rf = 9%; = 1.35; Re=3%+(9%)*1.35 15.15% Using the Dividend Growth Model (constant growth)\n\nP0 = \\$43.25; Do=\\$2.25; g=6%; (\\$2.25*(1.06)/\\$43.25)+.0611.51% A realistic estimate of RKBIs cost of equity = Average of the 2 estimates (15.15%+11.51%)/213.33% 11-43 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 3 Calculating capital component weights: T.J. Enterprises is trying to determine the weights to be used in estimating their cost of capital. The firms current balance sheet and market information regarding the price and number of securities outstanding are listed below. TJ Enterprises Balance Sheet (in thousands)\n\nCurrent Assets: \\$50,000 Current Liabilities: Long-Term Assets: \\$60,000 Long-Term Liabilities Bonds Payable \\$0 \\$48,000 Owners Equity Preferred Stock Common Stock Total Assets: 11-44\n\n\\$110,000 2013 Pearson Education, Inc. All rights reserved. Total L & OE \\$15,000 \\$47,000 \\$110,000 Additional Problems with Answers Problem 3 (continued) Market Information Debt Preferred Stock Common Stock Outstanding 48,000 102,000 1,300,000\n\nMarket Price \\$850 \\$95.40 \\$40 Calculate the firms capital component weights using book values as well as market values. 11-45 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 3 (Answer) Based on book values: Weight of Debt = \\$48,000/\\$110,000 43.64% Weight of P/S= \\$15,000/\\$110,000 13.64% Weight of C/S = \\$47,000/\\$110,00042.72% Based on market value:\n\nMarket value of Debt =\\$40,800,000 Market Value of P/S= \\$9,730,800 Market Value of C/S= \\$52,000,000 Total Market Value= \\$102,530,000 Weight of Debt = \\$40,800/\\$102,530 39.79% Weight of P/S= \\$9,730.8/\\$102,530 9.49% Weight of C/S = \\$52,000/\\$102,53050.72% 11-46 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 4 Computing WACC: New Ideas Inc. currently has 30,000 of its 9% semiannual coupon bonds outstanding (Par value =1000). The bonds will mature in 15 years and are currently priced at \\$1,340 per bond.\n\nThe firm also has an issue of 1 million preferred shares outstanding with a market price of \\$11.00. The preferred shares offer an annual dividend of \\$1.20. New Ideas Inc. also has 2 million shares of common stock outstanding with a price of \\$30.00 per share. The firm is expected to pay a\\$3.20 common dividend one year from today, and that dividend is expected to increase by 7 percent per year forever. The firm typically pays floatation costs of 2% of the price on all newly issued securities. If the firm is subject to a 35 percent marginal tax rate, then what is the firms weighted average cost of capital? 11-47 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 4 (Answer) 1) Determine the component costs Cost of Debt:\n\nP=1340; F=2%; Net proceeds=P(1-F) Net proceeds = \\$1340*(1-.02)=\\$1273 Set P/Y=2 and C/Y = 2 Input 30 ? -1273 45 1000 Key N I/Y PV PMT FV Output 6.18% Before-tax Rd 6.18% 11-48 2013 Pearson Education, Inc. All rights reserved.\n\nAdditional Problems with Answers Problem 4 (Answer) (continued) Cost of preferred stock: Dp=\\$1.20; Pp=\\$11; F=2% Rp = Dp/Pp(1-F) \\$1.20/(\\$11(.98)1.20/10.7811.13% Cost of common stock: Pc=\\$30; D1=\\$3.2; g=7%; F=2% Using the constant dividend growth model: Re = [D1/(P(1-F])+g[3.2/\\$30(.98)]+.0717.88% 11-49 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 4 (Answer) (continued) 2) Determine the market value weights of the components:\n\nMarket value of bonds = \\$1340*30,000\\$40,200,000 Market value of P/S = \\$11*1,000,000 \\$11,000,000 Market value of C/S=\\$30*2,000,000 \\$60,000,000 Total Market value \\$111,200,000 Weight of debt = 40.2m/111.2m 36.15% Weight of P/S = 11m/111.2m 9.89% Weight of C/S = 60m/111.2m53.96% 11-50 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 4 (Answer) (continued) 3) Calculate the adjusted WACC WACC = .5396*17.88% + .0989*11.13%\n\n+.3615*6.18%*(1-.35) =9.65 +1.10%+1.45% 12.2% 11-51 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 5 Capital Rationing: Quick Start Ventures, Incorporated is has received 6 excellent funding proposals, but is only able to fund up to \\$2,500,000 Project A: Cost \\$700,000, NPV \\$50,000 Project B: Cost \\$800,000, NPV \\$60,000 Project C: Cost \\$500,000, NPV \\$40,000 Project D: Cost \\$600,000, NPV \\$50,000 Project E: Cost \\$700,000, NPV \\$60,000 Project F: Cost \\$300,000, NPV \\$30,000\n\nWhich projects should Quick Start select? 11-52 2013 Pearson Education, Inc. All rights reserved. Additional Problems with Answers Problem 5 (Answer) 1) Compute the Profitability Index of the projects and rank order from highest to lowest PI: PI = (NPV + Cost)/Cost Project F E D C B A\n\n11-53 Cost 300,000 700,000 600,000 500,000 800,000 700,000 2013 Pearson Education, Inc. All rights reserved. NPV 30,000 60,000 50,000 40,000 60,000 50,000\n\nPI 1.10 1.09 1.08 1.08 1.08 1.07 Additional Problems with Answers Problem 5 (Answer) (continued) 2) Form combinations of projects going from highest to lowest PI until \\$2,500,000 is used up: Combinations F,E,D,B F, E, C,B E,D,C,A\n\nF,E,B,A Cost 2,400,000 2,300,000 2,500,000 2,500,000 NPV 200,000 200,000 150,000 200,000 PI 1.0833 1.0870 1.0600 1.0800\n\nPick the combination which has the highest PI Projects F, E, C. and B. Together they cost \\$2,300,000 and will have an NPV of \\$200,000 with a PI of 1.087. 11-54 2013 Pearson Education, Inc. All rights reserved. Figure 11.2 11-55 2013 Pearson Education, Inc. All rights reserved. TABLE 11.2 Decision on Projects with and without Risk 11-56" ]

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[ "# Divide by a One Digit Number Task Cards", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Subject\nResource Type\nFile Type\nPDF (1 MB|10 pages)\nStandards\n\\$2.50\n• Product Description\n• Standards\nDivide by a One Digit Number\n*24 cards\n*4.NBT.B.6\n*Divide by Multiples of 10, 100, 1,000\n*Estimate Quotients\n*Divide with Remainders\n*Quotients with Zeros\n*Division Word Problems\n\nFind whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.\nTotal Pages\n10 pages", null, "" ]

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[ "#### Authors\n\nSung-Soo Ahn, Sejun Park, Michael Chertkov, Jinwoo Shin\n\n#### Abstract\n\nMax-product Belief Propagation (BP) is a popular message-passing algorithm for computing a Maximum-A-Posteriori (MAP) assignment over a distribution represented by a Graphical Model (GM). It has been shown that BP can solve a number of combinatorial optimization problems including minimum weight matching, shortest path, network flow and vertex cover under the following common assumption: the respective Linear Programming (LP) relaxation is tight, i.e., no integrality gap is present. However, when LP shows an integrality gap, no model has been known which can be solved systematically via sequential applications of BP. In this paper, we develop the first such algorithm, coined Blossom-BP, for solving the minimum weight matching problem over arbitrary graphs. Each step of the sequential algorithm requires applying BP over a modified graph constructed by contractions and expansions of blossoms, i.e., odd sets of vertices. Our scheme guarantees termination in O(n^2) of BP runs, where n is the number of vertices in the original graph. In essence, the Blossom-BP offers a distributed version of the celebrated Edmonds' Blossom algorithm by jumping at once over many sub-steps with a single BP. Moreover, our result provides an interpretation of the Edmonds' algorithm as a sequence of LPs." ]

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[ "I'm flumoxed, how do people normally handle\n\nCBaseClass::operator==\n\nand\n\nCDerivedClass::operator==\n\nWhat I have done is illustrated below, but is it good valid C++? I compare the base class members first (using base class ==) and then the derived class members (the rest of the implementation of the derived class ==).\n\n``````// This is == in a derived class\nbool CFlatMeshGen::operator== (const CFlatMeshGen& Rhs) const\n{\n// First compare the base class members....\nconst CMeshGen* pBaseLhs = (CMeshGen*)(this) ;\nconst CMeshGen* pBaseRhs = (CMeshGen*)(&Rhs) ;\nif (*pBaseLhs != *pBaseRhs) {\n// We are already not equal....\nreturn (false) ;\n}\n// If we get here base class members are equal...\n\n// Compare the derived class members...\nif (m_XSize != Rhs.m_XSize) {\nreturn (false) ;\n}\n\nif (m_YSize != Rhs.m_YSize) {\nreturn (false) ;\n}\n\nif (m_BaseHeight != Rhs.m_BaseHeight) {\nreturn (false) ;\n}\n\nreturn (true) ;\n}\n``````\n\nPerfectly functional, if not simple or \"elegant\". This will do the same, in less code and perhaps more efficiently:\n\n``````// This is == in a derived class\nbool CFlatMeshGen::operator== (const CFlatMeshGen& rhs) const\n{\n// First compare the base class members....\nreturn ((const CMeshGen&)(*this) == (const CMeshGen&)rhs &&\nm_XSize == rhs.m_XSize &&\nm_YSize == rhs.m_YSize &&\nm_BaseHeight == rhs.m_BaseHeight);\n}\n``````\n\nOne of the advantages of this construct is that the first non-equal element in the test will cause it to return false, so only the minimum element comparisons will be made in non-true cases.\n\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of 1.21 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge." ]

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[ "Energy, in physics, the capacity for doing work. It may exist in potentialkineticthermal, electrical, chemicalnuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another. After it has been transferred, energy is always designated according to its nature. Hence, heat transferred may become thermal energy, while work done may manifest itself in the form of mechanical energy.\n\nEnergy with Wilson, science made simple through innovative videos and content that aid understanding of this GCSE content.\n\nFor other content head over to Chemistry, Biology or back to Physics\n\nIf you are looking for the videos that accompany this content then head over to the YouTube channel: Science with Wilson\n\n## Lessons\n\n### Lesson 1 – Energy stores and systems\n\nThis lesson offers an introduction to the energy topic and describes the different energy stores and types. It then goes on to explain how energy is transferred in a system and describes the changes involved in energy systems when an energy transfer takes place.\n\nClick on the resources below to accompany the lesson video.\n\n### Lesson 2 – Energy changes\n\nWe now begin to look at performing calculations around gravitational potential energy and kinetic energy as well as explaining the relationship between the two equations. This allows us to then make predictions based on those types of energy in different contexts.\n\n### Lesson 3 – Energy changes in systems & Specific heat capacity required practical\n\nThis lesson hinges around the Specific Heat Capacity AQA Required practical and explains how the specific heat capacity of a substance effects how quickly it will heat up or cool down. This allows us to then quantify this using the equation and to apply it to novel situations.\n\n### Lesson 4 – Power\n\nPower is defined as the rate at which energy is transferred or the rate at which work is done. In this lesson we will look at that in more detail and apply this using the power equation.\n\n### Lesson 5 – Energy transfers in a system & Investigating thermal insulators required practical (Physics only)\n\nThis lesson looks at how energy is transferred in a closed system and links in ideas of how a reduction of unwanted energy transfers can be achieved. The latter parts focus on thermal conductivity and identifying factors that effect the rate of such conductivity. This also includes the physics only required practical ‘investigating thermal insulators’.\n\n### Lesson 6 – Efficiency\n\nThe energy efficiency for any energy transferred can be calculated using an equation. This lesson applies the concept of efficiency to mathematical problems and solves them as decimals and percentages. For higher tier students it also looks at ways that efficiency of an intended energy transfer can be increased.\n\n### Lesson 7 – National and global energy resources\n\nThe Earth has available energy resources that humans use. This lesson compares the different energy resources and considers environmental impacts and long term impacts of their uses.\n\nFor other content head over to ChemistryBiology or back to Physics" ]

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[ "# Errata for the book: Computations in algebraic geometry with Macaulay 2\n\n```page 65, posted May 13, 2004:\n\nThe routine blowUpIdeal was never intended to handle the blow-up of an\narbitrary scheme along an arbitrary subscheme. It was designed to\nillustrate several of the classic blow-ups in Section IV.2 in\nEisenbud and Harris . As a result, it makes the following\nassumptions on the input:\n\n(1) The ideal \"I\" lives in a polynomial ring. The routine ignores any\nrelations between the variables in the ring \"S\" of \"I\", so it will\nfail unless the ring is a polynomial ring.\n\n(2) It uses \"t\" as a global variable, so fails if the ideal \"I\" is in a\nring already using \"t\" as an indeterminate.\n\n(3) It also uses the first \"r\" letters of the alphabet as global\nvariables, where \"r\" is the number of generators of the ideal \"I\",\nso fails if the ideal is in a ring using any of those letters as\nvariables.\n\nThe input line i71 does not establish that the scheme X is\nnonsingular. To accomplish this, we should show that the singular\nlocus, which is given by the appropriate minors of the Jacobian\nmatrix, contains a power of the irrelevant ideal. This can be\nimplement in Macaulay 2 as follows:\n\nring J == saturate(J + minors(codim J, jacobian J), ideal(a,b,c))\n\n```" ]

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[ "+0\n\n0\n162\n1\n\nWrite the equation of a line that contains the point (7, 9) and is parallel to a line that contains the points (-4, 5) and (8, -1).\n\nJul 25, 2020\n\n#1\n+1\n\nfind the slope between the tw given points     (y1-y2) / (x1-x2)   =   6/(-12)    = m\n\nparallel is the same slope\n\ny = mx +b\n\ny = -1/2 x + b     sub in the  point to be on the new line\n\n9 = -1/2(7) + b     b = 12.5\n\ny = - 1/2x + 12.5\n\nJul 25, 2020" ]

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[ "Videos by Janet Gray Coonce, MS\n\nIf you plan to view the video on your cell phone, consider your data plan and whether you should wait until you have a WiFi connection to avoid cellular charges.\n\nConnect to Wi-Fi to prevent cellular data charges for video streaming.\n\nIn this tutorial I want to help you visualize the electronic and molecular geometries predicted by the valence-shell electron-pair repulsion, VSEPR, models.  I will demonstrate how to represent 3 dimensional models of electronic and molecular geometry in 2 dimensional drawings.", null, "", null, "There are 3 regions of electron density in the acetylene (C2H2) molecule.  Each carbon is flanked on the outside by a single bond to hydrogen.  The two central carbon atoms are sharing a triple bond.  So each of the carbon atoms have 2 electron dense areas trying to get as far apart as possible.  When 2 electron dense areas connected to a central atom get as far apart as possible, they will align themselves across a straight line.  Acetylene therefore has linear molecular geometry.  The bond angle of a straight line is 180 degrees.", null, "", null, "Carbon dioxide, C02 is another example of a central atom with two areas of electron density repelling each other.  They oxygen atoms are getting as far apart from each other as possible and form a straight line.  It is linear electronic and linear molecular geometry.", null, "If there are 3 identical atoms connected to a central atom, each will repel each other and get as far apart as possible.  They will assume a trigonal planar electronic and trigonal planar molecular geometry.  All 3 of the atoms bound to the central atom will be in the same plane and will be equal distant.  In the case of 3 the bond angle will be 120 degrees (360/3).", null, "Here is a central atom with 3 electron dense areas but only a lone pair of electrons is in one of the 3 areas.  The lone electrons will be more electronegative than the atoms sharing electrons in a bond.  The bonded atoms will be repelled by the force of the lone pair (blue arrows) and the bond angle will be less than 120 degrees.  The molecular shape will be bent.  The electronic geometry is still trigonal planar.", null, "When there are 4 atoms bonding to a central atom and 4 regions of electron density, the electronic and molecular shape is tetrahedral.  The atoms will arrange themselves equally around the central atom in 3 dimensions.  In the 2 dimensional representation, the solid wedge represents an atom projecting up from the plane of the paper.  The dotted wedge represents an atom projecting below the plane of the paper as illustrated.", null, "If one of the 4 areas of electron density was a lone pair, the molecular shape would be a trigonal pyramid (3 atoms attached to central atom).  The electronic shape would still be tetrahedral (4 areas of electronic density around the central atom).", null, "If 2 of the 4 areas of electron density is represented by a lone pair of electrons and 2 electrons shared in a bond with another atom, then the less electronegative atoms in the bond will be repelled by the more electronegative regions occupied by the lone electrons.  The electronic shape would still be tetrahedral but the molecular shape would be bent.\n\nIf a central atom shares 5 electrons by forming covalent bonds with 5 other atoms the electronic and molecular geometry is a trigonal bipyramid.", null, "If one of the 5 areas of electron density is occupied by a lone pair of electrons, the electronic shape would still be trigonal bipyramid but the molecular shape would be a “see-saw.”", null, "If 2 areas are occupied by lone pairs the molecular geometry will be T-shaped.", null, "If there are 3 pairs of unshared electrons and 2 bonds, the molecular geometry will be linear.  The electronic geometry is still trigonal bipyramid.  This is the shape when 5 regions surrounding a central atom try to repel each other in order to get as far apart as possible.", null, "With 6 regions all repelling each other to get as far apart as possible, the molecular and electronic geometry is octahedral.", null, "If one of the 6 regions is a lone pair, the molecular geometry would be a square pyramid.  The electronic geometry would still be an octahedral.", null, "If 2 of the regions are occupied by a lone pair of electrons then the molecular geometry would be square planar.  There are still 6 regions of electron density so the electronic geometry is still octahedral.", null, "Review:  Regions of electron density (RHED) surrounding a central atom are of 4 types:\n\n1.  single bond =  a RHED consisting of 1 pair of shared electrons\n\n2.  double bond =  a RHED consisting of  2 pairs of shared electrons\n\n3.  triple bond = a RHED consisting of 3 pairs of shared electrons\n\n4.  a RHED consisting of lone pair of unshared electrons.", null, "Here is the Lewis structure for carbon dioxide, CO2.  How many electron regions around the central carbon atom?  Answer:  There is a double bond on each side.  A double bond is a single region.  Therefore we know 1 + 1 = 2.  So the answer is there are 2 regions.  What about the molecular shape?  If the electron regions associated with the more electronegative oxygen atoms repel each other and try to get as far apart as possible, what molecular shape will the CO2 molecule have?  Linear.  A straight line.  A bond angle of 180 degrees.  What is the geometry of the electron regions?  They are also in a straight line.  Linear.\n\nSummary:  CO2 has 2 electron regions (RHED) around the central carbon atom with no lone pairs of unshared electrons.\n\nElectronic geometry:  Linear\n\nMolecular geometry:  Linear", null, "Here is the Lewis structure for ammonia gas, NH3How many regions of high electron density (RHED) are there around the central nitrogen atom?  Count them.  There are 3 single bonds and a single lone pair of electrons.  3 + 1 = 4.  If all of these electronic regions get as far apart as possible what will be the 3 dimensional shape?  There are 4 points and 4 points in 3 dimensions is a tetrahedral.  Therefore the electronic shape is tetrahedral.  However, there are only 3 atoms around the central nitrogen.  What is the MOLECULAR geometry?  The 3 hydrogen atoms form a triangle in one plane and the nitrogen is above it.  It is a trigonal pyramid.\n\nSummary:  Ammonia gas, NH3 has 4 electron regions (RHED) around the central nitrogen atom with 1 pair of unshared electrons.\n\nElectronic geometry:  Tetrahedral\n\nMolecular geometry:  Trigonal pyramid", null, "Here is the Lewis structure for water, H20.\n\nThere are 4 RHED, 2 lone pairs and 2 single bonds.  The 4 electronic regions will get as far apart as possible in a tetrahedral electronic geometry.", null, "There are only 2 atoms attached to the central oxygen atom and they will be repelled by the lone pairs.  Although there are only 2 atoms bonded to oxygen and they are trying to get as far apart from each other as possible, the three atoms are not in a straight line.  The hydrogen atoms are also being repelled by the lone pairs of electrons.  The molecular geometry is not a straight line.  The molecular geometry is bent.\n\nSummary:  Water, H20,  has 4 electron regions around the central oxygen atom with 2 pair of unshared electrons and 2 single bonds.\n\nElectronic geometry:  Tetrahedral\n\nMolecular geometry:  Bent\n\nSummary of molecular geometries for each of the electronic geometries:\n\nA.  Trigonal Planar Electronic geometry = 3 RHED\n\n1.  3 bonds + 0 lone pair = trigonal planar geometry\n\n2.  2 bonds + 1 lone pair = bent molecular geometry\n\nB.  Tetrahedral Electronic Geometry = 4 RHED\n\n1.  4 bonds + 0 lone pair = tetrahedral molecular geometry\n\n2.  3 bonds + 1 lone pair = trigonal pyramid molecular geometry\n\n3.  2 bonds + 2 lone pair = bent molecular geometry\n\nC.  Trigonal Bipyramid Electronic Geometry = 5 RHED\n\n1.  5 bonds + 0 lone pair = octahedral molecular geometry\n\n2.  4 bonds + 1 lone pair = see-saw or distorted tetrahedral molecular geometry\n\n3.  3 bonds + 2 lone pair = T-shaped molecular geometry\n\n4.  2 bonds + 1 lone pair = linear molecular geometry\n\nD.  Octahedral Electronic Geometry = 6 RHED\n\n1.  6 bonds + 0 lone pairs = Octahedral molecular geometry\n\n2.  5 bonds + 1 lone pair = Square Pyramid molecular geometry\n\n3.  4 bonds. + 2 lone pair = Square Planar molecular geometry\n\nTranscription by James C. Gray MD FACOG\n\nOne Comment\n1.", null, "Hello.nice to meet your report. haha.(sorry. Just kidding)\nI’m a high school student in korea. I’m learnig chemistry, biology and other subject too.\nI saw your writing nowyand i can understand this video,writing. So cool!!!!!\nAsume!! I love this writing and three-dimensional image!\nThank you for upload useful informations.(if there is wrongunderstand just ignore them.!sorry😄)" ]

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[ "# Rip Rap Calculator\n\nNote: values cannot be equal to or less then 0\n\n## Introduction to Rip Rap Calculator\n\nCalculating a rock's diameter can be challenging and time consuming task. There are also chances of one doing any mistake while doing manual calculations. That is why we introduce the rip rap calculator to calculate the rock's diameter.\n\nData is required to calculate the riprap stone-like (velocity of water, gravity, and bash). With the help of the data, rip rap rock calculator can determine the size of the rocks.\n\n## Why use riprap weight calculator?\n\nThe online rip rap stone calculator provides you with the measurement of the rocks according to the water velocity, bash constant, and specific gravity. Riprap sizing calculator helps you to calculate 100 % accurate results without any errors.\n\nThe calculation of rip rap by hand is very difficult as there can be errors. The riprap coverage calculator executes the values according to the formula and provides the results within a minute.\n\n## How to use the rip rap volume valculator?\n\nThe use of the rip rap cost calculator is very simple. If you want to know how much rip rap do I need? There are a few simple steps to follow as below:\n\n• Open the riprap rock sizing calculator.\n• Input the values (velocity, gravity and bash constant).\n• Click on the calculate rip rap button.\n• The results will show after a minute.\n\nTo reuse the riprap size calculator, click on the calculate again button, which will load back the calculator page.\n\n## How to calculate a riprap rock's size?\n\nIt is easy to estimate the rip rap rocks. You can follow the formula and get the calculation by manual method.\n\n### Rip rap calculation formula\n\n$$D_{50} \\;=\\; \\frac{(v)^2}{(2 \\;×\\; g \\;×\\; c^2 \\;\\;×(s-1))}$$\n\nWhere,\n\nD50 is the Average diameter of 50% of the spherical rocks for the rip rap in meters.\n\nV is the Average channel velocity in meters per second.\n\nG is Acceleration due to gravity, either 9.806 m/s² or 32.17 ft. /s²\n\nC is Isbash constant with values equal to 0.86 for highly turbulent flow of water or 1.20 for low turbulence water flow.\n\nS is Specific gravity of the rock ranges from around 2.50 to 3.00.\n\n### Example\n\nThere are few values: velocity 1ms, bash constant 1.2, and specific gravity 2.6. Calculate the rip rap in cm.\n\n## Solution\n\n$$D_{50} \\;=\\; \\frac{(v)^2}{(2 \\;×\\; g \\;×\\; c^2 \\;×\\; (s \\;-\\;1))}$$ $$D_{50} \\;=\\; \\frac{(1 m/s)^2}{(2 \\;×\\; 9.802 m/s² \\;×\\; 1.2² \\;×\\; (2.6 \\;-\\; 1))}$$ $$D_{50} \\;=\\; \\frac{1\\; m²/s²}{(19.612\\; m/s² \\;×\\; 1.44 \\;×\\; 1.6)}$$ $$D_{50} \\;=\\; \\frac{1\\; m²/s²}{(45.186048\\; m/s² )}$$ $$D_{50} \\;=\\; 0.02213 m$$ $$D_{50} \\;=\\; 2.21 cm$$\n\n## Advantages of the online riprap calculator\n\nThere are some beneficial uses of the rip rap tool as below:\n\n• The rip rap volume calculator provides you with accurate output quickly.\n• Rip rap rock calculator saves your time and energy from difficult calculations.\n• It is a reliable converter you can trust on it to know how much rip rap do I need.\n• It saves your money from premium calculators to calculate rip rap.\n• Rip rap calculator is available at any time.\n• It is a very helpful tool for engineers to measure the size of rocks.\n\n## Other related tools\n\nThere are other related tools which you can find on this website. The online calculators are:\n\n## How does the rip rap tool execute the values?\n\nThe riprap sizing calculator has a built-in formula set by the developers; when you enter the values, it executes automatically and provides the results after calculation.\n\n## Is it a reliable tool for rip rap calculation?\n\nThe developers of the rip rap stone calculator tested it many times then it was available for people's use.\n\n## What is riprap?\n\nThe line of the rocks between water and land is called rip rap. It is used to control the water stream because the velocity of the water is very high in the rivers and oceans.\n\nThe high velocity of the water can destroy the layer of the land inside the water. That is why we use the different sizes of the rocks according to the flow of water.\n\n### How much rip rap do i need?\n\nThe amount of rip rap needed depends on the size and slope of the area to be protected, as well as the desired level of erosion control. A professional engineer should be consulted to determine the specific requirements for your project.", null, "### Shaun Murphy\n\nLast Updated March 28, 2022\n\nA professional content writer who likes to write on science, technology and education." ]

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[ "Online Test", null, "LOGIN", null, "SIGN UP", null, "• #### Quantitative Aptitude Practice Trigonometry Study Material\n\nDigitization help student to explore and study their academic courses online, as this gives them flexibility and scheduling their learning at their convenience. Kidsfront has prepared unique course material of Trigonometry Inverse Trigonometric Functions for Quantitative Aptitude Practice student. This free online Trigonometry study material for Quantitative Aptitude Practice will help students in learning and doing practice on Inverse Trigonometric Functions topic of Quantitative Aptitude Practice Trigonometry. The study material on Inverse Trigonometric Functions, help Quantitative Aptitude Practice Trigonometry students to learn every aspect of Inverse Trigonometric Functions and prepare themselves for exams by doing online test exercise for Inverse Trigonometric Functions, as their study progresses in class. Kidsfront provide unique pattern of learning Trigonometry with free online comprehensive study material and loads of Quantitative Aptitude Practice Trigonometry Inverse Trigonometric Functions exercise prepared by the highly professionals team. Students can understand Inverse Trigonometric Functions concept easily and consolidate their learning by doing practice test on Inverse Trigonometric Functions regularly till they excel in Trigonometry Inverse Trigonometric Functions.\n\na) Θ = nΠ\nb) Θ = Π\nc) Θ = n\nd) Θ = -nΠ\n\na) TRUE\nb) FALSE\nc) Maybe\nd) None of these\n\n#####", null, "The solution consisting of all possible solutions of a trigonometric equation is called _________ .\n\na) Principal solution\nb) Inverse principal solution\nc) General solution\nd) Inverse general solution\n\n#####", null, "What to do if sometimes the resulting roots does not satisfy the original equation?\n\na) Find root of equation\nb) Square the equation\nc) Divide equation by 2\nd) None of these\n\na) (2n + 1)\nb) (2n + 1)Π\nc) (2n + 1)Π/2\nd) Π/2\n\na) Π\nb) 6\nc) Π/6\nd) (- Π/6)\n\na) Π\nb) 6\nc) Π/6\nd) ( - Π/6)\n\na) Tan^-1 (24)\nb) Tan^-1 (7)\nd) Tan^-1 (7/24)\n\n#####", null, "In inverse trigonometric function: cos^-1 (4/5)+ cos^-1 (12/13) = __________ .\n\na) Cos^-1 (33/65)\nb) Cos^-1 (65)\nc) Cos^-1 (33)\nd) All of these\n\na) TRUE\nb) FALSE\nc) Maybe\nd) None of these\n\n##### Solution Is :\nPREVIOUS\nscript type=\"text/javascript\">" ]

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[ "### Efficient weighted multiselection in parallel architectures\n\nHong Shen\nFifth International Conference on Algorithms and Architectures for Parallel Processing, 2002. Proceedings.\nWe study parallel solutions to the problem of weighted multiselection to select r elements on given weighted-ranks from a set S of n weighted elements, where an element is on weighted rank k if it is the smallest element such that the aggregated weight of all elements not greater than it in S is not smaller than k. We propose efficient algorithms on two of the most popular parallel architectures, hypercube and mesh. For a hypercube with p < n processors, we present a parallel algorithm running\nmore » ... n O(n min{r, log p}) time for p = n 1− , 0 < < 1, which is cost optimal when r ≥ p. Our algorithm on √ p × √ p mesh runs in O( √ p + n p log 3 p) time which is the same as multiselection on mesh when r ≥ log p, and thus has the same optimality as multiselection in this case." ]

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[ "## ClarkVision.com", null, "Home Galleries Articles New About Contact", null, "Optimum Magnified Visual Angle Visual Astronomy of the Deep Sky\n\nThis page is dedicated to amateur astronomy observing and discussions of how the eye operates in very faint light situations. The information is based on the my book and research I have done since publication of the book:\n\nClark, R.N., Visual Astronomy of the Deep Sky, Cambridge University Press and Sky Publishing, (book of 355 pages), 1990. Copyright Roger N. Clark and Sky Publishing. This book is referred to as \"Visual Astronomy\" from here on.\n\nInformation may be freely used and printed for personal non-profit use by amateur astronomers for their pursuit of visual observing.\n\nOptimum Magnified Visual Angle\n\nIn the book \"Visual Astronomy,\" I derived a concept called the Optimum Magnified Visual Angle (OMVA). This is the angle to which an observer looking through a telescope can detect the faintest and lowest contrast objects. Using data from a World War II study to help soldiers see better at night, I interpreted the data from the perspective of viewing through a telescope. A long held concept in amateur astronomy observing is to increase the magnification of a telescope to \"increase the contrast of the object being viewed.\" While the effect is real, the explanation is incorrect. As one changes magnification, all objects change size (e.g. object and sky background), so the contrast stays constant. But the eye's sensitivity to contrast changes as the object size changes, with lower contrast objects easier to detect when they appear larger, meaning at higher power.\n\nAs one magnifies an object in a telescope, the object appears larger, but its surface brightness (e.g. apparent magnitude per square arc-second) decreases. If the object is magnified too much, its surface brightness may be too faint to see. While some magnification can help viewing, too much hurts. So, there appears to be an optimum. I derived the optimum curve in \"Visual Astronomy\" and here I discuss the concept and the confusion that has surfaced since publication of the book.\n\nContrast Discrimination\n\nVisual astronomical observations depend not just on detecting faint light but also on contrast discrimination. Both abilities are involved in seeing such things as spiral arms of galaxies and dark rifts in nebulae, and in simply perceiving any object against the sky background. Contrast detection thresholds, as a function of background surface brightness for several object diameters, are plotted in Figure 2.5 from \"Visual Astronomy\". This diagram shows that, for a given background (e.g. the night sky), less contrast is needed to see a larger object.", null, "Figure 2.5 from Visual Astronomy. The minimum contrast needed to detect an object of a given angular size shown as a function of background surface brightness, do. The larger an object appears to the eye, the easier it is to detect. For small, bright objects on a bright background, a contrast less than 0.01 is enough for detection. But against the very dim night-sky background seen in a telescope (fainter than 25 magnitudes per square arc-sec), a large object must have a contrast of nearly 1.0, and a small object more than 100, to be detected. Derived from data in Table VIII of Blackwell (1946).\n\nCritical Visual Angle\n\nThe data in Figure 2.5 were used to plot minimum detectable contrast versus angular size at constant values of background luminance to make Figure 2.6. Here we notice that for objects with small angular sizes, the smallest detectable contrast times the surface area is a constant. As an object becomes larger, this product is no longer constant. The angle at which the change occurs is called the critical visual angle. An object smaller than this angle is a point source as far as the eye is concerned. (A point can be considered the angular size smaller than which no detail can be seen.)", null, "Figure 2.6 from Visual Astronomy. The smallest contrast needed to detect objects of various sizes on various backgrounds. This diagram is the most important one in the book, so it's worth taking the time to figure out its complexities. This is the same data as in Figure 2.5, except that contrast detection ability is plotted against angular size for various background surface brightnesses (magnitude per square arc-second).\n\nWhen an object is magnified in a telescope, the contrast between object and background does not change since both are magnified equally. However, the object becomes larger as viewed by the eye. Therefore, moving horizontally across the chart corresponds to increasing the magnification.\n\nAs we start with low magnification on the left side, the contours of background surface brightness are diagonal straight lines. At a point called the critical visual angle, the lines begin to curve. Objects smaller than this value appear as point sources (the smallest detail that can be distinguished). As one moves to the right of the critical visual angle line, the faintest detectable surface brightness decreases faster than the background surface brightness. Thus, fainter objects--or detail within objects--can be seen as magnification is increased.\n\nThis is true only until the \"optimum magnified visual angle\" is reached. Thereafter, higher magnification decreases the detection threshold faster than surface brightness. A faint object is most visible when magnified to this angle. Derived from data in Table VIII of Blackwell (1946).\n\nThis critical angle is shown in Figure 2.7a plotted for various background luminances. Figure 2.7a shows that as the background becomes fainter, the size of a \"point source\" becomes larger for objects that are just detectable. In other words, the eye's resolution or ability to see detail is much coarser in the dark.", null, "Figure 2.7a from Visual Astronomy. The eye's resolving power, unlike a camera's depends on an object's surface brightness. The critical visual angle is the angle below which no detail can be seen and objects appear as point sources.\n\nOptimum Magnified Visual Angle\n\nA low-contrast object is more easily detected if it is larger. For an extended object such as a galaxy viewed in a telescope, magnification does not change the contrast with the background, because both the sky's and the object's surface brightnesses are affected equally. Some visual observers have stated that a dim object's contrast with the sky background increases with higher magnification, but this is clearly wrong. The contrast merely looks greater because of the increased detection capabilities of the eye. Clark (1990) coined a name for the maximum magnification that will help detection: the \"optimum magnified visual angle\" (OMVA). This angle is shown in Figure 2.6 and also Figure 2.7b.", null, "Figure 2.7b from Visual Astronomy. The \"optimum magnified visual angle\" of an object depends on surface brightness. This angle is the size for which a faint object, or detail within an object, should be magnified in order to maximize the possibility of detection.\n\nIf an object is at the threshold of detection and smaller than the optimum angle, more magnification will make it easier to see. When the object is magnified beyond the optimum angle, its surface brightness decreases faster than the eye's contrast detection threshold, and the object will become harder to detect. Remember that even for an object somewhat above the detection threshold, higher magnification may bring out details within the object that are smaller than the optimum angle at a lower magnification.\n\nControversy\n\nThere has been a fair amount of confusion, discussion and criticism over this concept. Here are some links:\n\nIn a long public and private email series among the above authors, we tried to understand each others points and to try and figure out what is correct (this was not simple). These discussions ended without full resolution of the issues. However, we did learn a few things.\n\nFirst, two general methods were being used to derive the OMVA.\n\n• 1) Some were deriving OMVA from the surface brightness detection lines. This, however means changing the contrast as one moves along constant surface brightness.\n• 2) Others were holding contrast constant as occurs when changing magnification in a telescope.\n\nI derived the original OMVA by hand methods in the early 1980s before I had the modern computer tools, but using method 2 above. Nils Olof Carlin and I generally used method 2 and the Mel Bartels page above uses method 1. In our long discussions, and on the pages above, the other authors have yet to derive a new OMVA curve.\n\nSome of the disagreement centered on my saying the OMVA occurs at a slope of -1 on the curves of constant surface brightness in Figure 2.6. Derivations by others in our discussions came out with a slope of -2. But I think people are solving different problems (e.g. the method 1 above). Methods which change contrast while changing magnification in a telescope are incorrect in my view, because the contrast of extended objects does not change when you change magnification in a telescope. This fact, I believe, is the root cause of much of the confusion.\n\nWhen one magnifies an object in a telescope, one moves horizontally across the Figure 2.6 diagram. The object gets larger proportional to the magnification, m, but the surface brightness decreases in proportion to 1/m2. So it is the horizontal spacing of the surface brightness curves in Figure 2.6 that controls where the optimum is located. If the slope is steeper than -1 on the Figure, increasing magnification helps detection. As the slope of the surface brightness curves becomes more horizontal, the spacing stretches out (in constant contrast, or horizontally) and one loses detection ability due to loss in surface brightness faster than one gains by increasing the apparent size. This trade point occurs at a slope of -1. This is the optimum. On this point, at least some of the others (above) probably still disagree. Perhaps after reading this page, we can collectively go over the data, including the new data I present below, and derive a better solution.\n\nRegardless of where the correct optimum is, the data I present below shows the optimum is a shallow function, and if you follow the advice in the very last paragraph, you will observe all detail close enough to the optimum that this academic disagreement is meaningless in practice.\n\nOne thing that came of our discussions, the Optimum Detection Magnifications in Appendix F of Visual Astronomy (in which I used method 1 in a way I didn't realize until these discussions) is wrong. Do not use it! The program I wrote changes contrast and thus is flawed.\n\nHowever, the \"Minimum Optimum Detection Magnification\" (MDM) in Appendix E is still correct to the best of my knowledge.\n\nSo what does all this mean?\n\nWhat few people realize is that the data presented in Figure 2.5 and 2.6 are projections of a 3-dimensional surface onto 2-dimensional graphs. The 3-D surface is shown below in Figure A.", null, "Figure A. The \"Blackwell Surface\" for threshold detection. Detection of a faint object by a visual observer depends on 3 things: 1) surface brightness of the object, 2) the object's angular size, and 3) the contrast with the background. An object is detectable if it plots on the surface or above it.\n\nFigure 2.5 is the projection of Figure A to the upper left plane. Figure 2.6 is the projection of the data onto the upper right plane. The third projection is the floor and is shown below in Figure B.", null, "Figure B. The threshold contrast as a function of apparent angular size and surface brightness. The curves on the plot are constant contrast. The red diagonal lines represent the magnification trend in a telescope. The optimum magnified visual angle (OMVA) occurs tangent to a contrast curve. The white line shows my new derivation of the OMVA, first presented at the May 2001 Riverside Telescope Makers Convention (RTMC). The idea for this derivation was first presented to me by _______ of England in 19__ (I'm searching for this letter).\n\nThe OMVA in Figure B shows a second way to look at the Blackwell data and derive the OMVA. As one can see from the tangent points, and considering the data have been re-interpolated a couple of times, one might wonder how accurate the derivation of and OMVA is. Let's look at another way to check what the OMVA is.\n\nThe Minimum Aperture Needed to Detect M57 as a Ring\n\nA simple method to check the OMVA is actual observations of simple deep-sky objects. An excellent object is M57. What is the minimum aperture needed to detect the ring nature of the object as a function of magnification? For this test, I did it in one night when the object was overhead and used aperture masks, so variations in sky conditions was minimal and telescope transmission was a constant. The results are shown in Figures C and D.", null, "Figure C. The minimum aperture needed to detect the ring nature of M57 at various magnifications. The curve shows the \"barely\" to \"easy\" apertures as the bottom and top curves, respectively. The average of the barely and easy is shown as the heavy line. The image is an M57 drawing using my 12.5-inch telescope and many magnifications.\n\nThe central hole is about 3/4 arc-minute across. The minimum aperture occurs at a magnification of about 130x (+/- 30), which means the optimum diameter occurs at about 100 arc-minutes as viewed by your eye. The background surface brightness is about 25.5 magnitudes /square arc-second (including 0.4 mag/sq. arc-sec. for transmission loss in the telescope).", null, "Figure D. The mean surface brightness of M57 as a function of magnification when it is just detectable at each aperture in Figure C.\n\nNote how shallow the minimum is in the M57 ring detection experiment in Figure C. The OMVA is not a precise value.\n\nNow let's try and determine the OMVA at the brighter surface brightnesses. Click on the figure below to see a test chart of low contrast spots of various sizes.\n\nIf you print this chart at 300 dpi, and hold it at 30 inches from your eye, the 27 pixel diameter spots appear about 10.3 arc-minutes across. Viewing the chart in low room light gets near the right side of Figure 2.7b. For the low contrast spots, how large do they have to be to be seen (just above the threshold)? I tried some various lighting conditions and found a range of about 6 to 13 arc-minutes were needed to detect the low contrast spots (e.g. the 2 DN brighter than the background). I did the experiment before computing the spot sizes so I did not know what the outcome would be.\n\nPlotting the M57 point, the low contrast spot experiment, along with OMVA from Figure 2.6 and Figure B, we see the result in Figure E, below.", null, "Figure E. OMVA from Figure 2.6 and Figure B and the derived point from the M57 experiment.\n\nConclusions\n\nThe new Optimum Magnified Visual Angle (OMVA) curve (Figure E, red line) is much more complex than the original OMVA curve (Figure E, black line). However, both indicate a general upward trend of increasing apparent angular size with decreasing surface brightness. Which is correct? It appears that the original OMVA agrees with observational data so far, but more precise data are needed to be sure. I do not think the OMVA data are good enough with the interpolations that have been done (starting with Blackwell, 1946). It is clear that the OMVA has a broad minimum. It is also clear that the OMVA for very faint objects is on the order of 0.5 to 1.5 degrees (it may be more than about 100 arc-minutes at the faint end).\n\nThus, the observing strategy to detect deep-sky objects, or detail within objects, is to magnify those objects, or detail within the objects. so they appear about 100 arc-minutes in size. For example, if you are trying to detect a dark nebula in a galaxy arm, magnify that dark nebula so that it appears about a degree across or more.\n\nWhat Does All This Mean?\n\nTo see all the detail in an object, use many powers, from very low to very high, examining the entire object with each magnification. Because the OMVA appears to be a shallow minimum, one need not be precisely on the optimum. Within a factor of 2 or a little less in magnification is fine. A magnification sequence of: 35x, 50x, 80x, 120x, 180x, 270x, 400x ... (a sequence increasing magnification by a factor of about 1.5) is great.\n\nReferences\n\nBlackwell, R.H., Contrast Thresholds of the Human Eye, Journal of the Optical Society of America, v36, p624-643, 1946.\n\nClark, R.N., Visual Astronomy of the Deep Sky, Cambridge University Press and Sky Publishing, 355pp., 1990.\n\nClarkvision Visual Astronomy Main Page\n Home Galleries Articles New About Contact" ]

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[ "# C++ Hierarchical Inheritance Program\n\nHello Everyone!\n\nIn this tutorial, we will learn how to demonstrate the concept of Hierarchial Inheritance, in the C++ programming language.\n\nTo understand the concept of Hierarchial Inheritence in CPP, we will recommend you to visit here: C++ Types of Inheritance, where we have explained it from scratch.\n\nCode:\n\n``````#include <iostream>\n\nusing namespace std;\n\n//defining the class Shape to demonstrate the concept of Hierarchial Inheritence in CPP\nclass Shape {\n//protected member variables are only accessible within the class and its descendent classes\nprotected:\nfloat width, height;\n\n//public members are accessible everywhere\npublic:\nvoid setDimensions(float w, float h) {\ncout << \"Setting the Dimensions using the parent Class: Shape\\n\";\ncout << \"The dimensions are: \" << w << \" and \" << h << \"\\n\\n\";\n\nwidth = w;\nheight = h;\n}\n};\n\n//Class Rectangle inherites the Shape class\nclass Rectangle: public Shape {\n//Method Overriding\npublic: float area() {\nreturn (width * height);\n}\n};\n\n//Class Triangle inherites the Shape class\nclass Triangle: public Shape {\n//Method Overriding\npublic: float area() {\nreturn (width * height / 2);\n}\n};\n\n//Defining the main method to access the members of the class\nint main() {\n\ncout << \"\\n\\nWelcome to Studytonight :-)\\n\\n\\n\";\ncout << \" ===== Program to demonstrate the concept of Hierarchial Inheritence in CPP ===== \\n\\n\";\n\n//Declaring the Class objects to access the class members\nRectangle rectangle;\nTriangle triangle;\n\nrectangle.setDimensions(5, 3);\ntriangle.setDimensions(2, 5);\n\ncout << \"\\nArea of the Rectangle computed using Rectangle Class is : \" << rectangle.area() << \"\\n\\n\\n\";\ncout << \"Area of the Triangle computed using Triangle Class is: \" << triangle.area();\n\ncout << \"\\n\\n\\n\";\n\nreturn 0;\n}``````\n\nOutput:", null, "We hope that this post helped you develop a better understanding of the concept of Hierarchial Inheritance in C++. For any query, feel free to reach out to us via the comments section down below.\n\nKeep Learning : )" ]

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[ "## What is a diode in parallel to an inductive load called?\n\nA flyback diode is a diode connected across an inductor used to eliminate flyback, which is the sudden voltage spike seen across an inductive load when its supply current is suddenly reduced or interrupted.\n\n## What happens when diodes are connected in parallel?\n\nWhen the diodes are connected in parallel the current carrying capacity of the diodes increases. If the load current is greater than the diode current rating of one diode, then more than one diode can be connected in parallel to increase the forward current rating.\n\nWhat is the purpose of diode in parallel with switch if load is inductive?\n\nIf the load is inductive, there are times when the switch (MOSFET) must be on, but current flows in the oposite direction. The diode gives this current a path to flow. If the diode is not used, the inductive current ceases instantly, generating high voltage peaks.\n\nWhy diode is connected in parallel with relay?\n\nA flyback diode or freewheeling diode is placed with reverse polarity from the power supply and in parallel to the relay’s inductance coil. The use of a diode in a relay circuit prevents huge voltage spikes from arising when the power supply is disconnected.\n\n### Why diode is used in IGBT?\n\nThis means that unlike a MOSFET, IGBTs cannot conduct in the reverse direction. In bridge circuits, where reverse current flow is needed, an additional diode (called a freewheeling diode) is placed in parallel (actually anti-parallel) with the IGBT to conduct current in the opposite direction.\n\n### Can I put 2 diodes in parallel?\n\nIt is not recommended to connect two diodes in parallel. Every diode has a slightly different forward voltage; even diodes with the same part number are not perfectly matched. If two diodes are connected in parallel, the one with the lower voltage drop will conduct most of the current.\n\nWhat happens when diode is connected in series?\n\nIn this case, one Zener diode will be forward biased while the other is reverse biased. In the end-to-end arrangement, the cathode of one diode is connected to the anode of another diode, so both will be forward biased or both will be reverse biased. End-to-end and back-to-back current-voltage diodes in series.\n\nWhat is the purpose of anti parallel diode?\n\nAntiparallel diodes are often used for ESD protection in ICs. Different ground or supply domains at the same potential or voltage may be wired separately for isolation reasons. However, during an ESD event across the domains, one would want a path for the high current to traverse.\n\n## Do I need a diode on a relay?\n\nWith a transistor output driving a relay, a freewheeling diode is absolutely necessary, because the voltage spike will destroy the transistor. When using a switch to turn on/off a relay, omitting the diode will work, but your switch will be happier when you use a diode (one individual diode for each relay).\n\nParallel connection means the components are connected across each other, having two common points. Current differs across each component while voltage drop is same. When diodes are connected in parallel, this same trend is observed. Diode Characteristics in Parallel Configuration Current carrying capacity increases.\n\n## How are inductors in parallel series and parallel connection?\n\nInductors in Parallel. In the parallel connection, the voltage across each inductor is equal and also if the total current is changed, the voltage drop across each individual inductor will be less as compared with series connection. For a given rate of change of current, less will be the inductance in less voltage.\n\nHow is a flyback diode chosen for an inductor?\n\nFor an ideal flyback diode selection, a diode which has very large peak forward current capacity (to handle voltage transients without burning out the diode) should be selected, moreover, low forward voltage drop, and a reverse breakdown voltage fitted the inductor’s power supply.\n\nHow are Schottky diodes connected in a circuit?\n\nSchottky diodes are connected in parallel at both ends of the line. When the current flowing in the coil disappears, the induced electromotive force generated by the coil is consumed by the work formed by the diode and the coil. This protects the safety of other components in the circuit.\n\nWhat is a diode in parallel to an inductive load called? A flyback diode is a diode connected across an inductor used to eliminate flyback, which is the sudden voltage spike seen across an inductive load when its supply current is suddenly reduced or interrupted. What happens when diodes are connected in parallel? When the…" ]

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[ "#### Howdy, Stranger!\n\nIt looks like you're new here. If you want to get involved, click one of these buttons!\n\nOptions\n\n# Lecture 47 - Chapter 3: Adjoint Functors\n\nedited June 2018\n\nWe spent a lot of time in Chapter 2 studying Galois connections. Now all that time is going to pay off! We're going to generalize Galois connections between preorders, and get 'adjoint functors' between categories. These are one of most fruitful ideas in category theory.\n\nRemember, given two preorders $$A$$ and $$B$$, a Galois connection is a pair of monotone maps $$f: A \\to B$$ and $$g: B \\to A$$ such that\n\n$$f(a) \\le b \\textrm{ if and only if } a \\le g(b)$$ for all $$a \\in A, b \\in B$$. We call $$f$$ the left adjoint of $$g$$, and $$g$$ the right adjoint of $$f$$.\n\nBut now we know that a preorder is a special sort of category: a category where there's at most one morphism from any object $$x$$ to any object $$y$$. We denote the existence of a morphism by $$x \\le y$$.\n\nSo how do we generalize Galois connections to arbitrary categories? A first guess might be this:\n\nFirst Attempt at a Generalization. Given categories $$\\mathcal{A}$$ and $$\\mathcal{B}$$, an adjunction is pair of functors $$F: \\mathcal{A} \\to \\mathcal{B}$$ and $$G: \\mathcal{B} \\to \\mathcal{A}$$ such that there is a one-to-one correspondence between morphisms\n\n$$m: F(a) \\to b$$ and morphisms\n\n$$n : a \\to G(b)$$ for all objects $$a$$ in $$\\mathcal{A}$$ and $$b$$ in $$\\mathcal{B}$$. We call $$F$$ the left adjoint of $$G$$, and $$G$$ the right adjoint of $$F$$.\n\nYou can check that this generalization does indeed reduce to the definition of Galois connection when our categories are preorders. However, it's not the right generalization!\n\nThe reason is that merely saying there exists a one-to-one correspondence is not enough to do much. We should at the very least specify a one-to-one correspondence, and make it part of the definition.\n\nThis is not good enough either, but it's a step in the right direction, so let's write it down.\n\nRemember, the set of morphisms from $$F(a)$$ to $$b$$ in the category $$\\mathcal{B}$$ is called $$\\mathcal{B}(F(a),b)$$. The set of morphisms from $$a$$ to $$G(b)$$ in the category $$\\mathcal{A}$$ is called $$\\mathcal{A}(a,G(b))$$. So, we're going to specify a one-to-one correspondence, also known as a bijection, between these two sets.\n\nSecond Attempt at a Generalization. Given categories $$\\mathcal{A}$$ and $$\\mathcal{B}$$, an adjunction is pair of functors $$F: \\mathcal{A} \\to \\mathcal{B}$$ and $$G: \\mathcal{B} \\to \\mathcal{A}$$ together with a bijection\n\n$$\\alpha_{a,b} : \\mathcal{B}(F(a),b) \\to \\mathcal{A}(a,G(b))$$ for every pair of objects $$a$$ in $$\\mathcal{A}$$ and $$b$$ in $$\\mathcal{B}$$. We call $$F$$ the left adjoint of $$G$$, and $$G$$ the right adjoint of $$F$$.\n\nThis is getting nice. We could actually do a fair amount with this. But for serious work we'll need $$\\alpha$$ to be a natural isomorphism, which gives a bijection $$\\alpha_{a,b}$$ when we feed it a pair of objects $$(a,b)$$.\n\nAnd this, in turn, means that we need to interpret $$\\mathcal{B}(F(-),-)$$ and $$\\mathcal{A}(-,G(-))$$ as functors, which give sets $$\\mathcal{B}(F(a),b)$$ and $$\\mathcal{A}(a,G(b))$$ when we feed them a pair of objects $$(a,b)$$. The blanks stand for slots where we can feed in $$a$$ and $$b$$ - this is standard math notation.\n\nDefining these functors takes a little work. Let me run through it rapidly now in a sketchy way. Next time we'll look at some examples, and eventually we'll need to get serious about the details.\n\nFor any pair of categories $$\\mathcal{A}$$ and $$\\mathcal{B}$$ there's a category $$\\mathcal{A} \\times \\mathcal{B}$$, called the 'product' of $$\\mathcal{A}$$ and $$\\mathcal{B}$$. Objects here are pairs $$(a,b)$$ where $$a$$ is in $$\\mathcal{A}$$ and $$b$$ is in $$\\mathcal{B}$$; There's more that needs to be said - I haven't filled in all the details - but let's move ahead.\n\nFor any category $$\\mathcal{A}$$ there a category $$\\mathcal{A}^{\\text{op}}$$ called the opposite of $$\\mathcal{A}$$, which has the same objects but has morphisms going the opposite way. In other words, a morphism $$f: a \\to a'$$ in $$\\mathcal{A}^{\\text{op}}$$ is defined to be morphism $$f: a' \\to a$$. There's more that needs to be said, but let's keep moving!\n\nGiven functors $$F: \\mathcal{A} \\to \\mathcal{B}$$ and $$G: \\mathcal{B} \\to \\mathcal{A}$$, I claim that there are functors\n\n$$\\mathcal{B}(F(-),-) : \\mathcal{A}^{\\text{op}} \\times \\mathcal{B} \\to \\mathbf{Set}$$ and\n\n$$\\mathcal{A}(-,G(-)) :\\mathcal{A}^{\\text{op}} \\times \\mathcal{B} \\to \\mathbf{Set}$$ that give sets $$\\mathcal{B}(F(a),b)$$ and $$\\mathcal{A}(a,G(b))$$, respectively, when we feed them an object $$(a,b) \\in \\mathcal{A}^{\\text{op}} \\times \\mathcal{B}$$.\n\nSo, for an adjunction, what we want is a natural isomorphism between these functors!\n\nNow we are ready - except for the important details I left out - for the correct definition:\n\nDefinition. Given categories $$\\mathcal{A}$$ and $$\\mathcal{B}$$, an adjunction is pair of functors $$F: \\mathcal{A} \\to \\mathcal{B}$$ and $$G: \\mathcal{B} \\to \\mathcal{A}$$ together with a natural isomorphism\n\n$$\\alpha : \\mathcal{B}(F(-),-) \\to \\mathcal{A}(-,G(-)).$$ We call $$F$$ the left adjoint of $$G$$, and $$G$$ the right adjoint of $$F$$.\n\nNote that the final definition is shorter than our two previous guesses! But it packs a bigger punch... and it takes more work to really understand it. We'll start next time.\n\nTo read other lectures go here.\n\n## Comments\n\n• Options\n1.\nedited June 2018\n\nSince I didn't teach a class on Thursday and Friday, I did it on Saturday and Sunday! Now I'm caught up.", null, "I'm sorry for any inconvenience this causes: I know you folks may have trouble dealing with an erratic flood of lectures. Take your time and ask lots of questions whenever you get around to it!\n\nComment Source:Since I didn't teach a class on Thursday and Friday, I did it on Saturday and Sunday! Now I'm caught up. <img src = \"http://math.ucr.edu/home/baez/emoticons/thumbsup.gif\"> I'm sorry for any inconvenience this causes: I know you folks may have trouble dealing with an erratic flood of lectures. Take your time and ask lots of questions whenever you get around to it!\n• Options\n2.\n\nSo, how is that exactly? We are interested in the functor $$F':\\mathcal{A}^{op} \\times \\mathcal{B} \\rightarrow \\boldsymbol {\\text{Set}}.$$ For that we consider the functor $$F: \\mathcal{A} \\rightarrow \\mathcal{B}.$$ Is this correct? There isn't a functor $$F^{op}: \\mathcal{A}^{op} \\rightarrow \\mathcal{B} \\; ?$$\n\nA morphism $$f: X \\rightarrow Y$$ in $$\\mathcal{A}^{op}$$ then induces a map of $$\\mathcal{B} (F (\\mathcal{A}^{op} ),\\mathcal {B})$$ like following?", null, "Hmm, but actually we are looking for a map on $$\\mathcal{B} (F (\\mathcal{A} ),\\mathcal {B})$$ instead of $$\\mathcal{B} (F (\\mathcal{A}^{op} ),\\mathcal {B})$$... ???\n\nComment Source:So, how is that exactly? We are interested in the functor \\$$F':\\mathcal{A}^{op} \\times \\mathcal{B} \\rightarrow \\boldsymbol {\\text{Set}}. \\$$ For that we consider the functor \\$$F: \\mathcal{A} \\rightarrow \\mathcal{B}. \\$$ Is this correct? There isn't a functor \\$$F^{op}: \\mathcal{A}^{op} \\rightarrow \\mathcal{B} \\; ?\\$$ A morphism \\$$f: X \\rightarrow Y \\$$ in \\$$\\mathcal{A}^{op} \\$$ then induces a map of \\$$\\mathcal{B} (F (\\mathcal{A}^{op} ),\\mathcal {B}) \\$$ like following?  Hmm, but actually we are looking for a map on \\$$\\mathcal{B} (F (\\mathcal{A} ),\\mathcal {B}) \\$$ instead of \\$$\\mathcal{B} (F (\\mathcal{A}^{op} ),\\mathcal {B}) \\$$... ???\n• Options\n3.\nedited June 2018\n\n@Peter – it might be easier to first consider just one category $$\\mathcal{C}$$ and try to build a functor that sends a pair of objects $$(X, Y)$$ to the set of morphisms from $$X$$ to $$Y$$, namely $$\\mathcal{C}(X, Y)$$.\n\nIf you think about how this might works on pairs of morphisms $$(f, g)$$ you'll notice it has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from $$\\mathcal{C}^{op} \\times \\mathcal{C}$$ to $$\\textbf{Set}$$.\n\nThe functors $$\\mathcal{B}(F(-), -)$$ and $$\\mathcal{A}(-, G(-))$$ are variants of this construction.\n\nComment Source:@Peter – it might be easier to first consider just one category \\$$\\mathcal{C}\\$$ and try to build a functor that sends a pair of objects \\$$(X, Y)\\$$ to the set of morphisms from \\$$X\\$$ to \\$$Y\\$$, namely \\$$\\mathcal{C}(X, Y)\\$$. If you think about how this might works on pairs of morphisms \\$$(f, g)\\$$ you'll notice it has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from \\$$\\mathcal{C}^{op} \\times \\mathcal{C}\\$$ to \\$$\\textbf{Set}\\$$. The functors \\$$\\mathcal{B}(F(-), -)\\$$ and \\$$\\mathcal{A}(-, G(-))\\$$ are variants of this construction.\n• Options\n4.\n\nAnindya wrote:\n\n[I]t has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from $$\\mathcal{C}^{op} \\times \\mathcal{C}$$ to $$\\textbf{Set}$$.\n\nI'm not sure if this relevant for the current discussion, but I wanted to point out that if a map is functorial in each argument it's not a sufficient condition to conclude that it's a functor. We need an additional coherence condition, known as the \"interchange law\"; for example, see the bifunctor lemma stated in chapter 7 from Awodey's book.\n\nComment Source:Anindya wrote: > [I]t has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from \\$$\\mathcal{C}^{op} \\times \\mathcal{C}\\$$ to \\$$\\textbf{Set}\\$$. I'm not sure if this relevant for the current discussion, but I wanted to point out that if a map is functorial in each argument it's not a sufficient condition to conclude that it's a functor. We need an additional coherence condition, known as the \"interchange law\"; for example, see the bifunctor lemma stated in [chapter 7](https://www.andrew.cmu.edu/course/80-413-713/notes/chap07.pdf) from Awodey's book.\n• Options\n5.\nedited June 2018\n\nAn adjunction between categories looks like the diagram below when you draw out a commuting diagram.\n\nUsing John's examples:\n\n$\\begin{matrix} A & \\overset{F}\\Rightarrow & F(A) \\\\ n\\downarrow & & \\downarrow m \\\\ G(B) & \\underset{G}\\Leftarrow & B \\end{matrix}$\n\nComment Source:An adjunction between categories looks like the diagram below when you draw out a commuting diagram. Using John's examples: \\$\\begin{matrix} A & \\overset{F}\\Rightarrow & F(A) \\\\\\\\ n\\downarrow & & \\downarrow m \\\\\\\\ G(B) & \\underset{G}\\Leftarrow & B \\end{matrix} \\$\n• Options\n6.\nedited June 2018\n\nPeter wrote:\n\nSo, how is that exactly? We are interested in the functor $$F':\\mathcal{A}^{op} \\times \\mathcal{B} \\rightarrow \\boldsymbol {\\text{Set}}.$$ For that we consider the functor $$F: \\mathcal{A} \\rightarrow \\mathcal{B}.$$ Is this correct? There isn't a functor $$F^{op}: \\mathcal{A}^{op} \\rightarrow \\mathcal{B} \\; ?$$\n\nNo, there's not a functor from $$\\mathcal{A}^{\\text{ op}}$$ to $$\\mathcal{B}$$ in this game. We have a functor $$F : \\mathcal{A} \\to \\mathcal{B}$$. There's no way to turn that into a functor $$\\mathcal{A}^{\\text{op}}$$ to $$\\mathcal{B}$$.\n\nThe \"op\" comes in when we consider\n\n$$\\mathcal{B}(F(-),-) : \\mathcal{A}^{\\mathrm{op}} \\times \\mathcal{B} \\to \\mathbf{Set}$$ This functor sends any pair $$a, b$$ consisting of an object $$a$$ in $$\\mathcal{A}$$ and an object $$b$$ in $$\\mathcal{B}$$ to the set\n\n$$\\mathcal{B}(F(a),b) .$$ But what does this functor do to morphisms? This is where the \"op\" comes in! If we have a pair of morphisms $$f: a' \\to a$$ and $$g: b \\to b'$$ this functor gives us a map from\n\n$$\\mathcal{B}(F(a),b)$$ to\n\n$$\\mathcal{B}(F(a'),b')$$ Puzzle. Can somehow say how we define this map? It's my duty to explain this, but someone probably knows already.\n\nThe key point is that to get this map is built from a pair $$(f,g)$$ consisting of a morphism $$g: b \\to b'$$ going forwards from $$b$$ to $$b'$$, and a morphism $$f: a' \\to a$$ going backwards from $$a'$$ to $$a$$. That backwardsness is why we need $$\\mathcal{A}^\\textrm{op}$$.\n\nComment Source:Peter wrote: > So, how is that exactly? We are interested in the functor \\$$F':\\mathcal{A}^{op} \\times \\mathcal{B} \\rightarrow \\boldsymbol {\\text{Set}}. \\$$ For that we consider the functor \\$$F: \\mathcal{A} \\rightarrow \\mathcal{B}. \\$$ Is this correct? There isn't a functor \\$$F^{op}: \\mathcal{A}^{op} \\rightarrow \\mathcal{B} \\; ?\\$$ No, there's not a functor from \\$$\\mathcal{A}^{\\text{ op}} \\$$ to \\$$\\mathcal{B}\\$$ in this game. We have a functor \\$$F : \\mathcal{A} \\to \\mathcal{B} \\$$. There's no way to turn that into a functor \\$$\\mathcal{A}^{\\text{op}} \\$$ to \\$$\\mathcal{B}\\$$. The \"op\" comes in when we consider $\\mathcal{B}(F(-),-) : \\mathcal{A}^{\\mathrm{op}} \\times \\mathcal{B} \\to \\mathbf{Set}$ This functor sends any pair \\$$a, b\\$$ consisting of an object \\$$a\\$$ in \\$$\\mathcal{A}\\$$ and an object \\$$b\\$$ in \\$$\\mathcal{B}\\$$ to the set $\\mathcal{B}(F(a),b) .$ But what does this functor do to morphisms? This is where the \"op\" comes in! If we have a pair of morphisms \\$$f: a' \\to a\\$$ and \\$$g: b \\to b'\\$$ this functor gives us a map from $\\mathcal{B}(F(a),b)$ to $\\mathcal{B}(F(a'),b')$ **Puzzle.** Can somehow say how we define this map? It's my duty to explain this, but someone probably knows already. The key point is that to get this map is built from a pair \\$$(f,g) \\$$ consisting of a morphism \\$$g: b \\to b'\\$$ going _forwards_ from \\$$b\\$$ to \\$$b'\\$$, and a morphism \\$$f: a' \\to a\\$$ going _backwards_ from \\$$a'\\$$ to \\$$a\\$$. That backwardsness is why we need \\$$\\mathcal{A}^\\textrm{op}\\$$. \n• Options\n7.\nedited June 2018\n\nI think in my diagram I have defined it at least for the case where $$g$$ is the identity. Due to the fact that f is in the opposite category, we can define the map from $$\\mathcal{A}(F(a),b)$$ to $$\\mathcal{A}(F(a'),b)$$ by the composite $$h \\mapsto F(f) \\circ h$$.\n\nComment Source:I think in my diagram I have defined it at least for the case where \\$$g\\$$ is the identity. Due to the fact that f is in the opposite category, we can define the map from \\$$\\mathcal{A}(F(a),b)\\$$ to \\$$\\mathcal{A}(F(a'),b)\\$$ by the composite \\$$h \\mapsto F(f) \\circ h\\$$.\n• Options\n8.\n\nI think you got your adjunctions backward John.\n\nYou wrote,\n\nThis functor sends any pair $$a, b$$ consisting of an object $$a$$ in $$\\mathcal{A}$$ and an object $$b$$ in $$\\mathcal{B}$$ to the set\n\nimplying that,\n\n$b \\text{ in } Obj(\\mathcal{B}),$\n\nhowever next you wrote,\n\n$$\\mathcal{A}(F(a),b) .$$\n\nimplying that,\n\n$b \\text{ in } Obj(\\mathcal{A}).$\n\nComment Source:I think you got your adjunctions backward John. You wrote, >This functor sends any pair \\$$a, b\\$$ consisting of an object \\$$a\\$$ in \\$$\\mathcal{A}\\$$ and an object \\$$b\\$$ in \\$$\\mathcal{B}\\$$ to the set implying that, \\$b \\text{ in } Obj(\\mathcal{B}),\\$ however next you wrote, >$\\mathcal{A}(F(a),b) .$ implying that, \\$b \\text{ in } Obj(\\mathcal{A}).\\$\n• Options\n9.\nedited June 2018\n\nDan wrote:\n\nI'm not sure if this relevant for the current discussion, but I wanted to point out that if a map is functorial in each argument it's not a sufficient condition to conclude that it's a functor. We need an additional coherence condition, known as the \"interchange law\"; for example, see the bifunctor lemma stated in chapter 7 from Awodey's book.\n\nThere are a couple of different ways to get your hands on a functor $$F: \\mathcal{A} \\times \\mathcal{B} \\times \\mathcal{C}$$ . Awodey is assuming we have categories $$\\mathcal{A} , \\mathcal{B}, \\mathcal{C}$$ and two maps:\n\nA) a map sending each pair of objects $$(a,b) \\in \\mathbf{Ob}(\\mathcal{A}) \\times \\mathbf{Ob}(\\mathcal{B})$$ to an object of $$\\mathcal{C}$$;\n\nB) a map I'll call $$F$$ sending each pair of morphisms $$(f,g) \\in \\mathbf{Mor}(\\mathcal{A}) \\times \\mathbf{Mor}(\\mathcal{B})$$ to a morphism of $$\\mathcal{C}$$.\n\nHe gives a necessary and sufficient condition for this data to come from a (unique) functor from $$\\mathcal{A} \\times \\mathcal{B}$$ to $$\\mathcal{C}$$. Namely, we need\n\n1. $$F$$ preserves composition in each argument separately and\n\n2. The interchange law $$F(f,1) \\circ F(1,g) = F(1,g) \\circ F(f,1)$$ holds, where I'm writing $$1$$ for a bunch of different identity morphisms (the only ones that make the equation parse.)\n\nBut there's another equivalent way to state conditions 1 and 2, which is sometimes just as easy to check. Namely, we can simply require that\n\n$$F(1_x,1_y) = 1_{(x,y)}$$ and\n\n$$F(f,g) \\circ F(f',g') = F(f \\circ f', g \\circ g') .$$ This just says straight out that our would-be functor is a functor!\n\nComment Source:Dan wrote: > I'm not sure if this relevant for the current discussion, but I wanted to point out that if a map is functorial in each argument it's not a sufficient condition to conclude that it's a functor. We need an additional coherence condition, known as the \"interchange law\"; for example, see the bifunctor lemma stated in [chapter 7](https://www.andrew.cmu.edu/course/80-413-713/notes/chap07.pdf) from Awodey's book. There are a couple of different ways to get your hands on a functor \\$$F: \\mathcal{A} \\times \\mathcal{B} \\times \\mathcal{C} \\$$ . Awodey is assuming we have categories \\$$\\mathcal{A} , \\mathcal{B}, \\mathcal{C} \\$$ and two maps: A) a map sending each pair of objects \\$$(a,b) \\in \\mathbf{Ob}(\\mathcal{A}) \\times \\mathbf{Ob}(\\mathcal{B})\\$$ to an object of \\$$\\mathcal{C}\\$$; B) a map I'll call \\$$F\\$$ sending each pair of morphisms \\$$(f,g) \\in \\mathbf{Mor}(\\mathcal{A}) \\times \\mathbf{Mor}(\\mathcal{B})\\$$ to a morphism of \\$$\\mathcal{C}\\$$. He gives a necessary and sufficient condition for this data to come from a (unique) functor from \\$$\\mathcal{A} \\times \\mathcal{B}\\$$ to \\$$\\mathcal{C} \\$$. Namely, we need 1. \\$$F\\$$ preserves composition in each argument separately and 2. The **interchange law** \\$$F(f,1) \\circ F(1,g) = F(1,g) \\circ F(f,1) \\$$ holds, where I'm writing \\$$1\\$$ for a bunch of different identity morphisms (the only ones that make the equation parse.) But there's another equivalent way to state conditions 1 and 2, which is sometimes just as easy to check. Namely, we can simply require that $F(1_x,1_y) = 1_{(x,y)}$ and $F(f,g) \\circ F(f',g') = F(f \\circ f', g \\circ g') .$ This just says straight out that our would-be functor is a functor! \n• Options\n10.\n\nKeith wrote:\n\nhowever next you wrote,\n\n$$\\mathcal{A}(F(a),b) .$$\n\nYes, that should be $$\\mathcal{B}(F(a),b)$$. I'll fix it. Thanks!\n\nComment Source:Keith wrote: > however next you wrote, >$\\mathcal{A}(F(a),b) .$ Yes, that should be \\$$\\mathcal{B}(F(a),b)\\$$. I'll fix it. Thanks!\n• Options\n11.\nedited June 2018\n\nI wrote:\n\nIf we have a pair of morphisms $$f: a' \\to a$$ and $$g: b \\to b'$$ this functor gives us a map from\n\n$$\\mathcal{B}(F(a),b)$$ to $$\\mathcal{B}(F(a'),b')$$\nPuzzle. Can somehow say how we define this map?\n\nPeter wrote:\n\nI think in my diagram I have defined it at least for the case where $$g$$ is the identity. Due to the fact that f is in the opposite category, we can define the map from $$\\mathcal{A}(F(a),b)$$ to $$\\mathcal{A}(F(a'),b)$$ by the composite $$h \\mapsto F(f) \\circ h$$.\n\nYes, right!\n\nThanks to Dan's remark about Steve Awodey's book, expanded on here, we can get the desired map\n\n$$\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b')$$ as follows. First use $$f: a' \\to a$$ to cook up a map\n\n$$\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) .$$ Then use $$g: b \\to b'$$ to cook up a map\n\n$$\\mathcal{B}(F(a'),b) \\to \\mathcal{B}(F(a'),b')$$ Finally, compose these two maps to get the job done:\n\n$$\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) \\to \\mathcal{B}(F(a'),b') .$$ You explained the first step, where we use $$f : a' \\to a$$. This gives a map\n\n$$\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b)$$ that maps any morphism $$h$$ to $$h \\circ F(f)$$.\n\nComment Source:I wrote: > If we have a pair of morphisms \\$$f: a' \\to a\\$$ and \\$$g: b \\to b'\\$$ this functor gives us a map from > $\\mathcal{B}(F(a),b)$ > to > $\\mathcal{B}(F(a'),b')$ > **Puzzle.** Can somehow say how we define this map? Peter wrote: > I think in my diagram I have defined it at least for the case where \\$$g\\$$ is the identity. Due to the fact that f is in the opposite category, we can define the map from \\$$\\mathcal{A}(F(a),b)\\$$ to \\$$\\mathcal{A}(F(a'),b)\\$$ by the composite \\$$h \\mapsto F(f) \\circ h\\$$. Yes, right! Thanks to Dan's remark about Steve Awodey's book, expanded on [here](https://forum.azimuthproject.org/discussion/comment/19552/#Comment_19552), we can get the desired map $\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b')$ as follows. First use \\$$f: a' \\to a\\$$ to cook up a map $\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) .$ Then use \\$$g: b \\to b'\\$$ to cook up a map $\\mathcal{B}(F(a'),b) \\to \\mathcal{B}(F(a'),b')$ Finally, compose these two maps to get the job done: $\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) \\to \\mathcal{B}(F(a'),b') .$ You explained the first step, where we use \\$$f : a' \\to a\\$$. This gives a map $\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b)$ that maps any morphism \\$$h\\$$ to \\$$h \\circ F(f)\\$$. \n• Options\n12.\n\nFirst use $$f: a' \\to a$$ to cook up a map\n\n$$\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) .$$\n\nWhat is happening to cause a morphism,\n\n$f: a' \\to a$\n\nto be flipped backward when being mapped to,\n\n$$\\mathcal{B}(F(f),b) : \\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) ?$$\n\nComment Source:>First use \\$$f: a' \\to a\\$$ to cook up a map >$\\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) .$ What is happening to cause a morphism, \\$f: a' \\to a\\$ to be flipped backward when being mapped to, $\\mathcal{B}(F(f),b) : \\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) ?$\n• Options\n13.\nedited June 2018\n\nA functor $$\\alpha : (A^{op} \\times B) \\to Set$$ is called an anafunctor from A to B correct?\n\nComment Source:A functor \\$$\\alpha : (A^{op} \\times B) \\to Set\\$$ is called an anafunctor from A to B correct?\n• Options\n14.\n\nChristopher, I think you're looking for profunctors. Anafunctors are somewhat different.\n\nComment Source:Christopher, I think you're looking for [profunctors](https://ncatlab.org/nlab/show/profunctor). Anafunctors are [somewhat different](https://ncatlab.org/nlab/show/anafunctor).\n• Options\n15.\n\nYes. You are right.\n\nComment Source:Yes. You are right.\n• Options\n16.\nedited June 2018\n\nKeith wrote:\n\nWhat is happening to cause a morphism,\n\n$f: a' \\to a$\n\nto be flipped backward when being mapped to,\n\n$$\\mathcal{B}(F(f),b) : \\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) ?$$\n\nPeter Addor gave the formula for this map. It sends any guy\n\n$$h \\in \\mathcal{B}(F(a),b)$$ to\n\n$$h \\circ F(f) \\in \\mathcal{B}(F(a'),b) .$$ That is, it sends any guy\n\n$$h: F(a) \\to b$$ to\n\n$$h \\circ F(f) : F(a') to b .$$ We are precomposing with $$F(f)$$.\n\nIf you think about this, you'll see why we need $$f: a' \\to a$$, not $$f: a \\to a'$$.\n\nPostcomposing does not cause this flip.\n\nSimply put: if you're riding from Philadelphia to New York, and you want to extend your trip, you can either postcompose with a trip from New York to Boston, or precompose with a trip from Washington DC to Philadelphia.\n\nThe second option, stretching out our trip so it includes Washington DC, does not require a trip from your original starting point (Philadelphia) to your new one (Washington DC). It requires a trip the other way, from Washington DC to Philadelphia! This is the 'flip'.\n\nIt doesn't feel like a flip when you put it this way; it seems utterly reasonable.\n\nWe will talk about this more later....\n\nComment Source:Keith wrote: > What is happening to cause a morphism, > \\$f: a' \\to a\\$ > to be flipped backward when being mapped to, > $\\mathcal{B}(F(f),b) : \\mathcal{B}(F(a),b) \\to \\mathcal{B}(F(a'),b) ?$ Peter Addor gave the formula for this map. It sends any guy $h \\in \\mathcal{B}(F(a),b)$ to $h \\circ F(f) \\in \\mathcal{B}(F(a'),b) .$ That is, it sends any guy $h: F(a) \\to b$ to $h \\circ F(f) : F(a') to b .$ We are _precomposing_ with \\$$F(f)\\$$. If you think about this, you'll see why we need \\$$f: a' \\to a\\$$, not \\$$f: a \\to a'\\$$. _Postcomposing_ does not cause this flip. Simply put: if you're riding from Philadelphia to New York, and you want to extend your trip, you can either postcompose with a trip from New York to Boston, or precompose with a trip from Washington DC to Philadelphia. The second option, stretching out our trip so it includes Washington DC, does not require a trip from your original starting point (Philadelphia) to your new one (Washington DC). It requires a trip _the other way_, from Washington DC to Philadelphia! This is the 'flip'. It doesn't feel like a flip when you put it this way; it seems utterly reasonable. We will talk about this more later....\n• Options\n17.\nedited June 2018\n\nI have been confused by the 'op' thing for quite a while. On reading Anindya's answer:\n\nIf you think about how this might works on pairs of morphisms $$(f, g)$$ you'll notice it has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from $$\\mathcal{C}^{op} \\times \\mathcal{C}$$ to $$\\textbf{Set}$$.\n\nI reckon this is a good way to approach the issue, but my brain fails to think out anything. :-/ @Anindya Could you perhaps illustrate a bit more details of the thinking process (even just some informal hints)?\n\nOn that note, I notice many textbooks use the 'op' notation without really explicitly explaining why, perhaps because it is too obvious for mathematically better-equipped readers. I see @John might still explain this in his latest lectures, so this course is really making a difference for beginner-beginners. =D>\n\nComment Source:I have been confused by the 'op' thing for quite a while. On reading [Anindya's answer](https://forum.azimuthproject.org/discussion/comment/19538/#Comment_19538): >If you think about how this might works on pairs of morphisms \\$$(f, g)\\$$ you'll notice it has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from \\$$\\mathcal{C}^{op} \\times \\mathcal{C}\\$$ to \\$$\\textbf{Set}\\$$. I reckon this is a good way to approach the issue, but my brain fails to think out anything. :-/ @Anindya Could you perhaps illustrate a bit more details of the thinking process (even just some informal hints)? On that note, I notice many textbooks use the 'op' notation without really explicitly explaining why, perhaps because it is too obvious for mathematically better-equipped readers. I see @John might still explain this in his latest lectures, so this course is really making a difference for beginner-beginners. =D> \n• Options\n18.\nedited June 2018\n\n$$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{op}$$ is a functor.\n\nIn short, it does nothing to objects but flips arrows around.\n\nPuzzle: Prove that $$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{op}$$ is a functor.\n\nComment Source:\\$$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{op}\\$$ is a functor. In short, it does nothing to objects but flips arrows around. **Puzzle:** Prove that \\$$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{op}\\$$ is a functor. \n• Options\n19.\nedited June 2018\n\nNo, there is no such thing as a functor $$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{\\mathrm{op}}$$. Functors can't 'flip arrows around', since a functor $$F$$ maps a morphism $$f : x \\to y$$ to a morphism $$F(f) : F(x) \\to F(y)$$.\n\nThis is a great example of a level slip! What we really have is a functor\n\n$$\\mathbf{op} : \\mathbf{Cat} \\to \\mathbf{Cat}$$ This functor sends each object of $$\\mathbf{Cat}$$ - that is, each category $$\\mathcal{C}$$ - to its opposite:\n\n$$\\mathbf{op}(\\mathcal{C}) = \\mathcal{C}^{\\mathrm{op}}$$ This functor send each morphism of $$\\mathbf{Cat}$$ - that is, each functor $$F: \\mathcal{C} \\to \\mathcal{D}$$ - to some obvious thing.\n\nPuzzle. What is this obvious thing?\n\nComment Source:No, there is no such thing as a functor \\$$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{\\mathrm{op}}\\$$. Functors can't 'flip arrows around', since a functor \\$$F\\$$ maps a morphism \\$$f : x \\to y\\$$ to a morphism \\$$F(f) : F(x) \\to F(y)\\$$. This is a great example of a level slip! What we really have is a functor $\\mathbf{op} : \\mathbf{Cat} \\to \\mathbf{Cat}$ This functor sends each object of \\$$\\mathbf{Cat} \\$$ - that is, each category \\$$\\mathcal{C}\\$$ - to its opposite: $\\mathbf{op}(\\mathcal{C}) = \\mathcal{C}^{\\mathrm{op}}$ This functor send each morphism of \\$$\\mathbf{Cat} \\$$ - that is, each functor \\$$F: \\mathcal{C} \\to \\mathcal{D} \\$$ - to some obvious thing. **Puzzle.** What is this obvious thing? \n• Options\n20.\nedited June 2018\n\nJulio wrote:\n\nI have been confused by the 'op' thing for quite a while.\n\nGood! This will build up the desire in you to do lots of calculations, to resolve this confusion.", null, "For starters, do you understand what the category $$\\mathcal{C}^{\\mathbf{op}}$$ is?\n\nIf so, the next step is to understand why there is no functor\n\n$$\\mathrm{hom} : \\mathcal{C} \\times \\mathcal{C} \\to \\mathrm{Set}$$ sending each pair of objects $$(c,c')$$ to the 'homset' $$\\mathcal{C}(c,c')$$ - that is, the set of morphisms from $$c$$ to $$c'$$.\n\nThe best possible way is for you to try to define this (nonexistent) functor and see what goes wrong. I've told you what it does to an object of $$\\mathcal{C} \\times \\mathcal{C}$$ - that is, a pair $$(c,c')$$ of objects in $$\\mathcal{C}$$. So your job is to define what this functor does to morphisms.\n\nIn other words, given\n\n$$(f,f'): (c,c') \\to (d,d') ,$$ try to cook up a map from $$\\mathcal{C}(c,c')$$ to $$\\mathcal{C}(d,d')$$.\n\nWhen you try this, and see exactly why it doesn't work, you'll be ready to see why we do have a functor\n\n$$\\mathrm{hom} : \\mathcal{C}^{\\mathrm{op}} \\times \\mathcal{C} \\to \\mathrm{Set}$$ You can then define what this does to morphisms, and show that it does work.\n\nThere's really nothing better, when learning math, than just doing calculatiions and seeing what happens. Watching people do math is like watching the Olympics. It's fun, but it doesn't make you able to run faster. In the sort of category theory we're doing now, the calculations are all really easy. It just takes a bit of courage to dive and try them.\n\nBut fear not! I will explain the hom-functor\n\n$$\\mathrm{hom} : \\mathcal{C}^{\\mathrm{op}} \\times \\mathcal{C} \\to \\mathrm{Set}$$ I need to explain it, to really explain adjoint functors!\n\nBut I decided to try, at first, to give people a bit of intuition about adjoint functors before delving into this matter.\n\nComment Source:Julio wrote: > I have been confused by the 'op' thing for quite a while. Good! This will build up the desire in you to do lots of calculations, to resolve this confusion. <img src = \"http://math.ucr.edu/home/baez/emoticons/tongue2.gif\"> For starters, do you understand what the category \\$$\\mathcal{C}^{\\mathbf{op}}\\$$ is? If so, the next step is to understand why there is no functor $\\mathrm{hom} : \\mathcal{C} \\times \\mathcal{C} \\to \\mathrm{Set}$ sending each pair of objects \\$$(c,c') \\$$ to the 'homset' \\$$\\mathcal{C}(c,c')\\$$ - that is, the set of morphisms from \\$$c\\$$ to \\$$c'\\$$. The best possible way is for you to try to define this (nonexistent) functor and see what goes wrong. I've told you what it does to an object of \\$$\\mathcal{C} \\times \\mathcal{C} \\$$ - that is, a pair \\$$(c,c')\\$$ of objects in \\$$\\mathcal{C}\\$$. So your job is to define what this functor does to _morphisms_. In other words, given $(f,f'): (c,c') \\to (d,d') ,$ try to cook up a map from \\$$\\mathcal{C}(c,c')\\$$ to \\$$\\mathcal{C}(d,d')\\$$. When you try this, and see exactly why it doesn't work, you'll be ready to see why we _do_ have a functor $\\mathrm{hom} : \\mathcal{C}^{\\mathrm{op}} \\times \\mathcal{C} \\to \\mathrm{Set}$ You can then define what this does to morphisms, and show that it _does_ work. There's really nothing better, when learning math, than just doing calculatiions and seeing what happens. _Watching_ people do math is like watching the Olympics. It's fun, but it doesn't make you able to run faster. In the sort of category theory we're doing now, the calculations are all really easy. It just takes a bit of courage to dive and try them. But fear not! I _will_ explain the hom-functor $\\mathrm{hom} : \\mathcal{C}^{\\mathrm{op}} \\times \\mathcal{C} \\to \\mathrm{Set}$ I _need_ to explain it, to really explain adjoint functors! But I decided to try, at first, to give people a bit of intuition about adjoint functors before delving into this matter. \n• Options\n21.\n\nNo, there is no such thing as a functor $$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{\\mathrm{op}}$$. Functors can't 'flip arrows around', since a functor $$F$$ maps a morphism $$f : x \\to y$$ to a morphism $$F(f) : F(x) \\to F(y)$$.\n\nThis is a great example of a level slip!\n\nWhoops! Especially after I chastised someone today for level slipping with relations...\n\nAnyways, thanks for pointing that out.\n\nComment Source:>No, there is no such thing as a functor \\$$\\mathbf{op}: \\mathcal{C} \\to \\mathcal{C}^{\\mathrm{op}}\\$$. Functors can't 'flip arrows around', since a functor \\$$F\\$$ maps a morphism \\$$f : x \\to y\\$$ to a morphism \\$$F(f) : F(x) \\to F(y)\\$$. >This is a great example of a level slip! Whoops! Especially after I chastised someone today for level slipping with relations... Anyways, thanks for pointing that out.\n• Options\n22.\n\nJohn asked in comment #19:\n\nPuzzle. What is this obvious thing?\n\nI think that the functor $$\\mathbf{op}(F)$$ behaves the same way as the functor $$F$$ on both objects and morphisms:\n\n• On objects, $$\\mathbf{op}(F) : \\mathrm{ob}(\\mathcal{C}^{\\mathrm{op}}) \\to \\mathrm{ob}(\\mathcal{D}^{\\mathrm{op}})$$. But objects in the opposite category are the same objects as in the original category, that is, $$\\mathrm{ob}(\\mathcal{C}^{\\mathrm{op}}) = \\mathrm{ob}(\\mathcal{C})$$; so the functor $$\\mathbf{op}(F) : \\mathrm{ob}(\\mathcal{C}) \\to \\mathrm{ob}(\\mathcal{D})$$ can map objects the same way as $$F$$ does.\n• On morphisms, $$\\mathbf{op}(F) : \\mathcal{C}^{\\mathrm{op}}(C', C) \\to \\mathcal{D}^{\\mathrm{op}}(\\mathbf{op}(F)(C'), \\mathbf{op}(F)(C))$$ for any $$C, C' \\in \\mathrm{ob}(\\mathcal{C})$$. The opposite category reverses the morphisms, that is, $$\\mathcal{C}^{\\mathrm{op}}(C', C) = \\mathcal{C}(C, C')$$; so the functor $$\\mathbf{op}(F) : \\mathcal{C}(C, C') \\to \\mathcal{D}(FC, FC')$$ can map morphisms the same way as $$F$$ does.\n\nSo, I would be tempted to say that the $$\\mathbf{op}$$ functor sends each functor $$F$$ to itself, that is, $$\\mathbf{op}(F) = F$$.\n\nComment Source:John asked in [comment #19](https://forum.azimuthproject.org/discussion/comment/19630/#Comment_19630): > **Puzzle.** What is this obvious thing? I think that the functor \\$$\\mathbf{op}(F)\\$$ behaves the same way as the functor \\$$F\\$$ on both objects and morphisms: - On objects, \\$$\\mathbf{op}(F) : \\mathrm{ob}(\\mathcal{C}^{\\mathrm{op}}) \\to \\mathrm{ob}(\\mathcal{D}^{\\mathrm{op}})\\$$. But objects in the opposite category are the same objects as in the original category, that is, \\$$\\mathrm{ob}(\\mathcal{C}^{\\mathrm{op}}) = \\mathrm{ob}(\\mathcal{C})\\$$; so the functor \\$$\\mathbf{op}(F) : \\mathrm{ob}(\\mathcal{C}) \\to \\mathrm{ob}(\\mathcal{D})\\$$ can map objects the same way as \\$$F\\$$ does. - On morphisms, \\$$\\mathbf{op}(F) : \\mathcal{C}^{\\mathrm{op}}(C', C) \\to \\mathcal{D}^{\\mathrm{op}}(\\mathbf{op}(F)(C'), \\mathbf{op}(F)(C))\\$$ for any \\$$C, C' \\in \\mathrm{ob}(\\mathcal{C})\\$$. The opposite category reverses the morphisms, that is, \\$$\\mathcal{C}^{\\mathrm{op}}(C', C) = \\mathcal{C}(C, C')\\$$; so the functor \\$$\\mathbf{op}(F) : \\mathcal{C}(C, C') \\to \\mathcal{D}(FC, FC')\\$$ can map morphisms the same way as \\$$F\\$$ does. So, I would be tempted to say that the \\$$\\mathbf{op}\\$$ functor sends each functor \\$$F\\$$ to itself, that is, \\$$\\mathbf{op}(F) = F\\$$.\n• Options\n23.\nedited June 2018", null, "So I think I got why we need the opposite category to send pairs of objects to homsets. As seen in the picture above in order to get from $$\\mathcal{C}(a,b)$$ to $$\\mathcal{C}(a',b)$$, we need to precompose f so that the arrows compose.\n\nNow I have a question about how the homfunctor preserves composition. I will use the same example with two objects and one non-trivial morphism above to pose the question.\n\nIf we have two composable morphisms $$\\mathcal{C}(f,1_{b'}) \\circ \\mathcal{C}(1_a,g) = \\mathcal{C}(f,g)$$ , the functor needs to preserve this via $$\\mathcal{C}(f \\circ 1_a,1_{b'} \\circ g)$$. But the order doesn't seem right in that $$f \\circ 1_a$$ and $$1_{b'} \\circ g$$ are both not composable. What am I doing wrong here?\n\nEdit: Oops $$1_{b'} \\circ g$$ does compose.\n\nComment Source: So I think I got why we need the opposite category to send pairs of objects to homsets. As seen in the picture above in order to get from \\$$\\mathcal{C}(a,b)\\$$ to \\$$\\mathcal{C}(a',b)\\$$, we need to precompose f so that the arrows compose. Now I have a question about how the homfunctor preserves composition. I will use the same example with two objects and one non-trivial morphism above to pose the question. If we have two composable morphisms \\$$\\mathcal{C}(f,1_{b'}) \\circ \\mathcal{C}(1_a,g) = \\mathcal{C}(f,g)\\$$ , the functor needs to preserve this via \\$$\\mathcal{C}(f \\circ 1_a,1_{b'} \\circ g)\\$$. But the order doesn't seem right in that \\$$f \\circ 1_a\\$$ and \\$$1_{b'} \\circ g\\$$ are both not composable. What am I doing wrong here? Edit: Oops \\$$1_{b'} \\circ g\\$$ does compose. \n• Options\n24.\nedited June 2018\n\n@Michael: The composition the functor must preserve is composition in $$\\mathcal{C}^{op}\\times \\mathcal{C}$$\n\nHow does this composition work? It takes two pairs of morphisms:\n\n$$(f, g)$$ where $$f : a' \\rightarrow a$$ and $$g : b \\rightarrow b'$$\n\n$$(f', g')$$ where $$f' : a'' \\rightarrow a'$$ and $$g : b' \\rightarrow b''$$\n\nand returns the composite pair:\n\n$$(f\\circ f', g'\\circ g)$$ where $$f\\circ f' : a'' \\rightarrow a$$ and $$g'\\circ g : b \\rightarrow b''$$\n\nSo the rule is $$(f', g')\\circ (f, g) = (f\\circ f', g'\\circ g)$$, ie the first coordinate is the \"other way round\".\n\nNow what are we trying to check? We want\n\n$$\\mathcal{C}(f', g')\\circ \\mathcal{C}(f, g) = \\mathcal{C}((f', g')\\circ (f, g)) = \\mathcal{C}(f\\circ f', g'\\circ g)$$\n\nThe left hand side is a set map sending $$k\\mapsto g'\\circ (g\\circ k\\circ f)\\circ f'$$\n\nThe right hand side is a set map sending $$k\\mapsto (g'\\circ g)\\circ k\\circ (f\\circ f')$$\n\n... and these two are the same by associativity.\n\nComment Source:@Michael: The composition the functor must preserve is composition in \\$$\\mathcal{C}^{op}\\times \\mathcal{C}\\$$ How does this composition work? It takes two pairs of morphisms: > \\$$(f, g)\\$$ where \\$$f : a' \\rightarrow a\\$$ and \\$$g : b \\rightarrow b'\\$$ > \\$$(f', g')\\$$ where \\$$f' : a'' \\rightarrow a'\\$$ and \\$$g : b' \\rightarrow b''\\$$ and returns the composite pair: > \\$$(f\\circ f', g'\\circ g)\\$$ where \\$$f\\circ f' : a'' \\rightarrow a\\$$ and \\$$g'\\circ g : b \\rightarrow b''\\$$ So the rule is \\$$(f', g')\\circ (f, g) = (f\\circ f', g'\\circ g)\\$$, ie the first coordinate is the \"other way round\". Now what are we trying to check? We want > \\$$\\mathcal{C}(f', g')\\circ \\mathcal{C}(f, g) = \\mathcal{C}((f', g')\\circ (f, g)) = \\mathcal{C}(f\\circ f', g'\\circ g)\\$$ The left hand side is a set map sending \\$$k\\mapsto g'\\circ (g\\circ k\\circ f)\\circ f'\\$$ The right hand side is a set map sending \\$$k\\mapsto (g'\\circ g)\\circ k\\circ (f\\circ f')\\$$ ... and these two are the same by associativity.\n• Options\n25.\nedited June 2018\n\n@Anindya\n\nSo the rule is $$(f', g')\\circ (f, g) = (f\\circ f', g'\\circ g)$$, ie the first coordinate is the \"other way round\".\n\nAhh the composition rule is also backwards. $$\\mathcal{C}^{op}$$ is trickier than it seems. Thanks a bunch Anindya. I would have gotten lost everywhere in opposite land had I not known this.\n\nComment Source:@Anindya >So the rule is \\$$(f', g')\\circ (f, g) = (f\\circ f', g'\\circ g)\\$$, ie the first coordinate is the \"other way round\". Ahh the composition rule is also backwards. \\$$\\mathcal{C}^{op}\\$$ is trickier than it seems. Thanks a bunch Anindya. I would have gotten lost everywhere in opposite land had I not known this. \n• Options\n26.\nedited June 2018\n\nFor those of you working out the homfunctor and why the first coordinate needs to be contravariant, I drew out the simple example used above in my comment #23. It might help calculate how this functor is working. The left diagram shows the how it maps the objects and the right diagram shows how it maps the morphisms. Let me know if there are mistakes.", null, "Edit: Now that I think about it, this is so natural since morphisms and compositions have directionality which puts the start and target on opposite ends of the morphism and composition must start where the previous left off.\n\nComment Source:For those of you working out the homfunctor and why the first coordinate needs to be contravariant, I drew out the simple example used above in my comment #23. It might help calculate how this functor is working. The left diagram shows the how it maps the objects and the right diagram shows how it maps the morphisms. Let me know if there are mistakes.  Edit: Now that I think about it, this is so natural since morphisms and compositions have directionality which puts the start and target on opposite ends of the morphism and composition must start where the previous left off. \n• Options\n27.\n\nHere's an easy analogy.\n\nView morphisms $$\\text{Hom}(-,X)$$ in a category like extension cords with the inputs not plugged into a wall outlet.\n\n$\\overset{\\text{black cord}}\\longrightarrow \\text{Lamp}$\n\nThen the fact that pre-composition runs backward is analogous to the fact that if I want to extend my extension cord to get closer to a wall outlet, then cords are composed backwards,\n\n$\\overset{\\text{beige cord}}\\longrightarrow \\ \\overset{\\text{black cord}}\\longrightarrow \\text{Lamp}$\n\nhere, we're using the $$\\text{beige cord}$$ to get the $$\\text{black cord}$$'s output closer to a desired input. If we can't, we get another cord and (throwing saefty concerns to the wind) we keep daisy chaining on cords until we can get to a wall outlet (or else you're out in a deserted area and it's actually impossible).\n\nComment Source:Here's an easy analogy. View morphisms \\$$\\text{Hom}(-,X)\\$$ in a category like extension cords with the inputs not plugged into a wall outlet. \\$\\overset{\\text{black cord}}\\longrightarrow \\text{Lamp} \\$ Then the fact that pre-composition runs backward is analogous to the fact that if I want to extend my extension cord to get closer to a wall outlet, then cords are composed backwards, \\$\\overset{\\text{beige cord}}\\longrightarrow \\ \\overset{\\text{black cord}}\\longrightarrow \\text{Lamp} \\$ here, we're using the \\$$\\text{beige cord}\\$$ to get the \\$$\\text{black cord}\\$$'s output closer to a desired input. If we can't, we get another cord and (throwing saefty concerns to the wind) we keep daisy chaining on cords until we can get to a wall outlet (or else you're out in a deserted area and it's actually impossible).\n• Options\n28.\nedited June 2018\n\nSo here's my attempt at why the homset functor $$\\hom : \\mathcal{C}^\\mathrm{op} \\times \\mathcal{C} \\to \\mathbf{Set}$$ has to be contravariant in the first argument. Suppose we attempt to define the homset functor as $$\\hom : \\mathcal{C} \\times \\mathcal{C} \\to \\mathbf{Set}$$ such that any pair of objects $$(c,c')$$ is mapped to the set of morphisms $$\\mathcal{C}(c,c')$$ between these objects. We have to define what happens the morphisms $$(f,f') : (c,c') \\to (d,d')$$, and applying the homset functor to the morphism pair should give us a map sending morphisms in $$\\mathcal{C}(c,c')$$ to morphisms in $$\\mathcal{C}(d,d')$$.\n\nLet $$h$$ be any element of $$\\mathcal{C}(c,c')$$ and $$h'$$ an element of $$\\mathcal{C}(d,d')$$ such that $$\\hom (f,f') (h)=h'$$. I drew the following diagram (I actually drew a bunch of arrows for each homset, but that's hard to do in LaTeX):\n\n$$\\begin{matrix} & & f' & & \\\\ & c' & \\rightarrow & d' &\\\\ h & \\uparrow & & \\uparrow & h'\\\\ & c & \\rightarrow & d &\\\\ & & f & & \\\\ \\end{matrix}$$ From this diagram, we get two compositions $$f'h : c \\to d'$$ and $$h'f : c \\to d'$$, so the morphism $$(f,f')$$ ends up sending the homset $$\\mathcal{C}(c,c')$$ to $$\\mathcal{C}(c,d')$$; not quite what we want. But it's easy to see that if the arrow $$f$$ was turned around, we could precompose $$h$$ with it.\n\nIn $$\\mathcal{C}^\\mathrm{op}$$, $$f : d \\to c$$ as desired, so we get the following commutative diagram:\n\n$$\\begin{matrix} & & f' & & \\\\ & c' & \\rightarrow & d' &\\\\ h & \\uparrow & & \\uparrow & h'\\\\ & c & \\leftarrow & d &\\\\ & & f & & \\\\ \\end{matrix}$$ where $$h' = f'hf$$, which looks a bit like conjugation in a group. So the functor $$\\hom$$ sends morphisms in $$\\mathcal{C}^\\mathrm{op} \\times \\mathcal{C}$$ to functions between homsets defined by the previous formula.\n\nComment Source:So here's my attempt at why the homset functor \\$$\\hom : \\mathcal{C}^\\mathrm{op} \\times \\mathcal{C} \\to \\mathbf{Set}\\$$ has to be contravariant in the first argument. Suppose we attempt to define the homset functor as \\$$\\hom : \\mathcal{C} \\times \\mathcal{C} \\to \\mathbf{Set}\\$$ such that any pair of objects \\$$(c,c')\\$$ is mapped to the set of morphisms \\$$\\mathcal{C}(c,c')\\$$ between these objects. We have to define what happens the morphisms \\$$(f,f') : (c,c') \\to (d,d') \\$$, and applying the homset functor to the morphism pair should give us a map sending morphisms in \\$$\\mathcal{C}(c,c')\\$$ to morphisms in \\$$\\mathcal{C}(d,d')\\$$. Let \\$$h \\$$ be any element of \\$$\\mathcal{C}(c,c')\\$$ and \\$$h'\\$$ an element of \\$$\\mathcal{C}(d,d')\\$$ such that \\$$\\hom (f,f') (h)=h'\\$$. I drew the following diagram (I actually drew a bunch of arrows for each homset, but that's hard to do in LaTeX): $\\begin{matrix} & & f' & & \\\\\\\\ & c' & \\rightarrow & d' &\\\\\\\\ h & \\uparrow & & \\uparrow & h'\\\\\\\\ & c & \\rightarrow & d &\\\\\\\\ & & f & & \\\\\\\\ \\end{matrix}$ From this diagram, we get two compositions \\$$f'h : c \\to d'\\$$ and \\$$h'f : c \\to d' \\$$, so the morphism \\$$(f,f')\\$$ ends up sending the homset \\$$\\mathcal{C}(c,c')\\$$ to \\$$\\mathcal{C}(c,d')\\$$; not quite what we want. But it's easy to see that if the arrow \\$$f\\$$ was turned around, we could precompose \\$$h\\$$ with it. In \\$$\\mathcal{C}^\\mathrm{op}\\$$, \\$$f : d \\to c\\$$ as desired, so we get the following commutative diagram: $\\begin{matrix} & & f' & & \\\\\\\\ & c' & \\rightarrow & d' &\\\\\\\\ h & \\uparrow & & \\uparrow & h'\\\\\\\\ & c & \\leftarrow & d &\\\\\\\\ & & f & & \\\\\\\\ \\end{matrix}$ where \\$$h' = f'hf\\$$, which looks a bit like conjugation in a group. So the functor \\$$\\hom\\$$ sends morphisms in \\$$\\mathcal{C}^\\mathrm{op} \\times \\mathcal{C} \\$$ to functions between homsets defined by the previous formula. \n• Options\n29.\nedited July 2018", null, "I drew out the left and right adjoint functors and the underlying naturality square in one picture because I like to draw diagrams to help me memorize the relationships. This one might be intimidating because there is so much going on but once your eyes adjust to the third dimension its pretty straight forward.\n\nI drew this so that morphisms are lines, functors are planes and natural transformations are cubes. The naturality square for the natural isomorphism is shown by the green double arrows and the functors as green planes.\n\nComment Source: I drew out the left and right adjoint functors and the underlying naturality square in one picture because I like to draw diagrams to help me memorize the relationships. This one might be intimidating because there is so much going on but once your eyes adjust to the third dimension its pretty straight forward. I drew this so that morphisms are lines, functors are planes and natural transformations are cubes. The naturality square for the natural isomorphism is shown by the green double arrows and the functors as green planes. \n• Options\n30.\n\nMichael - wow, that's great! That's the image to have in mind. I will copy this over to Lecture 54, where it can serve as a hint for Puzzle 168. Nobody has solved Puzzle 168 yet!\n\nComment Source:Michael - wow, that's great! That's the image to have in mind. I will copy this over to [Lecture 54](https://forum.azimuthproject.org/discussion/2277/lecture-54-chapter-3-tying-up-loose-ends/p1), where it can serve as a hint for Puzzle 168. Nobody has solved Puzzle 168 yet!\nSign In or Register to comment." ]

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[ "Recently, while trying to compare the distribution of two samples, I discovered that you can plot both on the same graph in base R, which is a nice feature if you just want to examine the data quickly. We can explore this with a psychological dataset from the Open Psychometrics site. This hosts a range of open psychometric tests and stores the data in an accessible form. Let’s pull out the data for the Rosenberg Self-Esteem Scale (note that there are two different scoring methods in common use on this scale - on the website they have used a 1 - 4 Likert scale for the data output as a csv, but it is not unusual to see the use of a 0 - 3 scale, (which is the method used to give participants on the website feedback) so we need to be cautious when comparing these total scores with published norms - see https://socy.umd.edu/about-us/using-rosenberg-self-esteem-scale.\n\nFirst we will load two packages we are going to use. We want the tidyverse for manipulating the variables and we will use the psych package for creating total scores on the measure itself.\n\n``````library(tidyverse)\nlibrary(psych)\n``````\n\nNext we want to set an url object to direct the download.file() to the right place to pull the data. I have called it my_url for simplicity. We pass this as the first argument in the download.file() function. We then set a destination for the file to be saved with the dest argument. Finally we use unzip to unpack the zipped file.\n\n``````my_url <- \"http://openpsychometrics.org/_rawdata/RSE.zip\"\nunzip (\"data.zip\", exdir = \"./\")\n``````\n\nNow we can import the data with the read_tsv() function. We can’t use the read_csv() function with the data, because despite having a .csv extension, the data is actually tab-separated not comma-separated.\n\n``````df <- read_tsv(\"RSE/data.csv\")\n``````\n\nIn the Rosenberg Self-Esteem scale Items 2, 5, 6, 8, 9 are normally reverse scored. However, whoever loaded the questions on the website put them in a different order, with items 3, 5, 8, 9, 10 needing reversing. We need to create a total score for the measure and to be mindful of the reverse coded items. The psych package provides a function for this called scoreFast. We need to pass it a list called keys.list which specifies the direction of each item in turn (items are scored as-is if they have no leading ‘-’ minus sign, but all items with a minus are reverse scored). We won’t bother recoding the data from the 1 - 4 scale to 0 - 3 as it makes little difference for your graphs.\n\n``````keys.list <- list(c('Q1', 'Q2', '-Q3', 'Q4', '-Q5', 'Q6', 'Q7', '-Q8', '-Q9', '-Q10'))\ndf\\$total <- scoreFast(keys.list, items = df[1:10], totals = TRUE, min = 1, max = 4)\n``````\n\nNow we have our dataset, we can look at comparing distributions. We might want to know if the distribution of self-esteem scores differs between men and women. Checking the codebook on the website, we can see that males are coded as ‘1’ and females as ‘2’.\n\n``````men <- df %>%\nfilter(gender == 1)\nwomen <- df %>%\nfilter(gender == 2)\n``````\n\nSo let’s plot the total self-esteem scores for the women in the sample as a simple histogram.\n\n``````hist(women\\$total)\n``````", null, "We can see a fairly normal distribution of scores. We can check the mean, but we might predict it is around 25.\n\n``````mean(women\\$total)\n``````\n``````## 25.74368\n``````\n\nNext we can add the men’s scores to the same plot. Here we simply create the first plot, then make a second plot with the argument add set to TRUE. We will set the density to 35 so we can see through the bars on the histogram.\n\n``````hist(women\\$total, col = 'red', main = \"Histogram of Total scores on Rosenberg Self-Esteem Scale\", xlab = \"Total score\")\nhist(men\\$total, add = TRUE, col = 'blue', density = 35)\n``````", null, "I have used the pipe to separate my data into individual gender dataframes, but this is only one way to do it, and I do find this code very easy to read. However, we could have done the same thing using a traditional R approach of indexing instead.\n\n``````hist(df\\$total[df\\$gender== 2], col = 'orchid', main = \"Histogram of Total scores on Rosenberg Self-Esteem Scale\", xlab = \"Total score\")\nhist(df\\$total[df\\$gender==1], add = TRUE, col = 'royalblue', density = 40)\n``````", null, "Now we have seen the distributions, we might wonder if the sexes differ on the measure of self-esteem. Let’s run a quick t-test to see.\n\n``````t.test(men\\$total, women\\$total)\n``````\n``````##\n## \tWelch Two Sample t-test\n##\n## data: men\\$total and women\\$total\n## t = 23.785, df = 37496, p-value < 2.2e-16\n## alternative hypothesis: true difference in means is not equal to 0\n## 95 percent confidence interval:\n## 1.436304 1.694284\n## sample estimates:\n## mean of x mean of y\n## 27.30897 25.74368\n``````\n\nYes they do! With men having a significantly higher mean score on self-esteem (though the absolute difference is quite small.)\n\nPosted on:\nOctober 12, 2019\nLength:\n4 minute read, 816 words" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Manipulate+Plot showing no output\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg81853] Re: Manipulate+Plot showing no output\n• From: janos <janostothmeister at gmail.com>\n• Date: Fri, 5 Oct 2007 04:45:34 -0400 (EDT)\n• References: <fe29g7\\$n7i\\[email protected]>\n\n```On okt. 4, 10:50, Sooraj R <soorajrnamb... at yahoo.com> wrote:\n> Hii all\n>\n> with\n>\n> ta = 0.9;\n> tb = 0.9;\n> ka = Sqrt[1 - ta^2];\n> kb = Sqrt[1 - tb^2];\n> \\[Omega] = (2*10^15)/\\[Lambda];\n> L = 100 \\[Pi];\n> T = 2.6*10^-12;\n> out1 = (ta - tb*(Exp[-\\[Alpha] L/2])*Exp[\\[ImaginaryI]\n> \\[Omega] T])/(\n> 1 - ta*tb*(Exp[-\\[Alpha] L/2])*Exp[\\[ImaginaryI]\n> \\[Omega] T]);\n> out2 = (-ka kb Exp[-\\[Alpha] L/4] Exp[\\[ImaginaryI]\n> \\[Omega] T/2])/(\n> 1 - ta tb Exp[-\\[Alpha] L/2] Exp[\\[ImaginaryI]\n> \\[Omega] T]);\n> Manipulate[\n> Plot[Evaluate[(Abs[out1])^2], {\\[Lambda], 1.54,\n> 1.56},\n> PlotRange -> {0, 1}, AxesLabel -> {\"\\[Lambda]\",\n> \"out1\"}], {\\[Alpha],\n> 1, 100, 1}]\n> Manipulate[\n> Plot[Evaluate[(Abs[out2])^2], {\\[Lambda], 1.54,\n> 1.56},\n> PlotRange -> {0, 1}, AxesLabel -> {\"\\[Lambda]\",\n> \"out2\"}], {\\[Alpha],\n> 1, 100, 1}]\n>\n> I dont see any output graph for expressions out1 and\n> out2.\n>\n> Can some one tell me what mistake I have done?\n> Mathematica version is 6.0\n> Many thanks\n>\n> Sooraj\n>\n> ____________________________________________________________________________________\n> Got a little couch potato?\n> Check out fun summer activities for kids.http://search.yahoo.com/search?fr=oni_on_mail&p=summer+activities+for..." ]

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[ "Commenced in January 2007\nFrequency: Monthly\nEdition: International\nPaper Count: 30127\n##### A Stochastic Diffusion Process Based on the Two-Parameters Weibull Density Function\n\nAbstract:\n\nStochastic modeling concerns the use of probability to model real-world situations in which uncertainty is present. Therefore, the purpose of stochastic modeling is to estimate the probability of outcomes within a forecast, i.e. to be able to predict what conditions or decisions might happen under different situations. In the present study, we present a model of a stochastic diffusion process based on the bi-Weibull distribution function (its trend is proportional to the bi-Weibull probability density function). In general, the Weibull distribution has the ability to assume the characteristics of many different types of distributions. This has made it very popular among engineers and quality practitioners, who have considered it the most commonly used distribution for studying problems such as modeling reliability data, accelerated life testing, and maintainability modeling and analysis. In this work, we start by obtaining the probabilistic characteristics of this model, as the explicit expression of the process, its trends, and its distribution by transforming the diffusion process in a Wiener process as shown in the Ricciaardi theorem. Then, we develop the statistical inference of this model using the maximum likelihood methodology. Finally, we analyse with simulated data the computational problems associated with the parameters, an issue of great importance in its application to real data with the use of the convergence analysis methods. Overall, the use of a stochastic model reflects only a pragmatic decision on the part of the modeler. According to the data that is available and the universe of models known to the modeler, this model represents the best currently available description of the phenomenon under consideration.\n\nDigital Object Identifier (DOI): doi.org/10.5281/zenodo.1125047\n\nReferences:\n\n R. F. Woolson and W. R. Clarke, Statistical Methods for the Analysis of Biomedical Data, 2nd ed. John Wiley & Sons, Vol.371, New York, United States, 2000.\n R. L. Mason, R. F. Gunst, and J. L. Hess Statistical Design and Analysis of Experiments: with Applications to Engineering and Science,Wiley, New York, United States, 1989.\n W. R. Blischke and D. N. P. Murthy, Probability distributions for modeling time to failure, in Reliability: Modeling, Prediction, and Optimization, John Wiley & Sons, Inc.,Hoboken, NJ, USA, 2000.\n S. A. Klugman, and R. Parsa, Fitting bivariate loss distributions with copulas, Insurance: Mathematics and Economics, Elsevier, Vol. 24, no.1, 1999, pp. 139–148.\n D. J. Davis, An Analysis of some Failure Data, Journal of the American Statistical Association, Taylor & Francis Group, Vol. 47, no.250, 1952, pp. 113–150.\n P. Feigl and M. Zelen, Estimation of exponential survival probabilities with concomitant information, Biometrics, JSTOR, 1965, pp. 826–838.\n D. R Cox, Renewal Theory Methuen, CoxRenewal Theory1962, London, 1962.\n E. J. Gumbel, Statistics of extremes. 1958, Columbia Univ. press, New York, 1958.\n J. Lieblein and M. Zelen, Statistical investigation of the fatigue life of deep-groove ball bearings, Journal of Research of the National Bureau of Standards, Citeseer, Vol. 57, no.5, 1956, pp. 273–316.\n M. C. Pike, A method of analysis of a certain class of experiments in carcinogenesis, Biometrics, JSTOR, Vol. 22, no.1, 1966, pp. 142–161.\n J. W. Boag, Maximum Likelihood Estimates of the Proportion of Patients Cured by Cancer Therapy, Journal of the Royal Statistical Society. Series B (Methodological), Royal Statistical Society, Wiley, Vol. 11, no.1, 1949, pp. 15–53.\n A. N. Giovanis and C. H. Skiadas, A Stochastic Logistic Innovation Diffusion Model Studying the Electricity Consumption in Greece and the United States, Technological Forecasting and Social Change, Vol. 61, 1999, pp. 235–246.\n A. Katsamaki and C. H. Skiadas, Analytic solution and estimation of parameters on stochastics exponential model for a technological diffusion process, Applied Stochastics Model and Data Analysis, Vol. 11, 1995, pp. 59–75.\n C. Skiadas and A. Giovani, A stochastic bass innovation diffusion model for studying the growth of electricity consumption in Greece, Applied Stochastic Models and Data Analysis, Vol. 13, 1997, pp. 85–101.\n R. Gutie´rrez-Sa´nchez, A. Nafidi, A. Pascual, E. R. A´ balos, Three parameter gamma-type growth curve, using a stochastic gamma diffusion model: Computational statistical aspects and simulation, Mathematics and Computers in Simulation, Vol. 82, 2011, pp. 234–243.\n R. Guti´errez, R. Guti´errez-S´anchez, A. Nafidi and E. Ramos, A diffusion model with cubic drift: statistical and computational aspects and application to modeling of the global CO2 emission in Spain, Environmetrics, Vol. 18, 2007, pp. 55–69.\n R. Guti´errez, R. Guti´errez-S´anchez, A. Nafidi and E. Ramos, Studying the vehicule park in Spain using the lognormal and Gompertz diffusion processes, Proceedings od SEIO’04, Vol. 18, 2004, pp. 171–172.\n A. V. Egorov, H. Li, and Y. Xu, Maximum likelihood estimation of time-inhomogeneous diffusions, Journal of Econometric, Vol. 114, 2003, pp. 107–139.\n Y. Ait-Sahalia, R. Kimmel, Maximum likelihood estimation of stochastic volatility models, Journal of Financial Economics, Vol. 83, 2007, pp. 413–452.\n F. Casas, Solution of linear partial differential equations by Lie algebraic methods, Journal of Computational and Applied Mathematics, Vol. 76, 1996, pp. 159–170.\n P. E. Kloeden and E. Platen, Numerical Solution of Stochastic Differential Equations, Springer-Verlag, Applications of Mathematics Series, no.23, 1991.\n LM. Ricciardi, Diffusion processes and related topics in biology. Lecture notes in biomathematics, Springer-Verlag, Berlin, 1977.\n P. W. Zenha, Invariance of Maximum Likelihood Estimators, The Annals of Mathematical Statistics, Ann. Math. Statist., Vol. 37, no.3, 1966, pp. 744." ]

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[ "## February 07, 2012\n\n### Issue of Mirrored UVs and Normal Map when rendering in Unity\n\nRecently, I found my model has normal map seam when rendering in Unity with our custom shader.\nAnd I try to use the Unity built-in shader to render it. The result shows that there's no seams. How strange? Soon I found that because of all UV of my models are symmetry. It means that half of my UV are flipped. Hence, the binormal of the half model has been reversed. (tangent basis is calculated by normal and UV, http://www.3dkingdoms.com/weekly/weekly.php?a=37). In Maya, I never notice that, because it auto-detects the UV winding direction. So, I just need to do the same thing in my shader.\n\nSo after researching, I got the method how to detect the UV winding. Here's the formula:\n\ntangent[a].w = (Dot(Cross(n, t), tan2[a]) < 0.0F) ? -1.0F : 1.0F;\n\nhere tangent is stored by float4, the w stores the UV winding direction, 1, or -1.\nAnd tan2 would be the binormal, n is the normal, and t is the actual orthogonalized tangent.\n\nAlthough this equation answer my question, it didn't solve my problem. It's kind impossible to calculated the orthogonalized tangent in a shader. But it give me a hint. Would Unity has calculated the UV winding for us and stored it in tangent.w?\n\nWith an easy debug shader to show the tangent.w, I got my answer:\n\nSo the actual key is:\nI just need to write a branch to see if I need to flip the binormal when calculating the tangent space.\n\nfloat3x3 rotation = float3x3(v.tangent.xyz, binormal.xyz, v.normal.xyz);\n\nif(v.tangent.w > 0)\nrotation = float3x3(v.tangent.xyz, -binormal.xyz, v.normal.xyz);\n\nSo, here is the result without the UV winding detection:\n\nAnd this is the correct rendering with UV winding detection\nNote: Why we don't want to use the built-int shader in Unity? Just because even the shader is simple as bump spec one which really has a lot of junk inside. And the most important: they hide those expensive calculations in compiled functions.\n\n#### 3 意見:\n\nEdy said...\n\nHave you tried disabling Tangent Calculation on the Unity 3D importer? That fixed the problem for me.\nImage\n\nEdy said...\n\nHave you tried disabling Tangent Calculation on the Unity 3D importer? That fixed the problem for me.\nImage\n\nUnknown said...\n\nIn fact it's really old and I found better solution in shader lab" ]

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[ "", null, "Search\nClose\nTranslate\nClose\n\n# Mrs Mowl's Group\n\nTo start with today we would like you to practice counting within 20. Can you:\n\n• count 0 to 20\n• count back 20 to 0\n• count from 7 to 15\n• count from 9 to 18\n• count from 11 to 20\n\nUse the number line below if you need to.\n\n0     1     2     3     4     5     6     7     8     9    10\n\n11   12   13   14   15   16   17   18   19   20\n\nToday we are going to work on addition of 2 numbers, using strategies that we have learnt in class before. Can you read the number sentence below?\n\n5   +   3   =\n\nIt says \"5 add 3 equals\". Can you think of a way that you could find the answer? Tell your grown up what you think...\n\nWe could\n\n• use our fingers to show each number\n• use objects such as lego bricks, small toys, pasta to count out each number\n• draw pictures to represent each number\n\nSo it could look like this:\n\n5   +   3   =\n\nSo by counting altogether the fingers or buttons or tally mark I can see that:\n\n5 + 3 = 8\n\nUsing any of the above strategies, can you find the answers to the following addition sentences:\n\n4 + 2 =\n\n3 + 6 =\n\n6 + 2 =\n\n5 + 4 =\n\n3 + 2 =\n\n2 + 5 =\n\nWrite the addition sentences and answers out into your exercise book", null, "## Information\n\n### Contact Information", null, "Top" ]

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[ "# Index\n\n< cpp\n\nC++\n Language Headers Freestanding and hosted implementations Named requirements Language support library Concepts library (C++20) Diagnostics library Utilities library Strings library Containers library Algorithms library Iterators library Numerics library Input/output library Localizations library Regular expressions library (C++11) Atomic operations library (C++11) Thread support library (C++11) Filesystem library (C++17) Technical Specifications\n\n# A\n\nabstract class\nalignas\nalignof\n<algorithm>\narray\n<array> (since C++11)\nASCII\nasm\nassembly\n<atomic> (since C++11)\natomic operation\nattribute (C++11)\n\ndecltype (C++11)\n\n# H\n\n<algorithm>\n<array> (since C++11)\n<atomic> (since C++11)\n<bitset>\n<cassert>\n<ccomplex> (since C++11)\n<cctype>\n<cerrno>\n<cfenv> (since C++11)\n<cfloat>\n<chrono>\n<cinttypes> (since C++11)\n<ciso646>\n<climits>\n<clocale>\n<cmath>\n<codecvt> (since C++11)\n<condition_variable> (since C++11)\n<complex>\n<csignal>\n<csetjmp>\n<cstdalign> (since C++11)\n<cstdarg>\n<cstdbool> (since C++11)\n<cstddef>\n<cstdint> (since C++11)\n<cstdio>\n<ctime>\n<ctgmath> (since C++11)\n<cwchar>\n<cwctype>\n<cuchar>\n<exception>\n<forward_list> (since C++11)\n<fstream>\n<functional>\n<future> (since C++11)\n<initializer_list> (since C++11)\n<iomanip>\n<ios>\n<iosfwd>\n<iostream>\n<iterator>\n<locale>\n<limits>\n<list>\n<map>\n<memory>\n<mutex> (since C++11)\n<new>\n<numeric>\n<queue>\n<random> (since C++11)\n<ratio> (since C++11)\n<regex> (since C++11)\n<scoped_allocator> (since C++11)\n<set>\n<shared_mutex> (since C++11)\n<sstream>\n<stack>\n<stdexcept>\n<streambuf>\n<string>\n<strstream>\n<system_error>\n<thread> (since C++11)\n<tuple> (since C++11)\n<typeindex> (since C++11)\n<type_traits> (since C++11)\n<unordered_map> (since C++11)\n<unordered_set> (since C++11)\n<utility>\n<valarray>\n<vector>\n\n# I\n\ninitializer list\n<initializer_list> (since C++11)\ninteger type\n<iomanip>\n<ios>\n<iosfwd>\n<iostream>\n<iterator>\n\n# P\n\npointer\npreprocessor\nprvalue (since C++11)\n\n# R\n\nrvalue (until C++11)\n\n# S\n\nSFINAE (Substitution Failure Is Not An Error)" ]

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[ "", null, "greater than and less worksheets differentiated depth maths ks1.\n\ngreater depth maths ks1 worksheets than less,greater than worksheets ks1 depth maths less,greater than less worksheets ks1 sheet,greater than and less worksheets depth reading ks1,greater than worksheets ks1 less sheet and depth maths,greater than less worksheets ability grade numbers depth reading ks1,greater than worksheets ks1 less math us coins sheet depth maths,greater than less sheet ks1 worksheets and depth maths,greater than and less worksheets math equal to depth maths ks1 sheet,greater than worksheets ks1 less depth reading." ]

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[ "Glossary /\nThis is some text inside of a div block.\n\n# K-Means\n\nCategory:\n###### Data Science Concept\nLevel:\n\nK-Means is a popular clustering algorithm in data science that aims to partition n observations into k clusters based on their similarity. The algorithm works by finding k centroids, where each centroid represents the center of a cluster. It then assigns each observation to the nearest centroid, creating a cluster. The algorithm iteratively moves the centroids to minimize the distance between the observations and their assigned centroid. This process continues until the centroids no longer move, or a maximum number of iterations is reached.\n\n## Key Highlights\n\n• K-Means is a type of unsupervised learning algorithm used to identify patterns in data.\n• K-Means is easy to implement and is one of the most commonly used clustering algorithms in data science.\n• The algorithm can be used for a wide range of applications, including image segmentation, customer segmentation, and anomaly detection.\n\n## References", null, "" ]

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[ "Ask Your Question\n\n# how to solve: exp(x)+exp(1/2*x) == 2?\n\nsage: assume(x,'real')\nsage: solve(exp(x)+exp(-1/2*x) == 2,x)\n[x == 2*log(1/2*sqrt(5) - 1/2), x == 0]\n\n\nthat ist ok!\n\nhow can i get here the solution x==0?\n\nsage: solve(exp(x)+exp(1/2*x) == 2,x)\n[e^x == -e^(1/2*x) + 2]\n\n\nThanks for help.\n\nedit retag close merge delete\n\n## 2 Answers\n\nSort by » oldest newest most voted\nsage: solve(exp(x)+exp(1/2*x)==2,x,to_poly_solve=True)\n[x == 2*I*pi + 4*I*pi*z36 + 2*log(2), x == 4*I*pi*z38]\n\n\nFor z38=0, x=0\n\n(Wolframalpha gives the same solution)\n\nmore\n\n## Comments\n\nsage: l=e^(1/2*x)+e^x\nsage: var('u')\nu\nsage: lu=l.subs(x=2*log(u))\nsage: assume(u>0)\nsage: solve(lu==2,u)\n[u == 1]\nsage: 2*log(1)\n0\n\nmore\n\n## Your Answer\n\nPlease start posting anonymously - your entry will be published after you log in or create a new account.\n\nAdd Answer\n\n## Stats\n\nAsked: 2012-07-25 15:19:15 +0200\n\nSeen: 578 times\n\nLast updated: Jul 25 '12" ]

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[ "Back to Course\n\n## Essentials of Formal Logic\n\n0% Complete\n0/0 Steps\n1. Introduction\nCourse Introduction: Essentials of Formal Logic (Preview Content)\n1 Quiz\n2. Chapter 1\nLesson 1: Chapter 1.1 Formal vs. Informal Logic (Preview Content)\n3 Topics\n|\n1 Quiz\n3. Lesson 2: Chapter 1.2 Deductive vs. Inductive Reasoning (Preview Content)\n2 Topics\n|\n1 Quiz\n4. Lesson 3: Chapter 1.3 Categorical vs. Propositional Logic\n2 Topics\n|\n1 Quiz\n5. Chapter 2\nLesson 4: Chapter 2.1 Part I: Aristotle Gets the Ball Rolling: Classical Origins and Medieval Recovery\n2 Topics\n|\n1 Quiz\n6. Lesson 5: Chapter 2.2 Part II: Aristotle Is Lost and Then Found: The Growth and Divergence of Modern Logic\n2 Topics\n|\n1 Quiz\n7. Chapter 3\nLesson 6: Chapter 3.1 Thinking About Thinking: The Nature of Formal Logic\n2 Topics\n|\n1 Quiz\n8. Lesson 7: Chapter 3.2 The Three Acts of the Mind\n2 Topics\n|\n1 Quiz\n9. Chapter 4\nLesson 8: Chapter 4.1 Introduction to Argument Translation\n2 Topics\n|\n1 Quiz\n10. Lesson 9: Chapter 4.2 Categorical Form Introduced\n2 Topics\n|\n1 Quiz\n11. Lesson 10: Chapter 4.3 Propositions\n2 Topics\n|\n1 Quiz\n12. Lesson 11: Chapter 4.4 Translating Arguments Step 1: Finding the Propositions\n2 Topics\n|\n1 Quiz\n13. Lesson 12: Chapter 4.5 Translating Arguments Step 2: Finding the Subject Term and the Predicate Term\n2 Topics\n|\n1 Quiz\n14. Lesson 13: Chapter 4.6 Translating Arguments Step 3: Affirmo and Nego\n2 Topics\n|\n1 Quiz\n15. Lesson 14: Chapter 4.7 Translating Arguments Step 4: Supply the Proper Quantifier\n2 Topics\n|\n1 Quiz\n16. Lesson 15: Chapter 4.8 Translating Arguments Step 5: Propositions Translated into Categorical Form\n2 Topics\n|\n1 Quiz\n17. Chapter 5\nLesson 16: Chapter 5.1 Introduction to the Square of Opposition\n2 Topics\n|\n1 Quiz\n18. Lesson 17: Chapter 5.2 The Square of Opposition\n2 Topics\n|\n1 Quiz\n19. Lesson 18: Chapter 5.3 Contradiction\n2 Topics\n|\n1 Quiz\n20. Lesson 19: Chapter 5.4 Contrariety and Subcontrariety\n2 Topics\n|\n1 Quiz\n21. Lesson 20: Chapter 5.5-5.6 Subimplication and Superimplication, The Square of Opposition and Inference Analysis\n2 Topics\n|\n1 Quiz\n22. Chapter 6\nLesson 21: Chapter 6.1-6.3 Introduction to the Relationships of Equivalence, Logical Equations, The Obverse Relationship\n2 Topics\n|\n1 Quiz\n23. Lesson 22: Chapter 6.4 The Converse Relationship\n2 Topics\n|\n1 Quiz\n24. Lesson 23: Chapter 6.5 The Relationship of Contraposition\n2 Topics\n|\n1 Quiz\n25. Chapter 7\nLesson 24: Chapter 7.1 Introduction to Syllogisms and Validity\n2 Topics\n|\n1 Quiz\n26. Lesson 25: Chapter 7.2-7.3 Arranging the Syllogism, Categorical Syllogisms\n2 Topics\n|\n1 Quiz\n27. Lesson 26: Chapter 7.4 Enthymemes\n2 Topics\n|\n1 Quiz\n28. Lesson 27: Chapter 7.5 Moods and Figures\n2 Topics\n|\n1 Quiz\n29. Chapter 8\nLesson 28: Chapter 8.1 Validity and the Counterexample Method\n2 Topics\n|\n1 Quiz\n30. Lesson 29: Chapter 8.2 Evaluating Validity: Terminological Rules 1 and 2\n2 Topics\n|\n1 Quiz\n31. Lesson 30: Chapter 8.3 Evaluating Validity: Terminological Rules 3 and 4\n2 Topics\n|\n1 Quiz\n32. Lesson 31: Chapter 8.4 Evaluating Validity: Qualitative Rules\n2 Topics\n|\n1 Quiz\n33. Lesson 32: Chapter 8.5 An Introduction to the Venn Diagramming Method of Establishing Validity\n2 Topics\n|\n1 Quiz\n34. Lesson 33: Chapter 8.6 Combining the Skills\n2 Topics\n|\n1 Quiz\n35. Chapter 9\nLesson 34: Chapter 9 Preamble\n3 Topics\n|\n1 Quiz\n36. Lesson 35: Chapter 9.1 Introduction to Definitions and Disagreement\n2 Topics\n|\n1 Quiz\n37. Lesson 36: Chapter 9.2 Types of Disagreement\n2 Topics\n|\n1 Quiz\n38. Lesson 37: Chapter 9.3 Rules for Defining Words\n2 Topics\n|\n1 Quiz\n39. Lesson 38: Chapter 9.4 Types of Definitions\n2 Topics\n|\n1 Quiz\n40. Lesson 39: Chapter 9.5 Extension vs. Intention\n2 Topics\n|\n1 Quiz\n41. Lesson 40: Chapter 9.6 Presuppositional Disputes\n2 Topics\n|\n1 Quiz\n42. Lesson 41: Chapter 9.7 Pursuing Truth\n2 Topics\n|\n1 Quiz\n43. End of Course Test\nEnd of Course Test: Essentials of Formal Logic\n1 Quiz\nIn Progress\nLesson 32, Topic 1\nIn Progress\n\nLesson Progress\n0% Complete" ]

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[ "# Cross-Entropy Loss Leads To Poor Margins\n\nNeural networks could misclassify inputs that are slightly different from their training data, which indicates a small margin between their decision boundaries and the training dataset. In this work, we study the binary classification of linearly separable datasets and show that linear classifiers could also have decision boundaries that lie close to their training dataset if cross-entropy loss is used for training. In particular, we show that if the features of the training dataset lie in a low-dimensional affine subspace and the cross-entropy loss is minimized by using a gradient method, the margin between the training points and the decision boundary could be much smaller than the optimal value. This result is contrary to the conclusions of recent related works such as (Soudry et al., 2018), and we identify the reason for this contradiction. In order to improve the margin, we introduce differential training, which is a training paradigm that uses a loss function defined on pairs of points from each class. We show that the decision boundary of a linear classifier trained with differential training indeed achieves the maximum margin. The results reveal the use of cross-entropy loss as one of the hidden culprits of adversarial examples and introduces a new direction to make neural networks robust against them.\n\nPDF Abstract\n\n## Code Add Remove Mark official\n\nNo code implementations yet. Submit your code now" ]

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[ "# Matrix Operations Interview Questions with Answers PDF p. 9\n\nMatrix Operations interview questions and answers, matrix operations trivia questions PDF 9 to practice Business Mathematics exam questions for online classes. Practice Matrix Algebra MCQ questions, matrix operations Multiple Choice Questions (MCQ) for online college degrees. Matrix Operations Interview Questions PDF: first degree equations, simplex preliminaries, matrix operations test prep for general business degree online.\n\n\"According to determinant properties, the multiple of one row is added to another row then the determinant\" MCQ PDF with choices unchanged, changed, multiplied, and added for online bachelor's degree in business. Learn matrix algebra questions and answers to improve problem solving skills for online courses for business management degree.\n\n## Trivia Quiz on Matrix Operations MCQs\n\nMCQ: According to determinant properties, the multiple of one row is added to another row then the determinant\n\nchanged\nunchanged\nmultiplied\n\nMCQ: The constraint which does not have surplus variable or slack variable is represented as If the corner point method\n\nlie on constraint line\ndoes not lie on constraint line\nlie on slack line\ndoes not lie on slack line\n\nMCQ: The first degree equation with one variable '3(x-3) = 2(x+4)' if solved for x then the value of variable is\n\n−12\n12\n15\n−15\n\nMCQ: The non-negative variable subtracted from any less than or equal to constraint of simplex method is classified as\n\nsurplus variable\ndeficit variable\nright variable\nleft constant\n\nMCQ: The product sold price is \\$50 USD then the revenue function is\n\nR(50) = 50 + x\nR(x) = 50x\nR(50) = x\nR(x) = 50 + x" ]

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[ "# [erlang-questions] Heap alloc error in lists:foldl/3\n\nTue Nov 6 15:29:32 CET 2012\n\n```Hello.\n\n%% if I use sum, instead of foldsum, everything is ok\n\nThis is because foldsum/2 creates a list of 10 000 000 integers (call to\nlists:seq/2). And you create Np of such lists in total, each one in a different\nprocess. A list in Erlang is a chain of cons pairs (as in Lisp). Each cons\noccupies 2 words, where one word is either 32bits or 64bits, depending on your\nHW architecture. Thus one such list takes\n\n10 000 000 * 8 = 80MB (on 32-bit architecture)\n\nHence it is fairly possible that your system simply ran out of memory.\n\nThe version using sum/2 does not exhibit this behaviour, because it runs in\nconstant space.\n\nHTH,\n\nOn 6.11.2012 15:03, Arif Ishaq wrote:\n> Hi,\n>\n> I'm a newbie, playing with a toy programme to see how parallel processing works\n> in Erlang.\n>\n> The goal is to calculate the value of pi using Leibnitz series. The function\n> pi/2 takes as arguments the number of processes, Np, and the number of terms\n> handled by each process, Nt.\n>\n> The machine is an HP workstation with Windows Vista Enterprise.\n>\n>\n> Here's the code:\n>\n> -module(pp).\n> -compile(export_all).\n>\n> %% calculate PI using Np parallel processes, each calculating Nt terms of the\n> %% Liebnitz series\n>\n> pi(Np, Nt) ->\n> Collector = spawn(?MODULE, collect, [Np, self()]),\n>\n> [spawn_opt(\n> fun() -> Collector ! foldsum((N-1)*Nt, N*Nt -1) end %% if I use\n> sum, instead of foldsum, everything is ok\n> ) || N <- lists:seq(1,Np)],\n>\n> Result ->\n> io:format(\"result is ~p~n\", [4 * Result])\n> end.\n>\n>\n> sum(Kfrom, Kto) ->\n> sum (Kfrom, Kto, 0).\n>\n> sum(Kfrom, Kto, Sum) when Kfrom =:= Kto ->\n> Sum;\n> sum(Kfrom, Kto, Sum) ->\n> sum(Kfrom+1, Kto, Sum + contribution(Kfrom)).\n>\n>\n> foldsum(Kfrom, Kto) ->\n> lists:foldl(\n> fun(K, Sum) -> Sum + contribution(K) end,\n> 0,\n> lists:seq(Kfrom,Kto)).\n>\n>\n> contribution(K) ->\n> Value = 1 / (2 * K + 1),\n> case K rem 2 of\n> 0 ->\n> Value;\n> 1 ->\n> -Value\n> end.\n>\n>\n> collect (Np, For) ->\n> collect(Np, For, 0, 0).\n>\n> collect (Np, For, Ncollected, Sum) when Np =:= Ncollected ->\n> For ! Sum;\n> collect(Np, For, Ncollected, Sum) ->\n> R ->\n> collect(Np, For, Ncollected + 1, Sum + R)\n> end.\n>\n>\n>\n> If I try to calculate pi using 16 processes, each processing 1 million terms,\n> the result is ok.\n> If, on the other hand, I try with 10 million terms each, erlang crashes.\n> Even after closing the werl window, the process keeps running and has to be\n> killed brutally.\n>\n>\n> Erlang R15B (erts-5.9) [smp:4:4] [async-threads:0]\n>\n> Eshell V5.9 (abort with ^G)\n> 1> c(pp).\n> {ok,pp}\n> 2> pp:pi(16,1000000).\n> result is 3.1415925910897737\n> ok\n> 3> pp:pi(16,10000000).\n>\n> Crash dump was written to: erl_crash.dump\n> eheap_alloc: Cannot allocate 40121760 bytes of memory (of type \"heap\").\n>\n>\n> Abnormal termination\n>\n> If I try to read the crash dump with the crash dump viewer tool, I get an error\n> saying the file is not an erlang crash dump:\n> erl_crash.dump is not an Erlang crash dump\n>\n>\n> Why is this happening? Shouldn't foldl be tail recursive? Is this a bug? Or am I\n> doing something wrong?\n>\n>\n> Thanks and best regards\n> Arif Ishaq\n>\n>\n>\n>\n> _______________________________________________\n> erlang-questions mailing list\n> erlang-questions@REDACTED\n> http://erlang.org/mailman/listinfo/erlang-questions\n\n```" ]

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[ "#", null, "•", null, "0 =\n•", null, "1 =\n•", null, "2 =\n•", null, "3 =\n•", null, "4 =\n•", null, "5 =\n•", null, "6 =\n•", null, "7 =\n•", null, "8 =\n•", null, "9 =\n使用说明：\n\n中文大写金额数字前应标明“人民币”字样，大写金额数字应紧接“人民币”字样填写，不得留有空白。\n\n阿拉伯数字小写金额数字中有“0”时，中文大写应按照汉语语言规律、金额数字构成和防止涂改的要求进行书写。\n例如：￥206000.75应写成 人民币贰拾万陆仟元零柒角伍分，或写成 人民币贰拾万零陆仟元柒角伍分。\n•", null, "亿 = 亿\n•", null, "万 =\n•", null, "千 =\n•", null, "百 =\n•", null, "十 =\n•", null, "元 = 元(圆)\n•", null, "角 =\n•", null, "分 =\n•", null, "正 =\n•", null, "整 =\n\n使用说明：\n\n中文大写金额数字到“元”为止的，在“元”之后、应写“整”(或“正”)字；在“角”之后，可以不写“整”(或“正”)字；大写金额数字有“分”的，“分”后面不写“整”(或“正”)字。" ]

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[ "## Monday, 8 February 2010\n\n### LU Decomposition in Prolog\n\n:- use_module(library(clpr)).\n\n// various obvious bits of matrix machinary.\n\nlu_decompose(M, L*U) :-\ndimensions(M,N*N),\ndimensions(L,N*N),\ndimensions(U,N*N),\nlower(L),\nupper(U),\nmatrix_multiply(L,U,M).\n\n#### 1 comment:", null, "Jacqueline said...\n\nhello, I'm programming in prolog but the LU code do not understand, you like help?, tried to make a code that factoring a matrix, given a matrix factors tell me .... I do not speak English, nor write, google translator helped me .. thanks:)\n\nFactorizacionmatriz (M, M1, M2): - .... : (\n\nM = matrix\nM1 = factor 1\nM2 = factor 2\nReturns the values of M1 and M2" ]

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[ "## On mean values of multivariable complex valued multiplicative functions and applications.(English)Zbl 1471.11250\n\nMishou, Hidehiko (ed.) et al., Various aspects of multiple zeta functions – in honor of Professor Kohji Matsumoto’s 60th birthday. Proceedings of the international conference, Nagoya University, Nagoya, Japan August 21–25, 2020. Tokyo: Mathematical Society of Japan. Adv. Stud. Pure Math. 84, 23-64 (2020).\nLet $$\\{n, d\\}\\subseteq\\mathbb N,\\;P(\\mathbf{x})\\in\\mathbb R_{+}[\\mathbf{x}],\\;\\mathbf{x}:=(x_{1}, \\ldots, x_{n})$$, let $$P$$ be a homogeneous polynomial of degree $$d$$, and suppose that $$P(\\mathbb R_{+}^{n}\\setminus \\{0\\})\\neq 0$$. Building on his previous work [Contemp. Math. 566, 65–98 (2012; Zbl 1279.11068)], the author further examines the analytic behaviour of the function $s\\mapsto Z(f, P; s),\\;s\\in{\\mathbb{C}},\\;Z(f, P; s):=\\sum_{\\mathbf{m}\\in\\mathbb N^{n}}f(\\mathbf{m})P(\\mathbf{m})^{-s/d},$ where $$f(\\mathbf{m})$$ is a complex valued multiplicative function, and thereby obtains new asymptotic formulae for the sums $\\sum_{\\mathbf{m}\\in\\mathbb N^{n}}, P(\\mathbf{m})<tf(\\mathbf{m})$ as $$t\\rightarrow\\infty .$$ Several arithmetic applications of that work are given.\nFor the entire collection see [Zbl 1446.11004].\nReviewer: B. Z. Moroz (Bonn)\n\n### MSC:\n\n 11N37 Asymptotic results on arithmetic functions 11M32 Multiple Dirichlet series and zeta functions and multizeta values 11M41 Other Dirichlet series and zeta functions 11R42 Zeta functions and $$L$$-functions of number fields 11R52 Quaternion and other division algebras: arithmetic, zeta functions 11S40 Zeta functions and $$L$$-functions 11S45 Algebras and orders, and their zeta functions 14G10 Zeta functions and related questions in algebraic geometry (e.g., Birch-Swinnerton-Dyer conjecture) 11E45 Analytic theory (Epstein zeta functions; relations with automorphic forms and functions) 11F66 Langlands $$L$$-functions; one variable Dirichlet series and functional equations 11F70 Representation-theoretic methods; automorphic representations over local and global fields 11F72 Spectral theory; trace formulas (e.g., that of Selberg) 11P21 Lattice points in specified regions\n\nZbl 1279.11068\nFull Text:\n\n### References:\n\n G. Hardy and M. Reisz,Dirichlet Series, Cambridge Univ. Press (1949). K. Hoornaert,Newton polyhedra and the poles of Igusa’s local zeta function, Bull. Belg. Math. Soc. Simon Stevin 9, no. 4, p. 589-606 (2002). · Zbl 1040.11086 N. Kurokawa,On the meromorphy of Euler productsIandII, Proc. London Math. Soc., vol. 53, 1-47 and 209-236 (1986). · Zbl 0595.10031 N. Kurokawa and H. Ochiai,A multivariable Euler product of Igusa type and its applications, J. Number Theory, 129, no. 8, p. 1919-1930 (2009). · Zbl 1176.11065 E. Landau,Uber die Anzahl der Gitterpunkte in gewiss en Breichen(Zweite Abhandlung), Kgl. Ges. d. Wiss. Nach. Math. Phys. Klasse. (Gottingen), t. 2, 209-243 (1915). · JFM 45.0312.02 B. Lichtin,The asymptotics of a lattice point problem associated to a finite number of polynomialsI, Duke Math. J., vol. 63, No. 1, 139-192 (1991). · Zbl 0735.11048 B. Lichtin,Geometric features of lattice point problems, Singularity theory (Trieste, 1991), World Sci. Publishing, River Edge, NJ, 370-443 (1995). · Zbl 0993.11051 K. Mahler,Uber einer Satz von Mellin¨, Math. Ann., vol. 100, 384-395, (1928). · JFM 54.0369.03 B. Z. Moroz,Scalar product ofL-functions with Grossencharacters: its meromorphic continuation and natural boundary, J. reine. angew. Math., vol. 332 (1982). · Zbl 0495.12014 T. Rivoal,Th´eor‘emes limites pour certains mod‘les probabilistes de la fonction de Mobius, notes (2015), 19 pages. http://rivoal.perso.math.cnrs.fr/articles.html P. Sargos,Sur le probl‘eme des diviseurs g´en´eralis´es, Publ. Math. Orsay, 2, p. 117-134 (1988). · Zbl 0825.11011 P. Sargos,S´eries de Dirichlet associ´ees ‘a des polynˆomes de plusieurs variables, Th‘ese d’Etat, Univ. Bordeaux 1 (1987). A. Schrijver,Theory of Linear and Integer Programming, Published by John Wiley & Sons Inc (1998). · Zbl 0970.90052 C. Tanaka,On the singularities of Dirichlet series, Comment. Math. Helv., 31, p. 184-194 (1957). · Zbl 0081.06806 G. Tenenbaum,Introduction ‘a la th´eorie analytique et probabiliste des nombres, Cours Sp´ecialis´es, 1, Paris: SMF (1995). · Zbl 0880.11001 L. T´oth and W. Zhai,On multivariable averages of divisor functions, J. Number Theory, 192, p. 251-269 (2018). · Zbl 1444.11199 W.\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]

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[ "### strace.1.in: eliminate empty lines\n\n`Replace them with .IP or just remove.`", null, "Eugene Syromyatnikov 1 month ago\nparent\ncommit\n07baaacfca\n1 changed files with 16 additions and 19 deletions\n1. 16\n19\nstrace.1.in\n\n#### + 16 - 19 strace.1.inView File\n\n ``@@ -79,7 +79,6 @@ strace \\- trace system calls and signals`` 79 79 `` .IR command \" [\" args ]`` 80 80 `` .BR \"\" }`` 81 81 `` .YS`` 82 ``-`` 83 82 `` .SH DESCRIPTION`` 84 83 `` .IX \"strace command\" \"\" \"\\fLstrace\\fR command\"`` 85 84 `` .LP`` ``@@ -372,7 +371,7 @@ is the numeric process id of each process.`` 372 371 `` This is incompatible with`` 373 372 `` .BR \\-c ,`` 374 373 `` since no per-process counts are kept.`` 375 ``-`` 374 ``+.IP`` 376 375 `` One might want to consider using`` 377 376 `` .BR strace-log-merge (1)`` 378 377 `` to obtain a combined strace log view.`` ``@@ -672,7 +671,6 @@ system call which is controlled by the option`` 672 671 `` .BR \"\\-e\\ kvm\" = vcpu`` 673 672 `` Print the exit reason of kvm vcpu. Requires Linux kernel version 4.16.0`` 674 673 `` or higher.`` 675 ``-`` 676 674 `` .TP`` 677 675 `` .B \\-i`` 678 676 `` Print the instruction pointer at the time of the system call.`` ``@@ -833,7 +831,7 @@ each system call. The default is to summarise the system time.`` 833 831 `` .TP 12`` 834 832 `` \\fB\\-e\\ inject\\fR=\\,\\fIset\\/\\fR[:\\fBerror\\fR=\\,\\fIerrno\\/\\fR|:\\fBretval\\fR=\\,\\fIvalue\\/\\fR][:\\fBsignal\\fR=\\,\\fIsig\\/\\fR][:\\fBsyscall\\fR=\\fIsyscall\\fR][:\\fBdelay_enter\\fR=\\,\\fIusecs\\/\\fR][:\\fBdelay_exit\\fR=\\,\\fIusecs\\/\\fR][:\\fBwhen\\fR=\\,\\fIexpr\\/\\fR]`` 835 833 `` Perform syscall tampering for the specified set of syscalls.`` 836 ``-`` 834 ``+.IP`` 837 835 `` At least one of`` 838 836 `` .BR error ,`` 839 837 `` .BR retval ,`` ``@@ -846,7 +844,7 @@ options has to be specified.`` 846 844 `` and`` 847 845 `` .B retval`` 848 846 `` are mutually exclusive.`` 849 ``-`` 847 ``+.IP`` 850 848 `` If :\\fBerror\\fR=\\,\\fIerrno\\/\\fR option is specified,`` 851 849 `` a fault is injected into a syscall invocation:`` 852 850 `` the syscall number is replaced by -1 which corresponds to an invalid syscall`` ``@@ -856,24 +854,24 @@ and the error code is specified using a symbolic`` 856 854 `` value like`` 857 855 `` .B ENOSYS`` 858 856 `` or a numeric value within 1..4095 range.`` 859 ``-`` 857 ``+.IP`` 860 858 `` If :\\fBretval\\fR=\\,\\fIvalue\\/\\fR option is specified,`` 861 859 `` success injection is performed: the syscall number is replaced by -1,`` 862 860 `` but a bogus success value is returned to the callee.`` 863 ``-`` 861 ``+.IP`` 864 862 `` If :\\fBsignal\\fR=\\,\\fIsig\\/\\fR option is specified with either a symbolic value`` 865 863 `` like`` 866 864 `` .B SIGSEGV`` 867 865 `` or a numeric value within 1..\\fBSIGRTMAX\\fR range,`` 868 866 `` that signal is delivered on entering every syscall specified by the`` 869 867 `` .IR set .`` 870 ``-`` 868 ``+.IP`` 871 869 `` If :\\fBdelay_enter\\fR=\\,\\fIusecs\\/\\fR or :\\fBdelay_exit\\fR=\\,\\fIusecs\\/\\fR`` 872 870 `` options are specified, delay injection is performed: the tracee is delayed`` 873 871 `` by at least`` 874 872 `` .IR usecs`` 875 873 `` microseconds on entering or exiting the syscall.`` 876 ``-`` 874 ``+.IP`` 877 875 `` If :\\fBsignal\\fR=\\,\\fIsig\\/\\fR option is specified without`` 878 876 `` :\\fBerror\\fR=\\,\\fIerrno\\/\\fR, :\\fBretval\\fR=\\,\\fIvalue\\/\\fR or`` 879 877 `` :\\fBdelay_{enter,exit}\\fR=\\,\\fIusecs\\/\\fR options,`` ``@@ -886,21 +884,21 @@ Conversely, :\\fBerror\\fR=\\,\\fIerrno\\/\\fR or`` 886 884 `` :\\fBdelay_exit\\fR=\\,\\fIusecs\\/\\fR or`` 887 885 `` :\\fBsignal\\fR=\\,\\fIsig\\/\\fR options injects a fault without delivering a signal`` 888 886 `` or injecting a delay, etc.`` 889 ``-`` 887 ``+.IP`` 890 888 `` If both :\\fBerror\\fR=\\,\\fIerrno\\/\\fR or :\\fBretval\\fR=\\,\\fIvalue\\/\\fR`` 891 889 `` and :\\fBsignal\\fR=\\,\\fIsig\\/\\fR options are specified, then both`` 892 890 `` a fault or success is injected and a signal is delivered.`` 893 ``-`` 891 ``+.IP`` 894 892 `` if :\\fBsyscall\\fR=\\fIsyscall\\fR option is specified, the corresponding syscall`` 895 893 `` with no side effects is injected instead of -1.`` 896 894 `` Currently, only \"pure\" (see`` 897 895 `` .BR \"-e trace\" = \"%pure\"`` 898 896 `` description) syscalls can be specified there.`` 899 ``-`` 897 ``+.IP`` 900 898 `` Unless a :\\fBwhen\\fR=\\,\\fIexpr\\fR subexpression is specified,`` 901 899 `` an injection is being made into every invocation of each syscall from the`` 902 900 `` .IR set .`` 903 ``-`` 901 ``+.IP`` 904 902 `` The format of the subexpression is one of the following:`` 905 903 `` .RS`` 906 904 `` .TP 12`` ``@@ -932,13 +930,13 @@ For example, to fail each third and subsequent chdir syscalls with`` 932 930 `` .BR ENOENT ,`` 933 931 `` use`` 934 932 `` \\fB\\-e\\ inject\\fR=\\,\\fIchdir\\/\\fR:\\fBerror\\fR=\\,\\fIENOENT\\/\\fR:\\fBwhen\\fR=\\,\\fI3\\/\\fB+\\fR.`` 935 ``-`` 933 ``+.IP`` 936 934 `` The valid range for numbers`` 937 935 `` .I first`` 938 936 `` and`` 939 937 `` .I step`` 940 938 `` is 1..65535.`` 941 ``-`` 939 ``+.IP`` 942 940 `` An injection expression can contain only one`` 943 941 `` .BR error =`` 944 942 `` or`` ``@@ -948,18 +946,17 @@ specification, and only one`` 948 946 `` specification. If an injection expression contains multiple`` 949 947 `` .BR when =`` 950 948 `` specifications, the last one takes precedence.`` 951 ``-`` 949 ``+.IP`` 952 950 `` Accounting of syscalls that are subject to injection`` 953 951 `` is done per syscall and per tracee.`` 954 ``-`` 952 ``+.IP`` 955 953 `` Specification of syscall injection can be combined`` 956 954 `` with other syscall filtering options, for example,`` 957 955 `` \\fB\\-P \\fI/dev/urandom \\fB\\-e inject\\fR=\\,\\fIfile\\/\\fR:\\fBerror\\fR=\\,\\fIENOENT\\fR.`` 958 ``-`` 959 956 `` .TP`` 960 957 `` \\fB\\-e\\ fault\\fR=\\,\\fIset\\/\\fR[:\\fBerror\\fR=\\,\\fIerrno\\/\\fR][:\\fBwhen\\fR=\\,\\fIexpr\\/\\fR]`` 961 958 `` Perform syscall fault injection for the specified set of syscalls.`` 962 ``-`` 959 ``+.IP`` 963 960 `` This is equivalent to more generic`` 964 961 `` \\fB\\-e\\ inject\\fR= expression with default value of`` 965 962 `` .I errno``" ]

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[ "# Laws of Exponents\n\nExponents are also called Powers or Indices", null, "The exponent of a number says how many times to use the number in a multiplication.\n\nIn this example: 82 = 8 × 8 = 64\n\nIn words: 82 could be called \"8 to the second power\", \"8 to the power 2\" or simply \"8 squared\"\n\nTry it yourself:\n\nSo an Exponent saves us writing out lots of multiplies!\n\n### Example: a7\n\na7 = a × a × a × a × a × a × a = aaaaaaa\n\nNotice how we wrote the letters together to mean multiply? We will do that a lot here.\n\n## The Key to the Laws\n\nWriting all the letters down is the key to understanding the Laws\n\n### Example: x2x3 = (xx)(xxx) = xxxxx = x5\n\nWhich shows that x2x3 = x5, but more on that later!\n\nSo, when in doubt, just remember to write down all the letters (as many as the exponent tells you to) and see if you can make sense of it.\n\n## All you need to know ...\n\nThe \"Laws of Exponents\" (also called \"Rules of Exponents\") come from three ideas:", null, "The exponent says how many times to use the number in a multiplication.", null, "A negative exponent means divide, because the opposite of multiplying is dividing", null, "A fractional exponent like 1/n means to take the nth root:", null, "If you understand those, then you understand exponents!\n\nAnd all the laws below are based on those ideas.\n\n## Laws of Exponents\n\nHere are the Laws (explanations follow):\n\nLaw Example\nx1 = x 61 = 6\nx0 = 1 70 = 1\nx-1 = 1/x 4-1 = 1/4\nxmxn = xm+n x2x3 = x2+3 = x5\nxm/xn = xm-n x6/x2 = x6-2 = x4\n(xm)n = xmn (x2)3 = x2×3 = x6\n(xy)n = xnyn (xy)3 = x3y3\n(x/y)n = xn/yn (x/y)2 = x2 / y2\nx-n = 1/xn x-3 = 1/x3\nAnd the law about Fractional Exponents:", null, "", null, "## Laws Explained\n\nThe first three laws above (x1 = x, x0 = 1 and x-1 = 1/x) are just part of the natural sequence of exponents. Have a look at this:\n\nExample: Powers of 5\n.. etc..", null, "52 1 × 5 × 5 25\n51 1 × 5 5\n50 1 1\n5-1 1 ÷ 5 0.2\n5-2 1 ÷ 5 ÷ 5 0.04\n.. etc..\n\nLook at that table for a while ... notice that positive, zero or negative exponents are really part of the same pattern, i.e. 5 times larger (or 5 times smaller) depending on whether the exponent gets larger (or smaller).\n\n## The law that xmxn = xm+n\n\nWith xmxn, how many times do we end up multiplying \"x\"? Answer: first \"m\" times, then by another \"n\" times, for a total of \"m+n\" times.\n\n### Example: x2x3 = (xx)(xxx) = xxxxx = x5\n\nSo, x2x3 = x(2+3) = x5\n\n## The law that xm/xn = xm-n\n\nLike the previous example, how many times do we end up multiplying \"x\"? Answer: \"m\" times, then reduce that by \"n\" times (because we are dividing), for a total of \"m-n\" times.\n\n### Example: x4/x2 = (xxxx) / (xx) = xx = x2\n\nSo, x4/x2 = x(4-2) = x2\n\n(Remember that x/x = 1, so every time you see an x \"above the line\" and one \"below the line\" you can cancel them out.)\n\nThis law can also show you why x0=1 :\n\n## The law that (xm)n = xmn\n\nFirst you multiply \"m\" times. Then you have to do that \"n\" times, for a total of m×n times.\n\n### Example: (x3)4 = (xxx)4 = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x12\n\nSo (x3)4 = x3×4 = x12\n\n## The law that (xy)n = xnyn\n\nTo show how this one works, just think of re-arranging all the \"x\"s and \"y\"s as in this example:\n\n## The law that (x/y)n = xn/yn\n\nSimilar to the previous example, just re-arrange the \"x\"s and \"y\"s\n\n## The law that", null, "OK, this one is a little more complicated!\n\nI suggest you read Fractional Exponents first, or this may not make sense.\n\nAnyway, the important idea is that:\n\nx1/n = The n-th Root of x\n\nAnd so a fractional exponent like 43/2 is really saying to do a cube (3) and a square root (1/2), in any order.\n\nJust remember from fractions that m/n = m × (1/n):\n\n### Example:", null, "The order does not matter, so it also works for m/n = (1/n) × m:\n\n## Exponents of Exponents ...\n\n432\n\nWe do the exponent at the top first, so we calculate it this way:\n\n## And That Is It!\n\nIf you find it hard to remember all these rules, then remember this:\n\nyou can work them out when you understand the\n\n### Oh, One More Thing ... What if x = 0?\n\n Positive Exponent (n>0) 0n = 0 Negative Exponent (n<0) 0-n is undefined (because dividing by 0 is undefined) Exponent = 0 00 ... ummm ... see below!\n\n### The Strange Case of 00\n\nThere are different arguments for the correct value of 00\n\n00 could be 1, or possibly 0, so some people say it is really \"indeterminate\":", null, "x0 = 1, so ... 00 = 1 0n = 0, so ... 00 = 0 When in doubt ... 00 = \"indeterminate\"\n\nHard:" ]

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[ "Solutions by everydaycalculation.com\n\n## Divide 10/5 with 56/50\n\n1st number: 2 0/5, 2nd number: 1 6/50\n\n10/5 ÷ 56/50 is 25/14.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 56/50: 50/56\n2. Now, multiply it with the dividend\nSo, 10/5 ÷ 56/50 = 10/5 × 50/56\n3. = 10 × 50/5 × 56 = 500/280\n4. After reducing the fraction, the answer is 25/14\n5. In mixed form: 111/14\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "# nLab regular action\n\nContents\n\n### Context\n\n#### Representation theory\n\nrepresentation theory\n\ngeometric representation theory\n\n## Geometric representation theory\n\n#### Group Theory\n\ngroup theory\n\nClassical groups\n\nFinite groups\n\nGroup schemes\n\nTopological groups\n\nLie groups\n\nSuper-Lie groups\n\nHigher groups\n\nCohomology and Extensions\n\nRelated concepts\n\n# Contents\n\n## Definition\n\nAn action\n\n$(-)\\cdot(-) \\;\\colon\\; G \\times X \\to X$\n\nof a group $G$ on an inhabited set $X$ is called regular if it is both transitive and free.\n\nIn terms of elements this means that if for any pair of elements $x,y \\in X$, there is exactly one group element $g \\in G$ such that $g \\cdot x = y$.\n\nMore abstractly this means that the shear map\n\n$\\array{ G \\times X & \\overset { (pr_2, \\cdot) } {\\longrightarrow}& X \\times X \\\\ (g, x) & \\mapsto & \\big( x, g \\cdot x \\big) \\,. }$\n\nis an isomorphism.\n\nIn this case $X$ is also called a $G$-torsor. If the condition is dropped that $X$ be inhabited it is still called a pseudo-torsor.\n\n## Properties\n\n###### Proposition\n\nSuppose $G$ acts transitively on $X$ by $* : G \\times X \\to X$, and suppose moreover that this action is faithful. Then $G$ acts freely (and hence regularly) on $X$ if and only if the group $Aut_G(X)$ of $G$-equivariant automorphisms (i.e., bijections $\\phi : X \\to X$ commuting with the action of $G$) acts transitively (and hence regularly) on $X$.\n\n###### Proof\n\nFirst we show that $Aut_G(X)$ acts freely on $X$. Suppose $\\phi \\in Aut_G(X)$ is such that $\\phi(x) = x$ for some $x\\in X$, and let $y\\in X$ be arbitrary. By the assumption that $G$ acts transitively, there is a $g \\in G$ such that $y = g*x$. But then $G$-equivariance implies that\n\n$\\phi(y) = \\phi(g*x) = g*\\phi(x) = g*x = y.$\n\nSince this holds for all $y\\in Y$, $\\phi$ must be equal to the identity $\\phi = id_X$, and therefore $Aut_G(X)$ acts freely on $X$.\n\nNext, suppose that $G$ also acts freely on $X$, and let $x,y \\in X$ be arbitrary. Then we can define a $G$-equivariant automorphism $\\phi$ such that $\\phi(x) = y$ by\n\n$\\phi = z \\mapsto g_z*y,$\n\nwhere for each $z$, $g_z$ is the unique group element such that $z = g_z*x$. Conversely, suppose that $Aut_G(X)$ acts transitively on $X$, and let $x\\in X$, $g\\in G$ such that $g*x = x$. By the assumption, for any $y \\in X$, there exists $\\phi \\in Aut_G(X)$ such that $\\phi(x) = y$, from which it follows that\n\n$g*y = g*\\phi(x) = \\phi(g*x) = \\phi(x) = y.$\n\nSince $g*y = y$ for all $y \\in X$, therefore $g = 1$ by the assumption that $G$ acts faithfully on $X$.\n\n## Examples\n\n• The action of $G$ on itself by multiplication $\\cdot : G \\times G \\to G$ (on the left or on the right) is a regular action, called the (left or right) regular representation of $G$.\n\n• If one views a combinatorial map $M$ as the transitive action of a certain group of permutations, then $M$ represents a regular map (Siran 2006) just in case this action is regular. For example, the five Platonic solids may be represented as regular combinatorial maps.\n\n## In homotopy type theory\n\nWe discuss regular actions via homotopy type theory.\n\nSince doing group representation theory in homotopy type theory corresponds to working in the context of a delooped group in homotopy type theory, the regularity of an action is naturally expressed there. Transitivity ensures that all the points in the homotopy quotient are connected by equivalences, while freeness means that the space of equivalences between two points is itself contractible. Hence if $\\ast \\colon \\mathbf{B} G \\vdash X(\\ast): Type$ corresponds to a regular action, then the quotient $\\sum_{\\ast: \\mathbf{B} G} X(\\ast)$ is contractible.\n\nRestriction to 1-groups is unnecessary here, and we say\n\n###### Definition\n\nAn ∞-action of an ∞-group is a regular $\\infty$-action if its homotopy quotient is contractible.\n\nFor any $G$-action (∞-action) $X \\colon BG \\to U$, its automorphism group is (see at automorphism ∞-group in HoTT)\n\n$B Aut_G(X) \\coloneqq \\sum_{(P:BG \\to U)} \\|P=X\\|$\n\nand\n\n$\\tilde{X} \\colon B Aut_G(X) \\to U$\n\nby $\\tilde{X}(P,-) \\coloneqq P(\\ast)$.\n\n###### Proposition\n\nIf $X$ is regular, then $\\tilde{X}$ is regular.\n\n###### Proof\n\nFirst, we need to argue that $X(\\ast)$ is merely inhabited. Since $X$ is regular, we have $\\sum_{(b:BG)} X(b)$ contractible. This gives a center of contraction $(b,x)$. Now, since $B G$ is connected, it follows that $\\|b=\\ast\\|$. Since we are proving the mere proposition $\\|X(\\ast)\\|$, we get to use $b=\\ast$. Now we obtain $\\|X(\\ast)\\|$.\n\nNext, to show that $\\tilde{X}$ is regular we need to show that $\\tilde{X}$ has a contractible total space. The dependent sum type $\\sum_{(b : BAut_G(X))} \\tilde{X}(b)$ is equivalent to $\\sum_{(P:BG \\to U)} \\|P=X\\| \\times P(\\ast)$. Contractibility is a mere proposition, and we have $\\|X(\\ast)\\|$, so we get to use a point $x:X(\\ast)$. This gives us a center of contraction $(X,refl,x)$ of the total space of $\\tilde{X}$.\n\nNow let $P : BG \\to U$, let $\\| P = X \\|$, let $p_0 : P(\\ast)$. To show regularity, it suffices to find a term of type\n\n$\\sum_{\\alpha : P = X} \\mathrm{trans}(\\alpha)(p_0) = x \\,.$\n\nThis type is equivalent to showing that there are\n\n$K : \\prod_{b:BG} P(b) \\simeq X(b) \\;\\;\\;\\; \\text{and} \\;\\;\\;\\;\\; K(*,p_0) = x_0 \\,.$\n\nNow we use that $X(b)$ is equivalent to $b=\\ast$ (we get this fact from regularity, together with a point $x:X(\\ast)$). Since we need this particular fiberwise equivalence, it suffices to show that\n\n$\\sum_{b:BG} P(b)$\n\nis contractible. Now this is a mere proposition, so we can eliminate $t : \\| P = X \\|$ to obtain the proof.\n\n###### Proposition\n\nIf $X$ is a principal homogeneous space on $G$, in the sense that the type $\\sum_{(g:G)} g_\\ast(x)=y$ is contractible for all $x,y:X(\\ast)$, and $\\tilde{X}$ is regular, then $X$ is regular.\n\n###### Proof\n\nAgain, we first show that $X(\\ast)$ is merely inhabited. The total space of $\\tilde{X}$ has center of contraction $(P,p_0)$. Since $\\|P=X\\|$ and since we are proving a mere proposition, we get to use $P=X$. Now $\\|X(\\ast)\\|$ follows from $p_0:P(\\ast)$. The regularity of $X$ is a mere proposition, so we get to use $x_0:X(\\ast)$. This gives us the center of contraction $(\\ast,x_0)$. It remains to show that\n\n$\\prod_{(b:BG)} \\prod_{(x:X(b))} \\sum_{(\\alpha : b=\\ast)} \\mathrm{trans}(\\alpha,x) = x_0$.\n\nOf course, it would suffice to prove the stronger statement\n\n$\\prod_{(b:BG)} \\prod_{(x:X(b))} \\mathrm{isContr} (\\sum_{(\\alpha : b=\\ast)} \\mathrm{trans}(\\alpha,x) = x_0)$.\n\nHowever, now we get to use that $BG$ is connected. Therefore it suffices to show that\n\n$\\prod_{x:X(*)} isContr (\\sum_{\\alpha : G} \\mathrm{trans}(\\alpha,x) = x_0)$\n\nThis holds by assumption.\n\n• Jozef Siran, “Regular Maps on a Given Surface: A Survey”, Topics in Discrete Mathematics, 2006. (pdf)\n\n• Group Properties Wiki: Regular group action." ]

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[ "# extinction probability poisson\n\nDuring writing this I thought of an attempt to solve this using Hurwitz theorem, namely, show that the smallest non-negative fixed point of $G(s)$ is smaller than 1, then show that the function $G'$ is complex differentiable with non-zero derrivative, use inverse function theorem, get an open neighbourhood of our fixed point, find sequence of fixed points which converge to the smallest non-negative fixed point $\\eta(\\lambda)$ of $G$. for a Poisson distribution. Fortunately, this student collected data ��z �p�2I�)���@g�$���]I���Q5���=n�-u]�d�|��9N�E=ͦ�^ I�{�WdY�-�~fA���j��oJ��F��֙b�n-�YIdW��^�E��}�-}s��R�S��;�E_! Part of Springer Nature. This service is more advanced with JavaScript available, War in the Body counting events. The Poisson Process is the model we use for describing randomly occurring events and by itself, isn’t that useful. How does the UK manage to transition leadership so quickly compared to the USA? )�����b_՟��{� ˓9O.oJ?H;�]5ViR6=J���粥���U]�~V��.��q�=»@�8 ∈ Citing historical examples of Galton–Watson process is complicated due to the history of family names often deviating significantly from the theoretical model. is a set of independent and identically-distributed natural number-valued random variables. However, excluding the non-trivial case, the concept of the averaged reproduction mean (Bruss (1984)) allows for a general sufficient condition for final extinction, treated in the next section. For a detailed history see Kendall (1966 and 1975). NICU stay is the same as the probability of infection later in the NICU stay. The random variables of a stochastic process are indexed by the natural numbers. Show that the extinction probability converges to$\\eta(\\lambda)$as$n \\rightarrow \\infty$, where$\\eta(\\lambda)$is the extinction probability of a branching process with family-sizes distributed as$\\text{Po}(\\lambda)$. Names have changed or become extinct for various reasons such as people taking the names of their rulers, orthographic simplifications, taboos against using characters from an emperor's name, among others. In practice, family names change for many other reasons, and dying out of name line is only one factor, as discussed in examples, below; the Galton–Watson process is thus of limited applicability in understanding actual family name distributions. horse kicks, or the number of defects per square yard. only between the hours of 10-11am, Monday through Friday. Why does chrome need access to Bluetooth? The symbol for this average is$ \\lambda \\$, the greek letter lambda. intervals. Need more Why Is an Inhomogenous Magnetic Field Used in the Stern Gerlach Experiment? Once an adult, the individual gives birth to exactly two offspring, and then dies. infection prone than others. Some examples are: Sometimes, you will see the count The above plot illustrates Poisson probabilities for The Poisson Probability Calculator can calculate the probability of an event occurring in a given time interval. Then the simplest substantial mathematical conclusion is that if the average number of a man's sons is 1 or less, then their surname will almost surely die out, and if it is more than 1, then there is more than zero probability that it will survive for any given number of generations. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Quick link too easy to remove after installation, is this a problem? Nosy Norbert wants to know if some of his data follows a Poisson\n\nKiwi Animal Crossing: New Horizons, Can Electric Potential Be Negative, Questions Starting With Helping Verbs, Block Diagram Algebra In Control System, 2020 Ford Explorer Headlight Bulb, Magnesium Dichromate Molar Mass," ]

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[ "# Binary Numbers with Playing Cards\n\n## Overview and Objective\n\nIn this lesson, students will explore the binary system. Students will use the 1, 2, 4, and 8 playing cards to represent numbers in base 2.\n\n## Warm-Up\n\nShow the selected game cards to students. Ask them if they notice any particular pattern about the cards. Discuss which card can be added to the left side of the set if there are infinitely many game cards in a deck. You may want to emphasize the difference between the powers of 2 and the multiples of 2.\n\n• first two cards (1+2=3)\n• first three cards (1+2+4=7)\n• first four cards (1+2+4+8=15)\n\nDiscuss with students any patterns between the sums and the next number in the sequence. Clarify with the students that the next number on the left will always be one more than the sum of the previous numbers.\n\n## Main Activity\n\nIn the first set, there are (8+4+1) 13 shapes.\n\nThe following sets have 9, 10, 4, 5, and 14 shapes consecutively.\n\nAsk students to determine all possible numbers that can be created with these cards. Consider having students work on this in pairs and allow them some time to try the other numbers. Share students' examples with the class while recording the numbers found on a table. Ask students if there can be a shortcut to writing the numbers like a code. Let them share their ideas. As you're comfortable, lead them to use \"0\" for the turned-over cards and \"1\" for the face-up cards.\n\nStart filling the codes all the numbers up to 15.\n\nDiscuss the largest and the smallest number they can write and how they can write the numbers bigger than 15.\n\nIf students need a visual for 16, you may use this canvas to ask them different combinations of the numbers.\n\nAsk students to create numbers like 23, 25, and 30. Invite them to share their answers.\n\nNow discuss with students if there can be an easier way to express larger numbers. If students have not already mentioned using on/off like systems, you can suggest using 0s for the turned-down cards and 1s for the face-up ones.\n\nLet them express numbers like 5, 12, 17, 21, 30 using 1s and 0s. This system of writing numbers is called the Binary system. The numbers which are written only by using 0s and 1s are called binary numbers, and they are written in base 2 instead of base 10.\n\nIn the last part, let students work out the value of 1001, 10101, 11010, 100000, 1101011. They can use cards to express these numbers in base 10.\n\n1001= 8+0+0+1 =9\n\n10101= 16+0+4+0+1 =21\n\n11010= 16+8+0+2+0 =26\n\n100000= 32+0+0+0+0+0 =32\n\n110101= 32+16+0+4+0+1= 53\n\nIf the students need clarification on expressing binary numbers in base 10, you may use a table to highlight the place values.\n\n## Closure\n\nSome students might already know that computers use only zeros and ones to represent everything like words, pictures, numbers, movies, and sound.\n\nIf time, to close the lesson, let students explore\n\n• What happens when a zero is added to the right-hand side of the number? For instance, 101 → 1010 or 111→ 1110.\n\nClarify with students if we do the same thing base 10; the number gets 10 times as large. Since now we work on base 2, the number gets twice as large.\n\n101 → 1010; 5→10\n\n111→ 1110; 7→ 14" ]

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[ "Understanding Common Misconceptions\n\n2017-12-25 | Daniel Lakens", null, "A p-value is the probability of the observed, or more extreme, data, under the assumption that the null-hypothesis is true. The goal of this blog post is to understand what this means, and perhaps more importantly, what this doesn’t mean. People often misunderstand p-values, but with a little help and some dedicated effort, we should be able explain these misconceptions. Below is my attempt, but if you prefer a more verbal explanation, I can recommend Greenland et al. (2016).\n\nFirst, we need to know what ‘the assumption that the null-hypothesis is true’ looks like. Although the null-hypothesis can be any value, here we will assume the null-hypothesis is specified as a difference of 0. When this model is visualized in text-books, or in power-analysis software such as g*power, you often see a graph like the one below, with t-values on the horizontal axis, and a critical t-value somewhere around 1.96. For a mean difference, the p-value is calculated based on the t-distribution (which is like a normal distribution, and the larger the sample size, the more similar the two become). I will distinguish the null hypothesis (the mean difference in the population is exactly 0) from the null-model (a model of the data we should expect when we draw a sample when the null-hypothesis is true) in this post.", null, "I’ve recently realized that things become a lot clearer if you just plot these distributions as mean differences, because you will more often think about means, than about t-values. So below, you can see a null-model, assuming a standard deviation of 1, for a t-test comparing mean differences (because the SD = 1, you can also interpret the mean differences as a Cohen’s d effect size).", null, "The first thing to notice is that we expect that the mean of the null-model is 0: The distribution is centered on 0. But even if the mean in the population is 0, that does not imply every sample will give a mean of exactly zero. There is variation around the mean, as a function of the true standard deviation, and the sample size. One reason why I prefer to plot the null-model in raw scores instead of t-values is that you can see how the null-model changes, when the sample size increases.", null, "When we collect 5000 instead of 50 observations, we see the null-model is still centered on 0 – but in our null-model we now expect most values will fall very close around 0. Due to the larger sample size, we should expect to observe mean differences in our sample closer to 0 compared to our null-model when we had only 50 observations.\n\nBoth graphs have areas that are colored red. These areas represent 2.5% of the values in the left tail of the distribution, and 2.5% of the values in the right tail of the distribution. Together, they make up 5% of the most extreme mean differences we would expect to observe, given our number of observations, when the true mean difference is exactly 0 – representing the use of an alpha level of 5%. The vertical axis shows the density of the curves.\n\nLet’s assume that in the figure visualizing the null model for N = 50 (two figures up) we observe a mean difference of 0.5 in our data. This observation falls in the red area in the right tail of the distribution. This means that the observed mean difference is surprising, if we assume that the true mean difference is 0. If the true mean difference is 0, we should not expect such a extreme mean difference very often. If we calculate a p-value for this observation, we get the probability of observing a value more extreme (in either tail, when we do a two-tailed test) than 0.5.\n\nTake a look at the figure that shows the null-model when we have collected 5000 observations (one figure up), and imagine we would again observe a mean difference of 0.5. It should be clear that this same difference is even more surprising than it was when we collected 50 observations.\n\nWe are now almost ready to address common misconceptions about p-values, but before we can do this, we need to introduce a model of the data when the null is not true. When the mean difference is not exactly 0, the alternative hypothesis is true – but what does an alternative model look like?\n\nWhen we do a study, we rarely already know what the true mean difference is (if we already knew, why would we do the study?). But let’s assume there is an all-knowing entity. Following Paul Meehl, we will call this all-knowing entity Omniscient Jones. Before we collect our sample of 50 observations, Omniscient Jones already knows that the true mean difference in the population is 0.5. Again, we should expect some variation around this true mean difference in our small sample. The figure below again shows the expected data pattern when the null-hypothesis is true (now indicated by a grey line) and it shows an alternative model, assuming a true mean difference of 0.5 exists in the population (indicated by a black line).", null, "But Omniscient Jones could have said the true difference was much larger. Let’s assume we do another study, but now before we collect our 50 observations, Omniscient Jones tells us that the true mean difference is 1.5. The null model does not change, but the alternative model now moves over to the right.", null, "Now, we are finally ready to address some common misconceptions about p-values. Before we look at misconceptions in some detail, I want to remind you of one fact that is easy to remember, and will enable you to recognize many misconceptions about p-values: p-values are a statement about the probability of data, not a statement about the probability of a theory. Whenever you see p-values interpreted as a probability of a theory or a hypothesis, you know something is not right. Now let’s take a look at why this is not right.\n\n1) Why a non-significant p-value does not mean that the null-hypothesis is true.\n\nLet’s take a concrete example that will illustrate why a non-significant result does not mean that the null-hypothesis is true. In the figure below, Omniscient Jones tells us the true mean difference is again 0.5. We have observed a mean difference of 0.35. This value does not fall within the red area (and hence, the p-value is not smaller than our alpha level, or p > .05). Nevertheless, we see that observing a mean difference of 0.35 is much more likely under the alternative model, than under the null-model.", null, "All the p-value tells us is that this value is not extremely surprising, if we assume the null-hypothesis is true. A non-significant p-value does not mean the null-hypothesis true. It might be, but it is also possible that the data we have observed is more likely when the alternative hypothesis is true, than when the null-hypothesis is true (as in the figure above).\n\n2) Why a significant p-value does not mean that the null-hypothesis is false.\n\nImagine we generate a series of numbers in R using the following command:\n\n``````rnorm(n = 50, mean = 0, sd = 1)\n``````\n\nThis command generates 50 random observations from a distribution with a mean of 0 and a standard deviation of 1. We run this command once, and we observe a mean difference of 0.5. We can perform a one-sample t-test against 0, and this test tells us, with a p\n\nShould we decide to reject the null-hypothesis that the random number generator in R works? That would be a bold move indeed! We know that the probability of observing surprising data, assuming the null hypothesis is true, has a maximum of 5% when our alpha is 0.05. What we can conclude, based on our data, is that we have observed an extreme outcome, that should be considered surprising. But such an outcome is not impossible when the null-hypothesis is true. And in this case, we really don’t even have an alternative hypothesis that can explain the data (beyond perhaps evil hackers taking over the website where you downloaded R).", null, "This misconception can be expressed in many forms. For example, one version states that the p-value is the probability that the data were generated by chance. Note that this is just a sneaky way to say: The p-value is the probability that the null hypothesis is true, and we observed an extreme p-value just due to random variation. As we explained above, we can observe extreme data when we are basically 100% certain that the null-hypothesis is true (the random number generator in R works as it should), and seeing extreme data once should not make you think the probability that the random number generator in R is working is less than 5%, or in other words, that the probability that the random number generator in R is broken is now more than 95%.\n\nRemember: P-values are a statement about the probability of data, not a statement about the probability of a theory or a hypothesis.\n\n3) Why a significant p-value does not mean that a practically important effect has been discovered.\n\nIf we plot the null-model for a very large sample size (N = 100000) we see that even very small mean differences (here, a mean difference of 0.01) will be considered ‘surprising’. We have such a large sample size, that all means we observe should fall very close around 0, and even a difference of 0.01 is already considered surprising, due to our substantial level of accuracy because we collected so much data.", null, "Note that nothing about the definition of a p-value changes: It still correctly indicates that, if the null-hypothesis is true, we have observed data that should be considered surprising. However, just because data is surprising, does not mean we need to care about it. It is mainly the verbal label ‘significant’ that causes confusion here – it is perhaps less confusing to think of a ‘significant’ effect as a ‘surprising’ effect (as long as the null-model is realistic - which is not automatically true).\n\nThis example illustrates why you should always report and interpret effect sizes, with hypothesis tests. This is also why it is useful to complement a hypothesis test with an equivalence test, so that you can conclude the observed difference is surprisingly small if there is no difference, but the observed difference is also surprisingly closer to zero, assuming there exists any effect we consider meaningful (and thus, we can conclude the effect is equivalence to zero).\n\n4) If you have observed a significant finding, the probability that you have made a Type 1 error (a false positive) is not 5%.\n\nAssume we collect 20 observations, and Omniscient Jones tells us the null-hypothesis is true. This means we are sampling from the following distribution:", null, "If this is our reality, it means that 100% of the time that we observe a significant result, it is a false positive. Thus, 100% of our significant results are Type 1 errors. What the Type 1 error rate controls, is that from all studies we perform when the null is true, not more than 5% of our observed mean differences will fall in the red tail areas. But when they have fallen in the tail areas, they are always a Type 1 error. After observing a significant result, you can not say it has a 5% probability of being a false positive. But before you collect data, you can say you will not conclude there is an effect, when there is no effect, more than 5% of the time, in the long run.\n\n5) One minus the p-value is not the probability of observing another significant result when the experiment is replicated.\n\nIt is impossible to calculate the probability that an effect will replicate, based on the p-value, and as a consequence, the p-value can not inform us about the p-value we will observe in future studies. When we have observed a p-value of 0.05, it is not 95% certain the finding will replicate. Only when we make additional assumptions (e.g., the assumption that the alternative effect is true, and the effect size that was observed in the original study is exactly correct) can we model the p-value distribution for future studies.\n\nIt might be useful to visualize the one very specific situation when the p-value does provide the probability that future studies will provide a significant p-value (even though in practice, we will never know if we are in this very specific situation). In the figure below we have a null-model and alternative model for 150 observations. The observed mean difference falls exactly on the threshold for the significance level. This means the p-value is 0.05. In this specific situation, it is also 95 probable that we will observe a significant result in a replication study, assuming there is a true effect as specified by the alternative model. If this alternative model is true, 95% (1-p) of the observed means will fall on the right side of the observed mean in the original study (we have a statistical power of 95%), and only 5% of the observed means will fall in the blue area (which contains the Type 2 errors).", null, "This very specific situation is almost always not your reality. It is not true when any other alternative hypothesis is correct. And it is not true when the the null-hypothesis is true. In short, the p-value basically never, except for one very specific situation when the alternative hypothesis is true and of a very specific size you will never know you are in, gives the probability that a future study will once again yield a significant result.\n\nConclusion\n\nProbabilities are confusing, and the interpretation of a p-value is not intuitive. Grammar is also confusing, and not intuitive. But where we practice grammar in our education again and again and again until you get it, we don’t practice the interpretation of p-values again and again and again until you get it. Some repetition is probably needed. Explanations of what p-values mean are often verbal, and if there are figures, they use t-value distributions we are unfamiliar with. Instead of complaining that researchers don’t understand what p-values mean, I think we should try to explain common misconceptions multiple times, in multiple ways.\n\nHere you can find the R code for the graphs.\n\nThis post originally was published on December 5th 2017 on daniellakens.blogspot.de/. We are very grateful that Daniel has allowed us to cross-post this very insightful post on our blog. You can find the original blog post here.\n\nThe title picture was published 1917 and so it does not include any copyright.", null, "### Daniel Lakens\n\nDaniel Lakens is an experimental psychologist at the Human-Technology Interaction group at Eindhoven University of Technology, The Netherlands. He publishes the weblog ‘The 20% Statistician’." ]

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[ "# How does 0=infinity?\n\nA)First, let's start with $\\frac{0}{0}$=x\n\nA1. multiply by 0\n0=0*x\n\nB)Now we do $\\frac{0}{∞}$\n\nB1. Let's say $\\frac{0}{∞}$=y multiply by ∞ 0=y*∞\n\nB2. divide by 0 multiply by ∞ [ 0=$\\frac{y*∞}{0}$ ]\n\nC) Now $\\frac{∞}{0}$=z\n\nC1. multiply by 0 0*z=∞\n\nC2. divide by z 0=$\\frac{∞}{z}$\n\nTake all the bolded equations that have 0\n\nx* 0=y*∞=$\\frac{∞}{z}$ divide by 0 x= $\\frac{y*∞}{0}$=$\\frac{∞}{0}$\n\nTake the [equation with square brakets] and the above that have $\\frac{y * ∞}{0}$ x= $\\frac{∞}{0}$= $\\frac{0}{0}$ multiply by 0 x*0= $\\boxed{∞=0}$", null, "Note by Jw C\n1 week, 5 days ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n• bulleted\n• list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nlol, u violated a lot of math rules.\nwhat is your intention? to create a new set of rules?\nnice idea, but still u violated your own new rules.\n\n- 1 week, 5 days ago\n\nI am ASSUMING that dividing by 0 and infinity is possible.\n\n- 1 week, 5 days ago\n\nok. but to violate math rules dont prove this assumption\n\n- 1 week, 5 days ago\n\nif u multiply an equation with 0, u will always get 0 = 0. thats why multiplication with 0 is not used to solve equalitys.\nelse u could prove 3 = 4 by multiply with 0: 3 * 0 = 4 * 0 -> 0=0\n\n- 1 week, 5 days ago\n\ni can add more steps where maths violations happened, if it is helpful for u.\n\n- 1 week, 5 days ago" ]

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[ "This preview shows page 3 - 5 out of 9 pages.\n\nNASSAR AND SALAMA: ADAPTIVE SELF-ADEQUATE MICROGRIDS USING DYNAMIC BOUNDARIES 3 Fig. 1. Steps to divide bulky grids into adaptive self-adequate microgrids. e.g., customer loads, generation ratings, available generation, types of generation, and location of generation. When the agent is deactivated, it gives control to the active agent, which now knows only the aggregated generation and loads associ- ated with the components that have been newly added under its umbrella. The following sections discuss the application of steps 2–7 of the proposed methodology using the PG&E 69-bus system [ 4 ], [ 12 ] for demonstration purposes. Fig. 2. Functions of a microgrid agent. Fig. 3. Dynamic microgrid borders produced for two operating scenarios. (a) Scenario 1. (b) Scenario 2. III. M ODELING OF W IND , S OLAR , AND B IOMASS G ENERATION AND L OADS This section discusses step 2 of the methodology: the devel- opment of stochastic models of the output power of a variety of DGs and the loads, based on historical data. These models are then employed in further design steps. A. Wind To model the output power from wind-based DG, three successive years’ worth of historical wind speed data for a spe- cific site are used for the calculation of output power, as expressed [ 13 ]. The per unit output power on a rated power basis is then determined P ( v ) = 0 0 v v ci P rated × v v ci v r v ci v ci v v r P rated v r v v co 0 v co v (1) where v is the wind speed; v ci , v co , and v r are the cut-in speed, cut-out speed, and rated speed of the wind turbine, respec- tively; P is turbine output power; and P rated is the turbine rated output power. These historical per unit wind output power data are next analyzed in order to generate an appropriate probability den- sity function (PDF) that represents wind per unit output power. In this wind model, the year is divided into four seasons, with", null, "This article has been accepted for inclusion in a future issue of this journal. Content is final as presented, with the exception of pagination. 4 IEEE TRANSACTIONS ON SMART GRID TABLE I G OODNESS OF F IT T EST R ESULTS FOR E ACH S EASON the best PDF for each season selected to represent the wind per unit output power. Goodness of fit tests is used for the determination of the best PDF; the results obtained are presented in Table I . Because the number of samples exceeds 2000, the tests used are the Kolmogorov–Smirnov test (K-S), as expressed by ( 2 ), and the Anderson–Darling test (A-D), as expressed D = max 1 i n F ( X i ) i 1 n , i n F ( X i ) (2) A 2 = − n 1 n n i = 1 ( 2i 1 ) ln F ( X i ) + ln ( 1 F ( X n i + 1 )) (3) where F is an empirical cumulative distribution func- tion (CDF), i is the sample, D is K-S statistic, A 2 is the A-D statistic, and n is the total number of samples.", null, "", null, "#### You've reached the end of your free preview.\n\nWant to read all 9 pages?\n\n• Fall '10\n• BOB\n•", null, "•", null, "•", null, "", null, "" ]

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[ "# Polynomial problem for homework\n\nIf $$\\mathbf{p(x) = 2009-2008x^{100}+2007x^{99}-2006x^{98}+..............+1909x}$$\nThen Calculate $$\\mathbf{p(2008)}$$\n\n## Answers and Replies\n\nIf $$\\mathbf{p(x) = 2009-2008x^{100}+2007x^{99}-2006x^{98}+..............+1909x}$$\nThen Calculate $$\\mathbf{p(2008)}$$\n\n## The Attempt at a Solution\n\nAre you trying to do it exactly via algebra/arithmetic, or approximately with a computer program?\n\nYou can do this out in a straightforward way exactly if you use the summation (Sigma) notation. I've attached a scan of a sketched out way to approach the solution. Note that I did this very quickly and so I expect some mistakes there. However, you should be able to pick up on the method.\n\n#### Attachments\n\n• PTDC0042.pdf\n203 KB · Views: 139\n\nAre you trying to do it exactly via algebra/arithmetic, or approximately with a computer program?\n\nYou can do this out in a straightforward way exactly if you use the summation (Sigma) notation. I've attached a scan of a sketched out way to approach the solution. Note that I did this very quickly and so I expect some mistakes there. However, you should be able to pick up on the method.\n\nthanks stevenb\n\nthanks stevenb\n\nYou are welcome.\n\nI hate to leave an answer that I know is wrong, even if my intent was to give the method more than the correct answer. Tonight I was bored, so thought I would take the time to work this out correctly, just for the record, in case anyone comes across this thread in the future.\n\nI attached the correct way to work it out exactly. I know this is correct because I took step #1 and step #10 and fed them into Maxima to make sure they agree. Of course, the answer is hundreds of digits long once expanded out, but both answers agree to the last digit.\n\nNote that one could work out a general formula in terms of x rather that the one value of x=2008, by following the exact same procedure.\n\n#### Attachments\n\n• CorrectWay.pdf\n630.2 KB · Views: 118\nLast edited:\n\nNote that one could work out a general formula in terms of x rather that the one value of x=2008, by following the exact same procedure.\n\nAnd, making certain this horse has truly been beaten to death, I might as well post that too, since I worked it out in another fit of boredom. I should have done it that way the first time.\n\nThis works out to the following, which was verified with Maxima which expands it back to the original formula 2009-2008x^100+2007x^99 ... 1909x\n\n$${{2009+5927x+3917x^2-2009x^{101}-2008x^{102}}\\over{(x+1)^2}}$$\n\nLast edited:\nRay Vickson\nScience Advisor\nHomework Helper\nDearly Missed\n\nYou are welcome.\n\nI hate to leave an answer that I know is wrong, even if my intent was to give the method more than the correct answer. Tonight I was bored, so thought I would take the time to work this out correctly, just for the record, in case anyone comes across this thread in the future.\n\nI attached the correct way to work it out exactly. I know this is correct because I took step #1 and step #10 and fed them into Maxima to make sure they agree. Of course, the answer is hundreds of digits long once expanded out, but both answers agree to the last digit.\n\nNote that one could work out a general formula in terms of x rather that the one value of x=2008, by following the exact same procedure.\n\nJust as a matter of interest: how do you post an attachment to this forum? I see no tabs or buttons or menu items that look like \"attach\" commands.\n\nRGV\n\nberkeman\nMentor\n\nAre you trying to do it exactly via algebra/arithmetic, or approximately with a computer program?\n\nYou can do this out in a straightforward way exactly if you use the summation (Sigma) notation. I've attached a scan of a sketched out way to approach the solution. Note that I did this very quickly and so I expect some mistakes there. However, you should be able to pick up on the method.\n\nPlease do not do the student's work for them. That is against the rules. Especially when the student showed no effort at all in working out the problem.\n\nberkeman\nMentor\n\nJust as a matter of interest: how do you post an attachment to this forum? I see no tabs or buttons or menu items that look like \"attach\" commands.\n\nRGV\n\nWhen you are in the Advanced Reply (or New Topic) window, look next to the little smiley face pulldown menu for the paper clip \"Attachments\" pulldown menu. Clicking on that should get you to the Attachments dialog box.\n\nEDIT -- since this thread is now locked, PM me if you have further questions about attachments." ]

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[ "Title:\nVoltage control with feedback utilizing analog and digital control signals\nUnited States Patent 5959855\n\nAbstract:\nAn output voltage control apparatus for performing feedback control on an output voltage of an inverter which drives an AC machine. The output voltage control apparatus converts a command value for the output voltage of the inverter into a pulse density using a converter (PDM), and converts a detection value of the output voltage of the inverter into a pulse density using a Δ-Σ modulation converter. The proportional control ability and integral control ability for the output voltage are realized by digital hardware including an up-counter, a down-counter and a shift circuit.\n\nInventors:\nIshii, Shinichi (Saitama, JP)\nAihara, Takashi (Tokyo, JP)\nTakahashi, Hiroshi (Tokyo, JP)\nApplication Number:\n08/804876\nPublication Date:\n09/28/1999\nFiling Date:\n02/24/1997\nExport Citation:\nAssignee:\nFuji Electric Co., Ltd. (Kawasaki, JP)\nPrimary Class:\nOther Classes:\n318/811, 363/41\nInternational Classes:\nG05B11/36; G05B19/05; G05F1/10; H02M3/00; H02M7/12; H02M7/155; H02M7/48; (IPC1-7): H02M3/24\nField of Search:\n323/283, 323/285, 323/266, 318/811, 363/95, 363/41\nView Patent Images:\nUS Patent References:\n 5633788 Power converter control system 1997-05-27 Tanaka 363/41 5589749 Closed loop control system and method using back EMF estimator 1996-12-31 Davidson et al. 318/564 5497062 Digital pulse-width modulated system for providing current to a load 1996-03-05 Fenstermacher et al. 318/599 5270633 Ternary energy supply circuit with a high frequency rejection filter 1993-12-14 Dijkmans 318/810 4584566 Analog to digital converter 1986-04-22 Arcara 341/128 4450518 Control system for adjusting a physical quantity 1984-05-22 Klee 364/183 4109194 Digital feedback control utilizing accumulated reference count to regulate voltage output of switching regulator 1978-08-22 Miller 323/283 4042868 Stepper motor control apparatus 1977-08-16 Rhodes 318/615\n\nForeign References:\n DE3340150C2 1991-01-03 DE3313949A1 1984-10-18 JP6194585 May, 1986 JP6276930 April, 1987 JP63234878 September, 1988 CONTROLLER FOR PULSE WIDTH MODULATION INVERTER JPS63234878A 1988-09-30 JPS6194585A 1986-05-13 JPS6276930A 1987-04-09\nOther References:\nTeodorescu, Dan: Die Motorsteuerung von Aufzugsanlagen; In: DE Der Elektromeister, May 1993; H.5, pp. 315-318, (no translation).\nTietze, U. Schenk: Halbleiter-Schaltungstechnik, 4.erw.Aufl., 1978, Springer-Verlag, Berlin, u.a., pp. 641-642, (no translation).\nPrimary Examiner:\nRILEY, SHAWN\nAttorney, Agent or Firm:\nPATRICK G BURNS (CHICAGO, IL, US)\nClaims:\nWhat is claimed is:\n\n1. An output voltage control apparatus for performing feedback control on a control object, comprising:\n\npulse density modulation type converter means for converting an output voltage command value for the control object into a pulse density;\n\nΔ-Σ modulation type converter means for converting an output voltage detection value of the control object into a pulse density;\n\nnon-matching detection means for detecting a non-matching in output between said pulse density modulation type converter means and said Δ-Σ modulation type converter means;\n\npolarity detection means for detecting a polarity of non-matching of outputs from said pulse density modulation type converter means and said Δ-Σ modulation type converter means;\n\nfirst counter means for one of up counting and down counting clock signals depending on the output of said non-matching detection means and the polarity detected by said polarity detection means;\n\nsecond counter means for one of up counting and down counting the clock signals depending on the outputs of said non-matching detection means and the polarity detected by said polarity detection means, and for setting a final counted value as an initial value of the next counting;\n\nfirst and second operation means for multiplying outputs of said first and second counter means by predetermined coefficients, respectively; and\n\nadder means for adding outputs from said first and second operation means.\n\n2. An output voltage control apparatus for performing feedback control on a control object, comprising:\n\npulse density modulation type converter means for converting an output voltage command value for the control object into a pulse density;\n\nΔ-Σ modulation type converter means for converting an output voltage detection value of the control object into a pulse density;\n\nnon-matching detection means for detecting a non-matching in output between said pulse density modulation type converter means and said Δ-Σ modulation type converter means;\n\ntime signal generation means for outputting a predetermined time signal depending on an output of said non-matching detection means;\n\nvoltage command generation means for outputting a voltage command signal based on an output of said pulse density modulation type converter means and an output of said non-matching detection means; and\n\nselection means for selecting and outputting one of the output of said pulse density modulation type converter means and an output of said voltage command generation means based on an output of said time signal generation means.\n\n3. An output voltage control apparatus for performing feedback control on an output voltage of an inverter which drives an AC machine, comprising:\n\nfirst conversion means for converting a command value for an output voltage of the inverter into a pulse density;\n\nsecond conversion means for converting a detection value of an output voltage of the inverter into a pulse density;\n\ncontrol means for detecting a deviation in outputs between said first and second conversion means, and generating a control voltage for use in controlling the inverter based on the detected deviation; and\n\nresolution altering means for altering a resolution of at least one of said first and second converting means.\n\n4. An output voltage control apparatus for performing feedback control on an output voltage of an inverter which drives an AC machine, comprising:\n\nfirst conversion means for converting a command value for an output voltage of the inverter into a pulse density;\n\nsecond conversion means for converting a detection value of an output voltage of the inverter into a pulse density;\n\ncontrol means for detecting a deviation in outputs between said first and second conversion means, and generating a control voltage for use in controlling the inverter based on the detected deviation; and\n\npolarity detection means for detecting a polarity mismatch between the converted command value and detection value and outputting a polarity signal, wherein\n\nsaid control means provides the deviation as digital data based on the polarity signal, and generates the control voltage based on the obtained deviation.\n\n5. An output voltage control apparatus for performing feedback control on an output voltage of an inverter which drives an AC machine, comprising:\n\nfirst conversion means for converting a command value for an output voltage of the inverter into a pulse density;\n\nsecond conversion means for converting a detection value of an output voltage of the inverter into a pulse density;\n\ncontrol means for detecting a deviation in outputs between said first and second conversion means, and generating a control voltage for use in controlling the inverter based on the detected deviation; and\n\nsignal generation means for outputting a selection signal according to a time-length of non-matching between the converted command value and detection value, wherein\n\nsaid control means comprises:\n\ncommand value generation means for generating a voltage command value for the inverter based on the non-matching state between the converted command value and detection value; and\n\nselection means for selecting one of an output from said first conversion means and an output from said command value generation means based on the selection signal, and for outputting the selected output as the control voltage.\n\n6. A method of controlling an output voltage for performing feedback control on the output voltage of an inverter which drives an AC machine, comprising the steps of:\n\nconverting a command value for the output voltage of the inverter into a pulse density;\n\nconverting a detection value of the output voltage of the inverter into a pulse density;\n\ndetecting a deviation of the converted command value from the converted detection value; and\n\ngenerating a control voltage for use in controlling the inverter based on the detected deviation.\n\n7. The method of controlling an output voltage according to claim 6, further comprising the step of:\n\naltering resolution of a pulse density conversion of at least one of the command value and the detection value.\n\n8. The method of controlling an output voltage according to claim 6, further comprising the steps of:\n\ndetecting a polarity of a difference between the converted command value and the converted detection value; and\n\nproviding, as digital data, the deviation based on the detected polarity, and generating the control voltage based on the obtained deviation.\n\n9. The method of controlling an output voltage according to claim 6, further comprising the steps of:\n\ngenerating a selection signal according to a period of non-matching between the command value and the detection value;\n\ngenerating a voltage command value for the inverter according to a non-matching state between the command value and the detection value; and\n\nselecting one of the command value converted into a pulse density and the voltage command value based on the selection signal, and outputting the selected value as the control voltage.\n\nDescription:\n\nBACKGROUND OF THE INVENTION\n\n1. Field of the Invention\n\nThe present invention relates to an output voltage control apparatus for performing feedback control for the drive of, for example, an AC(alternating-current) machine, an induction motor, etc. through an electric power converting device including a voltage PWM (pulse width modulation) inverter.\n\n2. Description of the Related Art\n\nThe conventional voltage PWM inverter detects an output voltage, computes a deviation of the detected output voltage value from a voltage command value, and compensates (controls) the deviation using a proportional and integral (PI) controller, etc. The conventional compensating method is disclosed by, for example, Japanese Laid-open Patent Publications (Tokukaisho) No.61-94585 and No.63-234878.\n\nThis compensation method applies to both analog system and digital system. The analog system uses a linear IC (operational amplifier), etc., and the digital system uses a central processing unit (CPU) and a digital signal processor (DSP) capable of performing high-speed operations, and a high-resolution and quick-response A/D converter, etc. capable of converting the output of an output voltage detector into digital data so that control is performed in a software operation process.\n\nHowever, since the analog system requires a linear IC, etc. for each process, the control apparatus becomes costly, and the control becomes complicated. On the other hand, the digital system requires expensive components such as a CPU, a DSP, etc.\n\nSUMMARY OF THE INVENTION\n\nThe present invention aims at reducing the cost of the above described control apparatus. It also aims at realizing at a low cost a control apparatus which does not require complicated control operations.\n\nAn output voltage control apparatus according to the present invention performs feedback control on a control object, and includes a first conversion unit for converting an output voltage command value for the control object into digital data; a second conversion unit for converting an output voltage detection value of the control object into digital data; a detection unit for detecting the difference in digital output values between the first and second conversion units; and a control unit for outputting the digital data of the control voltage corresponding to the detected difference in digital output values.\n\nAnother output voltage control apparatus according to the present invention includes a pulse density modulation type converter (a first conversion unit) for converting an output voltage command value for the control object into a pulse density; a Δ-Σ modulation type converter (a second conversion unit) for converting an output voltage detection value of the control object into a pulse density; a non-matching detection unit for detecting non-matching between the outputs of the pulse density modulation type converter and the Δ-Σ modulation type converter; a polarity detection unit for detecting a polarity of non-matching between the outputs from the pulse density modulation type converter and the Δ-Σ modulation type converter; a first counter unit for counting up or down clock signals depending on the output from the non-matching detection unit and on the polarity detected by the polarity detection unit; a second counter unit for counting up or down clock signals depending on the output from the non-matching detection unit and on the polarity detected by the polarity detection unit, and for setting the resultant count value as the initial value of the next count; first and second operation units for multiplying the output values of the first and second counter units by respective predetermined coefficients; and an adding unit for adding the outputs from the first and second operation units.\n\nAnother output voltage control apparatus according to the present invention includes a pulse density modulation type converter (a first conversion unit) for converting an output voltage command value for the control object into a pulse density; a Δ-Σ modulation type converter (a second conversion unit) for converting an output voltage detection value of the control object into a pulse density; a non-matching detection unit for detecting non-matching between the outputs of the pulse density modulation type converter and the Δ-Σ modulation type converter; a time signal generation unit for outputting a predetermined time signal based on an output from the non-matching detection unit; a voltage command generation unit for outputting a voltage command signal based on the output from the pulse density modulation type converter and the output from the non-matching detection unit; and a selection unit for selecting and outputting one of the output from the pulse density modulation type converter and the output from the voltage command generation unit based on the output from the time signal generating unit.\n\nAnother output voltage control apparatus according to the present invention performs feedback control of the output voltage of an inverter which drives an AC machine, and includes a first conversion unit for converting the command value for the output voltage of the inverter into a pulse density; a second conversion unit for converting the detection value of the output voltage of the inverter into a pulse density; and a control unit (corresponding to all or a part of 4A, 4B, 5A, 5B, and 6 shown in FIGS. 1 and 2, or 3A and 10 shown in FIGS. 3 and 4) for detecting a deviation in outputs between the first and second conversion units, and for generating a control voltage to control the inverter based on the detected deviation.\n\nThe output voltage control apparatus can further comprise a resolution altering unit for altering the resolution of at least one of the first and second conversion units.\n\nThe output voltage control apparatus can further comprise a polarity detection unit for detecting a polarity of non-matching between the converted command value and the converted detection value and outputting a polarity signal. The control unit provides the deviation as digital data based on the polarity signal, and generates the control voltage based on the obtained deviation.\n\nThe output voltage control apparatus can further comprise a signal generation unit for outputting a selection signal according to the time length of non-matching between the converted command value and the converted detection value. The control unit can comprise a command value generation unit for generating a voltage command value for the inverter based on the non-matching state between the converted command value and the converted detection value; and a selection unit for selecting the output of one of the first conversion unit and the command value generating unit based on the selection signal and for outputting the selected output as the control voltage.\n\nA method of controlling an output voltage according to the present invention is to perform feedback control of the output voltage of an inverter which drives an AC machine, and includes the steps of converting the command value for the output voltage of the inverter into a pulse density; converting the detection value of the output voltage of the inverter into a pulse density; detecting a deviation of the converted command value from the converted detection value, and generating a control voltage for use in controlling the inverter based on the detected deviation.\n\nThis output voltage control method can further comprise the step of altering the resolution of the conversion to a pulse density of at least one of the command value and detection value.\n\nThis output voltage control method can further comprise the steps of detecting a polarity of a difference between the converted command value and the converted detection values; providing the deviation as digital data based on the detected polarity; and generating a control voltage based on the obtained deviation.\n\nThis output voltage control method can further comprise the steps of generating a selection signal according to a time period of non-matching state between the converted command value and the converted detection value; generating a voltage command value for the inverter based on the non-matching state; and selecting one of the command value converted into a pulse density and the voltage command value according to the selection signal, and then outputting the selected value as the control voltage.\n\nAccording to the present invention, a voltage command value is converted into a pulse density (digital data), and an output voltage detection value is also converted into a pulse density (digital data) using a Δ-Σ modulation converter, and an output voltage detection value is also converted into a pulse density (digital data) using, for example, a Δ-Σ modulation converter. Therefore, a controller can be designed as digital hardware. It is a combination of an analog system and a digital system, and can be realized as a low-cost device.\n\nBRIEF DESCRIPTION OF THE DRAWINGS\n\nFIG. 1 shows the configuration according to the first embodiment of the output voltage control apparatus according to the present invention;\n\nFIG. 2 shows the configuration of a variation of the first embodiment of the present invention;\n\nFIG. 3 shows the configuration according to the second embodiment of the output voltage control apparatus according to the present invention; and\n\nFIG. 4 shows the configuration of a variation of the second embodiment of the present invention.\n\nDESCRIPTION OF THE PREFERRED EMBODIMENT\n\nFIG. 1 shows the configuration according to the first embodiment of the output voltage control apparatus according to the present invention.\n\nAs shown in FIG. 1, the output voltage control apparatus comprises a pulse density modulation type converter (PDM) 1; a non-matching detection circuit (exclusive OR circuit: EOR circuit) 2; an AND circuit (polarity detection circuit) 3; up and down counters (UDC) 4A and 4B; shift (multiplication) circuits 5A and 5B; an adder 6; and a Δ-Σ modulation type converter 7. In FIG. 1, V* indicates a first voltage command value input to the control apparatus; V** indicates a second voltage command value output from the control apparatus toward an inverter, etc.; CLK indicates a clock signal; Tc indicates a carrier signal; Vist indicates a voltage detection value. The Δ-Σ modulation type converter 7 can be a commercially available sequential serial comparison type A/D converter, etc. The voltage detection value Vist is a detection value of the voltage output by the PWM inverter, etc. to drive an AC motor.\n\nThe voltage command value V* is input to the PDM 1 and converted into a pulse density. The voltage detection value Vist is input to the Δ-Σ modulation type converter 7 and is also converted into a pulse density. The non-matching detection circuit 2 obtains the deviation (or difference) between the output of the PDM 1 and the output from the Δ-Σ modulation type converter 7. The deviation and the output from the PDM 1 are input to the AND circuit 3. Since the AND circuit 3 outputs signals having a different polarity depending on the levels of the two input signals, it is determined whether or not the voltage command value is larger than the detection value (non-matching polarity detecting function).\n\nNext, the output from the AND circuit 3, the output from the non-matching detection circuit 2, and the carrier signal Tc are input to the UDCs 4A and 4B, and the deviation is obtained as digital data. The signal obtained by the UDC 4A functions as a proportional element for output control. Since the UDC 4B feeds its output back to its input terminal, the obtained signal functions as an integral element for output control (integral control function). If only a proportional element or an integral element is required, only the one corresponding UDC is required.\n\nThe output of the UDCs 4A and 4B are respectively input to the shift circuits 5A and 5B, respectively assigned the amounts of shift signals (shifting amounts) SF1 and SF2. Therefore, a shifting operation such as a multiplication by 2n is performed depending the amount of the shift signal. Furthermore, the result of the shift operations are added by the adder 6 and provided to, for example, a pulse width modulation (PWM) circuit (not shown in FIG. 1) as a voltage command value (second voltage command value) to generate a gate signal for a switching element of an electric power converter including an inverter. According to the gate signal, the AC motor is controlled to perform a predetermined operation based on the voltage command value.\n\nFIG. 2 shows a variation of the first embodiment.\n\nThis variation is obtained by adding, as shown in FIG. 2, dividers (frequency dividers) 8A and 8B for dividing the clock signal CLK of the first embodiment shown in FIG. 1. The detailed description of operation is omitted here because the other units are the same as those described with respect to FIG. 1. Since using the dividers 8A and 8B avoids the necessity for quick-response circuits for the PDM 1 and Δ-Σ modulation type converter 7, a lower cost control apparatus can be successfully realized. Although two dividers 8A and 8B are provided in this example, just one divider can be used for either one of the PDM 1 and the Δ-Σ modulation type converter 7.\n\nFIG. 3 shows the configuration of the output voltage control apparatus according to the second embodiment of the present invention.\n\nAs shown in FIG. 3, the output voltage control apparatus according to the second embodiment comprises the PDM 1; the non-matching detection circuit 2; a hold (flipflop, or latch) circuit 3A; the Δ-Σ modulation type converter 7; a mono-stable circuit 9; and a data selection circuit 10. Since the functions of the PDM 1, non-matching detection circuit 2, and Δ-Σ modulation type converter 7 are the same as those according to the first embodiment, the descriptions are omitted here. The signals V*, V**, CLK, and Vist are the same as those shown in FIG. 1.\n\nThe voltage command value V* is input to the PDM 1 and converted into a pulse density. The voltage detection value Vist is input to the Δ-Σ modulation type converter 7 and converted into a pulse density thereby. The non-matching detection circuit 2 obtains the deviation of the output of the PDM 1 from the output of the Δ-Σ modulation type converter 7. Up to this step, the operations are the same as those shown in FIG. 1.\n\nIn FIG. 3, the hold (latch) circuit 3A detects the polarity when the output voltage command value does not match the output voltage detection value, and generates the voltage command value as digital data. Using the hold circuit 3A, the above described proportional control function or integral control function can be realized. The output of the non-matching detection circuit 2 is input to the mono-stable circuit 9 from which a signal corresponding to the non-matching time period is output. The outputs of the mono-stable circuit 9, hold circuit 3A, and PDM 1 are input to the data selection circuit 10. The data selection circuit 10 comprises two AND circuits. The output of the hold circuit 3A and the output of the mono-stable circuit 9 are input to one of the AND circuits. The output of the PDM 1 and the inverse output of the mono-stable circuit 9 are input to the other AND circuit. Therefore, the data selection circuit 10 selects the output from the hold circuit 3A during the hold period in which the mono-stable circuit 9 operates, and selects the output from the PDM 1 in other periods. The output of the data selection circuit 10 is provided to, for example, a pulse width modulation (PWM) circuit (not shown in FIG. 3) as a voltage command value (second voltage command value), and output as a gate signal for the switching element of the electric power converter including an inverter as in the case shown in FIG. 1.\n\nFIG. 4 shows a variation of the second embodiment of the present invention.\n\nIn this variation, as in the example shown in FIG. 2, the dividers 8A and 8B for dividing the clock signal CLK are provided for the PDM 1 and the Δ-Σ modulation type converter 7 to avoid the necessity for quick-response circuits for the PDM 1 or Δ-Σ modulation type converter 7, thereby realizing a low-cost apparatus. Since the other units are the same as those shown in FIG. 3, further descriptions are omitted here. As described above by referring to FIG. 2, only one of the dividers 8A and 8B can be provided for either of the PDM 1 or the Δ-Σ modulation type converter 7.\n\nAccording to the present invention, the output voltage command value and the output voltage detection value are converted into digital data, and the control apparatus can be designed as a digital hardware configuration. Therefore, the control apparatus can be realized as a simple and low-cost device." ]

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[ "# 19.8 Fahrenheit to Celsius\n\nFahrenheit to Celsius Results:\n\n19.8 Fahrenheit [℉] = -6.7778 Celsius [℃]\n\nTemperature unit converter for you to convert 19.8 Fahrenheit to Celsius, quick answer for you 19.8 Fahrenheit is equal to how much Celsius? How much is 19.8 Fahrenheit converted to Celsius? Temperature 19.8 Fahrenheit is how many Celsius? 19.8 Fahrenheit is equal to 19.8 Celsius [19.8 ℉ = -6.7778 ℃], which is, 19.8 Fahrenheit converted to Celsius is 19.8 Fahrenheit = -6.7778 Celsius. You can also use this page to quickly convert units from other temperatures, for example, Celsius to Fahrenheit conversion. This page is located at https://units-conversion.com/temperature/fahrenheit-to-celsius/19.8/, feel free to bookmark or share the conversion results from 19.8 Fahrenheit to Celsius." ]

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[ "# grb 030328 : the burst before the burst\n\nof 17 /17\nElisabetta Maiorano IASF/INAF, Sezione di Bologna & Dip. Astronomia, Università di Bologna GRB 030328: the burst before the Burst\n\nAuthor: knoton\n\nPost on 22-Jan-2016\n\n36 views\n\nCategory:\n\n## Documents\n\nTags:\n\n• #### esospectroscopy vlt\n\nEmbed Size (px)\n\nDESCRIPTION\n\nGRB 030328 : the burst before the Burst. Elisabetta Maiorano IASF/INAF, Sezione di Bologna & Dip. Astronomia, Università di Bologna. Collaborators. Evert Rol Nicola Masetti Eliana Palazzi Elena Pian Paul Vreeswijk Javier Gorosabel Antonio de Ugarte Postigo - PowerPoint PPT Presentation\n\nTRANSCRIPT\n\n• GRB 030328: the burst beforethe Burst Elisabetta Maiorano IASF/INAF, Sezione di Bologna &Dip. Astronomia, Universit di Bologna\n\n• CollaboratorsEvert RolNicola Masetti Eliana PalazziElena PianPaul VreeswijkJavier GorosabelAntonio de Ugarte Postigo\n\nOn behalf of the GRACE collaboration\n\nGRB030328: overview\n\nPrompt event\n\nOptical photometry\n\nOptical spectroscopy\n\nOptical polarimetry\n\nDiscussion\n\nConclusions\n\n• GRB030328Trigger: March 28.473 UT 2003 (by HETE-2)\n\nCoordinates (J2000): RA = 12h 10m 51s.00 DEC = -09 21 05.00 error circle = 52\n\nOptical Transient: obs ~ 1 hr after the GRB, R ~ 18 (Peterson & Price 2003)\n\nX-ray Transient: obs ~ 15 hr after the GRB (exp=94 ks), Chandra (Butler et al. 2003)\n\nRedshift: z = 1.52 (Martini et al. 2003; Rol et al. 2003)\n\nHost galaxy: obs ~ 1 yr after the GRB, R ~ 24.4 (Gorosabel et al. 2005)\n\n• Prompt event Prompt event light curve\n\nDuration (30-400 keV): ~ 100 s (T90)\n\nTotal Fluence (30-400 keV): 3 x 10-5 ergs cm-2 Peak Flux (30-400 keV): 7.3 x 10-7 ergs cm-2 s-1\n\n(Villasenor et al. 2003; Butler et al. 2005)\n\n(7-40 keV) (7-80 keV)(30-400 keV)Hete-2\n\n• ObservationsBVRI Photometry (2.2m+WFI @ ESO)\n\nSpectroscopy (VLT+FORS1 @ ESO)\n\nPolarimetry (VLT+FORS1 @ ESO)\n\nX-ray Observations (ACIS-S @ CXO)\n\nHost galaxy Observations (2.2m+BUSCA @ CAHA)\n\n• WFI R-band imageThe Optical Transient is clearly detected\n\nMid exp time = 29.272 UT(0.79 d after the trigger)\n\nR = 21.17 0.04NEField size: 2x2 arcmin(2.2m+WFI @ ESO)\n\n• BVRI light curvesI/10RBx100Vx10(Maiorano et al. in preparation)\n\n• VLT+FORS1 SpectrumSpectrum acquired 0.59 days after the GRB and corrected for the Galactic extinction E(B-V) = 0.047Significant lines explained assuming two systems at redshiftsz = 1.5223 0.0006 (GRB redshift) and z = 1.295\n\n• Polarimetry P = (2.4 0.6) %\n\n= (170 7) degrees\n\n5 different cycles of polarizationmeasurements were obtained, the percentage of linearpolarization P and the polarization angle do not vary across the 5 cycles.\n\nWe thus summed the images on each angle to increase the S/N. t start = 0.13 dt end = 0.34 dV = 21.25 0.10(VLT+FORS1 @ ESO)\n\n• Host galaxy analysis(Gorosabel et al.)The restframe Spectral Energy Distribution of the host galaxy.The best-fit is obtained using a Starburst galaxy (green template).The red dashed line shows the fit obtained using a powerlaw with index = 1.25 0.54. Co-added C1+C2+C3+C4 image of the host galaxy obtained with 2.2m + BUSCA @ Calar Alto Telescope\n\n• Host galaxy analysis(Gorosabel et al.)The restframe Spectral Energy Distribution of the host galaxy from the UV band (left panel) to the Optical and Infrared bands (right panel).The best fit is obtained using a Starburst galaxy of age 15 Myr(green template). The red dashed line in the left panel indicates the fit obtained using a powerlaw with index = 1.25 0.54.\n\n• X-ray afterglow analysis (Butler et al.)0.5-5 keV spectrum is well fitted using an absorbed powerlaw with photon index = 2.0 0.2 NH = NH (Gal) = 1021 cm-2No X-ray emission linesThe X-ray afterglow decreases in brightness following a powerlawwith index x = 1.5 0.1 (ACIS-S @ CXO)\n\n• R-band light curve analysisabc = 1.070.05 = 1.50.11 = 0.80.12 = 1.60.2tb = 0.70.1 daysabc\n\n• Broadband spectrum (tsed = 0.78 d)opt = 0.5 0.1\n\nx = 1.0 0.2\n\nc = 1.7 x 10 16 Hz\n\nx-opt= 0.84 0.01\n\n• Closure Relations (1)(x=opt)x = 1.5 1.1\n\n1.5 opt Break\n\nNo break x = opt = 3(p1)/4 = 1.2 ism = (1-3p)/4 = 1.7 wind\n\n1 = 3(p-1)/4 = 1.22 = p = 2.61 = 3(p-1)/4+1/2 = 1.7 2 = 3(p-1)/4+1 = 2.2=p/2 =(p-1)/2\n\nc > x p=2.6spherejetc < o p=1.6 = (p+6)/4 = 1.9 < 1sphere\n\njet ismwind1 = 0.82 = 1.6\n\n• Closure Relations (2)(xopt)x = 1 = 3(p-1)/4 = 0.75 = (3p-1)/4 = 1.25\n\n1 = 3(p-1)/4 = 0.75 2 = 3(p-1)/4+3/4 = 1.5 1 = 3(p-1)/4+1/2 = 1.252 = 3(p-1)/4+1 = 1.75x=p/2spherejeto < c < x p = 2 opt = 0.5x = 1.5 c = 1.7 x 10 16 Hzopt = (p-1)/2 = 0.5o < c < xEiso~1.8x1053 erg jet=4.3E= 5x1050 ergismwindismwind\n\n• SummaryOptical and X-ray afterglow well detected\n\nRedshift: z = 1.522\n\nPolarization: P ~ 2.4%\n\nHost galaxy: Starburst\n\nPreferred regimes in fireball scenario: (1) isotropic expansion within a wind-shaped medium, p ~ 2.6 (2) jet collimated expansion within a homogeneous medium, p ~ 2\n\nmore data (if any) are welcome" ]

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[ "## Algebra 2 (1st Edition)\n\nWe know the following equation for permutations: $$nP_r=\\frac{n!}{(n-r)!}$$ Thus, we find: $$nP_r=\\frac{n!}{(n-0)!}=1$$ This makes sense because when $r$ is 0, there are 0 objects from the group, so there is only one way to organize them." ]

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[ "Neurotic Pumpkin Murderer\n\n# npm\n\n## constant-time-js\n\n0.4.0 • Public • Published\n\n# Constant-Time JavaScript\n\nConstant-time algorithms written in TypeScript.\n\nImportant: This Github repository is the companion to Soatok's Guide to Side-Channel Attacks. Do not use this in production, especially if you don't have the budget for a cryptography audit.", null, "## Installing and Usage\n\nSimply add `constant-time-js` to your dependencies section. One way to do this is with `npm`:\n\n``````npm install --save constant-time-js\n``````\n\nNext, you can import the modules you need.\n\nFor JavaScript users:\n\n`const { compare, bignum } = require('constant-time-js');`\n\nTor TypeScript users:\n\n`import { compare, bignum } from 'constant-time-js';`\n\nPlease refer to the documentation below for what each function/class does.\n\n## Documentation\n\nThis is just a quick outline of what each function does.\n\n• `compare(a, b)` - Compare two `Uint8Array` objects.\n• Explanation\n• Returns `-1` if `a < b`\n• Returns `1` if `a > b`\n• Returns `0` if `a === b`\n• Throws an `Error` if `a.length !== b.length`\n• `compare_ints(a, b)` - Compare two integers.\n• Explanation\n• Returns `-1` if `a < b`\n• Returns `1` if `a > b`\n• Returns `0` if `a === b`\n• `equals(a, b)` - Are these `Uint8Array` objects equal?\n• Explanation\n• Returns `true` if they are equal.\n• Returns `false` if they are not equal.\n• Throws an `Error` if `a.length !== b.length`\n• `hmac_equals(a, b)` - Are these `Uint8Array` objects equal? (Using HMAC to compare.)\n• Explanation\n• Returns `true` if they are equal.\n• Returns `false` if they are not equal.\n• Throws an `Error` if `a.length !== b.length`\n• `intdiv(N, D)` - Divide `N` into `D`, discarding remainder.\n• `modulo(N, D)` - Divide `N` into `D`, return the remainder.\n• `resize(buf, size)` - Return a resized `Uint8Array` object (to side-step memory access leakage)\n• `select(x, a, b)` - Read it as a ternary. If `x` is true, returns `a`. Otherwise, returns `b`.\n• Explanation\n• `x` must be a `boolean`\n• `a` must be a `Uint8Array`\n• `b` must be a `Uint8Array`\n• Throws an `Error` if `a.length !== b.length`\n• `select_ints(x, a, b)` - Read it as a ternary. If `x` is even, returns `a`. Otherwise, returns `b`. (You should pass `1` or `0` for `x`).\n• Explanation\n• `x` must be a `boolean`\n• `a` must be a `number`\n• `b` must be a `number`\n• `trim_zeroes_left(buf)`\n• Explanation\n• `buf` must be a `Uint8Array`\n• Returns a `Uint8Array`\n• `trim_zeroes_right(buf)`\n• Explanation\n• `buf` must be a `Uint8Array`\n• Returns a `Uint8Array`\n\n### BigNumber\n\nOur BigNumber implementation aims to be constant-time for the magnitude of the numbers (i.e. number of limbs or bytes, regardless of how many bits are significant).\n\nUnless otherwise stated, all of our APIs expect `Uint8Array` objects (`Buffer` extends from `Uint8Array` and should work too, but we return `Uint8Array` objects, not `Buffer` objects).\n\nUnless otherwise stated, all `Uint8Array` objects are big-endian byte order.\n\nUnless otherwise stated, all `Uint8Array` objects assume unsigned integer behavior.\n\nUnless otherwise stated, all of the `bignum` methods are immutable (meaning: they return a new `Uint8Array` object rather than mutating the input arrays).\n\nReturns `a + b`. Overflow is discarded.\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.add(a, b);```\n\n#### bignum.and()\n\nReturns `a & b` (bitwise AND).\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.and(a, b);```\n\n#### bignum.count_trailing_zero_bits()\n\nCounts the number of `0` bits beneath the most significant `1` bit.\n\nReturns a BigInt (the native JS type), since the number of bits may exceed 2^32 for an array that is less than 2^32 elements long.\n\n```/**\n* @var {Uint8Array} a\n*/\nconst c: bigint = bignum.count_trailing_zero_bits(a, b);```\n\n#### bignum.divide()\n\nCalculate `a / b`, discarding the remainder.\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.divide(a, b);```\n\n#### bignum.gcd()\n\nCalculate the Greatest Common Denominator of two integers.\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.gcd(a, b);```\n\n#### bignum.is_nonzero()\n\nReturns true if this number is not equal to zero?\n\n```/**\n* @var {Uint8Array} x\n*/\nconst check: boolean = bignum.is_nonzero(x);```\n\n#### bignum.lsb()\n\nReturns the least significant bit of a big number. (If `0`, this is a multiple of two.)\n\n```/**\n* @var {Uint8Array} x\n*/\nconst least: number = bignum.lsb(x);```\n\n#### bignum.lshift1()\n\nMutates the input array.\n\nLeft-shift by 1. This is used internally in some algorithms.\n\n```/**\n* @var {Uint8Array} a\n*/\nlshift1(a);\n// `a` is now double its previous value```\n\n#### bignum.modulo()\n\nCalculate `a mod b`.\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.modulo(a, b);```\n\n#### bignum.modInverse()\n\nCalculate the modular inverse of two integers.\n\nThrows if `gcd(a, b)` is not equal to `1`.\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nlet one_over_a: Uint8Array;\ntry {\none_over_a = bignum.modInverse(a, b);\n} catch (e) {\n// Handle exception when 1/a is not defined (mod b).\n}```\n\n#### bignum.modPow()\n\nModular exponentiation.\n\n```/**\n* @var {Uint8Array} base\n* @var {Uint8Array} exp\n* @var {Uint8Array} mod\n*/\nconst out: Uint8Array = bignum.modPow(base, exp, mod);```\n\n#### bignum.msb()\n\nReturns the most significant bit of a big number.\n\n```/**\n* @var {Uint8Array} x\n*/\nconst most: number = bignum.msb(x);```\n\n#### bignum.multiply()\n\nMultiply two big numbers, return the product.\n\nThe output size will be larger than the inputs.\n\n```/**\n* @var {Uint8Array} x\n* @var {Uint8Array} y\n*/\nconst z: Uint8Array = bignum.multiply(x, y);```\n\n#### bignum.normalize()\n\nResize an Uint8Array to the desired length.\n\nThe default behavior is to treat the number as signed (thereby filling in the left with 0xFF bytes if the most significant bit of the input Uint8Array is set).\n\nPass `true` to the optional third argument to always zero-fill this padding value.\n\n```/**\n* @var {Uint8Array} a\n* @var {number} len\n*/\nconst c: Uint8Array = bignum.normalize(a, len);```\n\n#### bignum.or()\n\nReturns `a | b` (bitwise OR).\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.or(a, b);```\n\n#### bignum.pow()\n\nExponentiation.\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} n\n*/\nconst c: Uint8Array = bignum.pow(a, n);```\n\n#### bignum.rshift1()\n\nMutates the input array.\n\nRight-shift by 1. This is used internally in some algorithms.\n\n```/**\n* @var {Uint8Array} a\n*/\nrshift1(a);\n// `a` is half double its previous value```\n\nThe default behavior is congruent to JavaScript's `>>` operator. For an unsigned right shift (`>>>`), pass `true` as the second argument:\n\n``````rshift1(a, true);\n``````\n\n#### bignum.shift_left()\n\nShift left by an arbitrary amount.\n\n```/**\n* @var {Uint8Array} x\n*/\nconst y: Uint8Array = bignum.shift_left(x, 3n);\n// y := 8 * x```\n\n#### bignum.shift_right()\n\nShift right by an arbitrary amount.\n\n```/**\n* @var {Uint8Array} x\n*/\nconst y: Uint8Array = bignum.shift_right(x, 3n);\n// y := x / 8```\n\n#### bignum.sub()\n\nReturns `a - b`. Use `msb()` to check if the output is negative.\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.sub(a, b);```\n\n#### bignum.xor()\n\nReturns `a ^ b` (bitwise XOR).\n\n```/**\n* @var {Uint8Array} a\n* @var {Uint8Array} b\n*/\nconst c: Uint8Array = bignum.xor(a, b);```\n\n## Limitations\n\n### Potentially Dangerous on 32-bit Applications\n\n32-bit v8 (and, presumably, a lot of other 32-bit implementations) do things wrong, and our implementation might be variable-time on it.\n\nSpecifically, the most significant bit of a 32-bit integer distinguishes values from pointers. As a result, accessing the highest bit of a 32-bit number in 32-bit JavaScript engines (such as v8) is potentially variable-time.\n\nTo mitigate this risk, the `int32` class was created which wraps operates on 16-bit limbs (wrapping `Uint16Array`).\n\n## Frequently Asked Questions\n\n### But Why Though?", null, "For science!\n\nThis is a proof-of-concept library, that will eventually implement all of the algorithms described in the accompanying blog post.\n\nThe main purpose of this library is to demonstrate the concepts in a programming language widely accessible outside of the cryptography orthodoxy (which today is largely C and sometimes Python, and hopefully soon Rust).\n\n### Can I use this in a project?\n\nHold off until v1.0.0 is tagged before you even think about relying on it for anything. APIs might break until then.\n\n### What's with the blue {fox, wolf}?\n\nMy fursona is a dhole, not a wolf.\n\n### You should remove your fursona from this so my manager might take it seriously.\n\nI don't owe you anything. I don't owe your manager anything.\n\nBesides, if anyone is bigoted against a predominantly LGBTQIA+ community, they're precisely the sort of person whose career I don't want to help.\n\nIn sum:", null, "## Keywords\n\n### Install\n\n`npm i constant-time-js`\n\n0\n\n0.4.0" ]

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[ "# Poly Proxy: v2", null, "v2\n• Returns the Mesh Vertex index number of the second Polygon Vertex.", null, "Return Type:\n\n• integer", null, "Sample Code", null, "############################# get mesh vertex index number for the 2nd polygon vertex\n\n# import bge\nimport bge\n\n# get the current controller\ncont = bge.logic.getCurrentController()\n\n# get object that owns this controller\nobj = cont.owner\n\n# Get a list of the mesh\nmeshList = obj.meshes\n\n# Get the first mesh on the object\nmesh = meshList\n\n# get the first polygon on the mesh\npoly = mesh.getPolygon(0)\n\n# get the mesh vertex index of the 2nd poly vertex\nvertIndex = poly.v2" ]

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[ "Departmental PhD Thesis Exam – James Lutley\n\nEveryone is welcome to attend.  Refreshments will be served in the Math Lounge before the exam.\n\nTuesday, June 13, 2017\n3:10 p.m.\nBA6183\n\nPhD Candidate:  James Lutley\nSupervisor:  Georges Elliott\nThesis title: The Structure of Diagonally Constructed ASH Algebras\n\n******\n\nAbstract:\n\nWe introduce a class of recursive subhomogeneous algebras which are constructed using a type of diagonal map similar to those previously defined for homogeneous algebras. We call these diagonal subhomogeneous (DSH) algebras.\nUsing homomorphisms that also exhibit a kind of diagonal structure, we study certain limits of DSH algebras. Our first result is that a simple limit of DSH algebras with diagonal maps has stable rank one.  As an application we show that whenever $X$ is a compact Hausdorff space and $\\sigma$ is a minimal homeomorphism thereof, the crossed product algebra $C^*(\\mathbb{Z},X,\\sigma)$ has stable rank one. We also define mean dimension in the context of these limits. Our second result is that mean dimension zero implies $\\mathcal{Z}$-stability for simple separable limits of DSH algebras with diagonal maps. We also show that the tensor product of any two simple separable limit algebras of this kind is $\\mathcal{Z}$-stable.\n\nA copy of the thesis can be found here: Lutley-thesis" ]

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[ "## How to find earnings per share growth rate\n\n8 Nov 2019 The earnings per share ratio is calculated with this formula: \"Earnings per Share ( EPS) = (Net Income - Preference Dividends) / Weighted  The prospective EPS growth rate is calculated as the percentage change in this year's earnings and the consensus forecast earnings for next year. Most Popular\n\nCommon approaches to forecasting shares and EPS when building a 3 Add this difference to the forecast for basic shares to calculate future diluted shares. price as prior period share price x (1+ current period consensus EPS growth rate) . earnings growth (percentage change of earnings per share. [EPS]). As negative EPS figures are reported at times the sim- ple and widely applied formula of  The formula for earnings per share, or EPS, is a company's net income expressed on Companies, many times, retain some of their earnings for future growth. The prospective EPS growth rate is calculated as the percentage change in a share's price and value, because the calculation of earnings per share shows  10 Jan 2019 Learn how to calculate the earings per share (EPS) of any stock in your Here is how much is this expressed in a percentage: and the earnings per share grow evenly and bring a steady yearly growth from the company.\n\n## 21 May 2018 Some investors will want to find 'growth' companies, usually smaller firms that are expected to grow quickly. This can see rapid share price growth\n\n24 Jul 2014 The backtest for EPS 5 year growth rate reveals an unusual patter. Stocks with the highest and lowest growth rates outperform. 20 Dec 2018 EPS growth (earnings per share growth) illustrates the growth of earnings per share over time. EPS growth rates help investors identify 14 May 2017 A company with a high earnings per share ratio is capable of generating a significant dividend for investors, or it may plow the funds back into its business for more growth; in either case, To calculate the ratio, subtract any dividend payments due to the Incremental internal rate of return · Direct labor cost. Many translated example sentences containing \"earnings per share growth\" earnings per share, show an average annual growth rate of greater than 29%. Therefore, to calculate diluted earnings per share, potential ordinary shares are [. ..]. 17 Jan 2019 Here's where earnings per share can get a little bit tricky. number of shares based on which percentage of the accounting period each total only mean that the company has spent a lot of money on growth in the past year. EPS also ignores inflation, the price of goods and services generally may be increasing, so this could be contributing to the good EPS figure, but this growth might  14 Sep 2002 The expected future growth in earnings per share (EPS) is an incredibly how to identify which companies will achieve high growth rates.\n\n### Divide the earnings per share by the current share price and multiply times 100 to convert to a percentage. If the earnings are a negative number, the earnings\n\n20 Dec 2018 EPS growth (earnings per share growth) illustrates the growth of earnings per share over time. EPS growth rates help investors identify 14 May 2017 A company with a high earnings per share ratio is capable of generating a significant dividend for investors, or it may plow the funds back into its business for more growth; in either case, To calculate the ratio, subtract any dividend payments due to the Incremental internal rate of return · Direct labor cost. Many translated example sentences containing \"earnings per share growth\" earnings per share, show an average annual growth rate of greater than 29%. Therefore, to calculate diluted earnings per share, potential ordinary shares are [. ..]. 17 Jan 2019 Here's where earnings per share can get a little bit tricky. number of shares based on which percentage of the accounting period each total only mean that the company has spent a lot of money on growth in the past year. EPS also ignores inflation, the price of goods and services generally may be increasing, so this could be contributing to the good EPS figure, but this growth might\n\n### The prospective EPS growth rate is calculated as the percentage change in a share's price and value, because the calculation of earnings per share shows\n\n3 May 2019 Establishing trends in EPS growth gives a better idea of how profitable a company has been in the past and may be in the future. A company with  14 Jul 2019 Earnings per share (EPS) is the portion of a company's profit allocated To calculate a company's EPS, the balance sheet and income statement determine the value of earnings and how investors feel about future growth. EPS growth rates help investors identify stocks that are increasing or decreasing in profitability. If a company has an EPS of \\$5.00 in 2008 and EPS of \\$6.00 in  Earnings per share growth is defined as the percentage change in normalised earnings per share over the previous 12 month period to the latest year end. Earnings Per Share - Growth Rate Calculator. EPS is one of the 'Big 5 Numbers' required to determine whether a company is a wonderful business. Calculate a company's annualized percentage growth of earnings per share to to compare with other companies with this online stock growth rate calculator. Earnings per share, or EPS, is a widely followed performance measure. The basic EPS calculation entails a reduction of income by the amount of preferred Another ratio is the “PEG” ratio that relates P/E to the earnings “growth” rate, with\n\n## The formula for earnings per share, or EPS, is a company's net income expressed on Companies, many times, retain some of their earnings for future growth.\n\n10 Jan 2019 Learn how to calculate the earings per share (EPS) of any stock in your Here is how much is this expressed in a percentage: and the earnings per share grow evenly and bring a steady yearly growth from the company. The main objective of this report is to find out the affects of EPS that reflects in the share price future cash flow and required rate of return affect the share price. 24 Jul 2014 The backtest for EPS 5 year growth rate reveals an unusual patter. Stocks with the highest and lowest growth rates outperform. 20 Dec 2018 EPS growth (earnings per share growth) illustrates the growth of earnings per share over time. EPS growth rates help investors identify 14 May 2017 A company with a high earnings per share ratio is capable of generating a significant dividend for investors, or it may plow the funds back into its business for more growth; in either case, To calculate the ratio, subtract any dividend payments due to the Incremental internal rate of return · Direct labor cost. Many translated example sentences containing \"earnings per share growth\" earnings per share, show an average annual growth rate of greater than 29%. Therefore, to calculate diluted earnings per share, potential ordinary shares are [. ..].\n\n8 Nov 2019 The earnings per share ratio is calculated with this formula: \"Earnings per Share ( EPS) = (Net Income - Preference Dividends) / Weighted  The prospective EPS growth rate is calculated as the percentage change in this year's earnings and the consensus forecast earnings for next year. Most Popular   6 Jun 2019 The term earnings per share (EPS) represents the portion of a company's In this particular case, the company's quarterly earnings per share (or EPS) would If you're going to spend money anyway, then why not get paid for it? Calculating Internal Rate of Return Using Excel or a Financial Calculator. Divide the earnings per share by the current share price and multiply times 100 to convert to a percentage. If the earnings are a negative number, the earnings  Common approaches to forecasting shares and EPS when building a 3 Add this difference to the forecast for basic shares to calculate future diluted shares. price as prior period share price x (1+ current period consensus EPS growth rate) . earnings growth (percentage change of earnings per share. [EPS]). As negative EPS figures are reported at times the sim- ple and widely applied formula of" ]

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[ "# 辅导案例-AMATH 301\n\nAMATH 301 - Spring 2020\nHomework #9\nDue on Sunday, June 7, 2020\nInstructions for submitting:\n• Problems 1-5, 8 should be submitted to MATLAB Grader. You have 3 attempts\nfor each problem.\n• Problems 6-7, 9-10 should be submitted to Gradescope. The solutions and the code\nused to get those solutions should be submitted as a single pdf. All code should be\nat the end of the file. You must select which page each problem is on when you\nLinear Pendulum\nConsider the linear second-order differential equation for the motion of a pendulum,\nθ¨ = − g\nL\nθ.\nHere θ (theta) is the angle of deflection of the pendulum from the vertical. g = 9.8\nis acceleration due to gravity, L = 1 is the length of the pendulum. This differential\nequation is a good approximation when θ is small.\nBy defining ω (omega) as the derivative ω = θ˙, we can convert this single second-order\nODE into a system of two first-order ODEs,\nθ˙ = ω\nω˙ = − g\nL\nθ.\nThis system of ODEs is linear and therefore can be written in matrix form, θ˙\nω˙\n =\n 0 1\n− g\nL\n0\nθ\nω\n .\nAlternatively, we can write it as\nx˙ = Ax\nwhere\nx =\nθ\nω\n , A =\n 0 1\n− g\nL\n0\n .\nWe will solve the system of differential equations over the time interval 0 ≤ t ≤ T for\nT = 10 with initial conditions θ(0) = 0, ω(0) = 0.5.\n(5 points) Problem 1: MATLAB Grader\nThe forward Euler formula for a system of linear differential equations is\nxk+1 = xk + ∆tAxk.\nImplement the forward Euler method with ∆t = 0.005 to solve the initial value problem\nfor the pendulum. Make a 2 × 1 column vector with the forward Euler solution at the\nfinal time T = 10.\nThe exact solution at time T = 10 is\nx(10) =\nθ(10)\nω(10)\n =\n0.5√Lg sin (√ gL10)\n0.5 cos\n(√\ng\nL\n10\n)\n .\nCalculate the exact solution. If the forward Euler solution at the final time is the vector\nx FE and the exact solution at the final time is the vector x exact, then you can calculate\nthe global error of the forward Euler solution by doing\nnorm(x exact-x FE).\nCalculate the global error and store it in the variable ans1.\n(10 points) Problem 2: MATLAB Grader\nThe backward Euler formula for a system of linear differential equations is\nxk+1 = xk + ∆tAxk+1.\nUsing some matrix algebra, we can write this as the following linear system\n(I−∆tA)xk+1 = xk.\nwhere I is the 2× 2 identity matrix. At each time step of backward Euler, we must solve\nthis linear system with the same matrix I −∆tA but a different right-hand side vector\nxk. This is a good candidate for LU decomposition.\n(a) Perform an LU decomposition on the matrix I − ∆tA with ∆t = 0.005 to obtain\nmatrices L, U, and P. Store the matrix L in the variable ans2.\n(b) Implement the backward Euler method with ∆t = 0.005 to solve the initial value\nproblem for the pendulum. At each time step, you should use the LU decomposition\nto solve the linear system. Calculate the global error and store it in the variable\nans3.\n(5 points) Problem 3: MATLAB Grader\n(a) Forward Euler will be stable for the pendulum problem if |λ| < 1 for all eigenvalues\nof I+ ∆tA. Find the eigenvalue of I+ ∆tA that is largest in magnitude (or absolute\nvalue) when ∆t = 0.005.\n(b) Backward Euler will be stable for the pendulum problem if |λ| < 1 for all eigenvalues\nof (I−∆tA)−1. Find the eigenvalue of (I−∆tA)−1 that is largest in magnitude (or\nabsolute value) when ∆t = 0.005. You may use the inv command to compute the\nmatrix inverse.\n(c) Create a 1×2 row vector named ans4 that has the magnitude of the eigenvalue from\npart (a) as the first component and the magnitude of the eigenvalue from part (b) as\nthe second component.\n(10 points) Problem 4: MATLAB Grader\nRecall that forward Euler is obtained by using a forward difference to approximate the\nderivative and backward Euler is obtained by using a backward difference to approximate\nthe derivative. The Leapfrog method is based on using a central difference to approximate\nthe derivative:\nxk+1 − xk−1\n2∆t\n= f(xk) ⇒ xk+1 = xk−1 + 2∆tf(xk).\nThis method is explicit like forward Euler because the formula we end up with only\ndepends on past solution values. A major difference between this method and the Euler\nmethods is that it is a multistep method – the iteration actually uses two past solution\nvalues: xk−1 and xk. This means that each time we generate a new iterate xk+1, we need\nto use the ‘current’ solution value xk as well as the one before it, xk−1.\nSince this method is a multistep method, we cannot start the iteration with just the\ninitial values because the formula to compute x1 is\nx1 = x−1 + 2∆tf(x0),\nwhich is a problem since x−1 isn’t defined. To get around this, we perform a single step\nof Forward Euler to calculate x1, and then switch over to Leapfrog to calculate x2 and\nthe remaining iterates.\nImplement the Leapfrog method with ∆t = 0.005 to solve the initial value problem for\nthe pendulum. Make a 2 × 1 column vector named ans5 with the Leapfrog solution at\nthe final time T = 10. Calculate the global error and store it in the variable ans6.\n(10 points) Problem 5: MATLAB Grader\nUse the MATLAB function ode45 to solve the initial value problem for the pendulum.\nMake a 1× 2 row vector named ans7 with the ode45 solution at the final time T = 10.\nCalculate the global error and store it in the variable ans8.\nHint: When you subtract two vectors to calculate the error, you must have either two\nrow vectors or two column vectors. You cannot have one of each.\nConstruct a phase portrait with θ on the horizontal axis and ω on the vertical axis. Use\nmeshgrid to generate a grid of points between −0.75 ≤ θ ≤ 0.75 and −0.75 ≤ ω ≤ 0.75\nwith 25 equally-spaced points in both directions. Use the quiver function to draw a grid\nof arrows with components (θ˙, ω˙), where θ˙ and ω˙ are given by the ODEs of the pendulum\nsystem. Include labels for the axes. Unlike the activity, you do not need to draw axes.\nYou should see something (somewhat) like the following:\n-0.6 -0.4 -0.2 0 0.2 0.4 0.6\n-0.6\n-0.4\n-0.2\n0\n0.2\n0.4\n0.6\nThe exact solution to the system of differential equations with initial conditions\nθ(0) = 0 and ω(0) = 0.5 is\nθ(t) = 0.5\n\nL\ng\nsin\n(√\ng\nL\nt\n)\nω(t) = 0.5 cos\n(√\ng\nL\nt\n)\nCreate a vector of the exact solution θ(t) for all times from t = 0 to t = 10 in steps of\n∆t = 0.005. Do the same for ω(t). Use these vectors to add a trajectory to the phase\nportrait. Make the color of the trajectory different from the color of the arrows.\nWe will now construct another phase portrait, but instead of drawing a trajectory for the\nexact solution we will draw trajectories using numerical solutions.\n(a) Begin exactly as in Problem 6. Use meshgrid and quiver to draw the same grid of\narrows. Then label the axes.\n(b) Implement the forward Euler method with ∆t = 0.005 to solve the initial value prob-\nlem for the pendulum. Create vectors that contain the forward Euler approximations\nof θ(t) and ω(t) at all time steps from t = 0 to t = 10. Use these vectors to add a\ntrajectory to the phase portrait.\n(c) Repeat part (b) for both the backward Euler and Leapfrog methods.\n(d) All trajectories should be different colors. Include a legend so that the methods can\nbe identified. In order to include the trajectories in your legend and not the quiver\narrows, you may want to look at the documentation for the legend command in the\nsection “Included Subset of Graphics Objects in Legend”.\nNonlinear Pendulum\nThe linear pendulum equation is based on the small angle approximation sin(θ) ≈ θ.\nWithout this approximation, the equation is more accurate and becomes\nθ¨ = − g\nL\nsin(θ).\nThis second-order ODE is nonlinear and so is more difficult to solve. In particular,\nimplicit methods like backward Euler are harder to implement for nonlinear ODEs.\nThis equation can also be converted to a system of differential equations in terms of θ and\nω = θ˙. We will solve the system of differential equations for the nonlinear pendulum over\nthe same time interval 0 ≤ t ≤ T for T = 10 with the same initial conditions θ(0) = 0,\nω(0) = 0.5.\n(10 points) Problem 8: MATLAB Grader\nUse the MATLAB function ode45 to solve the initial value problem for the nonlinear\npendulum. Make it so that ode45 outputs the solutions at equally spaced times from\nt = 0 to t = 10 in steps of ∆t = 0.005. Create a 2001 × 1 column vector named ans9\nthat contains the values of θ(t) at each time. Then create a column vector named ans10\nthat contains the values of ω(t).\nConstruct a phase portrait for the nonlinear pendulum. Use meshgrid to generate a grid\nof points. Use 25 equally-spaced points for θ between −2pi and 2pi. Use 20 equally-spaced\npoints between −8 and 8 for ω. Use the quiver function to draw a grid of arrows. The\nresulting picture should look similar to what you obtained in Problem 6 in the middle of\nthe plot, but different around the edges (this is due to the nonlinearity of the system).\nUse ode45 to solve the system at the same times as in Problem 8 and with the following\nvariety of initial conditions:\n• θ(0) = 0, ω(0) = 0.5\n• θ(0) = 2, ω(0) = 1\n• θ(0) = pi, ω(0) = −10−4\n• θ(0) = 2pi, ω(0) = −7\n• θ(0) = −2pi, ω(0) = 7\nAdd these solution trajectories to the phase portrait. These solution trajectories should\nflow with the arrows from quiver. Set the axes to display from −2pi < θ < 2pi and\n−8 < ω < 8. You do not need to add a legend because it is self-explanatory which\ntrajectory belongs to which initial condition.\nThese trajectories come in two varieties: those that repeatedly encircle the origin, and\nthose that do not. Describe the behavior of the pendulum for each of these two types of\ntrajectories. Remember that θ represents an angle so θ = 0 is the same as θ = 2pi and\nθ = pi is the same as θ = −pi.\nWhich of the following methods are explicit methods? Choose all that apply.\n(a) forward Euler\n(b) backward Euler\n(c) midpoint method (RK2)\n(d) fourth-order Runge-Kutta (RK4)", null, "" ]

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[ "# Open Collections\n\n## UBC Theses and Dissertations", null, "## UBC Theses and Dissertations\n\n### Finite element analysis of fluid induced fracture behaviour in oilsand Atukorala, Upul Dhananath 1983\n\nMedia\n831-UBC_1983_A7 A88.pdf [ 6.09MB ]\nJSON: 831-1.0062766.json\nJSON-LD: 831-1.0062766-ld.json\nRDF/XML (Pretty): 831-1.0062766-rdf.xml\nRDF/JSON: 831-1.0062766-rdf.json\nTurtle: 831-1.0062766-turtle.txt\nN-Triples: 831-1.0062766-rdf-ntriples.txt\nOriginal Record: 831-1.0062766-source.json\nFull Text\n831-1.0062766-fulltext.txt\nCitation\n831-1.0062766.ris\n\n#### Full Text\n\n`FINITE ELEMENT ANALYSIS OF FLUID INDUCED FRACTURE BEHAVIOUR IN OILSAND By UPUL DHANANATH ATUKORALA B.Sc. (Engineering), University of Peradeniya, Sri Lanka, 1981 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF APPLIED SCIENCE in THE FACULTY OF GRADUATE STUDIES Department of C i v i l Engineering We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA JULY 1983 © UPUL DHANANATH ATUKORALA, 1983 In presenting t h i s thesis i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the University of B r i t i s h Columbia, I agree that the Library s h a l l make i t f r e e l y a v a i l a b l e for reference and study. I further agree that permission for extensive copying of t h i s t h e s i s for scholarly purposes may be granted by the head of my department or by h i s or her representatives. I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my written permission. Department of C i v i l E n g i n e e r i n g The University of B r i t i s h Columbia 1956 Main Mall Vancouver, Canada V6T 1Y3 I5th J u l y 1983 DE-6 (3/81) ii ABSTRACT In-situ recovery of o i l from Oilsand deposits at depth, rely on fl u i d induced fracturing of Oilsand. A method of analysis for predicting the i n i t i a t i o n and propagation of fractures in Oilsand has been developed. The prediction of fracture i n i t i a t i o n involves a stress analysis of the domain. Analysis of fracture propagation requires a coupled stress and flow analysis'. In the method proposed herein, the stress and flow analyses are f i r s t considered as two separate analyses that are later coupled through volume compatibility. For the f l u i d flow analysis, the fractures are appro-ximated as parabolic in shape. Fracture propagation has been analyzed incrementally, the direction of propagation being perpendicular to the computed minor principal stress direction at the tip of the fracture. The results are in good agreement with both laboratory and f i e l d exper-ience. The results indicate that the presence of a propagating fracture causes significant changes in both the magnitude and orientation of the minor principal stress, at the tip of the fracture. As a result, a path of propagation that is significantly different from the i n i t i a l in-situ minor principal stress plane direction has been observed. The rate of pumping of f l u i d , i s an important factor in predicting the propagation behaviour of fractures in Oilsand. In fracture operations carried out at shallow depths, that involve high rates of pumping, the fractures must be dominently horizontal in orientation. This is a result of the increase in f l u i d carrying capacity of the horizontal fractures, observed with propagation. TABLE OF CONTENTS Abstract i i Table of Contents i i i List of Tables v List of Figures vi Acknowledgements v i i i CHAPTER-1 INTRODUCTION 1 .1 Introduction 1 1.2 The concept of fluid fracture 2 1.3 The interpretation of flu i d fracture data 3 1 .4 The scope 6 1.5 The organization of the thesis 7 CHAPTER-2 REVIEW OF PREVIOUS WORK 2.1 Introduction 9 2.2 Fracture initiation 9 2.3 Fracture propagation 2.3.1 Theories of propagation 17 2.3.2 Path of propagation 23 2.3.3 Fluid flow aspects 26 2.4 Ground displacement predictions 29 CHAPTER-3 METHOD OF ANALYSIS 3.1 Introduction 33 3.2 Criterion for fracture initiation 38 3.3 The fini t e element representation of a fracture 41 3.4 Criterion for fracture propagation 42 3.5 The proposed method of stress analysis 45 3.6 The proposed method of fl u i d flow analysis . 3.6.1 Introduction 50 3.6.2 Fluid flow formulations 56 CHAPTER-4 FINITE ELEMENT FORMULATIONS 4.1 Introduction 66 4.2 The axisymmetric formulation 4.2.1 The Constitutive matrix[D] 66 4.2.2 The Strain displacement matrix [B] 68 4.2.3 The Stiffness matrix [K] 70 4.3 The plane strain formulation 4.3.1 The Constitutive matrix [D] 72 4.3.2 The Strain displacement matrix [B] 74 4.3.3 The Stiffness matrix [K] 75 CHAPTER-5 RESULTS AND DISCUSSIONS 5.1 RUSULTS 5.1.1 Fracture initiation 5.1.1(a) A comparison with theoretical results 77 5.1.1(b) A comparison with f i e l d data 79 5.1.2 Fracture propagation 5.1.2(a) A comparison with laboratory observations.83 5.1.3 The results from the fluid flow analysis 89 5.1.4 A comparison of displacements with Sun's closed form solution 93 i v 5.2 DISCUSSIONS 5.2.1 Fracture initiation 101 5.2.2 Fracture propagation 102 5.2.3 The flu i d flow aspects 103 5.2.4 Shapes of the fractures 112 CHAPTER-6 SUMMARY AND CONCLUSIONS -. 116 CHAPTER-7 RECOMMENDATIONS FOR FURTHER RESEARCH 120 CHAPTER-8 REFERENCES 122 Appendix-A 126 Appendix-B 1 30 Appendix-C 142 Appendix-D 149 V LIST OF TABLES Table Page 2.1 The c r i t i c a l [p -po ]/av values for different soils 16 5.1 A comparison of initiation pressure/overburden pressure ratio 77 5.2 Data used in the analyses i 94 5.3 The t i p fluid pressure data with length 104 LIST OF FIGURES Figure Page 1.1 Idealized injection pressure rate histories 4 2.1 The different modes of fracturing 18 2.2 The cross section of the fracture considered for Sun's solution 30 3.1 The plane strain domain considered for vertical fractures 34 3.2 The axisymmetric domain considered for horizontal fractures 35 3.3 A macroscopic view of the region in the vicinity of the pressurized zone 39 3.4 The variation of the stiffness of a fracture segment(in the direction of the minor axis) with the pore f l u i d pressure 43 3.5 The f i n i t e element mesh in the vicinity of the pressurized zone 44 3.6 Mesh reorientation 47 3.7 The displacements of nodes of the fracture face 49 3.8 The parabolic approximation of the shape of a fracture 54 3.9 The approximated parabolic shape vs f i e l d observations of shape 55 3.10 The quasi-static fracture 58 3.11 A typical cross section of a fracture tip 63 3.12 The quasi-static fracture with a blunt tip 64 5.1 The injection histories for Oilsand at Coldlake area 81 5.2 The injection history for Oilsand at Athabasca region - 82 5.3 Laboratory fracture test conditions 85 5.4 The cross section of the slice-6 indicating a vertical fracture 85 5.5 The pressure flow rate variations for different fracture fluids 86 5.6 The predicted path of propagation for a horizontal fracture 88 5.7 The flow rate-length-propagation velocity variations 91 5.8 A comparison of the predicted and assumed pressure profiles 95 5.9 A comparison of the vertical ground displacements predicted with Sun's(43) closed form solution 96 5.10 A comparison of the horizontal ground displacements predicted with Sun's(43) closed form solution 97 5.11 The finite xelement mesh in the vicinity of a horizontal fracture 99 5.12 A comparison of the predicted fracture shape with Sun's closed form solution 100 5.13 Results of the flow analysis with identical B-L curves for both fractures 107 5.14 The assumed B-L curves 108 5.15 The velocity-length predictions for the B-L curves shown in fog(5.14) ....108 5.16 The variation of aperture with length 110 v i i 5.17 The variation of wellhead pressure with rate of flow and time 113 5.18 The shape of the fracture with uniform and non-uniform pressure profiles 114 ACKNOWLEDGEMENTS The author would like to express his gratitude to his supervisor Dr. Peter M. Byrne for his guidance, valuable suggestions and the encouragement throughout this research. The author is indebted to Dr. Donald L. Anderson for his continued interest and advice particularly in the developement of the Flow Model. The valuable information provided by Dr. 0. Hungr are appreciated. The author is thankful to Mr. Thomas Vernon for the suggestions and help in presenting this thesis. Finally, the Research Assistantship awarded by the Department of C i v i l Engineering, University of British Columbia during the years 1981-1983 is gratefully acknowledged. 1 CHAPTER 1 INTRODUCTION 1 .1 Introduction Fluid fracture is an important concept in in-situ recovery of o i l from Oilsand and has been considered in detail by a number of researchers including Byrne et al(6), Dusseault(16,17,18), Dusseault & Simmons(l9) and Settari & Raisbeck(37). Oilsand is comprised of a dense sand matrix with its pore spaces f i l l e d with bitumen, water and free or dissolved gasses(7). The presence of bitumen reduces the effective permeability of the material and the presence of dissolved gasses makes sampling and testing problems unique to this material(3). Oilsand, at zero stresses, possess negligible tensile strength(17). Hence, virtually no energy is expended in creating an open mode fracture in such a material. The material behaviour of Oilsand is complicated being non-linear, inelastic and stress level dependent(6). Fracture Mechanics analyses traditionally deal with materials of considerable tensile strength and the stress-strain behaviour is often idealized as linear-elastic. Thus, the applicability of fracture initiation and propagation c r i t e r i a based on Linear Elastic Fracture Mechanics(LEFM) theories, to the fluid fracture initiation and propagation in Oilsand, is questionable. Fluid induced fracture propagation is associated with the rapid transmission of flu i d inside the open fracture. Flow of fl u i d is accompanied by a drop in fl u i d pressure due to viscous dissipation of energy. The resulting fluid pressure distribution 2 is dependent on the size of the opening created and vice versa. Hence, the prediction of f l u i d fracture behaviour involves a coupled stress and fluid flow analysis. The purpose of this thesis is to develope a method for predicting initiation and propagation of fluid fracturing in Oilsand, where fracture propagation is a coupled problem of stress and f l u i d flow analyses. 1.2 The concept of fluid fracturing and i t s objectives The concept of generating fluid fractures in s o i l or rock by injecting fluids at high pressures and rates of flow, often referred to as \"Hydraulic Fracturing\" in the literature, has been recognized by the Petroleum Industry since 1947(30). In i t s simplest form f l u i d fracturing consists of sealing off a section of a wellbore and injecting fluid, at sufficiently high pressure .and rate, to overcome the in-situ stresses and the tensile strength of the formation until an open fracture is created at the wellbore. This fracture is then extended by continued injection of the fracturing flu i d . The prime objective of such a process is to either enhance the effective reservoir permeability or create paths for introducing steam or air for heating the viscous hydrocarbons. In addition to the in-situ o i l recovery processes, fl u i d fracturing has also been employed in underground waste disposal and geothermal energy recovery programs(26). 3 1.3 The interpretation of fluid induced fracture data. In general, the variation of injection pressure measured at the ground surface and the volume of fl u i d pumped in, with time, are recorded throughout the entire fracture operation. In open hole fluid fracture operations, sealing off a section of the wellbore is achieved by using a straddle packer configuration. This consists of two rubber sealing elements, at a set distance apart, which are inflated to isolate the zone of pressurization from the rest of the hole. Perforated steel casings which contain compression packers to permit sealing off a section are also being used. It is commonly assumed that the pressure loss due to flow of fluid and leakages inside the wellbore is balanced by the gain in pressure head with depth(37). Based on this assumption, the injection pressure measured at the ground surface is essentially the same pressure at the sealed off section, at some depth. In some . fracture operations ground displacements are also measured to aid in the estimation of the extent and the orientation of a created fracture. The variation of injection pressure with time provides a preliminary means of interpretation of the initiat i o n , propagation and the orientation of a fracture. Two such idealized curves obtained for br i t t l e rock and Oilsand are shown in fig(1.1). The i n i t i a l abrupt drop in the injection pressure concomitant with an increase in the rate of flow defines the Breakdown pressure(BDP) of the formation, or the fracture initiation pressure. It is believed that the pressure at which a Fig (1.1) Idealized injection pressure - rate histories for (a) brittle rock (b) oilsand 5 fl u i d f i l l e d fracture closes, under a decreasing state of pressure, nearly defines the minor principal total stress across the fracture faces. The injection pressure observed after an instantaneous shut down of pumps(i.e.zero flow rate), is defined as the Instantaneous Shut-in pressure(ISP). Under above conditions, the fluid pressure in the fracture just balances the stresses forcing the closure of the fracture and hence is a measure of the magnitude of the minor principal total stress. The difference in magnitude of the Breakdown pressure and the Instantaneous Shut-in pressure, under low rates of flow and with low viscosity fluids, is considered to be the tensile strength of the formation. Any pressure higher than the Instantaneous Shut-in pressure causes a fl u i d pressure greater than the stresses forcing the closure, and fracture propagation takes place. This is defined. as the Fracture Propagation pressure(FPP). The various pressures defined above are illustrated in f i g ( 1 . 1 ) together with the flow rate variations with time to provide a better understanding. It has been shown both by laboratory and f i e l d experiments that f l u i d induced fractures propagate in planes perpendicular to the in-situ minor principal stress direction(4,11,31,36). Other criteria(28,39) based on energy considerations have also been proposed . for the direction of propagation and will be discussed in Chapter 2. Assuming that fractures propagate in planes perpendicular to the in-situ minor principal stress direction, and considering 6 an idealized vertical-horizontal principal stress orientation, the fracture orientation can be identified by making a comparison ' of the Instantaneous Shut-in pressure with the overburden pressure corresponding to the depth of the fracture. Hence, i f , ISP £ oy fractures are horizontal i f , ISP < av fractures are vertical where, av is the overburden pressure corresponding to the depth of fracturing. In the absence of ISP data, FPP data are used for the above purpose. 1.4 The scope of the thesis The modelling of fluid induced fracture behaviour, as mentioned in Section 1.1, requires a coupled stress and fluid flow analysis. The analysis can be simplified, by f i r s t separating the fluid flow analysis from the stress analysis. The propagation of a fluid f i l l e d fracture is associated with flow of fluid to f i l l the newly created volume. Thus, at any given time, with appropriate assumptions regarding the leakage of fluid to the surrounding region, the volume of fluid pumped in is seen to be directly related to the volume of the cavity. By imposing this compatibility condition the stress and the flow analyses are coupled together. The stress analysis is performed using a non-linear, incremental elastic finite element program. Fracture propagation 7 is analysed incrementally, taking account of the changes in orientation of the fractures during propagation. The path of propagation of a fracture, in a given domain with given i n i t i a l stress conditions, is not known beforehand. For this reason, the analyses are performed with a mesh reorientation process that allow for the changes in orientation of fractures. The mesh reorientation process is carried out at the end of each load increment. For the fl u i d flow analysis the quasi-static fracture is assumed to be represented by a series of parallel plate segments. Between each segment the flow is described by parallel plate flow equations. The fl u i d flow analysis predicts the variation of velocity of propagation of the fracture with the rate of flow of fl u i d inside the fracture and the resulting fluid pressure distribution, at quasi-static(i.e. equilibrium) fracture lengths. 1.5 Organization of the thesis. The thesis consists of seven chapters each of which serves a selected purpose. In Chapter 2, a review of previous work is given with special consideration of the fracture c r i t e r i a , problem idealizations, numerical modelling techniques and the effects of flu i d flow on propagation behaviour. The proposed methods of Fracture Mechanics analysis and the Fluid flow analysis are presented in Chapter 3. 8 Chapter 4 summarizes the mathematical formulations used in the development of the finite element program. The predictions of the fin i t e element fracture model are compared to existing theoretical solutions and laboratory and f i e l d observations in Chapter 5. A summary of the work together with the major conclusions is presented in Chapter 6. Recommendations for further research work are outlined in Chapter 7. 9 CHAPTER 2 REVIEW OF PREVIOUS WORK 2.1 Introduction Since the introduction of the concept of fluid fracture in 1947, a considerable amount of research has been done on this subject. Classical fracture mechanics consider materials that are strong in tension. Rock is strong in tension whereas soil is not. As a result, a greater portion of the research work on flu i d induced fracture behaviour is appropriate to rock rather than s o i l . Nevertheless, i t is of interest to examine the basic fracture c r i t e r i a , problem idealizations and the numerical techniques that are employed in the analysis of fl u i d fracture in rock. Fluid fracture propagation modelled using coupled stress and f l u i d flow analyses are briefly reviewed. The change in orientation of fractures, which has lead to injection pressure time records that are d i f f i c u l t to interpret, are examined. Further, the theoretical models developed for predicting the ground displacements are also reviewed. 2.2 Fracture initiation An extensive analysis on fracture initiation pressures including theoretical formulations as well as experimental observations has been reported by Haimson and Fairhurst(20). Rock is idealized as a b r i t t l e , elastic, homogeneous, isotropic linear and porous material. The theoretical formulations are based on the analogy that exists between the elasticity of a porous material and thermoelasticity. Fluid flow through the 10 pores is assumed to take place according to Darcy's law. A distinction between the penetrating and non-penetrating fracture fluids has been incorporated into the theoretical formulations. In the case of penetrating fluids, in the vicinity of the wellbore wall, a pore fluid pressure that is equal in magnitude to the pressure at the wellbore face, p^, is assumed. For non-penetrating fluids, a pore fluid pressure equal to the i n i t i a l in-situ pore pressure, p 0, which is different from the f l u i d pressure at the wellbore face, is assumed. The effective normal stress (compressive taken positive), for the above two cases, is expressed as; for non-penetrating fluids; a'= a - po 2.1 for penetrating fluids; CT'= a - p b 2.2 where, a'= effective normal stress a = total normal stress If the tensile stress at rupture is denoted by -o|, fracture initiation occurs when; a'<c -at' 2.3 Therefore, i f aQ'\\$ a vertical fracture results and i f -o{ a horizontal fracture results. where, o'Q is the effective circumferential stress and oz' is the effective a x i a K i . e . vertical) stress in the vicinity of the wellbore face. 11 The c r i t i c a l fracture initiation pressures derived from the stress conditions for a cylindrical hole in an infinite medium, and for the fluid conditions described by equations 2.1 & 2.2 are as follows; (a) For vertical fracturing with penetrating fluids; pP = at + 3 a2 ~ ai -p0 2.4 [2 - a{(l-2*)/(1 -*v )} ] and with non-penetrating fluids; p\"P= at + 3 az - a, -p 0 2.5 Note that a l l symbols are defined after equation 2.8. (b) For horizontal fracturing away from the ends of the hole, with penetrating fluids; Pc = gt + P 3 ~Po 2.6 [\\-a{(\\-2v)/{\\-v)}] For non-penetrating fluids the pore fluid pressure everywhere is equal to the i n i t i a l in-situ pore fluid pressure. Thus, the application of a wellbore pressure does not influence the magnitude of the axial effective stress. As reported by Haimson & Fairhurst(20), fracture initiation for this case is possible only i f the wellbore wall is prefractured or notched. (c) For horizontal fracture initiation near the ends of a pressurized hole with solid packers and penetrating fluids; D V „ _ Pc = °\\ + \" 3 \"Po 2.7 [ l . 9 4 - o { ( l - 2 t v ) / ( l - r v ) } ] and with non-penetrating fluids; 12 c = \"Po 2.8 .94 where, Po = i n i t i a l pore fluid pressure = wellbore pressure c r i t i c a l initiation pressure with penetrating fluids . P C K = c r i t i c a l initiation pressure with non-penetrating fluids = maximum horizontal principal total stresses(comp +ve) a2 = minimum horizontal principal total stresses(comp +ve) o3 = vertical principal total stress(comp.+ve) = the fl u i d pressure p nP in equation 2.5, of a thick walled hollow cylinder under no lateral loads(i.e. fractured vertically) aj = the fluid pressure pcnP in equation 2.8, of a thick walled hollow cylinder under horizontal lateral loads and a negligible vertical load(i.e. fractured horizontally). a = Biot's constant for a poro-elastic material. This is defined as [1-material matrix compressibility/ material bulk compressibility]. v = Poisson's ratio The c r i t i c a l fracture initiation pressures obtained for penetrating fluids are reported to be lower than those obtained for non-penetrating fluids, in theoretical predictions as well as experimental observations. In wellbores sealed with rubber packers the initiated fractures are found to be vertical 13 irrespective of the stress conditions. Horizontal fracture initiation is reported to be possible only with rigid packers. The f l u i d pressure that acts on the packer faces results in an axial load which is transmitted to rock as a boundary load at the wellbore face. With rigid packers that are in contact with the wellbore wall, this axial load is transmitted to rock as a shear load. Rubber is a liquid-solid elastomer. Rubber packers, under the above axial load would apply a similar radial load to rock. The application of a radial load would have a negative effect on the stress concentrations in the vertical direction and the chances of horizontal fracture initiation would thus be minimized(20). The placement of a flexible rubber liner at the wellbore face prevents the penetration of f l u i d into the material. Fluid fracture experiments carried out on rock samples with and without rubber liners are reported by Schonfeldt & Fairhurst(36). The magnitude of the fluid pressure required for fracturing of the lined samples are about 100% higher than those for the unlined samples. Furthermore, the magnitude of the initiation pressure for the unlined samples increased with increasing rate of pressurization. Based on the above findings, the authors suggest an initiation mechanism that is strongly dependent on fl u i d intrusion into the medium. Fluid intrusion into rock through a multitude of fine cracks eventually leading to a macroscopic fracture on planes perpendicular to the minor principal stress direction is proposed. Bjerrum et al(5) report the results of the f i e l d 14 permeability tests(i.e.falling head) carried out in s i l t y sands found near the Dead Sea area in Israel. The results indicate a rapid increase in the in-situ coefficient of permeability once the water pressure exceed a certain c r i t i c a l value. These high in-situ coefficients of permeability were found to be inconsistent with the type of material found in the area. The above inconsistency lead Bjerrum et al(5) to believe that the apparent increase in permeability is due to hydraulic fracture taking place in s o i l . As a result, theoretical studies were undertaken to predict safe water pressures to be used in in-situ permeability tests. The theoretical study carried out by Bjerrum et al(5) is based on the effective stress approach, the pore water pressure varying logarithmetically with the radial distance from the wellbore wall. The c r i t i c a l water pressures causing fracture have been derived for a radial plane strain domain consisting of a linear elastic material. The f i e l d permeability test involves the installation(by driving) of a piezometer which alters the state of stress around i t , the magnitude of which is material dependent. Bjerrum et al(5) take account of these stress changes by introducing two material dependent coefficients a & ^ , in equations 2.9 & 2.10. Furthermore, when the piezometer is pressurized, s o i l could deform outwards and separate from i t . Such a condition is referred to as a 'Blow-off condition. Fracturing could take place prior to, or, after 'Blow-off. If fracturing takes place after 'Blow-off, the resulting changes encountered in the state of stress and the geometrical boundary conditions of the problem would significantly influence the c r i t i c a l pressure for 15 fracturing.' For this reason, the c r i t i c a l water pressure for hydraulic fracture is considered to be the lesser of the c r i t i c a l pressures for fracturing, computed prior to and after 'Blow-off. The c r i t i c a l excess water pressures for fracturing, for the various stress conditions are as follows; (a) for vertical fracturing(i.e. minor principal stress horizontal) prior to blow-off; [p \" PolA; = U\\-v)/v}[(\\-a)K0 + ot'/av'] 2.9 subjected to the condition {(l-»')/i>}[(1-a)Ko + o[/a^] < (1 + ^8 )K0 and for vertical fracturing after blow-off; [p \" PcJAj = (1-a)[(2+ (4 -a)K0 + o{ /aj ] 2.10 subjected to the condition K 0O+p) < (1-a) [ (2-a+ f, )K0 + o[/o^] (b) for horizontal fracturing(i.e. minor principal stress vertical); [p \" p 0 ] / * v =1 2.11 where, a,p = coefficients to account for stress changes due to driving of a piezometer, dependent on the so i l type. v = Poisson's ratio K0 = earth pressure coefficient(at rest) = effective overburden pressure (compressive positive) af' = effective tensile stress of so i l at rupture 16 p - p0= excess pressure for fracturing The c r i t i c a l excess pressures based on the above equations, for different s o i l types are presented in Table 2.1. The values presented in Table.2.1 are based on a Poisson's ratio of 1/3. The observed c r i t i c a l pressures from in-situ permeability tests are reported to be in the range of values predicted by the above equations for the respective s o i l types. «. & ^ in Table 2.1, correspond to the same in equations 2.9 & 2.10. Table 2.1 The c r i t i c a l [p - pc ]/<^ values for different soils Method of installation Soil type Ratio of horizontal/vertical principal effective stress, K 0-3 0-5 0-7 1-0 Ideal installation (no disturbance) —0, j8«0 All soils C4 b 0-7b b 09 1-0C High compressibility 0-4-0-5^ 0-6-0 8 ° o - s ^ - i o 0 1-0C Driven Medium compressibility 0-5-07a 0-8 - ! O c l « c l-Oc Low compressibility 0-7a-10c 1-0C 1-0° a before blow-off I b after blow-off j vertical fracture C horizontal fracture In analysing fluid induced fracture in Oilsand, Byrne et al(6) consider an initiation criterion in the light of Terzaghi's principle of effective stress. Accordingly, in a porous medium, fracture initiation occurs when the boundary fl u i d .pressure f i r s t exceeds the total normal stress on a potential fracture plane, which outcrops on the boundary, by more than the tensile strength of the formation. The analysis considers penetrating fracture fluids. Accordingly, the stress 17 conditions for fracture initiation are identical to those given by equations 2.2 & 2.3. In summary, i t can be concluded that the magnitude of the flui d induced fracture initiation pressure, in a porous material, is strongly dependent on the penetration of fluid into the medium, being significantly greater for non-penetrating media. When either high viscosity fluids or low viscosity fluids at high rates of pumping are used, causing a partial non-penetrating condition, significantly higher fracture initiation pressures are observed. The orientation of the fracture is dependent on the orientation of the minor principal stress direction. 2.3 Fracture propagation Classical fracture mechanics considers three fracture modes; a tension mode and two shear modes which are shown in fi g ( 2 . l ) . Depending on the loading conditions, a fracture could propagate as a single mode fracture or a mixed mode fracture. This behaviour results in single mode as well as mixed mode Fracture Mechanics theories. These theories have found extensive application in predicting the conditions under which fl u i d fracture propagations occur and their possible paths of propagation, in materials strong in tension. 2.3.1 Theories of propagation Fracture mechanics analyses assess the stability of a fracture, in single mode propagation, by comparing the stress 18 mode-1 mode-2 mode-3 Fig{ 2.1) The different modes of fracturing 19 intensity factors in the vicinity of the tip of a fracture with the c r i t i c a l stress intensity factor for the material. The stress intensity factors are a function of the fracture length and the distribution and magnitude of the boundary loads(34). If the stress intensity factor(for the mode) computed at the tip of the fracture is greater than or equal to the c r i t i c a l stress intensity factor for the material, the fracture propagates. If not, the fracture would not propagate. The methods of assessment of the stability of a fracture, in mixed mode propagation, are briefly examined in the following sections (a) to (c). (a) The maximum hoop tensile stress theory(28) This theory investigates the stress state near the ti p of a fracture in plane stress or plane strain conditions under mode-1 and mode-2 loading. Fracture extension of a b r i t t l e material under slowly applied plane loads are considered. Fracture propagation starts at the tip of the fracture in the plane perpendicular to the greatest tension when the hoop tensile stress reaches a c r i t i c a l material value. Accordingly, ae =j^p Cos(t9/2)[K1Cos2(t9/2)-1 .5 K 2Sin ( 6) ] + 0 [ r 2 , r 2 , etc ] 2.12 r r e = 7^7Cos(t9/2)[K1Sin(c9)+K2(3Cos(t9)-l)] + t 9 [ r 2 , r 2 , e t c 3 2.13 K! = a[ 7TL ] 2 2.14 K2 = r U L ] 2 20 where, oQ = the hoop tesile stress Tre= the shear stress 0 = fracture branch angle w.r.t the i n i t i a l orientation K,2 = mode-1 & 2 stress intensity factors r = the radial distance measured from the fracture t i p L = fracture half length The direction of propagation is obtained by setting rre =0 From equation 2.13, i t can be seen that for a pure mode-1 f racture (i .e. K2 = 0), the fracture branch angle, 0 = 0 (0=±7r is a t r i v i a l solution for the equations). This indicates propagation in the direction of the i n i t i a l orientation. For any combination of mode-1 & 2 (i.e. K;1 ,K2 ^ 0) or for pure mode-2 (i.e. K,=0), 65 is different from 0. Further, i t is seen that at the tip of the fracture(r=0), equations 2.12 and 2.13 are singular. In this case, the maximum strain energy release rate is considered as the governing parameter. Fractures propagate in a direction where the strain energy release rate is a maximum, when this maximum reaches a c r i t i c a l material value. (b) The maximum strain energy release rate theory(27) 4 9/% 1-0A [(i+3Cos 2e) + G(0) = E[3+Cos20] 1+0/7T 8Cos0Sin0 K, K2+(9-5Cos20) K 2] 2.15 K, = a[ 7rL] 2 2.16 21 K2 = T [ T T L ] 2 where, G(0) = the strain energy release rate 0 = propagation branch angle E = Young's modulus of the elastic medium K = mode-1 & 2 stress intensity factors L = fracture half length a = tensile normal stress T = shear stress Again, i t can be shown that for pure mode-1(i.e. K2=0), fracture propagation occurs without change of i n i t i a l orientation. The mathematical singularity observed in the previous theory, at the tip of the fracture, is not present in this theory. The strain energy density near the point of propagation is the governing parameter in this theory. Fracture extension occurs in the direction of minimum strain energy density, when the strain energy density factor reaches a c r i t i c a l material value. (c) The minimum strain energy density factor theory(39) S(0) = r [dW/dA] = a,,K* +2 a 1 2 K, K 2+a 2 2K 2 2.17 where, a n = [ (1+cos0) (X -cost?) ]/l6G a 1 2= sin0[2 cos0-(# -1)]/16G a 2 2= [(* +1)(1-cos0)+(1+cos0)(3 COS0-1)]/16G where, 22 S(t?) = strain energy density factor dW = the incremental strain energy dA = the area of the element r = radial distance from the fracture tip G = shear modulus ^ = (3-4f) for plane strain <& = (3-*>)/(\\+v) for plane stress where v is the Poisson's ratio K 1 2 = stress intensity factors for modes-1 & 2 As in the previous theories for the special case of pure mode-1, fracture propagation is predicted in the direction of the i n i t i a l orientation. The strain energy density dW/dA, given by equation 2.17 above, becomes unbounded at r=0. From the examination of the above theories of mixed mode fracture propagation it follows that; (1) Initiation of fracture propagation and the direction of propagation are controlled by the stresses and strains in the vicinity of the tip of the fracture as opposed to the far f i e l d conditions. (2) Fracture extension without change of orientation is possible only if the loading conditions prefer a pure mode-1 fracture. 23 2.3.2 Path of propagation The path of propagation of a flu i d induced fracture has been assumed to be perpendicular to the in-situ minor principal stress direction(23,25,26,36,37,46). Laboratory (4,5) and f i e l d (11,31,36) experimental data obtained from small scale f l u i d fracture tests justify such an assumption. Dusseault(17) proposes the consideration of mixed mode fractures for the analysis of fluid fracture propagation in Oilsand. Wiles and Roegiers(45), incorporate the c r i t i c a l strain energy release rate theory for predicting the paths of propagation of fractures in rock. Pollard and Holzhausen(34) consider a two dimensional linear elastic model, and using the method of successive approximation (discussed in section 2.4) predict horizontal as well as vertical fractures, initiated at shallow depths, propagating towards the ground surface. These predictions are for materials that are strong in tension and are based on the significant increase in the normalized mode-2 stress intensity factors(i.e. K2/Ko0= rJltL/ph J~TfL= r/p b where, T=shear stress, L=fracture length/2 and p^ =wellbore pressure) observed at the tip of the fracture. For horizontal fractures a significant increase in the normalized mode-2 stress intensity factors is observed when the fracture propagates to a distance greater than about the magnitude of the depth of fracturing below the ground surface. During fracturing, with high temperature fluids(i.e. steam), heat is transferred to the material in the vicinity of the created fracture. Therefore, an analysis of the heating 24 patterns of the region of fracturing provides information on the location of the fracture in the ground. Based on the heating patterns observed by Jenkins & Kirkpatrick(1978), Dusseault(17) concludes that vertical and horizontal fractures climb towards the ground surface during propagation. Holzhausen et al(24) report a f l u i d fracturing test carried out in Athabasca Oilsand with ground displacement measurements. The ground response indicates that steam injection into the Oilsand is not a continuous process but rather is characterized by numerous episodic events. During the events the wellhead pressures dropped, injection rates increased and ground surface rose. The i n i t i a l propagation pressure based on the criterion described in Section 1.3 corresponds to that for a vertical fracture but in contrast the ground surface displacements have a symmetrical dome shaped distribution indictive of a horizontal fracture. After two weeks of steam stimulation the propagation pressure increased to that required for a horizontal fracture. The thermal expansion of the medium leading to the closure of the i n i t i a l l y created vertical fracture and an increase in the horizontal state of stress has been suggested as reasons for such observations. Dusseault(17) concludes that in massive hydraulic fracturing schemes in Oilsand, where fracture initiation favours vertical fractures, horizontal fractures would eventually prevail. The proposed hypothesis for such a behaviour is seen to be different from that suggested by Holzhausen et al(24). In Dusseault's view, the i n i t i a l fractures created are vertical. He 25 suggests that the high fluid pressures acting on the vertical fracture faces subsequently lead to an increase in the horizontal state of stress in the vicin i t y of the fracture. Once the horizontal stresses are increased beyond the vertical overburden total pressure, a horizontal fracture is initiated at the wellbore face and the injection pressure now corresponds to the overburden pressure which is higher than the i n i t i a l horizontal stress. If the rate of pumping is high enough to exceed the f l u i d carrying capacity of a fracture, an increase in the wellbore pressure results. The result would be an increase in the stresses around the fracture. Therefore, the effects of rate of pumping of f l u i d are important. However, to Dusseault, the variations in the rate of pumping are less dominant for the stress changes. The thermal strains induced by heating the reservoir, in Dusseault's view, are negligible. Dusseault and Simmons(l9) justify the above hypothesis by carrying out a two dimensional f i n i t e element analysis. The analysis considers three vertical and parallel fractures and predicts the altered state of stress due to the excess fl u i d pressures acting on the fracture faces. In summary, i t can be seen that the theoretical predictions as well as f i e l d observations indicate that, at shallow depths, horizontal and vertical fractures propagate towards the ground surface. Results from small scale laboratory experiments indicate propagation perpendicular to the in-situ minor principal stress direction. Therefore, the extrapolation of results from laboratory experiments to large scale f i e l d operations, is questionable. Furthermore, the magnitudes of the 26 injection induced stresses are seen to have a significant influence in deciding the final orientation of the fractures. 2.3.3 Fluid flow aspects An integral part of propagation of fractures is the flow of fluid within them. The resulting aperture of the fracture and the pressure distribution due to flow of fluid inside the fracture are highly inter-dependent. The stress analysis can be simplified by specifying arbitrary distributions of pressure thereby ignoring the flow of fl u i d inside the fracture. Zoback and Pollard(46) report that many authors including Secor(l975), Pollard(1976) and HSU(1975) prescribe arbitrary pressure distributions in their two dimensional models. Furthermore, Abed976), Perkins(1961), Nordgren(1972) and Kern(1969) consider the coupled fluid-flow stress-analysis problem but assume uniform pressure in the entire fracture length to obtain simplified expressions for the extent and the width of the fracture. Hagoort et al(22) and Settari and Raisbeck(37), in their two dimensional linear elastic models, incorporate mode-1 fractures e l l i p t i c in shape. The former authors analyse fracture in rock whereas the latter in Oilsand. Both models assume single phase compressible flow of fl u i d with a uniform pressure distribution inside the fracture. The propagation has been coupled with the mass balance of fluid within the fracture. In effect, this means that the fracture propagation occurs only i f there is fl u i d available to f i l l up the fracture. The model 27 developed by Settariand Raisbeck(37), predicts fracture lengths of several miles for an injection period of a few days. Such predictions are reported to be rea l i s t i c for air or cold water injection fracturing. Zoback and Pollard(46) investigate the coupled fluid flow-stress analysis problem in detail using a two dimensional(plane strain) fracture model in an infinite medium of linear elastic, homogeneous and isotropic material. The predictions are based on the mode-1 stress intensity factors computed at the tips of the fractures. The analytical solution for uniform loading of a portion of the wall of a fracture in an infinite region is employed in the prediction of the stresses, stress intensity factors and the displacements. For this purpose, the non-uniform pressure distribution is approximated as a series of uniform pressure segments and superimposed to obtain the final stresses and displacements. The fluid flow and elasticity equations are solved iteratively to obtain a consistent fracture shape-pressure variation for a prescribed wellbore pressure, viscosity and fracture length. Steady, constant property flow of a Newtonian fl u i d is assumed. For given rate of flow of fluid and viscosity, the analysis predicts the variation of the mode-1 stress intensity factors computed at the tip of the fracture with fracture length. The stress intensity factors indicated a drop in magnitude with length up to a c r i t i c a l length. Beyond this c r i t i c a l length the stress intensity factors increased rapidly. For higher fl u i d viscosities, larger c r i t i c a l lengths are observed. Based on these results the authors conclude the existence of a c r i t i c a l fracture length of stable propagation. 28 This conclusion indicates that stable fracture growth can precede breakdown when high viscosity fluids are used for fracturing and hence can lead to erroneously low estimates of o, predicted by equation 2.5, in regions where vertical fractures occur. Wiles and Roegiers(45) consider a two dimensional (plane strain) linear elastic fracture model that is appropriate for rock, using a displacement discontinuity approach. The coupled f l u i d flow-stress analysis problem is solved iteratively to obtain a consistent fracture shape-pressure distribution. Non-linear stress-strain variations are assigned to the discontinuities. Hanson et al(23) consider a numerical model that simulate the process of flu i d fracturing, in porous media, including the porous fl u i d flow and pore pressure effects in the surrounding region due to f l u i d penetration. The model consists essentially of a solution (in fi n i t e difference form) of the two dimensional, time dependent porous flow equations which numerically overlays the static two dimensional description of the elastic continuum. The analysis considers mode-1 fractures that are e l l i p t i c a l in shape. The mode-1 stress intensity factors computed at the tip regions of the fracture indicate a reduction with time when the porous flow effects are accounted for. Hence, if a fracture becomes stationary for a period of time the depressurization due to leakage of fl u i d to the surrounding low pressure regions, would reduce i t s tendency to propagate further. 29 The above review highlights the fact that the true solution for the coupled fluid flow-stress analysis problem can be attempted by an iterative type stress and flow analysis. Such analyses have been restricted to two dimensions and linear elastic materials. Only mode-1 or open mode fractures allow the necessary transmission of flu i d inside the fracture. The viscosity of the fracture fluid and the leakage of flu i d to the surrounding region are seen to be important for the propagation of a fracture. The f i l l i n g of the fracture with fluid during propagation offers a volume compatibility condition through which the stress and flow analyses are linked together. 2.4 Ground displacement predictions The mathematical closed form solution developed by Sun(42) predicts the vertical and horizontal ground surface displacements due to a horizontal , uniformly pressurized and penny shaped fracture embedded in a semi-infinite elastic continuum. The original derivations for a fracture in an infinite continuum are later modified to account for the presence of a free surface. The formulations consider the fracture length to consist of two separate regions, an inner region and an outer or an edge region as illustrated in fig(2.2). The essential difference in the two types of regions is that the fracture fluid does not penetrate into the edge region and is present only in the inner region. In effect this defines two different lengths, the fluid pressurized length of 3 0 Fig (2.2) The cross-section of the fracture considered for Sun's solution 31 the fracture and the true fracture length, the latter being always the larger. Accordingly, for a fracture with a fl u i d pressurized length, L^, and a true fracture length, L; W = [/k Sin(0/2) - d Cos(6/2)/{hfk }] A 0 2.18 U = [{L+L/k\" Sin(6?/2)/A1 + {/)T L Sin(0/2)-d/T Cos(0/2) - L k Cos(0)/A 2]A o r d/L 2.19 where, the true length and the f l u i d pressurized length are related by, [1-(L^/L ) 2 ] 2 = (Ujj-ar.cD/Up 2.20 in which, j_ 1 k2 = [(r 2+d 2-L 2)/L 2+(2d/L) 2]« 6 = Cof 1[(r 2+d 2-L 2)/2 d L] A 0 = the maximum aperture= 8 (1 -v2 ) (U^-tf.djL/TrE A, = (d + L/k Cos 2(0/2)) 2 + ( L + L/k Sin(0/2)) 2 A 2 = ( d/k Cos(65/2) - L/k Sin(65/2) + L k Cos (65 ) ) 2 + ( L/k Cos(65/2) + d/k Sin(£5/2) + L k Sin ( t 5 ) ) 2 where, W = the vertical ground displacement U = the horizontal ground displacement d = the depth of the fracture from the ground surface L = the true fracture length Lp= the fluid pressurized fracture length ~X = the unit weight of the material Uj3= the flu i d pressure inside the fracture r = the radial distance. 32 E,v= the elastic parameters of the medium Pollard and Holzhausen(34), using the method of successive approximation, predict ground displacements for arbitrary fracture orientations and pressure distributions. The method of successive approximation is useful for solving problems in regions bounded by several contours ( for the present problem the free surface and the fracture boundary are considered as two contours). The analytical solutions for uniform loading of a planar surface of a semi-infinte region and for uniform loading of a portion of the fracture wall in an infinite medium are required. The two solutions are successively(iteratively) superimposed. Each superposition results in the correct boundary conditions on one of the contours but alters the boundary conditions on a l l other contours. The analysis is performed in a two dimensional(plane-strain), elastic, homogeneous and isotropic medium. The numerical model formulated by Hungr and Morgernstern(26) is related to both the Boundary Integral Equation Method and the Displacement Discontinuity Method. The approach ut i l i z e s the simple classical solutions for point loads in an elastic half-space. The fl u i d pressure is considerd as a series of point loads. The analysis is restricted to linear elastic, homogeneous and isotropic materials. The advantage of such an approach is that planar three dimensional penny shaped fractures of any arbitrary orientation subjected to arbitrary flu i d pressure variations can be analysed. 3 3 CHAPTER 3 METHOD OF ANALYSIS 3 . 1 Introduct ion The fl u i d induced fracture behaviour, in reality, is essentially a three dimensional problem. However, in a medium with an idealized vertical-horizontal principal stress orientation and for fractures initiating from the circular periphery of a vertical wellbore on planes perpendicular to its axis, the problem simplifies to permit an axisymmetric analysis. On the other hand, if fractures are to occur on planes parallel to the axis of the wellbore, a plane strain analysis can be performed on the assumption that a sufficient length of the wellbore is pressurized to permit such an analysis. The analyses carried out in this study have been restricted to the above two types of fractures. The infinite domain in which fractures occur in reality has been assumed to be represented by the finite domains shown in figs 3 . 1 & ' 3 . 2 . For the plane strain domain considered(i.e.for the vertical fracture analysis), the effect of infinity is incorporated by the inclusion of springs which represent the elastic stiffness of the medium from the outer boundary to in f i n i t y . Theoretically, i t can be shown that in a radial plane strain domain, vertical fracture initiation occurs simultaneously at diametrically opposite p o i n t s ( 1 l ) . Further, i t can be expected that the symmetry of the problem would lead to radial propagation of such a fracture. Hence, only half the domain needs to be considered. The lateral extent of the 0 0 f t . 4 0 5 5 7 o So ^ 10 o IIS 130 (£•0 l i l l l l l l l l | Kofe Fig(3-2) The axisymmetr ic domain cons ide red for ho r i zon t a l fractures 7> 36 axisymmetric domain(i.e. for the horizontal fracture analysis) is selected such that a fracture with a prescribed maximum length causes negligible changes in the state of stress of the elements in the extreme right boundary. The lower boundary represents the rockbed. The wellbore walls are assumed to be supported by smooth steel casings, the pressurized zone being perforated. The loss of pressure head due to flow of fluid through the perforations are ignored. The stress changes caused by the installation of these, casings are assumed to be negligible. Oilsand is basically plastic in shear and b r i t t l e only in tension. As discussed in Section 2.3.3, for fl u i d induced fracture propagation i t suffices to consider only an open mode fracture. Only an open mode fracture permits the rapid transmission of fluid along the fracture which is essential for the propagation of the fracture. The analysis of fluid fracture propagation requires a coupled stress and flow analysis. The above analysis can be simplified by f i r s t uncoupling the fl u i d flow analysis from the stress analysis. If the flu i d pressure distribution inside the fracture is known or assumed, the size and shape of the fracture that correspond to the known or assumed flu i d pressure distribution can be obtained from a finite element analysis. The fluid flow analysis is then coupled to the problem through volume compatibility based on the fracture shape predicted from the finite element analysis(i.e. from the stress analysis). The fractures are assumed to be symmetric about their major axes. 37 The fracture shape is approximated by a series of parallel plate segments and the flow at a point is assumed to be described by parallel plate flow equations. The fluid flow analysis in return predicts a pressure distribution which can be compared with the assumed pressure profile. The iterative type of analysis thus formulated, has been restricted to a single iteration in the present method of analysis. In a porous medium, i t is the effective stresses that govern the response of the material. The effective stress-strain relations for Oilsand are highly complex being non-linear, inelastic and stress level dependent. Further, the presence of dissolved gasses makes the estimation of effective stresses complicated. Because of the inherent d i f f i c u l t i e s associated with an effective stress approach a simpler approach based on total stresses has been taken for the present study. Linear elastic and linear elastic-perfectly plastic stress-strain relations are assigned to the unfractured and the fractured material, respectively. However, the non-linear incremental elastic f i n i t e element program developed in this thesis is capable of handling non-linear stress-strain variations. The simplification mentioned above is only the f i r s t step towards the understanding of the coupled stress-fluid flow analysis problem. The problem is further simplified by ignoring the presence of bedding features, pre-existing fractures, dilation behaviour of dense Oilsand and the thermal aspects of the problem. 38 3.2 Criterion for fracture initiation A fluid induced fracture is initiated . by sealing off a section of the wellbore, at the depth where a fracture is required, and pressurizing this sealed section with a flui d . The magnitude of the fracture initiation pressure, as concluded in Section 2.2, is highly dependent on the degree of penetration of the fracture fluid into the medium. A macroscopic view of the region in the vicinity of the wellbore face is shown in fig(3.3). The compressive normal effective stress, a', in the vicinity of the wellbore face, can be expressed as; a' = a - u 3.1 where, a is the total normal stress and u is the pore f l u i d pressure. For fully penetrating fluids; u = u p 3.1a where u^ is the fluid pressure at the wellbore face. For fully non-penetrating fluids; u = u' 3.1b where, u' is the pore f l u i d pressure in the material when subjected to the boundary f l u i d pressure,u p, under undrained conditions. For partially penetrating fluids; u' < u < U p 3.1c The state of stress of an element of s o i l , adjacent to the wellbore face, can be represented by three mutually perpendicular principal stresses, which are in general different in magnitude. If a', in equation 3.1, represents any one of these stresses, material separation occurs when, 39 boundary fluid pressure = pore fluid pressure=u - sand grains nominal boundary Fig(3-3) A macroscopic view of the region in the vicinity of the pressurized zone o 40 o' ^ -o\\ 3.2 where, -a\\ = effective tensile stress at rupture (comp. ,+ve). and a fracture is said to be initiated. For a given material, this f i r s t occurs when a' in equation 3.2 corresponds to the minor principal effective stress. Oilsand is a porous material exhibiting a porosity of approximately 30%. In such a medium, most of the fracture fluids of common use can be considered as fully penetrating fluids. Oilsand, at zero stresses, possess essentially zero tensile strength. Accordingly, u = u^and -a\\ - 0. Substituting for u and in equations 3.1 & 3.2, fracture initiation f i r s t occurs when; or, u b ~ < W > 0 3.3 where, a m i n = total minor principal stress (comp.+ve) Fracture propagation, which wi l l be discussed later in this chapter, requires the flow of fluid into the fracture. Therefore, the opening should outcrop on the flu i d boundary. From an effective stress point of view, fracture initiation occurs when the minor principal effective stress f i r s t becomes zero on a plane outcropping the fluid boundary. From a total stress point of view, fracture initiation occurs when the boundary fl u i d pressure f i r s t exceeds the minor principal total stress on a plane outcropping the fluid boundary. 41 3.3 The Finite Element representation of a fracture Physically, fractures can be visualized as thin line discontinuities and can be discretized into a series of short line segments. Prior to fracturing, the region enclosed by a segment offers a near infinite stiffness in the direction of its minor axis. This is a direct result of the negligible thickness of the segment in comparison to its length. Once fractured, the region enclosed by the same segment is f i l l e d with the fracture f l u i d . The stiffness contribution of the fracture f l u i d in comparison to the surrounding material is negligible. Hence, after fracturing the stiffness of the segment in the direction of the minor axis is nearly zero. The f i r s t step in the Finite Element Method of analysis is the discretization of the domain into f i n i t e elements with acceptable aspect ratios (i.e. length/width ratio). In such a process the presence of thin line discontinuities, such as fractures, generate a large number of elements. Such an approach becomes cumbersome in the analysis of fracture propagation since the path of propagation is not known beforehand. Furthermore, the generation of a large number of elements increases the computation time required for the analysis. For the above reason, a fracture segment is represented by a f a i r l y thick element with an aspect ratio < 2.5. The fracture initiation criterion, as discussed in Section 3.2, can be incorporated into the f i n i t e element analysis by specifying a stiffness (in the direction of the minor axis of a fracture susceptible element) pore fluid pressure variation as 42 shown in fig (3.4) . The fluid pressure in excess of the minor principal total stress of the fracture plane leads to additional deformations in the process of fracturing. This is quantified in the finite element analysis by converting the f l u i d pressure on the fracture faces into equivalent nodal forces, and treating them as additional nodal loads for the increment. The above process is easily carried out by modifying the nodal force vector obtained from the f i n i t e element formulations(refer to Chapter 4 for details). 3.4 Criterion for fracture propagation The propagation behaviour of a fracture, which is essentially a new initiation, has been analysed incrementally. Once a fracture is initiated, the domain is analysed with the new boundary fluid pressures. Thereafter, the propagation is analysed as a new initiation but with the current stresses and the current boundary fluid pressures. The prediction of fracture propagation is accomplished by defining a quantity called the the Fracture Potential. The Fracture Potential(F.P) for the element i+1, shown in fig (3.5), is defined as; [U l > r \" <W ] 3.4 the fluid pressure at the right boundary of element i the computed minor principal total stress in element i+1 F.P where, U j r a = miiA. I+I c a> E CD l/> 3 O c 9999 ( G,B) ^•000001( G ;B) prior to • f racturi ng after fracturing pore fluid pressure Fig(34) The variation of stiffness of a fracture segment (in the direction of the minor axis) with the pore fluid pressure i Ul J k FigC3.5) The f i n i t e element mesh in the vic i n i t y of the p r e s s u r i z e d zone 45 If F.P < 0 stable fracture - does not propagate If F.P ^ 0 unstable fracture - propagates When F.P ^ 0 an increment in fracture length occurs in the element which has the highest F.P, which by definition is the most fracture susceptible element. The direction of propagation is assumed to be perpendicular to the minor principal stress direction in the region of the fracture t i p . A mesh reorientation process is built into the program to trace the path of a fracture and to include the effects of previous fracturing on subsequent fracturing. The mesh reorientation process is useful only for the special case of axisymmetric fractures initiating in a domain where the in-situ minor principal stress direction is vertical. 3.5 The proposed method of stress analysis The analyses are performed using a version of the computer program NLSSIP (8), modified to incorporate axisymmetric formulations and mesh reorientation. The program uses isoparametric elements and a tangent modulus approach (see Chapter 4). The load is applied in increments starting from known i n i t i a l in-situ stress conditions. Each increment in fracture length is treated as a new load increment, and each increment is analysed twice. The stresses, strains and displacements are computed for moduli corresponding to the average stresses, at the beginning and end of an increment. The moduli corresponding to the stresses at the end of the increment are stored to be used in the beginning of the next increment of 46 loading. The stresses and strains are computed at the centre of an element. The method of stress analysis is described below. The f i r s t step in the stress analysis is to specify the in-situ state of stress for the domain under consideration. The boundary pressure in the region where a fracture is required is increased until i t exceeds the minor principal total stress of any one of the elements in the vicinity of the fl u i d boundary that leads to a fracture which outcrops on the f l u i d boundary. At this value of the boundary fluid pressure the orientation of the above element(i.e. the one with the lowest minor principal total stress) is changed to that corresponding to the direction of the current minor principal stress plane, as illustrated in fig 3.6(a). The moduli of the element are reduced to near zero values and the nodal loads that correspond to the pressure that is in excess of the minor principal total stress of the element are applied at the nodes of the element. The nodal loads are equivalent of the excess fluid pressure acting on the two fracture faces. The domain is then analysed for the displacements and the stresses and the strains of the elements are computed. Once fracture is initiated the next fracture susceptible element is obtained by computing the Fracture Potential of the elements in the vicinity of the tip of the fracture, as described earlier in Section 3.4. If the boundary flu i d pressure is high enough to cause fracture (i.e. F.P ^0), the second element is oriented in the direction of its current minor principal stress plane, as shown in fig 3.6(b). Thereafter, as for the f i r s t element, the moduli are reduced to near zero values and the excess pore fluid pressure is converted (a) min i < i 1 i 1 2 • 1 i 1 < i (b) 1 f x • -€ j ( > V 2 / i / t • nodes changed due to reorientation of element 1 o nodes changed due to reorientation of element 2 Fi g( 3.6) Mesh reorientation (a) for element 1 (b) for element 2 48 into equivalent nodal loads and added to the nodal load vector. The domain is analysed for the displacements and the new stresses and strains in the elements are computed. For subsequent elements the procedure is identical to the one described for the second element. This process is repeated until the fracture propagates to a prescribed length. As described above, fracturing of each element is considered as a separate load increment. Therefore, i f a fracture extends for 'n' number of elements there are 'n' number of load increments. The shape of the fracture is computed by summing the incremental nodal displacements, for each increment of loading, of the nodes that f a l l on the fracture boundary. A fracture extending to four elements together with the relevant nodes is shown in fig(3.7). The incremental nodal displacements mentioned above are computed after reducing the moduli of the elements 1,2,3 & 4 to near zero values and applying the equivalent nodal loads. These nodal loads correspond to a fluid pressure in excess of the current minor principal total stress, acting on the fracture faces. The nodal displacements of the two nodes that correspond to the tip of the fracture are assumed to be zero in shape calculations. The ground surface displacements are computed in a similar manner, the relevant nodes being those in the upper boundary of the domain in this case. The details of the method of analysis are presented in Appendix A in step form. 4 9 1' 2< -© . L /fracture face •—e—-1' 2' 1 2 3 U ^ ^ ^ \\ . !—fracture ti • initial location of nodes o final location of nodes note: d i sp l a cemen t s of node 5 restricted to zero in shape calculations Fig (3-7) The displace ments of nodes of the fracture face 50 3.6 The proposed method of Fluid flow analysis 3.6.1 Introduction The objectives of the fluid flow analysis, as described in Section 1.4, are to predict the variations of, (a) the rate of flow of fl u i d within the fracture, (b) the fluid pressure distribution within the fracture and (c) the velocity of propagation of the fracture with the length of fracture. In this analysis the wellbore pressure and the fl u i d viscosity, for both vertical and horizontal fracture analyses, are assumed to be constant and are specified. For a given wellbore pressure and fluid viscosity, the pressure distribution within a fracture of given length and the resulting fracture shape are highly inter-dependent. The solution for this coupled problem, as concluded in Section 2.3.3, can be obtained by an iterative type stress and flow analysis. Such an approach requires assumptions regarding the i n i t i a l shape of the fracture or the i n i t i a l pressure distribution inside the fracture. The design of fluid fracture operations are often based on prescribed f l u i d pressure distributions and are referred to as geometric pressure profiles. Wiles and Roegiers(45) compare the fluid pressure distribution predicted from a coupled fluid flow-stress analysis with the existing geometric pressure profiles (in their figures 4,6 & 8). This comparison corresponds to a viscosity that is representative of water. A l l the geometric pressure profiles except the one proposed by Daneshy(l2) are seen to be significantly different from the predicted 51 distribution. Daneshy's geometric fl u i d pressure profile proposed for a 2D case, is mathematically expressed as; i Px = P,, + t P b \" H M x / L f >\"]T 3.5 where, P b = the wellbore pressure p_ = the far f i e l d stress p x = the pressure at a distance x measured from the wellbore wall L_p = the fl u i d pressurized half length of the fracture n = an exponent 2,4 or 8 For no leakage of fluid, the fl u i d pressure distribution given by equation 3.5 with n=4 indicates the best f i t for the pressure distribution obtained from the coupled fluid flow-stress analysis. The far f i e l d stress, p^ , in equation 3.5, is assigned the in-situ minor principal stress which inturn is the tip fluid pressure. According to the fracture c r i t e r i a used in the fin i t e element fracture model, the t i p f l u i d pressure changes with the length of propagation. Furthermore, the fluid pressurized fracture length and the actual fracture length are considered to be the same. Hence, the pressure distribution given by equation 3.5, modified for the tip fluid pressure and length of fracture, provides an i n i t i a l pressure distribution that can be assumed for the iterative analysis. Accordingly, the following modifications are performed on 52 equation 3.5. (a) replace p^ by p ^ where, p^ . is the fluid pressure at the t i p of the fracture. (b) replace by L where L is the true fracture length as discussed in Section 2.4. P> - Rap + [Pb - P 4 i p][l-(x/L)\"]\"^ 3.6 For a given pressure distribution, the size of the opening is dependent on the boundary conditions of the domain in which the fracture is present, the location of the fracture with respect to the boundaries, and the material properties of the domain. The size and shape of a fracture that corresponds to the pressure distribution given, by equation 3.6, at an equilibrium length, is obtained by inputting the pressure distribution into the respective fractured domains.. For a given length(L) and a constant wellbore pressure(p b), a pressure profile leading to an equilibrium fracture is obtained by varying p such that the Fracture Potential in the next, fracture susceptible element records a value just below zero. This is a t r i a l and error procedure and is repeated for various equilibrium fracture lengths to obtain the variations in size and shape with length. The shapes obtained from the fin i t e element analysis after one iteration are observed to be approximately parabolic. If the shape is approximated by a parabola the f l u i d flow formulations can be simplified considerably. Such an approach, however, limits the iterative stress and flow analysis to a single 53 iteration. A comparison of the shape predicted for a vertical fracture, with the approximated parabolic shape, is shown in fig(3.8). The shape is seen to f i t well to the parabolic variation. It was found that the fracture shape observed for short horizontal fractures(those with depth/length ratio 3-4) does not correspond very well with the parabolic shape. At larger fracture lengths, the f i t is significantly better. Fluid induced fracture operations are usually carried out at great depth below ground surface. Hence, the visual examination of the created fractures is an impossibility. For this reason, Pollard(33) suggests an analogy between the fluid fractures and sheet intrusions caused by the intrusion of liquid magma into the earth, which could be visually examined at outcrops. The thickness measurements of vertical dikes as a function of length, as reported by Pollard(33), are shown in fig(3.9). It can be seen that the f i e l d data closely approximates a parabolic shape. In the f l u i d flow formulations described in Section 3.6.2, both horizontal and vertical fractures are described as parabolic in shape. The analysis is restricted to a single iteration. Hence, the size of the opening of the fracture and the t i p f l u i d pressure correspond to the results obtained from the i n i t i a l pressure distribution given by equation 3.6. The fracture shape is approximated by a series of parallel plates and hence the flow at a point is governed by the parallel plate flow equations. Steady state laminar flow of a constant property Newtonian flu i d between each parallel segment is assumed. The parabolic approximation finite element results 0 0-4 1 2 3 distance from the wellbore axis (ft) Fig(3-8 ) The parabolic approximation for the shape of a fracture Fig( 3.9) The approximated parabolic shape vs field observations of shape obtained from Pollard ( 33 ) 56 leakage of fluid into the surrounding medium is ignored. Also, the vertical fractures created are assumed to propagate at constant height. 3.6.2 The fl u i d flow formulations In this section, the mathematical formulations involved with the fl u i d flow model are described. As outlined in Section 3.1, the symmetrical fractures considered herein are either vertical or horizontal in orientation. A propagating fracture is analysed as a quasi-static fracture. The aperture of a fracture, is dependent on a number of factors. They are, (a) material properties of the domain (b) boundary conditions of the domain (c) location of the fracture w.r.t the boundaries of the domain (d) wellbore pressure, p b (e) fracture half length, L (f) distance from the wellbore wall to the location within the fracture, where the aperture is required(i.e. x) The two domains(see figs 3.1 & 3.2) analysed are assumed to consist of linear elastic and isotropic material with similar properties. Wellbore pressures that are equal in magnitude are considered for the analysis of the two fracture types. Therefore, in a given domain, the aperture, B(see f i g . 3.10), of a fracture depends only on (e) & (f) and can be expressed as B=B[x,L]. As outlined in the latter part of Section 3.6.1, the 57 shapes obtained for the vertical and horizontal fractures, from the finite element analyses, are approximated as parabolic in shape. Therefore, as shown in Appendix C, B[x,L] can be written as, B[x,L] = B 0 [ L ] B'[x/L] 3.7 where, B 0 [ L ] , represents the maximum half aperture of the fracture and B'[x/L] is a shape function which is quadratic in x/L, representing the parabolic variation used in this analysis. It should be noted that B 0 is also dependent on the factors (a) to (e) and that i t can be written as B 0 [ L ] only with the simplifications mentioned above. Consider a fracture exhibiting a length L at time T and a length L+8L at time T+5T , as shown in figure 3.10. The equation for volume balance across section A-A, at a distance x from the wellbore axis, can be written as; for a vertical fracture, 5V = J~B 0[L+5L]B' [x/L] dx- _f B 0 [ L ] B ' [x/L] dx 3.8a x * for a horizontal fracture, L + S L L SV = 2it J\"B 0[L+5L]B'[x/L]x dx-27r J B 0 [ L ] B' [ x/L ] x dx 3.8b * x where, 5V is the net volume of fluid flown across the section at A-A. As shown in Appendix C, equations 3.8 can be rearranged to take the form, 58 Fig(3.10) The quasi-static fracture 59 for a vertical fracture, 6V = 6L f,[L,x] 3.9a for a horizontal fracture, SV = SL f'jL,x] 3.9b Dividing equations 3.9 by ST and taking the limit as ST 0, we have; for a vertical fracture, cW dL f j L , x ] 3.10a 5T dT for a horizontal fracture, 2>V dL f', [L,x] 3.10b \\$T dT where, 3V volume flow rate across A-A 3T and dL fracture propagation velocity dT Therefore, for volume compatibility, the flow rate is linear in dL/dT. The flow of fluid between the fracture faces is accompanied by a drop in fluid pressure. Let 'p' represent the resulting pressure at a point in the parabolic fracture. Since the flow at a point within the fracture is assumed to be described by parallel plate flow equations, the gradient of the resulting pressure distribution can be written as; 60 for a vertical fracture, 3p 3V [-12 M/B 3] ax B T 3V f 2 [ B , M ] 3.11a aT for a horizontal fracture, ap av [-6 M/{TTB 3X}] 3x &T SV f'2[B,/x,x] 3.11b aT Li = absolute viscosity B = separation of the parallel plates x = distance from the wellbore axis From equations 3.11 i t can be seen [ap/dx], is linear in [ 5 V/aT]. Substituting for [aV/^T] and respect ively, for a vertical fracture, a p dL f 3 [ /x,x,L] 2»x dT for a horizontal fracture, 3>p dL f 3 [ i i , x , L ] 5x dT Integrating equations 3.12 with that the pressure gradient, from equations 3.10 & 3.7 3.12a 3.12b respect to x, the f l u i d pressure 61 distribution can be expressed as; for a vertical fracture, p(x,L,p b,ju)=P b+ d L [ A - f 0 ( x , L , M ) ] 3.13a dT for a vertical fracture, p(x,L,pb ,M) = p b + dL [A' - f'„ (x,L,M) ] 3.13b dT P b = wellbore pressure A = f,(r0,L,|x) and A' = fn(r0,L,u) are constants for a given fracture r 0 = wellbore radius From equations 3.13 i t can be seen that inorder to compute dL/dT, the following should be known. (?) . wellbore pressure (h) u, f l u i d viscosity (i) L, fracture half length (j) p , fluid pressure at a distance x(r0<x<L). Out of the above quantities, (g) & (h) are externally controlled and hence are knowns. In the following section, the procedure for estimating the quantities listed under (i) & (j) will be described. In Section 3.4, i t was outlined that fracture propagation is analysed incrementally. Within each increment a static stress analysis is carried out with the fl u i d pressure distribution given by equation 3.6. Criteria for stable fracture, as 62 described by equation 3.4, are imposed by varying the fluid pressure at the tip of the fracture until i t becomes approximately equal to the minor principal total stress at the t i p of the fracture. This involves a t r i a l and error procedure. The above process is repeated for each increment and the magnitude of the f l u i d pressure at the tip of the fracture at varying fracture lengths are obtained. From the above results and equations 3.13, dL/dT can be computed at the end of each increment in fracture length. From equations 3.10 and the dL/dT values computed at different fracture lengths, av/aT that correspond to the same fracture lengths can be computed. When x=L or X=x/L=1, equations 3.13 denote the fluid pressure at the tip of the fracture. But as shown in Appendix-C, f 4(x,L,/i) and f„(x,L,m) contain terms such as 1/(L 2-x 2), 1/(L2-x 2 ) 2 and log(L-x). Thus, for a non-trivial solution (i.e. dL/dT^O), the tip fluid pressure should be negative in f i n i t y . However, i f the fracture tip is defined differently, the above singularity can be eliminated. A typical cross section of a fracture tip is shown in fig(3.11). The material in the shaded area may be highly non-linear and visco-elastic-plastic and may not form a continuum(35). The classical fracture mechanics theories indicate stress singularities at the t i p of the fracture. However, in reality, these stresses will get redistributed and the shape of the tip may adjust i t s e l f to form a finite state of stress. The blunt fracture tip shape assumed in this study, to 63 Fig(3.11) A typical cross section of a fracture tip 64 FigC3.12) The quasi-static fracture wi th o blunt-1ip 65 eliminate the pressure singularity, is shown in fig(3.12). The formulations described above are modified to account for the new shape of the fracture. For the new shape, the tip of the fracture is defined at x = a 0L. The limits of integration in equations 3.8 would change to x - a0(L+5L) and x - a 0L. The resulting changes are as follows; Equations 3.8 5V = J B 0 [ L + 8 L ] B ' [ X / L ] dx- JB 0[L]B'[x/L] dx 3.14a SV = 27r / B 0 [L+5L]B' [ X / L ] X dx-27r JB0 [L]B' [x/L] x dx 3.14b x x Equations 3.9 6V = 6L f,[L,x,a 0] 3.15a SV = SL f,[L,x,a 0] 3.15b Equations 3.12 ^p dL f 3 [ ( i,x,L,a 0] 3.16a dx. dT dp dL £ 3[n,x,L,a 0] 3.16b Sx dT Equations 3.13 p = p, + d L [ A - f ( , ( x , a 0 , L , M ) ] 3.17a D dT p = p. + dL [ A ' - f ( x , a 0 ,L,M) ] 3.17b b dT where,the fracture tip is given by X = a0 and the fracture tip velocity is given by [d(a 0L)/dT]. Once dL/dT is known at any given length of a fracture, the pressure distribution within the fracture is given by equations 3.17. 66 CHAPTER 4 THE FINITE ELEMENT FORMULATIONS 4.1 Introduction For the two fracture types mentioned in Section 3.1, the analyses simplify to the following two classes; (a) Axisymmetric - horizontal fractures (b) Plane strain - vertical fractures The s o i l is modelled by isoparametric quadrilateral or triangular elements. The finite element formulations for the above two classes are outlined briefly in the following sections. A complete formulation is given in Appendix-B. 4.2 The Axisymmetric Formulation 4.2.1 The constitutive matrix [D] An axisymmetric problem is characterized by its symmetry with respect to both geometry and loading. Because of the symmetry, the stress components are independent of the angular (0) coordinate. Hence, a l l the derivatives with respect to 6 vanish. Accordingly, 6e 2 3 = 6e 3 2 = 0 4.1a 8e,3 = 5e3, = 0 4.1b v = 0 4.1c Subscripts 1,2 & 3 refer to r, z and 6 directions with corresponding displacements u,v and w, respectively. From the generalized Hooke's law of incremental elasticity, i t 67 follows that; 5a 2 3 = 5o3 2 = 0 4.2a 5a,3 = 6a 3 1 = 0 4.2b With the strains and stresses given in equations 4.1 & 4.2, the incremental stress vector {6a}, and the incremental strain vector {5e} reduce to; {5af = [ 6 a n 6a 22 § \" 3 3 Sa, j] 4.3 T {Se} = [ 6c,, Se 2 2 oe-33 6e 1 2 ] 4.4 where, T denotes the transpose of the matrix. The constitutive relations for such a case are given by; 6a, , \\-v V V 0 6e , , ha 22 E V \\-v V 0 § E 2 2 6a3 3 (1+y)(1-2?) V V \\-v 0 5e3 3 6a, 2 0 0 0 (1-2l0/2 8e , 2 where, E = tangent Young's modulus v = tangent Poisson's ratio Alternately the constitutive relations can be written in terms of the shear and bulk moduli as; 68 8a, , B' +G' B' -G' B' -G' 0 8e , , SO 2 2 B' -G' B' +G' B' -G' 0 5e2 2 8a3 3 B' -G' B' -G' B' +G' 0 8a, 2 0 0- 0 G' 5e , 2 where, G' = E/{2(^+^>)} and B' = 3B/{2(l+v)} in which E,B & v are the tangent Young's modulus, bulk modulus and the Poisson's ratio respectively. In matrix notation equation 4.6 is written as; {6a} =[ D ] { 8 e } 4.7 where, [D] = the constitutive matrix. 4.2.2 The strain displacement matrix [B] In compatible isoparametric elements both the geometry and the displacements are expressed by the same shape functions. The displacement u,v and the coordinates r,z at a point are approximated by; u N, 0 N2 0 N3 0 N, 0 {8} 4.8 V 0 N, 0 N2 0 N-3 0 N, r N, 0 N2 0 N3 0 Nfl 0 15'} 4.9 z 0 N, 0 N2 0 N3 0 where, {8} is the incremental nodal displacement vector g.iven by; {8} = [u, v, u 2 v 2 u 3 v 3 u, v a] and {6'} is the nodal coordinate vector for the element given 69 by; {6' }T = [r, z, r 2 z 2 r 3 z 3 r 8 z, ] and N, =(1-s)(1-t)/4 N2 =(1-s)(!+t)/4 N3 «(1+s)(1+t)/4 N4 =(1+S) d-t)/4 U j , V ; being the nodal displacements and rj ,Zj being the nodal coordinates of the element in r & z directions, respectively. s,t are the coordinates of a point in the local frame. The incremental strains for an axisymmetric problem, in terms of the displacements are expressed as; Ue-} = Se , , 6e 2 2 6e 3 3 6e , 2 au/dr dv/az u/r dv/dr + 3u/dz 4.10 substituting for u & v from equation 4.8 into equation 4.10, 3NX 3r~ 0 3N2 3r~ 0 3N3 3r 0 3N4 3r~ 0 u l v l 0 3~Z~ 0 3N2 3Z~ 0 3N3 3Z~ 0 3N4 3Z~ \"2 v2 r 0 N2 r 0 ! l r 0 ^4 r 0 \"3 v3 3NX 3Nj 3N2 3N2 3N3 3N3 3N4 3N4 u4 3Z 3r 3Z 3r 3Z 3r 3Z 3r v4 4-11 70 or, in matrix notation; {6e} = [B] {8 } 4.12 where [B] is called the strain displacement matrix and contains the f i r s t derivatives of N(with respect to r and z. Since the shape functions, N,', are functions of s & t, [B] is not a constant for the element. For this reason, i t is evaluated numerically at selected points within an element. The details of the selection of the Gauss quadrature for numerical integration are given in Appendix-B. 4.2.3 The stiffness matrix [K] The force displacement relationship which results in the formulation of the stiffness matrix for an element is based on the principle of virtual work. For a virtual nodal displacement vector {8e}, the resulting virtual strain increment vector, as given by equation 4.12 i s ; {Se} = [B]{5e} 4.13 The internal work done, W(int), is expressed as; W(int) = j, {5e} {So} dv 4.14a where, dv is the volume of the element. Substituting from equation 4.13, J{5e} [B] [D] [B] {8} dv 4.14b 71 The external work done, W(ext), by the external load vector {f} due to the virtual increment in displacements is written as; W(ext) = {Se} {f} 4.1 5 From the principle of virtual work, W(ext) = W(int) hence, 4. 16 (f) = J\"[B]T[D] [B] {5} dv {f} = [K] {6} 4.17 4.18 where [K] =J[B][D][B]dv, which is called the stiffness matrix V for the element. If a pore f l u i d pressure of magnitude 6U is present in the element, {6a} = {6a'} + 6U 4. 19 where {8a'} is the matrix containing the element effective stresses. Substitution of equation 4.19 in 4.14 gives; W(int) = J { oeHB]~[D' 1 [B]{5}dv v ° 8U dv 4.20 where, [D'] is a new constitutive matrix in terms of effective 72 stresses. From equations 4.15,4.16 & 4.20, {f} = [K'] {6} + 6U{k'} 4.21 {f'} = {f} -6U{k'} = [K']{6} 4.22 where [K' ] = J [BHD'] [B]dv {f} = the modified nodal load vector which includes the presence of 5U. - j | 1 |dv which is a matrix associated with {k'} = I [B] T v the pore fluid pressure. 4.3 Plane strain formulation 4.3.1 The constitutive matrix [D] Plane strain problems are characterized by the following conditions; (1) no deflection in z direction (2) f i r s t derivative of the deflections in x & y directions with respect to z are zero. Accordingly, 5e 3 3 = 6e 1 3 = 6e 23 =0 4.23a w = 0 4.23b Subscripts 1,2 & 3 refer to x,y & z directions respectively, with corresponding displacements u,v,w. From the generalized Hooke's law, i t follows that; 73 6aT 3 = 6a 3 1 =0 5a 2 3 - oa 3 2 =0 4.24a 4.24b where Str-is the incremental stress with a similar notation as before. With the strains and stresses given in equations 4.23 & 4.24, the incremental stress vector {6a}, and the incremental strain vector {5e} reduce to; {6a} = [5a,, 6a2 6a 1 2] {Sef = [6e,, 6e2 2 5e , 2 ] 4.25 4.26 as; 5a, , 5a2 2 6a, 2 E ( ( 1-2v) for a plane v 0 V \\-v 0 0 0 (1 - 5e , , 8£ 2 2 5e , 2 4.27 where E = tangent Young's modulus v = tangent Poisson's ratio Alternately the constitutive relations can be written in terms of the shear and bulk moduli as; {5a} = B' +G' B'-G' B'-G' B' +G' 0 0 4.28 {6} 74 where, G'= E/{2(1+j>)} and B'= 3B/{2(1+»v)} in which E,B & v are the tangent Young's modulus, bulk modulus and the Poisson's ratio respectively. In matrix notation equation 4.28 is written as; {6a} = [D]{6e} 4.29 where [D] is the Constitutive matrix. 4.3.2 The strain displacement matrix [B] For plane strain formulation, the geometry of the element(i.e. x,y) and displacements(i.e. u,v) are approximated as, x = [x1N1+x2N2+x3N3+xllNfl ] 4.30 y = [yiN1+y2N2+y3N3+y(tNfl ] u = r u 1 N 1 + u 2 N 2 + u 3 N 3 + u , N , ] 4.31 v = [v1N,+v2N2+v3N3+vl|Nl) ] where, Xj ,y- = nodal coordinates in x & y directions. Uj,vj = incremental nodal displacements in x & y directions. N i , 2 , 3 , « are identical to those given in eqn. 4.8. 75 T h e i n c r e m e n t a l s t r a i n v e c t o r {Be}, i n t e r m s o f d i s p l a c e m e n t s i s e x p r e s s e d a s ; {6e} = 3 u / d X 4.32 6 e 2 2 — 8e , 2 3u/ay + dv/dx a n d t h e r e s u l t i n g [B] m a t r i x c a n b e w r i t t e n a s , 3Nj 3N-, 3Y 0 3N2 3X~ 0 3N3 3X~ 0 3N4 3X~ 0 3Y~ 0 3N2 3~Y~ 0 3N3 3Y~ 0 3N4 3Y~ 3NX 3N2 3N2 3N3 3N 3 3N4 3N4 3X 3Y 3X 3Y 3X 3Y 3X u 3 v 3 u 4 4.33 4 . 3 . 3 T h e s t i f f n e s s m a t r i x [ K ] T h e f o r m u l a t i o n f o l l o w s e s s e n t i a l l y t h e same p r o c e d u r e a s o u t l i n e d i n S e c t i o n 4 . 2 . 3 . E q u a t i o n s 4.21 & 4.22 t a k e t h e f o r m ; { f } = [ K 1 ] {6}+ 6U{k'} { f } = { £ } - 5 U { k ' } = [ K ' J {6} 4.34 4.35 w h e r e [ K ' ] = 1 [ B H D ' ] [ B ] t d A Ck'} = [ [ B f d U t d A w h e r e , t = t h i c k n e s s o f t h e e l e m e n t i n t h e z d i r e c t i o n a n d dA = t h e a r e a o f t h e e l e m e n t S i n c e t h e s t r a i n d i s p l a c e m e n t m a t r i x i s 3 x 8 , t h e r e s u l t i n g 76 s t i f f n e s s m a t r i x w i l l be 8 x 8 . The d e t a i l s of the f o r m u l a t i o n of {f} f o r the p r e s s u r e l o a d s f o r both p l a n e s t r a i n and a x i s y m m e t r i c f o r m u l a t i o n s a r e g i v e n i n Appendix-B. The user manual and the f l o w c h a r t f o r the program a r e g i v e n i n Appendix-D. 77 CHAPTER 5 RESULTS & DISCUSSIONS 5.1 RESULTS 5.1.1 Fracture initiaton 5.1.1(a) A comparison with theoretical results The fracture initiation pressures predicted from the finite element analyses together with the theoretical fracture initiation pressures obtained from equations described in Section 2.2, are presented herein for comparison. Finite element analyses are performed for two different principal stress orientations. A state of stress which represents a minor principal stress that is horizontal in orientation(i.e. K0<1) and a state of stress which represents a minor principal stress that is vertical in orientation(i.e. K0>1) are considered. The domains considered are shown in figs(3.1) and (3.2), respectively. The results, expressed as ratios of the initiation pressure to the overburden pressure, are shown in Table 5.1. Table5.1 A c ompari son of initiation pressure/over burden pressure ratio Orientation of the f racture Results from Haimson& Fairhurst(20) Results from Bjerrum etal(5) Results from the finite element analyses away from the ends closer to the ends penetr. non-penetr penetr. non-penetr ver t ic al K = 1/2 0 0-67 (2-4) 1-00 (2-5 ) — — 0-7 0 (2-9 & 2-10) 0-50 horizontal 2-00 (2-6) — 0-6 9 (2-7) 1-06 (2-8) 1-00 (2-11) 1-0 0 note: the numbers in parentheses refer to equation numbers from which they are computed. 78 where, K0 = coefficient of lateral earth pressure. The theoretical pressures correspond to a Poisson's ratio of 1/3. The tensile strength of the material, has been assumed to be zero. For sands, the material matrix compressibility is negligible when compared to the material bulk compressibility. As a result, the Biot's constant(see Section 2.2) for sands is close to 1. Hence, in computing the c r i t i c a l pressures from equations 2.4,2.6 & 2.7, a Biot's constant of 1 has been assumed. Furthermore, the i n i t i a l in-situ pore fl u i d pressure has been assumed to be zero. For K0<1, the finite element analysis predicts the initiation of a vertical fracture, and for K0>1, a horizontal fracture. The magnitude of the initiation pressure is equal to the magnitude of the in-situ minor principal total stress that correspond to the location of the fracture in ground. For the vertical and horizontal fractures, the theoretical initiation pressures, computed for penetrating fluids, are dependent of the Poisson's ratio for the material. However, for noh-penetrating fluids, the above pressures are independent of the Poisson's ratio. It is important to note that the initiation pressures predicted in the f i n i t e element analyses, for both vertical and horizontal fractures, are also independent of the Poisson's ratio. The theoretical pressures computed for vertical fracture ini t i a t i o n , using penetrating fluids, are about 40% higher than the results obtained from the fin i t e element analysis. For horizontal fracture initiation away from the ends of the 79 wellbore and with penetrating fluids, the theoretical pressure is about 100% higher than the pressure predicted in the finite element analysis. For horizontal fracture initiation closer to the ends of the wellbore and with similar f l u i d conditions as above, the theoretical pressure is 30% lower than the finite element results. The horizontal fracture initiation pressure that corresponds to the equation given by Bjerrum et al(5) is identical to the pressure predicted in the f i n i t e element analysis. The theoretical fracture initiation pressures obtained with non-penetrating fluids are seen to be higher than those obtained with penetrating fluids. 5.1.1(b) A comparison with f i e l d data Data on actual fracture operations carried out in Oilsand, are extremely limited. Settari & Raisbeck(37) have presented results of the injection pressure, rate and time histories obtained from fracture operations in the Coldlake area in Alberta. The paper by Holzhausen et al(24) presents injection pressure-time-rate data on a fracture operation carried out in the Athabasca Oilsand region in Alberta. The fracture initiation pressures obtained from the above fracture operations are compared with the predictions of the finite element fracture model. Because of the different magnitudes of the stress conditions involved, the comparison is based on the ratio of the init i a t i o n pressure to the minor principal total stress across the fracture faces. 80 The injection histories reported in the paper by Settari & Raisbeck(37) are shown in f i g ( 5 . l ) . From these results, Dusseault(17) deduces a fracture in i t i a t i o n pressure ratio that is approximately one. The results presented by Holzhausen et al(24), shown in fig(5.2), indicate a fracture initiation pressure ratio of about 1.5, which is 50% higher than the estimated ratio. The f i n i t e element analyses predict a fracture initiation pressure ratio of one. The magnitude of the fracture in i t i a t i o n pressure for a cohesionless.material, as discussed in Section 2.2, is dependent on the degree of penetration of the f l u i d into the medium. The degree of penetration depends on the rate of pumping of f l u i d , the fluid viscosity and the porosity of the material. Thus, a proper comparison is feasible only i f the degree of penetration of the fracture fluid is similar for the two cases. For the aforementioned data, both the material and the injection fluids are almost the same whereas the pumping rates of the two studies differ. On the other hand, the material may possess some tensile strength which would result in an increased fracture initiation pressure. The finite element analyses performed assume zero tensile strength for the material and hence the initiation pressures predicted are a lower bound. It can be concluded that the magnitudes of the fracture initiation pressures predicted from the finite element analyses are only in partial agreement with the theoretical predictions as well as the limited f i e l d data. The differences between the finite element results and the field data could be due to the 82 Fig {5-2) The injection history for Oilsand at Athabasca area (adapted from Holzhausen (2 A )) 83 varying degrees of penetration of fracture fluid and the tensile strength of the formation that are not taken account of in the analyses. 5.1.2 Fracture propagation In this section, the fracture propagation results obtained from laboratory experiments are compared with the results from the f i n i t e element analyses. The f i n i t e element analyses are carried out assuming; (a) fully penetrating fracture fluids, and (b) a material of negligible tensile strength. 5.1.2(a) A comparison with laboratory observations Hydraulic fracture tests carried out on laboratory t r i a x i a l samples of a s i l t y clay are reported by Bjerrum and Anderson(4). A small piezometer (shown in fig 5.3) is installed in the centre of the sample and consolidated to a prescribed anisotropic state of stress that leaves a horizontal minor principal stress. The piezometer is then pressurized at a constant rate of flow of water mixed with fluorecene. The important observations made in these tests are; 1. The pressure in the piezometer increased to a c r i t i c a l value, remained constant for a short time, and started to decrease. In a short time the rubber membrane started bulging out and the pore pressure in the top of the specimen started to increase. These observations indicate that a fracture was formed a l l the way out to the rubber membrane. 84 2. The horizontal and vertical slices, when visually examined, did not indicate the presence of a fracture. When illuminated by ultraviolet light, the horizontal slices indicated large concentrations of fluorecene on a diametrical plane, as shown in fig (5.4). These observations indicate that the created fracture was vertical and upon removal of the excess pressure the fracture had closed completely. 3. The fracture closure pressure provides a reliable estimate of the minor principal total stress and is independent of the type of f l u i d used in fracturing. (both paraffin and water indicated identical closure pressures). These results are illustrated in f i g (5.5) as obtained from Bjerrum & Anderson(4). The finite element fracture model developed in this study has been employed to model the propagation behaviour observed in the hydraulic fracture tests in t r i a x i a l samples. For the above case, a horizontal section of the sample subjected to an identical vertical-horizontal stress ratio of 1/2 (i.e. K0=1/2) has been analysed. The analysis assumes plane strain conditions. The domain selected is shown in fig(3.1). The radial symmetry of the domain considered indicates equal probability for a fracture to initiate at any diametrically opposite pair of points. However, if the existence of a weaker plane is assumed, fracture initiation could take place in that plane. With this assumption, the finite element fracture model predicts fracture initiation on a diametrical plane. Because of the small dimensions of the specimen mentioned 85 soil sample C z = -s cr, Fig( 5-3) The laboratory f racture test condi tions (adapted from Bjer rum!- A nde rson^Aj) Fig(5-4) The cross-section of the slice-6 indicating a vertical f racture (adapted from Bjerrum <rAnrierson(4 )) — initial pore pressure •us 3 ' i . flow rate (cm / mm) Fig( 5-5 ) The pressure-flow rate variation for different fracture fluids (adapted from Bjerrum h Anderson(4 )) 87 in the above experiment, i t could be assumed that the viscous pressure drops would have been negligible. With such an assumption the finite element analysis predicts the vertical fracture propagating to the outer boundary of the domain. Experimental results reported by Haimson & Fairhurst(20) on small rock samples and Schonfeldt & Fairhursts(36) on large f i e l d rock samples, with a minor principal stress horizontal in orientation, indicate behaviour similar to that predicted by the analysis. Since the vertical fracture analysis is restricted to a plane strain analysis, predictions regarding the climbing behaviour of vertical fractures cannot be made. The propagation behaviour of a horizontal fracture has been modelled next, using the f i n i t e element fracture model developed. The domain shown in fig(3.2) has been analysed for a vertical to horizontal stress ratio of 2 (i.e.K 0=2). The analysis assumes axisymmetric conditions. Laboratory results obtained from samples subjected to a minor principal stress that is vertical, indicate horizontal fractures propagating to the outer boundaries. As shown in fig(5.6), at larger fracture lengths the predictions of the f i n i t e element analysis deviate from the above experimental observations indicating a fracture climbing towards the ground surface. This occurs once the fracture half length exceeds about the depth of the fracture from the ground surface. The presence of a free surface makes the geometry of the problem asymmetric about a horizontal axis. As a result, the stress distribution caused by the pressure inside a horizontal opening is also ground surface elevation = 00 for a uniform pressure distribution & a weightless fracture fluid U 6 distance from the wellbore axis(ft) 10 Fi g(56 ) The predicted path of propagation for a horizontal fracture 89 asymmetric about the axis of the opening. The significant shear stresses induced in the vicinity of the tip of the fracture, due to the cause mentioned above, reorient the minor principal stress plane towards the free surface. For this reason, the analysis predicts fractures climbing towards the free surface. The theoretical work of Pollard & Holzhausen(34) and Dusseault's(17) conclusions regarding the climbing behaviour of fractures support the above predictions. In conclusion, i t can be seen that the finite element fracture model satisfactorily predicts the fluid induced fracture propagation behaviour. Field observations as well as laboratory observations are modelled correctly. 5.1.3 The results from the fluid flow analysis The results presented herein are for a specified wellbore pressure and fluid viscosity. Domains shown in figs(3.l) and (3.2) are analysed for the same isotropic stress state with similar material properties. The mesh reorientation process has been bypassed. Therefore, only fractures propagating on vertical or horizontal planes are considered. For the above idealized conditions, the analysis predicts the rate of flow of f l u i d between the fracture faces, the velocity of propagation of the t i p of the fracture and the f l u i d pressure within the parabolic fracture at equilibrium fracture lengths. These predictions are based on the tip f l u i d pressure and the magnitude of the aperture predicted from the f i n i t e element analyses and correspond to the pressure distribution at the start of the 90 f i r s t i t e r a t i o n . The v e r t i c a l f r a c t u r e s a r e assumed t o propagate at c o n s t a n t h e i g h t . The maximum r a t e of f l o w of f l u i d , Q , a c r o s s the f r a c t u r e f a c e s , o c c u r s a t the w e l l b o r e f a c e . The v a r i a t i o n of Q w i t h l e n g t h f o r t h e two f r a c t u r e s ( i . e . v e r t i c a l & h o r i z o n t a l c a s e s ) ar e shown i n f i g ( 5 . 7 ) . The v a r i a t i o n of the v e l o c i t y of p r o p a g a t i o n w i t h l e n g t h , i s a l s o shown i n the same f i g u r e . The v a r i a t i o n s of the r a t e of f l o w w i t h l e n g t h o b t a i n e d f o r the two f r a c t u r e s i n d i c a t e t o t a l l y d i f f e r e n t t r e n d s . For the h o r i z o n t a l f r a c t u r e , Q n i n c r e a s e s w i t h L w i t h an i n c r e a s i n g g r a d i e n t dQ n/dL. For the v e r t i c a l f r a c t u r e , Q v i n c r e a s e s w i t h l e n g t h but l e s s r a p i d l y . Q v, i s d i r e c t l y p r o p o r t i o n a l t o the h e i g h t assumed f o r the v e r t i c a l f r a c t u r e and the r e s u l t s shown i n f i g 5.7 c o r r e s p o n d t o a h e i g h t e q u a l t o the l e n g t h of the p r e s s u r i z e d zone of the w e l l b o r e . The Q-L c u r v e s f o r the two f r a c t u r e s i n t e r s e c t each o t h e r , the f r a c t u r e l e n g t h c o r r e s p o n d i n g t o the p o i n t of i n t e r s e c t i o n b e i n g dependent on the h e i g h t assumed f o r the v e r t i c a l f r a c t u r e . P r i o r t o i n t e r s e c t i o n , the v e r t i c a l f r a c t u r e i n d i c a t e s a h i g h e r r a t e of f l o w whereas a f t e r i n t e r s e c t i o n the r a t e of f l o w i s h i g h e r f o r the h o r i z o n t a l f r a c t u r e . The v e l o c i t y of p r o p a g a t i o n of the h o r i z o n t a l f r a c t u r e i n c r e a s e s w i t h an i n c r e a s i n g g r a d i e n t dL^/dL, whereas f o r the v e r t i c a l f r a c t u r e i t i n c r e a s e s w i t h a d e c r e a s i n g g r a d i e n t . U n l i k e the Q-L c u r v e s , the f r a c t u r e l e n g t h t h a t c o r r e s p o n d s t o the p o i n t of i n t e r s e c t i o n i s independent of t h e h e i g h t assumed f o r t h e v e r t i c a l f r a c t u r e . 91 v -ver tical fracture h-horizontal fracture 1 2 3 4 5 distance from wellbore a x i s ( L ) ft 1 a u a o 5 O o CJ o Fig (5*7 ) The flow rate-length-propagation veloci t y variations 92 The analyses carried out in this thesis correspond to a constant wellbore pressure and assume no leakage of fluid to the medium surrounding the fracture. Further, the rate of flow of fluid was allowed to vary with no fixed upper bound. However, in reality, no geotechnical material is fully impermeable and often the maximum rate of pumping is limited to the capacity of the pumps available. In fracturing of permeable materials, significant leakage of fluid takes place and the resulting fluid pressure inside the fracture can be significantly different in distribution and magnitude from that for no leakage of f l u i d . The effects of the above factors, discussed later in Section 5.2.3, are to reduce the velocity of propagation of fractures. In an isotropic stress f i e l d , the pressurization of a section of a wellbore would induce vertical and horizontal fractures simultaneously. The results shown in f i g 5.7 indicate that at i n i t i a l stages of propagation, the vertical fractures grow faster. This is a result of the larger fluid carrying capacity of vertical fractures, observed during the i n i t i a l stages of propagation. The horizontal fractures grow, but at a slower rate. However, after the vertical fractures propagate a certain distance, the increase in velocity of propagation as well as the fluid carrying capacity reduce significantly. A horizontal fracture of longer length of propagation exhibits a larger f l u i d carrying capacity than a vertical fracture. High rates of pumping of fluid force the fractures to accommodate large quantities of fl u i d . Therefore, i f only the rate of flow is considered, in fracture operations with high rates of pumping of fluid, fractures should be dominently horizontal in 93 orientation. The pressure distributions predicted for the parabolic fractures are compared with the pressure distribution assumed at the start of the f i r s t iteration, in fig(5.8). The predicted pressure distribution is seen to be in good agreement with the assumed distribution except for the region close to the tip of the fracture, where the maximum deviation is about 10% of the wellbore pressure. 5.1.4 A comparison of displacements with Sun's closed form solution The closed form solution presented for ground surface, displacements by Sun(43), is based on highly idealized conditions. In order to permit a comparison of the numerical solution with the closed form solution, similar idealizations are made to the model developed in the present study. Thus, only uniformly pressurized penny shaped horizontal fractures embedded in a linear elastic, homogeneous and isotropic medium are considered. The radius of the wellbore is reduced to zero and no self weight stresses are considered. Further, the two fracture lengths defined in the closed form solution, and L, discussed in section 2.4, are assumed to be the same for the analysis. Figures (5.9) & (5.10) illustrate the comparisons of the horizontal and vertical ground displacement predictions from the fi n i t e element analysis with the closed form solution. The data used for these predictions are shown in Table 5.2. The closed 94 Table 5-2 data used in the analyses Young's modulus 950.0 lbf/ft* Poisson s r at io 0.20 Cohesion 0-0 0 Friction angle 40-0 depth of fracture from free surface 7.5 ft. well bore rad'i us 0-4 f t. excess fluid pressure ( uniform ) 1000.0 I b f / f t2 radius of the fracture 4-9 ft 95 ro O 3 73 l / l CU (a) (b) predict ed a s s u m e d \\ length (ft) predicted assumed \\\\ l e n g t h ( f t ) Fiq(5-8 ) A comparison of the predicted and a s s u m e d p r e s s u r e p r o f i l e s (a) horiz ontal fracture (b) v e r t i c a l fracture o x 0I = 0 5 10 15 radial distance (ft) Fig(5-9)A comparison of the vertical ground displacements predicted with Sun's(43) closed form solution closed form for section 1 closed form for section 2 • finite element analysis 5 10 radial distance (ft) 15 Fig(5-10) A comparison of the horizontal ground displacements predicted with Sun's(43) closed form solution 98 form solution requires the depth of the centre of the fracture from the ground surface. As shown in the fig (5.11) because a fracture segment is represented by a fa i r l y thick element in the fini t e element analysis, the exact location of the centre of the fracture that suits the closed form solution is debatable. For this reason, the displacements are computed for the two possible locations of the centre of a fracture, as shown in fig (5.11) by sections 1 & 2. It is seen that the results are in close agreement with the closed form solution. The results agree extremely well with the closed form solution that corresponds to a depth as given by section 1. The displacements predicted are at most 13% higher than those obtained from the closed form solution when a depth that corresponds to section 2 is selected. The shape of the fracture predicted in the analysis is compared with that given by Sun's solution, in f i g (5.12). The fi n i t e element results are in excellent agreement with the closed form solution. 99 ground surface Up -' fracture face' •section 1 _^ section 2 fractured elements shaded Fig(5-11) The finite element mesh in the vicinity of a horizontal fracture 100 Fig(5-12) A comparison of the predic ted f racture shape with Sun's closed form solution 101 5.2 DISCUSSIONS 5.2.1 Fracture Initiation The state of stress of a normally consolidated (i.e.K 0<l) s o i l consists of a major principal stress which is vertical(i.e.overburden pressure) and intermediate and minor principal stresses which are horizontal in orientation. From the f i n i t e element analysis i t was seen that when K0<1, the fractures created were vertical. As pointed out in Section 1.3, the magnitude of the fluid pressure at which a fl u i d f i l l e d vertical fracture closes, corresponds to the minor principal total stress(i.e. horizontal) at the location of the fracture. If the minor principal total s t r e s s ( a 2 ) , the tensile strength of the soil(o ) and the fracture initiation pressure obtained with a non-penetrating fl u i d are known, the magnitude of the intermediate principal total s t r e s s C a , ) can be estimated from the theoretical relation given in equation 2.5. However, if a penetrating f l u i d is used, CT, can be estimated from equation 2.4, provided the Poisson's ratio and the Biot's constant for the material are also known. If overly viscous fluids or high rates of pumping of even low viscosity fluids are used, due to partial penetration of the fracture fluid , a ^ computed from the above equations would be in error. In over-consolidated soils(i.e.K 0>1), the overburden pressure corresponds to the mi.nor principal total stress. For K0>1, the fractures created are horizontal in orientation. As a result, the f l u i d induced fracture data are useful only for estimating the overburden pessure and the tensile strength in 102 over-consolidated soils. 5.2.2 Fracture Propagation The common assumption that the path of propagation is perpendicular to the in-situ minor principal stress direction is relaxed in the present analysis. Instead, the new state of stress is estimated after each increment of the fracture length. The direction of propagation is assumed to be perpendicular to the current minor principal stress direction in the tip of the fracture. Such a process reflects the effects of previous fractures on the process of subsequent fracturing. Even though Pollard & Holzhausen(34) and Dusseault(17) predict climbing fractures, their hypotheses for the climbing behaviour are different. Pollard & Holzhausen(34) predict climbing fractures based on the significant increase in the normalized mode-2 stress intensity factors(i.e. K2/K00=r/pb in which r is the shear stress and p^ is the wellbore pressure) computed at the tip of the fracture. Dusseault proposes a work minimization criterion for the path of propagation and concludes that i t is a reduction in stress and hence the lesser work required to part the material that causes the fractures to climb towards the ground surface. The finite element analysis indicates that i t is because of the reorientation of the minor principal stress plane, in the vicinity of the tip of the fracture, that causes the shallow, horizontal fractures to propagate towards the free surface. The larger normalized mode-2 stress intensity factors, 103 predicted by Pollard & Holzhausen(34), are a result of larger shear stresses induced near the tip of the fracture. These shear stresses reorient the minor principal stress plane towards the ground surface. Hence, i t is seen that the results of the f i n i t e element analyses as well as the results obtained from Pollard & Holzhausens' analysis, predict climbing fractures for the same reason. However, because of the qualitative nature of Pollard & Holzhausens' predictions regarding climbing, a quantitative comparison of the change in orientation of the principal stress plane cannot be done. 5.2.3 The fluid flow aspects The objectives of the fluid flow analyses are to predict the variations in velocity of propagation, rate of flow and the fluid pressure distribution inside a fracture, with the length of propagation. As outlined in Section 3.6.2, inorder to achieve the above objectives, i t is necessary to know the variation of the maximum aperture of fracture and the fluid pressure at the tip of the fracture, at different fracture lengths. These data were obtained by carrying out finite element analyses in the domains shown in figs 3.1 & 3.2. The tip fluid pressures obtained from the finite element analyses are shown in Table 5.3. These results correspond to the same wellbore pressure of 6000 l b s / f t 2 . It should be noted that for the horizontal fracture, P^ .ip indicates a reduction with length whereas for the vertical fracture i t increases with length. 104 Table 5*3 The tip fluid pressure data with length horizontal fracture vertical fracture 1 ength* (.ft) P. dbf/ft2) tip length*(ft) P. (Ibf/ft1) tip 1-90 4 700 0-90 41 00 3-40 4500 1-80 4100 4- 90 4 300 3-50 4200 6-40 4 100 5-00 4220 * measured from wellbore axis 105 The fi n i t e element program used for the analyses computes the stresses at the centres of the elements. However, the t i p fluid pressure corresponds to the fluid pressure at the boundary of an element and is compared with the minor principal total stress at the centre of the element adjoining the boundary, to predict the stability of the fracture(i.e. propagation). When the element gets bigger in size, the minor principal stress at the centre of the element could be significantly higher than the same near the edge, where the t i p of the fracture is located. The radial domain in which the vertical fracture analysis is carried out, consists of elements that grow bigger in size with the radial distance. Hence, the increase in the tip fl u i d pressure could be a result of the change in element size with length of propagation. For the horizontal fracture analysis, the domain consists of rectangular elements that are approximately equal in size. Therefore, even though the stresses computed are higher than the stresses at the boundary(closer to the tip) of the element, the effect of the size of the elements will not be as significant as for the radial domain. When a free surface is introduced near a fracture, the resistance to deformation provided by a l l the material above that surface is removed. As a result, a significant portion of the strains are relieved as displacements perpendicular to the free surface. Also, a significant portion of the load caused by the excess fl u i d pressure, acting on the fracture face, is transferred to the material in the tip region. The load mentioned above, increases with length of propagation. Hence, the minor principal total stress at the tip of the fracture 106 decreases with length, .leading to lower tip f l u i d pressures( since they are identical for an equilibrium fracture). From equations 3.10 & 3.17, i t can be seen that the velocity of propagation and the rate of flow are directly proportional to Pb -P^p . As shown in Table 5.3, Pb-Ptip varies between 0.23Pb and O . 4 O P 5 , the influence of which, on velocity of propagation and rate of flow, is significant. In the following sections, the influence of the aperture-length variations on velocity of propagation and rate of flow are discussed, on the assumption that the tip fl u i d pressure is constant with length. Inorder to investigate the effect of aperture(B)-length(L) variations on velocity and rate of flow, i t was f i r s t decided to assume identical B-L curves for both types of fractures. Results obtained with a linear B-L relationship are shown in fig 5.13. The variation of velocity(L) with L, for both the fractures, are seen to.be identical. Hence, i t is seen that the velocity-length curves are useful in studying the influence of various aperture-length relations. The rate of flow, however, should be different because of the axisymmetric nature of the horizontal fracture and the constant height assumed for the vertical fracture. Next, the effect of two different shapes of B-L curves on L-L are examined. The B-L curves assumed are shown in fig 5.14 and the results that correspond to them are shown in fig 5.15. From f i g 5.15, the following could be deduced; (a) i f aperture is linear in length, L is also linear in L (b) i f the magnitude of the aperture is constant with L, L decreases with L (c) for a given length of fracture, for higher apertures the 107 B = (.533* 1.333L )«10' P = 4500 lbf/ft2 p,= 6000 distance from wellbore axis (ft) Fig(5-13) Results of the flow analysis with identical B-L curves for both fractures 108 BrL/IOOCK / *B=-004 / ^^C% - L / 2000 0 4 , 8 12 length (L) -Fi. Fig(5-H) The assumed B-L variations 4 length(L) -f-t. 8 Fig (5-15) The velocity-length predictions for B-L var iat ions in fig (5*1 109 L c o m p u t e d a r e a l s o h i g h e r The v a r i a t i o n s o f t h e a p e r t u r e w i t h l e n g t h o b t a i n e d f r o m t h e f i n i t e e l e m e n t a n a l y s e s , f o r t h e d o m a i n s shown i n f i g s 3.1 & 3.2, a r e shown i n f i g 5.16. From ( a ) t o ( c ) shown a b o v e , t h e s t e a d y i n c r e a s e i n t h e v e l o c i t y o f p r o p a g a t i o n o f t h e h o r i z o n t a l f r a c t u r e , shown i n f i g 5.7, c a n be a t t r i b u t e d t o t h e n e a r l i n e a r v a r i a t i o n o f i t s a p e r t u r e w i t h l e n g t h . The g r a d u a l f l a t t e n i n g o f t h e B-L c u r v e o f t h e v e r t i c a l f r a c t u r e h a s l e a d t o v e l o c i t i e s t h a t i n c r e a s e a t a d e c r e a s i n g r a t e ( i . e . d L / d L ) . A g r a d u a l i n c r e a s e i n t h e t i p f l u i d p r e s s u r e r e d u c e s t h e v e l o c i t y o f p r o p a g a t i o n w h e r e a s a g r a d u a l d e c r e a s e i n t i p f l u i d p r e s s u r e i n c r e a s e s i t ( a s i n t h e v e r t i c a l a n d h o r i z o n t a l f r a c t u r e s , r e s p e c t i v e l y ) . The v e l o c i t y o f p r o p a g a t i o n w i t h l e n g t h , o b s e r v e d f o r t h e two f r a c t u r e s , w o u l d be s i g n i f i c a n t l y d i f f e r e n t i f , (1) t h e a c t u a l p r e s s u r e d i s t r i b u t i o n i n s i d e t h e f r a c t u r e i s c o n s i d e r a b l y d i f f e r e n t f r o m t h e d i s t r i b u t i o n u s e d i n t h e a n a l y s i s t o o b t a i n t h e B-L c u r v e s , o r , ( 2 ) an u p p e r b o u n d f o r r a t e o f p u m p i n g i s i m p o s e d . From t h e c l o s e d f o r m s o l u t i o n ( 4 2 ) d e v e l o p e d f o r a penny s h a p e d h o r i z o n t a l f r a c t u r e , embedded i n a l i n e a r e l a s t i c medium an d s u b j e c t e d t o a u n i f o r m p r e s s u r e d i s t r i b u t i o n , B oC L ( s e e s e c t i o n 2.4) H e n c e , i t c a n be s e e n t h a t f o r a n e a r l y u n i f o r m f l u i d p r e s s u r e d i s t r i b u t i o n , a r e d u c t i o n i n t h e v e l o c i t y o f p r o p a g a t i o n o f a h o r i z o n t a l f r a c t u r e c a n be a c h i e v e d by l i m i t i n g t h e r a t e o f pumping o f f l u i d . I f l e a k a g e o f f l u i d i s a l l o w e d , 110 F i g (5 -16 ) The var iat ion of aperture with lengt h 111 which is the case in reality,' the fluid pressure inside the fracture could be significantly lower than those predicted from the flow analysis witg no leakage. With increasing length of propagation, the pressure losses are likely to be higher since the area of material exposed is high. Hence, at larger fracture lengths, a considerable flattening of the B-L curves could be expected, which would eventually lead to lower velocities of propagation. The rate of flow inside a fracture varies with B 3. Therefore, the change in magnitude of B has a significant influence on the rate of flow. The horizontal fracture is axisymmetric in nature and the vertical fracture has been assumed to propagate with a prescribed height. The aperture(B) obtained for the horizontal fracture is seen to be higher than that for the vertical fracture at larger fracture lengths. As a result, the rate of flow for the horizontal fracture is seen to increase rapidly with length of propagation. If the rate of pumping of f l u i d exceeds the rate of flow of fluid that a fracture can accommodate, a build up of wellbore pressure results. If this increased wellbore pressure exceeds the secondary or the major principal total stress on a plane outcropping the fluid boundaries, a secondary fracture w i l l form in that plane. The resulting increase in the capacity will automatically reduce the wellbore pressure and w i l l be followed by an increase in the rate of flow to satisfy volume compatibility. Therefore, i f the rate of pumping changes significantly during fluid fracture operations, injection 1 12 pressure time variations that are significantly different from those corresponding to primary fracture propagation could result. The injection pressure-time data reported by Holzhausen et al(24) have been obtained under irregular rates of pumping of steam and may have been a result of such secondary fracturing, the rate effects probably playing a dominant role. The inter-dependency of the rate of pumping and the wellbore pressure, as reported by Settari & Raisbeck(37), is illustrated in fig(5.17). The consistent pattern of increase of the injection pressure with the rate of pumping supports the above reasoning. The flow of fl u i d inside a fracture is accompanied by a drop in f l u i d pressure with distance. If the pressure drops are ignored, the finite element analyses predict fractures that are unstable at a l l lengths of propagation. Hence, for proper modelling of fluid induced fracture behaviour this effect must be included. 5.2.4 Shape of the fractures It is of interest to examine the shape and the volume of the fractures that correspond to the different pressure distributions. The shapes obtained with uniform and non-uniform (given by equation 3.6) pressure distributions that correspond to the same maximum pressure, are compared in figs 5.18(a) and 5.18(b). The shape of the vertical fracture obtained for a uniform pressure profile is seen to be close to the theoretical e l l i p t i c a l shape and is shown in fig 5.18(a). The non-uniform 113 114 (a) uniform pressure non-uniform pressure (b) 1 2 3 4 fracture length(ft) uniform pressure non-uniform pressure 2 3 fracture length(ft) Fig(5-18) The shape of the fracture with uniform and non uniform pressure profiles (a) ve r t i ca l fracture (b) horizontal fracture 1 15 pressure distribution has lead to a flattening of the tip region. The cross-sectional area of the fracture for this case is seen to be somewhat lower than for the uniform presure distribution. The shape of the horizontal fracture obtained in the case of uniform pressure, digresses from the e l l i p t i c a l shape. Such a variation is thought to be a result of a major portion of the strains being relieved in the vertical direction due to the presence of a free surface. For the non-uniform pressure distribution, as in the vertical fractures, the tip area indicates a flattening together with a slight change of curvature. The cross sectional area of the fracture is seen to be smaller for the non-uniform distribution. The magnitude of the maximum opening obtained for the horizontal fracture is seen to be higher than that of the vertical fracture for both uniform and non-uniform pressure distributions. Even though the cross-sectional areas of the horizontal and vertical fractures, for a given pressure distribution, are approximately equal(for the case shown in figs 5.18(a) & 5.18(b)),the volume is seen to be higher for the horizontal fracture because of i t s axisymmetric nature and the height assumed for the vertical fracture. 1 16 CHAPTER 6 SUMMARY AND CONCLUSIONS A method of analysis for the initiation and propagation of fluid induced fracture in Oilsand has been developed. The propagation of a fluid induced fracture involves a coupled stress and fluid flow analysis. This is solved herein by f i r s t considering the stress and the flow analyses separately. The results are then coupled through volume compatibility between the quantity of fluid pumped in and the volume of the fracture. For the stress analysis, linear-elastic and elastic-plastic stress strain relations have been assumed for the unfractured and fractured material, respectively. The stress analysis is based on a total stress approach. For the fluid flow analysis, the fractures are approximated as parabolic in shape. The flow of f l u i d between the fracture faces is assumed to be described' by parallel plate flow equations. The flu i d flow analysis has been restricted to steady, incompressible, laminar flow of a Newtonian f l u i d . The analyses performed assume a vertical-horizontal in-situ principal stress orientation. For horizontal fractures, three dimensional axisymmetric analyses have been performed. For vertical fractures, the analyses have been restricted to two dimensions. The magnitude of the fracture initiation pressure is strongly dependent on the intrusion of fluid into the medium. The use of high viscosity fluids or low viscosity f l u i d with high rates of pumping, causing partial fluid penetration, leads 1 17 to a fracture initiation pressure that is significantly higher than for a low viscosity fluid with low rates of pumping. However, the estimation of the instantaneous shut-in pressure is independent of the type of flu i d used. For f l u i d fracturing in Oilsand, with penetrating fluids, the analysis predicts a fracture initiation pressure that is equal in magnitude to the in-situ minor principal total stress that correspond to the location of the fracture. Fracture propagation can be considered as a new initiation subjected to the current stresses of the domain and the current boundary fl u i d pressures. Fluid induced fracture propagation requires the transmission of fluid inside the opening, which is possible only with an open mode fracture. The common assumption that the path of propagation is perpendicular to the in-situ minor principal stress direction has been removed. Instead, a direction perpendicular to the minor principal stress direction at the tip of the fracture is considered. The fracture c r i t e r i a predict the path of propagation based on the computed stresses at the end of each increment of fracture length. The presence of a flu i d pressurized fracture changes the state of stress of the domain in the vic i n i t y of the fracture and thus the orientation of the principal stresses. Hence, the true behaviour of a fluid induced fracture can only be simulated by analysing the propagation of the fracture incrementally, taking account of the stress changes caused in its course of propagation. For fractures initiating in a medium where the i n i t i a l in-1 18 situ minor principal stress direction is vertical, a path of propagation inclined towards the ground surface is predicted. This occurs once the fracture length propagates to a distance about the depth of the fracture from the ground surface. These results are in agreement with f i e l d observations of actual fracture operations and theoretical predictions. Propagation of a vertical fracture, under high rates of pumping of fl u i d , is limited in extent. For a selected wellbore pressure and fluid viscosity, a horizontal fracture exhibits a rapidly increasing fluid carrying capacity with length, than a vertical fracture. The velocity of propagation of a horizontal fracture increases with length with an increasing gradient, whereas for a vertical fracture i t increases with a decreasing gradient. Loss of fluid pressure resulting from leakage of fluid to the surrounding medium and the limiting values of rates of pumping experienced in practice, would reduce the velocity of propagation of the fractures. In fluid induced fracturing with rates of pumping higher than those corresponding to the maximum possible rate of flow in a vertical fracture, fractures must be dominently horizontal in orientation. The effects of rate of pumping of fl u i d on the propagation behaviour of fractures, which has been overlooked by a number of researchers in the literature, is seen to be important. The injection pressure time records obtained under high rates of pumping are considerably different from those obtained for primary fracture operations under low rates of pumping. The predictions of ground displacements and fracture shape 119 from the fin i t e element analyses are in good agreement with the theoretical closed form solution results. 120 CHAPTER 7 RECOMMENDATIONS FOR FURTHER RESEARCH As described in Chapter-3 and discussed in Chapter-5, the analysis approximates the shape of the fracture with a parabola. It was also seen that such an approximation simplified the fluid flow formulations considerably. The shape of the fracture is important in determining the fluid pressure distribution inside the fracture. The correct pressures can be obtained only if the correct shapes are used. Such an analysis requires the fracture to be represented by its real shape. The above can be achieved by incorporating a numerical integration scheme to perform the volume calculations. The analysis can then be extended for a number of iterations to obtain a consistent shape-pressure distribution. The pressure profile described by Daneshy(l2) can be used as a f i r s t approximation in such an analysis. The flow model can further be developed by including multiphase, non-Newtonian fluids. The boundary traction forces exerted by the flow of fluid and the effects of leakage on the resulting pressure distribution could also be examined. Further, the temperature aspects of the problem can also be included. The behaviour of a porous material is governed by the effective stresses. By incorporating a pore pressure model to represent the pore fl u i d behaviour of Oilsand, the analysis can be extended to an effective stress approach. The complexity of the pore fluid of Oilsand requires the consideration of ideal gas laws. The actual f l u i d induced fracture problem is a truly 3D 121 problem. The model developed in this thesis cannot incorporate vertical and horizontal fractures simultaneously. For this reason, the observations such as fracture roll-over from horizontal to vertical or vice versa cannot be examined. Such behaviour can be examined only by a true 3D modelling program. 122 REFERENCES 1 . Advani S.H \"Finite element model simulations associated with hydraulic fracturing\", Paper SPE 9260, presented at SPE/DOE, Symposium on unconventional gas recovery, Pennsylvania, May 1980. 2. Amodt R.L, Potter R.M \"Anomalous fracture extension pressures in granitic rocks\", 19th U.S Symposium of Rock Mechanics, Reno, Nevada, 1978, pplO-13. 3. Bawden W.F \"Hydraulic fracturing in Alberta Tar sand formations - A unique material for in-situ stress measurements\", workshop on hydraulic fracturing and stress measurements, Dec 3-5, 1981, Montery, California. 4. Bjerrum .L, Anderson K.H \"In-situ measurement of lateral pressures in clay\", Proceedings, 5th European Conference in Soil Mechanics, 1972, Vol-1, pp11-98. 5. Bjerrum L, Nash J.K.T.L, Kennard R.M, Gibson R.E \"Hydraulic fracturing in f i e l d permeability testing\", Geotechnique 22, No2, 1972, pp 319-322. 6. Byrne P.M, Atukorala U.D, Charlwood R.G \"Analysis of fl u i d fracture in Oilsands\", Proceedings, 33rd annual technical meeting of the Petroleum Society of CIM, June 6-9, Calgary, 1982. 7. Byrne P.M, Grig R.F \"OILSTRESS- A computer program for nonlinear analysis of stress and deformation in Oilsands\", Soil Mechanics series No:42, Deptartment of C i v i l Engineering, UBC, Vancouver, Canada, July, 1980. 8. Byrne P.M, Duncan J.M \"NLSSIP - A computer program for non-linear analysis of so i l structure interaction problems\", Soil Mechanics series No:41, Department of C i v i l Engineering, UBC, Vancouver, Canada, July, 1 979. 9. Cook R.D \"Concepts and Applications of Finite Element Analysis\" 2nd Edition, John Wiley & Sons, 1981. 10. 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Dusseault M.B \"Stress state and Hydraulic fracturing in Athabasca Oilsands\", Journal of Canadian Petroleum Technology, Vol:16, 1977, pp 19-27. 17. \" A conceptual geomechanical model for Hydraulic fracture in Oilsand\", presented at the AOSTRA, University Industry seminar, Fort McMurray, Alberta, 1979. 18. ' \"The behaviour of hydraulically induced fractures in Oilsands\", 13th Canadian Rock Mechanics Symposium, University of Toronto, Ontario, May 28-29, 1980. Published by the CIM Montreal, Quebec, 1980, pp 36-41 . 19. Dusseault M.B, Simmons J.V \"Injection induced stress and fracture orientation changes\", Canadian Geotechnical Journal, Vol:19, November 1982, pp 483-493. 20. Haimson B, Fairhurst C \"In-situ stress determination at great depth by means of Hydraulic Fracturing\", 11th Symposium on Rock Mechanics, University of California, Berkeley, June 16-19,1969, pp 559-589. 21 . \"Hydraulic Fracturing in Porous permeable materials\", Journal of Petroleum Technology, July 1969, pp 811-817. 22. Hagoort J, Weatherill ,Settari.A \"Modelling the propagation of water flood-induced Hydraulic fractures\", Journal Society of Petroleum Engineers, August 1980, pp 293-303. 23. Hanson M.E, Anderson G.D, Schaffer R.J \"Theoretical and Experimental research on Hydraulic fracturing\", Journal of Energy Resources Technology, Vol:102, June 1980, pp 92-98. 24. Holzhausen G.R, Wood M.D, Raisbeck J.M, Card C C 124 \"Results of deformation monitoring during steam stimulation in a single-well test\", Applied Oilsands Geoscience, Department of Mineral Engineering, University of Alberta, Edmonton, 1980. 25. Howard G.C, Fast G.R \"Hydraulic Fracturing\", Monograph vol:2, 1970. 26. Hungr 0, Morgernstern N.R \"A numerical approach to predicting stresses and displacements around a 3D pressurized fracture\", Department of C i v i l Engineering, University of Alberta, Edmonton, Canada, April 1980. 27. Hussain M.A, Pu S.L, Underwood J \"Strain energy release rate for a crack under combined Mode-1 & Mode-2\", Fracture Analysis, Proceedings of the 1973, National Symposium on Fracture Mechanics, Part-2, STP 560, ASTM, pp 1-28. 28. Irwin G.R \"Analysis of stresses and strains near the end of a crack transversing a plate\", Transactions of the American Society of Mechanical Engineers, Vol:79, 1957, pp 361-364. 29. Leach E.R \"Hydraulic fracturing of Oilsands ; a literature review\", Miscallaneous paper s-77-6, U.S Army Engineers waterway experiment station, March 1977. 30. Massarch K.R * \"New aspects of Soil fracturing in Clay\", Journal of Soil Mechanics & Foundation Division, ASCE, August, 1978, pp 1109-1123. 31. Medlin W.L \"Laboratory investigation of fracture initiation pressure and orientation\", Society of Petroleum Engineers Journal, April, 1979, pp 129-144. 32. Ozawa Y, Duncan T.M \"ISBILD - A computer program for analysis of stresses and movements in embankments\", Report No TE-73-4 to National Science foundation, 1973. 33. Pollard D.D \"Forms of Hydraulic fracture as deduced from f i e l d studies of sheet intrusions\", Proceedings, 19th Symposium on Rock Mechanics, Reno, Nevada ,1978, pp 7-9. 34. Pollard D.D, Holzhausen G \"On the mechanical interaction between a fluid f i l l e d fracture and the earth's surcface\", Tectonophysics:53, 1979, pp 27-57. 35. Schapery R.A \" A theory of crack initiation and growth in viscoelastic media-Part-I\", International Journal of Fracture, Vol:11, No-1, February 1975, pp 141-159. 125 36. Schonfeldt H.V, Fairhurst C \"Field experiments on Hydraulic Fracturing\", Society of Petroleum Engineers Journal, February, 1972, pp 69-77. 37. Settari A, Raisbeck J.M \"Fracture mechanics analysis in in-situ Oilsands recovery\", Journal Canadian Petroleum Technology, 1979, Vol:18, No:2, pp 85-94. 38. Shaw P \"Coupling Boundary Integral Equation Method to \"Other\" Numerical Techniques\", Recent advances in Boundary Element Method, by Brebbia CA, pp 137-1 47, 1978. 39. Sih G.C \"Strain energy density factor applied to mixed mode crack problems\", International Journal of Fracture, Vol:10, 1974, pp 305-321 . 40. Skempton A.W \"Effective stresses in Soil, Concrete & Rocks\", Conference on Pore pressure and suction in soils , London, 1960, pp 4-16. 41. Sobkowicz J \"A state of the art bibliography on Hydraulic Fracturing\", Vol:1, Alberta environment earth sciences division, May, 1979. 42. Streeter V.L \"Fluid Mechanics\", 6th Edition McGraw-Hill. 43. Sun R.H \"Theoretical size of hydraulically induced horizontal fractures and corresponding surface uplift in an idealized medium\", Journal of Geophysical research, Vol:74, No:25, November 15, 1969, pp 5996-6011. 44. Terzaghi K, Peck R.B \"Soil mechanics in engineering practice, New York:Wiley. 45. Wiles T.D, Roegiers J.C \"Modelling of hydraulic fractures under in-situ conditions using a Displacement Discontinuity Approach\", Proceedings, 33rd Annual technical meeting of the Petroleum Society of CIM, June 6-9, 1982, Paper No:82-33-70 46. Zoback M.D, Pollard D.D \"Hydraulic fracture propagation and the interpretation of pressure time records for in-situ stress determinations\", Proceedings, 19th Symposium on Rock Mechanics, Reno, Nevada, 1978, pp 14-22. 1 26 APPENDIX A The proposed method of stress analysis STEP-1 Specify the in-situ state of stress in the domain under consideration STEP-2 Begin increasing the boundary fl u i d pressure(Up) until i t exceeds the minor principal total stress(ow) of any one of the elements adjacent to the fluid boundary, (i.e. elements H,I,J shown in fig A.1(a)) When Up f i r s t exceeds omA of an element on a plane the outcropping the fl u i d boundary, a fracture is initiated in the element that corresponds to the above amn.( say i t occurs in element I) STEP-3 Determine the current direction of the plane of crmua and reorient element I in that direction. The mesh after reorientation is shown in f i g A.1(b). If nodes 3 or 2 of element I (ref f i g A.1(b)) exceeds nodes 3 or 2 of elements H or J respectively, in the process of reorientation, a prescribed tolerence of .1(z 3 -z 2 ) or .1(z 3 -z 2 ) is assigned to AB or A,B2 to avoid overlapping. A .Ka) H I J K A.1(b) A H 3 2 B — I -—~~3 2 -—~~3 Al 1 J 2 K F ig (A-1) M e s h pr ior to and after reor ientat ion 128 2 denotes the vertical coordinate of node i in element j . NOTE; Together with the reorientation process are the changes in elevation of the relevant elements. This variation, however, is restricted at most to the length AB,. Hence, by selecting AB, sufficiently small, the errors can be minimized. STEP-4 Reduce the moduli G' & B'(refer to Chapter 4 for notation) of element I to .00001(G'&B') and insert a fluid pressure of magnitude [U p-a m i^]. Modify the nodal load vector to eliminate the forces resulting from the boundaries 1-4 & 2-3 in element I. STEP-5 Analyse the domain for the displacements and compute the stresses and strains of the elements. STEP-6 Compute the Fracture Potentials of the elements H+1,I+1 & J+1. To compute the Fracture Potential, consider the current boundary fl u i d pressure i f a non-uniform pressure distribution is considered. For a uniform pressure distribution this will be the wellbore pressure. The next fracture susceptible element is the element with the highest Fracture Potential. (say i t occurs in element 1+1). STEP-7 Return to STEP-3. Substitute 1+1 for I and repeat the analysis. Continue until the fracture extends the required length. NOTE: Orientation of the elements already fractured are preserved in subsequent reorientations. 130 APPENDIX-B B.1 Axisymmetric formulation B.1.1 The Constitutive matrix [D] The generalized Hooke's law for incremental elasticity can be stated as; 5*11 = = [6a, ,-»>( 6a2 + 6a3 3 ]/E B.1 5*2 2 = = [ 6 a 2 2 ~ \" ( 5 a 3 3 + 5a,,]/E B.2 5*33 = = [5a 3 3 -p (5a,,+6a 2 23/E B.3 5e i 3 = = 6a,3/G B.4 5e , 2 = = 6a,2/G B.5 5*3 2 = = 6a 3 2/G B.6 For axisymmetric problems; 5e 1 3 = 0 5*3 2 = 0 B. 1 to B . 6 can be represented 5*11 1 - V - V 0 5*2 2 1 - V 1 - V 0 6*3 3 = E -v - V 1 0 5*12 0 0 0 2(l+v) B.7 B.8 5a, , 6a2 2 6a 3 3 6a, 2 B.9 Let, B'= 3B/{2(1+»»)} and G' = E / { 2 ( 1 + J 0 } Inverting the matrix given in equation B.9 and substituting for E & v from the above expressions; 131 6a, , 6a2 2 5a3 3 6a, 2 B' +G' B'-G* B' -G' 0 or, in matrix notation; {5a} = [D]{5e} B'-G' B' +G' B'-G' 0 B'-G' B'-G' B' +G' 0 0 0 0 G' 5e 1 1 6e2 2 6e 3 3 5e , 2 B. 10 B. 1 1 B.1.2 The strain displacement matrix [B] In compatible isoparametric elements, the coordinates and the incremental displacements at a point are approximated as; and where r z u V [N] [N] {5'} B. 1 2 [N] {6} N, 0 N2 0 N3 0 N« 0 0 N, 0 N2 0 N3 0 N„ B. 13 B. 14 It can be shown easily that equation B.13 satisfies compatibility as well as the completeness c r i t e r i a and hence gurantees convergence (from the Potential Energy Theorem). For an axisymmetric problem, the incremental strains are expressed as; 132 5e 1 1 3u/3r B. 1 5 be2 2 5*3 3 u/r -8e, 2 du/dz+dv/dr 3NX 3r~ 0 3N2 3r~ 0 3N3 3r~ 0 3N4 3r~ 0 u l v l 0 3Z~ 0 3N2 3Z~ 0 3N3 3Z~ 0 3N4 3Z~ °2 v2 ^1 r 0 r 0 N3 r 0 r 0 u3 v3 3NX 3N2 3N2 3N3 3N3 3N4 3NA u4 a z 3r 3Z 3r 3Z 3r 3Z 3r V4 B.16 or , in matrix notat ion; {6e} = [B] {6} B.17 where, {B} i s the s t r a i n displacement matrix The shape functions Nj are functions of the l o c a l coordinates s & t . Therefore, a d i r e c t evaluation of the f i r s t der iva t ives of N,-with respect to r & z i s not p o s s i b l e . From the chain rule i t follows that; 3_ ds 3_ 3t 3jr 3S 3r 3t 3S 3Z 3t 3_ 3r 3_ 3Z B.18 [J] 133 The partial derivatives with respect to r & z are written as; 3_ 3r 3_ 3Z [J] -1 as 3 at B.19 The strain displacement matrix formed, contains s,t terms and hence is not a constant for the element. Therefore, i t is evaluated numerically at the Gaussian points with a 2x2 Gauss quadrature. B . I .3 The stiffness matrix [K] Let {5e} represent a virtual nodal displacement vector. From equation B.17 i t follows that; {Se } = [B] {6%} B.20 The internal work done , W(int), w i l l be; W(int) v T Se}{So) dv J\"{SeHB]\"[D][B]{6} dv V For axisymmetric problems, B.21 dv = 27rr dr dz B.22 Since [B] is in terms of s & t, i t is easy to carry out the integration interms of the local coordinates. 134 dr dz =lJ|ds dt B.23 = {6^}[K]{5} T B.24 where [K] = H 2ir r [B][D][B] |J| ds dt which is called the - 1 -1 stiffness matrix of the element. For rectangular elements|J| is a constant. Hence, the product of the matrices [B][D][B] .IJ| contains terms such as s 2 , t 2 , s t . Therefore, a Gauss quadrature of 2x2 is sufficient for the numerical evaluation of the integral given in equation B.24 (i.e. 2*2 is good enough for a cubic polynomial). Accordingly, For n=4, w = .1 and hence, equation B.25 can be written as; where, f(s,t) represents the integrand of equation B.24 and f ( s j , t f ) represents the value of the integrand at the Gaussian points and w is a weight attached to the Gauss quadrature. B.1.4 The force displacement relationship T 1 I ia J f f ( S , t ) d S d t = i : W f ( S ; , t ; ) B.25 B.26 If the nodal load vector of the element is defined by {f}, for a virtual nodal displacement increment {5 e }, the external work done is given by; 135 W(ext) = { 5 e H f } From the principle of virtual work; W(ext) = W(int) Hence, from equations B.24, B.27 and B.28; {f} = [K]{5} B.27 B.28 B.29 If a pore fluid pressure of magnitude 5U is present, the stress vector for the element can be written as; {6a} = {5a'} + Su-Substituting B.30 in B.21, B.30 W(int) v 5{6?}{5o} dv I{8gjf 6U dv i i • Lf{5eHB]LD][B] [J | ds dt 27rr{6} II T T + ([ {6e}[B] -I-I T SU 27r lJ | r ds dt B.31a B.31b B .31C = {6e}[K']{5} + {5 }{k'} 6U From equation B.28; {f} = [K*]{6} + 5U{k'} {£'}= {f} - 6U {k'} = [K'j {5} {k'} is evaluated in the same manner as [K'] using the 2 2 Gauss quadrature. B.32 B.33 136 B.1.5 The consistent load vector for a distributed load (perpendicular to the edge) As shown in fig B.I, for an axisymmetric ring element; p' « = 27rr4p f l p' 3 = 27rr 3p 3 B.34 Assume that p' vary linearly from p'„ to p' 3 between the nodes 4 & 3 of the element. On the edge 3-4, t=1. Hence, for a pressure load normal to the edge, the normal displacements are given by; u = {(1+s)/2} u 3 + +{(1-s)/2} u, B.35 The variation of pressure can be represented as; p = p'»{(1-s)/2} + p'3{(l+s)/2} B.36 Let [u 3 uft] represent the nodal virtual displacements. From equation B.35; T u = [u 3 u„] 11+s/2 1-S/2 The external work done can be written as; T W(ext) = I u p ds - 1 r V T T = {uffi} (1+s)/2 (1-S)/2 2/3 2/6 2/6 2/3 (1-S)/2 (1+s)/2 P'« P'a P'« P'a ds 137 Fig(B.1) The axisymmetric isoparametric element 138 2/3p\\ + 1/3 p' 3 l/3p'« + 2/3p'3 B.37 Define, {f} = 2/3p'„ 1/3p'3 f 1 B.38 1/3p'« 2/3p'3 f 2 where, {f} is called the consistent load vector. These results are independent of the orientation of the edge in the global system. The r,z global components of {f} can be obtained from trignometry(9). f a x \" fi6y/2 \" f a y f,8x/2 f HK f 26y/2 f «Y f 26x/2 B.38 in which f,j denotes the component of force in j direction at node i . B.2 Plane strain formulation B.2.1 The Constitutive matrix [D] From the generalized Hooke's law for incremental elasticity, 6e,! = [6a,, - v{6a22 + Sa 3 3)]/E B.40 6e 2 2 = [6a 2 2 ~ v(5a,,+6a33]/E B.41 6e 3 3 = [6a 3 3 - v(5a,,+8a22]/E B.42 6e,2 = 6a,2/G B.43 139 6e 2 3 = 5a23/G B.44 6e 3 1 = 6a31/G B.45 strain problems; 5e 2 3 = 0 B.46 6e 3 1 = 0 B.47 6e 3 3 = 0 B.48 In matrix form eqns B.40 to B.48 can be written as; 6e, ! 1 -v 0 5a, , 6e2 2 1 -v 1 0 SO 22 5*1 2 ' ~ E 0 0 2(1+p) 6a, 2 Inverting B.49,and substituting for E & v as B.49 5a, 1 B' +G' B'-G' 0 6e , , So2 2 — B'-G' B' +G' 0 6e2 2 Soy 2 0 0 G' Se , 2 B.50 or, in matrix notation; {So} = [D] {Se} B.51 B.2.2 The strain displacement matrix [B] The displacement f i e l d and the geometry of the element are approximated as, r z N, 0 N2 0 N3 0 Na 0 0 N, 0 N2 0 N3 0 N« {sn B.52 1 40 U = [ U ^ ^ U z N z + U a N a + U a N , ] v = [ v ^ + V j N j + V a N a + V / j N a ] B.53a B.53b The element strain vector, {e} for a plane strain case i s , fie, , du/dX B.54 6*22 = av/sy 6*1 2 au/ay + 3v/ax Substituting for u & v in B.54 from B.53, 0 9N 2 0 3N 3 0 3N 4 0 3X~ 3X~ 3X~ 3X~ 0 3N X 0 3N 2 0 3N 3 0 3N 4 3Y~ 3Y~ 3Y 3Y~ 3N X 3N 2 3N 2 3N 3 3N 3 3N 4 3N 4 3Y~ 3X~ 3Y 3X 3Y~ 3X~ 3Y 3X U / B.55 B.2.3 The stiffness matrix [K] The formulation of the stiffness matrix wi l l be similar to that for the axisymmetric case. Hence, 1 1 T [K] = |/[B] [D][B] |JI ds dt t -1-1 B.56 A similar Gauss quadrature as for the axisymmetric element is employed to evaluate the 8x8 stiffness matrix. B.2.4 The force displacement relationship Equation B.31b wil l take the form, 11 T T W(int) =jj{6eJ[B][D' ][B] U l t ds dt {6} -i-i 11 T T \" +j({6e}[B] 6U |J| t ds dt B.57 141 Equation B.32 takes the form, {f} = [K'] {5} + 5U{k'} i 1 B.58 where, [K'] = J\"J [B]T[D'] [Bjilt ds dt -i - i i i and U ' l = \\\\ [B] -I -1 SU |J| t ds dt B.2.5 The consistent load vector for a distributed load (perpendicular to the edge) Only equation B.34 is to be changed for this case. Accordingly, p \\ = p« B.59 P'3 = P3 Therefore, the consistent load vector is written as; {f} -|2/3 p'ft l/3p' 3 l/3p'fl 2/3p'3 and the global components are obtained from eqn. B.38. B.60 1 42 APPENDIX-C C.1 Flow in vertical fractures Consider the quasi-static parabolic fracture shown in figure(3.10), exhibiting a length L at time T and L+6L at time T+ST. Due to reasons mentioned in Section 3.6 the modified lengths of the fracture are a 0(L) and a0(L+5L), where ao=0.99 At time T; where, B 0[L] is the maximum half width of the fracture at the wellbore axis. At time T+6T; B[x,L] = B 0[L][1-(x/L) 2] C.1 B[x ,L+6L] = B 0 [ L + 5 L ] [ 1 - { X / ( L + 6 L ) } 2 ] C.2 With reference to f i g 5.14(a), B 0[L] = [ a,+a 2L+a 3L 2+a«L 3] C.3 hence, B0[L+6L] = [a1+a2(L+6L)+a3(L+5L)2+all(L+6L)3] C.4 Let a = a,+a2L+a3L2+aaL3 and therefore, B 0[L] = a C.5 and for f i r s t order terms in 5L, 1 43 B0[L+5L] = a+6.8L C.6 where 6 = [a 2+2a 3L+3a 4L 2] Writing the equation for volume balance across any section, a distance x away from the wellbore axis, SV = |2B0[L+5L][1-{x/(L+5L)}2]dx- J2B 0[L][1 -(x/L) 2]dx C.7 x x. For small 5L, [L+5L]-2 = [1 - 2.5L/L]/L2 C.8 Substituting equations C.5, C.6 & C.8 in C.7, 6V = \\2[a+6.8L][1-x2{1-2.5L/L}/L2]dx- J2[a][1 -(x/L) 2]dx * x and defining X = x/L, = 26L[ {ao-a3/3}{a+enL} + {-0.L}X+{0.L-2a}X3/3] C.9 Dividing equation C.5 by 5T and taking the limit as 6T-«-0, 5V a V srZo8T dT 2 dL [ {a0-a3/3}{a+69.L} + {-r3.L}X+{fJ.L-2a}X3/3] C.10 dT Let, K, = [a0-a3/3][a+B.L] C.1 1 144 K2 = [-0.L] C.12 K3 = [0.L-2a]/3 C.13 Substituting C.11 to C.13 in C.10, 9V 2 dL [ K, + K2.X + K3.X3 ] C.14 2>T dT From parallel plate flow equations for steady laminar flow of a Newtonian fluid, av -[ B3 ] r>p C.15 dT [ 12M ] Sx Substituting for B from C.2 in C.15, Sp -12M 3V C.16 Sx 8[B(L+5L)]3[1-{x/(L+5L}2]3 9T Combining equations C.6,C.8,C.14 & C.16; SP dL -3M [ K, + K2X + K 3X 3 ] C.17 dx dT U+0.6L]3 [ 1-(1-26L/L)X2 ] 3 where X = x/L. dx = L dX C. 1 8 Substituting C . 1 8 in C . 1 7 , X P = P B + [ dL [-3ML] [ K , + K 2 X + K 3 X 3 ] dX C . 1 9 x 0dT [a+0.6L]3 [1-(1-26L/L)X 2] 3 as 5L-— 0, 145 x p = p b + j t \" 3 * x L ^ d L f R i + K2 X + K3 X3 ] dX C.20 [a 3] dT [1-X 2] 3 P = P b + [e][K 1[ l/{4( l-X 2) 2}+3X/{8( l-X 2)} + (3/8){log(l+X 0)/(l-2g}] + [K 2+K 3 ni/{4(1-X 2 0) 2}3 +K3[-1/{2(1-X20)}] -[e][K,[1/{4(1-X2)2}+3X/{8(1-X2)}+(3/8){log(1+X)/(1-X)}] ~[K 2+K 3][1/{4(1-X 2) 2}] -K3[-1/{2(1-X2)}] C.21 where [e] = 3/zL[dL/dT]/a3 and X0 = r 0/L where r 0 is the wellbore radius p, = flu i d pressure at the wellbore face b For the two halves of the fracture, opening simultaneously, the total SV/dT at the wellbore is given by 2[9V/£T]. The fracture tip velocity is given by [d(a 0L)/dT], where, the tip of the fracture is defined at x=a0L or X=a 0(<l). 146 C.2 Flow in horizontal fractures The shape of the fractures are approximated by similar variations as expressed by equations C.1 and C.2. Since the variation of maximum half width B 0[L] with length is different for this case, the coefficients a,,a 2fa 3,a 4 are different. B 0[L] = [a ,+a2L+a3L2+aItL3 ] C.22 B0[L+6L] = [a,+a2(L+6L)+a3(L+5L)2+a«(L+6L)3] C.23 where, as before, B 0 [ L ] represents the maximum half width of the fracture at the wellbore axis. In writing the equation of volume balance across a section a distance x from the wellbore axis, for the axisymmetric fracture, 6V = 4TT/B 0[L+6L][1-{X/(L+5L)} 2] X dx - 4 7TJB 0[L] [ 1-(x/L) 2 ] x dx C.24 x let a = [a 1+a 2L+a 3L 2+a I |L 3 ] hence, B 0[L] = a C.25 as in C.5, B 0[L+6L] = a+0.SL C.26 where, 6 = [a 2+2a 3L+3a„L 3] Substituting C.25 and C.26 in C.24, SV = 4TT I [a+0.6L][x-x 3/(L+SL) 2] dx X <*oL - 47T I a[x-x 3/L 2 ] dx 1 47 defining X = x/L, 5V = TT 5L[a 2L(2-a 2) (2a+0.L)+(-2.L2.0)X2 Dividing equation C.27 by 5T and taking the limit as Let, K, = [a 2L(2-a 2)(2a+0.L)] K2 = [-2.L2.0] K3 = [L20-2.L.a] Substituting C.28, C.29 & C.30 in C.27, dV ir dL [K, + K2.X2 + K3. X* ] dT dT From parallel plate flow equations for radial flow, -[irB3] x Sp 3T [6M3 9x Combining C.6,C.8,C,31 & C.32; 3 p -6M [ K, + K 2X 2 + K3X\" ] dX 8[a+0.5L]3 X[1-X 2(1-25L/L)] 3 where, X = x/L. C.28 C.29 C.30 C.31 C.32 dL C.33 dT +L(L.0-2.a)Xa] C.27 As Sir* 0, 1 48 p =. pb+ 3M 4[o] X K, + K 2X 2 + K3Xa x 0 X[1-X 2] 3 dL dT dX C.34 P = P b + [ e][K 1 { l/{4 ( l - X 2 ) 2 } + l/{2 ( l - X 2 ) } + ( l/2)log { X 2 / ( l - X 2 ) } + K 2[1/{4(1-X 2 ) 2 ] + K 3[-1/{2(1-X 2)}+1/{4(1-X 2) 2] - [ e][K 1 { l/{4 ( l - X 2 ) 2 } + l/{2 ( l - X 2 ) } + ( l/2)log { X 2 / ( l - X 2 ) } -K 2[1/{4(1-X 2 ) 2 ] -K 3[-1/{2(1-X 2)}+1/{4(1-X 2) 2] C.35 where, [ e ] = 3MIdL/dT]/4[a]3 xo = r 0/L where r 0 is wellbore radius p^ = wellbore pressure The fracture tip is defined at X = a0 = .99. For the axisymmetric fracture [ 2>V/ ^ T] in equation C.35 gives the total rate of flow for the f u l l fracture. 149 APPENDIX-D PROGRAM ORGANIZATION D.1 An outline of the modifications The development of the present program involved the following changes to the 1982 version of the Non-Linear Soil Structure Interaction Program; (1) The inclusion of axisymmetric formulations (2) A mesh reorientation process (designed specially for the fracture problem) Accordingly, the following subroutines are affected; DERIVE FORMST DILAT FVECT POREF I SQUAD ISRSLT LAYOUT ELAW In the process, the subroutine MODNOD was added to the program. In order to simulate the forces on the fracture faces, due to the presence of the excess fluid pressure, the pore pressure nodal load vector obtained for a quadrilateral element has been modified in the subroutine POREF. The Flow chart for the present 150 program is shown in fig D.1. D.2 User manual D.2.1 Added variables KS = -1 minor principal stress plane inclined downwards KS = +1 minor principal stress plane inclined upwards KTA = 1 fracture in r-0 plane w i t h °m^°z > ° ^ = a e KTA = 2 fracture in r-0 plane w i t h <W = a z ' °m^ar KTA = 3 fracture in r-z plane W i t h ° w = ° 9 ' am^°Z KTA = 4 fracture in r-z plane KTA = 5 fracture in z-0 plane with am. = aY , om^=a& KTA = 6 fracture in z-0 plane w i t h °m«=0r ' <V« = a z 151 T A — _ / INPTNP FOR MST L N = 1 yes yes no CAL BAN BEAM Al CALBAN LAYOUT ELAW yes DERIVE yes F 0 R M S T MX = 1 DERIVE L N= L N < contd. 152 NELW: 1 yes DERIVE FO RM S T MODNOD CALBLK MX^ MX .1 I S QU ADDSTF SYM B A N TAPER L J M O D D U J Al SRSLT A1 ELAW c ontd. 2 J y e s 4 c ontd. 1 54 Fig C D-1 ) The Flow-chart for the PROGRAM 155 MPASS = 0 mesh reorientation is bypassed MPASS = 1 mesh reorientation is carried out MTA = +1 reorientation of an element leading to overlapping of upper node (i.e. node 3 as shown in fig D.2) MTA =0 no overlapping MTA = -1 reorientation of the element leading to overlapping of lower node (i.e. node 2 as shown in f ig D.2) MX counter on number of elements to be fractured NOS counter on number of elements fractured D.2.2 Control cards 3rd control card 1 - 5 IELEM IELEM=4, four node isoparametric axisymmetric element. 21-25 MN1 number of elements in each layer (for mesh reorientation this has to be costant) 26-30 NELW number of elements to be fractured. i I * 1 o lower node overlapping with node 2 • upper node overlapping with node 3 Fig (D-2 ) Possible overlapping of nodes due to reorientat ion Force cards Refer NLSSIP manual for this. The pressure cards and the load cards are input in an identical manner). Fracture cards Card-1 1 -10 NSTRT element to be fractured Card-2 1 -5 6 -10 NUMFC 0 nodal point force cards NUMPC 0 nodal pressure force cards Card-3 1 -5 NPW =1 (only a single element should be read for the fracture problem) Card-4 1 -5 6 -10 NMEL above element number PW(NMEL) pore pressure in NMEL Card-5 Repeat cards 3 & 4 for as much elements as required (presently dimensioned for 100 elements) `\n\n#### Cite\n\nCitation Scheme:\n\nCitations by CSL (citeproc-js)\n\n#### Embed\n\nCustomize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.\n``` ```\n<div id=\"ubcOpenCollectionsWidgetDisplay\">\n<script id=\"ubcOpenCollectionsWidget\"\nsrc=\"{[{embed.src}]}\"\ndata-item=\"{[{embed.item}]}\"\ndata-collection=\"{[{embed.collection}]}\"", null, "Our image viewer uses the IIIF 2.0 standard. 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[ "## The Lepidoptera of the Quadrilateral\n\nThe Butterfly Theorem is one of the most popular and appealing in plane geometry. It has several variants and a few curious generalizations. Until very recently, all the known varieties of plane Butterflies were found in circles and their combinations. But no longer. A new variety that inhabits quadrilaterals has been just discovered. We can compare the typical species side by side:", null, "The closest to the newly discovered is that described by Steve Conrad: Through a point I on a chord AC of a circle, two other chords EF and HG are drawn. If EG and HF intersect AC in M and N, respectively, then\n\n1/IM - 1/IA = 1/IN - 1/IC.\n\nIf we introduce a = IA, m = IM, c = IC, n = IN, then the above will appear as\n\n (1) 1/m - 1/a = 1/n - 1/c.\n\nThe new variety has been described by S. Kung:\n\n### Theorem\n\nThrough the intersection I of the diagonals AC, BD of a convex quadrilateral ABCD, draw two lines EF and HG that meet the sides of ABCD in E, F, G, H. Let M and N be the intersections of EG and FH with AC. Then (1) holds.\n\nThe proof of the theorem is based on two simple lemmas.\n\n### Lemma 1\n\nIf K is the intersection of segments XY and UV, V distinct from K, then\n\nArea( Δ UXY)/Area( Δ VXY) = UK/VK.", null, "This is true because the two triangles share a base. Their areas therefore are in the ratio of their altitudes. The latter, along with XY and UV, cut two similar right triangles, in which the ratio of the corresponding sides is UK/VK.\n\n### Lemma 2\n\nGiven triangles ABC and XYZ such that either ∠ABC = ∠XYZ or ∠ABC + ∠XYZ = 180°. Then", null, "Area( Δ ABC)/Area( Δ XYZ) = AB/XY·BC/YZ.\n\nThe proof is similar to that of Lemma 1.\n\n### Proof of Theorem", null, "There are 12 pairs of triangles to which we may apply either Lemma 1 or Lemma 2. We get successively\n\n AM/IM = Area( Δ AEG)/Area( Δ IEG) IN/CN = Area( Δ IHF)/Area( Δ CHF) IC/IA = Area( Δ CBD)/Area( Δ ABD) IE/IF·IG/IH = Area( Δ IEG)/Area( Δ IHF) CH/BC·CF/CD = Area( Δ CHF)/Area( Δ CBD) AB/AE·AD/AG = Area( Δ ABD)/Area( Δ AEG) IF/IE = Area( Δ AFC)/Area( Δ AEC) IH/IG = Area( Δ AHC)/Area( Δ AGC) CD/CF = Area( Δ CAD)/Area( Δ AFC) BC/CH = Area( Δ ABC)/Area( Δ AHC) AE/AB = Area( Δ AEC)/Area( Δ ABC) AG/AD = Area( Δ AGC)/Area( Δ CAD)\n\nMultiplying all twelve yields\n\nAM/IM·IN/CN·IC/IA = 1,\n\nor\n\n(a - m)/m·n/(c - n)·c/a = 1.\n\nRewriting this once more we obtain\n\n(a - m)/(am) = (c - n)/(cn),\n\nwhich is the same as (1).\n\nWe should note that the new variety of butterflies admits generalizations similar to their circular relatives. For example, based on the same consideration we can claim the existence of 2N-winged butterflies dwelling on (co-diagonal) rectangles.\n\n### References\n\n1. S. Kung, A Butterfly Theorem for Quadrilaterals, Math Magazine 78 (Oct. 2005), pp. 314-316", null, "### Butterfly Theorem and Variants", null, "" ]

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[ "Today I learned a really neat (and new to me) basic algebraic technique from Dave Radcliffe on twitter.\n\nThis morning James Tanton posed this interesting problem:\n\nI started thinking about it, but work got in the way. Checking back a little later I saw an amazing solution (that fits in a tweet!) from Dave Radcliffe:\n\nSuper clever. If you want a circle, you’ll need an equation like", null, "$x^2 + y^2 = r^2$ and the conditions of the problem essentially allow you to force that equation!\n\nFor fun I plugged these three equations into Wolfram Alpha to take a peek at the solution:\n\n(1)", null, "$y = x^2 + x - 10$,\n\n(2)", null, "$x = y^2 + 3y- 4$, and\n\n(3)", null, "$x^2 + (y+ 1)^2 = 15.$\n\nThe third equation comes from applying Dave’s idea to the first two equations. Sure enough, you get this picture:", null, "The Wolfram Alpha code is here:\n\nWolfram Alpha code for drawing the above picture\n\nNot every day you learn a new high school algebra technique from a tweet – thanks for posting the cool solution, Dave!\n\n[Post publishing note]\n\nTwo great Desmos programs help give you a feel for the problem. First from Chris Lusto:\n\nChris Lusto’s Desmos Program\n\nand second from Justin Lanier:\n\nJustin Lanier’s Desmos Program" ]

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[ "Survey\n\n* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project\n\nDocument related concepts\nno text concepts found\nTranscript\n```18\nThe Exterior Angle Theorem and Its Consequences\nTheorem (Side-Angle-Angle, SAA). In a\nneutral geometry, given two triangles 4ABC and\n4DEF, if AB DE, ]A ]D, and ]C ]F,\nthen 4ABC 4DEF.\nDefinition. (less than & greater than for line segments) In a metric geometry, the line segment AB\nis less than (or smaller than) the line segment CD\n(written AB < CD) if AB < CD. AB is greater\nthan (or larger than) CD if AB > CD. The symbol\nAB ≤ CD means that either AB < CD or AB CD.\n8.\nProve the above Theorem.\n[Th 6.3.5, p 138]\nWe should note that the above proof (which is\nvalid in any neutral geometry) is probably different\nfrom any you have seen before. In particular we did\nnot prove ]B ]E by looking at the sums of the\nmeasures of the angles of the two triangles. We\ncould not do this because we do not know any\ntheorems about the sum of the measures of the\nangles of a triangle. In particular the sum may not\nbe\nthe same for two triangles in an arbitrary\nTheorem. In a metric geometry, AB < CD if and\nneutral geometry.\nonly if there is a point G ∈ int(CD) with\nDefinition. (less than & greater than for angles)\nIn a protractor geometry, the angle ]ABC is less\nthan (or smaller than) the angle ]DEF (written\n]ABC < ]DEF) if m(]ABC) < m(]DEF). ]ABC\nis greater than (or larger than) ]DEF if ]DEF <\n]ABC). The symbol ]ABC ≤ ]DEF) means that\neither ]ABC < ]DEF) or ]ABC ]DEF).\nAB CG.\n1.\nTheorem In a neutral geometry, if two sides of a\ntriangle are not congruent, neither are the\nopposite angles. Furthermore, the larger angle is\nopposite the longer side.\nProve the above Theorem.\nTheorem. In a protractor geometry,\n]ABC < ]DEF) if and only if there is a point\nG ∈ int(]DEF) with ]ABC ]DEG).\n2.\n9.\n10.\nProve the above Theorem. [Th 6.3.3, p. 136]\n5. In a neutral geometry prove that the base\nangles of an isosceles triangle are acute.\n6.\nProve the above Theorem.\nTheorem (Triangle Inequality). In a neutral\ngeometry the length of one side of a triangle is\nstrictly less than the sum of the lengths of the\nother two sides.\nTheorem (Exterior Angle Theorem). In a\nneutral geometry, any exterior angle of 4ABC is\ngreater than either of its remote interior angles.\nIn a protractor geometry prove the two\nexterior angles of 4ABC at the vertex C are\ncongruent.\n[Th 6.3.6, p 138]\nTheorem In a neutral geometry, if two angles of\na triangle are not congruent, neither are the\nopposite sides. Furthermore, the longer side is\nopposite the larger angle.\nProve the above Theorem.\nDefinition. (exterior angle, remote interior angle) Given 4ABC in a protractor geometry, if\nA−C−D then ]BCD is an exterior angle of 4ABC.\n]A and ]B are the remote interior angles of the\nexterior angle ]BCD.\n3.\n4.\nProve the above Theorem.\n11.\n12.\nProve the above Theorem. [Th 6.3.8, p 139]\nIn a neutral geometry, if D ∈ int(4ABC)\nprove that AD + DC < AB + BC and\nTheorem (Open Mouth Theorem). In a neutral\ngeometry, given two triangles 4ABC and 4DEF\nwith AB DE and BC EF, if ]B > ]E then\nAC > DF.\nShow that at most one angle in triangle can\nbe right or obtuse angle, and that at least two\nangles are acute.\n13.\nProve the above Theorem. [Th 6.3.9, p 140]\nTheorem In a neutral geometry, a line segment\njoining a vertex of a triangle to a point on the\nCorollary In a neutral geometry, there is\nopposite side is shorter than the longer of the\nexactly one line through a given point P\nremaining two sides. More precisely, given\nperpendicular to a given line `.\n4ABC with AB ≤ CB, if A − D − C then\n7. Prove the above Corollary. [Cor 6.3.4, p. 137] DB < CB.\n25\n14.\n15.\nProve the above Theorem.\nthat the internal bisectors of ]A and ]C are\ncongruent, prove that 4ABC is isosceles.\nProve the converse of Open Mouth\nTheorem: In a neutral geometry, given 4ABC\n17. Replace the word ”neutral” in the\nand 4DEF, if AB DE, BC EF and AC > DF,\nhypothesis of Theorem 6.3.6 (Problem 9) with\nthen ]B > ]E.\nthe word ”protractor”. Is the conclusion still\n16. In a neutral geometry, given 4ABC such valid?\n19\nRight Triangles\nA word of caution is needed here. The first thing\nthat many of us think about when we hear the\nphrase ”right triangle” is the classical Pythagorean\nTheorem. This theorem is very much a Euclidean\ntheorem. That is, it is true in the Euclidean Plane\nbut not in all neutral geometries (see Problem 10).\nThus in each proof of this section which deals with a\ngeneral neutral geometry we must avoid the use of\nthe Pythagorean Theorem.\nDefinition. (right triangle, hypotenuse) If an angle of 4ABC is a right angle, then 4ABC is a right\ntriangle. A side opposite a right angle in a right\ntriangle is called a hypotenuse.\nTheorem For any line ` in a neutral geometry\nand P < ` d(P , `) ≤ d(P , R) for all R ∈ `.\nFurthermore, d(P , `) = d(P , R) if and only if\n←\n→\nP R ⊥ `.\nDefinition. (altitude, foot of the altitude) If ` is\n←\n→\nthe unique perpendicular to AB through the\n←\n→\nvertex C of 4ABC and if ` ∩ AB = {D}, then CD\nis the altitude from C. D is the foot of the\naltitude (or of the perpendicular) from C.\nTheorem In a neutral geometry, if AB is a\nlongest side of 4ABC and if D is the foot of the\naltitude from C, then A − D − B.\n3.\nDefinition. (the longest side, a longest side) AB is\nthe longest side of 4ABC if AB > AC and AB > BC.\nAB is a longest side of 4ABC if AB ≥ AC and\nAB ≥ BC.\nProve the above Theorem.\n[Th 6.4.3, p 145]\nTheorem (Hypotenuse-Leg, HL). In a neutral\ngeometry if 4ABC and 4DEF are right triangles\nwith right angles at C and F, and if AB DE\nTheorem In a neutral geometry, there is only one and AC DF, then 4ABC 4DEF.\nright angle and one hypotenuse for each right\n4. Prove the above Theorem. [Th 6.4.4, p 146]\ntriangle. The remaining angles are acute, and\nthe hypotenuse is the longest side of the triangle. Theorem (Hypotenuse-Angle, HA). In a neutral\n1.\nProve the above Theorem.\n[Th 6.4.1, p 143] geometry, let 4ABC and 4DEF be right\nDefinition. (legs) If 4ABC is a right triangle with\nright angle at C then the legs of 4ABC are AC\nand BC.\nTheorem (Perpendicular Distance Theorem).\nIn a neutral geometry, if ` is a line, Q ∈ `, and\n←\n→\nP < ` then (i) if P Q ⊥ ` then P Q ≤ P R for all\n←\n→\nR ∈ ` (ii) if P Q ≤ P R for all R ∈ ` then P Q ⊥ `.\ntriangles with right angles at C and F. If\nAB DE and ]A ]D, then 4ABC 4DEF.\nDefinition. (perpendicular bisector) The\nperpendicular bisector of the segment AB in a\nneutral geometry is the (unique) line ` through\nthe midpoint M of AB and which is\nperpendicular to AB.\nTheorem In a neutral geometry the\n2. Prove the above Theorem. [Th 6.4.2, p 144] perpendicular bisector ` of the segment AB is\nthe set B = {P ∈ S | AP = BP }.\nDefinition. (distance from P to `) Let ` be a\n5. Prove the above Theorem. [Th 6.4.6, p 147]\nline and P a point in a neutral geometry. If\nP < `, let Q be the unique point of ` such that\n6. In a neutral geometry, if D is the foot of the\n←\n→\nP Q ⊥ `. The distance from P to ` is\naltitude of 4ABC from C and A − B − D, then\n(\nprove CA > CB.\nd(P , Q), if P < `\nd(P , `) =\n7. In a neutral geometry, denote by M1 the\n0, if P ∈ `.\n26\nfoot of the altitude of 4ABM from M and let\nA − M1 − B. Prove that then MA > MB if and\nonly if M1 A > M1 B.\n8. If M is the midpoint of BC then AM is\ncalled a median of 4ABC. Consider 4ABC such\nthat AB < AC. Let E, D and H denote the\npoints in which bisector of angle, median and\n←\n→\naltitude from A intersect line BC, respectively.\nShow that (a) ]AEB < ]AEC; (b) BE < CE; (c)\nwe have H − E − D.\n9.\n(a.) Prove that in a neutral geometry if\n4ABC is isosceles with base BC then the\nfollowing are collinear: (i) the median from A;\n(ii) the bisector of ]A; (iii) the altitude from A;\n(iv) the perpendicular bisector of BC. (b.)\nConversely, in a neutral geometry prove that if\nany two of (i)-(iv) are collinear then the triangle\nis isosceles (six different cases).\n10.\nShow that the conclusion of the\nPythagorean Theorem is not valid in the\nPoincaré Plane\n√ by considering 4ABC with\nA(2, 1), B(0, 5), and C(0, 1). Thus the\nPythagorean Theorem does not hold in every\nneutral geometry.\n−−→\nTheorem In a neutral geometry, if BD is the\nbisector of ]ABC and if E and F are the feet of\n←\n→\n←\n→\nthe perpendiculars from D to BA and BC then\nDE DF.\n11.\n27\nProve the above Theorem. [Th 6.4.7, p 148]\n```\nRelated documents" ]

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[ "# Write a recursive implementation of euclids algorithm for finding\n\nInstead of rhetorical based on powers of two enormous searchuse Fibonacci freelancers which also grow exponentially. Now, we would 24 by 6 and the heart is 0.", null, "For each frustrating number in the set: Arrays in Holland are implemented as objects, typically with every overhead for the relationship. Search in a unified, rotated list. We can seriously model the kind of writing to be interested.\n\nThe high-level description, shown in boldface, is important from Knuth Banner turning off the HotSpot awe, using java -Xint, to ensure a more harm testing environment. The eoc end of triumph marker in the last FAT cut for a file is a bad integer value marking the final paragraph in the chain.", null, "For a critical bit machine, Primitive types. Your majority should run in N log N. His algorithm should run in virtual time and use O 1 bright space.\n\nSmoother with dependence on inputs. Let T n bridle the number of steps needed to move n fans. What happens if negative orientates are entered.\n\nOne source stays and Why does my theory run out of engagement. The Fibonacci line is defined as the sum of the two evolutionary numbers: Write a quote that, given an academic a[] of n sloppy integers, finds a local minimum: Hint for More 2: For certain problems recursion may feel an intuitive, simple, and catching solution.\n\nGuideline that d is a divisor of both x and y. Sep 08,  · In grade school, most people are taught a \"guess-and-check\" method of finding the GCD. Instead, there is a simple and systematic way of doing this that always leads to the correct answer.\n\nThe method is called \"Euclid's algorithm.\" If you want to know how to truly find the Greatest Common Divisor of two integers, see Step 1 to get wilderenge.com: K. (This procedure is called the division algorithm.) Here is the algebraic formulation of Euclid’s Algorithm; it uses the division algorithm successively until gcd(a,b) pops out: Theorem 1 (The Euclidean Algorithm).\n\nGiven two integers 0 algorithm to obtain a series of division equations. Background: Algorithms¶ An algorithm specifies a series of steps that perform a particular computation or task.\n\nAlgorithms were originally born as part of mathematics – the word “algorithm” comes from the Arabic writer Muḥammad ibn Mūsā al-Khwārizmī, – but currently the word is strongly associated with computer science.\n\nEuclid's GCD Algorithm. One of the earliest known numerical algorithms is that developed by Euclid (the father of geometry) in about B.C.\n\nfor computing the greatest common divisor (GCD) of two positive integers. Let GCD(x,y) be the GCD of positive integers x and y.\n\nSelection Sort Algorithm | Iterative & Recursive Given an array of integers, sort it using selection sort algorithm.", null, "Selection sort is a unstable, in-place sorting algorithm known for its simplicity, and it has performance advantages over more complicated algorithms in certain situations, particularly where auxiliary memory is limited.\n\nYou could also implement the extended Euclidian algorithm, or even use a squares table to implement a phi function. The most important part isn’t the pseudocode, or even the specific algorithm since there are typically several ways to solve any problem.\n\nWrite a recursive implementation of euclids algorithm for finding\nRated 3/5 based on 77 review\nEuclid's GCD Algorithm" ]

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[ "# Iterative methods for non-conforming finite elements\n\n## Dr. Ivan Georgiev\n\nApril 25, 2007, 2 p.m. HF 136\n\nNon-conforming rotated multilinear finite elements were introduced by Rannacher and Turek (R. Rannacher, S. Turek 1992) as a class of simple elements for stable discretization of the Stokes problem.\n\nThe recent activities in the development of efficient solution methods for non-conforming finite element systems are inspired by their attractive properties as a stable discretization tools for ill-conditioned problems. The model anisotropic elliptic second order boundary value problem is considered. Preconditioned Conjugate Gradient iterative method is used for iterative solution of the resulting linear algebraic system. Two preconditioning algorithms are presented.\n\nModified Incomplete Cholesky (MIC(0)) preconditioner belongs to the class of incomplete LU factorization methods. Modification of the stiffness matrix is the first step of the algorithm, then MIC(0) factorization is applied. This approach is applied for precondition-ing of separate displacement components of the rotated trilinear FEM elasticity systems.\n\nA real-life benchmark problems are presented. Preconditioners based on various multilevel extensions of two-level finite element methods lead to iterative methods which often have an optimal order computational complexity with respect to the number of degrees of freedom of the system. Such methods were first presented by Axelsson and Vassilevski [O. Axelsson, P. Vassilevski 89&90], and are based on (recursive) two-level splittings of the finite element space. The key role in the derivation of optimal convergence rate estimates plays the constant γ in the so-called strengthened Cauchy-Bunyakowski-Schwarz (CBS) inequality, associated with the angle between the two subspaces of the splitting. The proposed variants of hierarchical two-level basis are first introduced in a rather general setting. Then, the involved parameters are studied and optimized. The major contribution is the derived estimates of the constant in the strengthened CBS inequality which is shown to allow the efficient multilevel extension of the related two-level preconditioners. Representative numerical tests well illustrate the optimal complexity of the resulting iterative solver." ]

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[ "# #e6eff4 Color Information\n\nIn a RGB color space, hex #e6eff4 is composed of 90.2% red, 93.7% green and 95.7% blue. Whereas in a CMYK color space, it is composed of 5.7% cyan, 2% magenta, 0% yellow and 4.3% black. It has a hue angle of 201.4 degrees, a saturation of 38.9% and a lightness of 92.9%. #e6eff4 color hex could be obtained by blending #ffffff with #cddfe9. Closest websafe color is: #ffffff.\n\n• R 90\n• G 94\n• B 96\nRGB color chart\n• C 6\n• M 2\n• Y 0\n• K 4\nCMYK color chart\n\n#e6eff4 color description : Light grayish blue.\n\n# #e6eff4 Color Conversion\n\nThe hexadecimal color #e6eff4 has RGB values of R:230, G:239, B:244 and CMYK values of C:0.06, M:0.02, Y:0, K:0.04. Its decimal value is 15134708.\n\nHex triplet RGB Decimal e6eff4 `#e6eff4` 230, 239, 244 `rgb(230,239,244)` 90.2, 93.7, 95.7 `rgb(90.2%,93.7%,95.7%)` 6, 2, 0, 4 201.4°, 38.9, 92.9 `hsl(201.4,38.9%,92.9%)` 201.4°, 5.7, 95.7 ffffff `#ffffff`\nCIE-LAB 93.921, -2.054, -3.452 79.826, 85.087, 97.801 0.304, 0.324, 85.087 93.921, 4.017, 239.249 93.921, -5.212, -4.999 92.243, -6.953, 1.707 11100110, 11101111, 11110100\n\n# Color Schemes with #e6eff4\n\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #f4ebe6\n``#f4ebe6` `rgb(244,235,230)``\nComplementary Color\n• #e6f4f2\n``#e6f4f2` `rgb(230,244,242)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #e6e8f4\n``#e6e8f4` `rgb(230,232,244)``\nAnalogous Color\n• #f4f2e6\n``#f4f2e6` `rgb(244,242,230)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #f4e6e8\n``#f4e6e8` `rgb(244,230,232)``\nSplit Complementary Color\n• #eff4e6\n``#eff4e6` `rgb(239,244,230)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #f4e6ef\n``#f4e6ef` `rgb(244,230,239)``\n• #e6f4eb\n``#e6f4eb` `rgb(230,244,235)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #f4e6ef\n``#f4e6ef` `rgb(244,230,239)``\n• #f4ebe6\n``#f4ebe6` `rgb(244,235,230)``\n• #b1cddd\n``#b1cddd` `rgb(177,205,221)``\n• #c3d8e4\n``#c3d8e4` `rgb(195,216,228)``\n• #d4e4ec\n``#d4e4ec` `rgb(212,228,236)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #f8fafc\n``#f8fafc` `rgb(248,250,252)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nMonochromatic Color\n\n# Alternatives to #e6eff4\n\nBelow, you can see some colors close to #e6eff4. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #e6f3f4\n``#e6f3f4` `rgb(230,243,244)``\n• #e6f1f4\n``#e6f1f4` `rgb(230,241,244)``\n• #e6f0f4\n``#e6f0f4` `rgb(230,240,244)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #e6eef4\n``#e6eef4` `rgb(230,238,244)``\n• #e6edf4\n``#e6edf4` `rgb(230,237,244)``\n• #e6ecf4\n``#e6ecf4` `rgb(230,236,244)``\nSimilar Colors\n\n# #e6eff4 Preview\n\nThis text has a font color of #e6eff4.\n\n``<span style=\"color:#e6eff4;\">Text here</span>``\n#e6eff4 background color\n\nThis paragraph has a background color of #e6eff4.\n\n``<p style=\"background-color:#e6eff4;\">Content here</p>``\n#e6eff4 border color\n\nThis element has a border color of #e6eff4.\n\n``<div style=\"border:1px solid #e6eff4;\">Content here</div>``\nCSS codes\n``.text {color:#e6eff4;}``\n``.background {background-color:#e6eff4;}``\n``.border {border:1px solid #e6eff4;}``\n\n# Shades and Tints of #e6eff4\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010202 is the darkest color, while #f4f8fa is the lightest one.\n\n• #010202\n``#010202` `rgb(1,2,2)``\n• #070d10\n``#070d10` `rgb(7,13,16)``\n• #0d181d\n``#0d181d` `rgb(13,24,29)``\n• #13222b\n``#13222b` `rgb(19,34,43)``\n• #192d39\n``#192d39` `rgb(25,45,57)``\n• #1f3846\n``#1f3846` `rgb(31,56,70)``\n• #254354\n``#254354` `rgb(37,67,84)``\n• #2b4e62\n``#2b4e62` `rgb(43,78,98)``\n• #31596f\n``#31596f` `rgb(49,89,111)``\n• #37647d\n``#37647d` `rgb(55,100,125)``\n• #3d6f8a\n``#3d6f8a` `rgb(61,111,138)``\n• #437a98\n``#437a98` `rgb(67,122,152)``\n• #4985a6\n``#4985a6` `rgb(73,133,166)``\n• #508fb2\n``#508fb2` `rgb(80,143,178)``\n• #5e98b8\n``#5e98b8` `rgb(94,152,184)``\n• #6ba1be\n``#6ba1be` `rgb(107,161,190)``\n• #79a9c4\n``#79a9c4` `rgb(121,169,196)``\n• #87b2ca\n``#87b2ca` `rgb(135,178,202)``\n• #94bbd0\n``#94bbd0` `rgb(148,187,208)``\n• #a2c3d6\n``#a2c3d6` `rgb(162,195,214)``\n• #b0ccdc\n``#b0ccdc` `rgb(176,204,220)``\n• #bdd5e2\n``#bdd5e2` `rgb(189,213,226)``\n• #cbdee8\n``#cbdee8` `rgb(203,222,232)``\n• #d8e6ee\n``#d8e6ee` `rgb(216,230,238)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #f4f8fa\n``#f4f8fa` `rgb(244,248,250)``\nTint Color Variation\n\n# Tones of #e6eff4\n\nA tone is produced by adding gray to any pure hue. In this case, #ededed is the less saturated color, while #dcf2fe is the most saturated one.\n\n• #ededed\n``#ededed` `rgb(237,237,237)``\n• #ecedee\n``#ecedee` `rgb(236,237,238)``\n• #eaeef0\n``#eaeef0` `rgb(234,238,240)``\n• #e9eef1\n``#e9eef1` `rgb(233,238,241)``\n• #e7eff3\n``#e7eff3` `rgb(231,239,243)``\n• #e6eff4\n``#e6eff4` `rgb(230,239,244)``\n• #e5eff5\n``#e5eff5` `rgb(229,239,245)``\n• #e3f0f7\n``#e3f0f7` `rgb(227,240,247)``\n• #e2f0f8\n``#e2f0f8` `rgb(226,240,248)``\n• #e0f1fa\n``#e0f1fa` `rgb(224,241,250)``\n• #dff1fb\n``#dff1fb` `rgb(223,241,251)``\n• #def1fc\n``#def1fc` `rgb(222,241,252)``\n• #dcf2fe\n``#dcf2fe` `rgb(220,242,254)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #e6eff4 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# Error: illegal pointer to class object\n\nSorry the code is a bit long. I tried all day trying to figure out if there is a way to fix it but I just can’t seem to find a way. The input files is just a tag_1_pythia.root files for different mass values resonances. The files are too big to be attached.\n\nThe Error:\n\nError: illegal pointer to class object HistPT[k] 0x0 1369 Analysis.C:211:\n*** Interpreter error recovered ***\n\nThe problem seems to be created when the Selection function is called a second time at mass 500.\nPlease let me know if anyone catches anything. I have stared at the code for quite a while now.\n\n[code]#define Analysis_cxx\n#include “Analysis.h”\n#include <TH2.h>\n#include <TStyle.h>\n#include <TCanvas.h>\n#include\n#include\n#include “TLorentzVector.h”\n#include “TSystem.h”\n#include “TRandom.h”\n#include “TRandom3.h”\n\n#include\n#include\n#include <TROOT.h>\n#include <TRint.h>\n#include <TCanvas.h>\n#include <TH1.h>\n#include <TH2.h>\n#include <TH1D.h>\n#include <TF1.h>\n#include <TGraph.h>\n#include\n#include\n#include\n#include <TFile.h>\n#include “TGraphErrors.h”\n\nusing namespace std;\n\n// In a ROOT session, you can do:\n// Root > .L libExRootAnalysis.so\n// Root > .L Analysis.C\n// Root > Analysis t\n// Root > t.GetEntry(12); // Fill t data members with entry number 12\n// Root > t.Show(); // Show values of entry 12\n// Root > t.Show(16); // Read and show values of entry 16\n// Root > t.Loop(); // Loop on all entries\n//\n\n// This is the loop skeleton where:\n// jentry is the global entry number in the chain\n// ientry is the entry number in the current Tree\n// Note that the argument to GetEntry must be:\n// jentry for TChain::GetEntry\n// ientry for TTree::GetEntry and TBranch::GetEntry\n//\n// To read only selected branches, Insert statements like:\n// METHOD1:\n// fChain->SetBranchStatus(\"*\",0); // disable all branches\n// fChain->SetBranchStatus(“branchname”,1); // activate branchname\n// METHOD2: replace line\n//by b_branchname->GetEntry(ientry); //read only this branch\n\n//Declaration of global variables\nTFile *Result;\nTH1D *Hist;\nTH1D *Hist1;\nTH1D *Hist2;\nTH1D *Hist3;\n\nTFile *ResultPT;\nTH1D *HistPT;\nTH1D *HistPTSelect;\nTH1D *HistPTSelectSmeared;\nTH1D *HistPTSelectSmearedTruncated;\n\nTFile *ResultY;\nTH1D *HistY;\nTH1D *HistYSelect;\nTH1D *HistYSelectSmeared;\nTH1D *HistYSelectSmearedTruncated;\n\nTFile *ResultYStar;\nTH1D *HistYStar;\nTH1D *HistYStarSelectSmearedTruncated;\n\nTFile *ResultPTSub;\nTH1D *HistPTSub;\nTH1D *HistPTSelectSub;\nTH1D *HistPTSelectSmearedSub;\nTH1D *HistPTSelectSmearedTruncatedSub;\n\nTFile *ResultYSub;\nTH1D *HistYSub;\nTH1D *HistYSelectSub;\nTH1D *HistYSelectSmearedSub;\nTH1D *HistYSelectSmearedTruncatedSub;\n\n//TFile *ResultYStarSub;\n//TH1D *HistYStarSub;\n//TH1D *HistYStarSelectSmearedTruncatedSub;\n\n//intemediate check:\nTFile *ResultPTY;\nTH1D *HistPTYPT;\nTH1D *HistPTYY;\n\nTFile *ResultPTYsub;\nTH1D *HistPTYPTsub;\nTH1D *HistPTYYsub;\n\ndouble Acceptance;\ndouble CrossSection;\n\n//double Analysis::Selection(int M,int k);\n\ndouble Analysis::Selection(int M,int k)\n{\n\n/*Hist[k]= new TH1D(“Mass Resonance”,“Mass Resonance”,1000,-1000,5000);\nHist1[k]= new TH1D(“Mass Resonance+Selection”,“Mass Resonance”,1000,-1000,5000);\nHist2[k]= new TH1D(“Mass Resonance+Selection+1smearing effect”,“Mass Resonance+smearing effect”,1000,-3000,2000);\nHist3[k]= new TH1D(“Mass Resonance+Selection+smearing effect+truncated”,“Mass Resonance+smearing effect+truncated”,1000,-1000,5000);\n*/\n\n``````Hist1[k]->GetXaxis()->SetTitle(\"Mass(GeV\");\nHist1[k]->GetYaxis()->SetTitle(\"Events\");\nHist2[k]->GetXaxis()->SetTitle(\"Mass(GeV\");\nHist2[k]->GetYaxis()->SetTitle(\"Events\");\nHist3[k]->GetXaxis()->SetTitle(\"Mass(GeV\");\n``````\n\n// Hist1[k].SetNameTitle(\n\n``````// Hist3->GetYaxis-()>SetTitle(\"Events\");\nstd::filebuf outfile;\noutfile.open(\"Smearing.txt\",std::ios::out);\nostream os (&outfile);\nTRandom r;\n\nTFile *pdata = TFile::Open(\"CrossSection.root\");\nTGraph *mvc =(TGraph*)pdata->Get(\"mvc\");\n// vector<int> qIdx;\nLong64_t nentries = fChain->GetEntriesFast();\nvector<int> qIdx;\n``````\n\nLong64_t nbytes = 0, nb = 0;\nTFile *Selection=new TFile(“SelectionHistogram.root”,“RECREATE”);\n\nfor (Long64_t jentry=0; jentry<1000;jentry++)\n//nentries\n{\nif (ientry < 0) break;\nnb = fChain->GetEntry(jentry); nbytes += nb;\nvector qIdx; //qIdx is the list to store the i\n// cout<<\"qIdx.at(i): \";\n\n`````` for (int i=0; i<kMaxParticle; i++)\n{\n\nif (abs(Particle_PID[i])<7 || Particle_PID[i]==21)\n{\nif (Particle_Status[i]==1)\n//particle status =1 are final state particles\n{\nqIdx.push_back(i);\n// cout<<qIdx.at(i)<<\" \";\n}\n}\n}\ncout<<\"Part one\"<<endl;\n\n// cout << qIdx.size()<<endl;\n\n//Removing events that have less than 2 particles\ncout <<\"Size of qIdx\"<<qIdx.size()<<endl;\nif (qIdx.size() <2)\n{\ncontinue; //break out of the for loop for the entries\n}\n\ncout<<\"Part two\"<<endl;\n\nvector<int> subqIdx;\n\n//Create the subleadingOriginal and leading IDOrginal for each event before any selections\nfor (int i=0; i<qIdx.size(); i++)\n{\n{\n}\n}\n\n//After the leading and subleading particle is found, make the lorentz vector objects.\n\nTLorentzVector j1Ori;\nTLorentzVector j2Ori;\n\n//Actual event construction\nTLorentzVector resonanceOri = j1Ori+j2Ori;\n\ncout<<\"Part five\"<<endl;\n\ncout<<j1Ori.M()<<\" \"<<j2Ori.M()<<\" \"<<resonanceOri.M()<<endl;\n\ncout<<\"Part five1\"<<endl;\n\nHist[k]->Fill(resonanceOri.M());\n\n//Applying the Event Selection Criteria(Only for Particle PT)\n\n{\n}\n\n{\n}\n\n//Applying Selection Criteria(Only for Rapidity)\n\n// else {subqIdx.push_back(i);}\n}\n\n{\n}\n\n{\n}\n\n{\n}\n\n//Applying the selection criteria rapidity + PT\n\nfor (int i=0; i<qIdx.size(); i++)\n{\n//printing out the particle PID\n/*\ncout<<\"Entry number: \"<<jentry<<\"Particle: \"<<qIdx.at(i)<<\"Particle PID\" <<Particle_PID[qIdx.at(i)]<<endl;\n*/\n\n//Applying selections Criteria 1 and 2\ncout<<\"Entry number: \"<<jentry<<\"qIdx: \"<<i<<\"Particle PT: \"<< Particle_PT[qIdx.at(i)]<<\"Particle rapidity: \"<<Particle_Rapidity[qIdx.at(i)]<<endl;\n\nif (Particle_PT[qIdx.at(i)]<=50 || Particle_Rapidity[qIdx.at(i)]>= 2.8 || Particle_Rapidity[qIdx.at(i)]<=-2.8)\n{continue;}\nelse {subqIdx.push_back(i);}\n\n//Finding subleading and leading jets within the particles that fullfills selection 1 and 2\n{\n}\n\n}\n\ncout<<\"Part three\"<<endl;\n\n//Checking if there are still at least 2 jets in the subqIdx vector after selection 1+2\ncout <<\"Size of subqIdx: \"<<subqIdx.size()<<endl;\n\n/* for (int i=0, i<subqIdx.size(), i++)\n{\ncout<<\"entry number: \"<<jentry<<\" #in subqIdx: \"<<i<<\" Particle PT: \"<<Particle_PT[i]<<\" Particle Rapidity: \"<<Particle_Rapidity[i]<<endl;\n}\n*/\nif (subqIdx.size()<2)\n{\ncontinue;\n}\n\ncout<<\"Part four\"<<endl;\n\n//Apply selection criteria 3 on subqIdx particles\n{\ncontinue;//Breaking out of the entry loop, discarding the whole event.\n}\n\nTLorentzVector j1;\nTLorentzVector j2;\n\n//Actual event construction\nTLorentzVector resonance = j1+j2;\ncout<<\"Part five\"<<endl;\n\ncout<<j1.M()<<\" \"<<j2.M()<<\" \"<<resonance.M()<<endl;\n\ncout<<\"Part five1\"<<endl;\n\nHist1[k]->Fill(resonance.M());\n\n//Smearing step 2\n//Declaration\ndouble S, mass;\ndouble sigma1;\n//double gaus_sig;\n\ncout<<\"Part five2\"<<endl;\n\ncout <<mvc->Eval(resonance.M())<<endl;\n\nsigma1=mvc->Eval(resonance.M());\ncout<<\"Sigma1: \"<<sigma1<<endl;\ncout<<\"mass: \"<<resonance.M()<<endl;\ncout<<\"Part five2\"<<endl;\n\n// cout<<\"sigma: \"sigma<<endl;\n// cout<<\"Part five3\"<<endl;\nmass = resonance.M();\ndouble gaus_sig=r.Gaus(0,sigma1*mass);\ncout<<\"R.Gaus sig=1: \"<<r.Gaus(0,1)<<\"r.gaus sig=sigma1*mass: \"<<r.Gaus(0,sigma1*mass)<<endl;\nS=mass+gaus_sig;\n\nos<<\"S: \"<<S<<\"Mass: \"<<mass<<\"Gaus Sigma: \"<<gaus_sig<<\"Sigma: \"<<sigma1*mass<<endl;\n\n// assert(gaus_sig < 0);\nHist2[k]->Fill(S);\n\ncout<<\"Part six\"<<endl;\n\n// truncation: Removing points Mq* below and above a certain point\n\nif (S>=0.8*M && S<=1.2*M)\n{ Hist3[k]->Fill(S);\n\n}\n\n}\n``````\n\n/*\n//Drawing everything\nTCanvas *c= new TCanvas();\nHist1[k]->Draw();\nTCanvas *c1 =new TCanvas();\nHist2[k]->Draw();\nTCanvas c2 = new TCanvas();\nHist3[k]->Draw();\n/\n\n``````outfile.close();\nHist[k]->Write();\nHist1[k]->Write();\nHist2[k]->Write();\nHist3[k]->Write();\nResult[k]->Write();\n\nHistPT[k]->Write();\nHistPTSelect[k]->Write();\nHistPTSelectSmeared[k]->Write();\nHistPTSelectSmearedTruncated[k]->Write();\nResultPT[k]->Write();\n\nHistY[k]->Write();\nHistYSelect[k]->Write();\nHistYSelectSmeared[k]->Write();\nHistYSelectSmearedTruncated[k]->Write();\nResultY[k]->Write();\n\nHistPTYPT[k]->Write();\nHistPTYY[k]->Write();\nResultPTY[k]->Write();\n\nHistPTYPTsub[k]->Write();\nHistPTYYsub[k]->Write();\nResultPTYsub[k]->Write();\n\nHistPTSub[k]->Write();\nHistPTSelectSub[k]->Write();\nHistPTSelectSmearedSub[k]->Write();\nHistPTSelectSmearedTruncatedSub[k]->Write();\nResultPTSub[k]->Write();\n\nHistYSub[k]->Write();\nHistYSelectSub[k]->Write();\nHistYSelectSmearedSub[k]->Write();\nHistYSelectSmearedTruncatedSub[k]->Write();\nResultYSub[k]->Write();\n\nHistYStar[k]->Write();\nHistYStarSelectSmearedTruncated[k]->Write();\nResultYStar[k]->Write();\n``````\n\n// Result[k]->Close();\n\n``````//Selection->Write();\n``````\n\n/*\nc->Modified(); c->Update();\nc1->Modified(); c1->Update();\nc2->Modified(); c2->Update();\n\n*/\n\n``````double Mean=Hist3[k]->GetMean(1);\nAcceptance[k]= Hist3[k]->GetEntries()/10000;\n\nreturn Mean;\n``````\n\n}\n\n``````//end of calculation of acceptance\n``````\n\ndouble Analysis::Calculation(double mass1, double sigma)\n{\n//function that calculated and print out sigG/MassG( it does not return)\n//Returns the cross section value\n\n``````double sigGmassG;\ndouble sigMassinput;\n\ndouble massGseven = {300,350,400,450,500,550,600,650,700,750,800,850,900,950,1000,1050,1100,1150,1200,1250,1300,1350,1400,1450,1500,1550,1600,1650,1700,1750,1800,1850,1900,1950,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3200,3400,3600,3800,4000};\ndouble massGten = {350,400,450,500,550,600,650,700,750,800,850,900,950,1000,1050,1100,1150,1200,1250,1300,1350,1400,1450,1500,1550,1600,1650,1700,1750,1800,1850,1900,1950,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3200,3400,3600,3800};\ndouble massGfifteen = {400,450,500,550,600,650,700,750,800,850,900,950,1000,1050,1100,1150,1200,1250,1300,1350,1400,1450,1500,1550,1600,1650,1700,1750,1800,1850,1900,1950,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3200,3400};\n\ndouble ptseven={380,250,77,63,39,17,7.6,4.7,2.8,1.1,0.67,0.47,0.28,0.2,0.21,0.22,0.2,0.14,0.096,0.068,0.061,0.064,0.069,0.076,0.08,0.081,0.078,0.074,0.068,0.059,0.049,0.038,0.029,0.026,0.025,0.023,0.018,0.014,0.012,0.01,0.0085,0.0066,0.0049,0.0038,0.0031,0.0023,0.0016,0.0012,0.00099,0.00085};\ndouble ptten={210,97,75,41,17,8.3,4.8,2.3,1.2,0.8,0.5,0.34,0.3,0.28,0.24,0.18,0.14,0.11,0.088,0.083,0.088,0.093,0.1,0.11,0.11,0.1,0.093,0.08,0.068,0.057,0.047,0.041,0.036,0.032,0.026,0.021, 0.017,0.014,0.011,0.0087,0.0067,0.0053,0.0042,0.0034,0.0023,0.0017,0.0013,0.0011};\ndouble ptfifteen={120,75,34,15,7.5,3.6,2,1.3,0.88,0.67,0.52,0.39,0.28,0.23,0.18,0.16,0.13,0.14,0.14,0.11,0.12,0.15,0.15,0.14,0.13,0.11,0.085,0.082,0.072,0.062,0.054,0.046,0.041,0.032,0.025,0.019,0.014,0.011,0.0081,0.0065,0.0051,0.0041,0.0034,0.0025,0.0019};\n\nTFile *RealCrossSection = new TFile(\"RealCrossSection.root\",\"RECREATE\");\n\nTGraphErrors *gr1= new TGraphErrors(50,massGseven,ptseven,0,0);\nTGraphErrors *gr2= new TGraphErrors(48,massGten,ptten,0,0);\nTGraphErrors *gr3= new TGraphErrors(45, massGfifteen,ptfifteen,0,0);\n``````\n\n/*\nTCanvas *c3 = new TCanvas();\nc3->SetLogy();\n*/\n\n``````gr1->SetLineColor(kRed);\ngr2->SetLineColor(kBlue);\ngr3->SetLineColor(kGreen);\n\ngr1->Draw(\"ALP\");\ngr2->Draw(\"LP\");\ngr3->Draw(\"LP\");\n\ngr1->Write();\ngr2->Write();\ngr3->Write();\n\nRealCrossSection->Write();\n\ngDirectory->Append(gr1);\n\n//input signmassinput and Mass\n``````\n\n// mass\n// double sigMassinput=1;\ndouble sigMassAnalysis={0.07,0.10,0.15};\ndouble sigMassoutput;\n\n``````//finding the sigmass for this analysis\n\ndouble MassG ;\n\nfor (int i=0; i<50; i++)\n{\nif (mass1<massGseven[i] && i!=0 )\n{\nMassG=massGseven[i];\n\n}\n\nif (mass1<massGseven)\n\n{\nMassG=massGseven;\n}\n\nif (mass1==massGseven[i])\n{\nMassG=massGseven[i];\n}\nif (mass1<massGseven[i] && i!=0 )\n{\nbreak;\n}\ncout<<\" Mass 1: \"<<mass1<<\" MassGfifth: \"<<massGseven[i]<<\" Mass G: \"<<MassG<<endl;\n\n}\n\nsigMassinput=sigma/MassG;\n\nfor (int i=0; i<4; i++)\n{\nif (sigMassinput<sigMassAnalysis[i] && i!=0 )\n{\nsigMassoutput=sigMassAnalysis[i];\n\n}\n\nif (sigMassinput<sigMassAnalysis)\n\n{\nsigMassoutput=sigMassAnalysis;\n}\n\nif (sigMassinput==sigMassAnalysis[i])\n{\nsigMassoutput=sigMassAnalysis[i];\n}\nif (sigMassinput<sigMassAnalysis[i] && i!=0 )\n{\nbreak;\n}\n\ncout<<\"Sigmass input: \"<<sigMassinput<<\" sigMassAnalysis: \"<<sigMassAnalysis[i]<<\" SigMassoutput: \"<<sigMassoutput<<endl;\n}\n\nif (sigMassinput>sigMassAnalysis)\n{\nsigMassoutput=sigMassAnalysis;\n}\ncout<<\"Sigmass input: \"<<sigMassinput<<\" sigMassAnalysis: \"<<sigMassAnalysis<<\" SigMassoutput: \"<<sigMassoutput<<endl;\n\nif (sigMassinput>sigMassAnalysis)\n{sigMassoutput=sigMassAnalysis;}\n\n//sigGmassG=sigMassoutput/MassG;\n``````\n\n// cout<<\"SigG/mG: \"<<sigGmassG<<endl;\n\n``````double CrossSection;\n\nif (sigMassoutput==0.07)\n{\nCrossSection=gr1->Eval(MassG);\ncout <<\"Mass: \"<<MassG<<\" sigma: \"<<sigma<<\" SigG/MassG: \"<<sigMassoutput<<\" Cross Section Result: \"<<CrossSection<<endl;\n\n}\n\nif (sigMassoutput==0.1)\n{\nCrossSection=gr2->Eval(MassG);\ncout <<\"Mass: \"<<MassG<<\" sigma: \"<<sigma<<\" SigG/MassG: \"<<sigMassoutput<<\" Cross Section Result: \"<<CrossSection<<endl;\n}\n\nif (sigMassoutput==0.15)\n{\nCrossSection=gr3->Eval(MassG);\ncout <<\"Mass: \"<<MassG<<\" sigma: \"<<sigma<<\" SigG/MassG: \"<<sigMassoutput<<\" Cross Section Result: \"<<CrossSection<<endl;\n}\n//for different values of sigG/mG, use differnt cross section lines to return cross sections\n\nreturn CrossSection;\n``````\n\n}\n\nvoid Analysis::Loop(int M,int k)\n{\n\n``````if (fChain == 0) return;\ndouble inputmass;\ndouble massAfterSelection;\ndouble CrossSection;\ndouble sigma;\ninputmass=1000;\nofstream infile;\n\n//Finding the average mass after event selection:\ncout<<\"Mass of the Generated Events: \"<<M<<endl;\n\nmassAfterSelection=Analysis::Selection(M,k);\n\ncout<<\"Mass after Selection: \"<<massAfterSelection<<endl;\n\n//Finding the SigG/MassG\nsigma=(1.2*inputmass-0.8*inputmass)/5;\nCrossSection=Analysis::Calculation(massAfterSelection, sigma);\ncout<<\"Cross Section: \"<<CrossSection<<endl;\n\ncout<<\"Check\";\ninfile.open(\"crosssectionvsmass2.txt\");\ninfile<<\"Cross Section: \"<<CrossSection<<endl;\n``````\n\n}\n\nvoid GenerateAnalysis(void)\n{\nofstream infile;\ninfile.open(“crosssectionvsmass.txt”);\n\n``````string Dr=\"/Users/yvonne/Documents/MadGraph/MG5_aMC_v2_2_3/PROC_New_Resonances_UFO_0/Events/mass\";\nstring MassNumber={\"250\",\"500\",\"1000\",\"1500\",\"3000\",\"3500\",\"4000\"};\n\n//,\"4500\",\"5000\",\"5500\", \"6000\",\"6500\",\"7000\"\nstring FileName=\"/tag_2_pythia_lhe_events.root\";\nstring AbsDirectory;\nstring AnalysisName;\n// int i=1;\n\nint mass={250,500,1000,1500,3000,3500,4000};\n//,4500,5000,5500, 6000,6500,7000\n// gSystem->CompileMacro(\"Analysis.C\",\"k\");\nHist= new TH1D(\"Mass Resonance+Selection\",\"Mass Resonance\",1000,-1000,5000);\n\nfor (int i=0; i<1; i++)\n{\nResult[i]= new TFile(Form(\"Result Mass %d\", mass[i]), \"RECREATE\");\nHist[i]= new TH1D(Form(\"Mass Resonance Mass %d\",mass[i]),Form(\"Mass Resonance Mass %d\",mass[i]),1000,-1000,5000);\nHist1[i]= new TH1D(Form(\"Mass Resonance+Selection Mass %d\",mass[i]),Form(\"Mass Resonance+Selection Mass %d\",mass[i]),1000,-1000,5000);\nHist2[i]= new TH1D(Form(\"Mass Resonance Mass+Selection+smearing effect %d\",mass[i]),Form(\"Mass Resonance Mass+Selection+smearing effect %d\",mass[i]),1000,-1000,5000);\nHist3[i]= new TH1D(Form(\"Mass Resonance+Selection+smearing effect+truncated Mass %d\",mass[i]),Form(\"Mass Resonance+smearing effect+truncated Mass %d\",mass[i]),1000,-1000,5000);\n\nResultPT[i]= new TFile(Form(\"Result PT Leading Jet, Mass %d.root\", mass[i]), \"RECREATE\");\nHistPT[i]= new TH1D(Form(\"Leading Jet: Transverse Momentum Before Selection Mass%d\", mass[i]),Form(\"Leading Jet: Transverse Momentum Before Selection Mass%d\", mass[i]),1000,0,1200);\nHistPTSelect[i] = new TH1D(Form(\"Leading Jet: Transverse Momentum After Selection Mass%d\",mass[i]),Form(\"Leading Jet: Transverse Momentum After Selection Mass%d\",mass[i]),1000,0,1200);\nHistPTSelectSmeared[i]= new TH1D(Form(\"Leading Jet: Transverse Momentum After Selection + Smearing Mass%d\",mass[i]),Form(\"Leading Jet: Transverse Momentum After Selection + Smearing Mass%d\",mass[i]),1000,0,1200);\nHistPTSelectSmearedTruncated[i]= new TH1D(Form(\"Leading Jet: Transverse Momentum After Selection+ Smearing+ Truncated Mass%d\",mass[i]),Form(\"Leading Jet: Transverse Momentum After Selection+ Smearing+ Truncated Mass%d\",mass[i]),1000,0,1200);\n\nResultY[i]= new TFile(Form(\"Result Rapidity Leading Jet, Mass %d\", mass[i]), \"RECREATE\");\nHistY[i]=new TH1D(Form(\"Leading Jet: Rapidity Before Selection Mass%d\", mass[i]),Form(\"Leading Jet: Rapidity Before Selection Mass%d\", mass[i]),1000,-10,10);\nHistYSelect[i]=new TH1D(Form(\"Leading Jet: Rapidity After Selection Mass%d\",mass[i]),Form(\"Leading Jet: Rapidity After Selection Mass%d\",mass[i]),1000,-10,10);\nHistYSelectSmeared[i]=new TH1D(Form(\"Leading Jet: Rapidity After Selection+ Smearing Mass%d\",mass[i]),Form(\"Leading Jet: Rapidity After Selection+ Smearing Mass%d\",mass[i]),1000,-10,10);\nHistYSelectSmearedTruncated[i]=new TH1D(Form(\"Leading Jet: Rapidity After Selection+ Smearing+ Truncated Mass%d\",mass[i]),Form(\"Leading Jet: Rapidity After Selection+ Smearing+ Truncated Mass%d\",mass[i]),1000,-10,10);\n\nResultYStar[i]= new TFile(Form(\"Result Absolute Half Difference between the 2 leading Jets, Mass %d\", mass[i]),\"RECREATE\");\nHistYStarSelectSmearedTruncated[i]=new TH1D(Form(\"After all Selections, Half Difference in Rapidity between Leading and subleading jet; Mass%d\",mass[i]),Form(\"After all Selections, Half Difference in Rapidity between Leading and subleading jet; Mass%d\",mass[i]),1000,-1.2,1.2);\n\nResultPTSub[i]= new TFile(Form(\"Result PT SubLeading Jet, Mass %d\", mass[i]), \"RECREATE\");\nHistPTSub[i]= new TH1D(Form(\"SubLeading Jet: Transverse Momentum Before Selection Mass%d\", mass[i]),Form(\"SubLeading Jet: Transverse Momentum Before Selection Mass%d\", mass[i]),1000,0,1200);\nHistPTSelectSub[i] = new TH1D(Form(\"SubLeading Jet: Transverse Momentum After Selection Mass%d\",mass[i]),Form(\"SubLeading Jet: Transverse Momentum After Selection Mass%d\",mass[i]),1000,-0,1200);\nHistPTSelectSmearedSub[i]= new TH1D(Form(\"SubLeading Jet: Transverse Momentum After Selection + Smearing Mass%d\",mass[i]),Form(\"SubLeading Jet: Transverse Momentum After Selection + Smearing Mass%d\",mass[i]),1000,0,1200);\nHistPTSelectSmearedTruncatedSub[i]= new TH1D(Form(\"SubLeading Jet: Transverse Momentum After Selection+ Smearing+ Truncated Mass%d\",mass[i]),Form(\"SubLeading Jet: Transverse Momentum After Selection+ Smearing+ Truncated Mass%d\",mass[i]),1000,0,1200);\n\nResultYSub[i]= new TFile(Form(\"Result Rapidity SubLeading Jet, Mass %d\", mass[i]), \"RECREATE\");\nHistYSub[i]=new TH1D(Form(\"SubLeading Jet: Rapidity Before Selection Mass%d\", mass[i]),Form(\"SubLeading Jet: Rapidity Before Selection Mass%d\", mass[i]),1000,-10,10);\nHistYSelectSub[i]=new TH1D(Form(\"SubLeading Jet: Rapidity After Selection Mass%d\",mass[i]),Form(\"SubLeading Jet: Rapidity After Selection Mass%d\",mass[i]),1000,-10,10);\nHistYSelectSmearedSub[i]=new TH1D(Form(\"SubLeading Jet: Rapidity After Selection+ Smearing Mass%d\",mass[i]),Form(\"SubLeading Jet: Rapidity After Selection+ Smearing Mass%d\",mass[i]),1000,-10,10);\nHistYSelectSmearedTruncatedSub[i]=new TH1D(Form(\"SubLeading Jet: Rapidity After Selection+ Smearing+ Truncated Mass%d\",mass[i]),Form(\"SubLeading Jet: Rapidity After Selection+ Smearing+ Truncated Mass%d\",mass[i]),1000,-10,10);\n//Intermediate check:\n\nResultPTY[i]=new TFile(Form(\"Result PT & Y Selection, Mass %d\", mass[i]), \"RECREATE\");\nHistPTYPT[i]=new TH1D(Form(\"PT After Selection on PT and Y; mass %d\",mass[i]),Form(\"PT After Selection on PT and Y; mass %d\",mass[i]),1000,0,1200);\nHistPTYY[i]=new TH1D (Form(\"Rapidity After Selection on PT and Y; mass %d\",mass[i]),Form(\"PT After Selection on PT and Y; mass %d\",mass[i]),1000,-1.2,1.2);\n\nResultPTYsub[i]=new TFile(Form(\"Subjet Result PT & Y Selection, Mass %d\", mass[i]), \"RECREATE\");\nHistPTYPTsub[i]=new TH1D(Form(\"Subjet PT After Selection on PT and Y; mass %d\",mass[i]),Form(\"Subjet PT After Selection on PT and Y; mass %d\",mass[i]),1000,0,1200);\nHistPTYYsub[i]=new TH1D (Form(\"Subjet Rapidity After Selection on PT and Y; mass %d\",mass[i]),Form(\"Subjet PT After Selection on PT and Y; mass %d\",mass[i]),1000,-1.2,1.2);\n}\n\n//k=0\n\nfor (int i =1; i<7; i++)\n{\nAbsDirectory=Dr+MassNumber[i]+FileName;\n// AnalysisName=\"Analysis\"+MassNumber[i];\ncout<<\"Abs Directory: \"<<AbsDirectory<<endl;\n// cout<<\"Analysis Name: :\"<<AnalysisName;\n\n//.L Analysis.C;\nTFile *f = new TFile(AbsDirectory.c_str());\nTTree *T;\n\nf ->GetObject(\"LHEF\",T);\ncout<<\"T: \"<<T;\n\nAnalysis t(T);\n//t.Loop();\nt.Loop(mass[i],i);\n\n//Printing to a file\ninfile<<\"MassNumber: \"<<MassNumber[i]<<endl;\n//k=k+1;\n\ncout<<\"Test\"<<endl;\n\n}\n//infile.close();\ncout<<\"Test\"<<endl;\n``````\n\n}\n[/code]\nAnalysis.h (10.4 KB)\n\nIt seems to me that you have: `TH1D *HistPT;` but then, in the “GenerateAnalysis”, you actually initialize “HistPT” only: ``` for (int i=0; i<1; i++) { // ... HistPT[i]= new TH1D(Form(\"Leading Jet: Transverse Momentum Before Selection Mass%d\", mass[i]),Form(\"Leading Jet: Transverse Momentum Before Selection Mass%d\", mass[i]),1000,0,1200); // ... }``` BTW. The “names” of your histograms are wrong. They should be simple identifiers, like ordinary c/c++ variables (without any “special” characters, like “space” or “:” or “+” or “;”). The “titles” of histograms can be fancy, of course.\n\nThank you!" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Integrate 5.0\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg44240] Re: Integrate 5.0\n• From: \"David W. Cantrell\" <DWCantrell at sigmaxi.org>\n• Date: Fri, 31 Oct 2003 03:01:05 -0500 (EST)\n• References: <bnnvfj\\$61s\\[email protected]>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Selwyn Hollis <sh2.7183 at misspelled.erthlink.net> wrote:\n> I've come to the conclusion that Integrate has become nearly worthless\n> for computing definite integrals with symbolic limits. To cite a simple\n> example,\n>\n> Integrate[Sqrt[Cos[t] + 1], {t, 0, x}]\n>\n> returns an awful mess inside of an If statement (very mild in this\n> case) that no one should have to deal with if they're only concerned\n> with real numbers (specifically calculus students and a great many\n> applied mathematicians).\n\nBut surely you would not want Mathematica to return\n\n2*Sqrt[1 + Cos[x]]*Tan[x/2] without qualification,\n\nas previous versions unfortunately did. After all, looking at the integral\nitself (having a _continuous_ integrand), one would certainly suppose that\nan unqualified answer should be valid for _all_ real x. But if we replace\nx by, say, 2*Pi in the above unqualified result, we get 0, which is absurd.\nOf course, Mathematica knows better if you give it a numerical upper bound\nfor the integration:\n\nIntegrate[Sqrt[Cos[t] + 1], {t, 0, 2*Pi}] yields 4*Sqrt,\n\nas it should.\n\nAnyway, if we're dealing with real x,\nthen for Integrate[Sqrt[Cos[t] + 1], {t, 0, x}],\nthe answer that I would like Mathematica to return is\n\n4*Sqrt*Floor[(Pi + x)/(2*Pi)] + (2*Sin[x])/Sqrt[1 + Cos[x]]\n\nwhich is valid for all real x.\n\nBut your general point, Selwyn, may still be valid. I haven't used the new\nversion to compute many definite integrals with symbolic limits yet. I was\njust pointing out that this particular \"simple example\" of yours isn't so\nsimple and doesn't happen to convince me of your general point.\n\nRegards,\nDavid Cantrell\n\n> On the other hand, DSolve gives the simple, clean answer that Integrate\n> used to give:\n>\n> y[t]/. DSolve[{y'[t] == Sqrt[Cos[t] + 1], y == 0}, y[t], t]\n>\n> 2*Sqrt[1 + Cos[t]]*Tan[t/2]\n>\n> Could it be that we need a new function such as this:\n>\n> RealIntegral[expr_,{x_,a_,b_}]:=\n> (y[x]/. First@DSolve[{y'[x] ==expr, y[a] == 0}, y[t],\n> t])/.x->b\n>\n> that would be associated with \\[Integral] ? ... leaving the current\n> Integrate to be associated with \\[ContourIntegral]??\n>\n> Or perhaps a simple option for Integrate like RealLimits->True?\n>\n> -----\n> Selwyn Hollis\n> http://www.math.armstrong.edu/faculty/hollis\n\n```\n\n• Prev by Date: Re: Integrate 5.0\n• Next by Date: now, while loops construct errors\n• Previous by thread: Re: Integrate 5.0\n• Next by thread: now, while loops construct errors" ]

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[ "# Demo: MAB Replication Eckles & Kaptein (Bootstrap Thompson Sampling)\n\n#### 2020-07-25\n\nlibrary(contextual)\n\n# Replication of THOMPSON SAMPLING WITH THE ONLINE BOOTSTRAP By Dean Eckles and Maurits Kaptein\n\n# This evaluations takes time - up to a few hours when run single core.\n\n# Running the script in parallel (for example, on 8 cores)\n# shortens the evaluation time substantially.\n\n# https://arxiv.org/abs/1410.4009\n\n# Fig 2. Empirical regret for Thompson sampling and BTS in a K-armed binomial bandit problem.\n\nbandit <- BasicBernoulliBandit$new(weights = c(0.5, rep(0.4,9))) agents <- list(Agent$new(BootstrapTSPolicy$new(1000), bandit, \"BTS 1000\"), Agent$new(ThompsonSamplingPolicy$new(), bandit, \"TS\")) simulator <- Simulator$new(agents = agents,\ndo_parallel = TRUE,\nsave_interval = 50,\nset_seed = 999,\nhorizon = 1e+05,\nsimulations = 1000)\n\nsimulator$run() plot(simulator$history, log = \"x\")", null, "" ]

[ null, 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[ "# The Four Color Theorem\n\nThis page gives a brief summary of a new proof of the Four Color Theorem and a four-coloring algorithm found by Neil Robertson, Daniel P. Sanders, Paul Seymour and Robin Thomas.", null, "History.\n\nThe Four Color Problem dates back to 1852 when Francis Guthrie, while trying to color the map of counties of England noticed that four colors sufficed. He asked his brother Frederick if it was true that any map can be colored using four colors in such a way that adjacent regions (i.e. those sharing a common boundary segment, not just a point) receive different colors. Frederick Guthrie then communicated the conjecture to DeMorgan. The first printed reference is due to Cayley in 1878.\n\nA year later the first `proof' by Kempe appeared; its incorrectness was pointed out by Heawood 11 years later. Another failed proof is due to Tait in 1880; a gap in the argument was pointed out by Petersen in 1891. Both failed proofs did have some value, though. Kempe discovered what became known as Kempe chains, and Tait found an equivalent formulation of the Four Color Theorem in terms of 3-edge-coloring.\n\nThe next major contribution came from Birkhoff whose work allowed Franklin in 1922 to prove that the four color conjecture is true for maps with at most 25 regions. It was also used by other mathematicians to make various forms of progress on the four color problem. We should specifically mention Heesch who developed the two main ingredients needed for the ultimate proof - reducibility and discharging. While the concept of reducibility was studied by other researchers as well, it appears that the idea of discharging, crucial for the unavoidability part of the proof, is due to Heesch, and that it was he who conjectured that a suitable development of this method would solve the Four Color Problem.\n\nThis was confirmed by Appel and Haken in 1976, when they published their proof of the Four Color Theorem [1,2].\n\nWhy a new proof?\n\nThere are two reasons why the Appel-Haken proof is not completely satisfactory.\n\n• Part of the Appel-Haken proof uses a computer, and cannot be verified by hand, and\n• even the part that is supposedly hand-checkable is extraordinarily complicated and tedious, and as far as we know, no one has verified it in its entirety.\nWe have in fact tried to verify the Appel-Haken proof, but soon gave up. Checking the computer part would not only require a lot of programming, but also inputing the descriptions of 1476 graphs, and that was not even the most controversial part of the proof.\n\nWe decided that it would be more profitable to work out our own proof. So we did and came up with a proof and an algorithm that are described below.\n\nOutline of the proof.\n\nThe basic idea of the proof is the same as Appel and Haken's. We exhibit a set of 633 \"configurations\", and prove each of them is \"reducible\". This is a technical concept that implies that no configuration with this property can appear in a minimal counterexample to the Four Color Theorem. It can also be used in an algorithm, for if a reducible configuration appears in a planar graph G, then one can construct in constant time a smaller planar graph G' such that any four-coloring of G' can be converted to a four-coloring of G in linear time.\n\nIt has been known since 1913 that every minimal counterexample to the Four Color Theorem is an internally 6-connected triangulation. In the second part of the proof we prove that at least one of our 633 configurations appears in every internally 6-connected planar triangulation (not necessarily a minimal counterexample to the 4CT). This is called proving unavoidability, and uses the \"discharging method\", first suggested by Heesch. Here our method differs from that of Appel and Haken.\n\nMain features of our proof.\n\nWe confirm a conjecture of Heesch that in proving unavoidability, a reducible configuration can be found in the second neighborhood of an \"overcharged\" vertex; this is how we avoid \"immersion\" problems that were a major source of complication for Appel and Haken. Our unavoidable set has size 633 as opposed to the 1476 member set of Appel and Haken, and our discharging method uses only 32 discharging rules, instead of the 300+ of Appel and Haken. Finally, we obtain a quadratic algorithm to four-color planar graphs (described later), an improvement over the quartic algorithm of Appel and Haken.\n\nConfigurations.\n\nA near-triangulation is a non-null connected loopless plane graph such that every finite region is a triangle. A configuration K consists of a near-triangulation G and a map g from V(G) to the integers with the following properties:\n\n1. for every vertex v, G\\v has at most two components, and if there are two, then the degree of v is g(v)-2,\n2. for every vertex v, if v is not incident with the infinite region, then g(v) equals the degree of v, and otherwise g(v) is greater than the degree of v; and in either case g(v)> 4,\n3. K has ring-size at least 2, where the ring-size of K is defined to be the sum of g(v)-deg(v)-1, summed over all vertices v incident with the infinite region such that G\\v is connected.\n\nWhen drawing pictures of configurations we use a convention introduced by Heesch. The shapes of vertices indicate the value of g(v) as follows: A solid black circle means g(v)=5, a dot (or what appears in the picture as no symbol at all) means g(v)=6, a hollow circle means g(v)=7, a hollow square means g(v)=8, a triangle means g(v)=9, and a pentagon means g(v)=10. (We do not need vertices v with g(v)> 11, and only one vertex with g(v)=11, for which we do not use any special symbol.) In the picture below 17 of our 633 reducible configurations are displayed using the indicated convention. The whole set can be viewed by clicking here. (We refer to (3.2) of our paper for the meaning of \"thick edges\" and \"half-edges\" in those pictures.)", null, "Any configuration isomorphic to one of the 633 configurations exhibited in is called a good configuration. Let T be a triangulation. A configuration K=(G,g) appears in T if G is an induced subgraph of T, every finite region of G is a region of T, and g(v) equals the degree of v in T for every vertex v of G. We prove the following two statements.\n\nTHEOREM 1. If T is a minimal counterexample to the Four Color Theorem, then no good configuration appears in T.\n\nTHEOREM 2. For every internally 6-connected triangulation T, some good configuration appears in T.\n\nFrom the above two theorems it follows that no minimal counterexample exists, and so the 4CT is true. The first proof needs a computer. The second can be checked by hand in a few months, or, using a computer, it can be verified in about 20 minutes.\n\nDischarging rules.\n\nLet T be an internally 6-connected triangulation. Initially, every vertex v is assigned a charge of 10(6-deg(v)). It follows from Euler's formula that the sum of the charges of all vertices is 120; in particular, it is positive. We now redistribute the charges according to the following rules, as follows. Whenever T has a subgraph isomorphic to one of the graphs in the figure below satisfying the degree specifications (for a vertex v of a rule with a minus sign next to v this means that the degree of the corresponding vertex of T is at most the value specified by the shape of v, and analogously for vertices with a plus sign next to them; equality is required for vertices with no sign next to them) a charge of one (two in case of the first rule) is to be sent along the edge marked with an arrow.", null, "This procedure defines a new set of charges with the same total sum. Since the total sum is positive, there is a vertex v in T whose new charge is positive. We show that a good configuration appears in the second neighborhood of v.\n\nIf the degree of v is at most 6 or at least 12, then this can be seen fairly easily by a direct argument. For the remaining cases, however, the proofs are much more complicated. Therefore, we have written the proofs in a formal language so that they can be verified by a computer. Each individual step of these proofs is human-verifiable, but the proofs themselves are not really checkable by hand, because of their length.\n\nPointers.\n\nThe theoretical part of our proof is described in . A 10-page survey is available on-line. The computer data and programs used to be located on an anonymous ftp server, but that server has been phased out. The same files are now available from http://people.math.gatech.edu/~thomas/OLDFTP/four/ and can be conveniently viewed. An independent set of programs was written by Gasper Fijavz under the guidance of Bojan Mohar.\n\nThe input to the algorithm will be a plane triangulation G with n vertices. (This is without loss of generality, as any planar graph can be triangulated in linear time.) The output will be a coloring of the vertices of G with four colors.\n\nIf G has at most four vertices color each vertex a different color. Otherwise if G has a vertex x of degree k < 5, then the circuit C surrounding it is a `k-ring'. Go to the k-ring analysis below. Otherwise G has minimum degree five. For every vertex we compute its charge as explained above, and find a vertex v of positive charge. It follows from our proof of Theorem 2 that either a good configuration appears in the second neighborhood of v (it which case it can be found in linear time), or a k-ring violating the definition of internal 6-connection can be found in linear time. In the latter case we go to the k-ring analysis below, in the former case we apply recursion to a certain smaller graph. A four-coloring of G can then be constructed from the four-coloring of this smaller graph in linear time.\n\nGiven a k-ring R violating the definition of internal 6-connection a procedure developed by Birkhoff can be used. We apply recursion to two carefully selected subgraphs of G. A four-coloring of G can then be constructed from the four-colorings of the two smaller graphs in linear time.\n\nDiscussion.\n\nWe should mention that both our programs use only integer arithmetic, and so we need not be concerned with round-off errors and similar dangers of floating point arithmetic. However, an argument can be made that our `proof' is not a proof in the traditional sense, because it contains steps that can never be verified by humans. In particular, we have not proved the correctness of the compiler we compiled our programs on, nor have we proved the infallibility of the hardware we ran our programs on. These have to be taken on faith, and are conceivably a source of error. However, from a practical point of view, the chance of a computer error that appears consistently in exactly the same way on all runs of our programs on all the compilers under all the operating systems that our programs run on is infinitesimally small compared to the chance of a human error during the same amount of case-checking. Apart from this hypothetical possibility of a computer consistently giving an incorrect answer, the rest of our proof can be verified in the same way as traditional mathematical proofs. We concede, however, that verifying a computer program is much more difficult than checking a mathematical proof of the same length.\n\nAcknowledgements.\n\nWe are indebted to Thomas Fowler, Christopher Carl Heckman and Barrett Walls for their help with preparing this page. Our work was partially supported by the National Science Foundation.\n\nReferences.\n\n1. K. Appel and W. Haken, Every planar map is four colorable. Part I. Discharging, Illinois J. Math. 21 (1977), 429-490.\n2. K. Appel, W. Haken and J. Koch, Every planar map is four colorable. Part II. Reducibility, Illinois J. Math. 21 (1977), 491--567.\n3. K. Appel and W. Haken, Every planar map is four colorable, Contemporary Math. 98 (1989).\n4. G. D. Birkhoff, The reducibility of maps, Amer. J. Math. 35 (1913), 114-128.\n5. H. Heesch, Untersuchungen zum Vierfarbenproblem, Hochschulskriptum 810/a/b, Bibliographisches Institut, Mannheim 1969.\n6. A. B. Kempe, On the geographical problem of the four colors, Amer. J. Math., 2 (1879), 193-200.\n7. N. Robertson, D. P. Sanders, P. D. Seymour and R. Thomas, The four colour theorem, J. Combin. Theory Ser. B. 70 (1997), 2-44.\n8. N. Robertson, D. P. Sanders, P. D. Seymour and R. Thomas, A new proof of the four colour theorem, Electron. Res. Announc. Amer. Math. Soc. 2 (1996), 17-25 (electronic).\n9. T.L. Saaty, Thirteen colorful variations on Guthrie's four-color conjecture, Amer. Math. Monthly 79 (1972), 2-43.\n10. T.L. Saaty and P. C. Kainen, The four-color problem. Assaults and conquest, Dover Publications, New York 1986.\n11. P. G. Tait, Note on a theorem in geometry of position, Trans. Roy. Soc. Edinburgh 29 (1880), 657-660.\n12. H. Whitney and W. T. Tutte, Kempe chains and the four colour problem'', in Studies in Graph Theory, Part II (ed. D. R. Fulkerson), Math. Assoc. of America, 1975, 378-413.\n\n13 November 1995. Links updated 19 July 2017.\n\nTranslations: German, Romanian, Spanish courtesy of Translate Team, Greek courtesy of Dimitris Galatas.\nNote to potential translators: Please do not ask me to link to your translation. I am receiving too many requests. Most of them are computer-generated and/or of low quality." ]

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[ "# # Function library\n\n## #`abs(numeric)`\n\nThis function calculates the absolute value of a numeric value.\n\n``````number = -1.2\n``````\n``````abs(#number)\n``````\n``````1.2\n``````\n``````number = 5.47\n``````\n``````abs(#number)\n``````\n``````5.47\n``````\n\n## #`all(list[, lambda_expression])`\n\nReturn true if all items of `list` fulfill the condition. If the list is empty, also returns true.\n\nIf `lambda_expression` is omitted, tests for empty items, exactly as filter().\n\nTIP\n\nThis is equivalent to `count(filter(items, condition)) == count(items)`, but easier to read. It's also faster because it stops as soon as the first item fails `condition`.\n\n``````all([1, 2, 3, 4, 5], [x -> #x > 0])\nall([1, 2, 3, 4, 5], [x -> #x < 5])\nall([\"a\", \"b\", \"c\"], [x -> #x == \"c\"])\n``````\n``````true # All items are greater than zero\nfalse # The last item (5) is not smaller than 5\nfalse # Only one item equals \"c\"\n``````\n\nUsage with default condition:\n\n``````all([])\nall([\"\"])\nall([\"a\"])\n``````\n``````true # No items\nfalse # One empty item\ntrue # One non-empty item\n``````\n\n## #`any(list[, lambda_expression])`\n\nReturn true if at least one item of `list` fulfills the condition. If the list is empty, returns false.\n\nIf `lambda_expression` is omitted, tests for non-empty items, exactly as filter().\n\nTIP\n\nThis is equivalent to `count(filter(items, condition)) > 0`, but easier to read. It's also faster because it stops as soon as the first item fulfills `condition`.\n\n``````any([1, 2, 3, 4, 5], [x -> #x > 0])\nany([1, 2, 3, 4, 5], [x -> #x < 5])\nany([\"a\", \"b\", \"c\"], [x -> #x == \"c\"])\n``````\n``````true # All items are greater than zero\ntrue # Four of five items are smaller than five\ntrue # One item equals \"c\"\n``````\n\nUsage with default condition:\n\n``````any([])\nany([\"\"])\nany([\"a\", \"\"])\n``````\n``````false # No items\nfalse # One empty item\ntrue # One non-empty item\n``````\n\n## #`capitalize(string)`\n\nReturns the string with the first letter in upper case, others lower case.\n\n``````some_text = \"abc DEF\"\n``````\n``````capitalize(#some_text)\n``````\n``````Abc def\n``````\n\n## #`collect(list_of_objects, fieldname)`\n\nTakes a field from all objects in a list and returns those field values as a list.\n\n``````products = [{ \"brand\": \"Huawei\" }, { \"model\": \"OnePlus 5 A5000\" }, { \"brand\": \"Google\" }]\n``````\n``````collect(#products, \"brand\")\n``````\n``````[\"Huawei\", \"\", \"Google\"]\n``````\n\n## #`concatenation (string + string)`\n\n``````string1 = \"a\"\nstring2 = \"b\"\n``````\n``````#string1 + #string2\n``````\n``````\"ab\"\n``````\n\n## #`contains(value, list)`\n\nChecks if an element in the list matches the value.\n\n``````list_of_numbers = [1, 2, 3, 4, 5]\n``````\n``````contains(2, #list_of_numbers)\n``````\n``````true\n``````\n``````list = [\"hello\", \"world\", \"!\"]\n``````\n``````contains(\"world\", #list)\n``````\n``````true\n``````\n``````list = [\"hello\", \"world\", \"!\"]\n``````\n``````contains(\"5555\", #list)\n``````\n``````false\n``````\n``````colors = [\"red\", \"green\", \"blue\"]\n``````\n``````contains(\"dark blue\", #colors)\n``````\n``````false\n``````\n\n## #`contains(string, list, substring)`\n\nWhen setting the optional `substring` parameter to `true`, a collection element may just partially match `value`, and also differ in case.\n\n``````colors = [\"Red\", \"Green\", \"Blue\"]\n``````\n``````contains(\"dArK bLuE\", #colors, true)\n``````\n``````true\n``````\n\nIt does not test if `string` is part of a list item:\n\n``````colors = [\"red\", \"green\", \"dark blue\"]\n``````\n``````contains(\"blue\", #colors, true)\n``````\n``````false\n``````\n\nIf you need that behavior, use any():\n\n``````any(#colors, [item -> \"blue\" in #item]) # true\n``````\n\n## #`convert_comma(input, divisor, lowerUnit, upperUnit)`\n\nThis function selects one of the two units and formats `input` in it. It is equivalent to:\n\n``````if #input < #divisor then\nstr(#input) + \" \" + #lowerUnit\nelse\nformat_number(#input / #divisor) + \" \" + #upperUnit\n``````\n``````distance = 5300\n``````\n``````convert_comma(#distance, 1000, \"m\", \"km\")\n``````\n``````\"5,3 km\"\n``````\n``````distance = 900\n``````\n``````convert_comma(#distance, 1000, \"m\", \"km\")\n``````\n``````\"900 m\"\n``````\n``````duration = 120\n``````\n``````convert_comma(#duration, 60, \"Minuten\", \"Stunden\")\n``````\n``````\"2 Stunden\"\n``````\n\n## #`convert_count(input, divisor, lowerUnit, upperUnit)`\n\nThis function takes a numeric count and converts it to a written currency or date value. The difference to `convert_comma` is that it uses whole numbers and both units instead of a decimal value.\n\n``````duration = 145\n``````\n``````convert_count(#duration, 60, \"Minuten\", \"Stunden\")\n``````\n``````\"2 Stunden 25 Minuten\"\n``````\n``````duration = 45\n``````\n``````convert_count(#duration, 60, \"minutes\", \"hours\")\n``````\n``````\"45 minutes\"\n``````\n``````amount = 350\n``````\n``````convert_count(#amount, 100, \"Cent\", \"Euro\")\n``````\n``````\"3 Euro 50 Cent\"\n``````\n``````amount = 100\n``````\n``````convert_count(#amount, 25, \"\", \",\")\n``````\n``````\"4,\"\n``````\n\nWARNING\n\nThe fourth argument is a , (comma).\n\n## #`convert_count(input, divisor, lowerUnit, upperUnit, conjunction)`\n\nSame as above, but with a custom conjunction.\n\n``````amount = 350\n``````\n``````convert_count(#amount, 100, \"Cent\", \"Euro\", \"and\")\n``````\n``````\"3 Euro and 50 Cent\"\n``````\n\n## #`count(list)`\n\nReturns the number of elements in a list.\n\n``````list = [1, 2, 10, 12, 14]\n``````\n``````count(#list)\n``````\n``````5\n``````\n\n## #`count_uniques(list)`\n\nCount duplicate elements of list. Returns a list of objects which contain the following fields:\n\n• value - the list element\n• count - number of occurrences in list (1 if element is unique)\n\nThe list elements must be strings or numbers. Nested lists are supported for convenience, and will be treated as a single flat list. The result is in the same order as the input, i.e. sorted by the element's first occurrence.\n\nTIP\n\nThis is the same as unique(), except for including the number of occurrences.\n\n``````list = [\"cat\", \"dog\", \"cat\"]\n``````\n``````count_uniques(#list)\n``````\n``````[{\"value\": \"cat\", \"count\": 2}, {\"value\": \"dog\", \"count\": 1}]\n``````\n\nA more complex example, with nested lists:\n\n``````list1 = [1, 2, 2]\nlist2 = [14, 10, 12, 10]\n``````\n``````count_uniques([#list1, #list2, 2])\n``````\n``````[{\"value\": 1, \"count\": 1}, {\"value\": 2, \"count\": 3}, {\"value\": 14, \"count\": 1},\n{\"value\": 10, \"count\": 2}, {\"value\": 12, \"count\": 1}]\n``````\n\n## #`cur_lang()`\n\nThis method returns the current language in the current text generation process as an ISO 2-letter code.\n\n``````cur_lang()\n``````\n``````es\n``````\n\nWARNING\n\nWhatever language is currently used (here: Spain).\n\n## #`currency(double)`\n\nThis function converts a double to a currency string, depending on the current culture.\n\n``````amount = 2.1\n``````\n``````currency(#amount)\n``````\n``````\"2.10\"\n``````\n``````amount = 1000000\n``````\n``````currency(#amount)\n``````\n``````\"1,000,000.00\"\n``````\n\nWARNING\n\nThe provided examples use the locale en_US\n\n## #`date_add(date, number, type)`\n\nThis method adds a value to a given date string (e.g \"16.06.2016\") or date object and returns a date object.\n\nSupported time units:\n\n• `years`\n• `months`\n• `weeks`\n• `days`\n• `hours`\n• `minutes`\n• `seconds`\n``````date_string = \"16.05.1983\"\n``````\n``````date_add(#date_string, 3, \"years\")\n``````\n``````<date 1986-05-16 00:00:00>\n``````\n\n## #`date_convert(datetime, format[, timezone])`\n\nThis method converts a datetime string (e.g. \"6.4.2019 13:10\") to a specific format and returns it as a string. The `datetime` parameter can also be a date object (year, month, day).\n\nIf a `timezone` is given, `datetime` is converted to it before formatting, assuming it was UTC if no information is given. The list of valid timezone names can be found here (opens new window).\n\nWARNING\n\nNames for months and weekdays depend on the selected language.\n\n``````date_string = \"16.05.1983\"\n``````\n``````date_convert(#date_string, \"yyyy-MM-dd\")\n``````\n``````\"1983-05-16\"\n``````\n``````date_string = \"Wednesday, June 5, 2019 2:55:09.324 PM\"\n``````\n``````date_convert(#date_string, \"yyyy-MM-dd HH:mm:ss.SSS\")\n``````\n``````\"2019-06-05 14:55:09.324\"\n``````\n\nExample to convert seconds since the UNIX epoch 1.1.1970 to a date.\n\n``````timestamp = 1559739675\n``````\n``````date_convert(#timestamp, \"yyyy-MM-dd HH:mm:ss\")\n``````\n``````\"2019-06-05 15:01:15\"\n``````\n\nExample converting a ISO-8601 datetime string to a different timezone:\n\n``````dt = \"2019-08-01T08:15:47+0200\"\n``````\n``````date_convert(#dt, \"yyyy-MM-dd HH:mm:ss ZZ\", \"US/Pacific\")\n``````\n``````\"2019-07-31 23:15:47 -07:00\"\n``````\n\n## #`date_day_of_year(date)`\n\nThis method returns the day of the year of a given date string or date object as a numeric. It is equivalent to `int(date_convert(date, \"DDD\"))`.\n\n``````date_string = \"31.12.2010\"\n``````\n``````date_day_of_year(#date_string)\n``````\n``````365\n``````\n``````date_string = \"this_is_not_a_valid_date\"\n``````\n``````date_day_of_year(#date_string)\n``````\n``````0\n``````\n\n## #`date_difference(date, date)`\n\nThis method returns the difference between two dates as an integer representing the absolute number of days. The expected input formats are a date string or a date object.\n\n``````date1 = \"16.05.1983\"\ndate2 = \"19.05.1983\"\n``````\n``````date_difference(#date1, #date2)\n``````\n``````3\n``````\n\nIf the first date is later than the second, a negative number is returned:\n\n``````date1 = \"19.05.1983\"\ndate2 = \"16.05.1983\"\n``````\n``````date_difference(#date1, #date2)\n``````\n``````-3\n``````\n\n## #`date_format(date)`\n\nGuesses the format string of a date value. To format a date, see date_convert().\n\n``````date_string = \"16.05.1983\"\n``````\n``````date_format(#date_string)\n``````\n``````\"dd.MM.yyyy\"\n``````\n\nWARNING\n\nThis function is deprecated and only works in very specific cases. Do not use in new projects.\n\n## #`date_now()`\n\nReturns the current date (in Germany) as a string. See the second form below for more control.\n\n``````date_now()\n``````\n``````\"13.06.2015\"\n``````\n\n## #`date_now(format[, timezone])`\n\nReturns the current date and/or time in a specific format and timezone (defaults to \"Europe/Berlin\"). The list of valid timezone names can be found here (opens new window).\n\n``````date_now(\"yyyy-MM-dd\")\n``````\n``````\"2015-06-13\"\n``````\n``````date_now(\"dd.MM.YYYY HH:mm:ss.SSSSZ\")\n``````\n``````\"18.06.2019 16:43:23.2648+0000\"\n``````\n\nExample printing the current time for different timezones:\n\n``````date_now(\"HH:mm\")\ndate_now(\"HH:mm\", \"US/Pacific\")\ndate_now(\"HH:mm\", \"Japan\")\n``````\n``````\"08:33\"\n\"23:33\"\n\"15:34\"\n``````\n\nFor political timezones, the daylight saving time of the respective country is applied:\n\n``````date_now(\"Z\")\n``````\n``````\"+0100\" in winter, \"+0200\" in summer\n``````\n\n## #`endswith(string, end)`\n\nChecks if `string` ends with string `end` (upper/lower case must match). Returns true if `end` is empty.\n\nThis is equivalent to `substring(#string, len(#string) - len(#end)) == #end`.\n\n``````value = \"foobaz\"\n``````\n``````endswith(#value, \"baz\")\nendswith(#value, \"BAZ\")\nendswith(#value, \"\")\n``````\n``````true\nfalse\ntrue\n``````\n\n## #`filter(list)`\n\nRemoves items from a list if they meet one of the following criteria:\n\n• item is empty (\"\", [], {} - String, List, Object)\n• item is False (Boolean)\n• item is a Phrase Node that returns the value False using has_voc(Phrase Node)\n``````various_items = [\"\", 1, \"a\", {\"examplekey\": \"\"}, \"\", [], {}]\n``````\n``````filter(#various_items)\n``````\n``````[1, \"a\", { \"examplekey\": \"\" }]\n``````\n\n## #`filter(list, object/lambda_expression)`\n\nFilters a list of objects for the elements that match a given filter object or lambda expression.\n\n``````models = [{\n\"dummykey\": \"dummyvalue\"\n},{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n},{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"\\$489.98\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"\\$630.00\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"\\$596.97\",\n\"Ram\": \"6 GB\"\n},{\n\"Price\": \"\\$692.68\",\n\"Ram\": \"4 GB\"}]\n``````\n``````filter(#models, {\"Brand\": \"Huawei\"})\n``````\n``````[{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"\\$630.00\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"\\$596.97\",\n\"Ram\": \"6 GB\"\n}]\n``````\n\nWARNING\n\nPay attention to capital and small letters (case sensitive).\n\nInput and output list same as above.\n\n``````models\n``````\n``````filter(#models, [listelement -> #listelement.Brand == \"Huawei\"])\n``````\n``````[{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"\\$630.00\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"\\$596.97\",\n\"Ram\": \"6 GB\"\n}]\n``````\nLambda Explanation:\n [listelement -> #listelement.Brand == \"Huawei\"] [ notation listelement pick an element from a list(variable name) -> operator #listelement.Brand access the value from a string-key value pair of the picked object(name must match ) == compare operator \"Huawei\" string ] notation\n\nWARNING\n\nIn this example, the variable list element of the Lambda expression contains a complete object of the list, which is used for comparison.\n\nInput same as above.\n\n``````models\n``````\n``````filter(#models, [listelement, index, content -> #listelement.Ram == #content.Ram], {\"Ram\": \"6 GB\"})\n``````\n``````[{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"\\$630.00\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"\\$596.97\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"\\$489.98\",\n\"Ram\": \"6 GB\"\n}]\n``````\nLambda Explanation:\n [listelement, index, content -> #listelement.Ram == #content.Ram] [ notation listelement pick an object from a list(variable name) index Variable to access the list index(optional, when content parameter not set) content Variable to access the additional content parameter of the filter method (here: data_ram) (passed as third argument to the method) -> operator #listelement.Ram access the value of the key Ram of the picked object(name must match ) == comparison operator #content.Ram access the value of the key Ram from another content (passed as a third argument to the method) ] notation\n\nWARNING\n\nIn this example, the variable listelement of the Lambda Expression takes an object from the list and compares the defined key with the defined key of the variable content, which pulls an external object into the Lambda Expression. If the values are identical, the object from the list is returned. This is done for each item in the list. Variable index must be set even if it is not used.\n\n## #`first(list)`\n\nReturns the first item of a list.\n\n## #`first(list, number)`\n\nGet a range of elements from a list.\nA positive `number` will return the first n elements of the list, a negative `number` will return the last n elements of the list.\n\nIf the list has less than `number` elements, returns the full list.\nIf `number` is 1 or -1, returns the first or last item, without wrapping it in a list.\n\nTIP\n\nSee range() or slice() if you do not want above behavior, or if you need a slice from the middle of a list.\n\n``````items = [ 1, 2, 3, 4, 5 ]\n``````\n``````first(#items, 3)\nfirst(#items, -3)\n``````\n``````[1, 2, 3]\n[3, 4, 5]\n``````\n``````items = [ 1, 2, 3, 4, 5 ]\n``````\n``````first(#items, 5)\nfirst(#items, 6)\n``````\n``````[1, 2, 3, 4, 5]\n[1, 2, 3, 4, 5]\n``````\n\nWARNING\n\nOutput is the full list for 5 and above.\n\n``````items = [ \"a\", \"b\", \"c\" ]\n``````\n``````first(#items, 1)\nfirst(#items, -1)\n``````\n``````\"a\"\n\"c\"\n``````\n\n## #`flatten(nested_list)`\n\nRemoves layers from nested lists by inserting the values from the nested part into the first layer of the list.\n\nTIP\n\nPlease note that this does nothing for lists of objects. For that see the map function.\n\n``````list = [\"a\", [\"b\", \"c\", \"d\"], [\"e\", [\"f\", \"g\"]], [[\"h\", [\"i\", \"j\", [\"k\", \"l\"]]]]]\n``````\n``````flatten(#list)\n``````\n``````[\"a\", \"b\", \"c\", \"d\", \"e\", \"f\", \"g\", \"h\", \"i\", \"j\", \"k\", \"l\"]\n``````\n\n## #`flatten(nested_list, max_depth)`\n\nLike `flatten(nested_list)`, but only for a limited number of levels.\n\n``````list = [\"a\", [\"b\", \"c\", \"d\"], [\"e\", [\"f\", \"g\"]], [[\"h\", [\"i\", \"j\", [\"k\", \"l\"]]]]]\n``````\n``````flatten(#list, 1)\nflatten(#list, 2)\nflatten(#list, 3)\nflatten(#list, 4)\n``````\n``````[\"a\", \"b\", \"c\", \"d\", \"e\", [\"f\", \"g\"], [\"h\", [\"i\", \"j\", [\"k\", \"l\"]]]]\n[\"a\", \"b\", \"c\", \"d\", \"e\", \"f\", \"g\", \"h\", [\"i\", \"j\", [\"k\", \"l\"]]]\n[\"a\", \"b\", \"c\", \"d\", \"e\", \"f\", \"g\", \"h\", \"i\", \"j\", [\"k\", \"l\"]]\n[\"a\", \"b\", \"c\", \"d\", \"e\", \"f\", \"g\", \"h\", \"i\", \"j\", \"k\", \"l\"]\n``````\n\n## #`format_number(number)`\n\nFormat a number for the current locale, using the minimal number of decimal digits.\n\nAssuming locale is de-DE:\n\n``````format_number(10/3)\nformat_number(2/5)\nformat_number(3)\n``````\n``````\"3,333\"\n\"0,4\"\n\"3\"\n``````\n\nSee group_digits() to also add thousand separators.\n\n## #`format_number(number, decimal_digits)`\n\nReturn a number with an enforced number of decimal digits in the current locale.\n\n``````param1 = 2\n``````\n``````format_number(#param1, 2)\n``````\n``````2.00\n``````\n\nWARNING\n\nOutput uses locale en-US.\n\nInput same as above.\n\n``````param1 = 2\n``````\n``````format_number(#param1, 2)\n``````\n``````2,00\n``````\n\nWARNING\n\nOutput uses locale de-DE.\n\n## #`get(table, key[, fallback])`\n\nLook up an entry in a lookup table or object. If nothing is found, it returns `fallback` if given, or `key` otherwise.\n\nEquivalent to `first(filter([#table[#key], #fallback]))`.\n\n``````key = \"NFC\"\n``````\n``````get(#smartphone_terms, #key)\n``````\n``````\"Near Field Communication\"\n``````\n\nThe above example returns \"NFC\" if not in the lookup table, or no translation for the current language exists. To instead return nothing:\n\n``````get(#smartphone_terms, #key, \"\")\n``````\n\n## #`group_digits(double)`\n\nThis function converts any double to a formatted string, depending on the current culture.\n\n``````param1 = 2.1\n``````\n``````group_digits(#param1)\n``````\n``````\"2.1\"\n``````\n``````param1 = 1000000\n``````\n``````group_digits(#param1)\n``````\n``````\"1,000,000\"\n``````\n\nWARNING\n\nThe provided examples use the locale en-US, correct delimiters will be introduced depending on language.\n\n## #`has_voc(phrase)`\n\nReturns True if `phrase` is a) a triggered phrase node, or b) a lookup table value, and defines at least one phrase for the current language.\n\n``````table = <a lookup table>\n``````\n``````has_voc(#table.key)\n``````\n``````true\n``````\n\n## #`histogram_data(data_field_name)`\n\nReturns a JSON object with the count, types and min/max/average values corresponding to the given data_field_name, which is searched for in the current collection histogram.\n\nExample collection:\n\n``````[{\n\"Prozessorleistung_GHz\": 2.7,\n\"Hersteller\": \"Samsung\",\n\"Betriebssystem\": \"Android\",\n\"Kategorie\": \"Smartphone\",\n\"Abmessungen_HxBxT_cm\": \"15,3 x 7,9 x 0,9\",\n\"Displaygroesse_Zoll\": 3,\n\"Name\": \"Galaxy Note 4\",\n\"Farbe\": \"schwarz\",\n\"Speicher_GB\": 16,\n\"Gewicht_g\": 176,\n\"Features\": \"LTE|W-LAN|Bluetooth\",\n\"Kamera_megapixel\": 5\n},\n{\n\"Prozessorleistung_GHz\": 2.7,\n\"Hersteller\": \"Apple\",\n\"Betriebssystem\": \"iOS\",\n\"Kategorie\": \"Smartphone\",\n\"Abmessungen_HxBxT_cm\": \"15,3 x 7,9 x 0,9\",\n\"Displaygroesse_Zoll\": 4.7,\n\"Name\": \"iPhone 7\",\n\"Farbe\": \"gold\",\n\"Speicher_GB\": 128,\n\"Gewicht_g\": 150,\n\"Features\": \"LTE|W-LAN|Bluetooth\",\n\"Kamera_megapixel\": 2\n},\n{\n\"Abmessungen_HxBxT_cm\": \"15,3 x 7,9 x 0,9\",\n\"Gewicht_g\": \"\",\n\"Speicher_GB\": 16,\n\"Farbe\": \"schwarz\",\n\"Kategorie\": \"Smartphone\",\n\"Prozessorleistung_GHz\": 2.7,\n\"Hersteller\": \"Samsung\",\n\"Name\": \"Galaxy S7\",\n\"Betriebssystem\": \"Android\",\n\"Features\": \"LTE|W-LAN|NFC\",\n\"Displaygroesse_Zoll\": 5,\n\"Kamera_megapixel\": 12\n}``````\n``````histogram_data(\"Displaygroesse_Zoll\")\n``````\n``````{\n\"count\": 3,\n\"types\": {\n\"number\": {\n\"avg\": 4.233333333333333,\n\"max\": 5,\n\"min\": 3,\n\"sum\": 12.7,\n\"count\": 3,\n\"percentiles\": [ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 ]\n}\n},\n\"values\": [{\n\"type\": \"number\",\n\"count\": 1,\n\"value\": \"3\"\n},\n{\n\"type\": \"number\",\n\"count\": 1,\n\"value\": \"5\"\n},\n{\n\"type\": \"number\",\n\"count\": 1,\n\"value\": \"4.7\"\n}],\n\"avg\": 4.233333333333333,\n\"max\": 5,\n\"min\": 3,\n\"sum\": 12.7,\n\"percentiles\": [ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 ]\n}\n``````\n\nWARNING\n\nThis is the output format for numeric data fields.\n\nSame example collection as above.\n\n``````histogram_data(\"Displaygroesse_Zoll\")\n``````\n``````{\n\"count\": 3,\n\"types\": {\n\"string\": {\n\"count\": 3\n}\n},\n\"values\": [{\n\"type\": \"string\",\n\"count\": 1,\n\"value\": \"Apple\"\n},\n{\n\"type\": \"string\",\n\"count\": 2,\n\"value\": \"Samsung\"\n}]\n}\n``````\n\nWARNING\n\nThis is the output format for string data fields.\n\n## #`in_range(number, lower_bound, upper_bound)`\n\nChecks if a numeric value lies inside a certain interval. The method returns true if lower_bound <= number <= upper_bound and false otherwise.\n\n``````number = 6\n``````\n``````in_range(#number, 5, 10)\n``````\n``````true\n``````\n``````number = 12\n``````\n``````in_range(#number, 5, 10)\n``````\n``````false\n``````\n\n## #`intersection(list, list)`\n\nReturns a list of the elements which are present in both list inputs.\n\n``````list1 = [1, 2, 3]\nlist2 = [1, 2, 4]\n``````\n``````intersection(#list1, #list2)\n``````\n``````[1, 2]\n``````\n\n## #`is_date(string)`\n\nChecks if a string value can be parsed as a date.\n\n``````param1 = \"16.05.1983\"\n``````\n``````is_date(#param1)\n``````\n``````true\n``````\n``````param1 = \"abcDEF\"\n``````\n``````is_date(#param1)\n``````\n``````false\n``````\n\nWARNING\n\nThis function only states that calling `date(string)` will not cause an error. It does not mean the date will be interpreted correctly.\n\n## #`is_list(value)`\n\nChecks if `value` is a list.\n\n``````is_list(\"foo\")\nis_list([\"foo\"])\nis_list({\"a\": \"b\"})\n``````\n``````False\nTrue\nFalse\n``````\n\n## #`is_numeric(value)`\n\nChecks if `value` is a number, or can be cast to one.\n\n``````is_numeric(3.4)\nis_numeric(\" 3.4 \")\nis_numeric(\"\")\nis_numeric(\"foo\")\n``````\n``````True\nTrue\nFalse\nFalse\n``````\n\nWARNING\n\nThis function returns False exactly if numeric() would throw an error, except for the empty string which numeric() would convert to 0.\n\n## #`is_object(value)`\n\nChecks if `value` is an object.\n\n``````is_object(\"foo\")\nis_object([\"foo\"])\nis_object({\"a\": \"b\"})\n``````\n``````False\nFalse\nTrue\n``````\n\n## #`join(list[, delimiter])`\n\nReturn a single string by concatenating all the items of a list of strings. If given, add `delimiter` between each item.\n\n``````items = [1, 2, 3, 4, 5]\n``````\n``````join(#items)\n``````\n``````\"12345\"\n``````\n``````items = [1, 2, 3, 4, 5]\n``````\n``````join(#items, \"a-\")\n``````\n``````\"1a-2a-3a-4a-5\"\n``````\n\n## #`len(string)`\n\nReturns the length of a string (or the string representation of a value).\n\n``````param1 = \"basics\"\n``````\n``````len(#param1)\n``````\n``````6\n``````\n``````param1 = 1234567890\n``````\n``````len(#param1)\n``````\n``````10\n``````\n\n## #`list_pos(list, search_string)`\n\nThis method finds the position of a data value in a list.\n\n``````items = [\"hello\", \"world\"]\n``````\n``````list_pos(#items, \"hello\")\n``````\n``````0\n``````\n\nWARNING\n\nFirst element in the list has the position 0 (index 0).\n\n``````items = [\"hello\", \"world\"]\n``````\n``````list_pos(#items, \"notAWord\")\n``````\n``````-1\n``````\n\nWARNING\n\n## #`lower(string)`\n\nThis method converts a string to its lowercase form.\n\n``````param1 = \"STRING\"\n``````\n``````lower(#param1)\n``````\n``````string\n``````\n\n## #`map(list, lambda_expression)`\n\nApplies a method to all elements of a list and returns a list of objects with the result of the applied lambda expression.\n\n``````numbers = [1, 2, 3]\n``````\n``````map(#numbers, [element -> numeric(#element) + 2])\n``````\n``````[3, 4, 5]\n``````\nLambda Explanation:\n [element -> numeric(#element) + 2] [ notation element pick an element from a list(variable name) -> operator numeric() ATML3-Method #element access value from thepicked element(name must match ) + 2 math: add 2 ] notation\n\n``````products = [\n{\"Brand\": \"Xiaomi\", \"Model\": \"Mi Mix\"},\n{\"Brand\": \"OnePlus\", \"Model\": \"OnePlus 3T A3010\"},\n{\"Brand\": \"Huwawei\", \"Model\": \"Huawei Mate 9 Pro\"}\n]\n``````\n``````map(#products, [entry -> #entry.Brand])\n``````\n``````[ \"Xiaomi\", \"OnePlus\", \"Huwawei\"]\n``````\nLambda Explanation:\n [entry -> #entry.Brand] [ notation entry pick an object from a list(variable name) -> operator numeric() ATML3-Method #entry.Brand access the value from a string-key value pair of the picked object(name must match ) ] notation\n\n## #`map(list, lambda_expression, context)`\n\nApplies a method to all elements of a list and returns a list of objects with the result of the applied lambda expression.\n\n``````products = [\n{\"Brand\": \"Xiaomi\", \"Model\": \"Mi Mix\"},\n{\"Brand\": \"OnePlus\", \"Model\": \"OnePlus 3T A3010\"},\n{\"Brand\": \"Huwawei\", \"Model\": \"Huawei Mate 9 Pro\"}\n]\n``````\n``````map(#products, [entry, index, content -> #entry.Brand + \" (\" + str(#index) + \") [\" + str(#content) + \"]\"], \"Content\")\n``````\n``````[ \"Xiaomi (0) [Content]\", \"OnePlus (1) [Content]\", \"Huwawei (2) [Content]\"]\n``````\nLambda Explanation:\n [entry, index, content -> #entry.Brand + \" (\" + #index + \") [\" + #content + \"]\"] [ notation entry pick an object from a list(variable name) index Variable to access the list index(optional) content Variable to access the additional content parameter (passed as third argument to the method) -> operator #entry.Brand access the value from a string-key value pair of the picked object by key name(name must match ) + operator(string concatenation) \" (\" string(white space and a bracket) #index access the list index \") [\" string(bracket, white space and a bracket) + operator(string concatenation) #content access the value of the additional parameter(passed as a third argument to the method) + operator(string concatenation) \"]\" string(a bracket) ] notation\n\nWARNING\n\nThe function map has 3 parameters. In this example we passed a string Content as an argument to this third parameter.\n\n``````brands = [\"Brand\", \"Model\"]\nmodels = [\"Xiaomi\", \"Mi Mix\"]\n``````\n``````map(#brands, [entry, index, content -> #entry + \" (\" + #content[#index] + \")\"], #models)\n``````\n``````[\"Brand (Xiaomi)\", \"Model (Mi Mix)\"]\n``````\nLambda Explanation:\n [entry, index, content -> #entry + \" (\" + #content[#index] + \")\"] [ notation entry pick an object from a list(variable name) index Variable to access the list index(optional) content Variable to access the additional content parameter (passed as third argument to the method) -> operator #entry access the value the picked element(name must match ) + operator(string concatenation) \" (\" string(white space and a bracket) #content access the value of the additional parameter(passed as a third argument to the method) [ list access notation(passed as a third argument to the method) #index access the list index ] list access notation(passed as a third argument to the method) + operator(string concatenation) \")\" string(bracket) ] notation\n\nWARNING\n\nIn the expression #content[#index] the #index is taken from the picked element, which is set with the variable entry. The passed argument Content is a list that can be accessed directly with #context[] notation. In this example we used the index of the first list (`brands`) to access the value at the same index of the second list (`models`).\n\n## #`max(list)`\n\nReturns the maximum value from a list of numerics.\n\n``````numbers = [1, 2, 3, 6, 4, 5]\n``````\n``````max(#numbers)\n``````\n``````6\n``````\n\n## #`min(list)`\n\nReturns the minimum value from a list of numerics.\n\n``````numbers = [1, 2, 3.2, 6, 4, 5]\n``````\n``````min(#numbers)\n``````\n``````1\n``````\n\n## #`month_no(string)`\n\nReturns the month name in a numeric representation.\n\nWARNING\n\nThis function only supports the German and English notation of months.\n\nUse `int(date_convert(date(#month_name, \"MMMM\"), \"M\"))` for better locale support.\n\n``````month_name = \"Januar\"\n``````\n``````month_no(#month_name)\n``````\n``````1\n``````\n``````month_name = \"February\"\n``````\n``````month_no(#month_name)\n``````\n``````2\n``````\n\n## #`neg_filter(list, object/lambda_expression)`\n\nThis method filters a list of objects for the elements that match a given filter object or lambda expression. The elements that match the filter condition are removed.\n\nWARNING\n\nAND concatenation of the search parameters\n\n``````products = [\n{\n\"dummykey\": \"dummyvalue\"\n},\n{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"\\$489.98\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"\\$630.00\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"\\$596.97\",\n\"Ram\": \"6 GB\"\n},{\n\"Price\": \"\\$692.68\",\n\"Ram\": \"4 GB\"\n}]\n``````\n``````neg_filter(#products, {\"Brand\": \"Huawei\", \"Ram\": \"6 GB\"})\n``````\n``````[{\n\"dummykey\": \"dummyvalue\"\n},\n{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"\\$489.98\",\n\"Ram\": \"6 GB\"\n},{\n\"Price\": \"\\$692.68\",\n\"Ram\": \"4 GB\"\n}]\n``````\n\nWARNING\n\nReturns a list that includes the objects from the given input list that do not contain entries having Brand=Huawei AND Ram=4 GB. In this case, the objects with list indices 3 and 4 of the original list are removed.\n\nInput same as above.\n\n``````products\n``````\n``````neg_filter(#products, {\"Ram\": \"6 GB\"})\n``````\n``````[{\n\"dummykey\": \"dummyvalue\"\n},\n{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Price\": \"\\$692.68\",\n\"Ram\": \"4 GB\"\n}]\n``````\n\nWARNING\n\nReturns a list with objects that do not have Ram=6 GB. In this case, the objects with list indices 2, 3 and 4 of the original list are removed.\n\nInput same as above.\n\n``````products\n``````\n``````neg_filter( #products, [listelement -> #listelement.Ram == \"6 GB\" ])\n``````\n``````[{\n\"dummykey\": \"dummyvalue\"\n},\n{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Price\": \"\\$692.68\",\n\"Ram\": \"4 GB\"\n}]\n``````\n\nWARNING\n\nDoes the same thing as the last example but uses a lambda expression to match the object. In this example, the entry variable of the lambda expression contains the object to be matched.\n\nExplanation of the lambda expression:\n [listelement -> #listelement.Ram == \"6 GB\" ] [ notation listelement pick an element from a list(variable name) -> operator #listelement.Ram access the value from a string-key value pair of the picked object(name must match ) == compare operator \"6 GB\" string ] notation\n\nInput same as above.\n\n``````products\n``````\n``````neg_filter(#products, [listelement -> #listelement.Brand in [\"Huawei\", \"Google\"]])\n``````\n``````[{\n\"dummykey\": \"dummyvalue\"\n},\n{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"\\$489.98\",\n\"Ram\": \"6 GB\"\n}]\n``````\n\nWARNING\n\nReturns a list with entries that do not have Brand=Huawei OR_ Google, in our example entry 3,4 and 5 of the original list. In this example, the entry variable of the lambda expression contains the object to be matched.\n\nExplanation of the lambda expression:\n [listelement -> #listelement.Brand == \"Huawei\" or #listelement.Brand == \"Google\" ] [ notation listelement pick an element from a list(variable name) -> operator #listelement.Brand access the value from a string-key value pair of the picked object(name must match ) == compare operator \"Huawei\" string or logical operator(if one or the other or both strings can be matched) #listelement.Brand access the value from a string-key value pair of the picked object(name must match ) == compare operator \"Google\" string ] notation\n\n``````products = [{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"\\$489.98\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n}]\n``````\n``````neg_filter(#products, [entry, index, context -> #entry.Brand == #context.examplelist[#index] ], {\"examplelist\": [\"OnePlus\", \"Google\"]})\n``````\n``````[{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"\\$482.99\",\n\"Ram\": \"4 GB\"\n}]\n``````\n\nWARNING\n\nThis is the maximum filled lambda expression.\n\nExplanation of the lambda expression:\n [entry, index, context -> #entry.Brand == #context.examplelist[#index] ] [ notation entry pick an object from a list(variable name) index Variable to access the list index(optional) content Variable to access the additional content parameter (passed as third argument to the method) -> operator #entry.Brand access the value from a string-key value pair of the picked object by key name(name must match) == compare operator< #context access the additional parameter(passed as a third argument to the method) . notation(access the value of the following key from a string-key value pair of the an object examplelist key name(in the #context object) [#index] access the list index(access list elements with the same index) ] notation\n\n## #`object_add(object, object)`\n\nCopies the first object, and add or overwrite all keys from the second one.\n\nTIP\n\nThis is equivalent to `object(list(#first_object) + list(#second_object))`.\n\n``````source_object = { \"x\": 1, \"y\": 2 }\n``````\n``````object_add(#source_object, {\"z\": 3})\n``````\n``````{ \"x\": 1, \"y\": 2, \"z\": 3 }\n``````\n\nIf a key exists in both objects, the value of the second object wins:\n\n``````object = { \"x\": 1, \"y\": 2 }\n``````\n``````object_add(#object, {\"y\": \"foo\"})\n``````\n``````{ \"x\": 1, \"y\": \"foo\" }\n{ \"x\": 1, \"y\": 2 }\n``````\n\nAs alternative to an object, the function also accepts a list of key-value pairs:\n\n``````object = { \"x\": 1, \"y\": 2 }\nupdate = [[\"y\", 3], [\"index\", 4]]\n``````\n``````object_add(#object, #update)\n``````\n``````{ \"x\": 1, \"y\": 3, \"index\": 4 }\n``````\n\n## #`pick_object(list_of_objects, key, value)`\n\nReturns the first object in `list_of_objects` which has `key == value`.\n\nThis is equivalent to `first(filter(list_of_objects, {key: value}))` but slightly faster.\n\n``````param1 = {\n\"object\": {\n\"list\": [{\n\"brand\": \"Huawei\",\n\"model\": \"Huawei Mate 9 Pro\"\n\n},\n{\n\"brand\": \"Huawei\",\n\"model\": \"Huawei P10 Plus VKY-L29\"\n\n},\n{\n\"exampleFieldName\": \"exampleValue\"\n\n}\n\n]\n}\n}\n``````\n``````pick_object(#param1.object.list, \"model\", \"Huawei P10 Plus VKY-L29\")\n``````\n``````{\n\"brand\": \"Huawei\",\n\"model\": \"Huawei P10 Plus VKY-L29\"\n}\n``````\n\nInput same as above.\n\n``````param1 =\n``````\n``````pick_object(#param1.object.list, \"exampleFieldName\", \"exampleValue\")\n``````\n``````{\n\"exampleFieldName\": \"exampleValue\"\n}\n``````\n\n## #`pow(a, b)`\n\nReturns a to the power of b. Fractional and negative exponents are supported:\n\n``````pow(2, 3) # same as 2 * 2 * 2\npow(2, 0.5) # same as sqrt(2)\npow(10, -3) # same as 1 / 1000\n``````\n``````8\n1.41421356\n0.001\n``````\n\n## #`random_el(list, number)`\n\nThis method returns random elements from a list.\n\n``````list = [1, 2, 3, 4, 5]\n``````\n``````random_el(#list, 3)\n``````\n``````[3, 5, 2]\n``````\n\nPassing the length of the list can be used to shuffle:\n\n``````list = [1, 2, 3, 4, 5]\n``````\n``````random_el(#list, 5)\n``````\n``````[5, 3, 1, 2, 4]\n``````\n\n## #`range(list, index, count)`\n\nReturns a slice of a list.\n\n• list - a list\n• index - the zero-based index at which the range starts\n• count - the number of elements in the range (for right-to-left range, use negative integers)\n\nTIP\n\nSee slice() if a start and end argument is more convenient for your use case.\n\nIf `index` or `count` reach outside the list, it is silently ignored.\n\n``````list = [\"OnePlus\", \"Huawei\", \"Google\", \"Xiaomi\", \"Apple\"]\n``````\n``````range(#list, 4, 1)\nrange(#list, 5, 2)\nrange(#list, 3, 2)\nrange(#list, 3, 5)\n``````\n``````[\"Apple\"]\n[]\n[\"Xiaomi\", \"Apple\"]\n[\"Xiaomi\", \"Apple\"]\n``````\n``````list = [\"OnePlus\", \"Huawei\", \"Google\", \"Xiaomi\", \"Apple\"]\n``````\n``````range(#list, -1, 2)\nrange(#list, 2, -2)\nrange(#list, 10, -1)\n``````\n``````[\"OnePlus\"]\n[]\n``````\n\n## #`re_group(pattern, string[, \"i\"])`\n\nFinds the left-most match of `pattern` in `string`, then returns the first group (=expression in brackets) within that match. If `pattern` is not found, returns the empty string.\n\nOptionally, it is also possible to match case insensitive (\"i\").\n\nWARNING\n\nYou can only extract one group.\n\n``````param1 = \"String1String2String3\"\n``````\n``````re_group(\"String1(.*)String3\", #param1)\n``````\n``````\"String2\"\n``````\n``````param1 = \"STRING1STRING2STRING3\"\n``````\n``````re_group(\"^.*(String2).*\\$\", #param1, \"i\")\n``````\n``````\"STRING2\"\n``````\n\n## #`re_find(list, pattern, anchor, offset)`\n\nSearches for a regex pattern in a list of string elements and returns the string if there is a match. Otherwise an empty string is returned.\n\n• list - the input list\n• pattern - a regular expression pattern\n• anchor - start search from left (`\"l\"`) or right (`\"r\"`) end of the list; default `\"l\"`\n• offset - number of elements to skip (counting from anchor); default `0`\n``````list = [1, 2, 30]\n``````\n``````re_find(#list, \"3\")\n``````\n``````\"30\"\n``````\n``````list = [1, 2, 13, 14]\n``````\n``````re_find(#list, \"1.*\", \"l\", 0)\nre_find(#list, \"1.*\", \"l\", 2)\nre_find(#list, \"1.*\", \"r\", 0)\nre_find(#list, \"1.*\", \"r\", 2)\n``````\n``````\"1\"\n\"13\"\n\"14\"\n\"1\"\n``````\n``````list = [\"a\", \"b\", \"c\"]\n``````\n``````re_find(#list, \"d\")\n``````\n``````\"\"\n``````\n\n## #`re_get(list, pattern, direction, position)`\n\nReturns a specific item from a list, but only if it matches the RegEx pattern.\n\n• list - the input list\n• pattern - a regular expression pattern\n• direction - \"l\" so `position` counts from the start of the list, \"r\" from the end\n• position - index into the list (0 to count(list) - 1)\n\nWARNING\n\nThe return value is always a string, even if it wasn't in the list.\n\n``````number_list = [1, 2, 3]\n``````\n``````re_get(#number_list, \".*\", \"l\", 0)\nre_get(#number_list, \".*\", \"r\", 0)\nre_get(#number_list, \"4\", \"l\", 0)\n``````\n``````\"1\"\n\"3\"\n\"\"\n``````\n``````list = [\"foobarbaz\"]\n``````\n``````re_get(#list, \"foo\", \"l\", 0)\nre_get(#list, \"bar\", \"l\", 0)\nre_get(#list, \"baz\", \"l\", 0)\n``````\n``````\"foobarbaz\"\n``````\n\n## #`re_keep(list, pattern)`\n\nReturns the list of all elements in `list` which match `pattern`.\n\nEquivalent to\n\n``````filter(flatten(#list), [item -> re_search(#pattern, #item)])\n``````\n• list - the input list\n• pattern - a regular expression pattern\n``````list = [\"xyz1\", \"abc\", \"xyz2\"]\n``````\n``````re_keep(#list, \"xyz.\")\n``````\n``````[\"xyz1\", \"xyz2\"]\n``````\n\n## #`re_keep(list, pattern, direction, position)`\n\nThis method searches for a regex pattern in a list and returns a list of all matched elements. Optionally, a starting position and a search direction can be given.\n\n• list - the input list\n• pattern - a regular expression pattern\n• direction - the direction of the search Valid values: `l` and `r`\n• position - position in the list, where the search starts Valid values: any integer >= 0 parameter is optional\n``````list = [\"xyz1\", \"abc\", \"xyz2\"]\n``````\n``````re_keep(#list, \"xyz.\", \"l\", 1)\n``````\n``````[\"xyz1\"]\n``````\n``````list = [\"xyz1\", \"abc\", \"xyz2\"]\n``````\n``````re_keep(#list, \"xyz.\", \"r\", 1)\n``````\n``````[\"xyz2\"]\n``````\n\n## #`re_match(pattern, string[, \"i\"])`\n\nTests whether a string starts with a defined pattern. Use re_search() to allow matches anywhere in the string.\n\nOptionally, it is also possible to match case insensitive (`\"i\"`).\n\n``````param1 = \"String1String2\"\n``````\n``````re_match(\"String\\d\", #param1)\nre_match(\"String2\", #param1)\n``````\n``````true\nfalse\n``````\n``````param1 = \"STRING1STRING2\"\n``````\n``````re_match(\"String\\d\", #param1, \"i\")\n``````\n``````true\n``````\n\n## #`re_remove(list, pattern)`\n\nReturns the list of all elements in `list` which do not match `pattern`.\n\nEquivalent to\n\n``````filter(flatten(#list), [item -> not re_search(#pattern, #item)])\n``````\n• list - the input list\n• pattern - a regular expression pattern\n``````list = [\"xyz1\", \"abc\", \"xyz2\"]\n``````\n``````re_remove(#list, \"abc\")\n``````\n``````[\"xyz1\", \"xyz2\"]\n``````\n\n## #`re_remove(list, pattern, direction, position)`\n\nThis method searches for a regex pattern in a list and returns a list of all elements that do not contain matched elements. Optionally, a starting position and a search direction can be given.\n\n• list - the input list\n• pattern - a regular expression pattern\n• direction - the direction of the search Valid values: `l` and `r`\n• position - position in the list, where the search starts Valid values: any integer >= 0 parameter is optional\n``````list = [\"xyz1\", \"abc\", \"xyz2\"]\n``````\n``````re_remove(#list, \"abc\", \"l\", 1)\n``````\n``````[\"xyz1\", \"xyz2\"]\n``````\n``````list = [\"xyz1\", \"abc\", \"xyz2\"]\n``````\n``````re_remove(#list, \"xyz.\", \"r\", 1)\n``````\n``````[\"xyz1\", \"abc\"]\n``````\n\n## #`re_replace(string, old, new)`\n\nReturns a new string in which all occurrences conforming to the RegEx pattern old in string have been replaced by new.\n\n``````value = \"old Old old Old\"\n``````\n``````re_replace(#value, \"(o|O)ld\", \"new\")\n``````\n``````\"new new new new\"\n``````\n\nThe replacement string can include the following special replacement patterns:\n\nPattern Effect\n`\\$\\$` Inserts a literal \"\\$\" (only needed if followed by a number)\n`\\$0` or `\\$&` Inserts the matched substring\n`\\$1``\\$9` Inserts the nth parenthesized submatch string\n``````value = \"abc 1234 def 5678\"\n``````\n``````re_replace(#value, \"([0-9])([0-9]+)\", \"+\\$1\\$1 \\$2\")\n``````\n``````\"abc +11 234 def +55 678\"\n``````\n\n## #`re_search(pattern, string[, \"i\"])`\n\nTests whether a string matches a pattern. See re_match() to only allow matches at the start of `string` (or start pattern with `^`).\n\nOptionally, it is also possible to match case insensitive (`\"i\"`).\n\n``````param1 = \"String1String2\"\n``````\n``````re_search(\"String\\d\", #param1)\nre_search(\"String2\", #param1)\n``````\n``````true\ntrue\n``````\n``````param1 = \"STRING1STRING2\"\n``````\n``````re_search(\"String\\d\", #param1, \"i\"))\n``````\n``````true\n``````\n\n## #`reduce(list, lambda_expression, initial_value)`\n\nExecutes a reducer function on each element of the list, resulting in a single output value.\n\nFor example, `reduce(#list, [a, b -> #a + #b], 0)` is equivalent to `sum()`.\n\nThis works with an internal value called the accumulator, which is first set to `initial_value`. Then the given lambda is called with arguments `[accumulator, item]` for each item of the list. The accumulator is set to the lambda's return value, and its final value is returned.\n\nIf the list is empty, initial_value will be returned without calling the function.\n\nAn example execution of `reduce([2, 3, 4], [acc, x -> #acc * x], 1)`:\n\n callback iteration accumulator item return value first call 1 2 1 * 2 = 2 second call 2 3 2 * 3 = 6 third call 6 4 6 * 4 = 24\n``````reduce([2, 3, 4], [acc, x -> #acc + #x], 0) # 9\nreduce([2, 3, 4], [acc, x -> #acc * #x], 1) # 24\n``````\n\nIts true flexibility is that it can be used with any operator or function:\n\n``````data = [{\"a\": 1}, {\"b\": 2}, {\"c\": 34}]\n``````\n``````reduce(#data, [acc, x -> object_add(#acc, #x)])\n``````\n``````{\"a\": 1, \"b\": 2, \"c\": 34}\n``````\n\nWARNING\n\nContrary to `reduce()` in e.g. Javascript, omitting the initial_value argument will not default it to the first item of the list, but rather `\"\"`. We recommend that you always give an initial value.\n\n## #`replace(string, old, new)`\n\nThis method replaces all occurrences of a substring with another.\n\n``````param1 = \"old old old old\"\n``````\n``````replace(#param1, \"old\", \"new\")\n``````\n``````\"new new new new\"\n``````\n\n## #`replace(string, old, new, max)`\n\nThis method replaces all occurrences of a substring with a new stated substring.\n\nOptional parameter max sets the number of times of replacements.\n\n``````param1 = \"old old old old\"\n``````\n``````replace(#param1, \"old\", \"new\", 2)\n``````\n``````\"new new old old\"\n``````\n\n## #`replace_last(string, old, new)`\n\nReturns a new string in which the last occurrences of old in string have been replaced with new.\n\n``````param1 = \"1, 2, 3, 4\"\n``````\n``````replace_last(#param1, \",\", \" and\")\n``````\n``````\"1, 2, 3 and 4\"\n``````\n\n## #`reverse(list)`\n\nReverses the order of elements in a list.\n\n``````list = [1, 2, 3, 4, 5]\n``````\n``````reverse(#list)\n``````\n``````[5, 4, 3, 2, 1]\n``````\n\n## #`rnd_dbl()`\n\nReturns a random number between 0 and 1.\n\n``````rnd_dbl()\n``````\n``````0.599179536848878\n``````\n\n## #`rnd_dbl(lower_limit, upper_limit)`\n\nThis function returns a random number in the interval limited by lower_limit and upper_limit.\n\n``````lower = 0.0\nupper = 10.0\n``````\n``````rnd_dbl(#lower, #upper)\n``````\n``````7.73589949018131\n``````\n\n## #`rnd_int(lower_limit, upper_limit)`\n\nThis function returns a random integer in the interval limited by lower_limit and upper_limit (inclusive).\n\n``````param1 = 0\nparam2 = 10\n``````\n``````rnd_int(#param1, #param2)\n``````\n``````3\n``````\n\n## #`round(number)`\n\nRound `number` to the next whole number. In case of .5, round towards the next even number (\"banker's rounding\").\n\n``````round(0.6)\nround(2.5)\nround(3.5)\n``````\n``````1\n2\n4\n``````\n\n## #`round(number, decimals[, mode])`\n\nRound `number` to the given number of decimal places (0 to round to a whole number).\nThe optional parameter mode selects a rounding behavior:\n\n• `\"up\"`: Round away from zero\n• `\"down\"`: Round towards zero\n• `\"half_up\"`: Round to nearest digit, with .5 going away from zero.\n• `\"half_down\"`: Round to nearest, with .5 going towards zero.\n• `\"half_even\"`: Round to nearest, with .5 going towards next even number (\"banker's rounding\", the default)\n``````#num = 0.144743575\n``````\n``````round(#num, 3)\nround(#num, 2)\nround(3.85, 1)\nround(3.75, 1)\n``````\n``````0.145\n0.14\n3.8\n3.8\n``````\n\nRounding modes:\n\n``````round(3.2, 0, \"up\")\nround(3.8, 0, \"down\")\nround(2.5, 0, \"half_up\")\nround(3.5, 0, \"half_down\")\nround(3.85, 1, \"half_even\")\n``````\n``````4\n3\n3\n3\n3.8\n``````\n\n## #`slice(list, start[, end])`\n\nGets a slice of a list by providing a start- and endpoint.\n\nTIP\n\nSee range() if an index and count argument is more convenient for your use case.\n\n`start` and `end` are zero-based indices (0 meaning the first element, 1 the second, and so on), but can also be negative to count from the end of the list (-1 being the last element, and so on).\n\nThe function returns all list elements from `start` to just before `end` (which is not included). If `end` is omitted, returns all elements to the end of the list (the same could be done by supplying `count(#list)`).\n\n``````items = [1, 2, 3, 4, 5]\n``````\n``````slice(#items, 1, 4) # select index 1, 2, and 3\nslice(#items, 2, -1) # select from third to second-to-last\nslice(#items, -2) # select the last two elements\nslice(#items, 1) # omit the first element\nslice(#items, -3, -1) # select third-to-last and second-to-last\n``````\n``````[2, 3, 4]\n[3, 4]\n[4, 5]\n[2, 3, 4, 5]\n[3, 4]\n``````\n\nIn case `start` is later than `end`, or there are not enough elements, a shorter or empty list is returned:\n\n``````items = [1, 2, 3, 4, 5]\n``````\n``````slice(#items, 2, 2)\nslice(#items, -10)\nslice(#items, 2, 10)\n``````\n``````[]\n[1, 2, 3, 4, 5]\n[3, 4, 5]\n``````\n\n## #`sort(list)`\n\nSorts a list numerically or alphabetically (lowest to highest or A to Z).\n\n``````unsorted_numbers = [4, 5, 3, 2, 1]\n``````\n``````sort(#unsorted_numbers)\n``````\n``````[1, 2, 3, 4, 5]\n``````\n\n## #`sort(list, fieldname/lambda_expression)`\n\nSorts a list of objects by a given field, or by a custom key from a lambda expression.\n\n``````products = [{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"489.98\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"630.00\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"596.97\",\n\"Ram\": \"6 GB\"\n},{\n\"Price\": \"692.68\",\n\"Ram\": \"4 GB\"\n}]\n``````\n``````sort(#products, \"Price\")\n``````\n``````[{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"489.98\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"596.97\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"630.00\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Price\": \"692.68\",\n\"Ram\": \"4 GB\"\n}\n]\n``````\n\nWARNING\n\nReturns a sorted list with original elements sorted by the \"Price\" key in them (compare Huawei entries).\n\nExample list of objects:\n\n``````products = [{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"489.98\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"630.00\",\n\"Ram\": \"6 GB\"\n},{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"596.97\",\n\"Ram\": \"6 GB\"\n},{\n\"Price\": \"692.68\",\n\"Ram\": \"4 GB\"\n}]\n``````\n``````sort(#products, [item -> int(#item.Price)])\n``````\n``````[{\n\"Brand\": \"Xiaomi\",\n\"Model\": \"Mi Mix\",\n\"Price\": \"482.99\",\n\"Ram\": \"4 GB\"\n},\n{\n\"Brand\": \"OnePlus\",\n\"Model\": \"OnePlus 3T A3010\",\n\"Price\": \"489.98\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei P10 Plus VKY-L29\",\n\"Price\": \"596.97\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Brand\": \"Huawei\",\n\"Model\": \"Huawei Mate 9 Pro\",\n\"Price\": \"630.00\",\n\"Ram\": \"6 GB\"\n},\n{\n\"Price\": \"692.68\",\n\"Ram\": \"4 GB\"\n}]\n``````\n\nFor each element, the field `Price` is interpreted as integer (i.e. `[482, 489, 630, 596, 692]`), then the list is sorted according to that.\n\nWhen testing, you can replace `sort` with `map` to see what your key function is doing.\n\n## #`split(string[, delimiter])`\n\nSplits a string at every occurrence of `delimiter`, returning a list. If omitted, `delimiter` defaults to \" \" (whitespace).\n\nIf `delimiter` is not found, this function still returns a list, with `string` as the only item. If `delimiter` occurs multiple times in a row, you'll get empty elements in between.\n\n``````string_with_spaces = \"string1 string2 string3\"\n``````\n``````split(#string_with_spaces)\n``````\n``````[\"string1\", \"string2\", \"string3\"]\n``````\n``````string_with_commas = \"string1, string2, string3\"\n``````\n``````split(#string_with_commas, \", \")\n``````\n``````[\"string1\", \"string2\", \"string3\"]\n``````\n\n## #`sqrt(number)`\n\nReturns the square root of `number`.\n\n``````sqrt(2)\n``````\n``````1.414213562373\n``````\n\n## #`startswith(string, start)`\n\nChecks if `string` begins with string `start` (upper/lower case must match). Returns true if `start` is empty.\n\nThis is equivalent to `substring(#string, 0, len(#start)) == #start`.\n\n``````value = \"foobaz\"\n``````\n``````startswith(#value, \"foo\")\nstartswith(#value, \"FOO\")\nstartswith(#value, \"\")\n``````\n``````true\nfalse\ntrue\n``````\n\n## #`substring(string, index, length)`\n\nThis method extracts a substring from a string with a starting given index and length.\n\n``````param1 = \"Hello world\"\n``````\n``````substring(#param1, 0, 1)\n``````\n``````\"H\"\n``````\n\nWARNING\n\nThe substring from index 0 with a length of 1 character.\n\n``````param1 = \"Hello world\"\n``````\n``````substring(#param1, 5, 3)\n``````\n``````\"wor\"\n``````\n\nWARNING\n\nThe substring from index 5 (the w) with a length of 3 characters.\n\n``````param1 = \"Hello world\"\n``````\n``````substring(#param1, 5)\n``````\n``````\"world\"\n``````\n\nWARNING\n\nThe substring from index 5 to the end of the string.\n\n## #`sum(list)`\n\nSums numeric elements in a list. It does not work with non-numeric elements.\n\n``````numbers = [1, 2, 3]\n``````\n``````sum(#numbers)\n``````\n``````6\n``````\n\n## #`traverse_json_path(object, path)`\n\nThis function allows the navigation through complex JSON structures and the search in lists and sublists with a search mask.\n\n`path` may consist of the following components:\n\n• `.key`: A key in an object. Contrary to the normal form available with parameters, this may contain whitespace and any character except `.`, `[`, or `]`.\n• `[i]`: An element of a list, by index.\n• `[key=values;otherKey=values]`: Finds the first element of the list which\n• is an object\n• has all of the specified keys\n• each key has one of the `|`-separated values (exact match).\n``````object = {\n\"data\": [{\n\"type\": \"a\",\n\"color\": \"red\",\n\"price\": 345\n}, {\n\"type\": \"b\",\n\"color\": \"red\",\n\"price\": 239\n}, {\n\"type\": \"b\",\n\"color\": \"blue\",\n\"price\": 499\n}]\n}\n``````\n``````traverse_json_path(#object, \"data.price\")\ntraverse_json_path(#object, \"data[type=b].price\")\ntraverse_json_path(#object, \"data[type=b;color=blue|green].price\")\n``````\n``````345\n239\n499\n``````\n\n## #`trim(string)`\n\nThis method cuts away trailing and preceding whitespaces from a string.\n\n``````param1 = \"hello world\"\n``````\n``````trim(#param1)\n``````\n``````\"hello world\"\n``````\n\nWARNING\n\nNothing to trim\n\n``````param1 = \" hello world\"\n``````\n``````trim(#param1)\n``````\n``````\"hello world\"\n``````\n\nWARNING\n\nStripped whitespace from the beginning\n\nExample data param1:\n\n``````param1 = \"hello world \"\n``````\n``````trim(#param1)\n``````\n``````\"hello world\"\n``````\n\nWARNING\n\nstripped whitespace from the end\n\n``````param1 = \" hello world \"\n``````\n``````trim(#param1)\n``````\n``````\"hello world\"\n``````\n\nWARNING\n\nStripped whitespace from the beginning and the end\n\n## #`unique(list)`\n\nFinds the unique elements of a list.\n\nThe list elements must be strings or numbers. Nested lists are supported for convenience, and will be treated as a single flat list. The result is in the same order as the input, i.e. sorted by the element's first occurrence.\n\nTIP\n\nIf you also need the number of repeated elements, see count_uniques().\n\n``````list = [1, 2, 3, 1, 2]\n``````\n``````unique(#list)\n``````\n``````[1, 2, 3]\n``````\n\nUsing a nested list:\n\n``````list = [\"cat\", \"dog\", \"cat\"]\n``````\n``````unique([#list, 14, 10, 12, 10])\n``````\n``````[\"cat\", \"dog\", 14, 10, 12]\n``````\n\n## #`upper(string)`\n\nConverts a string to uppercase.\n\n``````param1 = \"string\"\n``````\n``````upper(#param1)\n``````\n``````\"STRING\"\n``````\n\n## #`weekday_int(days)`\n\nThis method takes a numeric n between 1 and 7. It returns a numeric representation of the weekday of today + n days. The returned numeric is to be interpreted as follows:\n\nWeekday Numeric\nMonday 1\nTuesday 2\nWednesday 3\nThursday 4\nFriday 5\nSaturday 6\nSunday 7\n``````weekday_int(1)\n``````\n``````2 # on Monday\n``````\n\nWARNING\n\nThe function returns the weekday of tomorrow: Assuming today is Monday (1), it gives 2 (Tuesday)\n\n``````days_to_add = 6\n``````\n``````weekday_int(#days_to_add)\n``````\n``````1 # on Tuesday\n``````\n\nWARNING\n\nAssuming today is Tuesday (2), in six days it will be Monday (1).\n\n## #`weekday_no(day)`\n\nThis method can be used to convert a weekday into a numerical representation.\n\nWARNING\n\nThis method only supports the German and English notation of weekdays.\n\n``````weekday = \"Tuesday\"\n``````\n``````weekday_no(#weekday)\n``````\n``````2\n``````\n\nWARNING\n\nTuesday is the second day in a week.\n\n``````weekday = \"Monday\"\n``````\n``````weekday_no(#weekday)\n``````\n``````1\n``````\n\nWARNING\n\nMonday is the first day of the week." ]

[ null ]

[ "• :00Days\n• :00Hours\n• :00Mins\n• 00Seconds\nA new era for learning is coming soon", null, "Suggested languages for you:\n\nEurope\n\nAnswers without the blur. Sign up and see all textbooks for free!", null, "Q35E\n\nExpert-verified", null, "Found in: Page 324", null, "### Linear Algebra With Applications\n\nBook edition 5th\nAuthor(s) Otto Bretscher\nPages 442 pages\nISBN 9780321796974", null, "# Show that similar matrices have the same eigenvalues. Hint:$\\stackrel{\\mathbf{\\to }}{\\mathbf{v}}$ If is an eigenvector of ${{\\mathbf{S}}}^{\\mathbf{-}\\mathbf{1}}{\\mathbf{AS}}$, then role=\"math\" localid=\"1659529994406\" ${\\mathbf{S}}\\stackrel{\\mathbf{\\to }}{\\mathbf{v}}{\\mathbf{}}$ is an eigenvector of A.\n\nWe have proved that similar matrices have the same eigenvalues.\n\nSee the step by step solution\n\n## Step 1: Definition of the Eigenvectors\n\nEigenvectors are a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector.\n\n## Step 2: Find eigenvalue\n\nAssume, that is an Eigen vector for${\\mathrm{S}}^{1}\\mathrm{AS}$ .\n\nTherefore, by definition:\n\n${S}^{-1}A{S}^{r}v=\\lambda {v}^{r}$\n\nNow manipulate the above equation as shown below:\n\n${\\mathrm{S}}^{-1}\\mathrm{AS}\\stackrel{^}{\\mathrm{v}}={\\mathrm{\\lambda v}}_{\\mathrm{r}}^{\\mathrm{r}}{\\mathrm{SS}}^{-1}\\phantom{\\rule{0ex}{0ex}}\\mathrm{ASv}={\\mathrm{S\\lambda }}_{\\mathrm{r}}^{\\mathrm{r}} \\left(\\mathrm{Multiplyby}\\mathrm{S}\\mathrm{fromLeft}\\right)\\phantom{\\rule{0ex}{0ex}}\\mathrm{ASv}=\\mathrm{\\lambda S}\\stackrel{\\mathrm{r}}{\\mathrm{r}} \\left(\\mathrm{M}\\right)$\n\nFrom the above, the Eigen value of $A$ is $\\lambda$and ${S}_{v}^{r}$ is the Eigenvector.\n\nSimilarly, Eigen vector of A is ${S}_{w}^{u}$ then ${{S}^{-1}}_{w}^{u}$ is an Eigen vector for ${S}^{-1}AS.$\n\nHence, similar matrices have same eigen values.", null, "### Want to see more solutions like these?", null, "" ]

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[ "# GSEB Solutions Class 7 Maths Chapter 10 Practical Geometry Ex 10.2\n\nGujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Textbook Questions and Answers.\n\n## Gujarat Board Textbook Solutions Class 7 Maths Chapter 10 Practical Geometry Ex 10.2", null, "Question 1.\nConstruct ∆XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.\nSolution:\nSteps of construction:\nI. Draw a line segment YZ = 5 cm.\nII. With centre Y and radius 4.5 cm, draw an arc.\nIII. With centre Z and radius 6 cm, draw another arc to cut the previous arc at X.\nIV. Join YX and ZX.", null, "Thus, ∆XYZ is the required triangle.\n\nQuestion 2.\nConstruct an equilateral triangle of side 5.5 cm.\nSolution:\nSteps of construction:\nI. Draw a line segment BC = 5.5 cm.\nII. With B as centre and radius 5.5 cm, draw an arc.\nIII. With centre C and radius 5.5 cm, draw another arc to cut the previous arc at A.\nIV. Join BA and CA.", null, "Thus, ∆ABC is the required equilateral triangle.", null, "Question 3.\nDraw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?\nSolution:\nSteps of construction:\nI. Draw a line segment QR = 3.5 cm.\nII. With Q as centre and radius 4 cm, draw an arc.\nIII. With R as centre and radius 4 cm, draw another arc such that it cuts the previous arc at R\nIV. Join QP and RP.\nThus, ∆PQR is the required triangle.\nSince, PQ = PR = 4 cm", null, "Therefore, PQR is an isosceles triangle.", null, "Question 4.\nConstruct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.\nSolution:\nSteps of construction:\nI. Draw a line segment BC = 6 cm.\nII. With centre as B and radius 2.5 cm, draw an arc.\nIII. With centre C and radius 6.5 cm, draw another arc such that it cuts the previous arc at A.\nIV. Join BA and CA.", null, "Thus, ∆ABC is the required triangle. On measuring, we find that ∠B = 90°." ]

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[ "# Equivalence Of Thevenin's And Norton's Theorems\n\nAccording to Thevenin's theorem a two terminal power supply can be represented like this:", null, "And According to Norton's theorem the same power supply can be represented like this:", null, "From the Thevenin circuit we have\n\n```IL = VT/(Rint+RL)\n```\n\nFrom the Norton circuit we have\n\n```VAB = ISC(RintRL/(Rint+RL))\n= ILRL\nILRL = ISC(RintRL/(Rint+RL))\nIL = VAB/RL\n= ISCRint/(Rint+RL)\n```\n\nVAB is the same for both circuits, so we can substitute what we stated IL was for the Thevenin circuit:\n\n```ISCRint/(Rint+RL) = VT/(Rint+RL)\nISCRint = VT\n```\n\nBoth theorems give us a way to represent complicated circuits as simpler ones, either as a simple series circuit, or a simple parallel circuit.\n\n## References\n\nFischer-Cripps. A.C., The Electronics Companion. Institute of Physics, 2005." ]

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[ "# ارزیابی رفتار ماسه‌ی زهکشی‌نشده با استفاده از پارامتر حالت\n\nنوع مقاله : پژوهشی\n\nنویسندگان\n\n1 دانشکده مهندسی عمران، دانشگاه تبریز\n\n2 دانشکده مهندسی عمران، دانشگاه صنعتی امیرکبیر\n\nچکیده\n\nمطالعات تجربی نشان داده‌اند که رفتار خاک‌های دانه‌یی از نسبت تخلخل و تنش همه‌جانبه‌ی تحکیم اولیه تأثیر می‌پذیرند. بنابراین لزوم تدوین روابطی برای تعریف تأثیر آنها در رفتار ماسه در چگالی‌ها، تنش‌های همه‌جانبه و برشی مختلف دیده می‌شود. در این نوشتار با معرفی ۴ پارامتر حالت: نسبت تنش همه‌جانبه، نسبت تنش همه‌جانبه‌ی ابتدایی، نسبت تنش برشی ابتدایی و شاخص فشار اولیه، دو فضای تنش جدید جهت تعریف رفتار ماسه تحت بارگذاری زهکشی‌نشده و یکنواخت در تمامی محدوده‌های تنش همه‌جانبه و چگالی پیشنهاد و دو رابطه بر اساس این فضاها ارائه شده است. در ادامه، با استفاده از داده‌های آزمایشگاهی دیگر پژوهشگران، تمامی مسیرهای تنش و مسیرهای تنشٓـ کرنش برای وضعیت‌های مختلف ماسه از حالت‌های شل تا متراکم تحت بارگذاری زهکشی‌نشده و یکنواخت شبیه‌سازی شده است. در نهایت، جهت صحت‌سنجی روابط پیشنهادی، نتایج حاصل با داده‌های آزمایشگاهی مربوط به آزمایش‌های سه محوری زهکشی‌نشده‌ی فشاری مقایسه و مورد بحث و ارزیابی قرار گرفته‌اند.\n\nکلیدواژه‌ها", null, "20.1001.1.26764768.1394.312.22.3.2\n\nعنوان مقاله [English]\n\n### E‌V‌A‌L‌U‌A‌T‌I‌N‌G S‌A‌N‌D B‌E‌H‌A‌V‌I‌O‌R T‌H‌R‌O‌U‌G‌H S‌T‌A‌T‌E P‌A‌R‌A‌M‌E‌T‌E‌R‌S\n\nنویسندگان [English]\n\n• T. A‌k‌h‌l‌a‌g‌h‌i 1\n• N. V‌a‌f‌a‌e‌i 2\n• H. K‌a‌t‌e‌b‌i 1\n1 D‌e‌p‌t. o‌f C‌i‌v‌i‌l E‌n‌g‌i‌n‌e‌e‌r‌i‌n‌g U‌n‌i‌v‌e‌r‌s‌i‌t‌y o‌f T‌a‌b‌r‌i‌z\n2 D‌e‌p‌t. o‌f C‌i‌v‌i‌l E‌n‌g‌i‌n‌e‌e‌r‌i‌n‌g A‌m‌i‌r‌k‌a‌b‌i‌r U‌n‌i‌v‌e‌r‌s‌i‌t‌y o‌f T‌e‌c‌h‌n‌o‌l‌o‌g‌y\nچکیده [English]\n\nI‌t h‌a‌s b‌e‌e‌n u‌n‌d‌e‌r‌s‌t‌o‌o‌d f‌r‌o‌m e‌x‌p‌e‌r‌i‌m‌e‌n‌t‌a‌l s‌t‌u‌d‌i‌e‌s t‌h‌a‌t g‌r‌a‌n‌u‌l‌a‌r s‌o‌i‌l b‌e‌h‌a‌v‌i‌o‌r c‌a‌n b‌e h‌i‌g‌h‌l‌y a‌f‌f‌e‌c‌t‌e‌d b‌y d‌e‌n‌s‌i‌t‌y a‌n‌d m‌e‌a‌n p‌r‌i‌n‌c‌i‌p‌a‌l e‌f‌f‌e‌c‌t‌i‌v‌e s‌t‌r‌e‌s‌s. T‌h‌e e‌f‌f‌e‌c‌t o‌f t‌h‌e‌s‌e f‌a‌c‌t‌o‌r‌s o‌n s‌a‌n‌d b‌e‌h‌a‌v‌i‌o‌r i‌s o‌f g‌r‌e‌a‌t s‌i‌g‌n‌i‌f‌i‌c‌a‌n‌c‌e, a‌n‌d d‌e‌f‌i‌n‌i‌n‌g\nt‌h‌e n‌a‌t‌u‌r‌e o‌f t‌h‌e l‌i‌n‌k b‌e‌t‌w‌e‌e‌n t‌h‌e‌m a‌n‌d t‌h‌e‌i‌r c‌o‌n‌t‌r‌o‌l‌s o‌n t‌h‌e b‌e‌h‌a‌v‌i‌o‌r o‌f s‌a‌n‌d, i‌s c‌a‌r‌r‌i‌e‌d o‌u‌t i‌n t‌h‌i‌s s‌t‌u‌d‌y.\nS‌a‌n‌d, i‌n i‌t‌s l‌o‌o‌s‌e o‌r d‌e‌n‌s‌e s‌t‌a‌t‌e, b‌e‌f‌o‌r‌e a‌p‌p‌r‌o‌x‌i‌m‌a‌t‌i‌n‌g i‌t‌s c‌r‌i‌t‌i‌c‌a‌l s‌t‌a‌t‌e, o‌b‌t‌a‌i‌n‌s a c‌o‌n‌s‌t‌a‌n‌t s‌t‌r‌e‌s‌s r‌a‌t‌i‌o. T‌h‌e‌r‌e‌f‌o‌r‌e, c‌h‌a‌n‌g‌e‌s i‌n t‌h‌e s‌t‌r‌e‌s‌s r‌a‌t‌i‌o o‌f s‌h‌e‌a‌r s‌t‌r‌e‌s‌s t‌o m‌e‌a‌n e‌f‌f‌e‌c‌t‌i‌v‌e s‌t‌r‌e‌s‌s m‌a‌y b‌e i‌g‌n‌o‌r‌e‌d. I‌t i‌s w‌o‌r‌t‌h m‌e‌n‌t‌i‌o‌n‌i‌n‌g t‌h‌a‌t i‌n l‌o‌o‌s‌e s‌a‌n‌d, i‌t i‌s c‌o‌r‌r‌e‌s‌p‌o‌n‌d‌e‌d t‌o p‌h‌a‌s‌e t‌r‌a‌n‌s‌f‌o‌r‌m‌a‌t‌i‌o‌n, b‌u‌t, i‌t h‌a‌p‌p‌e‌n‌s n‌e‌x‌t t‌o p‌h‌a‌s‌e t‌r‌a‌n‌s‌f‌o‌r‌m‌a‌t‌i‌o‌n i‌n d‌e‌n‌s‌e s‌a‌n‌d. T‌h‌e p‌o‌i‌n‌t a‌t w‌h‌i‌c‌h t‌h‌e c‌o‌n‌s‌t‌a‌n‌t r‌a‌t‌i‌o i‌n‌i‌t‌i‌a‌t‌e‌s i‌s o‌f g‌r‌e‌a‌t i‌m‌p‌o‌r‌t‌a‌n‌c‌e a‌n‌d i‌s c‌a‌l‌l‌e‌d I‌n‌i‌t‌i‌a‌l F‌a‌i‌l‌u‌r‌e P‌o‌i‌n‌t. T‌h‌e p‌r‌e‌s‌e‌n‌t s‌t‌u‌d‌y t‌r‌i‌e‌s t‌o o‌u‌t‌l‌i‌n‌e t‌w‌o n‌o‌b‌l‌e s‌t‌a‌t‌e p‌a‌r‌a‌m‌e‌t‌e‌r‌s; e‌a‌r‌l‌y c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s r‌a‌t‌i‌o, $R_{p‌e}$, a‌n‌d e‌a‌r‌l‌y s‌h‌e‌a‌r s‌t‌r‌e‌s‌s r‌a‌t‌i‌o, $R_{s‌e}$. E‌a‌r‌l‌y c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s r‌a‌t‌i‌o a‌n‌d e‌a‌r‌l‌y s‌h‌e‌a‌r s‌t‌r‌e‌s‌s r‌a‌t‌i‌o a‌r‌e d‌e‌f‌i‌n‌e‌d a‌s t‌h‌e s‌t‌r‌e‌s‌s r‌a‌t‌i‌o o‌f t‌h‌e c‌u‌r‌r‌e‌n‌t\nm‌e‌a‌n e‌f‌f‌e‌c‌t‌i‌v‌e s‌t‌r‌e‌s‌s t‌o t‌h‌e m‌e‌a‌n e‌f‌f‌e‌c‌t‌i‌v‌e s‌t‌r‌e‌s‌s o‌f t‌h‌e e‌a‌r‌l‌y p‌o‌i‌n‌t o‌f t‌h‌e f‌a‌i‌l‌u‌r‌e l‌i‌n‌e, a‌n‌d a‌s t‌h‌e s‌t‌r‌e‌s‌s r‌a‌t‌i‌o o‌f t‌h‌e c‌u‌r‌r‌e‌n‌t s‌h‌e‌a‌r s‌t‌r‌e‌s‌s t‌o t‌h‌e s‌h‌e‌a‌r s‌t‌r‌e‌s‌s o‌f t‌h‌e e‌a‌r‌l‌y p‌o‌i‌n‌t o‌f t‌h‌e f‌a‌i‌l‌u‌r‌e l‌i‌n‌e. F‌u‌r‌t‌h‌e‌r‌m‌o‌r‌e, a‌n‌o‌t‌h‌e‌r s‌t‌a‌t‌e p‌a‌r‌a‌m‌e‌t‌e‌r, c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s r‌a‌t‌i‌o, $R_{p0}$, i‌s d‌e‌f‌i‌n‌e‌d a‌s t‌h‌e s‌t‌r‌e‌s‌s r‌a‌t‌i‌o o‌f t‌h‌e c‌u‌r‌r‌e‌n‌t s‌t‌r‌e‌s‌s t‌o t‌h‌e i‌n‌i‌t‌i‌a‌l m‌e‌a‌n e‌f‌f‌e‌c‌t‌i‌v‌e s‌t‌r‌e‌s‌s.\nF‌i‌n‌a‌l‌l‌y, t‌h‌r‌o‌u‌g‌h i‌n‌t‌r‌o‌d‌u‌c‌i‌n‌g f‌o‌u‌r s‌t‌a‌t‌e p‌a‌r‌a‌m‌e‌t‌e‌r‌s i‌n‌c‌l‌u‌d‌i‌n‌g: i‌n‌i‌t‌i‌a‌l s‌t‌a‌t‌e p‌r‌e‌s‌s‌u‌r‌e i‌n‌d‌e‌x, $I_{p0}$, c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s r‌a‌t‌i‌o, $R_{p0}$ , e‌a‌r‌l‌y c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s r‌a‌t‌i‌o, $R_{p‌e}$, e‌a‌r‌l‌y s‌h‌e‌a‌r s‌t‌r‌e‌s‌s r‌a‌t‌i‌o, $R_{s‌e}$, t‌w‌o s‌t‌r‌e‌s‌s s‌p‌a‌c‌e‌s o‌f e‌a‌r‌l‌y c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s r‌a‌t‌i‌o-e‌a‌r‌l‌y s‌h‌e‌a‌r s‌t‌r‌e‌s‌s r‌a‌t‌i‌o a‌n‌d a‌x‌i‌a‌l s‌t‌r‌a‌i‌n-e‌a‌r‌l‌y c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s r‌a‌t‌i‌o, a‌r‌e i‌n‌t‌r‌o‌d‌u‌c‌e‌d t‌o d‌e‌f‌i‌n‌e s‌a‌n‌d b‌e‌h‌a‌v‌i‌o‌r u‌n‌d‌e‌r u‌n‌d‌r‌a‌i‌n‌e‌d m‌o‌n‌o‌t‌o‌n‌i‌c l‌o‌a‌d‌i‌n‌g i‌n a‌l‌l r‌a‌n‌g‌e‌s o‌f c‌o‌n‌f‌i‌n‌i‌n‌g s‌t‌r‌e‌s‌s a‌n‌d d‌e‌n‌s‌i‌t‌y, f‌r‌o‌m l‌o‌o‌s‌e s‌t‌a‌t‌e t‌o d‌e‌n‌s‌e s‌t‌a‌t‌e. S‌u‌b‌s‌e‌q‌u‌e‌n‌t‌l‌y, a‌c‌c‌o‌r‌d‌i‌n‌g t‌o t‌h‌e‌s‌e s‌p‌a‌c‌e‌s, t‌w‌o r‌e‌l‌a‌t‌i‌o‌n‌s a‌r‌e o‌f‌f‌e‌r‌e‌d. T‌h‌u‌s, u‌n‌d‌r‌a‌i‌n‌e‌d m‌o‌n‌o‌t‌o‌n‌i‌c l‌o‌a‌d‌i‌n‌g‌s a‌r‌e s‌i‌m‌u‌l‌a‌t‌e‌d. E‌v‌e‌n‌t‌u‌a‌l‌l‌y, i‌n o‌r‌d‌e‌r t‌o e‌v‌a‌l‌u‌a‌t‌e t‌h‌e a‌c‌c‌u‌r‌a‌c‌y o‌f t‌h‌e p‌r‌o‌p‌o‌s‌e‌d r‌e‌l‌a‌t‌i‌o‌n‌s, t‌h‌e r‌e‌s‌u‌l‌t‌i‌n‌g s‌i‌m‌u‌l‌a‌t‌i‌o‌n‌s a‌r‌e c‌o‌m‌p‌a‌r‌e‌d w‌i‌t‌h o‌t‌h‌e‌r l‌a‌b‌o‌r‌a‌t‌o‌r‌y t‌e‌s‌t‌s i‌n t‌h‌e l‌i‌t‌e‌r‌a‌t‌u‌r‌e o‌f u‌n‌d‌r‌a‌i‌n‌e‌d t‌r‌i‌a‌x‌i‌a‌l c‌o‌m‌p‌r‌e‌s‌s‌i‌o‌n. T‌h‌e r‌e‌s‌u‌l‌t‌s s‌h‌o‌w v‌e‌r‌y g‌o‌o‌d a‌g‌r‌e‌e‌m‌e‌n‌t, i‌n‌d‌i‌c‌a‌t‌i‌n‌g t‌h‌a‌t t‌h‌e p‌r‌o‌p‌o‌s‌e‌d r‌e‌l‌a‌t‌i‌o‌n‌s a‌r‌e c‌a‌p‌a‌b‌l‌e o‌f s‌i‌m‌u‌l‌a‌t‌i‌n‌g u‌n‌d‌r‌a‌i‌n‌e‌d s‌a‌n‌d b‌e‌h‌a‌v‌i‌o‌r u‌n‌d‌e‌r m‌o‌n‌o‌t‌o‌n‌i‌c\nl‌o‌a‌d‌i‌n‌g.\n\nکلیدواژه‌ها [English]\n\n• s‌a‌n‌d\n• s‌t‌a‌t‌e p‌a‌r‌a‌m‌e‌t‌e‌r\n• s‌t‌r‌e‌s‌s r‌a‌t‌i‌o\n• c‌r‌i‌t‌i‌c‌a‌l s‌t‌a‌t‌e\n• u‌n‌d‌r‌a‌i‌n‌e‌d l‌o‌a‌d‌i‌n‌g\n• s‌t‌r‌e‌s‌s p‌a‌t‌h" ]

[ null, "https://sjce.journals.sharif.edu/images/dor.png", null ]

[ "", null, "CrossCorrelation2D - Maple Help\n\nSignalProcessing\n\n CrossCorrelation2D\n calculate the cross-correlation between a pair of one- or two-dimensional rtables", null, "Calling Sequence CrossCorrelation2D( A, B, options )", null, "Parameters\n\n A, B - two one- or two-dimensional rtables container - (optional) container to compute and store the cross-correlation", null, "Description\n\n • The CrossCorrelation2D command takes two 1-D or 2-D rtables and computes their 2-D cross-correlation.\n • The inputs are converted to Matrices of complex datatype, and an error will be thrown if this is not possible. For this reason, it is most efficient for the inputs to already be Matrices having datatype complex.\n • The cross-correlation is computed via Discrete or Fast Fourier Transforms (DFTs or FFTs), but the result is equivalent to the standard definition (see example below). However, there may be numerical artifacts in the result that are small in size.\n • Suppose Matrices A and B, respectively, have dimensions (a,b) and (c,d), and define p=a+c-1 and q=b+d-1. Then, the cross-correlation Matrix of dimension (p,q) is given by the formula\n\n${C}_{i,j}={\\sum }_{r=1}^{a}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\sum }_{s=1}^{b}{A}_{r,s}\\stackrel{&conjugate0;}{{B}_{r-i+c,s-j+d}}$\n\n where A is assumed to be zero outside of the range of indices (1..a,1..b), and B is assumed to be zero outside of the range of indices (1..c,1..d).\n • FFTs are used when p and q, where p and q are as in the definition above, are powers of 2 and no less than 4 in value. Consequently, if p and q are both a little smaller than a power of 2, the user may want to pad the original Matrices with zeros. See below for an example.\n • If the container=C option is provided, then the results are stored in C. The container must have dimensions (p,q), and datatype complex.\n • If A is empty with no elements, a and b are both taken to be zero. Similarly, c and d are both taken to be zero if B has no elements. If both A and B are empty, an error will be thrown because the cross-correlation would have no meaning. However, if, say, B is empty but A is not, then the cross-correlation would be a zero Matrix of size (a-1,b-1).\n • As the underlying implementation of the SignalProcessing package is a module, it is also possible to use the form SignalProcessing:-CrossCorrelation2D to access the command from the package. For more information, see Module Members.", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{SignalProcessing}\\right):$", null, "Simple Examples", null, "Example 1\n\n > $A≔\\mathrm{Vector}\\left[\\mathrm{row}\\right]\\left(\\left[1,2\\right]\\right)$\n $\\left[\\begin{array}{rr}1& 2\\end{array}\\right]$ (1)\n > $B≔\\mathrm{Matrix}\\left(\\left[\\left[3,-4\\right],\\left[5I,6\\right]\\right]\\right)$\n $\\left[\\begin{array}{cc}3& -4\\\\ 5{}I& 6\\end{array}\\right]$ (2)\n > $C≔\\mathrm{CrossCorrelation2D}\\left(A,B\\right)$\n $\\left[\\begin{array}{ccc}5.999999999999999-1.0877919644084146{}{10}^{-15}{}I& 12.0-5.0{}I& 0.0-9.999999999999998{}I\\\\ -3.999999999999999-3.6259732146947156{}{10}^{-16}{}I& -4.999999999999999+0.0{}I& 5.999999999999999+0.0{}I\\end{array}\\right]$ (3)", null, "Example 2\n\n > $A≔\\mathrm{Matrix}\\left(\\left[\\left[1,2\\right],\\left[3,4\\right]\\right]\\right)$\n $\\left[\\begin{array}{rr}1& 2\\\\ 3& 4\\end{array}\\right]$ (4)\n > $B≔\\mathrm{Matrix}\\left(\\left[\\left[1,2,3\\right],\\left[4,5,6\\right],\\left[7,8,9\\right]\\right]\\right)$\n $\\left[\\begin{array}{rrr}1& 2& 3\\\\ 4& 5& 6\\\\ 7& 8& 9\\end{array}\\right]$ (5)\n > $\\mathrm{C1}≔\\mathrm{CrossCorrelation2D}\\left(A,B\\right)$\n $\\left[\\begin{array}{cccc}9.0+0.0{}I& 26.0+0.0{}I& 23.0+0.0{}I& 14.0+0.0{}I\\\\ 33.0+0.0{}I& 77.0+0.0{}I& 67.0+0.0{}I& 36.0+0.0{}I\\\\ 21.0+0.0{}I& 47.0+0.0{}I& 37.0+0.0{}I& 18.0+0.0{}I\\\\ 9.0+0.0{}I& 18.0+0.0{}I& 11.0+0.0{}I& 4.0+0.0{}I\\end{array}\\right]$ (6)\n > $\\mathrm{C2}≔\\mathrm{CrossCorrelation2D}\\left(B,A\\right)$\n $\\left[\\begin{array}{cccc}4.0+0.0{}I& 11.0+0.0{}I& 18.0+0.0{}I& 9.0+0.0{}I\\\\ 18.0+0.0{}I& 37.0+0.0{}I& 47.0+0.0{}I& 21.0+0.0{}I\\\\ 36.0+0.0{}I& 67.0+0.0{}I& 77.0+0.0{}I& 33.0+0.0{}I\\\\ 14.0+0.0{}I& 23.0+0.0{}I& 26.0+0.0{}I& 9.0+0.0{}I\\end{array}\\right]$ (7)", null, "Example 3\n\n > $A≔\\mathrm{Vector}\\left[\\mathrm{row}\\right]\\left(\\left[I,5,3-I\\right]\\right)$\n $\\left[\\begin{array}{ccc}I& 5& 3-I\\end{array}\\right]$ (8)\n > $B≔\\mathrm{Vector}\\left[\\mathrm{row}\\right]\\left(\\left[6-0.5I,7\\right]\\right)$\n $\\left[\\begin{array}{cc}6.-0.5{}I& 7\\end{array}\\right]$ (9)\n > $C≔\\mathrm{CrossCorrelation2D}\\left(A,B\\right)$\n $\\left[\\begin{array}{cccc}0.0+7.0{}I& 34.5+6.0{}I& 51.0-4.5{}I& 18.5-4.5{}I\\end{array}\\right]$ (10)", null, "Example 4\n\n > $A≔\\mathrm{Matrix}\\left(\\left[\\left[1,2+7I\\right],\\left[3-I,4\\right]\\right]\\right)$\n $\\left[\\begin{array}{cc}1& 2+7{}I\\\\ 3-I& 4\\end{array}\\right]$ (11)\n > $B≔\\mathrm{Array}\\left(\\left[\\left[5,6\\right],\\left[7,8+3I\\right]\\right]\\right)$\n $\\left[\\begin{array}{cc}5& 6\\\\ 7& 8+3{}I\\end{array}\\right]$ (12)\n > $C≔\\mathrm{Matrix}\\left(3,3,\\mathrm{datatype}=\\mathrm{complex}\\left[8\\right]\\right)$\n $\\left[\\begin{array}{ccc}0.0+0.0{}I& 0.0+0.0{}I& 0.0+0.0{}I\\\\ 0.0+0.0{}I& 0.0+0.0{}I& 0.0+0.0{}I\\\\ 0.0+0.0{}I& 0.0+0.0{}I& 0.0+0.0{}I\\end{array}\\right]$ (13)\n > $\\mathrm{CrossCorrelation2D}\\left(A,B,\\mathrm{container}=C\\right):$\n > $'C'=C$\n ${C}{=}\\left(\\left[\\begin{array}{ccc}8.000000000000004-2.999999999999995{}I& 43.999999999999986+50.0{}I& 14.000000000000004+49.0{}I\\\\ 27.0-16.999999999999993{}I& 69.99999999999999+23.0{}I& 38.0+34.999999999999986{}I\\\\ 18.0-5.999999999999995{}I& 38.999999999999986-4.999999999999997{}I& 20.0+7.105427357601002{}{10}^{-15}{}I\\end{array}\\right]\\right)$ (14)", null, "Direct Computation\n\n • To compute the cross-correlation directly from the definition, first take:\n > $X≔\\mathrm{Matrix}\\left(3,2,\\left[\\left[2,-2+I\\right],\\left[-I,2-I\\right],\\left[0,2-2I\\right]\\right]\\right)$\n $\\left[\\begin{array}{cc}2& -2+I\\\\ -I& 2-I\\\\ 0& 2-2{}I\\end{array}\\right]$ (15)\n > $Y≔\\mathrm{Matrix}\\left(4,3,\\left[\\left[1+I,2+2I,-1-I\\right],\\left[-2-I,1-I,-2I\\right],\\left[2I,1+2I,1+I\\right],\\left[-1,-I,2-I\\right]\\right]\\right)$\n $\\left[\\begin{array}{ccc}1+I& 2+2{}I& -1-I\\\\ -2-I& 1-I& -2{}I\\\\ 2{}I& 1+2{}I& 1+I\\\\ -1& -I& 2-I\\end{array}\\right]$ (16)\n > $a,b≔\\mathrm{upperbound}\\left(X\\right)$\n ${a}{,}{b}{≔}{3}{,}{2}$ (17)\n > $c,d≔\\mathrm{upperbound}\\left(Y\\right)$\n ${c}{,}{d}{≔}{4}{,}{3}$ (18)\n > $p≔a+c-1$\n ${p}{≔}{6}$ (19)\n > $q≔b+d-1$\n ${q}{≔}{4}$ (20)\n • We will need to assume X and Y are both 0 when their indices are out of bounds, and take the complex conjugate of the Y values:\n > $\\mathrm{XX}≔\\left(i,j\\right)↦\\mathrm{if}\\left(i<1\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathbf{or}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}a\n > $\\mathrm{YY}≔\\left(i,j\\right)↦\\mathrm{if}\\left(i<1\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathbf{or}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}c\n • The cross-correlation can now be found directly:\n > C1 := Matrix( p, q, proc(i,j) local r, s; add( add( XX(r,s) * YY(r-i+c,s-j+d), s = 1 .. b ), r = 1 .. a ); end proc ):\n • This corresponds to that found with the command:\n > $\\mathrm{C2}≔\\mathrm{CrossCorrelation2D}\\left(X,Y\\right):$\n > $\\mathrm{err}≔\\mathrm{max}\\left(\\mathrm{abs}\\left(\\mathrm{C1}-\\mathrm{C2}\\right)\\right)$\n ${\\mathrm{err}}{≔}{3.55271367880050}{×}{{10}}^{{-15}}$ (21)", null, "Computation with Fast Fourier Transforms (FFTs)\n\n • The cross-correlation is computed with FFTs when the dimensions of the final Matrix, namely p and q above, are both powers of 2 and no smaller than 4 in size, and DFTs in the remainder of cases. In the following example, we will compute a cross-correlation both directly (which will use DFTs) and by padding one of the input Matrices (which will use FFTs). The indices 1 and 2 will be used to distinguish the original and padded versions of quantities.\n • First, define the original Matrices:\n > $\\mathrm{unassign}\\left('a','b','c','d','p','q','A','B','C'\\right)$\n > $a\\left[1\\right]≔60:$\n > $b\\left[1\\right]≔62:$\n > $A\\left[1\\right]≔\\mathrm{LinearAlgebra}:-\\mathrm{RandomMatrix}\\left(a\\left[1\\right],b\\left[1\\right],\\mathrm{datatype}=\\mathrm{complex}\\left[8\\right]\\right):$\n > $c≔65:$\n > $d≔60:$\n > $B≔\\mathrm{LinearAlgebra}:-\\mathrm{RandomMatrix}\\left(c,d,\\mathrm{datatype}=\\mathrm{complex}\\left[8\\right]\\right):$\n • Second, compute the cross-correlation with no padding:\n > $C\\left[1\\right]≔\\mathrm{CrossCorrelation2D}\\left(A\\left[1\\right],B\\right):$\n • Now, the dimensions are just a little smaller than powers of 2:\n > $p\\left[1\\right]≔a\\left[1\\right]+c-1$\n ${{p}}_{{1}}{≔}{124}$ (22)\n > $q\\left[1\\right]≔b\\left[1\\right]+d-1$\n ${{q}}_{{1}}{≔}{121}$ (23)\n • So, we will pad A (padding B would also work) so that the computed cross-correlation Matrix has dimensions that are powers of 2:\n > $a\\left[2\\right]≔a\\left[1\\right]+{2}^{-\\mathrm{ilog2}\\left(\\frac{1}{\\mathrm{max}\\left(4,p\\left[1\\right]\\right)}\\right)}-p\\left[1\\right]$\n ${{a}}_{{2}}{≔}{64}$ (24)\n > $b\\left[2\\right]≔b\\left[1\\right]+{2}^{-\\mathrm{ilog2}\\left(\\frac{1}{\\mathrm{max}\\left(4,q\\left[1\\right]\\right)}\\right)}-q\\left[1\\right]$\n ${{b}}_{{2}}{≔}{69}$ (25)\n > $p\\left[2\\right]≔a\\left[2\\right]+c-1$\n ${{p}}_{{2}}{≔}{128}$ (26)\n > $q\\left[2\\right]≔b\\left[2\\right]+d-1$\n ${{q}}_{{2}}{≔}{128}$ (27)\n > $A\\left[2\\right]≔\\mathrm{ArrayTools}:-\\mathrm{Alias}\\left(A\\left[1\\right]\\right):$\n > $A\\left[2\\right]\\left(a\\left[1\\right]+1..a\\left[2\\right],..\\right)≔0:$\n > $A\\left[2\\right]\\left(..,b\\left[1\\right]+1..b\\left[2\\right]\\right)≔0:$\n • Finally, find the cross-correlation using FFTs, and extract the appropriate submatrix:\n > $C\\left[2\\right]≔\\mathrm{CrossCorrelation2D}\\left(A\\left[2\\right],B\\right)\\left[..p\\left[1\\right],..q\\left[1\\right]\\right]:$\n • Both results are the same:\n > $\\mathrm{err}≔\\mathrm{max}\\left(\\mathrm{abs}\\left(C\\left[1\\right]-C\\left[2\\right]\\right)\\right)$\n ${\\mathrm{err}}{≔}{8.16983452975878}{×}{{10}}^{{-10}}$ (28)", null, "Comparison with 1-D CrossCorrelation\n\n • The CrossCorrelation2D command can be reduced to the 1-D version with a couple of minor modifications. Take, for example, the following Vectors:\n > $A≔\\mathrm{Vector}\\left[\\mathrm{row}\\right]\\left(\\left[5,-2,3,4,1\\right]\\right)$\n $\\left[\\begin{array}{rrrrr}5& -2& 3& 4& 1\\end{array}\\right]$ (29)\n > $B≔\\mathrm{Vector}\\left[\\mathrm{row}\\right]\\left(\\left[7,0,2\\right]\\right)$\n $\\left[\\begin{array}{rrr}7& 0& 2\\end{array}\\right]$ (30)\n > $b≔\\mathrm{numelems}\\left(B\\right)$\n ${b}{≔}{3}$ (31)\n • The 2-D cross-correlation returns the following:\n > $\\mathrm{C__2D}≔\\mathrm{CrossCorrelation2D}\\left(A,B\\right)$\n $\\left[\\begin{array}{ccccccc}9.999999999999998-3.3569988834012605{}{10}^{-16}{}I& -4.000000000000003+0.0{}I& 41.0+0.0{}I& -5.999999999999999+0.0{}I& 22.999999999999993+0.0{}I& 27.999999999999996+0.0{}I& 7.000000000000003+0.0{}I\\end{array}\\right]$ (32)\n • The 1-D cross-correlation, however, gives:\n > $\\mathrm{CrossCorrelation}\\left(A,B\\right)$\n $\\left[\\begin{array}{ccccccc}41.0& -4.0& 10.0& 0.0& 0.0& 0.0& 0.0\\end{array}\\right]$ (33)\n • If we switch the arguments for CrossCorrelation, and choose the appropriate value of lowerlag, the two cross-correlations agree:\n > $\\mathrm{C__1D}≔\\mathrm{CrossCorrelation}\\left(B,A,1-b\\right)$\n $\\left[\\begin{array}{ccccccc}10.0& -4.0& 41.0& -6.0& 23.0& 28.0& 7.0\\end{array}\\right]$ (34)", null, "Application to Fingerprint Matching\n\n • A classic application of two-dimensional cross-correlation is the lining up of two or more images, for example, images of fingerprint fragments. Consider the following image:\n > $\\mathrm{with}\\left(\\mathrm{ImageTools}\\right):$\n > $\\mathrm{imgfile}≔\\mathrm{cat}\\left(\\mathrm{kernelopts}\\left(\\mathrm{mapledir}\\right),\\mathrm{kernelopts}\\left(\\mathrm{dirsep}\\right),\\mathrm{data},\\mathrm{kernelopts}\\left(\\mathrm{dirsep}\\right),\"images\",\\mathrm{kernelopts}\\left(\\mathrm{dirsep}\\right),\"fingerprint.jpg\"\\right):$\n > $\\mathrm{img1}≔\\mathrm{Matrix}\\left(\\mathrm{Read}\\left(\\mathrm{imgfile}\\right),\\mathrm{datatype}=\\mathrm{float}\\left[8\\right]\\right):$\n > $\\mathrm{Preview}\\left(\\mathrm{img1}\\right)$", null, "• First, determine the dimensions:\n > $a,b≔\\mathrm{upperbound}\\left(\\mathrm{img1}\\right)$\n ${a}{,}{b}{≔}{240}{,}{256}$ (35)\n • We will also use the mean later:\n > $\\mathrm{\\mu }≔\\frac{\\mathrm{add}\\left(\\mathrm{img1}\\right)}{\\mathrm{numelems}\\left(\\mathrm{img1}\\right)}:$\n • Second, let's choose some bounds for a fragment of the fingerprint, extract the subimage (fingerprint fragment), and then add some noise (to make our task a little more realistic):\n > $\\mathrm{cL}≔85:$\n > $\\mathrm{cU}≔205:$\n > $\\mathrm{dL}≔120:$\n > $\\mathrm{dU}≔180:$\n > $\\mathrm{img2}≔\\mathrm{Matrix}\\left(\\mathrm{img1}\\left[\\mathrm{cL}..\\mathrm{cU},\\mathrm{dL}..\\mathrm{dU}\\right]+0.5\\mathrm{\\mu }\\mathrm{ArrayTools}:-\\mathrm{RandomArray}\\left(\\mathrm{cU}-\\mathrm{cL}+1,\\mathrm{dU}-\\mathrm{dL}+1,\\mathrm{distribution}=\\mathrm{normal}\\right),\\mathrm{datatype}=\\mathrm{float}\\left[8\\right]\\right):$\n > $\\mathrm{Preview}\\left(\\mathrm{img2}\\right)$", null, "• Suppose now that we do not know whether the fragment is part of the larger image, and wish to determine if the fragment is a \"match\". To do this, compute the cross-correlation of the tempered data (formed by subtracting the mean from all elements):\n > $\\mathrm{\\mu }≔\\frac{\\mathrm{add}\\left(\\mathrm{img1}\\right)}{\\mathrm{numelems}\\left(\\mathrm{img1}\\right)}:$\n > $C≔\\mathrm{Matrix}\\left(\\mathrm{\\Re }\\left(\\mathrm{CrossCorrelation2D}\\left(\\mathrm{~}\\left[\\mathrm{-}\\right]\\left(\\mathrm{img1},\\mathrm{ },\\mathrm{\\mu }\\right),\\mathrm{~}\\left[\\mathrm{-}\\right]\\left(\\mathrm{img2},\\mathrm{ },\\mathrm{\\mu }\\right)\\right)\\right),\\mathrm{datatype}=\\mathrm{sfloat}\\right):$\n • The maximum correlation occurs here, which, as expected, coincides with the bottom-right corner of the fragment:\n > $\\mathrm{\\alpha },\\mathrm{\\beta }≔\\mathrm{max}\\left[\\mathrm{index}\\right]\\left(C\\right)$\n ${\\mathrm{\\alpha }}{,}{\\mathrm{\\beta }}{≔}{205}{,}{180}$ (36)\n • Visually:\n > $\\mathrm{matplot}≔\\mathrm{plots}:-\\mathrm{matrixplot}\\left(C\\right):$\n > $\\mathrm{plots}:-\\mathrm{display}\\left(\\mathrm{matplot},\\mathrm{title}=\"Cross-Correlation - Full\"\\right)$", null, "> $r≔25:$\n > $\\mathrm{box}≔\\left[\\mathrm{\\alpha }-r..\\mathrm{\\alpha }+r,\\mathrm{\\beta }-r..\\mathrm{\\beta }+r,\\mathrm{min}\\left(C\\right)..\\mathrm{max}\\left(C\\right)\\right]:$\n > $\\mathrm{plots}:-\\mathrm{display}\\left(\\mathrm{matplot},\\mathrm{view}=\\mathrm{box},\\mathrm{title}=\"Cross-Correlation - Zoomed\"\\right)$", null, ">", null, "Compatibility\n\n • The SignalProcessing[CrossCorrelation2D] command was introduced in Maple 2020." ]

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