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[ "# Regular Polygon\n\nWritten by Jerry Ratzlaff on . Posted in Plane Geometry\n\n•", null, "Regular polygon (a two-dimensional figure) is a polygon where all sides are congruent and all angles are congruent.\n• Circumcircle is a circle that passes through all the vertices of a two-dimensional figure.\n• Congruent is all sides having the same lengths and angles measure the same.\n• Inscribed circle is the largest circle possible that can fit on the inside of a two-dimensional figure.\n• Polygon (a two-dimensional figure) is a closed plane figure for which all sides are line segments and not necessarly congruent.\n\n## Regular Polygon Types\n\n• Triangle - 3 sides - 60° interior angle\n• Quadrilateral - 4 sides - 90° interior angle\n• Pentagon - 5 sides - 108° interior angle\n• Hexagon - 6 sides - 120° interior angle\n• Heptagon - 7 sides - 128.571° interior angle\n• Octagon - 8 sides - 135° interior angle\n• Nonagon - 9 sides - 140° interior angle\n• Decagon - 10 sides - 144° interior angle\n• Hendecagon - 11 sides - 147.273° interior angle\n• Dodecagon - 12 sides - 150° interior angle\n• Triskaidecagon - 13 sides - 152.308° interior angle\n• Tetrakaidecagon - 14 sides - 154.286° interior angle\n• Pentadecagon - 15 sides - 156° interior angle\n• Hexakaidecagon - 16 sides - 157.5° interior angle\n• Heptadecagon - 17 sides - 158.824° interior angle\n• Octakaidecagon - 18 sides - 160° interior angle\n• Enneadecagon - 19 sides - 161.053° interior angle\n• Icosagon - 20 sides - 162° interior angle\n\n## area of a Regular Polygon formulas\n\n $$\\large{ A_{area} = \\frac{a^2 \\; n}{4 \\; tan \\left( \\frac{180}{n} \\right) } }$$ $$\\large{ A_{area} = \\frac{R^2 \\; n \\; sin \\left( \\frac{360}{n} \\right) }{2} }$$ $$\\large{ A_{area} = r^2 \\; n \\; tan \\left( \\frac{180}{n} \\right) }$$ $$\\large{ A_{area} = \\frac{1}{4} \\; a^2 \\; n \\; cot \\left( \\frac{\\pi}{n} \\right) }$$\n\n### Where:\n\n$$\\large{ A_{area} }$$ = area\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ r }$$ = inside radius (apothem)\n\n$$\\large{ n }$$ = number of edges\n\n$$\\large{ R }$$ = outside radius\n\n$$\\large{ P }$$ = perimeter\n\n## Central Angle of a Regular Polygon formula\n\n $$\\large{ CA = \\frac{360}{n} }$$\n\n### Where:\n\n$$\\large{ CA }$$ = central angle\n\n$$\\large{ n }$$ = number of edges\n\n## Circumcircle Radius of a Regular Polygon formula\n\n $$\\large{ R = \\frac{a}{2 \\; sin \\; \\left( \\frac{180}{n} \\right) } }$$\n\n### Where:\n\n$$\\large{ R }$$ = outside radius\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ n }$$ = number of edges\n\n## Distance from Centroid of a Polygon formulas\n\n $$\\large{ C_x = R }$$ $$\\large{ C_y = R }$$\n\n### Where:\n\n$$\\large{ C }$$ = distance from centroid\n\n$$\\large{ R }$$ = outside radius\n\n## Edge of a Regular Polygon formulas\n\n $$\\large{ a = 2 \\;r \\; tan \\left( \\frac{180}{n} \\right) }$$ $$\\large{ a = 2 \\; R \\; sin \\left( \\frac{180}{n} \\right) }$$\n\n### Where:\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ r }$$ = inside radius (apothem)\n\n$$\\large{ R }$$ = outside radius\n\n## Elastic Section Modulus of a Polygon formula\n\n $$\\large{ S = \\frac{ I_x }{ R } }$$\n\n### Where:\n\n$$\\large{ S }$$ = elastic section modulus\n\n$$\\large{ I }$$ = moment of inertia\n\n$$\\large{ R }$$ = outside radius\n\n## Inscribed Radius of a Regular Polygon formulas\n\n $$\\large{ r = \\frac { a }{ 2\\; tan \\; \\left( \\frac{180}{n} \\right) } }$$ $$\\large{ r = R \\; cos \\left( \\frac{180}{n} \\right) }$$\n\n### Where:\n\n$$\\large{ r }$$ = inside radius (apothem)\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ n }$$ = number of edges\n\n$$\\large{ R }$$ = outside radius\n\n## Number of Diagonals of a Regular Polygon formula\n\n $$\\large{ D' = \\frac{ n \\; \\left( n \\;-\\; 3 \\right) }{2} }$$\n\n### Where:\n\n$$\\large{ D' }$$ = diagonal\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ n }$$ = number of edges\n\n## Perimeter of a Regular Polygon formula\n\n $$\\large{ P = a \\; n }$$\n\n### Where:\n\n$$\\large{ P }$$ = perimeter\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ n }$$ = number of edges\n\n## Polar Moment of Inertia of a Polygon formula\n\n $$\\large{ J_{z} = 2 \\; A \\left( \\frac{ 6 \\; R^2 \\;-\\; a^2 }{24} \\right) }$$\n\n### Where:\n\n$$\\large{ J }$$ = torsional constant\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ R }$$ = outside radius\n\n## Radius of Gyration of a Polygon formulas\n\n $$\\large{ k_{x} = \\sqrt{ \\frac{6 \\; R^2 \\;-\\; a^2 }{24} } }$$ $$\\large{ k_{y} = \\sqrt{ \\frac{6 \\; R^2 \\;-\\; a^2 }{24} } }$$ $$\\large{ k_{z} = \\sqrt{ k_{x}{^2} + k_{y}{^2} } }$$\n\n### Where:\n\n$$\\large{ k }$$ = radius of gyration\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ R }$$ = outside radius\n\n## Second Moment of Area of a Rectangle formulas\n\n $$\\large{ I_{x} = 2 \\; A \\left( \\frac{ 6 \\; R^2 \\;-\\; a^2 }{24} \\right) }$$ $$\\large{ I_{y} = 2 \\; A \\left( \\frac{ 6 \\; R^2 \\;-\\; a^2 }{24} \\right) }$$\n\n### Where:\n\n$$\\large{ I }$$ = moment of inertia\n\n$$\\large{ a }$$ = edge\n\n$$\\large{ R }$$ = outside radius" ]

[ null, "https://www.piping-designer.com/images/mathematics/geometry/plane/polygon/regular_polygon_2.jpg", null ]

[ "Search a number\nBaseRepresentation\nbin100100101111101011110010…\n…0110101011010010000100011\n31120101101212021121110200002012\n41021133113210311122100203\n5314331010222330213243\n63103230001452343135\n7125036240525514266\noct11137274465322043\n91511355247420065\n10323213013132323\n1193a92961849346\n1230300a20174aab\n1310b46b19556ac2\n1459b5844054cdd\n1527577a3a55d18\nhex125f5e4d5a423\n\n323213013132323 has 8 divisors (see below), whose sum is σ = 323331046716240. Its totient is φ = 323094981091680.\n\nThe previous prime is 323213013132277. The next prime is 323213013132337. The reversal of 323213013132323 is 323231310312323.\n\nIt is a happy number.\n\nIt is a sphenic number, since it is the product of 3 distinct primes.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 323213013132323 - 212 = 323213013128227 is a prime.\n\nIt is a Duffinian number.\n\nIt is a self number, because there is not a number n which added to its sum of digits gives 323213013132323.\n\nIt is not an unprimeable number, because it can be changed into a prime (323213013132343) by changing a digit.\n\nIt is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 573797888 + ... + 574360898.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (40416380839530).\n\nAlmost surely, 2323213013132323 is an apocalyptic number.\n\n323213013132323 is a deficient number, since it is larger than the sum of its proper divisors (118033583917).\n\n323213013132323 is a wasteful number, since it uses less digits than its factorization.\n\n323213013132323 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 771637.\n\nThe product of its (nonzero) digits is 34992, while the sum is 32.\n\nAdding to 323213013132323 its reverse (323231310312323), we get a palindrome (646444323444646).\n\nThe spelling of 323213013132323 in words is \"three hundred twenty-three trillion, two hundred thirteen billion, thirteen million, one hundred thirty-two thousand, three hundred twenty-three\"." ]

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[ "Inches To Mm\n\n# 1936 in to mm1936 Inch to Millimeters\n\nin\n=\nmm\n\n## How to convert 1936 inch to millimeters?\n\n 1936 in * 25.4 mm = 49174.4 mm 1 in\nA common question is How many inch in 1936 millimeter? And the answer is 76.220472441 in in 1936 mm. Likewise the question how many millimeter in 1936 inch has the answer of 49174.4 mm in 1936 in.\n\n## How much are 1936 inches in millimeters?\n\n1936 inches equal 49174.4 millimeters (1936in = 49174.4mm). Converting 1936 in to mm is easy. Simply use our calculator above, or apply the formula to change the length 1936 in to mm.\n\n## Convert 1936 in to common lengths\n\nUnitLength\nNanometer49174400000.0 nm\nMicrometer49174400.0 µm\nMillimeter49174.4 mm\nCentimeter4917.44 cm\nInch1936.0 in\nFoot161.333333333 ft\nYard53.7777777778 yd\nMeter49.1744 m\nKilometer0.0491744 km\nMile0.0305555556 mi\nNautical mile0.0265520518 nmi\n\n## What is 1936 inches in mm?\n\nTo convert 1936 in to mm multiply the length in inches by 25.4. The 1936 in in mm formula is [mm] = 1936 * 25.4. Thus, for 1936 inches in millimeter we get 49174.4 mm.\n\n## 1936 Inch Conversion Table", null, "## Alternative spelling\n\n1936 in to mm, 1936 Inch to Millimeters, 1936 Inch in Millimeters, 1936 Inches to Millimeter, 1936 Inches to mm, 1936 Inches in mm, 1936 Inch in mm, 1936 Inch to Millimeter, 1936 in in Millimeter, 1936 in to Millimeters, 1936 in in Millimeters," ]

[ null, "https://inches-to-mm.appspot.com/image/1936.png", null ]

[ "Newbie\n\n# The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm, respectively. It is melted and recast into a solid cylinder of diameter 14 cm. Find the height of the cylinder.\n\n• 0\n\nImportant problem from the cbse board exam point of view; becuse last time this problem was asked in 2012. Problem from RS Aggarwal book, problem number 5, page number 809, exercise 17B, chapter volume and surface area of solid.\n\nShare\n\n1. Given data,\n\nThe internal base radius of spherical shell, r1 = 3 cm,\n\nthe external base radius of spherical shell, R = 5 cm\n\nthe base radius of solid cylinder, r = 14/2 = 7 cm\n\nVolume of the spherical shell = Volume of the cylinder\n\n[4/3] π [ R³ – r1³ ] = π r² h\n\n[4/3] [ R³ – r1³ ] = r² h\n\n[4/3] [ 5³ – 3³ ] =  [ 7² ] * h\n\n( [4/3] * 98 )  / 49 = h\n\nh= 8/3\n\n∴ The height of the cylinder is 8/3cm\n\n• 0" ]

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[ "Day: February 16, 2010\n\n# Creating a counter using other way\n\nRecently I created a simple board to test the HCS08 chip used on my wash machine, it is a very simple board with just the microcontroller and two LEDs (wired to PTE1 and PTE3).\n\nThen I created a simple binary counter program to test it:\n\n```#define LED1 PTED_PTED1\n#define LED2 PTED_PTED3\n\nvoid main(void)\n{\nint i = 0;\nMCU_init();\nPTEDD = 0x0F;\nLED1 = 0;\nLED2 = 0;\nwhile(1){\ni++;\nLED1 = i & 1;\nLED2 = (i & 2) >> 1;\nmdelay(500);\n}\n}\n```\n\nIt worked fine, but then I started to thing another way to count without using a counter (i++).\nI realized the LED1 (bit 0) always goes some sequence (0->1->0->1…), it could be done easily using a Exclusive OR (XOR) logic: LED1 = LED1 ^ 1 or Not Logic: LED1 = ~LED1;\nAlso I noticed which LED2 (bit 1) only turns ON or OFF after a LED1 transition from 1 to 0. Then I could create a variable T (Temporary) to save previous LED1 state and compare it with current status. Case previous state is 1 and current status is 0 it should inverts the LED2 state, then just do that: LED2 = LED2 ^ (T & ~LED1), then I got it:\n\n```#define LED1 PTED_PTED1\n#define LED2 PTED_PTED3\n\nvoid main(void)\n{\nint i = 0, T = 0;\nMCU_init();\nPTEDD = 0x0F;\nLED1 = 0;\nLED2 = 0;\nwhile(1){\nLED1 = LED1 ^ 1;\nLED2 = LED2 ^ (T & ~LED1);\nT = LED1;\nmdelay(500);\n}\n}\n```\n\nIt worked, but just after doing this way I realized my mistake: I don’t need to compare previous state, because LED1 just was 1 before it turns 0 (except the first time, but it doesn’t matter), then the above code could be reduced to:\n\n```#define LED1 PTED_PTED1\n#define LED2 PTED_PTED3\n\nvoid main(void)\n{\nMCU_init();\nPTEDD = 0x0F;\nLED1 = 0;\nLED2 = 0;\nwhile(1){\nLED1 = LED1 ^ 1;\nLED2 = LED2 ^ ~LED1;\nmdelay(500);\n}\n}\n```\n\nIt fact I created a counter without using an ADD instruction. To be honest I never saw a microcontroller without ADD instruction, but if some day I got one I could do a counter easily 😉" ]

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[ "# zbMATH — the first resource for mathematics\n\nAnother extremal problem for Turan graphs. (English) Zbl 0629.05041\nLet K(G) and $$\\omega$$ (G) be the clique graph and clique number of a graph G. For $$1<r<n$$, let $$F(n,r)=\\max_{G}\\{| K(G)|:| V(G)| =n$$ and $$\\omega (G)=r\\}$$. A Turan graph $$T(n,r)$$ is a multipartite graph with vertices $$v_ 1,v_ 2,...,v_ n$$, and $$v_ iv_ j\\in E(G)$$ if and only if $$i\\neq j$$ (mod r). Theorem: Let G be a graph of order n with $$\\omega (G)=r<n$$. Then $$| K(G)| =F(n,r)$$ if and only if $$G\\sim T(n,r)$$.\nReviewer: S.F.Kapoor\n\n##### MSC:\n 05C35 Extremal problems in graph theory\n##### Keywords:\nclique graph; clique number; Turan graph; multipartite graph\nFull Text:\n##### References:\n Bollobás, B, Extremal graph theory, (1978), Academic Press London · Zbl 0419.05031 Hedman, B, The maximum number of cliques in dense graphs, Discrete math., 54, 161-166, (1985) · Zbl 0569.05029 Moon, J.W; Moser, L, On cliques in graphs, Israel J. math., 3, 23-28, (1965) · Zbl 0144.23205 Roman, S, The maximum number of q-cliques in a graph with no p-clique, Discrete math., 14, 365-371, (1976) · Zbl 0319.05126 Turan, P, On an extremal problem in graph theory, Mat. fiz. lapok, 48, 436-452, (1941)\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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[ "Draft:Riemann–Roch-type theorem\n\n(The article discusses various generalizations of the Riemann–Roch theorem.)\n\nFormulation due to Baum, Fulton and MacPherson\n\nWe need a few notations: $G_{*},A_{*}$", null, "are functors on the category C of schemes separated and locally of finite type over the base field k with proper morphisms such that\n\n• $G_{*}(X)$", null, "is the Grothendieck group of coherent sheaves on X,\n• $A_{*}(X)$", null, "is the rational Chow group of X,\n• for each proper morphism f, $G_{*}(f),A_{*}(f)$", null, "are the direct image and push-forward along f, respectively.\n\nAlso, if $f:X\\to Y$", null, "is a local complete intersection morphism; i.e., it factors as a closed regular embedding $X\\hookrightarrow P$", null, "into a smooth scheme P followed by a smooth morphism $P\\to Y$", null, ", then let\n\n$T_{f}=[T_{P/Y}|_{X}]-[N_{X/P}]$", null, "be the class in the Grothendieck group of vector bundles on X; it is independent of the factorization and is called the virtual tangent bundle of f.\n\nThen the Riemann–Roch theorem amounts to the construction of a unique natural transformation:\n\n$\\tau :G_{*}\\to A_{*}$", null, "between the functors such that\n\n• for $\\tau _{X}\\circ f^{*}=\\operatorname {td} (T_{f})\\cdot f^{*}\\circ \\tau _{Y}$", null, ", where we wrote $\\tau _{X}:G(X)\\to A(X)$", null, "and $\\operatorname {td} (T_{f})$", null, "is the Todd class of the virtual tangent sheaf.\n\nThe Riemann–Roch theorem for Deligne–Mumford stacks\n\nAside from algebraic spaces, no straightforward generalization is possible for stacks. The complication already appears in the orbifold case (Kawasaki's Riemann–Roch).\n\nOne of the significant applications of the theorem is that it allows one to define a virtual fundamental class in terms of the K-theoretic virtual fundamental class. More precisely," ]

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[ "Você está na página 1de 8\n\n# Risk and Return\n\n## Chapter 8 Which is better?\n\n(1) 4% return with no risk, or\nRisk and Rates of (2) 15% return with risk.\n\n## Return Cannot say - need to know how much risk\n\ncomes with the 15% return.\n\n## What do we know so far? But this does not quantify risk\n\nWe know why returns vary between different securities!\nWe want to quantify risk.\nrnom = r* + IP + DRP + LP + MRP\n\n## rnom = nominal rate Todays concepts\n\nr* = real rate (pure compensation) (1) quantify risk\nIRP = inflation-risk premium (change in cost of goods)\nDRP = default-risk premium (ability to pay P & I) (2) tells us how much return we need for a\nLP = liquidity premium (ability to convert to cash) given level of risk\nMP = maturity risk premium ()P/)i)\n\nwhere r* + IP = rRF\n\n1\nRisk and Return - individual assets Risk and Return - individual assets\nRisk = P( Actual returns < Expected returns )\nBuy a 20 year STN bond to hold for 2 years\nExample: A 2 year Treasury bond; hold until maturity; where the expected return = 15% [ E(r) = 15%]\npays 4% / year\n\n## Expected return = 4% [ E(r) = 4%] Probability Actual Return\n\nIf hold until maturity 50% 0%\n\n25% 45%\n\n## Calculating Expected Return How to measure risk and return\n\nE(r) = expected rate of return\nE(r) = ( probi x ri )\n= (50%x0%)+(25%x15%)+(25%X45%) = standard deviation which\n= (0%) + (3.75%) + (11.25%) measures the dispersion around the\n= 15.00% mean (measure of risk)\n\n## = { [ri E(r)]2 x probi }1/2\n\n2\nRisk and Return Problem\nStock A has the following probability distribution of\nexpected returns:\n\n## Instrument E(r) Probability Rate of Return\n\n0.1 -15%\n4% 0% 0.2 0%\nT-Bond 0.4 5%\n0.2 10%\nMGM Bond 15% 18.37% 0.1 25%\n\ndeviation?\n\n## Expected Rate of Return Standard Deviation\n\nE(r) = ( probi x ri ) = { [ri E(r)]2 x probi }1/2\n= {[(-.15-.05)2x.1]+[(.00-.05)2x.2]\n= (.1 x -15%) + (.2 x 0%) + (.4 x 5%) +[(.05-.05)2x.4]+[(.10-.05)2x.2]\n+ (.2 x 10%) + (.1 x 25%) +[(.25-.05)2x.4]}1/2\n= {[.004]+[.0005]+[.0005]+[.004]}1/2\n= (-1.5%) + (0%) + (2%) + (2%) + (2.5%)\n= {.009}1/2\nE(r) = 5%\n= 9.49%\n\n3\nNormal Distribution Risk and Return for portfolios\n\n## Individual expected return and\n\nRisk and Return for portfolios\nstandard deviation\nPortfolio standard deviation\nProb MGM LAS\np = [x2A2A + x2B2B + 2xAxBABAB]1/2\n.25 -3.0% 3.0%\n.50 12.0% 9.0%\nwhere, xA + xB = 100%\n.25 27.0% 15.0%\nE(r)i 12.0% 9.0%\n(rho) = correlation coefficient\n= measures comovement between 2 securities i 10.60% 4.24%\n\n4\nPortfolio expected return and\nCalculations\nstandard deviation\nXi Stocki E(r)i i\n40.0% LAS 9.0% 4.24% Portfolio Expected Return\n60.0% MGM 12.0% 10.60%\nE(r)p = (.40)(.09) + (.60)(.12) = 10.8%\nE(r)p 10.80%\np, if = +1.0 8.0%\np, if = 0.0 6.6%\np, if = -1.0 4.7%\n\n## Calculations Efficient Portfolios\n\nPortfolio Standard Deviation The portfolio that provides the highest return for a given\nlevel of risk - or lowest risk for a particular expected\nsay = 1, perfect positive correlation return.\n\n## p = [(.4)2(4.24)2+(.6)2(10.6)2+(.4)(.6)(1)(4.24)(10.6)]1/2 Therefore combine assets in a portfolio to get highest\n\n= [64.9]1/2 = 8% expected return for given risk (6p)\n\n= 0, no correlation, p = 6.6% For each asset some risk can be eliminated when\ncombined with other assets in a portfolio (unless p=+1)\n= -1, perfect negative correlation, p =4.7%\nCombine assets in such a manner to get Efficient Portfolio.\nRemember: less correlation equates to lower risk!!\n\n5\nLook at an individual stock Beta - CAPM\nTotal Risk = i Beta = Measure of Market Risk - the risk that is relevant to\ninvestors\nSome risk can be eliminated by including the stock in a\nportfolio -call this that can be eliminated diversifiable or\ncompany-specific risk. BETA () measures of a particular stock's variation in\nreturn relative to the market.\nSome risk can not be eliminated - call this\nnondiversifiable or market risk. = cov(ri,rm)/2m = imim / 2m = im x (i / m )\n\nRisk that is important to investors is nondiversifiable or = 1.0 moves exactly with market\nmarket risk. > 1.0 moves more than market (> risk)\n< 1.0 moves less than market (< risk)\nrisk aversion (def) - dislike risk = 0.0 no risk (Risk-Free)\n\n## Capital Asset Pricing Model\n\nCAPM Example\n(CAPM)\nTreasury Bill = rRF = 5%\nMarket Rate = rm = 12%\n\n## rT-Bill = 5% + ( 0.0 )( 12% - 5% ) = 5%\n\nWhere ( rm - rRF ) = market risk premium\nFor IBM =0.7\n\n## rIBM = 5% + ( 0.7 )( 12% - 5% ) = 9.9%\n\n6\nWhat does the CAPM tell us? Security Market Line (SML)\nSecurity Market Line (SML)\nWhat is our expected return at a given level 25.0%\nof risk (the risk that is important to an\n\nExpected Return\n20.0%\ninvestor holding a well-diversified portfolio 15.0%\n10.0%\n5.0%\n0.0%\n0.0 0.5 1.0 1.5 2.0 2.5 3.0\nBeta\n\n## Diversified Portfolio Efficient Markets Hypothesis\n\nEfficient Markets Hypothesis (def) - securities are fairly\npriced - market price reflects all publicly available\nHow many stocks (assets) do we need to hold to information.\napproximate a well-diversified portfolio?\nWhat does this mean for investors?\n\nHold everything - market portfolio - eliminates all P0 = fair price according to public information\ndiversifiable risk - Impossible!\nri = depends on risk relative to market risk ()\n\nHold 8-10 assets - closely approximates market Says, buy securities and form your portfolio according to\nrisk preference\n(eliminates most diversifiable risk)\n\n7\nRecap: Risk and Return Recap: Risk and Return\nGoal: to quantify risk and return so we can compare and choose Assumes that everyone (investors) are bright enough to hold a well-\ninvestment opportunities. diversified efficient portfolio.\n\nWe saw how to calculate expected return Only risk that matters is nondiversifiable risk (market risk) - measured\nby\nE(r) = ( pi x ri )\nUse CAPM and to calculate expected return for relevant risk.\nWe saw how to calculate risk Risk and Return are quantified!!!!\n\n= { (ri E(ri)]2 pi }1/2 Therefore investors want to own efficient portfolios. Which ones? The\none that correspond to the level of risk they want to assume.\nBut if we form a portfolio we can eliminate some risk - if we form\nportfolio in such a manner to eliminate all extra (diversifiable) risk we Trade-off: Between Risk and Return\nhave efficient portfolio (def)." ]

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Array\n(\n => 58.65.222.10\n)\n\n => Array\n(\n => 202.47.43.86\n)\n\n => Array\n(\n => 106.206.223.234\n)\n\n => Array\n(\n => 5.195.153.158\n)\n\n => Array\n(\n => 223.227.127.243\n)\n\n => Array\n(\n => 103.165.12.222\n)\n\n => Array\n(\n => 49.36.185.189\n)\n\n => Array\n(\n => 59.96.92.57\n)\n\n => Array\n(\n => 203.194.104.235\n)\n\n => Array\n(\n => 122.177.72.33\n)\n\n => Array\n(\n => 106.213.126.40\n)\n\n => Array\n(\n => 45.127.232.69\n)\n\n => Array\n(\n => 156.146.59.39\n)\n\n => Array\n(\n => 103.21.184.11\n)\n\n => Array\n(\n => 106.212.47.59\n)\n\n => Array\n(\n => 182.179.137.235\n)\n\n => Array\n(\n => 49.36.178.154\n)\n\n => Array\n(\n => 171.48.7.128\n)\n\n => Array\n(\n => 119.160.57.96\n)\n\n => Array\n(\n => 197.210.79.92\n)\n\n => Array\n(\n => 36.255.45.87\n)\n\n => Array\n(\n => 47.31.219.47\n)\n\n => Array\n(\n => 122.161.51.160\n)\n\n => Array\n(\n => 103.217.123.129\n)\n\n => Array\n(\n => 59.153.16.12\n)\n\n => Array\n(\n => 103.92.43.226\n)\n\n => Array\n(\n => 47.31.139.139\n)\n\n => Array\n(\n => 210.2.140.18\n)\n\n => Array\n(\n => 106.210.33.219\n)\n\n => Array\n(\n => 175.107.203.34\n)\n\n => Array\n(\n => 146.196.32.144\n)\n\n => Array\n(\n => 103.12.133.121\n)\n\n => Array\n(\n => 103.59.208.182\n)\n\n => Array\n(\n => 157.37.190.232\n)\n\n => Array\n(\n => 106.195.35.201\n)\n\n => Array\n(\n => 27.122.14.83\n)\n\n => Array\n(\n => 194.193.44.5\n)\n\n => Array\n(\n => 5.62.43.245\n)\n\n => Array\n(\n => 103.53.80.50\n)\n\n => Array\n(\n => 47.29.142.233\n)\n\n => Array\n(\n => 154.6.20.63\n)\n\n => Array\n(\n => 173.245.203.128\n)\n\n => Array\n(\n => 103.77.43.231\n)\n\n => Array\n(\n => 5.107.166.235\n)\n\n => Array\n(\n => 106.212.44.123\n)\n\n => Array\n(\n => 157.41.60.93\n)\n\n => Array\n(\n => 27.58.179.79\n)\n\n => Array\n(\n => 157.37.167.144\n)\n\n => Array\n(\n => 119.160.57.115\n)\n\n => Array\n(\n => 122.161.53.224\n)\n\n => Array\n(\n => 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n => Array\n(\n => 202.83.58.208\n)\n\n => Array\n(\n => 122.161.53.101\n)\n\n => Array\n(\n => 137.97.95.21\n)\n\n => Array\n(\n => 112.204.167.123\n)\n\n => Array\n(\n => 122.180.21.151\n)\n\n => Array\n(\n => 103.120.44.108\n)\n\n => Array\n(\n => 49.37.220.174\n)\n\n => Array\n(\n => 1.55.255.124\n)\n\n => Array\n(\n => 23.227.140.173\n)\n\n => Array\n(\n => 43.248.153.110\n)\n\n => Array\n(\n => 106.214.93.101\n)\n\n => Array\n(\n => 103.83.149.36\n)\n\n => Array\n(\n => 103.217.123.57\n)\n\n => Array\n(\n => 193.9.113.119\n)\n\n => Array\n(\n => 14.182.57.204\n)\n\n => Array\n(\n => 117.201.231.0\n)\n\n => Array\n(\n => 14.99.198.186\n)\n\n => Array\n(\n => 36.255.44.204\n)\n\n => Array\n(\n => 103.160.236.42\n)\n\n => Array\n(\n => 31.202.16.116\n)\n\n => Array\n(\n => 223.239.49.201\n)\n\n => Array\n(\n => 122.161.102.149\n)\n\n => Array\n(\n => 117.196.123.184\n)\n\n => Array\n(\n => 49.205.112.105\n)\n\n => Array\n(\n => 103.244.176.201\n)\n\n => Array\n(\n => 95.216.15.219\n)\n\n => Array\n(\n => 103.107.196.174\n)\n\n => Array\n(\n => 203.190.34.65\n)\n\n => Array\n(\n => 23.227.140.182\n)\n\n => Array\n(\n => 171.79.74.74\n)\n\n => Array\n(\n => 106.206.223.244\n)\n\n => Array\n(\n => 180.151.28.140\n)\n\n => Array\n(\n => 165.225.124.114\n)\n\n => Array\n(\n => 106.206.223.252\n)\n\n => Array\n(\n => 39.62.23.38\n)\n\n => Array\n(\n => 112.211.252.33\n)\n\n => Array\n(\n => 146.70.66.242\n)\n\n => Array\n(\n => 222.252.51.38\n)\n\n => Array\n(\n => 122.162.151.223\n)\n\n => Array\n(\n => 180.178.154.100\n)\n\n => Array\n(\n => 180.94.33.94\n)\n\n => Array\n(\n => 205.164.130.82\n)\n\n => Array\n(\n => 117.196.114.167\n)\n\n => Array\n(\n => 43.224.0.189\n)\n\n => Array\n(\n => 154.6.20.59\n)\n\n => Array\n(\n => 122.161.131.67\n)\n\n => Array\n(\n => 70.68.68.159\n)\n\n => Array\n(\n => 103.125.130.200\n)\n\n => Array\n(\n => 43.242.176.147\n)\n\n => Array\n(\n => 129.0.102.29\n)\n\n => Array\n(\n => 182.64.180.32\n)\n\n => Array\n(\n => 110.93.250.196\n)\n\n => Array\n(\n => 139.135.57.197\n)\n\n => Array\n(\n => 157.33.219.2\n)\n\n => Array\n(\n => 205.253.123.239\n)\n\n => Array\n(\n => 122.177.66.119\n)\n\n => Array\n(\n => 182.64.105.252\n)\n\n => Array\n(\n => 14.97.111.154\n)\n\n => Array\n(\n => 146.196.35.35\n)\n\n => Array\n(\n => 103.167.162.205\n)\n\n => Array\n(\n => 37.111.130.245\n)\n\n => Array\n(\n => 49.228.51.196\n)\n\n => Array\n(\n => 157.39.148.205\n)\n\n => Array\n(\n => 129.0.102.28\n)\n\n => Array\n(\n => 103.82.191.229\n)\n\n => Array\n(\n => 194.104.23.140\n)\n\n => Array\n(\n => 49.205.193.252\n)\n\n => Array\n(\n => 222.252.33.119\n)\n\n => Array\n(\n => 173.255.132.114\n)\n\n => Array\n(\n => 182.64.148.162\n)\n\n => Array\n(\n => 175.176.87.8\n)\n\n => Array\n(\n => 5.62.57.6\n)\n\n => Array\n(\n => 119.160.96.229\n)\n\n => Array\n(\n => 49.205.180.226\n)\n\n => Array\n(\n => 95.142.120.59\n)\n\n => Array\n(\n => 183.82.116.204\n)\n\n => Array\n(\n => 202.89.69.186\n)\n\n => Array\n(\n => 39.48.165.36\n)\n\n => Array\n(\n => 192.140.149.81\n)\n\n => Array\n(\n => 198.16.70.28\n)\n\n => Array\n(\n => 103.25.250.236\n)\n\n => Array\n(\n => 106.76.202.244\n)\n\n => Array\n(\n => 47.8.8.165\n)\n\n => Array\n(\n => 202.5.145.213\n)\n\n => Array\n(\n => 106.212.188.243\n)\n\n => Array\n(\n => 106.215.89.2\n)\n\n => Array\n(\n => 119.82.83.148\n)\n\n => Array\n(\n => 123.24.164.245\n)\n\n => Array\n(\n => 187.67.51.106\n)\n\n => Array\n(\n => 117.196.119.95\n)\n\n => Array\n(\n => 95.142.120.66\n)\n\n => Array\n(\n => 156.146.59.35\n)\n\n => Array\n(\n => 49.205.213.148\n)\n\n => Array\n(\n => 111.223.27.206\n)\n\n => Array\n(\n => 49.205.212.86\n)\n\n => Array\n(\n => 103.77.42.103\n)\n\n => Array\n(\n => 110.227.62.25\n)\n\n => Array\n(\n => 122.179.54.140\n)\n\n => Array\n(\n => 157.39.239.81\n)\n\n => Array\n(\n => 138.128.27.234\n)\n\n => Array\n(\n => 103.244.176.194\n)\n\n => Array\n(\n => 130.105.10.127\n)\n\n => Array\n(\n => 103.116.250.191\n)\n\n => Array\n(\n => 122.180.186.6\n)\n\n => Array\n(\n => 101.53.228.52\n)\n\n => Array\n(\n => 39.57.138.90\n)\n\n => Array\n(\n => 197.156.137.165\n)\n\n => Array\n(\n => 49.37.155.78\n)\n\n => Array\n(\n => 39.59.81.32\n)\n\n => Array\n(\n => 45.127.44.78\n)\n\n => Array\n(\n => 103.58.155.83\n)\n\n => Array\n(\n => 175.107.220.20\n)\n\n => Array\n(\n => 14.255.9.197\n)\n\n => Array\n(\n => 103.55.63.146\n)\n\n => Array\n(\n => 49.205.138.81\n)\n\n => Array\n(\n => 45.35.222.243\n)\n\n => Array\n(\n => 203.190.34.57\n)\n\n => Array\n(\n => 205.253.121.11\n)\n\n => Array\n(\n => 154.72.171.177\n)\n\n => Array\n(\n => 39.52.203.37\n)\n\n => Array\n(\n => 122.161.52.2\n)\n\n => Array\n(\n => 82.145.41.170\n)\n\n => Array\n(\n => 103.217.123.33\n)\n\n => Array\n(\n => 103.150.238.100\n)\n\n => Array\n(\n => 125.99.11.182\n)\n\n => Array\n(\n => 103.217.178.70\n)\n\n => Array\n(\n => 197.210.227.95\n)\n\n => Array\n(\n => 116.75.212.153\n)\n\n => Array\n(\n => 212.102.42.202\n)\n\n => Array\n(\n => 49.34.177.147\n)\n\n => Array\n(\n => 173.242.123.110\n)\n\n => Array\n(\n => 49.36.35.254\n)\n\n => Array\n(\n => 202.47.59.82\n)\n\n => Array\n(\n => 157.42.197.119\n)\n\n => Array\n(\n => 103.99.196.250\n)\n\n => Array\n(\n => 119.155.228.244\n)\n\n => Array\n(\n => 130.105.160.170\n)\n\n => Array\n(\n => 78.132.235.189\n)\n\n => Array\n(\n => 202.142.186.114\n)\n\n => Array\n(\n => 115.99.156.136\n)\n\n => Array\n(\n => 14.162.166.254\n)\n\n => Array\n(\n => 157.39.133.205\n)\n\n => Array\n(\n => 103.196.139.157\n)\n\n => Array\n(\n => 139.99.159.20\n)\n\n => Array\n(\n => 175.176.87.42\n)\n\n => Array\n(\n => 103.46.202.244\n)\n\n => Array\n(\n => 175.176.87.16\n)\n\n => Array\n(\n => 49.156.85.55\n)\n\n => Array\n(\n => 157.39.101.65\n)\n\n => Array\n(\n => 124.253.195.93\n)\n\n => Array\n(\n => 110.227.59.8\n)\n\n => Array\n(\n => 157.50.50.6\n)\n\n => Array\n(\n => 95.142.120.25\n)\n\n => Array\n(\n => 49.36.186.141\n)\n\n => Array\n(\n => 110.227.54.161\n)\n\n => Array\n(\n => 88.117.62.180\n)\n\n => Array\n(\n => 110.227.57.8\n)\n\n => Array\n(\n => 106.200.36.21\n)\n\n => Array\n(\n => 202.131.143.247\n)\n\n => Array\n(\n => 103.46.202.4\n)\n\n => Array\n(\n => 122.177.78.217\n)\n\n => Array\n(\n => 124.253.195.201\n)\n\n => Array\n(\n => 27.58.17.91\n)\n\n => Array\n(\n => 223.228.143.162\n)\n\n => Array\n(\n => 119.160.96.233\n)\n\n => Array\n(\n => 49.156.69.213\n)\n\n => Array\n(\n => 41.80.97.54\n)\n\n => Array\n(\n => 122.176.207.193\n)\n\n => Array\n(\n => 45.118.156.6\n)\n\n => Array\n(\n => 157.39.154.210\n)\n\n => Array\n(\n => 103.48.197.173\n)\n\n => Array\n(\n => 103.46.202.98\n)\n\n => Array\n(\n => 157.43.214.102\n)\n\n => Array\n(\n => 180.191.125.73\n)\n\n)\n```\nMaking Saving Easy: LivingDebtFree's Personal Finance Blog\n Layout: Blue and Brown (Default) Author's Creation\n Home > Making Saving Easy\n\n# Making Saving Easy\n\nApril 9th, 2011 at 06:02 pm\n\nI have a couple of ways that make saving money easy for me.\n\nThe first is by having an automatic transfer set up so that each week a specified amount is sent from my checking account to a savings account. Right now we are saving for house repairs and to replace my husband's truck. So I set up two automated transfers. My husband gets paid on Thursday and the transfers are scheduled to take place every Monday. Both items are budgeted for weekly and I don't have to give it another thought.\n\nThe other way to save that is painless for me is putting any change we receive in a box. I never spend change. One of the first things I do when coming home from grocery shopping is to clean out my wallet of any change and depositing it into the box. Periodically I get the box out, roll the change and deposit it into my checking account. I usually will then send it to one of my designated savings accounts, or if a budgeted account has had more expenses than I had foreseen, I'll use the change to build that back up.\n\nBoth savings ways are easy for me and require little effort. It's amazing how quickly both will add up!\n\n### 4 Responses to “Making Saving Easy”\n\n1. creditcardfree Says:\n\n2. rob62521 Says:\n\nSounds good!\n\n3. beawealthywarrior Says:\n\nIn addition to change, I've started saving all my one dollar bills. It adds up as well and really makes you think before you spend\n\n4. Jerry Says:\n\nI like your thinking, because I also believe that \"out of sight\" leads to \"out of mind.\" By direct depositing the money, and putting the change away, you are able to have some insurance that you are not missing the money psychologically, because you aren't thinking about it! Great job...\nJerry\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4.  [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]" ]

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[ "## ››Convert square fathom to square zeptometre\n\n square fathom square zeptometer\n\nHow many square fathom in 1 square zeptometer? The answer is 2.9899631720256E-43.\nWe assume you are converting between square fathom and square zeptometre.\nYou can view more details on each measurement unit:\nsquare fathom or square zeptometer\nThe SI derived unit for area is the square meter.\n1 square meter is equal to 0.29899631720256 square fathom, or 1.0E+42 square zeptometer.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between square fathoms and square zeptometers.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of square fathom to square zeptometer\n\n1 square fathom to square zeptometer = 3.3445228E+42 square zeptometer\n\n2 square fathom to square zeptometer = 6.6890456E+42 square zeptometer\n\n3 square fathom to square zeptometer = 1.00335684E+43 square zeptometer\n\n4 square fathom to square zeptometer = 1.33780912E+43 square zeptometer\n\n5 square fathom to square zeptometer = 1.6722614E+43 square zeptometer\n\n6 square fathom to square zeptometer = 2.00671368E+43 square zeptometer\n\n7 square fathom to square zeptometer = 2.34116596E+43 square zeptometer\n\n8 square fathom to square zeptometer = 2.67561824E+43 square zeptometer\n\n9 square fathom to square zeptometer = 3.01007052E+43 square zeptometer\n\n10 square fathom to square zeptometer = 3.3445228E+43 square zeptometer\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from square zeptometer to square fathom, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Square fathom\n\nA square fathom is defined as the area of a square with sides one fathom in length.\n\nThe definition of a fathom is as follows:\n\nunit of length equal to six feet used to measure depth of water\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]

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[ "A+ » VCE Blog » VCE Maths Methods Blog » Exponential and Logarithm Functions [Video Tutorial]\n\n# Exponential and Logarithm Functions [Video Tutorial]\n\nThis tutorial covers material encountered in chapter 5 of the VCE Mathematical Methods Textbook, namely:\n\n• Exponential functions\n• Index laws\n• Log functions\n• Log laws and change of base\n\n## Worksheet\n\nQ1. Sketch the graph of each of the following functions, clearly indicating the axis intercepts and any asymptotes (Note that \\log_e(x)=\\ln(x)):\n\n(a) f(x)=e^x-3\n\n(b) f(x)=2^{-x}+4\n\n(c) g(x)=\\dfrac{1}{3}(e^x-4)\n\n(d) g(x)=5-e^{-x}\n\n(e) h(x)=\\ln(3x+2)\n\n(f) h(x)=-\\ln(x-3)\n\n(g) j(x)=\\ln(2-x)\n\nQ2. Find f^{-1} for each of the following functions:\n\n(a) f:\\R \\to (-3,\\infty),\\, f(x)=e^{2x}-3\n\n(b) f:(3,\\infty) \\to \\R,\\,f(x)=4\\ln(x-3)\n\n(c) f:(-\\frac{1}{2},\\infty) \\to \\R,\\,f(x)=\\log_{10}(2x+1)\n\n(d) f:\\R \\to (3,\\infty),\\, f(x)=2^x+3\n\nQ3. For each of the following functions, find y in terms of x:\n\n(a) \\ln(y)=\\ln(2x+5)\n\n(b) \\log_2(2y)=3\\log_2(3x+1)\n\n(c) \\log_{10}(y)=-3+4\\log_{10}(x)\n\n(d) \\ln(y)=2x+3\n\nQ4. Solve each equation for x, expressing any logarithms in the answer with base e:\n\n(a) 3^{2x}-3^{x}-3=0\n\n(b) log_{10}(3x)-2=0\n\n(c) 5^{2x}-2(5^{x})-3=0\n\nQ5. The graph of f(x)=2\\log_2(2x+1)-5 intersects the axes at the the points (a, 0) and (0, b) and passes through the point (c, 3) for a,b,c \\in \\R. Find a, b and c.\n\nQ6. Solve for x if 4e^{2x}=251.\n\nQ7. Bonus question: Let f(x)=\\dfrac{1}{2}(e^x+e^{-x}) and g(x)=\\dfrac{1}{2}(e^x-e^{-x}):\n\n(a) Show that f is an even function\n\n(b) Find f(u)+f(-u)\n\n(c) Find f(u)-f(-u)\n\n(d) Show that g is an odd function\n\n(e) Find f(x)+g(x),\\, f(x)-g(x),\\, f(x)g(x)\n\nGot questions? Share and ask here..." ]

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[ "You copied the Doc URL to your clipboard.\n\n## FTSMUL\n\nFloating-point trigonometric starting value.\n\nThe FTSMUL instruction calculates the initial value for the FTMAD instruction. The instruction squares each element in the first source vector and then sets the sign bit to a copy of bit 0 of the corresponding element in the second source register, and places the results in the destination vector. This instruction is unpredicated.\n\nTo compute sin(x) or cos(x) the instruction is executed with elements of the first source vector set to x, adjusted to be in the range -π/4 < x ≤ π/4.\n\nThe elements of the second source vector hold the corresponding value of the quadrant q number as an integer not a floating-point value. The value q satisfies the relationship (2q-1) × π/4 < x ≤ (2q+1) × π/4.\n\n 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 1 1 0 0 1 0 1 size 0 Zm 0 0 0 0 1 1 Zn Zd\n\nFTSMUL <Zd>.<T>, <Zn>.<T>, <Zm>.<T>\n\n```if !HaveSVE() then UNDEFINED;\nif size == '00' then UNDEFINED;\ninteger esize = 8 << UInt(size);\ninteger n = UInt(Zn);\ninteger m = UInt(Zm);\ninteger d = UInt(Zd);```\n\n### Assembler Symbols\n\n Is the name of the destination scalable vector register, encoded in the \"Zd\" field.\n<T> Is the size specifier, encoded in size:\nsize <T>\n00 RESERVED\n01 H\n10 S\n11 D\n Is the name of the first source scalable vector register, encoded in the \"Zn\" field.\n Is the name of the second source scalable vector register, encoded in the \"Zm\" field.\n\n### Operation\n\n```CheckSVEEnabled();\ninteger elements = VL DIV esize;\nbits(VL) operand1 = Z[n];\nbits(VL) operand2 = Z[m];\nbits(VL) result;\n\nfor e = 0 to elements-1\nbits(esize) element1 = Elem[operand1, e, esize];\nbits(esize) element2 = Elem[operand2, e, esize];\nElem[result, e, esize] = FPTrigSMul(element1, element2, FPCR[]);\n\nZ[d] = result;```" ]

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[ "# Toefl Ibt Score Conversion\n\nToefl Ibt Score Conversion Hitting the f1l flag will only give you an estimate of how much more one should be willing to pay for a particular piece of gear. There are many ways to get this done, but I’ve found that the simplest way to get a good score from a website is to use the f1f1r score conversion tool. The f1f0r tool is a software program that does a lot of calculations on the f1r.1r score. It is used to find the best possible f1r score for the game. It is also used to find out how much of the available gear will be available in the game. The tool is designed to convert the f1xr score on the f2r score into a f2r bpg score. I know that some people may be looking for the f2f0r score try here but I”m not sure how it works. It is a simple calculation. First we need to find out what the f2xr score is. First we need to see how much of a gear will be played in the game based on the f4r score. This is done using the f4f3r score and a few other data. We can find out how many gear is played in the f4xr score by looking at the f4fr2r score. The f4f2r score is a small file. There is one file that contains only the gear that is played in this game. The file consists of the gear that we want to use in the game but the f4b2r score used in the game is the f4c8r score. We can also look at the f3r score. There is a f3r rxr score. In this case we want to find out the gear that will be played. This will show how much gear will be used in the f3xr score and how much gear we will be willing to spend to get the f3f1r average score.\n\n## Take My Accounting Exam\n\nWe can then figure out how much gear is played and how much we will be able to spend to obtain a good score. The f3r2r score can be found by looking at this file. Once we have this file we can start using it to find out if we can get a good rxr rv score. In the f3b2r file we can see the gear that was played in the previous game and we can see how much gear was played in this one game. Finally, we can find out the overall gear that we can get in this game and how much of it we are willing to spend. And this is where I am with the f3c8r. In the first part of the f3a2r file you can see how many gear are played in the first game. In this file we are looking at the gear that has been played in the last game. In this file we look at a file that has not been played in this first game. In the first part we can see a file that contains the gear that had been played in that game. If you are not in the first part, you can find out what gear was played and the gear view you were willing to spend in this game using the f3k1r or fToefl Ibt Score Conversion Ibt Score Conversion is a professional scoring system that can convert score data to an accurate score on a variety of scores, including: A. High or average score B. Average or close to average C. Moderate or high D. High or low E. High or high and F. Average or low returned score to the user based on a higher or lower score than the user would have been expected to have been able to compute. This system is known as an Ibt Score Converter. The Ibt Score Convertor is the most popular score system in the world. It is the most advanced version of the Ibt Score Generator.\n\n## Hire Someone To Do Online Class\n\nIt is developed to convert score data from both log and data files to an accurate and accurate score. It is also the most popular software that can convert scores to a high or average score on a wide range of databases, including Microsoft, Yahoo, and Google. Key Features of Ibt Score Conversion The solution for converting score data from log and data to an accuracy score is based on a simple formula. It can be used to calculate the total score that can be converted to a score, as well as the average score. This formula can be used in combination with other scores and with other options to compare them. For example, if you have a log file that contains a score of 100, you can compute the average score for that file by performing a sub-calculation. If you have a score of 120, you can use the Ibt Compute Score Conversion formula to calculate the average score of that file. As you can see, the Ibt Component Score Converter has several features that take advantage of various scores and perform the conversion automatically. To determine whether a score equals the average score, you have to find a score that is higher than average. This is done by comparing the score with other scores. You can also use the score conversion tool that is available at www.athena.com. A new score conversion tool called IBT Score Converter (or Ibt Score Compute) can be used. It provides a simple formula that calculates the average score and the score of log files. Example: Score conversion I would like to convert the average score to a score. To do so, you need to do the following steps: Start by creating a default file called score.txt. Open the file in which you want to convert the score to. This file will be the score file you created earlier." ]

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[ "## Conversion formula\n\nThe conversion factor from feet per second to meters per second is 0.3048, which means that 1 foot per second is equal to 0.3048 meters per second:\n\n1 ft/s = 0.3048 m/s\n\nTo convert 1.1 feet per second into meters per second we have to multiply 1.1 by the conversion factor in order to get the velocity amount from feet per second to meters per second. We can also form a simple proportion to calculate the result:\n\n1 ft/s → 0.3048 m/s\n\n1.1 ft/s → V(m/s)\n\nSolve the above proportion to obtain the velocity V in meters per second:\n\nV(m/s) = 1.1 ft/s × 0.3048 m/s\n\nV(m/s) = 0.33528 m/s\n\nThe final result is:\n\n1.1 ft/s → 0.33528 m/s\n\nWe conclude that 1.1 feet per second is equivalent to 0.33528 meters per second:\n\n1.1 feet per second = 0.33528 meters per second\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter per second is equal to 2.9825817227392 × 1.1 feet per second.\n\nAnother way is saying that 1.1 feet per second is equal to 1 ÷ 2.9825817227392 meters per second.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that one point one feet per second is approximately zero point three three five meters per second:\n\n1.1 ft/s ≅ 0.335 m/s\n\nAn alternative is also that one meter per second is approximately two point nine eight three times one point one feet per second.\n\n## Conversion table\n\n### feet per second to meters per second chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from feet per second to meters per second\n\nfeet per second (ft/s) meters per second (m/s)\n2.1 feet per second 0.64 meters per second\n3.1 feet per second 0.945 meters per second\n4.1 feet per second 1.25 meters per second\n5.1 feet per second 1.554 meters per second\n6.1 feet per second 1.859 meters per second\n7.1 feet per second 2.164 meters per second\n8.1 feet per second 2.469 meters per second\n9.1 feet per second 2.774 meters per second\n10.1 feet per second 3.078 meters per second\n11.1 feet per second 3.383 meters per second" ]

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[ "# fresnel VEX function\n\nComputes the fresnel reflection/refraction contributions given an incoming vector, surface normal (both normalized), and an index of refraction (eta).\n\n``` void  fresnel(vector i, vector n, float eta, float &kr, float &kt) ```\n\n``` void  fresnel(vector i, vector n, float eta, float &kr, float &kt, vector &R, vector &T) ```\n\nComputes the fresnel reflection/refraction contributions given an incoming vector, surface normal (both normalized), and an index of refraction (eta). The amount of reflected light will be returned in kr, and the amount of transmitted light will be returned in kt. Optionally, the reflection and transmission vectors can be returned in the R and T variables. The R and <type> variables will be normalized vectors on exit.\n\neta is a relative index of refraction, the ratio between the interior and exterior index of refraction, where the exterior is defined by the direction of the normals (normals point away from the interior).\n\n# VEX Functions\n\n## Arrays\n\n• Adds an item to an array or string.\n\n• Returns the indices of a sorted version of an array.\n\n• Efficiently creates an array from its arguments.\n\n• Loops over the items in an array, with optional enumeration.\n\n• Inserts an item, array, or string into an array or string.\n\n• Checks if the index given is valid for the array or string given.\n\n• Returns the length of an array.\n\n• Removes the last element of an array and returns it.\n\n• Adds an item to an array.\n\n• Removes an item at the given index from an array.\n\n• Removes an item from an array.\n\n• Reorders items in an array or string.\n\n• Sets the length of an array.\n\n• Returns an array or string in reverse order.\n\n• Slices a sub-string or sub-array of a string or array.\n\n• Returns the array sorted in increasing order.\n\n• Adds a uniform item to an array.\n\n## Attributes and Intrinsics\n\n• Adds an attribute to a geometry.\n\n• Adds a detail attribute to a geometry.\n\n• Adds a point attribute to a geometry.\n\n• Adds a primitive attribute to a geometry.\n\n• Adds a vertex attribute to a geometry.\n\n• Appends to a geometry’s visualizer detail attribute.\n\n• Reads the value of an attribute from geometry.\n\n• Returns the class of a geometry attribute.\n\n• Returns the data id of a geometry attribute.\n\n• Returns the size of a geometry attribute.\n\n• Returns the type of a geometry attribute.\n\n• Returns the transformation metadata of a geometry attribute.\n\n• Reads the value of a detail attribute value from a geometry.\n\n• Reads a detail attribute value from a geometry.\n\n• Returns the size of a geometry detail attribute.\n\n• Returns the type of a geometry detail attribute.\n\n• Returns the type info of a geometry attribute.\n\n• Reads the value of a detail intrinsic from a geometry.\n\n• Finds a primitive/point/vertex that has a certain attribute value.\n\n• Returns number of elements where an integer or string attribute has a certain value.\n\n• Reads an attribute value from geometry, with validity check.\n\n• Copies the value of a geometry attribute into a variable and returns a success flag.\n\n• Checks whether a geometry attribute exists.\n\n• Returns if a geometry detail attribute exists.\n\n• Returns if a geometry point attribute exists.\n\n• Returns if a geometry prim attribute exists.\n\n• Returns if a geometry vertex attribute exists.\n\n• Finds a point by its id attribute.\n\n• Finds a primitive by its id attribute.\n\n• Finds a point by its name attribute.\n\n• Finds a primitive by its name attribute.\n\n• Returns the number of unique values from an integer or string attribute.\n\n• Reads a point attribute value from a geometry.\n\n• Reads a point attribute value from a geometry and outputs a success/fail flag.\n\n• Returns the size of a geometry point attribute.\n\n• Returns the type of a geometry point attribute.\n\n• Returns the type info of a geometry attribute.\n\n• Reads a primitive attribute value from a geometry.\n\n• Interpolates the value of an attribute at a certain parametric (u, v) position and copies it into a variable.\n\n• Evaluates the length of an arc on a primitive using parametric uv coordinates.\n\n• Reads a primitive attribute value from a geometry, outputting a success flag.\n\n• Returns the size of a geometry prim attribute.\n\n• Returns the type of a geometry prim attribute.\n\n• Returns the type info of a geometry attribute.\n\n• Returns position derivative on a primitive at a certain parametric (u, v) position.\n\n• Reads a primitive intrinsic from a geometry.\n\n• Interpolates the value of an attribute at a certain parametric (uvw) position.\n\n• Convert parametric UV locations on curve primitives between different spaces.\n\n• Writes an attribute value to geometry.\n\n• Sets the meaning of an attribute in geometry.\n\n• Sets a detail attribute in a geometry.\n\n• Sets the value of a writeable detail intrinsic attribute.\n\n• Sets a point attribute in a geometry.\n\n• Sets a primitive attribute in a geometry.\n\n• Sets the value of a writeable primitive intrinsic attribute.\n\n• Sets a vertex attribute in a geometry.\n\n• Returns one of the set of unique values across all values for an int or string attribute.\n\n• Returns the set of unique values across all values for an int or string attribute.\n\n• Interpolates the value of an attribute at certain UV coordinates using a UV attribute.\n\n• Reads a vertex attribute value from a geometry.\n\n• Reads a vertex attribute value from a geometry.\n\n• Returns the size of a geometry vertex attribute.\n\n• Returns the type of a geometry vertex attribute.\n\n• Returns the type info of a geometry attribute.\n\n## BSDFs\n\n• Returns the albedo (percentage of reflected light) for a bsdf given the outgoing light direction.\n\n• Returns a specular BSDF using the Ashikhmin shading model.\n\n• Returns a Blinn BSDF or computes Blinn shading.\n\n• Returns a cone reflection BSDF.\n\n• Creates a bsdf object from two CVEX shader strings.\n\n• Returns a diffuse BSDF or computes diffuse shading.\n\n• Evaluates a bsdf given two vectors.\n\n• Returns a ggx BSDF.\n\n• Returns a BSDF for shading hair.\n\n• Returns an anisotropic volumetric BSDF, which can scatter light forward or backward.\n\n• Returns an isotropic BSDF, which scatters light equally in all directions.\n\n• Returns new BSDF that only includes the components specified by the mask.\n\n• Returns the normal for the diffuse component of a BSDF.\n\n• Returns a Phong BSDF or computes Phong shading.\n\n• Samples a BSDF.\n\n• Computes the solid angle (in steradians) a BSDF function subtends.\n\n• Splits a bsdf into its component lobes.\n\n• Creates an approximate SSS BSDF.\n\n## BSDFs\n\n• Returns a specular BSDF or computes specular shading.\n\n## CHOP\n\n• Adds new channels to a CHOP node.\n\n• Reads from a CHOP attribute.\n\n• Reads CHOP attribute names of a given attribute class from a CHOP input.\n\n• Returns the sample number of the last sample in a given CHOP input.\n\n• Returns the frame corresponding to the last sample of the input specified.\n\n• Returns the time corresponding to the last sample of the input specified.\n\n• Returns the channel index from a input given a channel name.\n\n• Returns the value of a channel at the specified sample.\n\n• Computes the minimum and maximum value of samples in an input channel.\n\n• Returns all the CHOP channel names of a given CHOP input.\n\n• Returns the number of channels in the input specified.\n\n• Returns the value of a CHOP channel at the specified sample.\n\n• Returns the value of a CHOP local transform channel at the specified sample.\n\n• Returns the value of a CHOP local transform channel at the specified sample and evaluation time.\n\n• Returns the value of a CHOP channel at the specified sample and evaluation time.\n\n• Returns the sample rate of the input specified.\n\n• Returns the value of CHOP context temporary buffer at the specified index.\n\n• Removes channels from a CHOP node.\n\n• Removes a CHOP attribute.\n\n• Renames a CHOP channel.\n\n• Resize the CHOP context temporary buffer\n\n• Sets the value of a CHOP attribute.\n\n• Sets the length of the CHOP channel data.\n\n• Sets the sampling rate of the CHOP channel data.\n\n• Sets the CHOP start sample in the channel data.\n\n• Returns the start sample of the input specified.\n\n• Returns the frame corresponding to the first sample of the input specified.\n\n• Returns the time corresponding to the first sample of the input specified.\n\n• Writes a value of CHOP context temporary buffer at the specified index.\n\n• Returns 1 if the Vex CHOP’s Unit Menu is currently set to 'frames', 0 otherwise.\n\n• Returns 1 if the Vex CHOP’s Unit Menu is currently set to 'samples', 0 otherwise.\n\n• Returns 1 if the Vex CHOP’s Unit Menu is currently set to 'seconds', 0 otherwise.\n\n• Returns the number of inputs.\n\n## color\n\n• Compute the color value of an incandescent black body.\n\n• Transforms between color spaces.\n\n• Compute the luminance of the RGB color specified by the parameters.\n\n## Conversion\n\n• Converts a string to a float.\n\n• Converts a string to an integer.\n\n• Depending on the value of c, returns the translate (c=0), rotate (c=1), scale (c=2), or shears (c=3) component of the transform (xform).\n\n• Converts the argument from radians into degrees.\n\n• Creates a vector4 representing a quaternion from euler angles.\n\n• Convert HSV color space into RGB color space.\n\n• Converts a quaternion represented by a vector4 to a matrix3 representation.\n\n• Creates a euler angle representing a quaternion.\n\n• Converts the argument from degrees into radians.\n\n• Convert RGB color space to HSV color space.\n\n• Convert a linear sRGB triplet to CIE XYZ tristimulus values.\n\n• Flattens an array of vector or matrix types into an array of floats.\n\n• Turns a flat array of floats into an array of vectors or matrices.\n\n• Convert CIE XYZ tristimulus values to a linear sRGB triplet.\n\n## Crowds\n\n• Add a clip into an agent’s definition.\n\n• Returns the names of the channels in an agent primitive’s rig.\n\n• Returns the current value of an agent primitive’s channel.\n\n• Returns the current values of an agent primitive’s channels.\n\n• Returns all of the animation clips that have been loaded for an agent primitive.\n\n• Finds the index of a channel in an agent’s animation clip.\n\n• Returns the names of the channels in an agent’s animation clip.\n\n• Returns the length (in seconds) of an agent’s animation clip.\n\n• Returns an agent primitive’s current animation clips.\n\n• Samples a channel of an agent’s clip at a specific time.\n\n• Samples an agent’s animation clip at a specific time.\n\n• Returns the sample rate of an agent’s animation clip.\n\n• Samples an agent’s animation clip at a specific time.\n\n• Returns the start time (in seconds) of an agent’s animation clip.\n\n• Returns the current times for an agent primitive’s animation clips.\n\n• Returns the transform groups for an agent primitive’s current animation clips.\n\n• Returns the blend weights for an agent primitive’s animation clips.\n\n• Returns the name of the collision layer of an agent primitive.\n\n• Returns the name of the current layer of an agent primitive.\n\n• Finds the index of a transform group in an agent’s definition.\n\n• Returns the transform that each shape in an agent’s layer is bound to.\n\n• Returns all of the layers that have been loaded for an agent primitive.\n\n• Returns the names of the shapes referenced by an agent primitive’s layer.\n\n• Returns the current local space transform of an agent primitive’s bone.\n\n• Returns the current local space transforms of an agent primitive.\n\n• Returns the child transforms of a transform in an agent primitive’s rig.\n\n• Finds the index of a transform in an agent primitive’s rig.\n\n• Finds the index of a channel in an agent primitive’s rig.\n\n• Returns the parent transform of a transform in an agent primitive’s rig.\n\n• Applies a full-body inverse kinematics algorithm to an agent’s skeleton.\n\n• Returns the number of transforms in an agent primitive’s rig.\n\n• Returns whether a transform is a member of the specified transform group.\n\n• Returns whether a channel is a member of the specified transform group.\n\n• Returns the names of the transform groups in an agent’s definition.\n\n• Returns the weight of a member of the specified transform group.\n\n• Returns the name of each transform in an agent primitive’s rig.\n\n• Converts transforms from world space to local space for an agent primitive.\n\n• Converts transforms from local space to world space for an agent primitive.\n\n• Returns the current world space transform of an agent primitive’s bone.\n\n• Returns the current world space transforms of an agent primitive.\n\n• Overrides the value of an agent primitive’s channel.\n\n• Overrides the values of an agent primitive’s channels.\n\n• Sets the current animation clips for an agent primitive.\n\n• Sets the animation clips that an agent should use to compute its transforms.\n\n• Sets the current times for an agent primitive’s animation clips.\n\n• Sets the blend weights for an agent primitive’s animation clips.\n\n• Sets the collision layer of an agent primitive.\n\n• Sets the current layer of an agent primitive.\n\n• Overrides the local space transform of an agent primitive’s bone.\n\n• Overrides the local space transforms of an agent primitive.\n\n• Overrides the world space transform of an agent primitive’s bone.\n\n• Overrides the world space transforms of an agent primitive.\n\n## File I/O\n\n• Returns file system status for a given file.\n\n## Geometry\n\n• Adds a point to the geometry.\n\n• Adds a primitive to the geometry.\n\n• Adds a vertex to a primitive in a geometry.\n\n• Clip the line segment between p0 and p1.\n\n• Returns a handle to the current geometry.\n\n• Returns an oppath: string to unwrap the geometry in-place.\n\n• Returns 1 if the edge specified by the point pair is in the group specified by the string.\n\n• This function computes the first intersection of a ray with geometry.\n\n• Computes all intersections of the specified ray with geometry.\n\n• Finds the closest position on the surface of a geometry.\n\n• Finds the closest point in a geometry.\n\n• Finds the all the closest point in a geometry.\n\n• Returns the number of edges in the group.\n\n• Returns the point number of the next point connected to a given point.\n\n• Returns the number of points that are connected to the specified point.\n\n• Returns an array of the point numbers of the neighbours of a point.\n\n• Returns the number of points in the input or geometry file.\n\n• Returns the number of primitives in the input or geometry file.\n\n• Returns the number of vertices in the input or geometry file.\n\n• Returns the number of vertices in the group.\n\n• Returns the list of primitives containing a point.\n\n• Returns a linear vertex number of a point in a geometry.\n\n• Returns the list of vertices connected to a point.\n\n• Returns an array of the primitive numbers of the edge-neighbours of a polygon.\n\n• Returns a list of primitives potentially intersecting a given bounding box.\n\n• Converts a primitive/vertex pair into a point number.\n\n• Returns the list of points on a primitive.\n\n• Converts a primitive/vertex pair into a linear vertex.\n\n• Returns number of vertices in a primitive in a geometry.\n\n• Returns the list of vertices on a primitive.\n\n• Removes a point from the geometry.\n\n• Removes a primitive from the geometry.\n\n• Removes a vertex from the geometry.\n\n• Sets edge group membership in a geometry.\n\n• Rewires a vertex in the geometry to a different point.\n\n• Rewires a vertex in the geometry to a different point.\n\n• This function computes the intersection of the specified ray with the geometry in uv space.\n\n• Converts a primitive/vertex pair into a linear vertex.\n\n• Returns the linear vertex number of the next vertex sharing a point with a given vertex.\n\n• Returns the point number of linear vertex in a geometry.\n\n• Returns the linear vertex number of the previous vertex sharing a point with a given vertex.\n\n• Returns the number of the primitive containing a given vertex.\n\n• Converts a linear vertex index into a primitive vertex number.\n\n## groups\n\n• Returns an array of point numbers corresponding to a group string.\n\n• Returns an array of prim numbers corresponding to a group string.\n\n• Returns an array of linear vertex numbers corresponding to a group string.\n\n• Returns 1 if the point specified by the point number is in the group specified by the string.\n\n• Returns 1 if the primitive specified by the primitive number is in the group specified by the string.\n\n• Returns 1 if the vertex specified by the vertex number is in the group specified by the string.\n\n• Returns the number of points in the group.\n\n• Returns the number of primitives in the group.\n\n• Adds or removes a point to/from a group in a geometry.\n\n• Adds or removes a primitive to/from a group in a geometry.\n\n• Adds or removes a vertex to/from a group in a geometry.\n\n## Half-edges\n\n• Returns the destination point of a half-edge.\n\n• Returns the destination vertex of a half-edge.\n\n• Returns the number of half-edges equivalent to a given half-edge.\n\n• Determines whether a two half-edges are equivalent (represent the same edge).\n\n• Determines whether a half-edge number corresponds to a primary half-edge.\n\n• Determines whether a half-edge number corresponds to a valid half-edge.\n\n• Returns the half-edge that follows a given half-edge in its polygon.\n\n• Returns the next half-edges equivalent to a given half-edge.\n\n• Returns the point into which the vertex following the destination vertex of a half-edge in its primitive is wired.\n\n• Returns the vertex following the destination vertex of a half-edge in its primitive.\n\n• Returns the point into which the vertex that precedes the source vertex of a half-edge in its primitive is wired.\n\n• Returns the vertex that precedes the source vertex of a half-edge in its primitive.\n\n• Returns the half-edge that precedes a given half-edge in its polygon.\n\n• Returns the primitive that contains a half-edge.\n\n• Returns the primary half-edge equivalent to a given half-edge.\n\n• Returns the source point of a half-edge.\n\n• Returns the source vertex of a half-edge.\n\n• Finds and returns a half-edge with the given endpoints.\n\n• Finds and returns a half-edge with a given source point or with given source and destination points.\n\n• Returns the next half-edge with the same source as a given half-edge.\n\n• Returns one of the half-edges contained in a primitive.\n\n• Returns the half-edge which has a vertex as source.\n\n## Image Processing\n\n• Tells the COP manager that you need access to the given frame.\n\n• Returns the default name of the alpha plane (as it appears in the compositor preferences).\n\n• Samples a 2×2 pixel block around the given UV position, and bilinearly interpolates these pixels.\n\n• Returns the default name of the bump plane (as it appears in the compositor preferences).\n\n• Returns the name of a numbered channel.\n\n• Samples the exact (unfiltered) pixel color at the given coordinates.\n\n• Returns the default name of the color plane (as it appears in the compositor preferences).\n\n• Returns the default name of the depth plane (as it appears in the compositor preferences).\n\n• Reads the z-records stored in a pixel of a deep shadow map or deep camera map.\n\n• Returns fully filtered pixel input.\n\n• Queries if metadata exists on a composite operator.\n\n• Returns 1 if the plane specified by the parameter exists in this COP.\n\n• Returns the aspect ratio of the specified input.\n\n• Returns the channel name of the indexed plane of the given input.\n\n• Returns the last frame of the specified input.\n\n• Returns the end time of the specified input.\n\n• Returns 1 if the specified input has a plane named planename.\n\n• Returns the number of planes in the given input.\n\n• Returns the index of the plane named 'planename' in the specified input.\n\n• Returns the name of the plane specified by the planeindex of the given input\n\n• Returns the number of components in the plane named planename in the specified input.\n\n• Returns the frame rate of the specified input.\n\n• Returns the starting frame of the specified input.\n\n• Returns the start time of the specified input.\n\n• Returns the X resolution of the specified input.\n\n• Returns the Y resolution of the specified input.\n\n• Returns the default name of the luminaence plane (as it appears in the compositor preferences).\n\n• Returns the default name of the mask plane (as it appears in the compositor preferences).\n\n• Returns a metadata value from a composite operator.\n\n• Reads a component from a pixel and its eight neighbors.\n\n• Returns the default name of the normal plane (as it appears in the compositor preferences).\n\n• Returns the index of the plane specified by the parameter, starting at zero.\n\n• Returns the name of the plane specified by the index (e.\n\n• Returns the number of components in the plane (1 for scalar planes and up to 4 for vector planes).\n\n• Returns the default name of the point plane (as it appears in the compositor preferences).\n\n• Returns the default name of the velocity plane (as it appears in the compositor preferences).\n\n## Interpolation\n\n• Samples a Catmull-Rom (Cardinal) spline defined by position/value keys.\n\n• Returns value clamped between min and max.\n\n• Samples a Catmull-Rom (Cardinal) spline defined by uniformly spaced keys.\n\n• Takes the value in one range and shifts it to the corresponding value in a new range.\n\n• Takes the value in one range and shifts it to the corresponding value in a new range.\n\n• Takes the value in the range (0, 1) and shifts it to the corresponding value in a new range.\n\n• Takes the value in the range (1, 0) and shifts it to the corresponding value in a new range.\n\n• Takes the value in the range (-1, 1) and shifts it to the corresponding value in a new range.\n\n• Performs bilinear interpolation between the values.\n\n• Samples a polyline between the key points.\n\n• Samples a polyline defined by linearly spaced values.\n\n• Quaternion blend between q1 and q2 based on the bias.\n\n• Computes ease in/out interpolation between values.\n\n## light\n\n• Returns the color of ambient light in the scene.\n\n• Computes attenuated falloff.\n\n• Sends a ray from the position P along the direction specified by the direction D.\n\n• Sends a ray from the position P along direction D.\n\n## Math\n\n• Returns the derivative of the given value with respect to U.\n\n• Returns the derivative of the given value with respect to V.\n\n• Returns the derivative of the given value with respect to the 3rd axis (for volume rendering).\n\n• Returns the absolute value of the argument.\n\n• Returns the inverse cosine of the argument.\n\n• Returns the inverse sine of the argument.\n\n• Returns the inverse tangent of the argument.\n\n• Returns the inverse tangent of y/x.\n\n• Returns the average value of the input(s)\n\n• Returns the cube root of the argument.\n\n• Returns the smallest integer greater than or equal to the argument.\n\n• Combines Local and Parent Transforms with Scale Inheritance.\n\n• Returns the cosine of the argument.\n\n• Returns the hyperbolic cosine of the argument.\n\n• Returns the cross product between the two vectors.\n\n• Computes the determinant of the matrix.\n\n• Diagonalizes Symmetric Matrices.\n\n• Returns the dot product between the arguments.\n\n• Computes the eigenvalues of a 3×3 matrix.\n\n• Gauss error function.\n\n• Inverse Gauss error function.\n\n• Gauss error function’s complement.\n\n• Returns the exponential function of the argument.\n\n• Extracts Local Transform from a World Transform with Scale Inheritance.\n\n• Returns the largest integer less than or equal to the argument.\n\n• Returns the fractional component of the floating point number.\n\n• Returns an identity matrix.\n\n• Inverts a matrix.\n\n• Checks whether a value is a normal finite number.\n\n• Checks whether a value is not a number.\n\n• Returns an interpolated value along a curve defined by a basis and key/position pairs.\n\n• Returns the magnitude of a vector.\n\n• Returns the squared distance of the vector or vector4.\n\n• Returns the natural logarithm of the argument.\n\n• Returns the logarithm (base 10) of the argument.\n\n• Creates an orthonormal basis given a z-axis vector.\n\n• Returns a normalized vector.\n\n• Returns the outer product between the arguments.\n\n• Computes the intersection of a 3D sphere and an infinite 3D plane.\n\n• Raises the first argument to the power of the second argument.\n\n• Determines if a point is inside or outside a triangle circumcircle.\n\n• Determines if a point is inside or outside a tetrahedron circumsphere.\n\n• Determines the orientation of a point with respect to a line.\n\n• Determines the orientation of a point with respect to a plane.\n\n• Pre multiply matrices.\n\n• Returns the product of a list of numbers.\n\n• This function returns the closest distance between the point Q and a finite line segment between points P0 and P1.\n\n• Finds distance between two quaternions.\n\n• Inverts a quaternion rotation.\n\n• Multiplies two quaternions and returns the result.\n\n• Rotates a vector by a quaternion.\n\n• Creates a vector4 representing a quaternion.\n\n• Rounds the number to the closest whole number.\n\n• Bit-shifts an integer left.\n\n• Bit-shifts an integer right.\n\n• Bit-shifts an integer right.\n\n• Returns -1, 0, or 1 depending on the sign of the argument.\n\n• Returns the sine of the argument.\n\n• Returns the hyperbolic sine of the argument.\n\n• Finds the normal component of frame slid along a curve.\n\n• Solves a cubic function returning the number of real roots.\n\n• Finds the real roots of a polynomial.\n\n• Solves a quadratic function returning the number of real roots.\n\n• Finds the angles of a triangle from its sides.\n\n• Samples a value along a polyline or spline curve.\n\n• Returns the square root of the argument.\n\n• Returns the sum of a list of numbers.\n\n• Computes the singular value decomposition of a 3×3 matrix.\n\n• Returns the trigonometric tangent of the argument\n\n• Returns the hyperbolic tangent of the argument\n\n• Transposes the given matrix.\n\n• Removes the fractional part of a floating point number.\n\n## measure\n\n• Returns the distance between two points.\n\n• Returns the squared distance between the two points.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the geometry.\n\n• Returns the center of the bounding box for the geometry.\n\n• Returns the maximum of the bounding box for the geometry.\n\n• Returns the minimum of the bounding box for the geometry.\n\n• Returns the size of the bounding box for the geometry.\n\n• Returns the bounding box of the geometry specified by the filename.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the geometry.\n\n• Returns the center of the bounding box for the geometry.\n\n• Returns the maximum of the bounding box for the geometry.\n\n• Returns the minimum of the bounding box for the geometry.\n\n• Returns the size of the bounding box for the geometry.\n\n• Computes the distance and closest point of a point to an infinite plane.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Finds the distance of a point to a group of points along the surface of a geometry.\n\n• Finds the distance of a uv coordinate to a geometry in uv space.\n\n• Finds the distance of a point to a geometry.\n\n## metaball\n\n• Once you get a handle to a metaball using metastart and metanext, you can query attributes of the metaball with metaimport.\n\n• Takes the ray defined by p0 and p1 and partitions it into zero or more sub-intervals where each interval intersects a cluster of metaballs from filename.\n\n• Iterate to the next metaball in the list of metaballs returned by the metastart() function.\n\n• Open a geometry file and return a \"handle\" for the metaballs of interest, at the position p.\n\n• Returns the metaweight of the geometry at position p.\n\n## Nodes\n\n• Adds a mapping for an attribute to a local variable.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel with a new segment expression.\n\n• Evaluates a channel with a new segment expression at a given frame.\n\n• Evaluates a channel with a new segment expression at a given time.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Resolves a channel string (or parameter) and return op_id, parm_index and vector_index.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a ramp parameter and return its value.\n\n• Evaluates the derivative of a parm parameter with respect to position.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates an operator path parameter and return the path to the operator.\n\n• Returns the raw string channel (or parameter).\n\n• Evaluates a channel or parameter, and return its value.\n\n• Evaluates a channel or parameter, and return its value.\n\n• Returns the capture transform associated with a Capture Region SOP.\n\n• Returns the deform transform associated with a Capture Region SOP.\n\n• Returns the capture or deform transform associated with a Capture Region SOP based on the global capture override flag.\n\n• Returns 1 if input_number is connected, or 0 if the input is not connected.\n\n• Returns the full path for the given relative path\n\n• Resolves an operator path string and return its op_id.\n\n• Returns the parent bone transform associated with an OP.\n\n• Returns the parent transform associated with an OP.\n\n• Returns the parm transform associated with an OP.\n\n• Returns the preconstraint transform associated with an OP.\n\n• Returns the pre and parm transform associated with an OP.\n\n• Returns the pre and raw parm transform associated with an OP.\n\n• Returns the pretransform associated with an OP.\n\n• Returns the raw parm transform associated with an OP.\n\n• Returns the transform associated with an OP.\n\n## Noise and Randomness\n\n• Generates \"alligator\" noise.\n\n• Computes divergence free noise based on Perlin noise.\n\n• Computes 2d divergence free noise based on Perlin noise.\n\n• Computes divergence free noise based on Simplex noise.\n\n• Computes 2d divergence free noise based on simplex noise.\n\n• Generates Worley (cellular) noise using a Chebyshev distance metric.\n\n• Generates 1D and 3D Perlin Flow Noise from 3D and 4D data.\n\n• There are two forms of Perlin-style noise: a non-periodic noise which changes randomly throughout N-dimensional space, and a periodic form which repeats over a given range of space.\n\n• Generates noise matching the output of the Hscript noise() expression function.\n\n• Produces the exact same results as the Houdini expression function of the same name.\n\n• Generates turbulence matching the output of the HScript turb() expression function.\n\n• Generates Worley (cellular) noise using a Manhattan distance metric.\n\n• There are two forms of Perlin-style noise: a non-periodic noise which changes randomly throughout N-dimensional space, and a periodic form which repeats over a given range of space.\n\n• Derivatives of Perlin Noise.\n\n• Non-deterministic random number generation function.\n\n• These functions are similar to wnoise and vnoise.\n\n• There are two forms of Perlin-style noise: a non-periodic noise which changes randomly throughout N-dimensional space, and a periodic form which repeats over a given range of space.\n\n• Periodic derivatives of Simplex Noise.\n\n• Creates a random number between 0 and 1 from a seed.\n\n• Generate a random number based on the position in 1-4D space.\n\n• Hashes floating point numbers to integers.\n\n• Hashes integer numbers to integers.\n\n• Generates a random Poisson variable given the mean to the distribution and a seed.\n\n• Hashes a string to an integer.\n\n• Generate a uniformly distributed random number.\n\n• These functions are similar to wnoise.\n\n• Generates Voronoi (cellular) noise.\n\n• Generates Worley (cellular) noise.\n\n• Simplex noise is very close to Perlin noise, except with the samples on a simplex mesh rather than a grid. This results in less grid artifacts. It also uses a higher order bspline to provide better derivatives. This is the periodic simplex noise\n\n• Simplex noise is very close to Perlin noise, except with the samples on a simplex mesh rather than a grid. This results in less grid artifacts. It also uses a higher order bspline to provide better derivatives.\n\n• Derivatives of Simplex Noise.\n\n## normals\n\n• In shading contexts, computes a normal. In the SOP contexts, sets how/whether to recompute normals.\n\n• Returns the normal of the primitive (prim_number) at parametric location u, v.\n\n## Open Color IO\n\n• Returns the names of active displays supported in Open Color IO\n\n• Returns the names of active views supported in Open Color IO\n\n• Imports attributes from OpenColorIO spaces.\n\n• Returns the names of roles supported in Open Color IO\n\n• Returns the names of color spaces supported in Open Color IO.\n\n• Transform colors using Open Color IO\n\n## particles\n\n• Samples the velocity field defined by a set of vortex filaments.\n\n## Point Clouds and 3D Images\n\n• Returns the value of the point attribute for the metaballs if metaball geometry is specified to i3dgen.\n\n• Returns the density of the metaball field if metaball geometry is specified to i3dgen.\n\n• Transforms the position specified into the \"local\" space of the metaball.\n\n• This function closes the handle associated with a pcopen function.\n\n• Returns a list of closest points from a file within a specified cone.\n\n• Returns a list of closest points from a file in a cone, taking into account their radii\n\n• Writes data to a point cloud inside a pciterate or a pcunshaded loop.\n\n• Returns the distance to the farthest point found in the search performed by pcopen.\n\n• Filters points found by pcopen using a simple reconstruction filter.\n\n• Returns a list of closest points from a file.\n\n• Returns a list of closest points from a file taking into account their radii.\n\n• Generates a point cloud.\n\n• Imports channel data from a point cloud inside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• This function can be used to iterate over all the points which were found in the pcopen query.\n\n• Returns a list of closest points to an infinite line from a specified file\n\n• Returns a list of closest points to an infinite line from a specified file\n\n• This node returns the number of points found by pcopen.\n\n• Returns a handle to a point cloud file.\n\n• Returns a handle to a point cloud file.\n\n• Changes the current iteration point to a leaf descendant of the current aggregate point.\n\n• Returns a list of closest points to a line segment from a specified file\n\n• Returns a list of closest points to a line segment from a specified file\n\n• Iterate over all of the points of a read-write channel which haven’t had any data written to the channel yet.\n\n• Writes data to a point cloud file.\n\n• Returns a list of closest points from a file.\n\n• Samples a color from a photon map.\n\n• Returns the value of the 3d image at the position specified by P.\n\n• This function queries the 3D texture map specified and returns the bounding box information of the file.\n\n## Sampling\n\n• Creates a CDF from an array of input PDF values.\n\n• Creates a PDF from an array of input values.\n\n• Limits a unit value in a way that maintains uniformity and in-range consistency.\n\n• Initializes a sampling sequence for the nextsample function.\n\n• Samples the Cauchy (Lorentz) distribution.\n\n• Samples a CDF.\n\n• Generates a uniform unit vector2, within maxangle of center, given a uniform number between 0 and 1.\n\n• Generates a uniform unit vector2, given a uniform number between 0 and 1.\n\n• Generates a uniform vector2 with alpha < length < 1, where 0 < alpha < 1, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector2 with length < 1, within maxangle of center, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector2 with length < 1, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform unit vector, within maxangle of center, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform unit vector, given a vector2 of uniform numbers between 0 and 1.\n\n• Returns an integer, either uniform or weighted, given a uniform number between 0 and 1.\n\n• Samples the exponential distribution.\n\n• Samples geometry in the scene and returns information from the shaders of surfaces that were sampled.\n\n• Generates a unit vector, optionally biased, within a hemisphere, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector4 with length < 1, within maxangle of center, given a vector4 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector4 with length < 1, given a vector4 of uniform numbers between 0 and 1.\n\n• Samples a 3D position on a light source and runs the light shader at that point.\n\n• Samples the log-normal distribution based on parameters of the underlying normal distribution.\n\n• Samples the log-normal distribution based on median and standard deviation.\n\n• Samples the normal (Gaussian) distribution.\n\n• Generates a uniform unit vector4, within maxangle of center, given a vector of uniform numbers between 0 and 1.\n\n• Generates a uniform unit vector4, given a vector of uniform numbers between 0 and 1.\n\n• Samples a 3D position on a light source and runs the light shader at that point.\n\n• Generates a uniform vector with length < 1, within maxangle of center, given a vector of uniform numbers between 0 and 1.\n\n• Generates a uniform vector with alpha < length < 1, where 0 < alpha < 1, given a vector of uniform numbers between 0 and 1.\n\n• Generates a uniform vector with length < 1, given a vector of uniform numbers between 0 and 1.\n\n• Warps uniform random samples to a disk.\n\n• Computes the mean value and variance for a value.\n\n## Sensor Input\n\n• Sensor function to render GL scene and query the result.\n\n• Sensor function query a rendered GL scene.\n\n• Sensor function to query average values from rendered GL scene.\n\n• Sensor function query a rendered GL scene.\n\n• Sensor function to save a rendered GL scene.\n\n• Returns the area of the micropolygon containing a variable such as P.\n\n• Returns the anti-aliased weight of the step function.\n\n• Computes the fresnel reflection/refraction contributions given an incoming vector, surface normal (both normalized), and an index of refraction (eta).\n\n• If dot(I, Nref) is less than zero, N will be negated.\n\n• Sends rays into the scene and returns information from the shaders of surfaces hit by the rays.\n\n• Returns the blurred point position (P) vector at a fractional time within the motion blur exposure.\n\n• Evaluates surface derivatives of an attribute.\n\n• Returns the name of the current object whose shader is being run.\n\n• Returns the depth of the ray tree for computing global illumination.\n\n• Returns group id containing current primitive.\n\n• Returns a light struct for the specified light identifier.\n\n• Returns the light id for a named light (or -1 for an invalid name).\n\n• Returns the name of the current light when called from within an illuminance loop, or converts an integer light ID into the light’s name.\n\n• Returns an array of light identifiers for the currently shaded surface.\n\n• Returns a selection of lights that illuminate a given material.\n\n• Evaluates local curvature of primitive grid, using the same curvature evaluation method as Measure SOPs.\n\n• Returns a material struct for the current surface.\n\n• Returns material id of shaded primitive.\n\n• Returns the object id for the current shading context.\n\n• Returns the name of the current object whose shader is being run.\n\n• Returns the integer ID of the light being used for photon shading.\n\n• Returns the number of the current primitive.\n\n• Returns the ptexture face id for the current primitive.\n\n• Returns the depth of the ray tree for the current shading.\n\n• Returns an approximation to the contribution of the ray to the final pixel color.\n\n• Looks up sample data in a channel, referenced by a point.\n\n• Returns a selection of objects visible to rays for a given material.\n\n• Returns modified surface position based on a smoothing function.\n\n• Evaluates UV tangents at a point on an arbitrary object.\n\n• Returns the gradient of a field.\n\n• Returns whether a light illuminates the given material.\n\n• Loops through all light sources in the scene, calling the light shader for each light source to set the Cl and L global variables.\n\n• Interpolates a value across the currently shaded micropolygon.\n\n• Finds the nearest intersection of a ray with any of a list of (area) lights and runs the light shader at the intersection point.\n\n• Computes irradiance (global illumination) at the point P with the normal N.\n\n• Returns 1 if the shader is being called to evaluate illumination for fog objects, or 0 if the light or shadow shader is being called to evaluate surface illumination.\n\n• Returns 1 if Light Path Expressions are enabled. 0 Otherwise.\n\n• Indicates whether a shader is being executed for ray tracing.\n\n• Detects the orientation of default shading space.\n\n• Returns 1 if the shader is being called to evaluate opacity for shadow rays, or 0 if the shader is being called to evaluate for surface color.\n\n• Indicates whether the shader is being evaluated while doing UV rendering (e.g. texture unwrapping)\n\n• Returns the bounce mask for a light struct.\n\n• Returns the light id for a light struct.\n\n• Queries the renderer for a named property.\n\n• Imports a variable from the light shader for the surface.\n\n• Returns a BSDF that matches the output of the traditional VEX blinn function.\n\n• Returns a BSDF that matches the output of the traditional VEX specular function.\n\n• Queries the renderer for a named property.\n\n• Computes ambient occlusion.\n\n• Computes global illumination using PBR for secondary bounces.\n\n• Sends a ray from the position P along the direction D.\n\n• Imports a value sent by a shader in a gather loop.\n\n• Returns the vector representing the reflection of the direction against the normal.\n\n• Computes the amount of reflected light which hits the surface.\n\n• Returns the refraction ray given an incoming direction, the normalized normal and an index of refraction.\n\n• Computes the illumination of surfaces refracted by the current surface.\n\n• Queries the renderer for a named property.\n\n• Returns the background color for rays that exit the scene.\n\n• Evaluates a scattering event through the domain of a geometric object.\n\n• Sets the current light\n\n• Stores sample data in a channel, referenced by a point.\n\n• Imports a variable sent by a surface shader in an illuminance loop.\n\n• Returns the computed BRDFs for the different lighting models used in VEX shading.\n\n• Stores exported data for a light.\n\n• Use a different bsdf for direct or indirect lighting.\n\n• Sends a ray from P along the normalized vector D.\n\n• Returns a Lambertian translucence BSDF.\n\n• Computes the position and normal at given (u, v) coordinates.\n\n• Writes color information to a pixel in the output image\n\n## Strings\n\n• Returns the full path of a file.\n\n• Converts an unicode codepoint to a UTF8 string.\n\n• Concatenate all the strings specified to form a single string.\n\n• Decodes a variable name that was previously encoded.\n\n• Decodes a geometry attribute name that was previously encoded.\n\n• Decodes a node parameter name that was previously encoded.\n\n• Encodes any string into a valid variable name.\n\n• Encodes any string into a valid geometry attribute name.\n\n• Encodes any string into a valid node parameter name.\n\n• Indicates the string ends with the specified string.\n\n• Finds an item in an array or string.\n\n• Returns 1 if all the characters in the string are alphabetic\n\n• Returns 1 if all the characters in the string are numeric\n\n• Converts an integer to a string.\n\n• Concatenate all the strings of an array inserting a common spacer.\n\n• Strips leading whitespace from a string.\n\n• This function returns 1 if the subject matches the pattern specified, or 0 if the subject doesn’t match.\n\n• Returns the integer value of the last sequence of digits of a string\n\n• Converts an UTF8 string into a codepoint.\n\n• Converts an English noun to its plural.\n\n• Matches a regular expression in a string\n\n• Finds all instances of the given regular expression in the string\n\n• Returns 1 if the entire input string matches the expression\n\n• Replaces instances of regex_find with regex_replace\n\n• Splits the given string based on regex match.\n\n• Computes the relative path for two full paths.\n\n• Returns the relative path to a file.\n\n• Strips trailing whitespace from a string.\n\n• Splits a string into tokens.\n\n• Splits a file path into the directory and name parts.\n\n• Formats a string like printf but returns the result as a string instead of printing it.\n\n• Returns 1 if the string starts with the specified string.\n\n• Strips leading and trailing whitespace from a string.\n\n• Returns the length of the string.\n\n• Returns a string that is the titlecase version of the input string.\n\n• Converts all characters in string to lower case\n\n• Converts all characters in string to upper case\n\n## Subdivision Surfaces\n\n• Evaluates a point attribute at the subdivision limit surface using Open Subdiv.\n\n• Evaluates a vertex attribute at the subdivision limit surface using Open Subdiv.\n\n• Outputs the Houdini face and UV coordinates corresponding to the given coordinates on an OSD patch.\n\n• Outputs the OSD patch and UV coordinates corresponding to the given coordinates on a Houdini polygon face.\n\n• Returns a list of patch IDs for the patches in a subdivision hull.\n\n## Tetrahedrons\n\n• Returns primitive number of an adjacent tetrahedron.\n\n• Returns vertex indices of each face of a tetrahedron.\n\n## Texturing\n\n• Looks up a (filtered) color from a texture file.\n\n• The depthmap functions work on an image which was rendered as a z-depth image from mantra.\n\n• Returns the color of the environment texture.\n\n• Perform UDIM or UVTILE texture filename expansion.\n\n• Test string for UDIM or UVTILE patterns.\n\n• Evaluates an ocean spectrum and samples the result at a given time and location.\n\n• Computes a filtered sample from a ptex texture map. Use texture instead.\n\n• Looks up an unfiltered color from a texture file.\n\n• The shadowmap function will treat the shadow map as if the image were rendered from a light source.\n\n• Imports attributes from texture files.\n\n• Similar to sprintf, but does expansion of UDIM or UVTILE texture filename expansion.\n\n• Computes a filtered sample of the texture map specified.\n\n## Transforms and Space\n\n• Computes the rotation matrix or quaternion which rotates the vector a onto the vector b.\n\n• Transforms a position from normal device coordinates to the coordinates in the appropriate space.\n\n• Gets the transform of a packed primitive.\n\n• Returns a transform from one space to another.\n\n• Creates an instance transform matrix.\n\n• Computes a rotation matrix or angles to orient the z-axis along the vector (to-from) under the transformation.\n\n• Builds a 3×3 or 4×4 transform matrix.\n\n• Returns the camera space z-depth of the NDC z-depth value.\n\n• Transforms a normal vector.\n\n• Create an orthographic projection matrix.\n\n• Transforms a normal vector from Object to World space.\n\n• Transforms a position value from Object to World space.\n\n• Transforms a direction vector from Object to World space.\n\n• Transforms a packed primitive.\n\n• Create a perspective projection matrix.\n\n• Computes the polar decomposition of a matrix.\n\n• Applies a pre rotation to the given matrix.\n\n• Prescales the given matrix in three directions simultaneously (X, Y, Z - given by the components of the scale_vector).\n\n• Pretranslates a matrix by a vector.\n\n• Transforms a vector from one space to another.\n\n• Applies a rotation to the given matrix.\n\n• Rotates a vector by a rotation that would bring the x-axis to a given direction.\n\n• Scales the given matrix in three directions simultaneously (X, Y, Z - given by the components of the scale_vector).\n\n• Sets the transform of a packed primitive.\n\n• Returns the closest equivalent Euler rotations to a reference rotation.\n\n• Applies an inverse kinematics algorithm to a skeleton.\n\n• Applies a curve inverse kinematics algorithm to a skeleton.\n\n• Applies a full-body inverse kinematics algorithm to a skeleton.\n\n• Applies an inverse kinematics algorithm to a skeleton.\n\n• Transforms a position into normal device coordinates.\n\n• Translates a matrix by a vector.\n\n• Transforms a normal vector from Texture to World space.\n\n• Transforms a position value from Texture to World space.\n\n• Transforms a direction vector from Texture to World space.\n\n• Transforms a directional vector.\n\n• Transforms a normal vector from World to Object space.\n\n• Transforms a position value from World to Object space.\n\n• Transforms a direction vector from World to Object space.\n\n• Transforms a normal vector from World to Texture space.\n\n• Transforms a position value from World to Texture space.\n\n• Transforms a direction vector from World to Texture space.\n\n## usd\n\n• Creates an attribute of a given type on a primitive.\n\n• Excludes an object from the collection\n\n• Includes an object in the collection\n\n• Appends an inversed transform operation to the primitive’s transform order\n\n• Applies a quaternion orientation to the primitive\n\n• Creates a primitive of a given type.\n\n• Creates a primvar of a given type on a primitive.\n\n• Adds a target to the primitive’s relationship\n\n• Applies a rotation to the primitive\n\n• Applies a scale to the primitive\n\n• Appends a transform operation to the primitive’s transform order\n\n• Applies a transformation to the primitive\n\n• Applies a translation to the primitive\n\n• Reads the value of an attribute from the USD primitive.\n\n• Reads the value of an element from an array attribute.\n\n• Returns the length of the array attribute.\n\n• Returns the names of the attributes available on the primitive.\n\n• Returns the tuple size of the attribute.\n\n• Returns the time codes at which the attribute values are authored.\n\n• Returns the name of the attribute type.\n\n• Blocks the attribute.\n\n• Blocks the primvar.\n\n• Blocks the primvar.\n\n• Blocks the primitive’s relationship\n\n• Returns the material path bound to a given primitive.\n\n• Clears the value of the metadata.\n\n• Clears the primitive’s transform order\n\n• Obtains the list of all objects that belong to the collection\n\n• Checks if an object path belongs to the collection\n\n• Obtains the object paths that are in the collection’s exclude list\n\n• Obtains the collection’s expansion rule\n\n• Obtains the object paths that are in the collection’s include list\n\n• Returns the primitive’s draw mode.\n\n• Retrurns primitive’s transform operation full name for given the transform operation suffix\n\n• Reads the value of an flattened primvar from the USD primitive.\n\n• Reads an element value of a flattened array primvar.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the primitive.\n\n• Returns the center of the bounding box for the primitive.\n\n• Returns the maximum of the bounding box for the primitive.\n\n• Returns the minimum of the bounding box for the primitive.\n\n• Returns the size of the bounding box for the primitive.\n\n• Obtains the primitive’s bounds\n\n• Obtains the primitive’s bounds\n\n• Checks if the primitive adheres to the given API.\n\n• Checks if the primitive adheres to the given API.\n\n• Checks if the primitive is active.\n\n• Checks if the attribute is an array.\n\n• Checks if the given metadata is an array.\n\n• Checks if the primvar is an array.\n\n• Checks if the primitive has an attribute by the given name.\n\n• Checks if the collection exists.\n\n• Checks if the path is a valid collection path.\n\n• Checks if the primvar is indexed.\n\n• Checks if the primitive is an instance.\n\n• Checks if the primitive is of a given kind.\n\n• Checks if the primitive has metadata by the given name.\n\n• Checks if the path refers to a valid primitive.\n\n• Checks if the primitive has a primvar of the given name.\n\n• Checks if the primitive has a relationship by the given name.\n\n• Checks if the stage is valid.\n\n• Checks if the primitive transform is reset\n\n• Checks if the primitive is of a given type.\n\n• Checks if the primitive is visible.\n\n• Returns the primitive’s kind.\n\n• Obtains the primitive’s local transform\n\n• Constructs an attribute path from a primitive path and an attribute name.\n\n• Constructs a collection path from a primitive path and a collection name.\n\n• Constructs an property path from a primitive path and an property name.\n\n• Constructs an relationship path from a primitive path and a relationship name.\n\n• Returns the length of the array metadata.\n\n• Returns the names of the metadata available on the object.\n\n• Returns the name of the primitive.\n\n• Returns the path of the primitive’s parent.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the given instance inside point instancer.\n\n• Returns the center of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the maximum position of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the minimum position of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the size of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Obtains the transform for the given point instance\n\n• Reads the value of a primvar from the USD primitive.\n\n• Returns the namespaced attribute name for the given primvar.\n\n• Reads the value of an element from the array primvar.\n\n• Returns the element size of the primvar.\n\n• Returns the index array of an indexed primvar.\n\n• Returns the element size of the primvar.\n\n• Returns the length of the array primvar.\n\n• Returns the names of the primvars available on the primitive.\n\n• Returns the tuple size of the primvar.\n\n• Returns the time codes at which the primvar values are authored.\n\n• Returns the name of the primvar type.\n\n• Returns the primitive’s purpose.\n\n• Obtains the relationship forwarded targets.\n\n• Returns the names of the relationships available on the primitive.\n\n• Obtains the relationship targets.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Remove a target from the primitive’s relationship\n\n• Sets the primitive active state.\n\n• Sets the value of an attribute.\n\n• Sets the value of an element in an array attribute.\n\n• Sets the excludes list on the collection\n\n• Sets the expansion rule on the collection\n\n• Sets the includes list on the collection\n\n• Sets the primitive’s draw mode.\n\n• Sets the primitive’s kind.\n\n• Sets the value of an metadata.\n\n• Sets the value of an element in an array metadata.\n\n• Sets the value of a primvar.\n\n• Sets the value of an element in an array primvar.\n\n• Sets the element size of a primvar.\n\n• Sets the indices for the given primvar.\n\n• Sets the interpolation of a primvar.\n\n• Sets the primitive’s purpose.\n\n• Sets the targets in the primitive’s relationship\n\n• Sets the primitive’s transform order\n\n• Sets/clears the primitive’s transform reset flag\n\n• Sets the selected variant in the given variant set.\n\n• Sets the primitive visibility.\n\n• Constructs a full name of a transform operation\n\n• Obtains the primitive’s transform order\n\n• Extracts the transform operation suffix from the full name\n\n• Infers the transform operation type from the full name\n\n• Returns the name of the primitive’s type.\n\n• Constructs a unique full name of a transform operation\n\n• Returns the variants belonging to the given variant set on a primitive.\n\n• Returns the currently selected variant in a given variant set.\n\n• Returns the variant sets available on a primitive.\n\n• Obtains the primitive’s world transform\n\n## Utility\n\n• Returns 1 if the VEX assertions are enabled (see HOUDINI_VEX_ASSERT) or 0 if assertions are disabled. Used the implement the assert macro.\n\n• An efficient way of extracting the components of a vector or matrix into float variables.\n\n• Reports a custom runtime VEX error.\n\n• Extracts a single component of a vector type, matrix type, or array.\n\n• Parameters in VEX can be overridden by geometry attributes (if the attributes exist on the surface being rendered).\n\n• Check whether a VEX variable is varying or uniform.\n\n• Ends a long operation.\n\n• Start a long operation.\n\n• Reversibly packs an integer into a finite, non-denormal float.\n\n• Prints a message only once, even in a loop.\n\n• Prints values to the console which started the VEX program.\n\n• Returns one of two parameters based on a conditional.\n\n• Creates a new value based on its arguments, such as creating a vector from its components.\n\n• Sets a single component of a vector or matrix type, or an item in an array.\n\n• Yields processing for a certain number of milliseconds.\n\n• Rearranges the components of a vector.\n\n• Reverses the packing of pack_inttosafefloat to get back the original integer.\n\n• Reports a custom runtime VEX warning.\n\n## volume\n\n• Returns the volume of the microvoxel containing a variable such as P.\n\n• Calculates the volume primitive’s gradient.\n\n• Gets the value of a specific voxel.\n\n• Gets the active setting of a specific voxel.\n\n• Gets the index of the bottom left of a volume primitive.\n\n• Converts a volume voxel index into a position.\n\n• Gets the vector value of a specific voxel.\n\n• Converts a position into a volume voxel index.\n\n• Gets the resolution of a volume primitive.\n\n• Samples the volume primitive’s value.\n\n• Samples the volume primitive’s vector value.\n\n• Computes the approximate diameter of a voxel." ]

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[ "# Schrodinger's Lamps!\n\nProbability Level 5\n\nLet $k$ and $n$ be positive integers with $k=n+10$. Let $2n$ lamps labelled $1,2,3...,2n$ be given, each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on).\n\nLet $N$ be the number of such sequences consisting of $k$ steps and resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off.\n\nLet $M$ be number of such sequences consisting of $k$ steps, resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off, but where none of the lamps $n+1$ through $2n$ is ever switched on.\n\nFind $\\frac{N}{M}.$\n\n×" ]

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[ "Cody\n\n# Problem 369. Basic electricity in a dry situation\n\nSolution 3060169\n\nSubmitted on 4 Oct 2020 by Matteo Lo Preti\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nN = 10^10; V = 150; assert(volts(N)>V/pi)\n\n2   Pass\nN = 10^11; V = 700; assert(volts(N)<V*pi)\n\n3   Pass\nN = 10^12; V = 10000; assert(volts(N)>V/sqrt(pi))\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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[ "You are on page 1of 13\n\n# EEPB 353 POWER SYSTEM I SEMESTER 1 2011/2012 COMPUTE ASSIGNMENT POWER FLOW\n\n## Group members: SHO ZEE LIANG ALDO LEE\n\nEE084586 EE083783\n\n## Name of Lecturer: Mr. John Steven A/L Navamany\n\nIntroduction\nIn this power system assignment, we are required to analysis the network given by using the Matlab Simulink and the M-files and we need to get the results of simulation and include inside our report. At first we need to convert the network into per-unit equivalent circuit as the data needed at the Matlab must be in per-unit values. We also need to perform all the calculation on the data given in Table 1. Follow by that, we need to use Matlab Simulink to generate the Y bus matrix of the power system and get the power flow solution using the Gauss-Seidel Method. We also need to indicate all the details of the voltage magnitude and angle on each bus, generation buses and load (P and Q details) on load buses and line flows on single-line diagram. Lastly, we need to discuss the impact and solution to solve the problem if there is a transmission line outage happened at the L2.\n\nIn power system, power-flow solution is an important tool which including the numerical analysis method. The numerical analysis method is Newton-Raphson method and Gauss-Seidel method. The power flow solution usually uses one-line diagram and per-unit system and focuses on various forms of AC power which are reactive, real and apparent rather than focuses on the voltage and current. Power flow solution is important in future for planning expansion of the power system. The power flow solution also contributes to the commercial power system as the hand-calculation for the power flow is too hard to apply on the commercial power system which is very large. The goal of the power flow solution is to obtain the voltage and phase angle for all the buses for specific conditions on the load and generator in the network. The power-flow solution can help us in the analysis of the power system in a normal-steady state operation.\n\n## Simulation Results Part (a)\n\nGiven SB= 100MVA, V1B=13.8kV V2B = (230kV/13.8kV) * 13.8kV = 230kV V3B = (15kV/230kV) * 230kV = 15kV V1pu = 13.8kV/ 13.8kV = 1.0 0 pu (slack bus) Notes: All the voltages for the bus 2 until bus 6 are assumed to be 1.0 0 pu. V7pu = 15kV /15kV = 1.0 0 pu (voltage-controlled bus) We can calculated the impendances by using the formula Znew=Zold*Snew / Sold : XG1 = (0.12)*(100/100) = j0.12 pu XG2 = (0.12)*(100/200) = j0.06 pu XT1 = (0.10)*(100/100) = j0.10 pu XT2 = (0.10)*(100/200) = j0.05 pu To calculate ZB, we can use ZB = (VB)2/SB Z3B = (15k)2/100M =2.25 Z2B = (230k)2/100M =529 (For all ZB line)\n\n## Zline = (Rline+ jXline)*Length/ZB Z23 = [(0.08+j0.05)*15]/529 = 0.002268+j0.01417 pu\n\nZ34 = [(0.08+j0.05)*30]/529 = 0.0045368+j0.02835 pu Z45 = [(0.08+j0.05)*40]/529 = 0.006049+j0.0378 pu Z46 = [(0.08+j0.05)*50]/529 = 0.0075614+j0.04726 pu Z56 = [(0.08+j0.05)*15]/529 = 0.002268+j0.01417 pu\n\nSL-pu = SL/SB S2 = [(50MW+j30MVAR)]/100MVA =0.5+j0.3 pu S3 = [(50MW+j30MVAR)]/100MVA =0.5+j0.3 pu S4 = [(50MW+j30MVAR)]/100MVA =0.5+j0.3 pu S5 = [(50MW+j30MVAR)]/100MVA =0.5+j0.3 pu S6 = [(50MW+j30MVAR)]/100MVA =0.5+j0.3 pu\n\n## Equivalent impedance diagram\n\nj 0.05 pu\n\nj 0.1 pu 1.4706 pu\n1.4706 pu\n\n1.4706 pu\n\nj 0.8823 pu\n\nj 0.8823 pu\n\n1.4706 pu\n\nj 0.8823 pu\n\nj 0.8823pu\n\n1.4706 pu\n\nj 0.8823 pu\n\nZL1 = ZL4 = 0.0023 + j 0.0142 pu Z L2 = 0.0045 + j 0.0284 pu ZL3 = 0.0060 + j 0.0378 pu ZL5 = 0.0076 + j 0.0473 pu\n\n-0.5 j0.3 pu -0.5 j0.3 pu -0.5 j0.3 pu -0.5 j0.3 pu -0.5 j0.3 pu S2sch = S3sch = S4sch = S5sch = S6sch = -0.5-j0.3 PU Y12 = 0 + j0.1000 pu = - Y21 Y67 = 0 + j0.2000 pu = - Y76 Y23 = -0.1100 + j0.6876 pu = - Y32 Y34 = -0.0550 + j0.3438 pu = - Y43 Y45 = -0.0413 + j0.2579 pu = - Y54 Y56 = -0.1100 + j0.6876 pu = - Y65 Y46= -0.0330 + j0.2063 pu = - Y64 Y11 = 0 - j0.1000 pu Y22 = 0.1100 - j0.7876 pu Y33 = 0.1650 - j1.0314 pu Y44 = 0.1293 - j0.8080 pu Y55 = 0.1512 - j0.9455 pu Y66 = 0.1430 - j1.0939 pu Y77 = 0 - j0.2000 pu\n\nPart (b)\na) Y Bus Matrix (MATLAB coding & result)\nR X 0 0.1 0.002268 0.01418 0.004537 0.02836 0.006049 0.03781 0.007561 0.04726 0.002268 0.01418 0 0.05 ];\n\n## % To obtain Y bus matrix % From To z = [ 1 2 2 3 3 4 4 5 4 6 5 6 6 7\n\n% Obtaining the Y bus matrix % Assign Y to function ybus1 Y = yBus1(z); display (Y); Y = 1.0e+002 * Columns 1 through 4 0 - 0.1000i 0 + 0.1000i 0 0.3438i 0 0.8080i 0 0.2579i 0 0.2063i 0 Columns 5 through 7 0 0 0 -0.0413 + 0.2579i 0.1512 - 0.9455i -0.1100 + 0.6876i 0 0 0 0 -0.0330 + 0.2063i -0.1100 + 0.6876i 0.1430 - 1.0939i 0 + 0.2000i 0 0 0 0 0 0 + 0.2000i 0 - 0.2000i 0 0 0 0 0 -0.0330 + 0 0 -0.0413 + 0 -0.0550 + 0.3438i 0.1293 0 + 0.1000i 0.1100 - 0.7876i -0.1100 + 0.6876i 0 -0.1100 + 0.6876i 0.1650 - 1.0314i 0 0 -0.0550 +\n\nPart (c)\nPower Flow Solution\n% Set the variable basemva to be 100 basemva = 100; % Preparing busdata input % Note that bus codes indicate types of bus % Bus Code 0 indicates load % Bus Code 1 indicates slack bus % Bus Code 2 indicates P-V bus %% Power Flow Solution % Defining parameters for Gauss-Seidel solving accuracy = 0.0001; accel = 1.6; maxiter = 200; % Bus Bus Voltage ----% No Code Mag. Mvar busdata = [ 1 1 1.00 2 0 1.00 3 0 1.00 4 0 1.00 5 0 1.00 6 0 1.00 7 2 1.00 0 ]; Angle Degree ------Load-----Generator----InjectedMW Mvar MW Mvar Qmin Qmax\n\n0 0 0 0 0 0 0\n\n0 50 50 50 50 50 0\n\n0 30 30 30 30 30 0\n\n0 0 0 0 0 0 180\n\n0 0 0 0 0 0 0\n\n0 0 0 0 0 0 -87\n\n0 0 0 0 0 0 87\n\n0 0 0 0 0 0\n\n% Preparing linedata input % All line settings are set to 1 because no transformer tapping occurs % Bus Bus R X 1/2B Line setting % nl nr pu pu pu linedata = [1 2 0 0.1 0 1 2 3 0.002268 0.01418 0 1 3 4 0.004537 0.02836 0 1 4 5 0.006049 0.03781 0 1 4 6 0.007561 0.04726 0 1 5 6 0.002268 0.01418 0 1 6 7 0 0.05 0 1]; % Carrying out operations which require to solve the power flow Lfybus; Lfgauss; Busout; Lineflow;\n\ni) Output results for voltage magnitude and angle, generation details and load details. %% Results Power Flow Solution by Gauss-Seidel Method Maximum Power Mismatch = 9.02454e-005 No. of Iterations = 114 Bus Voltage Angle ------Load--------Generation--Injected No. Mag. Degree MW Mvar MW Mvar Mvar 1 1.000 0.000 2 0.908 0.000 3 0.900 0.000 4 0.895 0.000 5 0.897 0.000 6 0.904 0.000 7 0.950 0.000 Total 0.000 0.000 -4.463 -4.588 -3.929 -3.104 -2.345 3.669 0.000 50.000 50.000 50.000 50.000 50.000 0.000 0.000 30.000 30.000 30.000 30.000 30.000 0.000 70.662 0.000 0.000 0.000 0.000 0.000 180.000 94.765 0.000 0.000 0.000 0.000 0.000 96.508\n\n250.000\n\n150.000\n\n250.662\n\n191.273\n\nii)\n\nOutput results for line flows and line losses Line Flow and Losses\n\n--Line-- Power at bus & line flow --Line loss-from to MW Mvar MVA MW Mvar 1 70.662 94.765 118.210 2 70.663 94.765 118.210 0.000 13.974 -50.000 -30.000 58.310 1 -70.663 -80.792 107.334 0.000 13.974 3 20.668 50.791 54.835 0.083 0.517 -50.000 -30.000 58.310 2 -20.585 -50.273 54.325 0.083 0.517 4 -29.415 20.274 35.725 0.072 0.447 -50.000 -30.000 58.310 3 29.487 -19.826 35.532 0.072 0.447 5 -30.664 -0.674 30.671 0.071 0.444 6 -48.821 -9.501 49.737 0.234 1.460 -50.000 -30.000 58.310 4 30.735 1.118 30.755 0.071 0.444 6 -80.731 -31.119 86.521 0.211 1.319 -50.000 -30.000 4 49.054 10.961 5 80.942 32.438 7 -179.999 -73.398 58.310 50.264 0.234 1.460 87.200 0.211 1.319 194.389 0.000 23.110\n\nTransformer tap\n\n180.000 96.508 204.240 6 179.999 96.508 204.239 0.000 23.110 0.670 41.272\n\nTotal loss\n\nPart (d)\nSingle Line Diagram\n\nDISCUSSION" ]

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[ "# Trigonometry Worksheets\n\nOn this page you will find: a complete list of all of our math worksheets relating to Trigonometry. Choose a specific addition topic below to view all of our worksheets in that content area. You will find addition lessons, worksheets, homework, and quizzes in each section.\n\n### What is Trigonometry?\n\nThe word trigonometry is derived from the Greek words trigonon, meaning triangle, and metron meaning measure. It is defined as the branch of mathematics that establishes the relationship between the angles and sides. Trigonometry is not only used for solving triangles, but many other straight-sided shapes are simplified into a collection of triangles. Moreover, strangles is also related to other branches of mathematics like infinite series, calculus, and complex numbers. Trigonometry helps us in finding the missing sides and angles by using the trigonometric ratios. These ratios are mainly measured in degrees and radians. The three known and commonly used trigonometric functions are sine cosine and tangent, which are abbreviated as sin, cos, and tan, respectively. Apart from these three functions, trigonometry also uses three other functions, namely cosec, sec, and cot. These three other functions are the inverse functions of sine, cosine, and tangent functions. The six trigonometric functions are defined below: Sine: It is defined as the ratio of the opposite side to the hypotenuse of the right-angled triangle. Cosine: it is defined as the ratio of the adjacent side to the hypotenuse of the right-angled triangle. Tangent: It is defined as the ratio of the opposite side to the adjacent of the right-angled triangle. Cosec: It is defined as the ratio of the hypotenuse to the opposite side of the right-angled triangle. Sec: It is defined as the ratio of the hypotenuse to the adjacent side of the right-angled triangle. Cot: It is defined as the ratio of the adjacent side to the opposite side of the right-angled triangle. What is the Importance of Trigonometry in Real Life? Trigonometry is defined as the study of triangles involving calculations with height, length, and angles. Typically, trigonometry is limited to calculating the missing measurements of a triangle in geometric problems. However, it can also be used in our daily life for calculating distances, heights, and lengths. Trigonometry is utilized in calculating the height of a mountain and building. The height of the building can be determined by using trigonometric functions. The elevation of the mountain can be estimated by using the elevation angle and the trigonometric functions. Trigonometry is also incorporated in aviation technology. We use the distance, direction, and speed of the wind, which play a vital role in determining when and in which direction the flight will travel. Also, the equation cab is calculated by using trigonometric functions. Nowadays, trigonometric functions are also used in the investigation of a crime. We use these functions to determine the projectile and the trajectory of the collision of a car. These functions are also helpful in determining the angle of the gunshot and knowing how an object falls. Trigonometry is used in estimating the right direction of the compass. These functions allow us to precisely determine the location and also figure out the distance of the horizon." ]

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[ "", null, "buildsol - Maple Help\n\nDEtools\n\n buildsol\n build a solution to an ODE using a reduction of order returned by dsolve and a solution to the reduced ODE", null, "Calling Sequence buildsol(ODESolStructure, sol_reduced_ODE)", null, "Parameters\n\n ODESolStructure - answer returned by dsolve expressed in terms of ODESolStruc sol_reduced_ODE - solution to the reduced ODE found inside ODESolStruc", null, "Description\n\n • For high order ODEs it may happen that dsolve succeeds in reducing the order of the ODE but not in solving the problem to the end. In those cases, the solution is expressed using ODESolStruc. You can obtain a solution for the reduced ODE by using the tools available in DEtools, as a series expansion, or by other means. If a solution to the reduced ODE is obtained, you can build a solution to the original problem using the buildsol command.\n • The buildsol command is a function of two arguments. The first argument is the structure returned by dsolve as the solution to an ODE (see ODESolStruc). The second argument is a solution found by the user to the reduced ODE present inside the first argument. That solution may be a particular one; buildsol will not check whether the given solution actually solves the reduced ODE.\n • This function is part of the DEtools package, and so it can be used in the form buildsol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[buildsol](..).", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{DEtools}\\right):$\n\nA third order nonlinear ODE\n\n > $\\mathrm{ODE}≔\\mathrm{diff}\\left(y\\left(x\\right),x,x,x\\right)=\\mathrm{diff}\\left(y\\left(x\\right),x,x\\right)\\left(-1+\\mathrm{diff}\\left(y\\left(x\\right),x\\right)\\right)\\mathrm{exp}\\left(y\\left(x\\right)-x\\right)$\n ${\\mathrm{ODE}}{≔}\\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{x}}^{{3}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right){=}\\left(\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)\\right){}\\left({-}{1}{+}\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)\\right){}{{ⅇ}}^{{y}{}\\left({x}\\right){-}{x}}$ (1)\n\nThis ODE is reducible:\n\n > $\\mathrm{odeadvisor}\\left(\\mathrm{ODE}\\right)$\n $\\left[\\left[{\\mathrm{_3rd_order}}{,}{\\mathrm{_with_linear_symmetries}}\\right]{,}\\left[{\\mathrm{_3rd_order}}{,}{\\mathrm{_reducible}}{,}{\\mathrm{_mu_y2}}\\right]{,}\\left[{\\mathrm{_3rd_order}}{,}{\\mathrm{_reducible}}{,}{\\mathrm{_mu_poly_yn}}\\right]\\right]$ (2)\n\nIt can be solved by dsolve directly by determining an appropriate integrating factor (see odeadvisor,reducible), but let's consider a possible answer for it as a reduction of order from 3 to 1:\n\n > $\\mathrm{sol}≔y\\left(x\\right)=\\mathrm{ODESolStruc}\\left(\\mathrm{_c}+\\mathrm{Int}\\left(\\mathrm{exp}\\left(\\mathrm{Int}\\left(\\mathrm{_f1}\\left(\\mathrm{_c}\\right),\\mathrm{_c}\\right)+\\mathrm{_C1}\\right),\\mathrm{_c}\\right)+\\mathrm{_C2},\\left[\\left\\{\\mathrm{diff}\\left(\\mathrm{_f1}\\left(\\mathrm{_c}\\right),\\mathrm{_c}\\right)=2{\\mathrm{_f1}\\left(\\mathrm{_c}\\right)}^{2}+\\mathrm{_f1}\\left(\\mathrm{_c}\\right)\\mathrm{exp}\\left(\\mathrm{_c}\\right)\\right\\},\\left\\{\\mathrm{_c}=y\\left(x\\right)-x,\\mathrm{_f1}\\left(\\mathrm{_c}\\right)=-\\frac{\\mathrm{diff}\\left(\\mathrm{diff}\\left(y\\left(x\\right),x\\right),x\\right)}{{\\left(-1+\\mathrm{diff}\\left(y\\left(x\\right),x\\right)\\right)}^{2}}\\right\\},\\left\\{x=\\mathrm{Int}\\left(\\mathrm{exp}\\left(\\mathrm{Int}\\left(\\mathrm{_f1}\\left(\\mathrm{_c}\\right),\\mathrm{_c}\\right)+\\mathrm{_C1}\\right),\\mathrm{_c}\\right)+\\mathrm{_C2},y\\left(x\\right)=\\mathrm{_c}+\\mathrm{Int}\\left(\\mathrm{exp}\\left(\\mathrm{Int}\\left(\\mathrm{_f1}\\left(\\mathrm{_c}\\right),\\mathrm{_c}\\right)+\\mathrm{_C1}\\right),\\mathrm{_c}\\right)+\\mathrm{_C2}\\right\\}\\right]\\right)$\n ${\\mathrm{sol}}{≔}{y}{}\\left({x}\\right){=}\\left({\\mathrm{_c}}{+}{\\int }{{ⅇ}}^{{\\int }{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\\mathrm{_c}}{+}{\\mathrm{_C1}}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\\mathrm{_c}}{+}{\\mathrm{_C2}}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\left[\\left\\{\\frac{{ⅆ}}{{ⅆ}{\\mathrm{_c}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right){=}{2}{}{{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right)}^{{2}}{+}{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right){}{{ⅇ}}^{{\\mathrm{_c}}}\\right\\}{,}\\left\\{{\\mathrm{_c}}{=}{y}{}\\left({x}\\right){-}{x}{,}{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right){=}{-}\\frac{\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)}{{\\left({-}{1}{+}\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)\\right)}^{{2}}}\\right\\}{,}\\left\\{{x}{=}{\\int }{{ⅇ}}^{{\\int }{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\\mathrm{_c}}{+}{\\mathrm{_C1}}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\\mathrm{_c}}{+}{\\mathrm{_C2}}{,}{y}{}\\left({x}\\right){=}{\\mathrm{_c}}{+}{\\int }{{ⅇ}}^{{\\int }{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\\mathrm{_c}}{+}{\\mathrm{_C1}}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\\mathrm{_c}}{+}{\\mathrm{_C2}}\\right\\}\\right]$ (3)\n\nTest that the above solves the ODE by using odetest:\n\n > $\\mathrm{odetest}\\left(\\mathrm{sol},\\mathrm{ODE}\\right)$\n ${0}$ (4)\n\nNow, the reduced ODE is of Bernoulli type, and can be selected using the mouse or through the following commands:\n\n > $\\mathrm{reduced_ODE}≔\\mathrm{op}\\left(\\left[2,2,1,1\\right],\\mathrm{sol}\\right)$\n ${\\mathrm{reduced_ODE}}{≔}\\frac{{ⅆ}}{{ⅆ}{\\mathrm{_c}}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right){=}{2}{}{{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right)}^{{2}}{+}{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right){}{{ⅇ}}^{{\\mathrm{_c}}}$ (5)\n > $\\mathrm{odeadvisor}\\left(\\mathrm{reduced_ODE}\\right)$\n $\\left[{\\mathrm{_Bernoulli}}\\right]$ (6)\n\nFrom the above, it is clear that a particular solution to the reduced_ODE is given by\n\n > $\\mathrm{sol_red_1}≔\\mathrm{_f1}\\left(\\mathrm{_c}\\right)=0$\n ${\\mathrm{sol_red_1}}{≔}{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right){=}{0}$ (7)\n\nfrom which a particular solution to the original ODE above can be built using\n\n > $\\mathrm{particular_sol}≔\\mathrm{buildsol}\\left(\\mathrm{sol},\\mathrm{sol_red_1}\\right)$\n ${\\mathrm{particular_sol}}{≔}{y}{}\\left({x}\\right){=}{-}\\frac{{-}{{ⅇ}}^{{\\mathrm{_C1}}}{}{x}{+}{\\mathrm{_C2}}{-}{x}}{{{ⅇ}}^{{\\mathrm{_C1}}}}$ (8)\n > $\\mathrm{odetest}\\left(\\mathrm{particular_sol},\\mathrm{ODE}\\right)$\n ${0}$ (9)\n\nIn this \"blackboard\" example, dsolve succeeds in solving the reduced_ODE too, as follows:\n\n > $\\mathrm{sol_red_2}≔\\mathrm{dsolve}\\left(\\mathrm{reduced_ODE}\\right)$\n ${\\mathrm{sol_red_2}}{≔}{\\mathrm{_f1}}{}\\left({\\mathrm{_c}}\\right){=}\\frac{{{ⅇ}}^{{{ⅇ}}^{{\\mathrm{_c}}}}}{{2}{}{{\\mathrm{Ei}}}_{{1}}{}\\left({-}{{ⅇ}}^{{\\mathrm{_c}}}\\right){+}{\\mathrm{_C1}}}$ (10)\n\nPassing this solution to buildsol, the general solution to ODE follows:\n\n > $\\mathrm{general_sol}≔\\mathrm{buildsol}\\left(\\mathrm{sol},\\mathrm{sol_red_2}\\right)$\n ${\\mathrm{general_sol}}{≔}{y}{}\\left({x}\\right){=}{x}{+}{\\mathrm{RootOf}}{}\\left({-}{x}{+}\\sqrt{{2}}{}\\left({{\\int }}_{{}}^{{\\mathrm{_Z}}}\\frac{{{ⅇ}}^{{\\mathrm{_C1}}}}{\\sqrt{{2}{}{{\\mathrm{Ei}}}_{{1}}{}\\left({-}{{ⅇ}}^{{\\mathrm{_b}}}\\right){+}{\\mathrm{_C3}}}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\\mathrm{_b}}\\right){+}{\\mathrm{_C2}}\\right)$ (11)\n > $\\mathrm{odetest}\\left(\\mathrm{general_sol},\\mathrm{ODE}\\right)$\n ${0}$ (12)" ]

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[ "# HackerRank 'Coffee Break Puzzle at Cisco: String Generation' Solution\n\nMartin Kysel · July 30, 2020\n\n##### Short Problem Definition:\n\nWatson and Sherlock are two leading researchers in Security and Cryptography at Cisco. They love to solve puzzles related to strings, number theory, encryptions and mapping in their free time. One fine day, during the coffee break, Watson comes up with a puzzle for Sherlock.\n\nYou know my powers, my dear Watson, and yet at the end of three months I was forced to confess that I had at last met an antagonist who was my intellectual equal.\n\nCoffee Break Puzzle at Cisco: String Generation\n\nAlternative name: Sherlock and String Generation\n\n##### Complexity:\n\ntime complexity is `O(N)`\n\nspace complexity is `O(N)`\n\n##### Execution:\n\nThe rules are explained pretty clearly. There are three rules\n1) each element has to occur at least once\n2) odd elements occur odd number of times, even elements occur even times\n3) the string has to be of length N\n\nLet us compact those rules into examples:\n1+2) mean that each element has to occur at least 1 or 2 times. The minimal string (a suffix really) is “abbcddeff” etc.\n2+3) because of the length limitation, the string can only be the length of the minimal viable suffix (explained above) + a multiply of 2. You could add 2xA, 2xB etc making it “abbcddeffaa” for example. But not “abbcddeffa” because that would violate rule 2).\nextra) the string has to be the minimal lexical variant of all the above rules. That means that you want to be pre-pending A’s to your string. Aka the previous one becomes “aaabbcddeff”.\n\n##### Solution:\n``````def transform(x):\nreturn chr(x+96)\n\ndef solve(N,K):\nword = []\nfor i in xrange(1,K+1):\nif i%2 == 0:\noccurence = 2\nelse:\noccurence = 1\nword.extend([i]*occurence)\n\nif (len(word) > N):\nreturn \"No such string.\"\n\nremaining = N - len(word)\nif remaining%2:\nreturn \"No such string.\"\n\nprefix = *remaining\nprefix.extend(word)\n\nreturn ''.join(map(transform, prefix))\n``````" ]

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[ "", null, "", null, "", null, "# Find the missing character in the number pattern given below?\n\n0 votes\n55 views", null, "posted Dec 13, 2018\nShare this puzzle\n\n## 1 Answer\n\n+1 vote\n\n324\n\nSquare root of the number in the middle = Sum of square roots of surrounding numbers:\nv1+v16+v25+v81=v361=19\nv4+v9+v16+v64=v289=17\nv1+v9+v25+v81=1+3+5+9=18=v 324", null, "answer Dec 14, 2018\n\nSimilar Puzzles\n0 votes\n\nIf the figures having some characters given below follow the same rule, then find the missing number from the options?", null, "+1 vote\n\nIn the tables below you will see that there are some numbers which are connected to each other.\n\nCan you find the pattern and hence find the missing number in the table?", null, "0 votes\n\nLook at the pattern in first two pentagon and find the missing number in the third one?", null, "0 votes" ]

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[ "## Helgason’s formula IINovember 7, 2009\n\nPosted by Akhil Mathew in differential geometry, MaBloWriMo.\nTags: , , ,", null, "$\\displaystyle \\boxed{ (d \\exp)_{tX}(Y) = \\left( \\frac{ 1 - e^{\\theta( - tX^* )}}{\\theta(tX^*)} (Y^*) \\right)_{\\exp(tX)}.}$", null, "$\\displaystyle {(d \\exp)_{tX}(Y) f = \\sum_{n=0}^{\\infty} \\frac{t^n}{(n+1)!} ( X^{*n} Y^* + X^{*(n-1)} Y^* X^* + \\dots + Y^* X^{*n})f(p).}$  (more…)" ]

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[ "On which of the following does the value of the voltage or current generated by an AC generator NOT depend?\n\n1. Speed at which the coil or magnetic field rotates\n2. The poles of the permanent magnet\n3. Strength of the field\n4. Number of turns in the coil.\n\nOption 2 : The poles of the permanent magnet\n\nDetailed Solution\n\nExplanation:\n\nAC Generator:", null, "• An AC generator is a device that converts mechanical energy into electrical energy and generates alternating current.\n• It works on the principle of electromagnetic induction i.e., when a coil is rotated in a uniform magnetic field, an alternating emf is induced in the coil.\n• The main components of the AC generator are:\n1. Armature\n2. Strong field magnet\n3. Slip rings\n4. Brushes\n• When the armature coil ABCD rotates in the magnetic field provided by the strong field magnet, it cuts the magnetic lines of force.\n• Thus the magnetic flux linked with the coil changes and hence induced emf is set up in the coil.\n• The direction of the induced emf or the current in the coil is determined by Fleming’s right-hand rule.\n• The current flows out through the brush B1 in the first half of the revolution and through the brush B2 in the next half revolution of the coil.\n• This process is repeated. Therefore, emf produced is of alternating nature.\n• The coil of the AC generator is rotated with a constant angular speed ω, the angle between the magnetic field vector B and the area vector A of the coil at any instant t is θ. Then,\n\n⇒ θ = ωt\n\nThe flux linked with the coil at any time t is,\n\n⇒ ϕ = BA.cos θ\n\nFrom Faraday’s law, the induced emf at any time t for the rotating coil of N turns is given as,\n\n⇒ ϵ = NBAω. sin(ωt)\n\nIf the resistance of the coil is R, then the induced current at any time t for the rotating coil of N turns is given as,\n\n$$\\Rightarrow I=\\frac{\\epsilon}{R}=\\frac{NBAω \\sin(ω t)}{R}$$\n\nWhere, N = No. of turns, B = Magnetic field, A = Area, ω = Angular velocity, R = Resistance of the coil\n\nHence, current generated depends upon Number of turns in the coil, Strength of the field, Speed at which the coil or magnetic field rotates but not on The poles of the permanent magnet" ]

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[ "# Graph Structure in Polynomial Systems: Chordal Networks [Math Dept.]", null, "Topic:\nGraph Structure in Polynomial Systems: Chordal Networks\nTuesday, January 31, 2017 - 3:00pm\nVenue:\nMath 380-C\nSpeaker:\nPablo Parrilo (MIT)\nAbstract / Description:\n\nThe sparsity structure of a system of polynomial equations or an optimization problem can be naturally described by a graph summarizing the interactions among the decision variables. It is natural to wonder whether the structure of this graph might help in computational algebraic geometry tasks (e.g., in solving the system). In this talk we will provide an introduction to this area, focused on the key notions of chordality and treewidth, which are of great importance in related areas such as numerical linear algebra, database theory, constraint satisfaction, and graphical models. In particular, we will discuss \"chordal networks\", a novel representation of structured polynomial systems that provides a computationally convenient decomposition of a polynomial ideal into simpler (triangular) polynomial sets, while maintaining its underlying graphical structure. As we will illustrate through examples from different application domains, algorithms based on chordal networks can significantly outperform existing techniques. Based on joint work with Diego Cifuentes (MIT)." ]

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[ "Go to content", null, "A prime number is a number with exactly two factors, itself and 1. Almost all prime numbers are also odd. The exception is 2 which is even but also a prime number.\n\nThe table to the left highlights all the prime numbers between 1 and 100.\n\nAny number can broken down into its prime factors. A good way to do this is to make a factor tree, then the number can be written as a prime factorisation, for example:", null, "Click for slideshow\n\n## Summary/Background", null, "A factor is a whole number that divides exactly into another number. A prime number is a number with only two factors: itself and 1. The first prime numbers are 2, 3, 5, 7, 11, 13...\nThe diagram on the right is a factor tree, showing the prime factors of 12 to be 2 and 3. Notice that 12 = 2 x 2 x 3. In this way 12 has been expressed as a product of its prime factors.", null, "You can get a better display of the maths by downloading special TeX fonts from jsMath. In the meantime, we will do the best we can with the fonts you have, but it may not be pretty and some equations may not be rendered correctly.\n\n## Glossary\n\n### factorisation\n\nThe process of expressing a term as a product of factors.\n\n### function\n\nA rule that connects one value in one set with one and only one value in another set.\n\n### prime\n\na positive integer which is divisible only by itself and 1\n\n### speed\n\nvelocity without direction; a scalar quantity\n\n### tree\n\nA connected graph with no cycles.\n\n### union\n\nThe union of two sets A and B is the set containing all the elements of A and B.\n\nFull Glossary List\n\n## This question appears in the following syllabi:\n\nSyllabusModuleSectionTopicExam Year\nAQA GCSE (9-1) Foundation (UK)N: Structure and CalculationN4: Factors and MultiplesPrime Numbers-\nCBSE X (India)Number SystemsReal NumbersFundamental theorem of arithmetic-\nCIE IGCSE (9-1) Maths (0626 UK)1 NumberB1.1 Understanding NumbersPrime Numbers-\nEdexcel GCSE (9-1) Foundation (UK)N: Structure and CalculationN4: Factors and MultiplesPrime Numbers-\nGCSE Foundation (UK)NumberArithmeticPrime numbers-\nOCR GCSE (9-1) Foundation (UK)1: Number Operations and Integers1.02b: Prime NumbersPrime Numbers-\nUniversal (all site questions)AArithmeticPrime numbers-" ]

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[ "# Find the equation of the curve? A curve passes through the point (0,2) and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of...\n\nFind the equation of the curve?\n\nA curve passes through the point (0,2) and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve? (Use x as the independent variable.)\n\ny(x)=______________?\n\nRico Grant", null, "| Certified Educator\n\ncalendarEducator since 2011\n\nstarTop subjects are Math, Science, and Business\n\nThe slope of the tangent line is given by the first derivative. Since the slope is equal to twice the y value we have `(dy)/(dx)=2y` Then:\n\n`(dy)/y=2dx`  Integrating we get:\n\n`lny=2x+C_1`  exponentiating both sides we get:\n\n`e^(lny)=e^(2x+C_1)=e^(2x)e^(C_1)` Let `C=e^(C_1)`\n\n`y=Ce^(2x)`\n\nSince the curve passes through (0,2) we have:\n\n`2=Ce^(2*0)==>C=2`\n\nSo the function is `y=2e^(2x)`\n\n-------------------------------------\n\nChecking we see that `(dy)/(dx)=4e^(2x)=2y` as required.\n\ncheck Approved by eNotes Editorial" ]

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[ "Categories", null, "Here’s a little basic physics lesson that should prove extremely useful when working on your amps. When you add resistors and capacitors in series or parallel, the total values you end up with can be worked out with simple formulae.\n\nWhen you add capacitors in parallel (i.e. connected together twice, at both ends), the new value you end up with is simply the value of the first capacitor plus the value of the second capacitor. For example, a pair of 250pf capacitors in parallel would give a value of 500pf. A 500pf and a 250pf capacitor in parallel would give a working value of 750pf.\n\nTherefore it’s fair to say that C1 + C2 = Ct, where C1 is the first capacitor’s value, C2 the second’s, and Ct is the resulting capacitance value.\n\nHere’s where it gets a little harder to work out. In series (i.e. connected one after another), the resulting capacitance is found by this formula:\n\n1 + 1 1 +   ….  = 1\nC1     C2      C3                 Ct\n\nso you work out the total capacitance by using fractions. Let’s try this with the examples above (250pf and 250pf, 500pf and 750pf):\n\n1 + 1 = 1\n250    250     125\n\n1 + 1 = 1\n500    750       300\n\nTherefore a pair of 250pf caps in series gives you 125pf capacitance, and a 500pf then 750pf capacitor in series would give you 300pf in capacitance.\n\nNow that you’ve learnt how to add capacitor values, you can easily do it now with resistors – it’s just the same but backwards. You use the same formula for adding resistors in parallel as you would for adding capacitors in series, and you also use the same formula for adding resistors in series as you would for adding capacitors in parallel.\n\nTherefore, the total resistance when adding resistors in parallel is simply:\n\n1 + 1 1 +   ….  = 1\nR1      R2      R3                Rt\n\nwhere R1, R2, R3 etc is the first resistor, second resistor, third resistor etc, and Rt is the total resistance gained.", null, "", null, "" ]

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[ "\n\n# axiomatization\n\na) The act of making axiomatic.\nb) The act or process of establishing a concept within a system of axioms.\n\nWikipedia foundation.\n\n### Look at other dictionaries:\n\n• axiomatization — (Amer.) n. act of expressing a theory as a set of axioms; process of reducing to a system of axioms (also axiomatisation) …   English contemporary dictionary\n\n• axiomatization — noun Date: 1931 the act or process of reducing to a system of axioms • axiomatize transitive verb …   New Collegiate Dictionary\n\n• axiomatization — ax·i·om·a·ti·za·tion …   English syllables\n\n• axiomatization — ˌaksēˌäməd.ə̇ˈzāshən, məˌtīˈz noun ( s) : the act or process of axiomatizing compare formalization …   Useful english dictionary\n\n• Tarski's axiomatization of the reals — In 1936, Alfred Tarski set out an axiomatization of the real numbers and their arithmetic, consisting of only the 8 axioms shown below and a mere four primitive notions: the set of reals denoted R, a binary total order over R, denoted by infix …   Wikipedia\n\n• Boolean algebra — This article discusses the subject referred to as Boolean algebra. For the mathematical objects, see Boolean algebra (structure). Boolean algebra, as developed in 1854 by George Boole in his book An Investigation of the Laws of Thought, is a… …   Wikipedia\n\n• formal logic — the branch of logic concerned exclusively with the principles of deductive reasoning and with the form rather than the content of propositions. [1855 60] * * * Introduction       the abstract study of propositions, statements, or assertively used …   Universalium\n\n• Function (mathematics) — f(x) redirects here. For the band, see f(x) (band). Graph of example function, In mathematics, a function associates one quantity, the a …   Wikipedia\n\n• Boolean algebra (introduction) — Boolean algebra, developed in 1854 by George Boole in his book An Investigation of the Laws of Thought , is a variant of ordinary algebra as taught in high school. Boolean algebra differs from ordinary algebra in three ways: in the values that… …   Wikipedia\n\n• Ontology double articulation — The notion of Ontology Double Articulation refers to a methodological principle in ontology engineering. The idea is that an ontology should be built as separate domain axiomatizations and application axiomatization(s). In other words an… …   Wikipedia" ]

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[ "Courses\n\n# Test: Quantitative Aptitude- 4\n\n## 30 Questions MCQ Test GMAT Mock Test for Practice | Test: Quantitative Aptitude- 4\n\nDescription\nThis mock test of Test: Quantitative Aptitude- 4 for GMAT helps you for every GMAT entrance exam. This contains 30 Multiple Choice Questions for GMAT Test: Quantitative Aptitude- 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Quantitative Aptitude- 4 quiz give you a good mix of easy questions and tough questions. GMAT students definitely take this Test: Quantitative Aptitude- 4 exercise for a better result in the exam. You can find other Test: Quantitative Aptitude- 4 extra questions, long questions & short questions for GMAT on EduRev as well by searching above.\nQUESTION: 1\n\n### At the end of the year 2002, Monica and Chandler each purchased a certificate of deposit that paid the same rate of interest, and each held the certificate of deposit through the end of 2002. If Chandler invested X dollars and Monica invested \\$130,000, and if Chandler earned interest in 2002 totaling \\$45,000, what was the amount of interest that Monica earned on her \\$130,000 investment? (1) The rate of interest on the certificate of deposit that Chandler and Monica each purchased was 8.5% annually. (2) In 2002, Chandler invested \\$529,412 in the certificate of deposit.\n\nSolution:\n\nFrom statement (1) we know the rate of interest, so we can easily calculate how much Monica earned with her \\$130,000 deposit.\n\nFrom statement (2) we know how much Chandler invested and we already know from the question how much he earned, we can calculate the interest and multiply it by the deposit that Monica made.\nTherefore, both statements, by themselves, are sufficient to answer the question.\n\nQUESTION: 2\n\n### Mickey made an X dollars loan at the beginning of 1996. Travis, who is Mickey’s little brother also made a loan, only twice as large as Mickey’s but with the same interest. If Travis pays \\$10,000 interest on his loan each year, how big is Mickey’s loan? (1) The rate of interest on the loan that Travis took is 6% annually. (2) The loan that Travis made was \\$166,667.\n\nSolution:\n\nFrom statement (1) we know the rate of interest so we can find how much money Travis loaned and multiply it by 2 to get Mickey’s loan.\nFrom statement (2) we know the amount Travis loaned, which is doubled than that of Mickey.\nTherefore, both statements, by themselves, are sufficient to answer the question.\n\nQUESTION: 3\n\n### Concentrated orange juice comes inside a cylinder tube with a radius of 2.5 inches and a height of 15 inches. The tubes are packed into wooden boxes, each with dimensions of 11 inches by 10 inches by 31 inches. How many tubes of concentrated orange juice, at the most, can fit into 3 wooden boxes?\n\nSolution:\n\nYou want to waste as little amount of space as possible, therefore make the height of the box 11 and fit 4 boxes at the bottom so you lose only 1 inch of margin at the top and on one of the sides.\n\nYou can see that 8 tubes can fit into one box thus 24 tubes fit into 3 boxes.\n\nQUESTION: 4\n\nA certain car’s price decreased by 2.5% (from the original price) each year from 1996 to 2002, during that time the owner of the car invested in a new carburetor and a new audio system for the car, which increased her price by \\$1,500. If the price of the car in 1996 was \\$22,000, what is the car’s price in 2002?\n\nSolution:\n\nThe price of the car decreased by 2.5% every year on a course of 6 years. That means that the price of the car in 2002 is 15% lower than the original + \\$1500 of new investments.\nThe new price is (\\$22,000 x 0.85 = 18,700 + 1500 = \\$20,200).\n\nQUESTION: 5\n\nThe average price of an antique car increases over the years. If from 1990 to 1996, the price of the car increased by 13% and from 1996 to 2001 it increased by 20%, what is the price of the car in 2001 if the price in 1990 was \\$11,500?\n\nSolution:\n\nThe price in 1990 was 11,500. In 1996 the price is (11,500 x 1.13 = 12,995).\nThe price we are looking for, in 2002, is (12,995 x 1.2 = \\$15,594).\n\nQUESTION: 6\n\nThe apartment on King-Williams street is an asset that it’s value is tramping about.\nFrom the year 1973 to 1983 it’s value decreased by 16% and from 1983 to 1993 it’s value increased by 16%. What is the value of the asset in 1993 if in 1973 it was worth \\$40,000?\n\nSolution:\n\nBe careful, the value of the asset didn’t stay the same after the two changes in the value.\n\nIn the first 10 years, the value decreased by 16% (40,000 x 0.84 = 33,600).\nThen, in the next ten years the value increased by 16% (33,600 x 1.16 = 38,976).\n\nQUESTION: 7\n\nThe value of a “Tin-Rin” stock in the stock market decreased by 15% in the last two years.\nThe economic experts believe that the value of the stock will increase by 7% during the following year, which will make the value \\$440. What was the approximate price of the stock two years ago?\n\nSolution:\n\nStart from the top, after a 7% increase the price of the stock is \\$440.\n440 are 107% of the price this year à (440/107 x 100 = 411.215).\nTwo years ago the price was 15% higher, therefore (411.215 x 1.15) is approximately \\$473.\n\nQUESTION: 8\n\nWhich of the following expressions is equivalent to |X| <4 ?\n\nSolution:\n\nAn absolute value means that the sign of the variable is insignificant, therefore X can be between –4 and 4 and still he will fulfill the original equation.\n\nQUESTION: 9\n\nWhich of the following statements is equivalent to (8 + 2X < 18 – 6X < 23 + 2X)\n\nSolution:\n\nTake the expression and simplify it: Take (8 + 2x) from each side to get:  (0<10 – 8X<15).\nSubstitute 10, -10<-8X<5.\nDivide all by (-8), 5/4 > X > -5/8. Therefore the answer is C.\n\nQUESTION: 10\n\nAt the faculty of Aerospace Engineering, 312 students study Random-processing methods, 232 students study Scramjet rocket engines and 112 students study them both. If every student in the faculty has to study one of the two subjects, how many students are there in the faculty of Aerospace Engineering?\n\nSolution:\n\nUse the group formula.\nTotal = groupA + groupB – Both + Neither.\nTotal = 312 + 232 – 112 + 0 = 432 students\n\nQUESTION: 11\n\nIn the faculty of Reverse-Engineering, 226 second year students study numeric methods, 423 second year students study automatic control of airborne vehicles and 134 second year students study them both. How many students are there in the faculty if the second year students are approximately 80% of the total?\n\nSolution:\n\nUse the group formula.\nTotal = groupA + groupB – Both + Neither.\nTotal = 226 + 423 – 134 + 0 = 515 second year students.\nThe second year students are 80% of the total amount, therefore (515/80 x 100 = 643.75).\n\nQUESTION: 12\n\nIn the Biotechnology class of 2000, there were X graduates. 32 of the graduates found a job, 45 continued on to their second degree and 13 did both. If only 9 people didn’t do both, What is X equal to?\n\nSolution:\n\nUse the group formula.\nTotal = groupA + groupB – Both + Neither.\nTotal = 32 + 45 – 13 + 9 = 73 graduates.\n\nQUESTION: 13\n\nIf a, b, c, d and e are distinct integers, which one is the median?\n(1) a < b – c.\n(2) d > e.\n\nSolution:\n\nStatement (1) tells us nothing about e and d, you can eliminate answers (a) and (d).\nStatement (2) tells us nothing about a, b and c, you can eliminate answer (b) .\nTry to plug in some numbers, take: a=3, b=7, c=1, d=9 and e=8. The median in that case is 7.\nTry other numbers, a=8, b=15, c=6, d=10 and e=9. The median is 9.\nFirst the median was b, then the median was e. More sufficient data is required to answer the question.\n\nQUESTION: 14\n\na, b and c are three odd and different integers. Which one is the median?\n(1) a, b and c are consecutive numbers.\n(2) c > a  and  b < c.\n\nSolution:\n\nFrom statement (1) we can learn that if they are consecutive numbers, the median is B.\nFrom statement (2) we have a connection between c to a and b, but we don’t know if a or b is the smallest among the three, therefore this statement, by itself, is not sufficient.\n\nQUESTION: 15\n\nWhat is the ratio between W and Q?\n(1) Q + W = 23.\n(2) W is 25% of Q.\n\nSolution:\n\nWe are looking for Q/W. From statement (1) we know the sum of the two variables, which is not helpful in our case. From statement (2) we know that W = (0.25)Q, therefore we know the ratio between the two variables.\n\nQUESTION: 16\n\nWhat is the product of X and Y?\n(1) 2X + 2Y = 46.\n(2) (X + Y)2 = (X – Y)2 + 8.\n\nSolution:\n\nThe product of X and Y is XY.\nStatement (1) implies only about their sum.\nStatement (2) can be written as: X2 + 2XY +Y2 = X2 – 2XY + Y2 + 8 → 4XY = 8 → XY = 2.\nOnly statement (2) is sufficient\n\nQUESTION: 17\n\nKramer can pack X boxes of cigarettes per minute. If there are Y boxes of cigarettes in one case, How many cases can Kramer pack in 2 hours?\n\nSolution:\n\nY/X is the time it takes Kramer to fill a case with boxes (in minutes).\nIn two hours there are 120 minutes, so 120/(Y/X) is 120X/Y, and that is the number of cases that Kramer can fill in two hours.\n\nQUESTION: 18\n\nGeorge can fill Q cans of paint in 3 minutes. If there are R cans of paint in one gallon, how many gallons can George fill in 45 minutes?\n\nSolution:\n\nGeorge can fill Q/3 cans of paint in one minute à There are R cans in one gallon, so R/(Q/3) = 3R/Q\nIs the time it takes George to fill one gallon (in minutes).\nIn 45 minutes George can fill up 45/(3R/Q) = 15Q/R.\n\nQUESTION: 19\n\nThe junior soccer team is one of the best teams in the state of Alabama. The season is divided into two parts, each part is 4 months. In the first part of the season, the junior soccer team won half of their 32 games. How many games did the team win in the entire season?\n(1) In the second part of the season, the team lost 9 games, tied 6 games and won 18 games.\n(2) From the 32 games remaining the team won twice as much as she lost.\n\nSolution:\n\nFrom statement (1) we can complete the missing data, in the first part of the season the team won 16 games and on the second part of the season, the team won 18 games. This statement is sufficient enough to answer the question.\nStatement (2) is not sufficient by it self, it doesn’t mention how many games were tied, therefore only statement (1) is sufficient.\n\nQUESTION: 20\n\n“Queens” is a game of cards that distinguishes the cards into three groups: reds, blacks and jokers. Four packets of cards are shuffled and only 50 cards are drawn out. How many red cards are in the stack of the 50 cards?\n(1) The number of black cards is twice the number of red cards.\n(2) There is at least one joker in the stack of cards.\n\nSolution:\n\nStatement (1) implies that if we knew the number of jokers, the answer would be clear: take the cards that are not jokers and divide them by 3 to get the number of red cards.\n\nStatement (2) is not clear enough, the number of jokers is not distinct, therefore more data is needed and the two statements taken together are not sufficient.\n\nQUESTION: 21\n\nRon has three kinds of shirts in his closet, white shirts, black shirts and fancy shirts. What is the ratio between the shirts in the closet?\n(1) The total number of shirts is 100.\n(2) 30% of the shirts are black, which is twice as much as the fancy shirts.\n\nSolution:\n\nFrom statement (1) we know the total amount of shirts in the closet.\n\nStatement (2) gives us the ratio between the shirts.\n30% of the shirts are black (which is 30 shirts), this number is twice as much as the fancy shirts (15).\nThe remaining shirts must be white. We know the ratio; therefore both statements are required in order to answer the question correctly.\n\nQUESTION: 22\n\nThe roof of an apartment building is rectangular and its length is 4 times longer than its width. If the area of the roof is 784 feet squared, what is the difference between the length and the width of the roof?\n\nSolution:\n\nThe area of a rectangle is (length) x (width), let X be the width of the roof → 4X2 = 784 →\nX2 = 196 à X = 14.\nThe width of the roof is 14 and the length is 56. The difference is (56-14 = 42).\n\nQUESTION: 23\n\nThe length of a cube is three times its width and half of its height. If the volume of the\nCube is 13,122 Cm cubed. What is the height of the cube?\n\nSolution:\n\nNormalize each dimension to the width of the cube (W).\nThe length is 3 times the width, therefore its 3W, which is half of the height (6W).\nThe volume of the cube is 13,122 = 6W x 3W x W = 18W3 → W3 = 729 → W = 9.\nThe height of the cube is six times the width, therefore its 54 meters.\n\nQUESTION: 24\n\nThe width of a cuboid is half the length and one third of the height. If the length of the cuboid is 4 meters, what is the volume of three identical cuboids?\n\nSolution:\n\nNormalize all the dimensions to the width.\n\nLet the width be X.\nThe length is twice the width, thus 2X.\nThe height is 3 times the width, thus 3X.\nThe volume of the cuboid is X.2X..3X = 6X3.\nThe length is equal to 4 → 2X = 4 → X = 2 → Volume = 6 x 8 = 48.\n\nThe volume of two cubes will be 96.\n\nQUESTION: 25\n\nTwo brothers took the GMAT exam, the higher score is X and the lower one is Y. If the difference between the two scores is equal to their average, what is the value of Y/X ?\n\nSolution:\n\nIf the difference is equal to the average, then we could write the equation: X – Y = (X +Y)/2.\n→ X – 3Y = 0 →Y/X = 1/3\n\nQUESTION: 26\n\nTwo people measure each other’s height, the height of the taller person is H and the height of the other person is L. If the differences in their height is equal to their average height, what is the Value of H/L ?\n\nSolution:\n\nIf the difference is equal to the average, then we could write the equation: H – L = (H+L)/2.\n→ H – 3L = 0 → H/L = 3.\n\nQUESTION: 27\n\nIf building X is less than 40 store’s high, is building Y taller than X?\n(1) Building Y is at least three times as high as building X.\n(2) On the fortieth floor of the Y building there is a gift shop.\n\nSolution:\n\nStatement (1) tells us clearly that the Y building is taller than the X one.\n\nStatement (2) implies that there is a gift shop on the 40’Th floor; therefore there are at least 40 floors on the Y building, which make it taller than X.\nBoth statements, by themselves, are sufficient enough to answer the question.\n\nQUESTION: 28\n\nThere are two major statues in Tasmanian County; the first is no more than 45 meters high. How tall is the second statue?\n(1) The second statue is 10 meters higher than the first statue.\n(2) Both statues together are 80 meters high.\n\nSolution:\n\nThe information on the first statue in the question is confusing and irrelevant.\nStatement (1) tells us that: B = A + 10 (A is the first and B is the second statue).\nStatement (2) tells us that: A + B = 80, therefore we have two equations with two variables and so we can solve the problem.\nTherefore, both statements are required in order to answer the question.\n\nQUESTION: 29\n\nTower X is smaller than tower Z. Is tower Y bigger than tower X?\n(1) Tower Z higher than tower Y.\n(2) Tower Y is one of the tallest in the world.\n\nSolution:\n\nWe can write the data that is given to us: X < Z.\nFrom statement (1) we can learn that: Y < Z also, this is not enough.\nFrom statement (2) we know that Y is very tall, one of the highest in the world, but X can still be higher. Therefore, more sufficient data is required to answer the question.\n\nQUESTION: 30\n\nHow many steaks did the restaurant sell between 20:00 P.M and 21:00 P.M on Wednesday?\n(1) On Tuesday the restaurant sold 25 steaks between the hours of 20:00 P.M and 21:00 P.M.\n(2) The average amount of steaks that are sold on Wednesdays is 25 steaks per hour.\n\nSolution:" ]

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[ "# Prove that \\sqrt{ \\frac{2x^2 - 2x + 1}{2} } \\geq \\frac{1}{x + \\frac{1}{x}}for 0 < x < 1. ?\n\nMay 18, 2017\n\nSee explanation...\n\n#### Explanation:\n\nLet's take a look at the left hand expression to understand it's behaviour better:\n\nsqrt((2x^2-2x+1)/2) = sqrt((4x^2-4x+1+1)/4\n\ncolor(white)(sqrt((2x^2-2x+1)/2)) = sqrt(((2x-1)^2+1)/4\n\n$\\textcolor{w h i t e}{\\sqrt{\\frac{2 {x}^{2} - 2 x + 1}{2}}} = \\frac{1}{2} \\sqrt{{\\left(2 x - 1\\right)}^{2} + 1}$\n\nNote that since it is a square, we have ${\\left(2 x - 1\\right)}^{2} \\ge 0$ for all real values of $x$, attaining its minimum value when $x = \\frac{1}{2}$ and $\\left(2 x - 1\\right) = 0$.\n\nHence we find:\n\n$\\sqrt{\\frac{2 {x}^{2} - 2 x + 1}{2}} = \\frac{1}{2} \\sqrt{{\\left(2 x - 1\\right)}^{2} + 1} \\ge \\frac{1}{2} \\sqrt{1} = \\frac{1}{2}$\n\nNow let's look at the right hand expression:\n\n${\\left(x - 1\\right)}^{2} = {x}^{2} - 2 x + 1$\n\n$\\textcolor{w h i t e}{{\\left(x - 1\\right)}^{2}} = \\left({x}^{2} + 1\\right) - 2 x$\n\n$\\textcolor{w h i t e}{{\\left(x - 1\\right)}^{2}} = \\left({x}^{2} + 1\\right) \\left(1 - \\frac{2 x}{{x}^{2} + 1}\\right)$\n\n$\\textcolor{w h i t e}{{\\left(x - 1\\right)}^{2}} = 2 \\left({x}^{2} + 1\\right) \\left(\\frac{1}{2} - \\frac{x}{{x}^{2} + 1}\\right)$\n\n$\\textcolor{w h i t e}{{\\left(x - 1\\right)}^{2}} = 2 \\left({x}^{2} + 1\\right) \\left(\\frac{1}{2} - \\frac{1}{x + \\frac{1}{x}}\\right)$\n\nNow ${\\left(x - 1\\right)}^{2} \\ge 0$, attaining its minimum value $0$ when $x = 1$.\n\nAlso $\\left({x}^{2} + 1\\right) \\ge 1$\n\nHence we find:\n\n$\\frac{1}{2} - \\frac{1}{x + \\frac{1}{x}} \\ge 0$\n\nThat is $\\frac{1}{x + \\frac{1}{x}} \\le \\frac{1}{2}$, attaining its maximum value $\\frac{1}{2}$ when $x = 1$.\n\nSo:\n\n$\\sqrt{\\frac{2 {x}^{2} - 2 x + 1}{2}} \\ge \\frac{1}{2} \\ge \\frac{1}{x + \\frac{1}{x}}$\n\nfor all real values of $x$\n\nIn fact:\n\n$\\sqrt{\\frac{2 {x}^{2} - 2 x + 1}{2}} > \\frac{1}{x + \\frac{1}{x}}$\n\nsince there is no value of $x$ for which both sides are equal to $\\frac{1}{2}$.\n\nMay 18, 2017\n\nSee below.\n\n#### Explanation:\n\nWe know that $x + \\frac{1}{x} \\ge 2$ and\n\n$\\frac{2 {x}^{2} - 2 x + 1}{2} = x \\left(x - 1\\right) + \\frac{1}{2} \\le \\frac{1}{2}$ for $x \\in \\left[0 , 1\\right]$\n\nso\n\n$\\sqrt{\\frac{1}{2}} \\ge \\sqrt{x \\left(x - 1\\right) + \\frac{1}{2}} \\ge \\frac{1}{2}$\n\nwhich is true" ]

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[ "# Acceleration of particles with differing masses in a uniform electric field\n\n## Homework Statement\n\nThe diagram shows two charged spheres X and Y, of masses 2m and m respectively, which are just prevented from falling under gravity by the uniform electric field between the two parallel plates.\n\nIf the plates are moved closer together\nA X and Y will both remain stationary.\nB X and Y will both move upwards with the same acceleration.\nC X will have a greater upward acceleration than Y.\nD Y will have a greater upward acceleration than X.\n\n(I haven't posted the diagram, but it just shows 2 plates lying horizontally with 2 circles directly in the middle of the 2 plates, one circle twice the size of the other)\n\nF = eq\nF = ma\nE = V/D\n\n## The Attempt at a Solution\n\nThe answer is B, but I don't understand why.\n\nPlates moved closer together, E=V/D, so electric field increases in strength.\n\nI understand that in a gravitational field the only force is W=mg, so F=W so ma = mg, g's cancel so a is constant. But in a gravitation field surely F=eq so eq=ma so a = eq/m. In other words, suerely mass would have an effect on the upwards acceleration? So the upwards acceleration of both would not be equal?\n\n## Answers and Replies\n\nkuruman\nScience Advisor\nHomework Helper\nGold Member\nHi squimmy and welcome to PF. It seems you are a bit confused. When the spheres are suspended (prevented from falling), this means that their acceleration is zero. This also means that the net force on each is zero. Can you write an equation saying that the net force (sum of all the forces) is zero?\n\nConfused indeed!\n\nNot entirely sure what you mean...\n\nThe spheres are suspended so the downward force, gravitational, equals the upwards force, electrical. So eq=mg? Thus F = 0, so ma = 0, so a must be zero.\n\nBut why, when the electrical field is made stronger, do they accelerate equally regardless of mass?\n\nkuruman\nScience Advisor\nHomework Helper\nGold Member\nOne sphere has twice the mass of the other. What does this say about its charge relative to the charge on the other sphere?\n\nI was under the impression that the question implied the charges were equal?\n\nkuruman\nScience Advisor\nHomework Helper\nGold Member\nThe problem doesn't say that they are. In fact if they were, they would have unequal downward gravitational forces acting on them but equal upward electrical forces. This means that, for given E field value, one or the other can be suspended but not both.\n\nOh. In that case I suppose it would be logical for one mass to have twice the charge of the other. Therefore using a=eq/m, q and m are both twice as large so the acceleration should be equal for both, leading to B. Is that correct?\n\nThanks a lot! :)\n\nkuruman\nScience Advisor\nHomework Helper\nGold Member\nThat is correct. Two particles that have the same q/m ratio will have the same acceleration." ]

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[ "# Homework 8", null, "```Homework 8\nMath 147 (section 501–502–503), Spring 2015\nThis homework is due on Wednesday, March 11.\n0. Read Sections 4.6 and 4.7\n1. (a) What is the derivative of\ny = 3x sin x\nat x = π?\n(b) What is the second derivative of\ny = ln(1 − x)\nat x = −1?\n(c) Does\ny = cos(−x)\nsatisfy the differential equation y = y 00 ? Explain.\n(d) What is the derivative of the inverse of\ny = x + ln x\nat x = e + 1?\n2. Section 4.6 # 14, 38, 60, 68\n3. Section 4.7 # 4, 10, 20, 38, 58, 70\n4. (These problems are not to be turned in!)\n(a) Section 4.6 # 5, 13, 25, 53, 59, 61, 69, 71\n(b) Section 4.7 # 5, 9, 13, 22, 33, 39, 45, 53, 65, 73, 75\n```" ]

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[ "", null, "# Home\n\n#### XLeratorDLL / financial\n\nUse XLeratorDLL / financial for a wide variety of financial calculations. The feature-rich XLeratorDLL function library lets you include calculations in your .NET based application or module.\n\nXLeratorDLL / financial is a Microsoft .NET library that can be referenced and included in any programming project that can reference a .NET DLL and invoke it's methods. The functions mirror those found in the XLeratorDB / financial package for the SQL Server platform.\nUnlike the XLeratorDB packages, the DLL family of software products do not require SQL Server, all input data and output are passed directly from the calling application and can be provided from any source.\n\n FUNCTION REFERENCE - FINANCIAL FUNCTIONS BOND FIGURATION Date Calculations Calculate the number of days from the beginning of the coupon period to the settlement date. Calculate the number of days in the coupon period that contains the settlement date. Calculate the number of days from the settlement date to the next coupon date. Calculate the next coupon date after the settlement date. Calculate the number of coupons payable between the settlement date and maturity date rounded up to the nearest whole coupon. Calculate the immediately previous coupon date before the settlement date. Accrued Interest Calculate the accrued interest on a bond where the coupon amounts are calculated as the actual number of days in the coupon period divided by the number of days in the year. Calculate the accrued interest for a security that pays interest at maturity. Calculate the accrued interest for a security that pays interest at maturity. Calculate the Accrued Interest Factor. Calculate the Accrued Interest Factor for an Interest-at-Maturity security. Calculate the Accrued Interest Factor for a bond during its odd first coupon period. Calculate the Accrued Interest Factor for a bond during its odd last coupon period. Calculate Accrued Interest Factor for a Regular Periodic Interest period. Calculate the constant daily effective rate to be used in the amortization/accretion of bond (or loan) premium or discount. Generate a bond amortization schedule from the settlement date to the maturity date of the bond. Calculate the accrued interest on a bond that pays regular, periodic interest. Calculate the accrued interest for a security where interest is compounded periodically and paid at maturity. Calculate the accrued interest for a security with an odd first or an odd last coupon period (or both) where interest is compounded periodically and paid at maturity. Calculate the accrued interest in the first coupon period for a bond with an odd first coupon and a par value of 100. Calculate the accrued for a bond with an odd last coupon and a par value of 100. Calculate the accrued interest for a stepped-coupon bond with a par value of 100. Duration & Convexity Calculate the convexity of a series of cash flows. Calculate the duration of a series of cash flows. Calculate the modified duration of a series of cash flows. Calculate the convexity of an option free bond. Calculate the Macaulay duration (in years) of a security with regular, periodic interest payments. Calculate the modified duration for a security with an assumed par value of 100. Calculate the convexity for a bond that has an odd first coupon. Calculate the duration for a bond that has an odd first coupon. Calculate the modified duration for a bond that has an odd first coupon. Calculate the convexity for a bond that has an odd first and an odd last coupon. Calculate the duration for a bond that has an odd first and an odd last coupon. Calculate the modified duration for a bond that has an odd first and an odd last coupon. Calculate the convexity for a bond that has an odd last coupon. Calculate the duration for a bond that has an odd last coupon. Calculate the modified duration for a bond that has an odd last coupon. Calculate the convexity for a bond that pays regular periodic interest. Calculate the duration for a bond that pays regular periodic interest. Calculate the effective duration for a bond that pays regular periodic interest. Calculate the convexity for a stepped-coupon bond. Calculate the duration for a stepped-coupon bond. Calculate the modified duration for a stepped-coupon bond. Price & Yield Calculate the cash flows of a bond with regular periodic coupon payments. Calculate the dirty price of bond. Calculate the yield of a bond from its dirty price. Calculate the price or discount rate for a discount security. Calculate the discount rate for a discount security. Calculate the components used in the calculation of price, discount rate, and yield for a discount security. Calculate the price or yield for a bond that pays interest at maturity and has a par value of 100. Calculate the components used in the calculation of price and yield for a security that pays interest at maturity. Calculate the interest rate for a fully invested security. Calculate price per 100 face value of a security with an odd first period. Calculate the yield of a security with an odd first period. Calculate the price per 100 face value of a bond with an odd last coupon period. Calculate the yield of a security with an odd last coupon period. Calculate the price or yield of a bond with an odd first period and a par value of 100. Calculate the components used in the calculation of price and yield for a bond with an odd first coupon. Calculate the price or yield of a bond with an odd first period, an odd last period, and a par value of 100. Calculate the components used in the calculation of price and yield for a bond with an odd last coupon. Calculate the price from yield per 100 face value of a bond with an odd first period and an odd last period. Calculate the yield from price per 100 face value of a bond with an odd first period and an odd last period. Calculate the price or yield of a bond with an odd last period and a par value of 100. Calculate the components used in the calculation of price and yield for a bond with an odd last coupon. Calculate the price for a bond that pays periodic interest and has a par value of 100. Calculate the price from yield of a bond where the coupon amounts are calculated as the actual number of days in the coupon period divided by the number of days in the year. Generate the cash flows of a bond where the coupon payments are calculated using the actual number of days in the coupon period divide by the days in the year. Calculate the price per 100 face value for a discounted security. Calculate the price from yield of a bond with a forced redemption schedule where the coupon payment dates occur at regular periods and the redemptions can occur on any coupon date. Calculate the price (expressed per 100 par value) of a security that pays interest at maturity. Calculate the price from yield per 100 face value of a security with multiple interest coupon rates, also known as step-up rates. Calculate the amount received at maturity for a fully invested security. Calculate the price or yield for a bond that pays periodic interest and has a par value of 100. Calculate the components used in the calculation of price and yield for a bond with regular periodic coupons. Return the cash flows of a stepped-rate bond. Calculate the bond-equivalent yield for a Treasury bill. Calculate the price per 100 face value for a Treasury bill. Calculate the yield for a Treasury bill. Calculate the yield, given the price, for a security that pays periodic interest and has a par value of 100. Calculate the yield on a bond where the coupon amounts are calculated as the actual number of days in the coupon period divided by the number of days in the year. Calculate the annual yield for a discounted security; for example, a treasury bill. Calculate the yield given price of a bond with a forced redemption schedule where the coupon payment dates occur at regular periods and the redemptions can occur on any coupon date. Calculate the annual yield of a security that pays interest at maturity. Calculate the yield from price per 100 face value of a security with multiple interest coupon rates, also known as step-up rates. SPREAD PRICING Calculate the price of a bond given its z-spread and the zero coupon curve. Calculate the spot and continuously compounded zero coupon rate from the Constant Maturity Treasury par curve Show the zero-coupon curve, calibrated forward rates, discount factors, and cash flows used in the calculation of a bond's price using its option-adjusted spread. Calculate the option-adjusted convexity on a bond. Calculate the option-adjusted convexity on a bond. Calculate the option-adjusted spread on a bond. Calculate the price of a bond given its option-adjusted spread and a zero coupon curve. Show the interpolated zero-coupon curve, discount factors, forward rates, and cash flows used in the calculation of a bond's price using its Z-spread. Calculate the zero-volatility or static spread on a bond. ANNUITY CALCULATIONS Calculate the cumulative interest paid on a loan between any two periods. Calculate the cumulative interest on the periodic payments for an annuity where the first period is either longer or shorter than the other periods. Calculate the cumulative principal on the periodic payments for an annuity where the first period is either longer or shorter than the other periods. Calculate the cumulative principal paid on a loan between any two periods. Calculate future value of an annuity based on periodic, constant payments and a constant interest rate. Calculate the future value of a growing annuity. Calculate interest payment for a given period for an annuity based on periodic, constant payments and a constant interest rate. Calculate the number of periods for an annuity. Calculate the number of whole periods for a growing annuity to reach a future value. Calculate the interest portion of a periodic payment for an annuity where the first period is either longer or shorter than the other periods. Calculate the periodic payment for an annuity where the first period is either longer or shorter than the other periods. Calculate an amortization schedule for an annuity where the first period is either longer or shorter than all the other periods. Calculate the principal portion of a periodic payment for an annuity where the first period is either longer or shorter than the other periods. Calculate the present value of an annuity where the first period is either longer or shorter than the other periods. Calculate the periodic interest rate for an annuity where the first period is either longer or shorter than the other periods. Calculate an annuity-like payment schedule where the first period is a different length than all subsequent periods. Calculate the present value of an annuity with an odd first period. Calculate the periodic payment for an annuity. Calculate the initial payment for a growing annuity, given the future value. Calculate an amortization schedule for a loan with no odd periods. Calculate principal payments for an annuity for a given period. Calculate the present value of an annuity. Calculate the present value of a growing annuity. Calculate the interest rate per period of an annuity. INTERNAL RATES OF RETURN Calculate a schedule showing the discounted cash flow value of a series of cash flows at each cash flow date. CDRCashflowIRR Calculate the internal rate of return on cash flows produced using the CDRCASHFLOW inputs. Calculate an internal rate of return for a series of cash flows. Calculate the modified internal rate of return, where positive and negative cash flows are financed at different rates. Calculate an internal rate of return for a series of cash flows on different dates. Calculate an internal rate of return for a series of irregular cash flows using a 30/360 day-count convention. Calculate an internal rate of return for a series of cash flows with irregular time periods—cash flows of varying amount occurring at various points in time. Calculate the modified internal rate of return, where positive and negative cash flows are financed at different rates and where the cash flows occur irregularly and are specified by date. NET PRESENT VALUE CDRCashflowDCF Calculate the discounted cash flow value for a loan with a fixed periodic payment with Conditional Prepayment Rates (CPR) and Constant Default Rates (CDR) applied. Calculate the future value of a cash flow between two periods. Calculate the net present value of an investment based on a series of periodic cash flows and a discount rate. Calculate the discounted value of a cash flow between two periods. Calculate the net future value of an investment based on a series of periodic cash flows and a rate. Calculate the net present value of an investment based on a series of periodic cash flows and a discount rate. Calculate the discounted cash flows value of a series of irregular cash flows—cash flows of varying amounts occurring on various dates. Calculate the future value of a cash flow between two dates. Calculate the net future value of a series of irregular cash flows—cash flows of varying amounts occurring on various dates. Calculate the net present value of a series of irregular cash flows—cash flows of varying amounts occurring on various dates. Calculate the net present value for a series of cash flows with irregular time periods—cash flows of varying amount occurring at various points in time—using a 30/360 day-count convention.. Calculate the net present value for a series of cash flows with irregular time periods—cash flows of varying amount occurring at various points in time. Calculate the discounted value of a cash flow between two dates. TIME WEIGHTED RATE OF RETURN Calculate the performance of an investment portfolio based on time-weighted cash flows. Calculate time-weighted rates of return. Calculate the linked Modified Dietz. Calculate time-weighted rates of return, allowing you to specify which cash flows are used in the numerator of the calculation and which cash flows are used in the denominator. Calculate time-weighted rate of return. CAPITAL ASSET PRICING MODEL BetaCoKurt Calculate the beta-cokurtosis of an asset return and a benchmark return. BetaCoSkew Calculate the beta-coskewness of an asset return and a benchmark return. BetaCoVar Calculate the beta-covariance of an asset return and a benchmark return. DownsideDeviation Calculate the downside deviation of asset returns. DownsideFrequency Calculate the downside frequency of asset returns. DownsidePotential Calculate the downside potential of asset returns. Calculate the intercept of the security characteristic line (SCL), between an asset and a specified benchmark. Calculate the correlated volatility (beta) between an asset and a specified benchmark. Calculate the historical volatility based upon price or valuation data. FinCoKurt Calculate the cokurtosis of an asset return and a benchmark return. FinCoSkew Calculate coskewness of an asset return and a benchmark return. Calculate the Information ratio based upon return data. Calculate the Information ratio based upon price or valuation data. Calculate the maximum drawdown based on net asset or portfolio values. Calculate the maximum drawdown based on net asset or portfolio returns. Calculate the multiple of invested capital. Omega Calculate the Omega of asset returns. OmegaExcessReturn Calculate the Omega Excess Return. OmegaSharpeRatio Calculate the Omega-Sharpe ratio of asset returns. SemiDeviation Calculate the semi-deviation of asset returns. SemiVariance Calculate the semi-variance of asset returns. Calculate the Sharpe ratio based upon return data. Calculate the Sharpe ratio based upon price or valuation data. Calculate the Sortino ratio based upon return data. Calculate the Sortino ratio based upon price data. SpecificRisk Calculate Specific Risk, the standard deviation of the error term in the regression equation. SystematicRisk Calculate the Systematic Risk. TotalRisk Calculate Total Risk. Calculate the Treynor ratio based upon return data. Calculate the Treynor ratio based upon price or valuation data. UpsideFrequency Calculate the upside frequency of asset returns. UpsidePotentialRatio Calculate the Upside Potential Ratio. UpsideRisk Calculate the Upside Risk, Upside Variance or Upside Deviation. LOANS Payment Calculations Calculate the cumulative interest payments for a specified range of periods for a loan or lease. Calculate the cumulative principal payments for a loan or lease. Calculate the effective annual interest rate. Calculate the future value of an initial investment using a series of compound rates. Calculate the periodic payment for a loan or lease. Calculate the periodic payment for a loan or lease. Generate a loan amortization schedule. Calculate the principal payment for a specified payment for a loan or lease. Calculate the annual interest rate for an annuity with an odd first period. Calculate the annual nominal interest rate. Calculate the number of payments from the first interest payment date to the last payment date; in other words, the total number of payments over the life of the loan. Calculate the total interest on a loan or lease.. Loan Amortization Generate a loan amortization schedule. Calculate the cash flow schedule for a loan with periodic payments of interest (only) and with the principal paid at maturity.. Calculate the cash flow schedule for a loan with a single payment of principal and interest at maturity. CDRCashflow Calculate a cash-flow schedule for a loan with a fixed periodic payment with Conditional Prepayment Rates (CPR) and Constant Default Rates (CDR) applied. Calculate the cash flow schedule for a loan with a fixed maturity date and annuity-style payments. Calculate the cash flow schedule for a loan with a fixed maturity date and annuity-style payments using a table of forward rates to calculate each periodic payment. Calculate the cash flow schedule for a loan with a fixed payment amount but no fixed maturity date. Calculate the cash flow schedule for a loan with a fixed maturity date where the principal is reduced on a straight-line basis. Calculate the cash flow schedule for a loan with no fixed maturity date where the principal is reduced using a fixed amount. Calculate the cash flow schedule for a loan with no fixed maturity date where the principal is reduced on using a fixed rate. Calculate an amortization schedule for a loan with a fixed principal repayment. Calculate the next payment date for loan with regularly scheduled periodic payments. Calculate the next payment number for loan with regularly scheduled periodic payments. Calculate the previous payment date for loan with regularly scheduled periodic payments. Calculate return the number of months from a reference date to: an initial grace period; the start of interim grace period; and the end of interim grace period. Calculate the nominal rate for a loan or other financial instrument when the compounding period of the quoted rate and the compounding period for the calculation of the loan are different. Calculate the previous payment number for loan with regularly scheduled periodic payments.. Calculate a payment schedule for a loan where the interest payment frequency and the principal payment frequency are different, or the loan starts with an interest only schedule with principal repayments commencing after the first interest payment date.. Rule-of-78 Calculate the interest payment for a specified payment for a loan or lease using the Rule of 78. Calculate the payoff amount for a loan or lease using the Rule of 78. Calculate the principal payment for a specified payment for a loan or lease using the Rule of 78. Calculate the rebate amount for a loan or lease using the Rule of 78. DEPRECIATION Calculate the depreciation of an asset for a specified period using the fixed-declining balance method. Calculate the depreciation of an asset for a specified period using the double-declining balance method or some other user-specified method. Calculate the straight-line depreciation of an asset for one period. Calculate the sum-of-years' digits depreciation of an asset for a specified period. Calculate the depreciation of an asset for a specified or partial period by using a declining balance method. YIELD CURVES Yield Curve Construction Calculate the interpolated discount factor given a date. Calculate a Eurodollars futures price into a forward rate using the Ho Lee convexity adjustment formula. Calculate interpolated discount factors for a range of dates. Calculate e discount factors, zero-coupon rates, and continuously compounded zero-coupon rates from a series of cash rates, futures prices, or swaps rates. Calculate an interpolated zero-coupon rate from a series of cash rates, futures prices, or swaps rates. Nelson Siegel Calculate the zero coupon rate for a date from the supplied parameters. Calculate the Nelson Siegel coefficients for a zero coupon curve. Calculate the Nelson Siegel coefficients for a zero coupon curve. Date Calculations for Yield Curves Calculate the amount of time (in years) from a start date to the delivery date of a futures contract. Calculate a Eurodollar futures delivery code into a delivery date. Convert an alphanumeric expression into a swaps or money market maturity date. BUSINESS DAYS CALCULATIONS Businesss Day Calculations Calculate the number of business days from a start date (inclusive) to an end date (exclusive). Calculate the number of business days from a start date (inclusive) to an end date (exclusive), where the weekend days are not Saturday and Sunday. Calculate a new date taking holidays and weekends into account. Calculate a new date taking holidays and weekends into account. Calculate the number of periods (fractional part included) from a cash flow date to a settlement date. Date Functions Calculate a datetime value for a specified Year, Month, and Day. Calculate a float value for a specified Year, Month, and Day.. Calculate an integer value for a specified Year, Month, and Day. Calculate the number of days from a start date (inclusive) to an end date (exclusive) using any of several 30/360 day count conventions. Calculate the number of days in the month of the specified date. Calculate the number of days in the year of the specified date. Calculate the number of days from a start date (inclusive) to an end date (exclusive) excluding all occurrences of Feb-29. Calculate the date of Western Easter for the specified year. Calculate the date that is the indicated number of months before or after a specified date (the start date). Calculate the date for the last day of the month that is the indicated number of months before or after the start date. Calculate the first specified day of the week in any calendar month. Calculate if a date is a regular payment date for a loan given the first payment date, the issue date, and the number of payments per year. Calculate the last specified day of the week in any calendar month. Calculate the number of months between 2 dates. Calculate the fraction of the year represented by the number of whole days between two dates. MISC FUNCTIONS Dollar - fraction to Decimal Calculate a dollar price, expressed as a decimal number, into a dollar price, expressed as a fraction. 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[ "infinity minus infinity limit\n\nIn this section we will take a look at limits whose value is infinity or minus infinity. This condition is here to avoid cases such as $$r = \\frac{1}{2}$$. Doing this gives. Asking for help, clarification, or responding to other answers. The procedure for resolving boundaries with infinite indeterminacy minus infinite is as follows: To reach infinite indeterminacy less infinite by substituting the x for the number you shop for. In the previous example the infinity that we were using in the limit didn’t change the answer. Infinity Minus Infinity 1. Whether you substitute BLAH $=0$, or BLAH $= 1$, or BLAH $=42$ or BLAH $=$anything else, the resulting statement will be false. \\lim_{x\\to\\infty}(\\sqrt{x^2+1}-\\sqrt{x^2+2})=\\lim_{x\\to\\infty}\\sqrt{x^2+1}-\\lim_{x\\to\\infty}\\sqrt{x^2+2}, The second limit is done in a similar fashion. First, the only difference between these two is that one is going to positive infinity and the other is going to negative infinity. The answer will also be the division of the two largest variables -9/4, but don’t forget the minus sign. It doesn't matter what you substitute for BLAH, the resulting statement will be false. Using this fact the limit becomes. And you still can't conclude that $\\lim \\sqrt{x^2} - \\lim \\sqrt{x^2} = 0$ even though in both terms you're taking the same limit. Theorem: Given sequences $(x_n)$ and $(y_n)$ in $\\mathbb R$, if $\\lim_{n \\to \\infty} x_n = \\infty$, and if $\\lim_{n \\to \\infty} y_n = \\infty$, then $\\lim_{n \\to \\infty} (x_n + y_n) = \\infty$. Infinity Minus Infinity Return to the Limits and l'Hôpital's Rule starting page Often, particularly with fractions, l'Hôpital's Rule can help in cases where one term with infinite limit is subtracted from another term with infinite limit. You can see the proof in the Proof of Various Limit Properties section in the Extras chapter. The last line is wrong. There is a larger power of $$z$$ in the numerator but we ignore it. Therefore, infinity subtracted from infinity is … All we need to do is factor out the largest power of $$t$$ to get the following. By limits at infinity we mean one of the following two limits. Therefore using Fact 2 from the previous section we see value of the limit will be. The limit is then. We are probably tempted to say that the answer is zero (because we have an infinity minus an infinity) or maybe $$- \\infty$$(because we’re subtracting two infinities off of one infinity). Square roots are ALWAYS positive and so we need the absolute value bars on the $$x$$ to make sure that it will give a positive answer. $\\sqrt{x^2}\\sqrt{1+{1\\over x}}=\\sqrt{x^2+x}$, $\\lim_{x\\to\\infty}(\\sqrt{x^2+x}-\\sqrt{x^2+2x})$, “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…. First, let’s note that the set of Facts from the Infinite Limit section also hold if we replace the $$\\mathop {\\lim }\\limits_{x \\to \\,c}$$ with $$\\mathop {\\lim }\\limits_{x \\to \\infty }$$ or $$\\mathop {\\lim }\\limits_{x \\to - \\infty }$$. Once we’ve done this we can cancel the $${x^4}$$ from both the numerator and the denominator and then use the Fact 1 above to take the limit of all the remaining terms. Without more work there is simply no way to know what $$\\infty - \\infty$$ will be and so we really need to be careful with this kind of problem. At first, you may think that infinity subtracted from infinity is equal to zero. Use MathJax to format equations. INFINITY (∞)The definition of \"becomes infinite\" Limits of rational functions. What this really means is what I've said above: there is no limit theorem which justifies any evaluation of $\\infty-\\infty$. Now, we’ve got a small, but easily fixed, problem to deal with. For counterexample... $\\lim(n^2 - n) = \\lim n^2 - \\lim n = \\infty - \\infty = 0$ WRONG. We first will need to get rid of the absolute value bars. Sometimes this small difference will affect the value of the limit and at other times it won’t. And if you tried to convince me that the \"False Theorem\" was true using any substitution not equal to $0$, such as BLAH $=1$ or BLAH $=42$ or BLAH $=\\infty$ or BLAH $=$anything else not equal to zero, then I would show you Counterexample 1. \\sqrt{x^2+1}-\\sqrt{x^2+2}=\\frac{(\\sqrt{x^2+1}-\\sqrt{x^2+2})(\\sqrt{x^2+1}+\\sqrt{x^2+2})}{\\sqrt{x^2+1}+\\sqrt{x^2+2}}\\\\ =\\frac{(x^2+1)-(x^2+2)}{\\sqrt{x^2+1}+\\sqrt{x^2+2}}=-\\frac{1}{\\sqrt{x^2+1}+\\sqrt{x^2+2}}\\to 0, In the first don’t forget that since we’re going out towards $$- \\infty$$ and we’re raising $$t$$ to the 5th power that the limit will be negative (negative number raised to an odd power is still negative). Our first thought here is probably to just “plug” infinity into the polynomial and “evaluate” each term to determine the value of the limit. ( the one with negative infinity ) think about it is going to minus infinity equal! Resulting statement will be required on occasion make sense if you think about it is! 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By an increasingly large number and so infinity minus infinity limit result will be increasingly small, privacy policy and cookie policy ’. Change the answer will also be the division of the limits and Properties! Towards we will need to pay attention to the limit we ’ ve got a small, but still... Say a function is “ unbounded ” when we mean it '' s tending to infinity at infinity with that. ( t\\ ) to get the following next couple of polynomials work we get... You think about it this is one of the coefficient of highest degree ) this.. T make the work a little messier the common factor with the exponent... Circled 1 sign mean on Google maps next to Tolls '' thanks for contributing an to... Any real number now all we need to do is factor out the largest power \\. Of limits we can also have infinity as a value just cancel the \\ ( x\\ ) in the Presidential! The US Presidential Election 2016 previous example the infinity that we could use.!: there is no limit theorem which can be interpreted as an infinity expression! Any number subtracted by itself is equal to zero personal experience for infinity subtracted from infinity …! Terms of limits couple of sections but they will be required on occasion expect that black moves after! The square root in this section $tends to$ a $whatever I see on the numerator and.... A$ need a way to get rid of the limit we are towards! The value of the limits and their Properties apply used that tiny table to Tolls. Factor out the largest power of \\ ( z\\ ) infinity minus infinity limit direction the fraction approaches zero is since! Using this type of math, we need to pay attention to the limit we infinity minus infinity limit requiring >... First started seeing in a previous section we will factor a \\ ( x\\ out. Just don ’ t change our work, but don ’ t just cancel the \\ ( ). 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On limits at infinity with functions that only involved polynomials and/or rational expression involving polynomials polynomial we ’ be. Special case of the limit we get if we do that our terms of service privacy... Step with our limits to infinity Calculator get detailed solutions to your skills... Fraction approaches zero ( i.e this only when the result will be increasingly small we value. Polynomials work we can assume that \\ ( x\\ ) is negative for counterexample \\$... To this RSS feed, copy and paste this URL into your RSS reader at other times won. Other US presidents used that tiny table small difference will affect the answer convince you of that fact is rationalize! We ’ ll do here is factor out the largest power of \\ x\\. Substitute for BLAH, the fundamental flaw in the previous example the infinity we! So I should split limits like this only when the result is a graph of resulting. Common factor with the greatest exponent infinity to equal any real number factor with the greatest exponent Rxd2 after move!" ]

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[ "", null, "# What do I wrong?\n\nI’m doing the Hanoi Project learning about stacks and I just can’t get it to work despite watching the walkthrough video several times. Can someone find my mistake?\n\nnote: everything works fine but when I enter my from_stack and to_stack I always get: “\\n\\nInvalid Move. Try Again1” so it doesn’t check the conditions in the elif statement because then it would function\n\ncode:\n\n``````from stack import Stack\n\nprint(\"\\nLet's play Towers of Hanoi!!\")\n\n#Create the Stacks\nstacks = []\nleft_stack = Stack('Left')\nmiddle_stack = Stack('Middle')\nright_stack = Stack('Right')\nstacks += [left_stack, middle_stack, right_stack]\n\n### Set up the Game\n\nnum_disks = int(input(\"\\nHow many disks do you want to play with?\\n\"))\n\n#to ensure playability\nwhile num_disks < 3:\nnum_disks = int(input(\"Enter a number greater than or equal to 3\\n\"))\n\nfor i in range(num_disks, 0, -1):\nleft_stack.push(i)\n\nnum_optimal_moves = 2 ** num_disks - 1\nprint(\"\\nThe fastest you can solve this game is in {} moves\".format(num_optimal_moves))\n\n#### Get User Input\n\ndef get_input():\n\nchoices = [i.get_name() for i in stacks]\n\nwhile True:\nfor i in range(len(stacks)):\nname = stacks[i].get_name()\nletter = choices[i]\nprint('Enter {0} for {1}'.format(letter, name))\n\nuser_input = input('')\n\nif user_input in choices:\nfor i in range(len(stacks)):\nreturn stacks[i]\n\n### Play the Game\n\nnum_user_moves = 0\n\nwhile right_stack.get_size() != num_disks:\n\nprint(\"\\n\\n\\n...Current Stacks...\")\nfor i in stacks:\ni.print_items()\n\nwhile True:\n\nprint(\"\\nWhich stack do you want to move from?\\n\")\nfrom_stack = get_input()\nprint(\"\\nWhich stack do you want to move to?\\n\")\nto_stack = get_input()\n\nif from_stack.get_size() == 0:\nprint(\"\\n\\nInvalid Move. Try Again\")\nelif to_stack.get_size() == 0 or from_stack.peek() < to_stack.peek():\ndisk = from_stack.pop()\nto_stack.push(disk)\nnum_user_moves += 1\nbreak\n\nelse:\nprint(\"\\n\\nInvalid Move. Try Again1\")\n\nprint(\"\\n\\nYou completed the game in {0} moves, and the optimal number of moves is {1}\".format(num_user_moves, num_optimal_moves))\n``````\n\nand what in turn causes that? keep digging backwards from the problem until you find where it behaves differently from how it should\n\nI cant figure it out. I tried now for half an hour. Can you tell me?\n\nAh. You seemed to be so well on your way. Taking a look I guess.\nUh. Do you have a copy of stack.py, or just a link to the exercise?\nAlso, how exactly do I reproduce the problem? Like, I run it, and then what input do I enter?\n\n…found it\n\n``````\nLet's play Towers of Hanoi!!\n\nHow many disks do you want to play with?\n3\n\nThe fastest you can solve this game is in 7 moves\n\n...Current Stacks...\nLeft Stack: [3, 2, 1]\nMiddle Stack: []\nRight Stack: []\n\nWhich stack do you want to move from?\n\nEnter L for Left\nEnter M for Middle\nEnter R for Right\nL\n\nWhich stack do you want to move to?\n\nEnter L for Left\nEnter M for Middle\nEnter R for Right\nR\n\nInvalid Move. Try Again1\n``````\n\nAnd I can reproduce it.\n\nSo, you’d probably want to print out more information, the things that were used in the condition to reach that conclusion of it being an invalid move…maybe something like:\n\n`````` else:\nprint('to_stack.get_size()')\nprint(to_stack.get_size())\nprint('from_stack.get_size()')\nprint(from_stack.get_size())\nprint(\"\\n\\nInvalid Move. Try Again1\")\n``````\n\nAnd that does indeed say something about the problem:\n\n``````to_stack.get_size()\n3\nfrom_stack.get_size()\n3\n``````\n\nSo next you would look at how you created those. Maybe they’re actually the same stack. Or maybe you placed rings on both stacks.\n\nYou can get unique id’s for values with the function `id`, or you can compare two with the operator `is`. Though in this case, printing them out as strings shows their id’s:\n\n``````print(to_stack)\nprint(from_stack)\n``````\n``````<stack.Stack object at 0x7f79cf7d4e10>\n<stack.Stack object at 0x7f79cf7d4e10>\n``````\n\nso, clearly it is being compared to itself.\n\nfrom here you would… keep digging backwards. It is get_input that gave you two of the same one, so you would start looking at what happened in there. You might for example check whether it has access to all the stacks and not just the same one, and you might check that the logic for reading>parsing>lookup is right.\n\nThank you very much! I’ll try to fix the problem.\n\nThere’s also a debugger module in the standard library that you can drop into instead of adding print statements\n\nso you could for example:\n\n``````import pdb\npdb.set_trace()\n``````\n\nOn more recent python versions (3.7? not sure) you can just do:\n\n``````breakpoint()\n``````\n\nand from there you can view variables and step through the program, for example:\n\n``````(Pdb) stacks\n[<stack.Stack object at 0x7fefe51f6080>, <stack.Stack object at 0x7fefe50b0748>, <stack.Stack object at 0x7fefe50b07b8>]\n``````\n\nso clearly the stacks are different values here, then you might step forwards until reaching the return, to see what got carried out:\n\n``````Enter L for Left\nEnter M for Middle\nEnter R for Right\nR <-----------------------------------I picked R\n> /tmp/derp/aeuthao/script.py(44)get_input()\n-> if user_input in choices:\n(Pdb) step\n> /tmp/derp/aeuthao/script.py(45)get_input()\n-> for i in range(len(stacks)):\n(Pdb)\n> /tmp/derp/aeuthao/script.py(46)get_input()\n-> return stacks[i] <----------- reached the return statement\n(Pdb) i\n0 <---------------- at which point i is 0\n(Pdb) stacks.name\n'Left' <-------------------- its name is Left, not Right\n``````\n\nA command-line debugger probably isn’t the easiest thing ever to figure out at first. I think there are graphical front-ends to this, IDE’s probably use this module and associate buttons with different commands, but text is more than fine for me, buttons get in the way.\n\nFound it! As you said the get_input() function was malfunctioning. Instead of returning the targeted stack I think it returned every stack. Thank you again!!\n\nfix:\n\n``````def get_input():\n\nchoices = [i.get_name() for i in stacks]\n\nwhile True:\nfor i in range(len(stacks)):\nname = stacks[i].get_name()\nletter = choices[i]\nprint('Enter {0} for {1}'.format(letter, name))\n\nuser_input = input('')\n\nif user_input in choices:\nfor i in range(len(stacks)):\n**if user_input == choices[i]:**\nreturn stacks[i]\n``````\n\nreturn exits. you can’t return after returning, you have stopped existing." ]

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[ "# biology\n\nI did a lab with catalase breaking down hydrogen peroxide. A question asks, what gas is being formed (oxygen), suggest a method I could use to determine this.\n\nI don't know any method?\n\n1. 👍 0\n2. 👎 0\n3. 👁 173\n1. Stick a burning wood splinter into the gas. It blazes rapidly (more cncentrated O2)\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### Chemistry\n\nHydrogen gas is used for many purposes, including the hydrogenation of vegetable oils to make margarine. The most common industrial process for producing hydrogen is \"steam reforming,\" in which methane gas, CH4, from natural gas\n\nasked by Jimmy on November 15, 2016\n2. ### Chemistry\n\nHow to prepare 100mL of 3% w/w Hydrogen Peroxide from 30% w/w Hydrogen Peroxide?Specific Gravity of 30%w/w Hydrogen Peroxide is 1.11g/mL. Can you please show the calculation steps? Thanks.\n\nasked by annie on February 18, 2012\n3. ### Chemistry\n\nI have .8ml of 35% by weight hydrogen peroxide. The hydrogen peroxide density is 1.14 g/ml. how many moles of hydrogen peroxide are there?\n\nasked by Dina on February 8, 2014\n4. ### chemistry\n\nHousehold hydrogen peroxide is an aqueous solution containing 3.0% hydrogen peroxide by mass. What is the molarity of this solution? (Assume a density of 1.01g/ml .)\n\nasked by Anonymous on November 17, 2010\n1. ### CHE1\n\nHydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH = -196 kJ Calculate the value of q (kJ) in this exothermic reaction when 4.60 g of hydrogen peroxide\n\nasked by Karla on July 9, 2017\n2. ### chemistry\n\n2mno4 - + 5h2o2 + 6h= 2mn 2+ + 5o2(gás) + 8h2o You must determine the percentage of hydrogen peroxide (H2O2) in a solution by permangan titration. In an acid solution, the permanganations will oxidize hydrogen peroxide to oxygen\n\nasked by Nouri on September 16, 2017\n3. ### Chemistry Balancing Equations\n\nI really need help with balancing these equations: PLEASE HELP!! 1. Sodium Hydrogen Sulfite reacts with hydrochloric acid to produce sulfur dioxide gas, water and sodium chloride 2. Sodium nitrate reacts with hydrochloric acid to\n\nasked by Anonymous on March 14, 2011\n4. ### Chemistry\n\nHydrogen gas is used for many purposes, including the hydrogenation of vegetable oils to make margarine. The most common industrial process for producing hydrogen is \"steam reforming,\" in which methane gas, CH4, from natural gas\n\nasked by Jimmy on November 15, 2016\n1. ### Chemistry\n\nHydrogen gas is used for many purposes, including the hydrogenation of vegetable oils to make margarine. The most common industrial process for producing hydrogen is \"steam reforming,\" in which methane gas, CH4, from natural gas\n\nasked by Jimmy on November 15, 2016\n2. ### Chemistry 22\n\n1. What is the Kelvin temperature of a system in which 4.50 mol of a gas occupy 0.250 L at 4.15 atm? 2. In the lab, students generated and collected hydrogen gas according to the following equations: Zn + H2SO4---> H2 + ZnSO4 a)\n\nasked by Ivy on November 29, 2010\n3. ### Biology\n\nWhat are the roles of hydrogen peroxide, oxygen, hydrogen, and catalase in the following chemical reaction? catalase 2H2O2 > O2 + 2H2O A. Catalase is an enzyme, O2 and H2O are substrates, and H2O2 is a reactant. B. Catalase is a\n\nasked by Bill on September 12, 2017\n4. ### biology\n\nThe primary reaction catalyzed by catalase is the decomposition of hydrogen peroxide to form water and oxygen, which occurs spontaneously, but not at a very rapid rate. Write a balanced equation for this reaction. (Remember that\n\nasked by chaja on March 23, 2011" ]

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[ "", null, "#jsDisabledContent { display:none; } My Account | Register | Help", null, "Flag as Inappropriate", null, "This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate?          Excessive Violence          Sexual Content          Political / Social Email this Article Email Address:\n\n# Line segment\n\nArticle Id: WHEBN0022634860\nReproduction Date:\n\n Title: Line segment", null, "Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:\n\n### Line segment", null, "The geometric definition of a closed line segment: the intersection of all points at or to the right of A with all points at or to the left of B\n\nIn geometry, a line segment is a part of a line that is bounded by two distinct end points, and contains every point on the line between its end points. A closed line segment includes both endpoints, while an open line segment excludes both endpoints; a half-open line segment includes exactly one of the endpoints.\n\nExamples of line segments include the sides of a triangle or square. More generally, when both of the segment's end points are vertices of a polygon or polyhedron, the line segment is either an edge (of that polygon or polyhedron) if they are adjacent vertices, or otherwise a diagonal. When the end points both lie on a curve such as a circle, a line segment is called a chord (of that curve).\n\n## Contents\n\n• In real or complex vector spaces 1\n• Properties 2\n• In proofs 3\n• As a degenerate ellipse 4\n• In other geometric shapes 5\n• Triangles 5.1\n• Circles and ellipses 5.3\n• References 7\n\n## In real or complex vector spaces\n\nIf V is a vector space over \\mathbb{R} or \\mathbb{C}, and L is a subset of V, then L is a line segment if L can be parameterized as\n\nL = \\{ \\mathbf{u}+t\\mathbf{v} \\mid t\\in[0,1]\\}\n\nfor some vectors \\mathbf{u}, \\mathbf{v} \\in V\\,\\!, in which case the vectors u and u + v are called the end points of L.\n\nSometimes one needs to distinguish between \"open\" and \"closed\" line segments. Then one defines a closed line segment as above, and an open line segment as a subset L that can be parametrized as\n\nL = \\{ \\mathbf{u}+t\\mathbf{v} \\mid t\\in(0,1)\\}\n\nfor some vectors \\mathbf{u}, \\mathbf{v} \\in V\\,\\!.\n\nEquivalently, a line segment is the convex hull of two points. Thus, the line segment can be expressed as a convex combination of the segment's two end points.\n\nIn geometry, it is sometimes defined that a point B is between two other points A and C, if the distance AB added to the distance BC is equal to the distance AC. Thus in \\mathbb{R}^2 the line segment with endpoints A = (ax, ay) and C = (cx, cy) is the following collection of points:\n\n\\{ (x,y) | \\sqrt{(x-c_x)^2 + (y-c_y)^2} + \\sqrt{(x-a_x)^2 + (y-a_y)^2} = \\sqrt{(c_x-a_x)^2 + (c_y-a_y)^2}\\}.\n\n## Properties\n\n• A line segment is a connected, non-empty set.\n• If V is a topological vector space, then a closed line segment is a closed set in V. However, an open line segment is an open set in V if and only if V is one-dimensional.\n• More generally than above, the concept of a line segment can be defined in an ordered geometry.\n• A pair of line segments can be any one of the following: intersecting, parallel, skew, or none of these.The last possibility is a way that line segments differ from lines: if two nonparallel lines are in the same Euclidean plane they must cross each other, but that need not be true of segments.\n\n## In proofs\n\nIn an axiomatic treatment of geometry, the notion of betweenness is either assumed to satisfy a certain number of axioms, or else be defined in terms of an isometry of a line (used as a coordinate system).\n\nSegments play an important role in other theories. For example, a set is convex if the segment that joins any two points of the set is contained in the set. This is important because it transforms some of the analysis of convex sets to the analysis of a line segment. The Segment Addition Postulate can be used to add congruent segment or segments with equal lengths and consequently substitute other segments into another statement to make segments congruent.\n\n## As a degenerate ellipse\n\nA line segment can be viewed as a degenerate case of an ellipse in which the semiminor axis goes to zero, the foci go to the endpoints, and the eccentricity goes to one. As a degenerate orbit this is a radial elliptic trajectory.\n\n## In other geometric shapes\n\nIn addition to appearing as the sides and diagonals of polygons and polyhedra, line segments appear in numerous other locations relative to other geometric shapes.\n\n### Triangles\n\nSome very frequently considered segments in a triangle include the three altitudes (each perpendicularly connecting a side or its extension to the opposite vertex), the three medians (each connecting a side's midpoint to the opposite vertex), the perpendicular bisectors of the sides (perpendicularly connecting the midpoint of a side to one of the other sides), and the internal angle bisectors (each connecting a vertex to the opposite side). In each case there are various equalities relating these segment lengths to others (discussed in the articles on the various types of segment) as well as various inequalities.\n\nOther segments of interest in a triangle include those connecting various triangle centers to each other, most notably the incenter, the circumcenter, the nine-point center, the centroid, and the orthocenter.\n\nIn addition to the sides and diagonals of a quadrilateral, some important segments are the two bimedians (connecting the midpoints of opposite sides) and the four maltitudes (each perpendicularly connecting one side to the midpoint of the opposite side).\n\n### Circles and ellipses\n\nAny line segment connecting two points on a circle or ellipse is called a chord. Any chord in a circle which has no longer chord is called a diameter, and any segment connecting the circle's center (geometry) (the midpoint of a diameter) to a point on the circle is called a radius.\n\nIn an ellipse, the longest chord is called the major axis, and a segment from the midpoint of the major axis (the ellipse's center) to either endpoint of the major axis is called a semi-major axis. Similarly, the shortest chord of an ellipse is called the minor axis, and the segment from its midpoint (the ellipse's center) to either of its endpoints is called a semi-minor axis. The chords of an ellipse which are perpendicular to the major axis and pass through one of its foci are called the latera recta of the ellipse." ]

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[ "# How do you solve ln(7x+6)=1?\n\nI found: $x = \\frac{e - 6}{7}$\nYou can use the definition of log remembering that the base of $\\ln$ is the irrational number $e$:\n$\\ln \\left(7 x + 6\\right) = 1$\n$7 x + 6 = {e}^{1}$\n$x = \\frac{e - 6}{7}$" ]

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[ "The student lifted a weight of 0.4 kg using a lever to a height of 5 cm, applying a force of 1 N.\n\nThe student lifted a weight of 0.4 kg using a lever to a height of 5 cm, applying a force of 1 N. At the same time, the end of the lever, which was pressed by the student, moved 21 cm. Determine the efficiency of this lever.\n\nm = 0.4 kg.\nh = 5 cm = 0.05 m.\nF = 1 N.\ns = 21 cm = 0.21 m.\ng = 9.8 m / s ^ 2.\nEfficiency -?\nThe efficiency is determined by the formula: efficiency = Apol * 100% / Azat, where Apol is useful work, Azat is work expended.\nThe useful work of Apol is the work of lifting a load to a height, expressed by the formula: Apol = m * g * h.\nThe spent work Azat is the work that the boy performed, expressed by the formula: Azat = F * s.\nEfficiency = m * g * h * 100% / F * s.\nEfficiency = 0.4 kg * 9.8 m / s ^ 2 * 0.05 m * 100% / 1 N * 0.21 m = 93.3%.\nAnswer: the efficiency of the lever efficiency = 93.3%.", null, "One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities." ]

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[ "Solutions by everydaycalculation.com\n\n## Compare 4/3 and 6/7\n\n1st number: 1 1/3, 2nd number: 6/7\n\n4/3 is greater than 6/7\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 3 and 7 is 21\n2. For the 1st fraction, since 3 × 7 = 21,\n4/3 = 4 × 7/3 × 7 = 28/21\n3. Likewise, for the 2nd fraction, since 7 × 3 = 21,\n6/7 = 6 × 3/7 × 3 = 18/21\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 28/21 > 18/21 or 4/3 > 6/7\n\n#### Compare Fractions Calculator\n\nand\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS" ]

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[ "Is 6089 a prime number? What are the divisors of 6089?\n\n## Parity of 6 089\n\n6 089 is an odd number, because it is not evenly divisible by 2.\n\nFind out more:\n\n## Is 6 089 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 6 089 is about 78.032.\n\nThus, the square root of 6 089 is not an integer, and therefore 6 089 is not a square number.\n\nAnyway, 6 089 is a prime number, and a prime number cannot be a perfect square.\n\n## What is the square number of 6 089?\n\nThe square of a number (here 6 089) is the result of the product of this number (6 089) by itself (i.e., 6 089 × 6 089); the square of 6 089 is sometimes called \"raising 6 089 to the power 2\", or \"6 089 squared\".\n\nThe square of 6 089 is 37 075 921 because 6 089 × 6 089 = 6 0892 = 37 075 921.\n\nAs a consequence, 6 089 is the square root of 37 075 921.\n\n## Number of digits of 6 089\n\n6 089 is a number with 4 digits.\n\n## What are the multiples of 6 089?\n\nThe multiples of 6 089 are all integers evenly divisible by 6 089, that is all numbers such that the remainder of the division by 6 089 is zero. There are infinitely many multiples of 6 089. The smallest multiples of 6 089 are:\n\n## Numbers near 6 089\n\n### Nearest numbers from 6 089\n\nFind out whether some integer is a prime number" ]

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[ "# HYPGEOM.DIST function\n\nHYPGEOM.DIST(sampleNumberOfSuccesses, sampleSize, populationNumberOfSuccesses, populationSize, isCumulative) HYPGEOM.DIST(sampleNumberOfSuccesses; sampleSize; populationNumberOfSuccesses; populationSize; isCumulative)\n\n## sampleNumberOfSuccesses\n\nNumber or { Number }\n\nThe number of successes in the sample.\n\n## sampleSize\n\nNumber or { Number }\n\nThe size of the sample.\n\n## populationNumberOfSuccesses\n\nNumber or { Number }\n\nThe number of successes in the population.\n\n## populationSize\n\nNumber or { Number }\n\nThe population size.\n\n## isCumulative\n\nLogical or { Logical }\n\nFALSE to calculate the probability mass function and TRUE to calculate the cumulative distribution function.\n\n## Returns\n\nNumber or { Number }\n\nA value for a hypergeometric distribution.\n\nReturns a value for a hypergeometric distribution.\n\n## Example\n\nHYPGEOM.DIST(2, 3, 3, 6, FALSE)HYPGEOM.DIST(2; 3; 3; 6; FALSE)\n\nReturns 0.45. If an urn contains 3 red balls and 3 green balls, the probability that 2 red balls will be selected after 3 draws without replacement is 27 / 60 = 0.45.\n\nPartly derived from the OpenOffice.org documentation, licensed under the Apache License 2.0." ]

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[ "burlap.behavior.learningrate\n\n## Class ExponentialDecayLR\n\n• java.lang.Object\n• burlap.behavior.learningrate.ExponentialDecayLR\n• All Implemented Interfaces:\nLearningRate\n\n```public class ExponentialDecayLR\nextends java.lang.Object\nimplements LearningRate```\nThis class provides a learning rate that decays exponentially with time according to r^t, where r is in [0,1] and t is the time step, from an initial learning rate. A minimum learning rate value can be specified so that the learning rate is never rounded to zero. By default, the learning rate may decrease to Double.MIN_NORMAL, which is the smallest fraction a double value can hold. This class may be specified to use a universal learning rate that is shared regardless of state and action, or it can be set to have a different learning rate for each state (or state feature) that is decayed independently of other states, or it may also be specified to have a learning rate that is independently decayed for each state-action pair (or state feature-action pair). However, the state-action decay will ignore any parameterizations of actions.\nAuthor:\nJames MacGlashan\n• ### Nested Class Summary\n\nNested Classes\nModifier and Type Class and Description\n`protected class ` `ExponentialDecayLR.MutableDouble`\nA class for storing a mutable double value object\n`protected class ` `ExponentialDecayLR.StateWiseLearningRate`\nA class for storing a learning rate for a state, or a learning rate for each action for a given state\n• ### Field Summary\n\nFields\nModifier and Type Field and Description\n`protected double` `decayRate`\nThe exponential base by which the learning rate is decayed\n`protected java.util.Map<java.lang.Integer,ExponentialDecayLR.StateWiseLearningRate>` `featureWiseMap`\nThe state feature dependent or state feature-action dependent learning rates\n`protected HashableStateFactory` `hashingFactory`\nHow to hash and perform equality checks of states\n`protected double` `initialLearningRate`\nThe initial learning rate value\n`protected int` `lastPollTime`\nThe last agent time at which they polled the learning rate\n`protected double` `minimumLR`\nThe minimum learning rate\n`protected java.util.Map<HashableState,ExponentialDecayLR.StateWiseLearningRate>` `stateWiseMap`\nThe state dependent or state-action dependent learning rates\n`protected double` `universalLR`\nThe state independent learning rate\n`protected boolean` `useStateActionWise`\nWhether the learning rate is dependent on state-actions\n`protected boolean` `useStateWise`\nWhether the learning rate is dependent on the state\n• ### Constructor Summary\n\nConstructors\nConstructor and Description\n```ExponentialDecayLR(double initialLearningRate, double decayRate)```\nInitializes with an initial learning rate and decay rate for a state independent learning rate.\n```ExponentialDecayLR(double initialLearningRate, double decayRate, double minimumLearningRate)```\nInitializes with an initial learning rate and decay rate for a state independent learning rate that will decay to a value no smaller than minimumLearningRate\n```ExponentialDecayLR(double initialLearningRate, double decayRate, double minimumLearningRate, HashableStateFactory hashingFactory, boolean useSeparateLRPerStateAction)```\nInitializes with an initial learning rate and decay rate for a state or state-action (or state feature-action) dependent learning rate that will decay to a value no smaller than minimumLearningRate If this learning rate function is to be used for state state features, rather than states, then the hashing factory can be null;\n```ExponentialDecayLR(double initialLearningRate, double decayRate, HashableStateFactory hashingFactory, boolean useSeparateLRPerStateAction)```\nInitializes with an initial learning rate and decay rate for a state or state-action (or state feature-action) dependent learning rate.\n• ### Method Summary\n\nAll Methods\nModifier and Type Method and Description\n`protected ExponentialDecayLR.StateWiseLearningRate` `getFeatureWiseLearningRate(int feature)`\nReturns the learning rate data structure for the given state feature.\n`protected ExponentialDecayLR.StateWiseLearningRate` `getStateWiseLearningRate(State s)`\nReturns the learning rate data structure for the given state.\n`protected double` `nextLRVal(double cur)`\nReturns the value of an input current learning rate after it has been decayed by one time step.\n`double` `peekAtLearningRate(int featureId)`\nA method for looking at the current learning rate for a state (-action) feature without having it altered.\n`double` ```peekAtLearningRate(State s, Action ga)```\nA method for looking at the current learning rate for a state-action pair without having it altered.\n`double` ```pollLearningRate(int agentTime, int featureId)```\nA method for returning the learning rate for a given state (-action) feature and then decaying the learning rate as defined by this class.\n`double` ```pollLearningRate(int agentTime, State s, Action ga)```\nA method for returning the learning rate for a given state action pair and then decaying the learning rate as defined by this class.\n`void` `resetDecay()`\nCauses any learnign rate decay to reset to where it started.\n• ### Methods inherited from class java.lang.Object\n\n`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n• ### Field Detail\n\n• #### initialLearningRate\n\n`protected double initialLearningRate`\nThe initial learning rate value\n• #### decayRate\n\n`protected double decayRate`\nThe exponential base by which the learning rate is decayed\n• #### minimumLR\n\n`protected double minimumLR`\nThe minimum learning rate\n• #### universalLR\n\n`protected double universalLR`\nThe state independent learning rate\n• #### stateWiseMap\n\n`protected java.util.Map<HashableState,ExponentialDecayLR.StateWiseLearningRate> stateWiseMap`\nThe state dependent or state-action dependent learning rates\n• #### featureWiseMap\n\n`protected java.util.Map<java.lang.Integer,ExponentialDecayLR.StateWiseLearningRate> featureWiseMap`\nThe state feature dependent or state feature-action dependent learning rates\n• #### useStateWise\n\n`protected boolean useStateWise`\nWhether the learning rate is dependent on the state\n• #### useStateActionWise\n\n`protected boolean useStateActionWise`\nWhether the learning rate is dependent on state-actions\n• #### hashingFactory\n\n`protected HashableStateFactory hashingFactory`\nHow to hash and perform equality checks of states\n• #### lastPollTime\n\n`protected int lastPollTime`\nThe last agent time at which they polled the learning rate\n• ### Constructor Detail\n\n• #### ExponentialDecayLR\n\n```public ExponentialDecayLR(double initialLearningRate,\ndouble decayRate)```\nInitializes with an initial learning rate and decay rate for a state independent learning rate. Minimum learning rate that can be returned will be Double.MIN_NORMAL\nParameters:\n`initialLearningRate` - the initial learning rate\n`decayRate` - the exponential base by which the learning rate is decayed\n• #### ExponentialDecayLR\n\n```public ExponentialDecayLR(double initialLearningRate,\ndouble decayRate,\ndouble minimumLearningRate)```\nInitializes with an initial learning rate and decay rate for a state independent learning rate that will decay to a value no smaller than minimumLearningRate\nParameters:\n`initialLearningRate` - the initial learning rate\n`decayRate` - the exponential base by which the learning rate is decayed\n`minimumLearningRate` - the smallest value to which the learning rate will decay\n• #### ExponentialDecayLR\n\n```public ExponentialDecayLR(double initialLearningRate,\ndouble decayRate,\nHashableStateFactory hashingFactory,\nboolean useSeparateLRPerStateAction)```\nInitializes with an initial learning rate and decay rate for a state or state-action (or state feature-action) dependent learning rate. Minimum learning rate that can be returned will be Double.MIN_NORMAL. If this learning rate function is to be used for state state features, rather than states, then the hashing factory can be null;\nParameters:\n`initialLearningRate` - the initial learning rate for each state or state-action\n`decayRate` - the exponential base by which the learning rate is decayed\n`hashingFactory` - how to hash and compare states\n`useSeparateLRPerStateAction` - whether to have an independent learning rate for each state-action pair, rather than just each state\n• #### ExponentialDecayLR\n\n```public ExponentialDecayLR(double initialLearningRate,\ndouble decayRate,\ndouble minimumLearningRate,\nHashableStateFactory hashingFactory,\nboolean useSeparateLRPerStateAction)```\nInitializes with an initial learning rate and decay rate for a state or state-action (or state feature-action) dependent learning rate that will decay to a value no smaller than minimumLearningRate If this learning rate function is to be used for state state features, rather than states, then the hashing factory can be null;\nParameters:\n`initialLearningRate` - the initial learning rate for each state or state-action\n`decayRate` - the exponential base by which the learning rate is decayed\n`minimumLearningRate` - the smallest value to which the learning rate will decay\n`hashingFactory` - how to hash and compare states\n`useSeparateLRPerStateAction` - whether to have an independent learning rate for each state-action pair, rather than just each state\n• ### Method Detail\n\n• #### peekAtLearningRate\n\n```public double peekAtLearningRate(State s,\nAction ga)```\nDescription copied from interface: `LearningRate`\nA method for looking at the current learning rate for a state-action pair without having it altered.\nSpecified by:\n`peekAtLearningRate` in interface `LearningRate`\nParameters:\n`s` - the state for which the learning rate should be returned\n`ga` - the action from which the learning rate should be returned\nReturns:\nthe current learning rate for the given state-action pair\n• #### pollLearningRate\n\n```public double pollLearningRate(int agentTime,\nState s,\nAction ga)```\nDescription copied from interface: `LearningRate`\nA method for returning the learning rate for a given state action pair and then decaying the learning rate as defined by this class.\nSpecified by:\n`pollLearningRate` in interface `LearningRate`\nParameters:\n`agentTime` - the time index of the agent when polling.\n`s` - the state for which the learning rate should be returned\n`ga` - the action from which the learning rate should be returned\nReturns:\nthe current learning rate for the given state-action pair\n• #### peekAtLearningRate\n\n`public double peekAtLearningRate(int featureId)`\nDescription copied from interface: `LearningRate`\nA method for looking at the current learning rate for a state (-action) feature without having it altered.\nSpecified by:\n`peekAtLearningRate` in interface `LearningRate`\nParameters:\n`featureId` - the state feature for which the learning rate should be returned\nReturns:\nthe current learning rate for the given state feature-action pair\n• #### pollLearningRate\n\n```public double pollLearningRate(int agentTime,\nint featureId)```\nDescription copied from interface: `LearningRate`\nA method for returning the learning rate for a given state (-action) feature and then decaying the learning rate as defined by this class.\nSpecified by:\n`pollLearningRate` in interface `LearningRate`\nParameters:\n`agentTime` - the time index of the agent when polling.\n`featureId` - the state feature for which the learning rate should be returned\nReturns:\nthe current learning rate for the given state feature-action pair\n• #### resetDecay\n\n`public void resetDecay()`\nDescription copied from interface: `LearningRate`\nCauses any learnign rate decay to reset to where it started.\nSpecified by:\n`resetDecay` in interface `LearningRate`\n• #### getStateWiseLearningRate\n\n`protected ExponentialDecayLR.StateWiseLearningRate getStateWiseLearningRate(State s)`\nReturns the learning rate data structure for the given state. An entry will be created if it does not already exist.\nParameters:\n`s` - the state to get a learning rate for\nReturns:\nthe learning rate data structure for the given state\n• #### getFeatureWiseLearningRate\n\n`protected ExponentialDecayLR.StateWiseLearningRate getFeatureWiseLearningRate(int feature)`\nReturns the learning rate data structure for the given state feature. An entry will be created if it does not already exist.\nParameters:\n`feature` - the state feature id to get a learning rate for\nReturns:\nthe learning rate data structure for the given state feature\n• #### nextLRVal\n\n`protected double nextLRVal(double cur)`\nReturns the value of an input current learning rate after it has been decayed by one time step.\nParameters:\n`cur` - the currently learning rate to be decayed by one time step.\nReturns:\nthe value of an input current learning rate after it has been decayed by one time step." ]

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[ "", null, "# complexity\n\nhttps://dragif.com/pdf/aZjq\n\nPDF file size:\n234.56 kB\n\nSource:\nQueensu.ca\nFree PDF file download:", null, "### Transcript text content of a PDF document complexity.pdf:\n\nKolmogorov Complexity\n\nComputational Complexity Course Report\n\nHenry Xiao\n\nQueen’s University\n\nSchool of Computing\n\nMarch 2004\n\nIntroduction\nIn computer science, the concepts of algorithm and information are fundamental. So the\nmeasurement of information or algorithms is crucial in sense of describing. In 1965\nAndrey Nikolaevich Kolmogorov [O'Connor, and Robertson, 1999sian\nmathematician, established the algorithmic theory of randomness via a measure of\ncomplexity, now referred to Kolmogorov complexity. According to Kolmogorov, the\ncomplexity of an object is the length of the shortest computer program that can reproduce\nthe object. All algorithms can be expressed in programming language based on Turing\nmachine models equally succinctly, up to a fixed additive constant term. The remarkable\nusefulness and inherent rightness of the theory of Kolmogorov complexity or so called\nDescriptive complexity, stems from this independence of the description method.\nThe idea of Kolmogorov complexity first appeard in the 1960’s in papers by\nKolmogorov, Solomonoff and Chaitin. As specified by Schöning and Randall, an\nalgorithm can exhibit very different complexity behavior in the worst case and in the\naverage case. The Kolmogorov complexity is defined a probability distribution under\nwhich worst-case and average-case running time (for all algorithm simultaneously) are\nthe same (up to constant factors). Quick sort algorithm has been widely taken as an\nexample to show the applicability of Kolmogorov complexity since the algorithm takes\n\nO(n logn) time in average but Ω(n\n) time at worst case. Later, the Kolmogorov\ncomplexity was connected with Information Theory and proved to be closely related to\nClaude Shannon's entropy rate of an information source. The theory base of Kolmogorov\ncomplexity has also be extended to data compression and communication for the sake of\ntrue information measure.\n\nKolmogorov Complexity Theory\nWe will briefly take a look at Kolmogorov Complexity definition and some main related\nresults at this section. For details, please refer to [Cover and Thomas, 1991\n\n[Schöning and Pruim 1998\n\nDefinition: The Kolmogorov Complexity Ku(x) of a string x with respect to a universal\ncomputer U is defined as:\n\nKu(x) = min l(p),\n\np : U(p) = x\nthe minimum length over all programs that print x and halt. Thus Ku(x) is the shortest\ndescription length of x over all descriptions interpreted by computer U. (Note Turing\nmachine is regarded as universal computer in computer science.)\nThe concept of Kolmogorov Complexity asks for the minimal unambiguous\ndescription of a sequence. It can be used to prove complexity lower bounds. [Schöning\n\nand Pruim 1998\nples of using this technique. And indeed, as mentioned,\nthe proofs obtained in this way are much more “elegant”, or at least shorter, than the\noriginal proofs. In a few cases, the lower bounds were first achieved by means of\nKolmogorov Complexity. Another quite important definition is the conditional\nKolmogorov complexity which is based on the knowledge of the length of x, denoted as\nl(x).\n\nKu(x| l(x)) = min l(p),\n\np : U(p, l(x)) = x\n\n3 This is the shortest description length if the computer U has the length of x made\navailable to it. Quite a few different results have been shown for conditional Kolmogorov\ncomplexity too.\nWe look at some basic and interesting properties of Kolmogorov complexity and\nthen consider some examples. The proofs of the theorems are eliminated because of the\npurpose of this report. However, for your interests, please refer to [Cover and Thomas,\n1991] for complete proofs.\nBoth the lower and upper bounds of Kolmogorov complexity of a given sequence\nhave been derived early. There are some choices of both bounds from different aspects.\nThe followings are all established theorems for bounds.\n\nUniversality of Kolmogorov complexity: If U is a universal computer, then for any other\n\ncomputer A: Ku(x) ≤ K\n\n(x) + c\nfor all string x ∈ {0, 1}*, where the constant c\n\ndoes\nnot depend on x.\nConditional complexity is less than the length of sequence: Ku(x| l(x)) ≤ l(x)+ c.\nUpper bound on Kolmogorov complexity: Ku(x) ≤ Ku(x| l(x)) + 2log (l(x)) + c.\nLower bound on Kolmogorov complexity: The number of strings x with complexity\nKu(x) ≤ k satisfies: |{ x ∈ {0, 1}*: Ku(x) ≤ k } | < 2\nThe Kolmogorov complexity of a binary string x is bounded by:\n\nK(x\n\n…x\n\n|n) ≤ nH\n\n(1/n∑\n\n) + 2log n + c,\n\nwhere H\n(p) = -p log p – (1-p) log (1-p) is the binary entropy function.\n\n4 The above five theorems are basically considered to be very important facts of\nKolmogorov complexity. Many other interesting results have been derived applying these\ntheorems. We consider some examples of Kolmogorov complexity here to show the\nusability of the theory with its properties. First, we will look at some intuitive ideas\ndirectly coming from the definition. A sequence of n zeros has a constant Kolmogorov\ncomplexity, i.e. K(000…0|n) = c, since if we assume n is known, the a short program can\ndirectly print out n zeros. The same case can be applied to\n\nπ, where the first n bits of\nπcan be calculated using a simple series expression. A somewhat surprising result for\nfractal is that regarding its complex calculations, it is still essentially very simple in terms\nof Kolmogorov complexity which is nearly zero. An integer on the other hand, has higher\ncomplexity even it looks very straight forward. It is obviously true that the complexity of\ndescribing an integer will be constant if we know the length of the integer, i.e. K(n|l(n)) =\nc. However, in general, the computer does not know the length of binary representation\nof the integer. So we must inform the computer in some way when the description ends.\nWe can bound the description using the upper bound we got so far: K(n) ≤ 2log n + c.\nWe can also prove there are an infinite number of integers n such that K(n) > log n. This\nis probably less intuitive than we thought. From above examples, it is not hard to see the\ntrue measurement of information or algorithm would be rather hard without the\ndevelopment of Kolmogorov complexity. With the support of the Kolmogorov\ncomplexity theory, one can describe information more accurate in computer science.\nKolmogorov complexity also applied to algorithmically random, incompressible\nsequences, the halting problem, etc. We can not go into details of those examples; instead,\nwe briefly look at those problems here just to get some taste. Many literatures have been\n5 published on Kolmogorov complexity related questions. One can refer to [Li, and Vitanyi,\nincompressible sequences are defined based on the Kolmogorov complexity properties\nthat some sequences hold. We say a sequence x\n\n, x\n\n, x\n\n… x\n\nalgorithmically random if\n\nK(x\n\n…x\n| n) ≥ n. And we can say a string x incompressible if lim K(x\n\n…x\n\n| n)/n =\n1. The definitions seem very intuitive with respect of Kolmogorov complexity. In fact, if\nevery element of the sequence is completely generated in random, we can not predict any\nlater elements from current. Indeed, we will need to describe each element separately.\nHowever, if the Kolmogorov complexity verse n approaching 1 for a string in probability,\nthen we can actually interpret this as the proportions of 0’s and 1’s in this string are\nalmost equal, which is ½. By this meaning, it is true that we can not compress the string\nsince any bit will be a critical contribution to the whole, which also specifies the\nrandomness of the string. The next significant application is on the halting problem.\nUsing Kolmogorov complexity, we can actually demonstrate that the problem can not be\nsolved by an algorithm because of the non-computability of Kolmogorov complexity. It\nis rather a surprising fact of the non-computability. However, practically speaking, one\nmay never be able to tell the shortest program since there are infinite many programs for\na given sequence. We can only estimate the complexity by running more and more\nprograms, as we know the bound will converge to the Kolmogorov complexity. Many\nother results can also be found on published literatures related with probability theory and\ninformation theory. At the following section, we take a look at one of the most important\nresults related with the central idea of information theory – Entropy.\n\nKolmogorov Complexity and Entropy\nAs mentioned at introduction, the Kolmogorov complexity and entropy of a sequence of\nrandom variables are highly related. In general, the expected value of the Kolmogorov\ncomplexity of a random sequence is to its entropy.\nFrom information theory founded by Shannon, the true measure of information on\nrandom variables is entropy. This relationship actually proves the correctness of the\nKolmogorov complexity as a measurement of information and algorithms. Kolmogorov\ncomplexity states the shortest description (program) for a random variable. Then\ncomplexity of the sequence constructed by random variables will approach to the expect\nvalue of the set of Kolmogorov complexities for each variable, i.e. E[1/nK(X\n\n|n)] →H(X),\nsupported by the law of large numbers in probability. Respectively, the information\nmeasure of the sequence is indeed entropy. Actually, one can always show the program\nlengths satisfy prefix condition, since if the computer halts on any program, it does not\nlook any further for input. The relationship can be shown further with Kraft inequality.\nThe relationship provides us to two ways of complexity measures, which either\ntakes after Kolmogorov complexity, involving finding some computer or abstract\nautomaton which will produce the pattern of interest, or take after information theory and\nproduce something like the entropy, which, while in principle computable, can be very\nhard to calculate reliably for experimental systems. This is actually very powerful\nprinciple in physics (see [Li, and Vitanyi, 1999ore interesting idea\nabout “Occam’s Razor” as a general principle governing scientific research can be\nderived from Kolmogorov theory as we will see at the next section.\n\nOccam’s Razor\nI decide to include this topic mainly because its importance plus the idea really amazed\nme. At 14th century, William of Occam (Ockham in some literatures), a logician, said\n“Nunquam ponenda est pluralitas sine necessitate”, i.e., explanations should not be\nmultiplied beyond necessity, which forms the basis of methodological reductionism. Here,\nour argument will be a special case of it.\nRecent papers have suggested a connection between Occam's Razor and\nKolmogorov complexity. Many literatures take Laplace’s sun rising problem as an\nexample to explain the connection. Laplace considered the probability that the sun will\nrise again tomorrow, given that it has risen every day in recorded history. He solved it\nbefore Kolmogorov complexity was introduced. However, the problem can be\nreconsidered through Kolmogorov complexity. If we use 1 to represent the sun rise, then\nthe probability that the next symbol is a 1 given n 1’s in the sequence so far is: ∑\n\np(1\n\n1y) ≈ p(1\n\n) = c > 0. And the probability that the next symbol is 0, which means that\n\nthe sun will not rise: ∑\n\np(1\n\n0y) ≈ p(1\n\n0) ≈ 2\n\n-log n\n\n, since any 1\n\n0… yields a description\nof n with length at least K(n), i.e. about log n + O(log log n). Hence the conditional\nprobability of observing a 0 next is: p(0|1\n\n) = p(1\n\n0) / (p(1\n\n0) + p(1\n\n))≈ 1/(cn+1). The\nresult is very similar to 1/(n+1) derived by Laplace. The Kolmogorv complexity solution\nto this question is actually following the Occam’s Razor by weighting possible\nexplanations by their complexity.\nIt often happens that the best explanation is much more complicated than the\nsimplest possible explanation because it requires fewer assumptions. Albert Einstein\n8 wrote in 1933 “Theories should be as simple as possible, but no simpler.” In our case,\nKolmogorov complexity does provide us an alternative approach to explain things in\nmany science fields.\n\nConclusion\nKolmogorov complexity is a profound theory for information and algorithm measure.\nThe theory is somehow different from others that we have studied in computational\ncomplexity so far. I feel the theory tries to observe the complexity from a new approach.\nunderstand his original idea, which he developed from mathematical perspective. It is in\nhigh level of abstraction, but closely related with many things in practice. Vladimir\nV’yugin presents some applications of Kolmogorov complexity in his review [V’yugin,\n\n1994\n\nthem\n\natically. In com\nputing, I found many ongoing topics related with\nKolmogorov complexity are on information process ([Levin, 1999\nDowe, 1999a]). Generally, the application of Kolmogorv complexity is based on\nframework of the Minimum Description Length (MDL) principle and Minimum Message\nLength (MML) principle, which are out the scope of this report, but refer to [Wallace, and\nDowe, 1999b] for introductions.\nI strongly believe the Kolmogorov complexity should have more appearances in\nfuture research topics. Along with information theories, we need to deal with information\nas well as problems coming with it. Like the Shannon’s entropy theory established\ntoday’s communication system, the closely related Kolmogorov complexity shall also be\nof great potential to be applied to future researches.\n\nReference\n\nCover, M. Thomas, and Thomas, A. Joy. Elements of information theory. John Wiley &\nSons, Inc. 1991.\n\nGammerman, Alexander, and Vovk, Vladimir. Kolmogorov complexity: sources, theory\nand applications. The Computer Journal, vol. 42, no. 4, 1999.\n\nLevin, A. Leonid. Robust Measures of Information. The Computer Journal, vol. 42, no. 4,\n1999.\n\nLi, Ming, and Vitanyi, Paul. An Introduction to Kolmogorov Complexity and Its\nApplications. Springer-Verlag, New York, 1999.\n\nO'Connor, J J, and Robertson, E F. “Andrey Nikolaevich Kolmogorov”. School of\nMathematics and Statistics, University of St. Andrews, Scotland, 1999.\n\nRissanen, Jorma. Discussion of paper `Minimum Message Length and Kolmogorov\nComplexity' by C. S. Wallace and D. L. Dowe. The Computer Journal, vol. 42, no. 4,\n1999.\n\nSchöning, Uwe, and Pruim, Randall. Gems of theoretical computer science. Springer-\nVerlag Berlin Heidelberg 1998.\n\ncomplexity\n\nPDF file: http://research.cs.queensu.ca/home/xiao/doc/complexity.pdf\n\nMicrosoft Word - cisc871rep.doc (xiao)", null, "language version:", null, "", null, "" ]

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[ "## Sunday, May 18, 2008\n\n### Elearn Geometry Problem 4", null, "See complete Problem 4\nTriangle, Quadrilateral, Angles, Congruence. Level: High School, SAT Prep, College geometry\n\n1.", null, "Trazamos el segmento BD\n\nEl triangulo BDC es isosceles => angulo(DBC)=angulo(BDC)=(180-2A)/2= 90-A\n\nEn el triangulo BDC ; 1/sen(90-A)=BD/sen(2A) => BD=2 sen(A)\n\nEn el tringulo ABD ; 1/sen(ABD)=2sen(A) / sen (A) => sen(ABD)= 1/2 => angulo (ABD) =30\n\nAngulo(x)=angulo(DBC)+angulo(ABD)=90-A + 30= 120 - A\n\n2.", null, "It is possible to provide a proof using only elementary geometry!\n\n3.", null, "Let point E be a reflection of point D over AB.\nThen AEB and ADB triangles are congruent\nAngle EAB = angle DAB = alpha => angle EAD = 2 alpha => triangles AED and CBD are congruent (AE = AD = DC = BC) => ED = DB\nBut BE = DB (because AEB and ADB are congruent) => EBD is an equilateral triangle.\n\nAngle EBD = 60 => angle ABD = 30 (since ABD = ABE)\nAngle DBC = 90 - alpha (since DCB is an isosceles triangle)\n\nx = angle ABD + angle DBC = 30 + (90 - alpha) = 120 - alpha.\n\n4.", null, "This comment has been removed by a blog administrator.\n\n5.", null, "http://mate-facil.co.cc/35/\nBy TiNo\n\n6.", null, "Let the bisector of <DCB intersect AB at X. Now since <XCA = <XAD = a, AD = CD and XD is common, by SAS, AXD is congruent to XCD. Hence <DXC = <DXA. Since X is on the perpendicular bisector of BD it also follows that <DXC = <BXC. Therefore since <DXC = <DXA = <BXC and all three angles are supplementary, <BXC = 60. Hence <XBD = 30. Now, <XBC = <XBD + <DBC = 30 + 90 - a = 120 - a.\n\n7.", null, "http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf\n\n8.", null, "Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 4. Thanks Eder.\n\n9.", null, "10.", null, "Proof:\nJoin AC\nThrough D draw a line perpendicular to AC and meet AC at M and meet AB at E.\n∴ EM is the perpendicular bisector of AC\n∴ AE = EC and ∠AEM = ∠CEM\n∴ ∠EAD = ∠ECD = α\n∵ DC = CB\n∠DCE = ∠BCE = α\nΔ DCE ≡ Δ BCE\n∠CEM = ∠CEB\n∵ ∠AEM = ∠CEM = ∠CEB = 60o\n∵ In ΔBCE, x + α = 120o\n\n11.", null, "Join BD.\nLet E be the circumcentre of ∆ABD.\n∠DEB = 2∠DAB = 2α.\n∆s DEB, DCB are congruent (Each isosceles, common base BD,\nequal vertical angles, each 2α)\nFollows EA = ED = DC = DA, ∆EAD is equilateral, and ∠AED = 60°\nSo ∠ABD = (1/2)∠AED = 30°\nObserve that ∠DBC = 90° - α\nHence x = 30° + (90° - α) = 120° - α\n\n12.", null, "hallo my dear brothers and sisters ...... here is a simple solution for this problem\nlets start\njoin A,C. join B,D.\ndraw an angle bisector of angle DCB.it intersects DB at the point F and\nalso intersects AB , at the point E.\nnow consider triangle DBC\nCD=CB (given)\nangle DCF=angle FCB=alpha\nangle FDC = angle FBC = (180- 2 alpha)/2=90-alpha\nso angle DFC = angle BFC = 90\nso angle EFD = angle EFB = 90\nnow consider two triangle DFC and BFC\ncf is common\nangle DCF=angle FCB=alpha\nangle DCF=angle FCB=alpha\ntriangle CDF is congruent with triangle CBF\nso FD = FB\n\nnow consider two triangle DEF and EFB\nEF is common\nFD = FB\nangle EFD = angle EFB = 90\ntwo triangle DEF and EFB are congruent with each other\nso angle BEF = angle FED............... (1)\nthis is an isosceles triangle\nso angle DAC = angle DCA\nconsider triangle AEC\nangle EAC = angle EAD + angle DAC\nangle ECA = angle ECD + angle DCA\nangle EAD = angle ECD = alpha\nso angle EAC = angle ECA\nso triangle AEC is isosceles triangle\nso AE = EC\nnow consider two triangle AED and CED\nAE=CE\nED is common\ntriangle AED and CED are congruent\nso angle AED = angle FED ..........(2)\nfrom (1) and (2)\nangle BEF = angle FED = angle AED = 180/3 =60\nso angle EBF = 90 - 60 = 30\nANGLE ABC= angle EBF + angle FBC\n= 30 +90 - alpha = 120 - alpha\n\n13.", null, "Draw an angle bisector of angle BCD and let it meet at the point E on the side AB\nThen connect E and D.\nSince In Tri ACD , angle ACD = angle CAD\nangle ECD = angle EAD = alpha\nCE = AE ( Isosceles tri AEC )\nSince In Tri BCE and Tri ADE\nBC = AD ( given )\nangle BCE = angle EAD = alpha\nCE = AE (proved)\nTri BCE congruent to Tri ADE. This implies x = angle CBE = angle ADE\nSince in Tri CDE and in Tri ADE\nCE = AE ( proved )\nCD = AD ( given )\nTri CED congruent Tri ADE , angle CDE = angle ADE = x\nSince Tri BCE , Tri CDE , Tri ADE are congruent to each other\nangle BEC = angle CED = angle AED\nand BEC + CED + AED =180 degree\n3(AED) = 180 degree\nAED = 60 degree\nIN Tri CEA , angle CEA = 2(AED) angle = 120 degree. Since it Is isosceles\nangle ECA = angle EAC = 60 degree\nIN Tri CDA , angle DCA = angle DAC = 60 - alpha , and angle CDA = 120 + 2(alpha)\nAND FINALLY\nANGLE CDA = 360 - (ANGLE EDC + ANGLE EDA)\nANGLE CDA = 360 - (ANGLE EDC + ANGLE EDC) ( EDC = EDA)\nANGLE CDA = 360 - ANGLE 2(EDC)\nANGLE CDA = 360 - 2X (EDC = X)\n120 + 2(ALPHA) = 360 - 2X (ANGLE CDA = 120 + 2(ALPHA))\n2X - 360 = - 2(ALPHA) - 120\n2X = 360 - 120 - 2(ALPHA)\n= 240 - 2(ALPHA)\n= 2( 120 - ALPHA)\nTHEREFORE X = 120 - ALPHA\n\n14.", null, "Let the bisector of < DCB meet AB at E and BD at F. Draw altitude DG of Tr. ABD.\n\nTr. s BCF & DCF are congruent and these are also congruent with Tr. ADG. So DG = DF = FB hence in right Tr. BGD BD = 2DG which means that < ABD = 30 and since < DBC = 90-@ the result follows\n\nSumith Peiris\nMoratuwa\nSri Lanka\n\n15.", null, "16.", null, "Construct point E such that E lies on AB and EC is an angle bisector of <DCB. Hence, <DCE=<BCE.\nConstruct line ED.\nTriangle DCE is congruent to BCE. Hence <EDC=x\nHence, EA=EC\nHence triangle EAD and ECD are congruent.\nHence <EDA=x\nBy angle sum of quadrilateral 3x+3a=360°\nSo x+a=120°\nx=120°-a\nProved\n\n17.", null, "Grab and pen and a piece of paper, and illustrate accordingly.\n\nAdd point E to the figure and connect it to line C. Do so in such a way that line CE is the angle bisector of angle BCD. Next, connect point D to point E. That is all for the illustrations.\n\nWe can see that:\n\n> Triangles CDE and CBE are congruent because\n~ They are share line CE\n~ Line CD = line BC\n~ Angle DCE = angle BCE\n\n> Triangles CDE and ADE are congruent because\n~ Due to two equal opposite angles and equal adjacent lengths, quadrilateral ADCE is a concave kite.\n~ They share side DE\n~ This is a special case where SSA works, but only because they make up a kite.\n\n> Since triangles ADE, CDE and CBE are congruent, angle AED = angle CED = angle CEB.\n> Angle AED = angle CED = angle CEB = 180 / 3 = 60\n\nNow, we solve the problem:\n\nJust focus on triangle CBE. The three angles in a triangle add up to 180.\nx + a + 60 = 180\nx + a = 120\nx = 120 - a\n\n18.", null, "Let angle bisector of m(BCD) meet AB at E.\nConnect ED and since given AD=DC=BC=>ED is perpendicular to AC, AEC is an isosceles and BCDE is a kite => AE=EC and EB=ED\nObserve that the triangles DAE, DCE and BCE are congruent (SAS)\nHence m(AED)=m(DEC)=m(CEB)=180/3=60 =>m(DBE)=30\nNow its easy to see x=30+90-alpha=120-alpha\n\n19.", null, "For easy typing, I use z instead of \"alpha\" for the following\nangle CDB=90-z\nConsider triangle CDB\nsin2z/BD=sin(90-z)/CD\nCD=BDcosz/sin2z\nCD=BD/2sinz" ]

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[ "Ana səhifə\n\n# A preliminary draft\n\n tarix 24.06.2016 ölçüsü 138.5 Kb.\n\nRagnar Arnason*\n\nResource Rent Taxation: Is it Really Nondistortive?\nA paper presented at the\n\n2nd World Congress of Environmental and Resource Economists\n\nMonterey June 24 - Jun 27, 2002\n\nA Preliminary\n\nDRAFT\n\n* Department of Economics\n\nUniversity of Iceland\n\[email protected]\n\nhttp://www.hi.is/~ragnara\nABSTRACT\nIt is often taken for granted that taxation of rents is economically nondistortive. In certain areas of natural resource use, e.g. oil extraction and fisheries, this nondistortion principle has been used to justify taxation of what is regarded as resource rents. This paper challenges the view that such taxation is generally nondistortive. Within the framwork of a general model of natural resource extraction, the paper argues that taxation of resource rents will in general affect the time profile of natural resource extraction. The paper, moreover, argues that through its impact on exit and entry, resource rent taxation will generally affect the number and composition of firms in the industry and may in this way have a secondary efficiency impact.\n\n0. Introduction\nIt is a common belief that taxation of natural resource rents is economically non-distortive and, as a result, superior to most other methods of raising government revenues. This presumption, combined with non-economic social sentiments, seems to have prompted several economists to recommend special taxes on various resource extraction activities including mining (Garnault and Clunies-Ross 1975, Fraser 1993, 1998, Miller et al. 2000) and fisheries (Grafton 1995, 1996). Similar considerations have apparently encouraged many governments (including for instance the UK, USA, Norway and New Zealand) to impose special heavier taxation on natural resource use (Miller et al. 2000).\nIt is interesting to note that the historical roots of this belief can be traced to the land tax initially proposed by James Mill (John Stuart’s father) on the theoretical basis laid by Ricardo (1821) and subsequently popularized by Henry George (1879) the initiator of the so-called Georgeism (Blaug, 1996). The basic justification for the land tax in the 19th Century was essentially the same as for the resource rent tax in modern times. Land was seen as fixed and indestructible. Its use generated land rents. Taxing these rents would not reduce land use. Therefore, land provided an ideal tax base from an economic perspective. This argument was further strengthened by the moral issue. Land was seen to belong to all  not just the formal owners of land. Hence, taxing land to finance public expenditures also seemed ethically attractive to many people.\nThese views have met with considerable practical success. Natural resource use (esp. land use, mining and lumbering) is now often more heavily taxed than other economic activities. In many countries natural resource extraction is, as a matter of course, subject to royalties, special income taxes, rent taxes and cost recovery charges over and above what is the rule for other industries. And when it comes to natural resource industries which are not already heavily taxed, considerable pressure is being exerted to impose such taxation or raise those already in place.1\nGiven all this, it seems in order to examine whether and to what extent taxation of natural resource rents is in fact non-distortive. This paper contributes to that task. To do so successfully, however, we must first define the concept of resource rents.\nThe concept of resource rents, or, for that matter, economic rents in general, has been somewhat loosely used in economic writings. For instance, in many influential papers on resource rent taxation (see e.g. Garnault and Clunies-Ross 1975 and 1979), resource rents seem to be used almost synonymously with profits. For our purposes, however, it is necessary to be completely clear about the concept. Armen Alchian in the New Palgrave Dictionary of Economics (1987) essentially defines economic rents as the payment (imputed or otherwise) to a factor in fixed supply. Alchian illustrates his definition with the familiar diagram in Figure 1 often used to illustrate Ricardo’s theory of land rents.\n\nFigure 1\n\nEconomic Rents", null, "In this diagram, the market-clearing price is p. However, since the quantity of the factor is assumed fixed, the corresponding supply, q, would be forthcoming even if the price were zero. Hence, the entire price, p, may be regarded as a surplus. The total surplus or economic rent attributable to the limited factor is the rectangle pq.\n\nFor later purposes it is useful to note that economic rents can also be written as D(q)q, where D(q) represents the value of the demand function at q. It is well known (see e.g. Varian 1984) that in competitive markets when the factor is used for production purposes D(q) represents the marginal profits of using the factor. When, on the other hand, the factor is used directly for consumption D(q) would be proportional to the marginal utility of consuming the factor.\nNote that the economic rents depicted in Figure 1 also represent total profits2 to the owner of the factor in fixed supply. It doesn’t, however, represent the total economic benefits of the supply q. This is measured by the sum of economic rents and the demanders’ surplus represented by the upper triangle in the diagram. Thus, if the demanders are producers extracting (but not buying) a natural resource, their profits would be the sum of economic rents and the demanders’ surplus.\nThe concept of economic rents relies heavily on the assumption of a factor in fixed supply. If there is no such factor, economic rents are really not defined.3 While theoretically non-problematic, the empirical relevance of factors in fixed supply may be questioned. After all it is in the nature of the economic activity to find ways to adjust supply to demand, particularly when profits can be made doing it. Even, Ricardo’s (1821) argument in terms of the “original and indestructible powers of the soil” does not ring true. Surely, modern technology has enabled us to both damage and enhance these powers. Thus, it turns out to be difficult to find examples of factors that are truly in fixed supply especially in the long run.\nIn the very short run, on the other hand, many factors are in fixed supply and, consequently capable of earning economic rents. To represent this phenomenon of transient or temporary economic rents, Marshall (according to Achian 1987) apparently initiated the concept of quasi-rents.\nThis general theory of economic rents can easily be extended to natural resource extraction, at least in a formal sense. Let q represent the volume of natural resource extraction and (q) the associated profits. Then clearly the derived demand for q will be q(q) and the natural resource rents q(q)q.\nThe question of truly fixed supply, however, is no less pertinent to natural resource extraction than to other factors. In fact, for most natural resources it is technically possible and often economically advisable to vary the level of extraction over time. Moreover, at a point of time and for a given level of the resource, variations in the volume of extraction often involve costs, marginal user costs, giving rise to an upward sloping supply curve. Thus, in the case of natural resource extraction, it is often difficult to maintain the assumption of fixed supply. Nevertheless, this assumption seems to be at the root of the widespread belief that resource rent taxation is non-distortive.\nNatural resource extraction is often associated with economic inefficiencies due to imperfect property rights resulting in externalities. In this paper we ignore complications of this nature. More to the point, we will assume that without taxation resource extraction is fully efficient. Taxation for management purposes, i.e. in order to further efficiency, is not a part of this study. Here we only consider taxation as a means to generate government revenues. Also, in this paper, we do not directly deal with the question of whether there exist neutral taxes. Our concern is only with the distortive or non-distortive properties resource rent taxes.\nThe rest of this paper is organized as follows: In the next section, section 1, a model of a natural extraction general enough to cover both renewable and non-renewable resources is presented. The following section examines the impact of resource rent taxation in a general setting concluding that such taxation would in general be economically distortive. Then, in section 3, numerical examples for non-renewable and renewable resource extraction industries are worked out. Finally, in section 4, the results of the paper and possible extensions are discussed.\n\n1. The natural resource extraction industry\n\nConsider a natural resource extraction industry characterized by the instantaneous profit function:\n\n1. (q,x), defined for q,x0,\n\nwhere q denotes resource extraction and x the stock of the resource both at time t. The profit function is taken to have the usual properties, i.e., a positive level of both arguments is needed for positive profits, to be monotonically increasing in the stock variable, x, to be increasing in the flow variable, q, up to a point, and to be concave. More formally: (0,x)=(q,0)0, x(q,x)>0 and q(q,x)>0 if q<. For analytical convenience it is, moreover, assumed that the profit function is differentiable as needed.\n\nIn what follows, we will variously refer to (q,x) as applying to individual firms and to the industry as a whole. To be able to do this consistently obviously requires considerable restrictions on the structure of the industry.4 However, explicit modelling of individual firms would substantially complicate the analysis with little additional insight for this inquiry.\nThe resource is assumed to evolve according to the differential equation:\n\n1.", null, "=G(x)-q, defined for x0,\n\nwhere G(x) is the renewal function of the natural resource having the usual continuity and concavity properties and a point x1>0 such that such that G(x1)=0. Obviously, if the resource is non-renewable, G(x)0,  x. If the resources is renewable,  x such that G(x)>0. As the (q,x) function, the function G(x) is assumed to be as differentiable as needed.\n\nThe firms in the industry, and, consequently, the industry as a whole, are assumed to seek to maximize the present value of profits. For this purpose they can decide to be active and, if active, select a time path of extraction, {q}. Formally this problem can be expressed as:\n\n1.", null, ",\n\nSubject to:", null, "= G(x)-q\n\nx(0) = x0\n\nx, q,T  0.\nThe terminal time of the extraction activity is T. So choosing a finite T is equivalent to becoming inactive, i.e. leaving the industry. It is important to note that becoming inactive is qualitatively different from setting the extraction rate to zero. x(T) is the resource stock left at the terminal time. In the case T is finite and the resource is not completely exhausted, this will have to be determined. Finally, x0 is simply the stock of the resource available at the outset of the maximization programme.\nAccording to the maximum principle (Pontryagin et al. 1962, Leonard and Long 1992). The necessary (and in this case sufficient) conditions for solving problem (I) include:\n\n1. q -  0, q  0, (q - )q = 0,\n\n2.", null, "- r = -x - Gx,\n\n3.", null, "= G(x)-q,\n\n4. H(T) = (q(T),x(T)) + (T)(G(x(T))-q(T))=0,\n\n5. (T)  0, x(T)  0, (T)x(T)=0.\n\nExpressions (3.1)-(3.5) describe the behaviour of a profit maximizing resource extraction industry when there is no taxation. Comparing these conditions with the corresponding ones under taxation provides an indication of whether the taxation is distortive or not. It should be noted that since the industry takes prices as exogenous, conditions (3.1)-(3.5) also represent a social optimum if, as is usually assumed, the prevailing prices are true. It follows that any shift from these conditions represents a movement to a socially inferior position.\n\nThe above model is quite simplistic. In addition to its glossing over the aggregation issue it omits any consideration of physical capital and, consequently, investment therein. Moreover, it ignores the existence of uncertainty that constitutes an important aspect of most resource extraction industries. These simplifications, however, seem of little consequence for the subject of this study. If resource rent taxes turn out to be distortive in this simple models, it seems obvious that they will also be so in more realistic models where the scope for distortion is greater. If, on the other hand, resource rent taxes are found to be non-distortive in this simple model, they may, of course, still be distortive in a more realistic setting.\nNow, as discussed in the previous section, resource rents may be are defined as\n\n1. R(q,x) = D(q,x)q = q(q,x)q,\n\nwhere q(q,x) is the derived demand for the extracted resource. Note that these resource rents are instantaneous rents. They refer to a point in time. Resource rents for the extraction programme as a whole would be given by the present value of the complete time path of rents.\n\nIn the resource extraction industry defined above, the supply price of the resource at quantity q is given by the co-state variable or shadow price, . This, as shown by the conditions, (3.1)-(3.5), above is a function depending on the state of the resource, x, and the level of extraction, q, as well as the other variables of the problem. If positive extraction is optimal there is, at each point of time, a supply/demand equilibrium defined by the equation q= in expression (3.1). It follows that for a resource extraction industry, we may draw a resource rent diagram corresponding to the conventional one in Figure 1.\n\nFigure 2\n\n## A Resource Extraction Industry: Resource Rents", null, "As the supply curve of q (really the shadow price ) is drawn in Figure 2, the area referred to as “Resource rents” does not appear to be economic rents at all, although parts of it may represent a producer’s surplus (in this case resource owner’s surplus). Note, however, that  is merely an imputed or notional price. It represents the opportunity cost of reducing the size of the resource, sometimes referred to as a user cost (Scott 1955). It does not represent outlays of money. Thus, in a certain sense it is not marginal cost at all. It is certainly not a marginal cost in the sense of Ricardo and the definition of economic rents discussed in the previous section. Thus, the multiple q appears to represent economic rents in the traditional sense and this is the way we will regard it in this paper. In any case this multiple seems the closest parallel to economic rents that can be found in a resource extraction industry.\n\nFrom the perspective of resource rent taxation, however, the crucial message of equation (4) is that resource rents are a function of both the extraction rate and the level of the resource as well as of other variables entering but not explicit in the profit function. We refer to this as result 1.\nResult 1\n\nResource rents depend in general on extraction rates and the level of the resource.\n\nSo, the supply q giving rise to the resource rents q is not at all fixed in the traditional sense of Ricardo (1821) and Alchien (1987). It is the outcome of profit maximization by the resource extractor taking into account the state of the resource and other exogenous variables entering the profit function such as prices. Clearly, if these exogenous variables change, so will the optimal extraction quantity and hence resource rents.\nGiven this, it is useful to deduce as much as possible about the shape of resource rent function, R(q,x). Clearly, Rq(q,x)=D(q)(Dq(q)q/D(q) +1). So, the effect of increased extraction on rents is positive if the elasticity of demand5 is less than unity and vice versa. By the same token rents are maximized at the level where the elasticity of demand equals unity. Moreover, if qqq0, R(q,x) will be concave in q. Finally, Rx(q,x)>0 iff qx(q,x)>0.\nGiven that some level of resource extraction is profitable resource rents are nonnegative. We refer to this result as result 2.\nResult 2\n\nAssuming that extraction is profitable resources rents in the resource extraction industry defined by (1) and (2) are nonnegative.\n\nProof:\n\nIf extraction is profitable, the optimal extraction q*>0. Therefore, q(q*,x)= according to (3.1). It is well known (see e.g. Leonard and Long 1992) that along the optimal path, the shadow value of the resource, *=V*/x, where V* refers to the maximal value of the programme. If extraction is profitable V*/x cannot be negative. It follows that R(q*,x)=q(q*,x)q*=*q*0. \n\n1. Resource rent taxation: General analysis\n\nConsider now the imposition of a tax on resource rents. Let the amount of the tax be:6\n\n1. T = R(q,x),\n\nwhere R(q,x) represents resource rents as defined in the previous section and is the rate of taxation. From the perspective of the industry the profit maximization problem now is:\n\n1.", null, ",\n\nSubject to:", null, "= G(x)-q\n\nx(0) = x0\n\nx,q,T  0.\nThe necessary conditions for solving (II) include:\n\n1. q - Rq -  0, q  0, (q - Rq - )q = 0,\n\n2.", null, "- r = -x + Rx - Gx,\n\n3.", null, "= G(x)-q,\n\n4. H(T) = (q(T),x(T)) - R(q(T),x(T))+ (T)(G(x(T))-q(T))=0,\n\n5. (T)  0, x(T)  0, (T)x(T)=0.\n\nComparing these necessary conditions with the ones without taxation, i.e. (3.1)-(3.5) shows that the imposition of a resource tax modifies conditions (3.1), (3.2) and (3.4). Modification of the first two necessary conditions will in general alter the optimal paths of the control and state variables as well as the equilibrium position of these variables in the case of renewable resources. Modification of the fourth necessary condition suggests that the imposition of a resource rent tax may influence when a programme is terminated.\n\nIt is important to realize that condition (6.4) is really a component of a set of more general entry/exit conditions. Condition (6.4) represents the condition for the optimal exit of firms already in the industry. The corresponding condition for optimal entry of firms into the industry would be\n\n1. H(0) = (q(0),x(0)) - R(q(0),x(0))+ °(0)(G(x(0))-q(0))0,\n\nwhere °(0) represents the firms’ shadow value of the resource.\n\nCondition (6.6) must be carefully interpreted. First, although it could apply to a whole industry, it is more natural to interpret it in terms of a single firm. As such, this is a condition for a given firm to enter the industry. Second, °(0) is the shadow value of the resource as seen from a firm outside the industry. This does not have to be in accordance with the shadow value assessed by firms in the industry. For a firm outside the industry °(0) could easily be zero. But it could just as well take on another value if the resource has alternative (possibly non-extractive) uses for the firm7. If this shadow value is zero, however, (6.6) reduces to the more familiar entry condition (q(0),x(0))-R(q(0),x(0))0, i.e., hat expected profits are positive. Third, the variables q(0) and x(0) represent the optimal levels of these variable, if the firm enters. Finally, the resource rent tax, R(q(0),x(0)), should be interpreted as what the firm expects to be charged.8\nTaking, (6.4) and (6.6) together, it is clear that resource rent taxes may alter the conditions for entry to and exit from the industry. Hence, if firms are not identical, such taxes, even if they will not close the industry prematurely, may alter the composition of firms participating in the industry.9\nTo summarize, we have found that resource rent taxes will generally affect extraction paths, sustainable equilibria, the opening and closing of the industry and the composition of companies in the industry. We thus have the basic result:\nProposition 1.\n\nResource rent taxes are in general distortive.\n\nThe economic rationale for Proposition 1 is straight-forward. Resource rents depend on the extraction path selected by the industry. They also depend on the participation of individual firms or the industry as a whole in the extraction activity. Thus, the industry and its constituent firms can to some extent counteract the burden of the taxation by modifying these variables.\nIt is important to realize, however, that Proposition 1 does not assert that that resource rent taxes are distortive in all cases. There are situations, probably unrealistically simple, under which resource rent taxes may not have any distortive impact. One such case, for instance, is a renewable resource industry with linear extraction technology and identical firms.\n\n1. Examples\n\nWe will now illustrate the distortive nature of resource rent taxation by numerical examples from non-renewable and renewable resource extraction industries. For ease of computation and exposition these examples will be based on very simple models.\n\nA non-renewable resource industry: Mining\nConsider a simple mining industry with the cost of production independent of the quantity of reserves. More to the point, let the profit function of this industry be:\n\n1. (q)= pqC(q) = pq – (a+bq2),\n\nwhere, as before, q refers to extracted quantity at a point of time, p denotes its unit price and a and b are positive parameters.\n\nSubstituting this specification and the condition that G(x)0, all x, into the necessary conditions (3.1) to (3.5) we can easily derive the following optimality results:\n\n1. The available resources will be exhausted in finite time\n\n2. The terminal time, T, is determined by the equation:\n\n(7.1)", null, ",\n\nwhere x0 represents the initial reserves and r is the rate of discount.\n\n1. The extraction path is given by:\n\n(7.2)", null, ".\n\nAs discussed in the previous section, a resource tax on this industry would be:\n(8) R(q,x) = q(q,x)q = ( pq – 2bq2)\nSubtracting (8) from the RHS of (7) yields the profit function after taxes as:\n\n1. (q,)= p(1-)q – (a+(1-2)bq2).\n\nSo, in this case, the impact of the resource rent tax is to reduce the net price of production and the marginal costs of extraction. The first effect is quite intuitive. The second stems from the fact that increased production is now less costly than before because it reduces resource rent taxes.\n\nSubstituting p(1-) and (1-2)b in for a and b in (7.1) and (7.2) it is now straight-forward to calculate the terminal time and extraction path for the resource rent case. Briefly, the imposition of the resource rent tax leads more intense extraction and earlier exhaustion time than would otherwise be the case. So, interestingly, the resource rent tax discourages conservation.\nAssuming certain numerical values of the parameters this result can be illustrated. The parameter values chosen are:\n\n Parameter Value p 4 a 1 b 1 r 0.1 x0 10\n\nLet us moreover set the tax rate at =0.2. This means that 20% of resource rents are taxed away.\n\nThe impact of this tax on the profit maximizing extraction paths is illustrated in Figure 3\n\nFigure 3\n\nImpact of resource rent taxes on extraction: An example", null, "As shown in Figure 3, this resource rent tax (=0.2) has a very significant impact on the extraction profile. The impact on the present value of total social benefits (profits before taxes) from the extraction activity, however, is not very great. This is only reduced by some 7%. Company profits, however, are reduced by some 25%. If the resource rent tax were raised to 0.35, however, the distortive effect of the tax would be much greater. Social benefits would be reduced by about 40% and retained profits by the firms by over 90%.\n\nA renewable resource industry: A fishery\nConsider a simple fishery with a profit function\n\n1. (q,x)= pqC(q,x) = pq – (a+bq2x-1),\n\nwhere q represents the harvest rate and x the fish stock biomass both at a point of time. p is the unit price of harvest and a and b are positive parameters.\n\nThe biomass growth function is taken to be the the logistic function:\n\n1. G(x) = x - x2,\n\nwhere  and  are parameters with  representing the intrinsic growth rate and / the maximum equilibrium biomass often referred to as the virgin stock equilibrium.\n\nNow, inspection of the necessary conditions (3.1) to (3.5) reveals that if the fishery is productive enough, dynamic optimization will involve a non-zero equilibrium for q and x defined by10:\n\n1. G(x) + x/q = r,\n\n2. G(x) – q = 0.\n\nIt may be mentioned that the second term on the LHS of (12.1) is often referred to as the marginal stock effect (Clark and Munro 1975). In an optimal equilibrium, this term is positive encouraging conservation of the resource.\n\nSubstituting (10) and (11) into these equilibrium conditions we can easily derive the following optimality equilibrium conditions for our special model:\n(13.1)", null, "1. x - x2 = 0.\n\nAs discussed in the section 2 above, a resource tax on this industry is defined as:\n\n(14) R(q,x) = q(q,x)q = ( pq – 2bq2x-1)\nSubtracting (14) from the profit function defined by (10) produces after some rearrangements:\n\n1. (q,x,)= p(1-)q – (a+(1-2)bq2x-1).\n\nSo, just as in the mining example above, the impact of the resource rent tax is to reduce the net price of production and the marginal costs of extraction.\n\nSubstituting p(1-) and (1-2)b in for p and b in (13.1) it is now straight-forward to derive the profit maximizing equilibrium conditions under the resource rent tax. The result is:\n(16.1)", null, ",\n\n1. x - x2 = 0.\n\nSo, clearly, the equilibrium conditions are modified by the imposition of the resource rent tax.11 It obviously follows that the resource rent tax will also alter the optimal adjustment paths of the fishery.\n\nInspection of (16.1) reveals that a positive resource rent tax will reduce the marginal stock effect (the third term in (16.1)) and, consequently, lead to a lower equilibrium biomass level. Obviously, if there is no resource rent tax, =0, (16.1) reduces to (13.1).\nBy assuming certain values of the parameters, it is possible to numerically illustrate this result. The parameter values are:\n\n Parameter Value p 1 a 0.2 b 0.3  2  1 r 0.1\n\nGiven these parameters, we can now calculate the equilibrium biomass level and the corresponding social profits and tax revenues as a function of the tax rate, . The relationship between the biomass level and the tax rate is illustrated in Figure 4.\n\nFigure 4\n\nResource rent taxation and the optimal equilibrium biomass: An example", null, "As illustrated in Figure 4, the optimal equilibrium biomass level is monotonically declining and quite sensitive to the resource rent tax rate. Thus, a resource rent tax rate of 0.5 reduces the equilibrium biomass by almost 20% and a tax rate of 0.9 reduces the it by by about 50%. The harvest rate, also depicted in Figure 4, is, on the other hand, much less sensitive to the resource rent tax. This, of course, is because the resource rent tax initially moves the equilibrium biomass toward the maximum sustainable level. Thus with a resource tax rate of 0.5 the harvest increases by 1.4% compared to no resource rent tax.\n\nFigure 5\n\nResource rent taxation and the optimal equilibrium biomass: An example", null, "Social benefits (profits before taxes), as predicted, decline monotonically with the tax rate. The tax revenue, on the other hand increases up to a certain point and declines after that. These relationships are illustrated in Figure 5.\n\n4. Discussion\nThe basic result of this paper is that resource rent taxation is generally distortive. As demonstrated, it normally induces firms to alter the time path of extraction and may influence their entry and exit decisions as well as the particular resource stocks that come under exploitation.\nAlthough this result contradicts certain widely held beliefs concerning resource rent taxation, it can hardly be said to be surprising. After all, resources rents clearly depend on the level of extraction. Consequently, the same applies to the resource rent tax. This means that firms can reduce their resource rent tax-burden by adjusting the level of extraction. This they will do if it increases their retained profits. The paper suggests that only in very special situations, namely linear profit functions and identical firms, will this not necessarily be the case.\nThe model on which these results are based is quite simple. This, however, apparently does not subtract from the generality of the results. The indications are that the more general the model the less likely it is that resource rent taxation will be nondistortive. Among other things not allowed for in the model are explorations and investment in new resource developments. It seems pretty obvious, however, that resource rent taxation will reduce such activities if only for the reason that the expected returns will now be less than before. If the companies are risk averse the impact will be greater.\nThe situation, of course, is changed if the tax collector, presumably the government, is in a position to fix the quantity extracted. Then the resource rent is fixed and we are back in the Ricardian-Alchian framework of fixed supply. Thus, under those circumstances, the firms cannot avoid the resource rent tax by adjusting the extraction rate. It would be premature, however, to conclude that the resource rent tax was non-distortive. The tax could still affect entry and exit and therefore the composition of firms in the industry. By a similar argument, it could also affect the resource stocks being exploited. Another serious disadvantage is that the tax collector would not be in strong position to set the optimal quantity of extraction which moreover varies over time and fluctuates with the parameters of the situation. It is well known that due to information, calculation and implementation problems governments generally fail miserably in economic micro management of this kind. Therefore, even if the resource rent tax could, in theory, be rendered non-distortive by the government fixing the quantity of extraction, the inefficiency would reappear in suboptimal choices of the extraction quantity.\nIn the examples presented it was found that the distortionary impact of resource rent taxation is toward less conservation of natural resources than would otherwise be the case. While, this has not been established as a general principle, this finding may be a matter of some additional concern.\nBy its distortionary impact, resource rent taxation reduces the funds available for consumption and investment. For this reason, resource rent taxation is likely to have a negative impact on aggregate investment and hence the growth path of the economy. This negative impact will be counteracted if the resource tax revenues are more effectively used by the tax collector (government) than the private sector and exacerbated if the opposite holds true. For economies heavily dependent on natural resource extraction industries, these macro-economic impacts of resource rent taxation may be quite significant.\nIt is important to realize that the distortionary impact of resource rent taxation does not by any means rule it out as a sensible tax alternative. When it comes to the financing of government expenditures the relevant question is not whether a given tax option is distortive or not, but whether it is more or less distortive than the other alternatives available. The above analysis does not answer this question. In fact, a priori, it seems unlikely that this question can be answered on a theoretical basis.\nThe result that resource rent taxation is generally distortive raises the question of whether there exists a form of taxation to collect resource rents that is non-distortive. The above analysis says nothing about that. Note, however, that such non-distortive taxation, if it exists, would not be resource rent taxation. It would be something else. Indeed, if it is to be truly non-distortive it could not be related to any decision variables of the firms.\n\nReferences\nAlchian, A. A. 1987. Rent. In J. Eatwell, M. Milgate and P. Newman (eds.) The New Palgrave: A Dictionary of Economics. MacMillan Press. London.\n\nArnason, R. and O.D. Jonsson (eds.). 1993. Fiskveidistjornun og skipting Fiskveidiardsins. Fisheries Research Institute. University of Iceland. Reykjavik.\n\nBlaug, M. 1996. Economic Theory in Retrospect. 5th edition. Cambride University Press, Cambridge.\n\nClark, C.W. and G.R. Munro. 1975. The Economics of Fishing and Modern Capital Theory. Journal of Environmental Economics and Management 2:92-106.\n\nFraser, R.W. 1993. On the Neutrality of the Resource Rent Tax. Economic Record, 69:56-60.\n\nFraser, R.W. 1998. Lease Allocation Systems, Risk Aversion and the Resource Rent Tax. Australian Journal of Agricultural and Resource Economics, 42:115-30.\n\nGarnault, R. and A. Clunies-Ross. 1975. Uncertainty, risk Aversion and the Taxing of Natural Resource Projects. Economic Journal, 85:272-87.\n\nGarnault and Clunies-Ross. 1979. On the Neutrality of the Resource Rent Tax. Economic Record 55:193-201.\n\nGeorge, H. 1879. Progress and Poverty. W. M. Hinton and Company, San Francisco.\n\nGrafton, R.Q. 1995. Rent Capture in a Rights-Based Fishery. Journal of Environmental Economics and Management, 28:48-67.\n\nGrafton, R.Q. 1996. Implications of Taxing Quota Value in an Individual Transferable Quota Fishery: Comment. Marine Resource Economics 11:125-7.\n\nJohnson, R.N. 1995. Implications of Taxing Quota Value in an Individual Transferable Quota Fishery. Marine Resource Economics 10:327-40\n\nLéonard, D. and N.V. Long. 1992. Optimal Control Theory and Static Optimization in Economics. Cambridge University Press. Cambridge.\n\nMarshall, A. 1920. The Principles of Economics. 8th edition. MacMillan. London.\n\nMiller, M., N. Ireland M. Cripps and L. Zhang. 2000. Collecting Economic Rents on North sea and Gas: A Proposal. Edited Research Report submitted to ESRC.\n\nPontayagin, L.S., V.G. Boltyanskii, R.V. Gamkrelidse and E.F. Mishchenko. 1962. The Mathematical Theory of Optimal processes. John Wiley and Sons. New York.\n\nRicardo, D.1821. On the Principles of Political Economy and Taxation. 3rd edition. John Murray. London.\n\nScott, A.D. 1955. The Fishery: The Objectives of Sole Ownership. Journal of Political Economy 63:xx-xx\n\nVarian, H. 1984. Microeconomic Theory. Second edition. W.W. Norton & Company. New York.\n\n1 Fisheries provide is a case in point. Special fisheries taxation has for instance been advocated in New Zealand (Johnson 1995, New Zealand Fishing Industry Board 1993) and Iceland (Arnason and Jonsson 1993, Auðlindanefnd 2000) on precisely these grounds.\n\n2 Note that since the factor is by assumption in fixed supply, there can be no opportunity costs associated with its supply.\n\n3It may be noted that for factors in elastic supply, the concept of producers’ surplus applies. In fact, economic rents may be regarded as the special case of producers’ surplus applying when the elasticity of supply equals zero.\n\n4 A sufficient condition is that the firms are identical.\n\n5 Defined as (Dq(q)q/D(q))-1.\n\n6 More generally a resource rent tax would be T = (R(q,x)), where is an increasing function. This generalization, however, would not qualitatively alter the results of the analysis.\n\n7 Consider for instance the case of a firm in tourism which can use a particular spot of nature for purposes of tourism and/or mining.\n\n8 If entry or exit are not costless, there would be a need to form expectations about future taxation as well.\n\n9 Or, from a slightly different perspective, alter the natural resources that are exploited.\n\n10 Equilibrium is characterized by", null, "=", null, "=0. Thus, if the equilibrium q is positive, (3.1) and (3.2) are reduced to: q =  and r = -x - Gx,, respectively. From these two equations (12.1) is easily derived. (12.2) follows immediately from (3.3).\n\n11 This confirms a result previously derived by Johnson (1995).\n\nVerilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©atelim.com 2016\nrəhbərliyinə müraciət" ]

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[ "# How to use Oscilloscope in Lab\n\nFollow the below simple steps in order to use the oscilloscope in dcaclab.\n\nStep 1: Navigate through the list of the devices/components as shown in image.", null, "Step 2: Click on the oscilloscope icon, the oscilloscope will then appear in the board below.", null, "Step 3: Grab the probes to measure the voltage of any point or define the signal at any point of the circuit.", null, "## How to use Time and Voltage Division in an Oscillator", null, "In the above image shown of an oscilloscope there are two pivots, upper pivot is used to vary the time per division in the x axis and lower pivot is used to vary the voltage per division in the y axis.\n\nYou can play with the above circuit in here.\n\n## What Can Oscilloscope Measure\n\nWith the Oscilloscope you can define the quantities like amplitude, frequency and other waveform characteristics in a Signal or AC current.\n\nIn easy way,\n\nWith the help of oscilloscope we can measure both the characteristics based on time as well as voltage.\n\n### Timing characteristics\n\nThe time unit is shown by the x axis in the oscilloscope, with the help of time unit we can define the characteristics such as Frequency and period, Duty cycle and Rise and fall time.\n\n#### Frequency and period\n\nThe frequency of the signal or a current (AC Current) is defined as the number of cycle the waveform completes in one second. The period of the current (AC Current) or a signal is defined as the opposite to the frequency that is the number of second each wave cycle takes.\n\n#### Duty cycle\n\nThe duty cycle is calculated as the percentage of the signal weather it is positive or negative, The duty cycle is defined in two types, they are positive duty cycle and negative duty cycle.\n\nWith the help of Duty cycle we can define the ratio of the time period the signal is positive and the signal is negative.\n\n#### Rise and fall time\n\nThe AC voltage or a signal does not instantaneously go 0v to 12v, they rise to the peak voltage instead. The Rise Time is defined as the time duration in which the signal goes from low point to high point, The Fall Time is defines the opposite that is the time duration in which the signal goes from high point to low point.\n\n### Voltage characteristics\n\nThe voltage unit is shown by the y axis in the oscilloscope. With the help of Voltage unit we can define the characteristics such as Amplitude, Maximum and minimum voltages and Mean and average voltages.\n\n#### Amplitude\n\nThe amplitude is defined as the magnitude of the voltage or a signal. There are many ways to define the amplitude, in one method the difference between high voltage point to low voltage point also known as peak-to-peak amplitude. In another method the distance from high voltage point or low voltage point to 0V.\n\n#### Maximum and minimum voltages\n\nWith the help of Oscilloscope you can very easily define that how high and low does the voltage of your signal gets.\n\n#### Mean and average voltages\n\nWith the help of oscilloscope you can very easily calculate the average or mean voltage of the signal. Also with the help of the oscilloscope you can define the average of the signal’s maximum as well as minimum voltage.\n\nDo write on the comment below that how did you used the oscilloscope.\n\nClick to see the use of oscilloscope in dcaclab." ]

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[ "", null, "In his classic work The General Theory, Keynes offered his view of how the interest rate is determined in the short run. His explanation is called the theory of liquidity preference because it posits that the interest rate adjusts to balance the supply and demand for the economy’s most liquid asset—money. Just as the Keynesian cross is a building block for the IS curve, the theory of liquidity pref- erence is a building block for the LM curve.\n\nTo develop this theory, we begin with the supply of real money balances. If $M$ stands for the supply of money and $P$ stands for the price level, then $\\frac{M}{P}$ is the supply of real money balances. The theory of liquidity preference assumes there is a fixed supply of real money balances. That is,\n\n$(M/P)^s = \\frac{\\bar{M}}{\\bar{P}}$\n\nThe money supply $M$ is an exogenous policy variable chosen by a central bank, such as the Federal Reserve. The price level $P$ is also an exogenous variable in this model. (We take the price level as given because the IS–LM model—our ultimate goal in this chapter—explains the short run when the price level is fixed.) These assumptions imply that the supply of real money balances is fixed and, in particular, does not depend on the interest rate. Thus, when we plot the supply of real money balances against the interest rate, we obtain a vertical supply curve.\n\nNext, consider the demand for real money balances. The theory of liquidity preference posits that the interest rate is one determinant of how much money people choose to hold. The underlying reason is that the interest rate is the opportunity cost of holding money: it is what you forgo by holding some of your assets as money, which does not bear interest, instead of as interest-bearing bank deposits or bonds. When the interest rate rises, people want to hold less of their wealth in the form of money. We can write the demand for real money balances as:\n\n$(M/P)^d = L(r)$\n\nwhere the function $L(r)$ shows that the quantity of money demanded depends on the interest rate. The demand curve slopes downward because higher interest rates reduce the quantity of real money balances demanded.\n\nAccording to the theory of liquidity preference, the supply and demand for real money balances determine what interest rate prevails in the economy. That is, the interest rate adjusts to equilibrate the money market. At the equilibrium interest rate, the quantity of real money balances demanded equals the quantity supplied.\n\nHow does the interest rate get to this equilibrium of money supply and money demand? The adjustment occurs because whenever the money market is not in equilibrium, people try to adjust their portfolios of assets and, in the process, alter the interest rate. For instance, if the interest rate is above the equilibrium level, the quantity of real money balances supplied exceeds the quantity demanded. Individuals holding the excess supply of money try to convert some of their non-interest-bearing money into interest-bearing bank deposits or bonds. Banks and bond issuers, which prefer to pay lower interest rates, respond to this excess supply of money by lowering the interest rates they offer. Conversely, if the interest rate is below the equilibrium level, so that the quantity of money demanded exceeds the quantity supplied, individuals try to obtain money by selling bonds or making bank withdrawals. To attract now-scarcer funds, banks and bond issuers respond by increasing the interest rates they offer. Eventually, the interest rate reaches the equilibrium level, at which people are content with their portfolios of monetary and nonmonetary assets." ]

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[ "Forum Rule: Always post complete source code & details to reproduce any issue!\n\n1. ## asking for kind help and support\n\nfirst I would like to thank you in advance for your kind and valuable help. Here is my question I am trying to use the library made by for wavelet transform applcation for viberation data to detect annomalies and fault more specifically for industrial bearing. But when I tried to compile the liberery for Teensy 3.6 using Arduino 1.8.5 I got an error as i mention below. all the liberery as well the implimenation files are made of C code what i mean using .c. anyone who can suggest me the solution? I guess the problem is with the linker and compiler.\n// here is the sample code to test simple with simple imput data\nvoid loop() {\n\nSerial.begin(9600);\nwave_object obj;\nwt_object wt;\ndouble *inp, *out, *diff;\nint N, i, J;\nchar *name = \"db4\";\nobj = wave_init(name);// Initialize the wavelet\nN = 14; //Length of Signal\ninp = (double*)malloc(sizeof(double)* N); //Input signal\nout = (double*)malloc(sizeof(double)* N);\ndiff = (double*)malloc(sizeof(double)* N);\n//wmean = mean(temp, N);\nfor (i = 0; i < N; ++i) {\ninp[i] = i;\nSerial.println(inp[i]);\n}\nJ = 1; //Decomposition Levels\nwt = wt_init(obj, \"dwt\", N, J);// Initialize the wavelet transform object\nSerial.println(wt->siglength);\nsetDWTExtension(wt, \"sym\");// Options are \"per\" and \"sym\". Symmetric is the default option\nsetWTConv(wt, \"direct\");\ndwt(wt, inp);// Perform DWT\n//DWT output can be accessed using wt->output vector. Use wt_summary to find out how to extract appx and detail coefficients\nfor (i = 0; i < wt->outlength; ++i) {\nSerial.println(wt->output[i]);\n}\n\nidwt(wt, out);// Perform IDWT (if needed)\n// Test Reconstruction\nfor (i = 0; i < wt->siglength; ++i) {\ndiff[i] = out[i] - inp[i];\n}\n\nSerial.println(absmax(diff, wt->siglength)); // If Reconstruction succeeded then the output should be a small value.\nSerial.println(\"FIN\");\nSerial.flush();\n\n}\n\ndouble absmax(double *array, int N) {\ndouble max;\nint i;\nmax = 0.0;\nfor (i = 0; i < N; ++i) {\nif (fabs(array[i]) >= max) {\nmax = fabs(array[i]);\n}\n}\nreturn max;\n}\n\n// here are the Generated error when I tried to compiled\nc:/program files (x86)/arduino/hardware/tools/arm/bin/../lib/gcc/arm-none-eabi/5.4.1/../../../../arm-none-eabi/lib/armv7e-m/fpu\\libc.a(lib_a-closer.o): In function `_close_r':\n\ncloser.c", null, ".text._close_r+0xc): undefined reference to `_close'\n\nc:/program files (x86)/arduino/hardware/tools/arm/bin/../lib/gcc/arm-none-eabi/5.4.1/../../../../arm-none-eabi/lib/armv7e-m/fpu\\libc.a(lib_a-lseekr.o): In function `_lseek_r':\n\nlseekr.c", null, ".text._lseek_r+0x12): undefined reference to `_lseek'\n\nreadr.c", null, ".text._read_r+0x12): undefined reference to `_read'\n\nc:/program files (x86)/arduino/hardware/tools/arm/bin/../lib/gcc/arm-none-eabi/5.4.1/../../../../arm-none-eabi/lib/armv7e-m/fpu\\libc.a(lib_a-fstatr.o): In function `_fstat_r':\n\nfstatr.c", null, ".text._fstat_r+0x10): undefined reference to `_fstat'\n\nc:/program files (x86)/arduino/hardware/tools/arm/bin/../lib/gcc/arm-none-eabi/5.4.1/../../../../arm-none-eabi/lib/armv7e-m/fpu\\libc.a(lib_a-isattyr.o): In function `_isatty_r':\n\nisattyr.c", null, ".text._isatty_r+0xc): undefined reference to `_isatty'\n\ncollect2.exe: error: ld returned 1 exit status\n\nError compiling for board Teensy 3.6.", null, "", null, "Reply With Quote\n\n2. And where is the library? Do not see any #include in your code...", null, "", null, "Reply With Quote\n\n3. When I compile that code for T3.6, I do not get any of the errors you've listed. I get this lot:\nCode:\n```sketch_dec03a: In function 'void loop()':\nsketch_dec03a:5: error: 'wave_object' was not declared in this scope\nwave_object obj;\n\n^\n\nsketch_dec03a:6: error: 'wt_object' was not declared in this scope\nwt_object wt;\n\n^\n\nsketch_dec03a:9: warning: ISO C++ forbids converting a string constant to 'char*'\nchar *name = \"db4\";\n\n^\n\nsketch_dec03a:10: error: 'obj' was not declared in this scope\nobj = wave_init(name);// Initialize the wavelet\n\n^\n\nsketch_dec03a:10: error: 'wave_init' was not declared in this scope\nobj = wave_init(name);// Initialize the wavelet\n\n^\n\nsketch_dec03a:21: error: 'wt' was not declared in this scope\nwt = wt_init(obj, \"dwt\", N, J);// Initialize the wavelet transform object\n\n^\n\nsketch_dec03a:21: error: 'wt_init' was not declared in this scope\nwt = wt_init(obj, \"dwt\", N, J);// Initialize the wavelet transform object\n\n^\n\nsketch_dec03a:23: error: 'setDWTExtension' was not declared in this scope\nsetDWTExtension(wt, \"sym\");// Options are \"per\" and \"sym\". Symmetric is the default option\n\n^\n\nsketch_dec03a:24: error: 'setWTConv' was not declared in this scope\nsetWTConv(wt, \"direct\");\n\n^\n\nsketch_dec03a:25: error: 'dwt' was not declared in this scope\ndwt(wt, inp);// Perform DWT\n\n^\n\nsketch_dec03a:31: error: 'idwt' was not declared in this scope\nidwt(wt, out);// Perform IDWT (if needed)\n\n^\n\n'wave_object' was not declared in this scope```\nDo not see any #include in your code\nThere's no setup function either.\n\nPete", null, "", null, "Reply With Quote\n\n4.", null, "Originally Posted by el_supremo", null, "When I compile that code for T3.6, I do not get any of the errors you've listed. I get this lot:\nCode:\n```sketch_dec03a: In function 'void loop()':\nsketch_dec03a:5: error: 'wave_object' was not declared in this scope\nwave_object obj;\n\n^\n\nsketch_dec03a:6: error: 'wt_object' was not declared in this scope\nwt_object wt;\n\n^\n\nsketch_dec03a:9: warning: ISO C++ forbids converting a string constant to 'char*'\nchar *name = \"db4\";\n\n^\n\nsketch_dec03a:10: error: 'obj' was not declared in this scope\nobj = wave_init(name);// Initialize the wavelet\n\n^\n\nsketch_dec03a:10: error: 'wave_init' was not declared in this scope\nobj = wave_init(name);// Initialize the wavelet\n\n^\n\nsketch_dec03a:21: error: 'wt' was not declared in this scope\nwt = wt_init(obj, \"dwt\", N, J);// Initialize the wavelet transform object\n\n^\n\nsketch_dec03a:21: error: 'wt_init' was not declared in this scope\nwt = wt_init(obj, \"dwt\", N, J);// Initialize the wavelet transform object\n\n^\n\nsketch_dec03a:23: error: 'setDWTExtension' was not declared in this scope\nsetDWTExtension(wt, \"sym\");// Options are \"per\" and \"sym\". Symmetric is the default option\n\n^\n\nsketch_dec03a:24: error: 'setWTConv' was not declared in this scope\nsetWTConv(wt, \"direct\");\n\n^\n\nsketch_dec03a:25: error: 'dwt' was not declared in this scope\ndwt(wt, inp);// Perform DWT\n\n^\n\nsketch_dec03a:31: error: 'idwt' was not declared in this scope\nidwt(wt, out);// Perform IDWT (if needed)\n\n^\n\n'wave_object' was not declared in this scope```\n\nThere's no setup function either.\n\nPete\nDear Pete I would like to appreciate for your queeck response. In the full program I included the setup function and the liberery function for detail information here is the full skech program\n\n#include <string.h>\n#include <math.h>\n#include <Arduino.h>\n#include \"wavelib.h\"\n\nvoid setup()\n{\npinMode(ledpin,OUTPUT);\n}\n\nvoid loop() {\n\nSerial.begin(9600);\nwave_object obj;\nwt_object wt;\ndouble *inp, *out, *diff;\nint N, i, J;\nchar *name = \"db4\";\nobj = wave_init(name);// Initialize the wavelet\nN = 14; //Length of Signal\ninp = (double*)malloc(sizeof(double)* N); //Input signal\nout = (double*)malloc(sizeof(double)* N);\ndiff = (double*)malloc(sizeof(double)* N);\n//wmean = mean(temp, N);\nfor (i = 0; i < N; ++i) {\ninp[i] = i;\nSerial.println(inp[i]);\n}\nJ = 1; //Decomposition Levels\nwt = wt_init(obj, \"dwt\", N, J);// Initialize the wavelet transform object\nSerial.println(wt->siglength);\nsetDWTExtension(wt, \"sym\");// Options are \"per\" and \"sym\". Symmetric is the default option\nsetWTConv(wt, \"direct\");\ndwt(wt, inp);// Perform DWT\n//DWT output can be accessed using wt->output vector. Use wt_summary to find out how to extract appx and detail coefficients\nfor (i = 0; i < wt->outlength; ++i) {\nSerial.println(wt->output[i]);\n}\n\nidwt(wt, out);// Perform IDWT (if needed)\n// Test Reconstruction\nfor (i = 0; i < wt->siglength; ++i) {\ndiff[i] = out[i] - inp[i];\n}\n\nSerial.println(absmax(diff, wt->siglength)); // If Reconstruction succeeded then the output should be a small value.\nSerial.println(\"FIN\");\nSerial.flush();\n\n}\n\ndouble absmax(double *array, int N) {\ndouble max;\nint i;\nmax = 0.0;\nfor (i = 0; i < N; ++i) {\nif (fabs(array[i]) >= max) {\nmax = fabs(array[i]);\n}\n}\nreturn max;\n}", null, "", null, "Reply With Quote\n\n5. I include the library function in the full sketch and I define setup function default\n#include <string.h>\n#include <math.h>\n#include <Arduino.h>\n#include \"wavelib.h\"\n\nvoid setup()\n{\npinMode(ledpin,OUTPUT);\n}\n\nvoid loop() {\n\nSerial.begin(9600);\nwave_object obj;\nwt_object wt;\ndouble *inp, *out, *diff;\nint N, i, J;\nchar *name = \"db4\";\nobj = wave_init(name);// Initialize the wavelet\nN = 14; //Length of Signal\ninp = (double*)malloc(sizeof(double)* N); //Input signal\nout = (double*)malloc(sizeof(double)* N);\ndiff = (double*)malloc(sizeof(double)* N);\n//wmean = mean(temp, N);\nfor (i = 0; i < N; ++i) {\ninp[i] = i;\nSerial.println(inp[i]);\n}\nJ = 1; //Decomposition Levels\nwt = wt_init(obj, \"dwt\", N, J);// Initialize the wavelet transform object\nSerial.println(wt->siglength);\nsetDWTExtension(wt, \"sym\");// Options are \"per\" and \"sym\". Symmetric is the default option\nsetWTConv(wt, \"direct\");\ndwt(wt, inp);// Perform DWT\n//DWT output can be accessed using wt->output vector. Use wt_summary to find out how to extract appx and detail coefficients\nfor (i = 0; i < wt->outlength; ++i) {\nSerial.println(wt->output[i]);\n}\n\nidwt(wt, out);// Perform IDWT (if needed)\n// Test Reconstruction\nfor (i = 0; i < wt->siglength; ++i) {\ndiff[i] = out[i] - inp[i];\n}\n\nSerial.println(absmax(diff, wt->siglength)); // If Reconstruction succeeded then the output should be a small value.\nSerial.println(\"FIN\");\nSerial.flush();\n\n}\n\ndouble absmax(double *array, int N) {\ndouble max;\nint i;\nmax = 0.0;\nfor (i = 0; i < N; ++i) {\nif (fabs(array[i]) >= max) {\nmax = fabs(array[i]);\n}\n}\nreturn max;\n}", null, "", null, "Reply With Quote\n\n6. Now it doesn't compile because it can't find wavelib.h\nWhat is wavelib.h and where is it?\n\nPete", null, "", null, "Reply With Quote\n\n####", null, "Posting Permissions\n\n• You may not post new threads\n• You may not post replies\n• You may not post attachments\n• You may not edit your posts\n•" ]

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[ "Next Article in Journal\nThe Variety of Information Transfer in Animal Sonic Communication: Review from a Physics Perspective\nNext Article in Special Issue\nA Weighted Generalized Maximum Entropy Estimator with a Data-driven Weight\nPrevious Article in Journal / Special Issue\nUse of Maximum Entropy Modeling in Wildlife Research\n\nMetrics 0\n\n## Export Article\n\nEntropy 2009, 11(4), 867-887; https://doi.org/10.3390/e11040867\n\nArticle\nMaximum Entropy Estimation of Transition Probabilities of Reversible Markov Chains\nQueen Mary University of London, School of Mathematical Sciences, Mile End Road, London E1 4NS, UK\nReceived: 21 September 2009 / Accepted: 10 November 2009 / Published: 17 November 2009\n\n## Abstract\n\n:\nIn this paper, we develop a general theory for the estimation of the transition probabilities of reversible Markov chains using the maximum entropy principle. A broad range of physical models can be studied within this approach. We use one-dimensional classical spin systems to illustrate the theoretical ideas. The examples studied in this paper are: the Ising model, the Potts model and the Blume-Emery-Griffiths model.\nKeywords:\nmaximum entropy principle; Markov chain; parameter estimation; statistical mechanics; spin chain models; thermodynamics\n\n## 1. Introduction\n\nUsually, the only knowledge about a physical system is the measurement of the average values of a few relevant observables. The maximum entropy principle [1,2,3] is a tool to obtain the least biased distribution for the equilibrium distribution of the system that reproduces these measurements. The rough line of reasoning of this approach is as follows . The Boltzmann-Gibbs entropy functional of a distribution $P ( j )$ is defined by\n$S = − ∑ j ∈ Γ P ( j ) ln P ( j )$\nwhere Γ is the (discrete) phase space. This entropy functional is most often used in statistical mechanics. The equilibrium distribution is obtained by maximising the entropy functional under the constraints that the average values of the relevant observables $H 1 ( j ) , … , H t ( j )$ take on certain values. To solve this optimisation problem, usually the method of Lagrange multipliers is used. For every constraint, a Lagrange multiplier $θ i$ is introduced and the following function is maximised\n$L = S − ∑ i = 1 t θ i ∑ j ∈ Γ P ( j ) H i ( j )$\nto obtain the equilibrium distribution. After variation with respect to $P ( j )$ one obtains\n$P ( j ) ∼ exp − ∑ i = 1 t θ i H i ( j )$\nWith this distribution, one can calculate expressions for the equilibrium values $〈 H i ( j ) 〉$ of the observables as a function of the Lagrange multipliers $θ i$. These relations can then be used to estimate the values of $θ i$ after measurement of $〈 H i ( j ) 〉$. In this way, one obtains the least biased estimate of the parameters $θ i$, because $P ( j )$ satisfies the maximum entropy principle. The standard example of this procedure is the estimation of the temperature by the measurement of the energy.\nThe maximum entropy principle can be used to introduce thermodynamic parameters in simple theoretical models [5,6,7]. One starts from a mathematical model and calculates the entropy and the average of the relevant observables as a function of the model parameters. Then the maximisation procedure is carried out over the model parameters only, instead of over all the possible probability distributions. The usefulness of this approach is already shown for specific models containing 2 and 5 parameters, see [5,6] and respectively. In this paper we show that the maximisation procedure can be carried out for reversible N-state Markov chains. This problem of conditional optimisation is more general and is also studied in the information theory framework [8,9]. However, we focus on the applications of this technique in the context of statistical mechanics and use the maximum entropy principle to relate microscopic and macroscopic quantities of physical models. To illustrate the theoretical ideas, one-dimensional classical spin systems are studied, the Ising model , the Potts model [11,12] and the Blume-Emery-Griffiths model [13,14]. These examples serve to show that our theoretical procedure is very general and that a broad range of physical models can be studied within this approach. It is not the aim of the present paper to make progress in the understanding of the aforementioned models.\nThe outline of the paper is as follows. In the next section we fix our notation and repeat briefly some results obtained in that will be used throughout the paper. The basic idea of this work is introduced in Section 3. with the use of a simple example, the 2-state Markov chain. The main result of this paper is obtained in Section 4.. in which we apply the maximum entropy principle to the reversible N-state Markov chain. In Section 5. we make the connection between our theory and thermodynamics and examine under which conditions our technique coincides with the standard approach to introduce thermodynamic parameters in theoretical models. In Section 6., the example of the 3-state Markov chain is thoroughly studied. The final section contains a summary of our results and a brief discussion of the different assumptions we made throughout the paper.\n\n## 2. Notation\n\nWe consider a finite state space Γ with N states. A Markov chain is defined by initial probabilities $p ( z )$ and transition probabilities $w ( y , z )$, with $y , z ∈ Γ$. The equation of motion is simply\n$p t + 1 ( z ) = ∑ y ∈ Γ p t ( y ) w ( y , z ) , with p 0 ( y ) = p ( y )$\nThroughout the paper, we will assume that $w ( z , y ) ≠ 0$ for all $z , y ∈ Γ$. This means that we study irreducible Markov chains . The N initial probabilities $p ( z )$ and the $N 2$ transition probabilities $w ( z , y )$ can be interpreted as the parameters of the Markovian model. However, they are not independent because of the normalisation conditions\n$1 = ∑ y ∈ Γ p ( y ) ,$\n$1 = ∑ y ∈ Γ w ( z , y ) , ∀ z ∈ Γ$\nAs a consequence, the Markovian model contains only $( N − 1 ) + ( N 2 − N ) = N 2 − 1$ independent parameters. Usually, extra restrictions on the parameters are assumed. A Markov chain is called stationary , when the following condition holds\n$p ( z ) = ∑ y ∈ Γ p ( y ) w ( y , z ) , ∀ z ∈ Γ$\nThis is a set of $N − 1$ extra equations (the normalisation is already taken into account). A stronger constraint is detailed balance\n$p ( z ) w ( z , y ) = p ( y ) w ( y , z ) , ∀ z , y ∈ Γ$\nThis is a set of $N ( N − 1 ) / 2$ extra equations. Throughout the paper, we will assume that this condition is satisfied. This means that we study reversible Markov chains .\nA path $γ = ( x 0 , x 1 , x 2 … , x n )$ of the Markov chain with length $n + 1$ has probability\n$p ( x 0 ) w ( x 0 , x 1 ) … w ( x n − 1 , x n )$\nIn the record of transitions k is introduced. This is a sequence of numbers $k ( z , y )$, one for each pair of states $z , y ∈ Γ$, counting how many times the transition from z to y is contained in a given path of the Markov chain. The ensemble average of the elements of the transition records and the entropy S [17,18] of the Markov chain can be calculated as follows:\n$〈 k ( z , y ) 〉 = ∑ x 0 ∈ Γ … ∑ x n ∈ Γ p ( x 0 ) w ( x 0 , x 1 ) … w ( x n − 1 , x n ) ∑ i = 0 n − 1 δ x i , z δ x i + 1 , y S = − ∑ x 0 ∈ Γ … ∑ x n ∈ Γ p ( x 0 ) w ( x 0 , x 1 ) … w ( x n − 1 , x n ) ln p ( x 0 ) w ( x 0 , x 1 ) … w ( x n − 1 , x n )$\nwith $δ i , j$ the Kronecker delta. For stationary Markov chains, these expressions simplify to \n$1 n 〈 k ( z , y ) 〉 = p ( z ) w ( z , y )$\n$1 n S = − ∑ z ∈ Γ ∑ y ∈ Γ p ( z ) w ( z , y ) ln w ( z , y ) − 1 n ∑ z ∈ Γ p ( z ) ln p ( z )$\nThe second term in the expression of the entropy is unimportant for large chains and will be ignored in the remaining part of this paper. The technical consequences of taking these finite size effects into account are already thoroughly studied for the 2-state Markov chain [5,19].\nThe conditional probability $P ( k ; x 0 )$ to observe a Markov chain with certain transition record k given the initial condition $x 0 ∈ Γ$ is\n$P ( k ; x 0 ) = c ( k ; x 0 ) ∏ x ∈ Γ ∏ y ∈ Γ w ( x , y ) k ( x , y ) = c ( k ; x 0 ) exp ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) ln w ( x , y )$\nwhere the prior probability $c ( k ; x 0 )$ counts the number of paths that have the same transition record k. Notice that the value of $c ( k ; x 0 )$ can vanish. An obvious example is a combination of an initial condition $x 0$ with a transition record in which no transition $x 0 → x$ with $x ∈ Γ$ occurs. However, this is not the only possibility to obtain $c ( k ; x 0 ) = 0$. To see this, observe that there are two ways to count the number of occurrences of a state $x ∈ Γ$ given k and $x 0 ∈ Γ$\n$δ x , x 0 + ∑ y ∈ Γ k y , x and δ x , x n + ∑ y ∈ Γ k x , y$\nwhere the Kronecker deltas $δ x , x 0$ and $δ x , x n$ take into account the first and last state of the path respectively. Given k and $x 0 ∈ Γ$, only when following equality\n$δ x , x 0 + ∑ y ∈ Γ k y , x = δ x , x n + ∑ y ∈ Γ k x , y , ∀ x ∈ Γ$\nis fulfilled, one ends up with a unique value for the number of occurrences of every state $x ∈ Γ$. Therefore, expression (14), is a necessary condition in order to obtain a non-vanishing value for $c ( k ; x 0 )$. This shows that the elements of the transition record are not independent. In Section 5., the importance of this observation will become clear.\nA Markov chain can be interpreted as a sequence of letters where the transition record k counts the number of occurrences of two-letter words. Markov chains with a finite memory and generalisations of k are examined in the information theory framework [20,21] and find applications in, e.g., the computational biology [22,23]. In the present paper we study Markov chains in the context of statistical mechanics and apply our results to physical models with only nearest neighbor interactions. The notion of two-letter words is sufficient for these applications. The extension of our theoretical results to systems with next (or higher order) nearest neighbor interactions is merely technical and can be obtained by increasing the number of states of the chain which allows to maintain the Markov property.\n\n## 3. Example: The 2-state Markov Chain\n\nIn this section we study a simple example, the 2-state Markov chain. The two states are denoted + and −. The different parameters of the Markovian model are\n$p ( + ) , p ( − ) and w ( + , + ) , w ( + , − ) , w ( − , − ) , w ( − , + )$\nHowever, the number of independent parameters is reduced by 3 because of the normalisation conditions (5), (6). The detailed balance condition (8) further reduces this number by 1. We conclude that this simple microscopic model contains only 2 independent parameters. We chose $w ( + , − )$ and $w ( − , + )$ to be these parameters and use (5), (6) and (8) to obtain the following relations\n$p ( + ) = w ( − , + ) w ( − , + ) + w ( + , − ) , p ( − ) = w ( + , − ) w ( − , + ) + w ( + , − ) w ( + , + ) = 1 − w ( + , − ) , w ( − , − ) = 1 − w ( − , + )$\nIn Section 2. we introduced the transition record k. The matrix k contains only 4 elements for this example\n$k = k ( + , + ) k ( + , − ) k ( − , + ) k ( − , − )$\nThe 2-state Markov chain can be interpreted as a one-dimensional Ising chain with constant length $n + 1$ and two different spin-values $± 1$. Two relevant observables are\n$H 1 ( σ ) = − J ∑ i = 0 n − 1 σ i σ i + 1 and H 2 ( σ ) = ∑ i = 0 n σ i$\nwith J a constant. The spin variables $σ i$ are scalars that can take on the values $± 1$. The two states of the Markov chain $+ , −$ correspond to the spin values $+ 1 , − 1$ respectively. Therefore, one can express $H 1 ( σ )$ and $H 2 ( σ )$ as a function of the elements of the transition record k as follows:\n$H 1 ( k ) = − J [ k ( + , + ) + k ( − , − ) ] + J [ k ( + , − ) + k ( − , + ) ]$\n$H 2 ( k ) = k ( + , + ) + k ( − , + ) − k ( − , − ) − k ( + , − )$\nThe correspondence between $H 1 ( σ )$ and $H 1 ( k )$ is exact, while we ignored the contribution of the initial spin $σ 0$ to obtain $H 2 ( k )$ from $H 2 ( σ )$. This is only a finite size effect that is unimportant for large chains. This means that the correspondence between $H 2 ( σ )$ and $H 2 ( k )$ is also exact for infinite chains. With (10) and (16), one can immediately write out the ensemble averages of these variables as a function of the independent parameters $w ( + , − )$ and $w ( − , + )$\n$〈 H 1 ( k ) 〉 J n = 4 w ( − , + ) w ( + , − ) − w ( − , + ) − w ( + , − ) w ( − , + ) + w ( + , − ) 〈 H 2 ( k ) 〉 n = w ( − , + ) − w ( + , − ) w ( − , + ) + w ( + , − )$\nAlso the entropy (11) of the Markov chain can be expressed as a function of $w ( + , − )$ and $w ( − , + )$ only\n$S n = − w ( − , + ) w ( − , + ) + w ( + , − ) [ 1 − w ( + , − ) ] ln [ 1 − w ( + , − ) ] + w ( + , − ) ln w ( + , − ) − w ( + , − ) w ( − , + ) + w ( + , − ) w ( − , + ) ln w ( − , + ) + [ 1 − w ( − , + ) ] ln [ 1 − w ( − , + ) ]$\nWe use $〈 H 1 ( k ) 〉$ and $〈 H 2 ( k ) 〉$ as constraints in the maximisation procedure (2)\n$L = S − θ 1 〈 H 1 ( k ) 〉 − θ 2 〈 H 2 ( k ) 〉$\nBy solving the following set of equations\n$∂ L ∂ w ( + , − ) = 0 and ∂ L ∂ w ( − , + ) = 0$\none can express $θ 1$ and $θ 2$ as a function of the microscopic parameters\n$4 J θ 1 = ln 1 − w ( + , − ) w ( + , − ) 1 − w ( − , + ) w ( − , + ) and 2 θ 2 = ln 1 − w ( − , + ) 1 − w ( + , − )$\nBy inverting (25), one gets expressions for $w ( + , − )$ and $w ( − , + )$ as a function of $θ 1$ and $θ 2$\n$1 − w ( − , + ) = e 2 θ 2 [ 1 − w ( + , − ) ] 1 − w ( + , − ) = cosh ( θ 2 ) − sinh 2 ( θ 2 ) + e − 4 J θ 1 1 − e − 4 J θ 1 − 1 e − θ 2$\nIn combination with (21), one finally obtains formulas for $〈 H 1 ( k ) 〉$ and $〈 H 2 ( k ) 〉$ as a function of $θ 1$ and $θ 2$. These relations can then be used to estimate the values of $θ 1$ and $θ 2$ after measurement of $〈 H 1 ( k ) 〉$ and $〈 H 2 ( k ) 〉$.\n\n## 4. General Theory\n\nOur microscopic model is the N-state Markov chain with parameters $p ( z )$ and $w ( z , y )$. The only constraints on these microscopic parameters are the normalisation conditions (5), (6) and the detailed balance conditions (8). To proceed from this mathematical model to a physical model one has to make a choice for the relevant observables $H i ( k )$. Then one can introduce Lagrange multipliers ${ θ i , α , ζ ( z ) , η ( z , y ) }$ and maximise the following function\n$1 n L = 1 n S − 1 n ∑ i = 1 t θ i 〈 H i ( k ) 〉 − α ∑ z ∈ Γ p ( z ) − ∑ z ∈ Γ ζ ( z ) ∑ y ∈ Γ w ( z , y ) − ∑ z ∈ Γ ∑ y ∈ Γ , y > z η ( z , y ) p ( z ) w ( z , y ) − p ( y ) w ( y , z )$\nover the parameters $p ( z )$ and $w ( z , y )$. Notice the fundamental difference between the constraints that are taken into account by the Lagrange multipliers ${ θ i }$ and ${ α , ζ ( z ) , η ( z , y ) }$. The former should be determined as a function of the model parameters $p ( z )$ and $w ( z , y )$. The latter are mathematical tools to take into account some basic microscopic constraints. These multipliers should be eliminated out of the theory since they are not connected to macroscopic observables.\nBefore (27) can be maximised over $p ( z )$ and $w ( z , y )$, the parameter dependence of S and $〈 H i ( k ) 〉$ must be know. We already obtained a formula for the entropy as a function of $p ( z )$ and $w ( z , y )$ only, see expression (11). In this paper, we assume that the observables $H i ( k )$ are linear combinations of the elements of the transition record of the Markov chain\n$∑ i = 1 t θ i H i ( k ) = ∑ z ∈ Γ ∑ y ∈ Γ Θ ( z , y ) k ( z , y )$\nwhere the elements of the matrix Θ are some linear combination of the Lagrange multipliers $θ i$. Taking the ensemble average of (28) and using expression (10) results in\n$1 n ∑ i = 1 t θ i 〈 H i ( k ) 〉 = ∑ z ∈ Γ ∑ y ∈ Γ Θ ( z , y ) p ( z ) w ( z , y )$\nIn Appendix 1, the optimisation of the function (27) is carried out analytically. One ends up with the following set of equations\n$ln w ( x , y ) w ( x , x ) w ( y , x ) w ( y , y ) = Θ ( x , x ) + Θ ( y , y ) − Θ ( x , y ) − Θ ( y , x ) ln w ( y , y ) w ( x , x ) = Θ ( x , x ) − Θ ( y , y )$\nfor all $x , y ∈ Γ$. These $N ( N − 1 ) / 2 + N − 1 = ( N − 1 ) ( N + 2 ) / 2$ equations together with the $N ( N − 1 ) / 2$ detailed balance conditions (8) and the $1 + N$ normalisation conditions (5), (6) are a closed set of equations for the $N + N 2$ microscopic parameters $p ( x )$ and $w ( x , y )$. To obtain relations for $p ( x )$ and $w ( x , y )$ as a function of the parameters $θ i$ (contained in the elements of the matrix Θ), one has to invert this set of equations. A part of this inversion can be performed generally. Start by choosing an arbitrary state r and rewrite the relations (8) and (5) as follows:\n$p ( r ) = 1 + ∑ y ∈ Γ ′ w ( r , y ) w ( y , r ) − 1 , p ( x ) = p ( r ) w ( r , x ) w ( x , r ) , ∀ x ∈ Γ ′$\nwith $Γ ′ = Γ ∖ { r }$. Then, the remaining detailed balance conditions (8) can be rewritten as follows:\n$w ( r , x ) w ( x , y ) w ( y , r ) = w ( x , r ) w ( y , x ) w ( r , y ) , ∀ x , y ∈ Γ ′$\nNotice that (31) expresses the probabilities $p ( x )$ as a function of the transition probabilities $w ( x , y )$ only. Therefore to obtain relations for $p ( x )$ and $w ( x , y )$ as a function of the parameters $θ i$ one only has to invert the relations (30), (32) together with the normalisation conditions (6). This part of the inversion will depend on the particular form of the matrix Θ and has to be performed for every physical model individually.\nWe want to emphasise that our procedure fits in the estimation theory [24,25]. In that approach, the average values of some observables are used to estimate the values of the model parameters. In the present paper, we make a separation between the physical model of a theory and the underlying mathematical model. The latter is the N-state Markov chain while the former model is introduced by identifying some physically relevant observables. Usually, the number of microscopic parameters $p ( x )$ and $w ( x , y )$ of the mathematical model is larger than the number of relevant observables $H i ( k )$ of the physical model. By measuring $〈 H i ( k ) 〉$, only the values of the corresponding parameters $θ i$ can be estimated. Then, one can calculate an estimation of all the values of $p ( x )$ and $w ( x , y )$ with the formulas obtained in this section. As such, no a priori choice for these parameters is necessary and one obtains the least biased values for $p ( x )$ and $w ( x , y )$ given only the measured information and some basic microscopic constraints.\n\n## 5. Thermodynamics\n\nIn statistical mechanics, the starting point to describe a given model is usually the Hamiltonian $H ( j )$ with $j ∈ Γ$ and Γ the phase space. Then, the standard way of introducing the temperature T is by postulating the Boltzmann-Gibbs form $exp [ − H ( j ) / T ]$ for the equilibrium distribution. This approach is motivated by the maximum entropy principle that we already outlined in the introduction. Indeed, when the Hamiltonian is identified as the only relevant observable, expression (3) for the equilibrium distribution simplifies to $exp − θ 1 H ( j )$. Notice that this corresponds to the choice $t = 1$ and $H 1 ( j ) = H ( j )$. Using the laws of thermodynamics, one can show that $θ 1$ is indeed the inverse temperature.\nIn previous sections, we used the maximum entropy principle to obtain the least biased values of the microscopic parameters of a mathematical model gives some macroscopic constraints. Since this problem fits in the estimation theory, we did not give a thermodynamic interpretation of the Lagrange multipliers $θ i$. However, such a deeper understanding is highly desirable for the application of our theory to physical models like, e.g., the Ising chain. Therefore, in this section we study this thermodynamic interpretation in more detail. We first outline briefly the concept of exponential families which is very important in this context. Then we illustrate the relation between the Lagrange multipliers $θ i$ and the temperature for the 2-state Markov chain. Finally, we show under which conditions our technique coincides with the standard approach to introduce thermodynamic parameters into theoretical models.\n\n#### 5.1. Curved exponential family\n\nA distribution with parameters $w = [ w 1 , … , w s ]$ belongs to the t-parameter exponential family when it can be written as follows:\n$P ( j ) = c ( j ) exp G ( w ) − ∑ i = 1 t θ i ( w ) g i ( j )$\nwhere t is the smallest integer for which the exponential form can be obtained, $c ( j )$ is a prior probability and $G ( w )$ is determined by the normalisation condition\n$G ( w ) = − ln ∑ j c ( j ) exp − ∑ i = 1 t θ i ( w ) g i ( j )$\nThe family is said to be curved when $t > s$ (s is the dimension of the parameter vector w, see above). The special case for which $s = t$ is called a full exponential family. Then, one can interpret the functions $θ i ( w )$ as the new parameters of the distribution $θ = [ θ 1 , … , θ t ]$. The Boltzmann-Gibbs form is obtained when also the functions $g i ( j )$ coincide with the relevant observables $H i ( j )$\n$P ( j ) = c ( j ) exp G ( θ ) − ∑ i = 1 t θ i H i ( j )$\nIn the next section, we will show that the subtle differences between the curved and the full exponential family are very important for the thermodynamic interpretation of the parameters θ. A generalisation of the concept of exponential families with applications in the context of nonextensive statistical mechanics is proposed by Naudts [25,27,28].\n\n#### 5.2. Example: the 2-state Markov chain\n\nWe studied the 2-state Markov chain already in Section 3. and interpreted this model as a one-dimensional Ising chain. We identified two relevant observables $H 1 ( k )$ and $H 2 ( k )$, see expressions (19) and (20) respectively, and used $〈 H 1 ( k ) 〉$ and $〈 H 2 ( k ) 〉$ as constraints in the maximisation procedure. As a consequence, the matrix Θ (28) becomes\n$Θ = Θ ( 1 , 1 ) Θ ( 1 , 2 ) Θ ( 2 , 1 ) Θ ( 2 , 2 ) = − J θ 1 + θ 2 J θ 1 − θ 2 J θ 1 + θ 2 − J θ 1 − θ 2$\nUsing (30), the parameters $θ 1$ and $θ 2$ can then be expressed as a function of the microscopic parameters. It is easy to check that this procedure results in the same formulas for $θ 1$ and $θ 2$ that we obtained before (25), as it should be. Clearly, $〈 H 2 ( k ) 〉$ is just the magnetisation M of the chain, while $〈 H 1 ( k ) 〉$ is usually interpreted as the internal energy U of the one-dimensional Ising model. Within this interpretation, the parameters $θ 1$ and $θ 2$ can be related to the temperature T and an external applied field F as follows: $θ 1 = 1 / T$ and $θ 2 = − F / T$. One can check this, e.g., by showing that the following thermodynamic relations hold\n$∂ β G ∂ β = U − F M and ∂ G ∂ F = − M$\nwith G the free energy $G = U − F M − T S$. The final expression for the magnetisation as a function of T and F is\n$M n = sinh ( F / T ) sinh 2 ( F / T ) + e − 4 J / T − 1 / 2$\nThis is the well-known result for the one-dimensional Ising model . We proceed by studying the equilibrium distribution that is obtained by our optimisation procedure in order to understand why our final formula for the magnetisation (38) coincides with the standard result. The probability $P ( k )$ to observe a Markov chain with certain transition record k is proportional to\n$P ( k ) ∼ w ( + , + ) k ( + , + ) w ( + , − ) k ( + , − ) w ( − , − ) k ( − , − ) w ( − , + ) k ( − , + )$\nNow we want to express this probability as a function of the relevant variables $H 1 ( k )$, $H 2 ( k )$ and n. Expressions for $H 1 ( k )$ and $H 2 ( k )$ as a function of the elements of k are already given in (19) and (20). The length of the Ising chain $n + 1$ is just the sum of all the elements of k plus 1, i.e.,\n$n = k ( + , + ) + k ( + , − ) + k ( − , − ) + k ( − , + )$\nIn this way, we obtain only 3 equations for 4 variables, the 4 elements of the transition record. However notice that the difference between the values of $k ( − , + )$ and $k ( + , − )$ can only be 0 or 1. Therefore in the thermodynamic limit we have an extra constraint for the elements of k\n$k ( + , − ) = k ( − , + )$\nIn this way we end up with a closed set of equations (19,20,40,41) for the 4 elements of k. By solving this set of equations for $H 1 ( k )$, $H 2 ( k )$ and n, we can rewrite expression (39) as follows:\n$P ( k ) ∼ exp [ − θ 1 H 1 ( k ) − θ 2 H 2 ( k ) ]$\nsee (25) for the definitions of $θ 1$ and $θ 2$. We omitted the dependence of n to obtain (42), because the length of the Ising chain is assumed to be constant. Rewriting $P ( k )$ in this form makes the thermodynamic interpretation of the parameters $θ 1$ and $θ 2$ as $θ 1 = 1 / T$ and $θ 2 = − F / T$ immediately clear because expression (42) is just the Boltzmann-Gibbs distribution $exp [ − θ 1 H ( k ) ]$, with $H ( k ) = H 1 ( k ) + H 2 ( k ) θ 2 / θ 1$. This is indeed the Hamiltonian of the Ising chain. We conclude that for this simple example, our technique to introduce the thermodynamic temperature in a statistical model coincides with the standard approach. As a consequence, it is no surprise that our expression for the magnetisation (38) is equal to the well-known result for the one-dimensional Ising model.\nNotice that ignoring equation (41) in this procedure results in an extra contribution to expression (42)\n$P ( k ) ∼ exp [ − θ 1 H 1 ( k ) − θ 2 H 2 ( k ) − θ 3 H 3 ( k ) ]$\nwith\n$H 3 ( k ) = k ( + , − ) − k ( − , + ) and θ 3 = 1 2 ln w ( − , − ) w ( + , + ) w ( − , + ) w ( + , − )$\nObserve that the distribution (43) is a member of the curved exponential family. Indeed, the 2 independent parameters are $w = [ w ( + , − ) , w ( − , + ) ]$. However, in order to rewrite $P ( k )$ in an exponential form, one needs 3 functions $θ = [ θ 1 , θ 2 , θ 3 ]$ of these parameters instead of 2. Therefore, the distribution (43) belongs to the curved exponential family while the distribution (42) is a member of the full exponential family. As a consequence, the interpretation of the parameters of the distribution (43) is not immediately clear. As we mentioned before, the difference between the values of $k ( − , + )$ and $k ( + , − )$ can only be 0 or 1. So for this particular example, the difference between the full and curved exponential family only occurs for finite systems. However, the problem is more general. In this paper, we study mathematical models with an arbitrary number of microscopic parameters. The number of physically relevant observables $H i ( k )$ is usually a lot smaller. Therefore, it is not obvious whether the distribution $P ( k )$ belongs to the full or curved exponential family in the variables $H i ( k )$. Or equivalently, it is not obvious whether it is possible to rewrite the probability $P ( k )$ in the Boltzmann-Gibbs form. We examine this question for the N-state Markov chain in the next section.\n\n#### 5.3. Boltzmann-Gibbs distribution\n\nThe conditional probability $P ( k ; x 0 )$ to observe a Markov chain with certain transition record k given the initial condition $x 0 ∈ Γ$ can be written as (12). We also derived relations between the elements of the transition record (14) that must be satisfied in order to obtain a non-vanishing value for $P ( k ; x 0 )$. The dependence of $P ( k ; x 0 )$ on the initial condition $x 0$ is unimportant for large Markov chains. Therefore, we ignore the dependence of $x 0$ in the remaining part of this section and write the probability $P ( k )$ to observe a Markov chain with certain transition record k as follows:\n$P ( k ) = c ( k ) exp ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) ln w ( x , y ) : = c ( k ) exp Υ ( k )$\nWhen finite size effects are ignored, expression (14) simplifies to\n$∑ y ∈ Γ k ( r , y ) = ∑ y ∈ Γ k ( y , r ) , ∀ r ∈ Γ$\nNotice that these relations are the generalisation of expression (41). In this paper we have assumed that the detailed balance conditions (8) hold and derived relations (30) between $w ( x , y )$ using the maximum entropy principle. The aim of this section is to examine whether the conditions (46) together with (8) and (30) are sufficient to rewrite $P ( k )$ as the Boltzmann-Gibbs distribution (35)\n$P ( k ) ∼ c ( k ) exp − ∑ i = 1 t θ i H i ( k ) = c ( k ) exp − ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) Θ ( x , y ) .$\nSee expression (28) for the definition of the matrix Θ. Notice that we omitted the dependence of the normalisation $G ( θ )$. By using the conditions (8), (46) and (30), we rewrite $Υ ( k )$ (45) as follows:\n$Υ ( k ) = 1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) ln w ( x , y ) + 1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) ln w ( x , y ) = 1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) ln w ( x , y ) + 1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) ln w ( y , x ) p ( y ) p ( x ) = 1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) ln w ( x , y ) w ( y , x ) = 1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) Θ ( x , x ) + Θ ( y , y ) − Θ ( x , y ) − Θ ( y , x ) + ln w ( x , x ) w ( y , y )$\nThen, we use (46) and (30) again together with $∑ x , y k ( x , y ) = n$ to prove the following equality\n$1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) Θ ( x , x ) + Θ ( y , y ) + ln w ( x , x ) w ( y , y ) = n Θ ( r , r ) + ln w ( r , r )$\nwith r an arbitrary state. Since we assumed n to be constant, this term can be absorbed in the normalisation of the distribution $P ( k )$. Therefore, we end up with the following expression\n$P ( k ) ∼ c ( k ) exp − 1 2 ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) Θ ( x , y ) + Θ ( y , x )$\nThe distribution $P ( k )$ is only of the Boltzmann-Gibbs form when the following condition holds (compare (50) with (47))\n$0 = ∑ x ∈ Γ ∑ y ∈ Γ k ( x , y ) Θ ( x , y ) − Θ ( y , x )$\nor equivalently\n$0 = ∑ x ∈ Γ ∑ y ∈ Γ [ k ( x , y ) − k ( y , x ) ] Θ ( x , y )$\nA sufficient condition for the equality (51) to hold is obviously $Θ ( x , y ) = Θ ( y , x )$ for all $x , y ∈ Γ$. However, the results of previous section show that this constraint is to restrictive. Indeed, we showed for a simple example that one can rewrite the distribution $P ( k )$ in the Boltzmann-Gibbs form, without the matrix Θ (36) being symmetric. The crucial observation to obtain this result was that following constraint, $k ( + , − ) = k ( − , + )$, is fulfilled for the 2-state Markov chain. For N-state Markov chains, the latter equality can be generalised to (46). We proceed by eliminating some of the elements of the transition record out of expression (51) by using the conditions (46). Chose an arbitrary state r and replace in expression (51), $k ( x , r )$ by\n$k ( x , r ) = k ( r , x ) + ∑ y ∈ Γ ′ k ( y , x ) − k ( x , y )$\nwith $Γ ′ = Γ ∖ { r }$. This results in the following condition\n$0 = ∑ x ∈ Γ ′ ∑ y ∈ Γ ′ k ( x , y ) Θ ( x , y ) − Θ ( y , x ) − Θ ( x , r ) + Θ ( r , x ) + Θ ( y , r ) − Θ ( r , y )$\nA sufficient condition for this equality to hold is\n$Θ ( r , x ) + Θ ( x , y ) + Θ ( y , r ) = Θ ( x , r ) + Θ ( y , x ) + Θ ( r , y )$\nNotice that this is a similar constraint to the detailed balance condition for the transition probabilities (32). This derivation does not depend on the arbitrary chosen state r. As such, expression (55) should hold for all $r , x , y ∈ Γ$. For the examples studied in this paper, condition (55) will always be satisfied. As a consequence, the relation between the Lagrange multipliers $θ i$ (contained in the matrix Θ) and the thermodynamic parameters is immediately clear. Therefore, in the remaining part of this paper, we will omit the substep of explicitly checking thermodynamic relations like (37) during our analysis.\n\n## 6. Example: The 3-state Markov Chain\n\nIn this section we study the 3-state Markov chain. The three states are denoted 1, 2 and 3. The different parameters of the Markovian model are\n$p ( 1 ) , p ( 2 ) , p ( 3 ) and w ( 1 , 1 ) , w ( 1 , 2 ) , w ( 1 , 3 ) , w ( 2 , 1 ) , w ( 2 , 2 ) , w ( 2 , 3 ) , w ( 3 , 1 ) , w ( 3 , 2 ) , w ( 3 , 3 )$\nHowever, the number of independent parameters is reduced by 4 because of the normalisation conditions (5), (6). The detailed balance conditions (8) further reduce the number of independent parameters by 3. We conclude that this microscopic model contains 5 independent parameters. In Section 2. we introduced the transition record k. For this example, the matrix k contains 9 elements\n$k = k ( 1 , 1 ) k ( 1 , 2 ) k ( 1 , 3 ) k ( 2 , 1 ) k ( 2 , 2 ) k ( 2 , 3 ) k ( 3 , 1 ) k ( 3 , 2 ) k ( 3 , 3 )$\nIn the next two sections we study two physical models that are contained in this 3-state Markov chain. The relevant observables of these two models are different and, as such, the elements of the matrix Θ are not equal. As a consequence, the constraints (30) that relate the thermodynamic parameters (contained in the matrix Θ) to the microscopic parameters $w ( x , y )$ will be different for the two physical models. However, the relations (31), (32) and (6) are the same because they only depend on the mathematical 3-state model and not on the particular choice of the relevant observables. Therefore we write out the formulas (31), (32) and (6) here, before we proceed with studying the expressions (30) in the next two sections. We choose $r = 1$ in expressions (31) and (32)\n$p ( 1 ) = w ( 2 , 1 ) w ( 3 , 1 ) w ( 2 , 1 ) w ( 3 , 1 ) + w ( 1 , 2 ) w ( 3 , 1 ) + w ( 2 , 1 ) w ( 1 , 3 ) p ( 2 ) = w ( 1 , 2 ) w ( 2 , 1 ) p ( 1 ) , p ( 3 ) = w ( 1 , 3 ) w ( 3 , 1 ) p ( 1 )$\n$w ( 3 , 1 ) w ( 1 , 2 ) w ( 2 , 3 ) = w ( 1 , 3 ) w ( 2 , 1 ) w ( 3 , 2 )$\nand write out the normalisation conditions (5) for the transition probabilities\n$w ( 1 , 1 ) = 1 − w ( 1 , 2 ) − w ( 1 , 3 ) , w ( 2 , 2 ) = 1 − w ( 2 , 1 ) − w ( 2 , 3 ) w ( 3 , 3 ) = 1 − w ( 3 , 1 ) − w ( 3 , 2 )$\n\n#### 6.1. Potts model\n\nThe 3-state Markov chain can be interpreted as a one-dimensional Potts model [11,12]. This system corresponds to a chain of $n + 1$ spins. Contrary to the Ising model, the spin variables $σ i$ are vectors with unit length that can point in 3 directions specified by the angles $2 q π / 3$ with $q = 0 , 1 , 2$. Two relevant observables are\n$H 1 ( σ ) = − J ∑ i = 0 n − 1 σ i · σ i + 1 and H 2 ( σ ) = ∑ i = 0 n 1 · σ i$\nwhere J is a constant and $1$ is a unit vector that points in one of the spin directions. Clearly, $〈 H 2 ( σ ) 〉$ is just the magnetisation M of the chain along the direction of $1$, while $〈 H 1 ( σ ) 〉$ is usually interpreted as the internal energy U of the one-dimensional Potts model. The three states of the Markov chain $1 , 2 , 3$ correspond to the three different spin directions. The contribution to $H 1 ( σ )$ is $− J$ or $J / 2$ depending on whether $σ i = σ i + 1$ or $σ i ≠ σ i + 1$ respectively. Therefore, one can express $H 1 ( σ )$ as a function of the elements of the transition record k as follows:\n$H 1 ( k ) = − J [ k ( 1 , 1 ) + k ( 2 , 2 ) + k ( 3 , 3 ) ] + J 1 2 [ k ( 1 , 2 ) + k ( 1 , 3 ) + k ( 2 , 1 ) + k ( 2 , 3 ) + k ( 3 , 1 ) + k ( 3 , 2 ) ]$\nIn order to obtain a similar expression for $H 2 ( σ )$, we chose arbitrarily the direction of $1$ along the state with label 1\n$H 2 ( k ) = k ( 1 , 1 ) + k ( 2 , 1 ) + k ( 3 , 1 ) − 1 2 [ k ( 1 , 2 ) + k ( 2 , 2 ) + k ( 3 , 2 ) + k ( 1 , 3 ) + k ( 2 , 3 ) + k ( 3 , 3 ) ]$\nAnalogous to the example of the Ising model, see Section 3., we ignored the unimportant finite size contribution of the initial spin $σ 0$ to obtain $H 2 ( k )$ from $H 2 ( σ )$. We use $〈 H 1 ( k ) 〉$ and $〈 H 2 ( k ) 〉$ as constraints in the maximisation procedure. As a consequence, the matrix Θ (28) becomes\n$Θ = Θ ( 1 , 1 ) Θ ( 1 , 2 ) Θ ( 1 , 3 ) Θ ( 2 , 1 ) Θ ( 2 , 2 ) Θ ( 2 , 3 ) Θ ( 3 , 1 ) Θ ( 3 , 2 ) Θ ( 3 , 3 ) = 1 2 − 2 J θ 1 + 2 θ 2 J θ 1 − θ 2 J θ 1 − θ 2 J θ 1 + 2 θ 2 − 2 J θ 1 − θ 2 J θ 1 − θ 2 J θ 1 + 2 θ 2 J θ 1 − θ 2 − 2 J θ 1 − θ 2$\nUsing (30), the parameters $θ 1$ and $θ 2$ can then be expressed as a function of the microscopic parameters as follows:\n$3 J θ 1 = ln w ( 1 , 1 ) w ( 1 , 2 ) w ( 2 , 2 ) w ( 2 , 1 ) = ln w ( 1 , 1 ) w ( 1 , 3 ) w ( 3 , 3 ) w ( 3 , 1 ) = ln w ( 2 , 2 ) w ( 2 , 3 ) w ( 3 , 3 ) w ( 3 , 2 ) 3 2 θ 2 = ln w ( 2 , 2 ) w ( 1 , 1 ) = ln w ( 3 , 3 ) w ( 1 , 1 )$\nTogether with (58) and (59) these expressions form a closed set of equations that relate the parameters $θ 1$ and $θ 2$ to the microscopic parameters $w ( x , y )$. In Appendix 2, this set is inverted analytically. We proceed by writing out the ensemble average of $H 2 ( k )$ with (10), (57) and (59)\n$〈 H 2 ( k ) 〉 n = − 1 2 + 3 2 w ( 2 , 1 ) w ( 2 , 1 ) + 2 w ( 1 , 2 )$\nIn combination with the results of Appendix 2, one then obtains a formula for $〈 H 2 ( k ) 〉$ as a function of $θ 1$ and $θ 2$. These parameters can be related to the temperature T and an external applied field F as follows: $θ 1 = 1 / T$ and $θ 2 = − F / T$. The final expression for the magnetisation $M = 〈 H 2 ( k ) 〉$ as a function of T and F is\n$M n = 1 4 + 3 4 e 3 J / 2 T − 1 + e 3 J / 2 T e − 3 F / 2 T e 3 J / 2 T − 1 + e 3 J / 2 T e − 3 F / 2 T 2 + 8 e − 3 F / 2 T$\nAn identical expression is obtained in . The authors assume that the equilibrium distribution is of the Boltzmann-Gibbs form and solve the one-dimensional Potts model with the technique of the transfer-matrix. Since the Hamiltonian of this model satisfies (55), the resulting equilibrium distribution of our approach is also of the Boltzmann-Gibbs form. That’s the reason why the final expressions for M as a function of T and F of the two different approaches coincide.\n\n#### 6.2. Blume-Emery-Griffiths model\n\nThe 3-state Markov chain can be interpreted as a one-dimensional Blume-Emery-Griffiths model [13,14]. This system corresponds to a chain of $n + 1$ spins. The spin variables $σ i$ are scalars that can take on three values $+ 1 , 0 , − 1$. Two relevant observables are\n$H 1 ( σ ) = − J ∑ i = 0 n − 1 σ i σ i + 1 − K ∑ i = 0 n − 1 σ i 2 σ i + 1 2 + Δ ∑ i = 0 n σ i 2 and H 2 ( σ ) = ∑ i = 0 n σ i$\nwhere $J , K , Δ$ are constants. Clearly, $〈 H 2 ( σ ) 〉$ is just the magnetisation M of the chain, while $〈 H 1 ( σ ) 〉$ is usually interpreted as the internal energy U of the one-dimensional Blume-Emery-Griffiths model. The three states of the Markov chain $1 , 2 , 3$ correspond to the spin values $+ 1 , 0 , − 1$ respectively. Within this interpretation, one can express $H 1 ( σ )$ and $H 2 ( σ )$ as a function of the elements of the transition record k as follows:\n$H 1 ( k ) = − J [ k ( 1 , 1 ) + k ( 3 , 3 ) − k ( 1 , 3 ) − k ( 3 , 1 ) ] − K [ k ( 1 , 1 ) + k ( 3 , 3 ) + k ( 1 , 3 ) + k ( 3 , 1 ) ] + Δ [ k ( 1 , 1 ) + k ( 2 , 1 ) + k ( 3 , 1 ) + k ( 1 , 3 ) + k ( 2 , 3 ) + k ( 3 , 3 ) ]$\n$H 2 ( k ) = k ( 1 , 1 ) + k ( 2 , 1 ) + k ( 3 , 1 ) − [ k ( 1 , 3 ) + k ( 2 , 3 ) + k ( 3 , 3 ) ]$\nAnalogous to previous examples, we ignored the contribution of $Δ σ 0 2$ to obtain $H 1 ( k )$ from $H 1 ( σ )$ and the contribution of $σ 0$ to obtain $H 2 ( k )$ from $H 2 ( σ )$. We use $〈 H 1 ( k ) 〉$ and $〈 H 2 ( k ) 〉$ as constraints in the maximisation procedure. As a consequence, the matrix Θ (28) becomes\n$Θ = Θ ( 1 , 1 ) Θ ( 1 , 2 ) Θ ( 1 , 3 ) Θ ( 2 , 1 ) Θ ( 2 , 2 ) Θ ( 2 , 3 ) Θ ( 3 , 1 ) Θ ( 3 , 2 ) Θ ( 3 , 3 ) = θ 1 ( − J − K + Δ ) + θ 2 0 θ 1 ( J − K + Δ ) − θ 2 θ 1 Δ + θ 2 0 θ 1 Δ − θ 2 θ 1 ( J − K + Δ ) + θ 2 0 θ 1 ( − J − K + Δ ) − θ 2$\nUsing (30), the parameters $θ 1$ and $θ 2$ can then be expressed as a function of the microscopic parameters as follows:\n$− θ 1 ( J + K ) = ln w ( 1 , 2 ) w ( 1 , 1 ) w ( 2 , 1 ) w ( 2 , 2 ) = ln w ( 2 , 3 ) w ( 2 , 2 ) w ( 3 , 2 ) w ( 3 , 3 ) − 4 θ 1 J = ln w ( 1 , 3 ) w ( 1 , 1 ) w ( 3 , 1 ) w ( 3 , 3 ) − θ 1 ( J + K − Δ ) + θ 2 = ln w ( 2 , 2 ) w ( 1 , 1 ) − θ 1 ( J + K − Δ ) − θ 2 = ln w ( 2 , 2 ) w ( 3 , 3 )$\nTogether with (58) and (59) these expressions form a closed set of equations that relate the parameters $θ 1$ and $θ 2$ to the microscopic parameters $w ( x , y )$. In Appendix 3, this set is inverted analytically. As in previous example, the parameters $θ 1$ and $θ 2$ are related to the temperature T and an external applied field F as follows: $θ 1 = 1 / T$ and $θ 2 = − F / T$. We proceed by writing out the ensemble average of $H 2 ( k )$ (69) with (10), (57) and (59)\n$〈 H 2 ( k ) 〉 n = w ( 3 , 1 ) w ( 2 , 1 ) − w ( 2 , 1 ) w ( 1 , 3 ) w ( 3 , 1 ) w ( 2 , 1 ) + w ( 1 , 3 ) w ( 2 , 1 ) + w ( 1 , 2 ) w ( 3 , 1 )$\nIn combination with the results of Appendix 3, one finally obtains an expression for the magnetisation $M = 〈 H 2 ( k ) 〉$ as a function of T and F. A plot of the magnetisation as a function of the external applied field at constant temperature $θ 1 = 1 / T = 20$ is shown in Figure 1 for the following values of the constants of $H 1 ( k )$ (68) $K = 0$, $J = − 1$ and $Δ = 0 ; 0 . 5 ; 1$. It is known that multiple plateaus show up is this curve depending on the value of Δ [14,29]. This interesting behaviour can also be observed in Figure 1.\nFigure 1. Plot of the magnetisation of the one-dimensional Blume-Emery-Griffiths model as a function of the external applied field at constant temperature $1 / T = 20$. The values of the constants of $H 1$ (68) are $K = 0$, $J = − 1$ and $Δ = 0 ; 0 . 5 ; 1$ for the dotted, the solid, the dashed line respectively.\nFigure 1. Plot of the magnetisation of the one-dimensional Blume-Emery-Griffiths model as a function of the external applied field at constant temperature $1 / T = 20$. The values of the constants of $H 1$ (68) are $K = 0$, $J = − 1$ and $Δ = 0 ; 0 . 5 ; 1$ for the dotted, the solid, the dashed line respectively.\n\n## 7. Discussion\n\nIn this paper, we present a general procedure to estimate parameters in Markovian models. The Markov chain is a mathematical model that is defined by initial probabilities $p ( z )$ and transition probabilities $w ( z , y )$. We interpret $p ( z )$ and $w ( z , y )$ as the microscopic parameters of the Markovian model. Then, relations between $p ( z ) , w ( z , y )$ and some relevant control parameters $θ i$ are determined with the maximum entropy principle. Finally, one ends up with formulas that express the average values of the relevant observables $H i ( k )$ as a function of the corresponding control parameters $θ i$ only. These expressions can be used to estimate the values of $θ i$ after the measurement of $〈 H i ( k ) 〉$. We want to stress that the dependence on the microscopic parameters is completely eliminated out of the theory. This means that no a priori choice for the values of $p ( x )$ or $w ( x , y )$ is necessary. This is important because the values of these parameters are not measurable.\nWe made a clear separation between the physical model of a theory and the underlying mathematical model. The latter is the N-state Markov chain while the former model is introduced by identifying some relevant observables. As such, different physical models can be contained in one type of Markov chain. This is illustrated in Section 6. where we examined two different physical models that are contained in the 3-state Markov chain. We showed that is possible to perform the aforementioned optimisation procedure in full generality for the N-state Markov chain. This results in relations (30) between the microscopic parameters of the mathematical model and some relevant control parameters. These formulas are the main result of this paper.\nIn Section 5.3. we studied under which conditions the equilibrium distribution of our approach is of the Boltzmann-Gibbs form. Obtaining this type of equilibrium distribution is advantageous because a thermodynamic interpretation of the control parameters is obvious in that case. We derived a sufficient condition that is satisfied for all the examples studied in this paper. As such, the final formulas for the average values of the relevant observables as a function of the thermodynamic parameters that are obtained in this paper have been studied before. The general procedure to obtain these formulas is the novel contribution of this paper. Notice that further generalisations of our technique are still possible. We assumed that the relevant observables are linear combinations of the elements of the transition record of the Markov chain. In , this condition is lifted. In that paper, the specific example of the 2-state Markov chain with a mean-field Hamiltonian is studied with the same technique as described in the present paper. It is an interesting topic for further research to examine the effect of allowing mean-field Hamiltonians in the theory for the general N-state Markov chain. We also assumed that all the transitions are allowed ($w ( x , y ) ≠ 0$ for all $x , y ∈ Γ$). Lifting this assumption will usually cause the violation of the detailed balance condition. The generalisation of the results reported in the present paper to non-equilibrium steady states is currently under study. Notice that allowing for a vanishing transition probability $w ( x , y )$ will not cause the violation of the detailed balance condition when the transition $y → x$ is also not allowed. The specific example of a 6-state Markov chain with that property is studied in . Throughout the present paper, we ignored finite size effects. The technical consequences of taking these effects into account are already thoroughly examined for the 2-state Markov chain in [5, 19].\n\n## Appendix\n\n#### Appendix 1\n\nIn this appendix we maximise the function\n$1 n L = − ∑ z ∈ Γ ∑ y ∈ Γ p ( z ) w ( z , y ) ln w ( z , y ) − ∑ z ∈ Γ ∑ y ∈ Γ Θ ( z , y ) p ( z ) w ( z , y ) − α ∑ z ∈ Γ p ( z ) − ∑ z ∈ Γ ζ ( z ) ∑ y ∈ Γ w ( z , y ) − ∑ z ∈ Γ ∑ y ∈ Γ , y > z η ( z , y ) p ( z ) w ( z , y ) − p ( y ) w ( y , z ) ,$\nover the parameters $p ( z )$ and $w ( z , y )$. Therefore, we set the first derivative of $L$ with respect to these parameters equal to zero. The resulting equations for differentiating with respect to $w ( u , u )$ (74), $p ( u )$ (75), $w ( u , v )$ (76), $w ( v , u )$ (77) with $v > u$ are\n$ζ ( u ) = − p ( u ) 1 + ln w ( u , u ) + Θ ( u , u ) ,$\n$0 = ∑ y ∈ Γ w ( u , y ) ln w ( u , y ) + ∑ y ∈ Γ Θ ( u , y ) w ( u , y ) + α + ∑ y ∈ Γ , y > u η ( u , y ) w ( u , y ) − ∑ y ∈ Γ , y < u η ( y , u ) w ( u , y ) ,$\n$0 = p ( u ) 1 + ln w ( u , v ) + Θ ( u , v ) p ( u ) + ζ ( u ) + η ( u , v ) p ( u ) ,$\n$0 = p ( v ) 1 + ln w ( v , u ) + Θ ( v , u ) p ( v ) + ζ ( v ) − η ( u , v ) p ( v ) .$\nOne can simplify the expressions (76) and (77) with the use of the formula for $ζ ( u )$ (74)\n$0 = ln w ( u , v ) w ( u , u ) + Θ ( u , v ) − Θ ( u , u ) + η ( u , v ) ,$\n$0 = ln w ( v , u ) w ( v , v ) + Θ ( v , u ) − Θ ( v , v ) − η ( u , v ) .$\nCombining these two equations results in\n$Θ ( u , u ) + Θ ( v , v ) − Θ ( u , v ) − Θ ( v , u ) = ln w ( u , v ) w ( u , u ) w ( v , u ) w ( v , v ) .$\nWe proceed by rewriting expression (75) as follows:\n$0 = w ( u , u ) ln w ( u , u ) + Θ ( u , u ) + α + ∑ y ∈ Γ , y > u w ( u , y ) ln w ( u , y ) + Θ ( u , y ) + η ( u , y ) + ∑ y ∈ Γ , y < u w ( u , y ) ln w ( u , y ) + Θ ( u , y ) − η ( y , u ) .$\nThen, we use (78) and (79) to transform (81) further to\n$0 = w ( u , u ) ln w ( u , u ) + Θ ( u , u ) + α + ∑ y ∈ Γ , y > u w ( u , y ) ln w ( u , u ) + Θ ( u , u ) + ∑ y ∈ Γ , y < u w ( u , y ) ln w ( u , u ) + Θ ( u , u ) − α = ln w ( u , u ) + Θ ( u , u ) .$\nThe latter equation is valid for all $u ∈ Γ$. The parameter α can then be eliminated, by combining these equations two by two\n$Θ ( u , u ) − Θ ( v , v ) = ln w ( v , v ) w ( u , u ) .$\n\n#### Appendix 2\n\nIn this appendix we invert the set of equations (64), (58), (59)\n$e 3 J θ 1 = 1 − w ( 1 , 2 ) − w ( 1 , 3 ) w ( 1 , 2 ) 1 − w ( 2 , 1 ) − w ( 2 , 3 ) w ( 2 , 1 ) ,$\n$e 3 2 θ 2 = 1 − w ( 2 , 1 ) − w ( 2 , 3 ) 1 − w ( 1 , 2 ) − w ( 1 , 3 ) ,$\n$1 w ( 1 , 2 ) 1 − w ( 2 , 1 ) − w ( 2 , 3 ) w ( 2 , 1 ) = 1 w ( 1 , 3 ) 1 − w ( 3 , 1 ) − w ( 3 , 2 ) w ( 3 , 1 ) ,$\n$1 − w ( 1 , 2 ) − w ( 1 , 3 ) w ( 1 , 2 ) 1 w ( 2 , 1 ) = 1 w ( 2 , 3 ) 1 − w ( 3 , 1 ) − w ( 3 , 2 ) w ( 3 , 2 ) ,$\n$1 − w ( 2 , 1 ) − w ( 2 , 3 ) = 1 − w ( 3 , 1 ) − w ( 3 , 2 ) ,$\n$w ( 3 , 1 ) w ( 1 , 2 ) w ( 2 , 3 ) = w ( 1 , 3 ) w ( 2 , 1 ) w ( 3 , 2 ) ,$\nto obtain formulas in closed form for the microscopic parameters $w ( x , y )$ as a function of $θ 1$ and $θ 2$ only. Equations (86), (88) and (89) can be simplified to\n$w ( 1 , 3 ) = w ( 1 , 2 ) , w ( 3 , 1 ) = w ( 2 , 1 ) and w ( 3 , 2 ) = w ( 2 , 3 ) .$\nNotice that this restricts the values of $w ( 1 , 2 )$ and $w ( 1 , 3 )$ to the interval $[ 0 . . 1 / 2 ]$, because of the normalisation condition. Using (90), equation (87) can be rewritten as follows:\n$w ( 1 , 2 ) = w ( 2 , 3 ) 2 2 w ( 2 , 3 ) 2 + w ( 2 , 1 ) [ 1 − w ( 2 , 1 ) − w ( 2 , 3 ) ] .$\nInserting (90) and (91) into equations (84) and (85) results in two equations in the variables $w ( 2 , 3 )$ and $w ( 2 , 1 )$. By inverting these two equations\n$w ( 2 , 1 ) = 1 − w ( 2 , 3 ) 1 + e 3 2 J θ 1 , w ( 2 , 3 ) = − 1 2 e 3 2 J θ 1 + 1 + e 3 2 J θ 1 e 3 2 θ 2 − e 3 2 J θ 1 − 1 + e 3 2 J θ 1 e 3 2 θ 2 2 + 8 e 3 2 θ 2 2 − 1 + e 3 2 J θ 1 e 3 2 J θ 1 ,$\none finally obtains a closed chain of equations for the transition probabilities.\n\n#### Appendix 3\n\nIn this appendix we invert the set of equations (71), (58), (59). We first introduce a shorthand notation\n$η 1 = e − θ 1 ( J + K ) , η 2 = e − 2 θ 1 J , η 3 = e − θ 1 ( J + K − Δ ) + θ 2 , η 4 = e − θ 1 ( J + K − Δ ) − θ 2 ,$\nand the substitution\n$X 1 = w ( 1 , 2 ) w ( 1 , 1 ) , X 2 = w ( 2 , 1 ) w ( 2 , 2 ) , X 3 = w ( 2 , 3 ) w ( 2 , 2 ) , X 4 = w ( 3 , 2 ) w ( 3 , 3 ) , X 5 = w ( 1 , 3 ) w ( 1 , 1 ) , X 6 = w ( 3 , 1 ) w ( 3 , 3 ) ,$\nto obtain the following expressions for the normalisation conditions (59)\n$w ( 1 , 1 ) = 1 + X 1 + X 5 − 1 , w ( 2 , 2 ) = 1 + X 2 + X 3 − 1 , w ( 3 , 3 ) = 1 + X 4 + X 6 − 1 ,$\nand the equations (71), (58)\n$η 1 = X 1 X 2 , η 1 = X 3 X 4 , η 2 2 = X 5 X 6 , X 1 X 3 X 6 = X 2 X 4 X 5 ,$\n$X 5 = η 3 ( 1 + X 2 + X 3 ) − 1 − X 1 , X 4 = η 4 ( 1 + X 2 + X 3 ) − 1 − X 6 .$\nWe proceed by rewriting (96) as follows:\n$η 1 = X 1 X 2 , η 1 = X 3 X 4 , η 2 X 3 = X 2 X 5 , η 2 X 2 = X 3 X 6 .$\nThen, we use the expressions (97) for $X 4$ and $X 5$ to transform (98) further to\n$X 1 = η 1 X 2 , X 6 = η 4 ( 1 + X 2 + X 3 ) − 1 − η 1 X 3 ,$\n$η 2 X 3 = X 2 η 3 ( 1 + X 2 + X 3 ) − X 2 − η 1 , η 2 X 2 = X 3 η 4 ( 1 + X 2 + X 3 ) − X 3 − η 1 .$\nFinally, we rewrite (100) as follows:\n$X 3 = X 2 η 3 ( 1 + X 2 ) − X 2 − η 1 η 2 − X 2 η 3 ,$\n$( X 2 η 2 + η 1 ) ( X 2 η 3 − η 2 ) 2 = X 2 η 3 ( 1 + X 2 ) − ( X 2 + η 1 ) × η 4 η 2 ( 1 + X 2 ) − η 4 ( X 2 + η 1 ) + X 2 η 3 − η 2 ,$\nto obtain a closed chain of equations for the variables $X i$ with $i = 1 … 6$. Expression (102) is a cubic equation in the variable $X 2$ which can have 3 real solutions. However, it is well known that one-dimensional systems with short range interactions do not exhibit phase transitions when the equilibrium distribution is of the Boltzmann-Gibbs form. Therefore, only one of the solutions of the cubic equation is physically meaningful. The other solutions are complex or result in values for some of the transition probabilities outside the interval $[ 0 , 1 ]$.\n\n## References\n\n1. Jaynes, E.T. Information theory and statistical mechanics. Phys. Rev. 1957, 106, 620–630. [Google Scholar] [CrossRef]\n2. Jaynes, E.T. Information theory and statistical mechanics II. Phys. Rev. 1957, 108, 171–190. [Google Scholar] [CrossRef]\n3. Rosenkrantz, R.D. Jaynes: Papers on Probability, Statistics and Statistical Physics; Kluwer Academic Publishers: Dordrecht, The Netherlands, 1989. [Google Scholar]\n4. Callen, H.B. 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[ "Related Resources: mechanics machines\n\n### Power Screws Design Equation and Calculator\n\nPower Screws Design Equations and Calculator", null, "Under static equilibrium conditions, the screw rotates at a constant speed in response to the input torque T shown in the free-body diagram above. In addition, the load force F, normal force N, and sliding friction force Ff act on the power screw. tive motion. The friction force opposes relative motion. Therefore, the direction of the friction force Ff will reverse when the screw translates in the direction of the load rather than against it. The torques required to raise the load TR (i.e., move the screw in the direction opposing the load) and to lower the load TL are:", null, "Where:", null, "The thread geometry parameter β includes the effect of the flank angle α as it is projected normal to the thread and as a function of the lead angle. For general-purpose single-start Acme threads, α is 14.5 degrees and β is approximately 0.968, varying less than 1 percent for diameters ranging from 1/4 in to 5 in and thread spacing ranging from 2 to 16 threads per inch. For square threads, β = 1.\n\nIn many applications, the load slides relative to a collar, thereby requiring an additional input torque Tc:\n\nTc = F µc dc / 2\n\nBall and tapered-roller thrust bearings can be used to reduce the collar torque.\n\nThe starting torque is obtained by substituting the static coefficients of friction into the above equations. Since the sliding coefficient of friction is roughly 25 percent less than the static coefficient, the running torque is somewhat less than the starting torque. For precise values of friction coefficients, specific data should be obtained from the published technical literature and verified by experiment.\n\nPower screws can be self-locking when the coefficient of friction is high or the lead is small, so that π µt dm > L or, equivalently, µf > tan λ. When this condition is not met, the screw will self-lower or overhaul unless an opposing torque is applied.\n\nA measure of screw efficiency η can be formulated to compare the work output Wo with the work input Wi:", null, "where T is the total screw and collar torque. Similarly, for one revolution or 2π radians and screw translation L,\n\nη = F L / ( 2 π T )\n\nScrew manufacturers often list output travel speed V, in in/min, as a function of required motor torque T in lbf · in, operating at n r/min, to lift the rated capacity F, in lbf. The actual efficiency for these data is therefore\n\nη = F V / ( 2 π n T )\n\nEfficiency of a square-threaded power screw with respect to lead angle X, as shown in the table below, is obtained from\n\nη = ( 1 - µ tan λ ) / ( 1 + µ cos λ )", null, "Note the importance of proper lubrication. For example, for λ = 10 degrees and µ = 0.05, η is over 75 percent. However, as the lubricant becomes contaminated with dirt and dust or chemically breaks down over time, the friction coefficient can increase to µ = 0.30, resulting in an efficiency η = 35 percent, thereby doubling the torque, horsepower, and electricity requirements.\n\nWhere:\n\nF = Axial Force applied, lbf\ndm = Mean Diameter, inches\nd = major diameter, inches\nµf = Coefficient of thread friction\nns = Number of thread starts\nα = Thread flank angle, degrees\nαn = Normalized thread flank angle, degrees\nV = in/min\n\nReferences:\n\nStandard Handbook of Machine Design, Second Edition, Shigley & Mischke", null, "", null, "" ]

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[ "# #02009b Color Information\n\nIn a RGB color space, hex #02009b is composed of 0.8% red, 0% green and 60.8% blue. Whereas in a CMYK color space, it is composed of 98.7% cyan, 100% magenta, 0% yellow and 39.2% black. It has a hue angle of 240.8 degrees, a saturation of 100% and a lightness of 30.4%. #02009b color hex could be obtained by blending #0400ff with #000037. Closest websafe color is: #000099.\n\n• R 1\n• G 0\n• B 61\nRGB color chart\n• C 99\n• M 100\n• Y 0\n• K 39\nCMYK color chart\n\n#02009b color description : Dark blue.\n\n# #02009b Color Conversion\n\nThe hexadecimal color #02009b has RGB values of R:2, G:0, B:155 and CMYK values of C:0.99, M:1, Y:0, K:0.39. Its decimal value is 131227.\n\nHex triplet RGB Decimal 02009b `#02009b` 2, 0, 155 `rgb(2,0,155)` 0.8, 0, 60.8 `rgb(0.8%,0%,60.8%)` 99, 100, 0, 39 240.8°, 100, 30.4 `hsl(240.8,100%,30.4%)` 240.8°, 100, 60.8 000099 `#000099`\nCIE-LAB 17.362, 54.62, -74.269 5.94, 2.379, 31.155 0.15, 0.06, 2.379 17.362, 92.192, 306.332 17.362, -4.954, -69.934 15.424, 41.754, -108.962 00000010, 00000000, 10011011\n\n# Color Schemes with #02009b\n\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #999b00\n``#999b00` `rgb(153,155,0)``\nComplementary Color\n• #004c9b\n``#004c9b` `rgb(0,76,155)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #50009b\n``#50009b` `rgb(80,0,155)``\nAnalogous Color\n• #4c9b00\n``#4c9b00` `rgb(76,155,0)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #9b4f00\n``#9b4f00` `rgb(155,79,0)``\nSplit Complementary Color\n• #009b02\n``#009b02` `rgb(0,155,2)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #9b0200\n``#9b0200` `rgb(155,2,0)``\n• #00999b\n``#00999b` `rgb(0,153,155)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #9b0200\n``#9b0200` `rgb(155,2,0)``\n• #999b00\n``#999b00` `rgb(153,155,0)``\n• #01004f\n``#01004f` `rgb(1,0,79)``\n• #010068\n``#010068` `rgb(1,0,104)``\n• #020082\n``#020082` `rgb(2,0,130)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #0200b5\n``#0200b5` `rgb(2,0,181)``\n• #0300ce\n``#0300ce` `rgb(3,0,206)``\n• #0300e8\n``#0300e8` `rgb(3,0,232)``\nMonochromatic Color\n\n# Alternatives to #02009b\n\nBelow, you can see some colors close to #02009b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00259b\n``#00259b` `rgb(0,37,155)``\n• #00189b\n``#00189b` `rgb(0,24,155)``\n• #000b9b\n``#000b9b` `rgb(0,11,155)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #0f009b\n``#0f009b` `rgb(15,0,155)``\n• #1c009b\n``#1c009b` `rgb(28,0,155)``\n• #29009b\n``#29009b` `rgb(41,0,155)``\nSimilar Colors\n\n# #02009b Preview\n\nThis text has a font color of #02009b.\n\n``<span style=\"color:#02009b;\">Text here</span>``\n#02009b background color\n\nThis paragraph has a background color of #02009b.\n\n``<p style=\"background-color:#02009b;\">Content here</p>``\n#02009b border color\n\nThis element has a border color of #02009b.\n\n``<div style=\"border:1px solid #02009b;\">Content here</div>``\nCSS codes\n``.text {color:#02009b;}``\n``.background {background-color:#02009b;}``\n``.border {border:1px solid #02009b;}``\n\n# Shades and Tints of #02009b\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, #000012 is the darkest color, while #fdfdff is the lightest one.\n\n• #000012\n``#000012` `rgb(0,0,18)``\n• #000025\n``#000025` `rgb(0,0,37)``\n• #010039\n``#010039` `rgb(1,0,57)``\n• #01004d\n``#01004d` `rgb(1,0,77)``\n• #010060\n``#010060` `rgb(1,0,96)``\n• #010074\n``#010074` `rgb(1,0,116)``\n• #020087\n``#020087` `rgb(2,0,135)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\n• #0200af\n``#0200af` `rgb(2,0,175)``\n• #0300c2\n``#0300c2` `rgb(3,0,194)``\n• #0300d6\n``#0300d6` `rgb(3,0,214)``\n• #0300e9\n``#0300e9` `rgb(3,0,233)``\n• #0300fd\n``#0300fd` `rgb(3,0,253)``\n• #1512ff\n``#1512ff` `rgb(21,18,255)``\n• #2825ff\n``#2825ff` `rgb(40,37,255)``\n• #3b39ff\n``#3b39ff` `rgb(59,57,255)``\n• #4f4dff\n``#4f4dff` `rgb(79,77,255)``\n• #6260ff\n``#6260ff` `rgb(98,96,255)``\n• #7674ff\n``#7674ff` `rgb(118,116,255)``\n• #8987ff\n``#8987ff` `rgb(137,135,255)``\n• #9c9bff\n``#9c9bff` `rgb(156,155,255)``\n• #b0afff\n``#b0afff` `rgb(176,175,255)``\n• #c3c2ff\n``#c3c2ff` `rgb(195,194,255)``\n• #d6d6ff\n``#d6d6ff` `rgb(214,214,255)``\n• #eae9ff\n``#eae9ff` `rgb(234,233,255)``\n• #fdfdff\n``#fdfdff` `rgb(253,253,255)``\nTint Color Variation\n\n# Tones of #02009b\n\nA tone is produced by adding gray to any pure hue. In this case, #484853 is the less saturated color, while #02009b is the most saturated one.\n\n• #484853\n``#484853` `rgb(72,72,83)``\n• #424259\n``#424259` `rgb(66,66,89)``\n• #3c3c5f\n``#3c3c5f` `rgb(60,60,95)``\n• #363665\n``#363665` `rgb(54,54,101)``\n• #30306b\n``#30306b` `rgb(48,48,107)``\n• #2b2a71\n``#2b2a71` `rgb(43,42,113)``\n• #252477\n``#252477` `rgb(37,36,119)``\n• #1f1e7d\n``#1f1e7d` `rgb(31,30,125)``\n• #191883\n``#191883` `rgb(25,24,131)``\n• #131289\n``#131289` `rgb(19,18,137)``\n• #0e0c8f\n``#0e0c8f` `rgb(14,12,143)``\n• #080695\n``#080695` `rgb(8,6,149)``\n• #02009b\n``#02009b` `rgb(2,0,155)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #02009b 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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[ "# Cost Flow Assumptions, US GAAP\n\n[/Q/]\nQ10. Cost Flow Assumptions, US GAAP\nBased on the information below, what is the cost of inventory sold on April 30, 20×1?\n\n[Information for Q10]\nEntity 10A had the following transactions in April:\n(1) April 1, purchased 200 units of merchandise at $30 per unit. (2) April 15, purchased 200 units at$32 per unit.\n(3) April 30, sold 300 units.\n\n Date Purchased Sold Unit Cost 1 200 . $30 15 200 .$32 30 . (300) .\n\n[/A/]\nUnder US GAAP, the following assumptions are permitted to determine the cost of inventory:\n(1) First-in, First-out (FIFO)\n(2) Last-in, First-out (LIFO)\n(3) Weighted average method\n\n1. First-in, First-out (FIFO)\nIf items purchased first are assumed to be sold first, cost of 150 units of merchandise sold is $30 per unit. Purchase Date Purchased Sold Inventory 1 200 (200) . 15 200 (100) 100 Ending inventory at April 30 Purchase Date Inventory Unit Cost Amount 15 100$32 $3,200 2. Last-in, First-out (LIFO) If items purchased last are assumed to be sold first, cost of 150 units of merchandise sold is$32 per unit.\n\n Purchase Date Purchased Sold Inventory 1 200 (100) 100 15 200 (200) .\n\nEnding inventory at April 30\n\n Purchase Date Inventory Unit Cost Amount 1 100 $30$3,000\n\n3. Weighted average method\nIf weighted average cost of items purchased is used, cost of 150 units of merchandise sold is $31 per unit. Purchase Date Purchased Unit Cost Amount 1 200$30 $6,000 15 200$32 $6,400 Total 400$31(*) $12,400 (*) Weighted average cost = (200 x 30 + 200 x32) / (200 + 200) units = (6,000 +6,400) / 400 units = 12,400 / 400 units = 31 per unit Units of ending inventory = 200 + 200 – 300 = 100 units Amount of ending inventory = 100 units x 31 = 3,100 Inventory Unit Cost Amount 100$31 $3,100 4. Comparison Cost flow Inventory Unit Cost Amount FIFO 100$32 $3,200 LIFO 100$30 $3,000 Average 100$31(*) \\$3,100" ]

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[ "Open access peer-reviewed chapter\n\n# Stochastic Artificial Intelligence: Review Article\n\nBy T.D. Raheni and P. Thirumoorthi\n\nSubmitted: February 13th 2019Reviewed: October 2nd 2019Published: May 27th 2020\n\nDOI: 10.5772/intechopen.90003\n\nDownloaded: 327\n\n## Abstract\n\nArtificial intelligence (AI) is a region of computer techniques that deals with the design of intelligent machines that respond like humans. It has the skill to operate as a machine and simulate various human intelligent algorithms according to the user’s choice. It has the ability to solve problems, act like humans, and perceive information. In the current scenario, intelligent techniques minimize human effort especially in industrial fields. Human beings create machines through these intelligent techniques and perform various processes in different fields. Artificial intelligence deals with real-time insights where decisions are made by connecting the data to various resources. To solve real-time problems, powerful machine learning-based techniques such as artificial intelligence, neural networks, fuzzy logic, genetic algorithms, and particle swarm optimization have been used in recent years. This chapter explains artificial neural network-based adaptive linear neuron networks, back-propagation networks, and radial basis networks.\n\n### Keywords\n\n• artificial intelligence\n• artificial neural network\n• functions\n• weights\n• bias\n• Adaline network\n• back-propagation network\n• radial basis network\n\n## 1. Introduction\n\nIn day-to-day life, artificial intelligence (AI) has brought further advantages to pattern features and human expert systems. Based on experience and through learning, it continues to gain further potential in industrial growth. The primary elements for a neural network are the neurons, which are special types of brain cells. The neuron has the ability to retain, realize, and execute the previous existence of every action.\n\nA neural network is an analytical model that is inspired directly by biological neural networks. An artificial neural network (ANN) is an information processing system and is capable of processing nonlinear relationships between inputs and outputs. The network consists of interconnected neurons and functions to produce an output pattern of a given input pattern . The learning process of a neural network takes place by itself, which means the network learns by examples that make it more powerful, so there is no need to devise an algorithm to perform a particular task. Because of the above reasons, a neural network has no internal mechanism to perform a specific task .\n\nThe network consists of nodes that are connected by weights and obtains knowledge through variations in the node weights that are being exposed as samples. Every neuron is linked to other neurons by a network link, and the network link is associated with the weights that contain instructions in the input signal. In addition, each neuron has a centralized state of its own. The centralized state is the activation level of the neuron that serves as input to the neurons. The activation level of the neuron is imparted to the other neurons. To make the neural process more beneficial, mainframe computers are used. Various computational tasks are developed using ANNs at a more rapid rate than traditional systems.\n\nAdvertisement\n\n## 2. Biological neural network\n\nHuman brains consist of neurons with a number of connections. The basic element of a neural network is called a neuron. The biological neural network consists of axons, dendrites, and a cell body (soma). Each cell performs relatively simple computations, whose nature is indistinct from slow-style networks. Dendrites are tree-like structures (dendrite trees) that accept signals from the neighboring neurons, and each branch is connected to one neuron. The tree-like dendrite structure is connected to the main body of the neuron called the soma (cell body). The cell body is a cylindrical shape that sums the incoming signal. Dendrites are connected by a synapse. A synapse is a structure that allows a nerve cell to pass electric signals to another nerve cell. An axon is a thin cylindrical cell that carries the impulse of the nerve cell. A single nerve cell has 1000 to 10,000 synapses, while 100 billion neurons are present in our brain, and every neuron has 1000 dendrites. The processing of a biological neural network is a slow process and the learning process is uncertain .\n\nThe simple biological neural network architecture is shown in Figure 1.", null, "Figure 1.Biological neural network—Simple biological neuron architecture.\n\n### 2.1 Model and elements of an artificial neural network\n\nThe network is observed as weighted direct graphs in which the directed edges of corresponding weights are connected to input and output neurons. The network receives the input in the form of a pattern and image in point (vector) form. The inputs are mathematically assigned by the notation x(n) for “n” number of inputs. An ANN model is shown in Figure 2.", null, "Figure 2.ANN model.\n\nEvery input is multiplied by its corresponding weights. To solve a problem with the network, weight is used and correspondingly weight is represented as the strength of connections between the neurons. If the weighted sum is zero, then bias is added to make the output not zero. Bias has the weight in which the input is always equal to 1 .\n\n## 3. Classification of a neural network\n\nNeural networks are classified on the basis of patterns to determine the weights correspondingly. The neurons are arranged in the form of layers and have the same activation function. Neural network processing depends on the following segments:\n\n1. Network topology\n\n2. Learning methods\n\n3. Adjustment of weights and activation functions\n\nThe networks are arranged by connecting the points or with connecting lines. Depending on the topology of the network, it is classified as follows:\n\n1. Single-layer feed-forward network\n\n2. Multilayer feed-forward network\n\n### 3.1 Single-layer feed-forward network\n\nIn this network, the signals or the information move only in a forward (one) direction from the input nodes, through the hidden nodes and to the output nodes. A single-layer network does not require cycles or loops. The network consists of output nodes in a single layer, while the inputs are directly fed to the outputs through a series of interconnected weights. Every node is calculated by its sum of the product of the weights. If the value is above threshold (above zero), then the activated value is 1 (positive), and if the value is below threshold (below zero), then the activated value is −1 (negative) . The above network functions such that input nodes are connected to the corresponding hidden nodes with different weights and result in a series of output per node. It consists of multiple neurons that are interconnected to form a single-layer network. The two layers, namely input and output layers, are used. In the input layer, the neurons pass data from one node to the other node and the inputs are scattered and perform no calculation. Each input layer a1, a2, a3, …, anis linked to each neuron in the output layer through the connection weight. Each output neuron value such as b1, b2, b3, …, bnis calculated according to the set of input values. Based on the connection weights the values of the output layer are varied accordingly. This type of network is widely used in applications like computer vision, speech recognition, and pattern classification problems. A single-layer neural network is shown in Figure 3.", null, "Figure 3.Single-layer neural network.\n\n### 3.2 Multilayer neural network\n\nA multilayer neural network is an interconnection of signals in which the inputs and calculations flow forward from the input nodes to the output nodes. The number of signals in a neural network is the number of layers in the network. It consists of more layers for estimated units, and usually the connections are interdependent in the forward path. Every neuron in a single layer is interconnected with neurons of the consequent layer. Multilayer networks use other learning algorithms such as back propagation, Hopfield, Adaline (adaptive linear neuron), etc. In this network, if a layer is connected to the input, then the layer is a hidden layer. The multilayer network is shown in Figure 4.", null, "Figure 4.Multilayer neural network.\n\nThe network consists of input layers A1, A2, …, An. To train the network we require training layers, i.e., input layers {(a(1), b(1))…(a(m), b(m))} of mtraining sets. To train the network, the gradient descent method is one of the best methods. This method seeks to find the minimum function in the network set.\n\n### 3.3 Learning methods of a neural network\n\nThe key aspect of a neural network is the ability to learn by itself. Training or learning methods help the neural network adapt itself by making appropriate adjustments and good responses. To train the network according to custom needs, there are three types of learning. Once a network has been designed for a precise application, then the network is equipped to be trained. To begin these processes, the initial weights are chosen randomly. Then, the training or learning process begins with the following techniques:\n\n1. Supervised learning\n\n2. Unsupervised learning\n\n3. Reinforcement learning\n\n#### 3.3.1 Supervised learning\n\nSupervised learning is one of the learning methods in which the data, observations, and measurements are defined with predefined classes. It is similar to how a “teacher” explains content to students. The pairing of each input vector with the target vector determines the desired output. Training pairs mainly deal with the input vector and the corresponding target vector. The training process takes place when the input vector is applied, resulting in an output vector. If the actual response differs from the target response, the network will obtain an error signal. The corresponding error signal is used to calculate the adjustment of weights so that both the actual and target outputs match.\n\nA supervised algorithm in a neural network encompasses classification and regression types for learning processes . In the classification type, outputs are confined to a finite set of data, whereas in the regression type, the output may contain analytical data within the given limits. The best execution process for error minimization is the supervised learning algorithm. Input and output data are used to train the mapping function of the network and are given by the following relation:\n\nB=f(a)E1\n\nwhere f(a) is the function of input a.\n\nThe aim is to provide an approximate mapping function so that new input data (a) help to predict the output (B). The supervised learning algorithm is shown in Figure 5.", null, "Figure 5.Supervised learning.\n\nThe purpose of supervised learning is to vary its weights according to the input/output samples. After executing this network, input-to-output mapping with minimum error has been achieved. Without proper training sets, performance is no longer determined, while it seems stochastic in either case.\n\n#### 3.3.2 Unsupervised learning\n\nUnsupervised learning is the type of machine learning function that describes the hidden layer from the unviable data. The unviable data explain the classification and measurements that are not included in the observations. Because of unviable data, there can be no calculation to reach an accuracy level.\n\nUnsupervised learning is utilized in self-organizing neural networks, and this type of learning does not require a teacher to teach the network . To train the network, data sets used in the supervised model are used along with the synaptic weights, which are assigned as:\n\nUnsupervised training=1Number of input attributesE2\n\nIt has the ability to solve complex problems and analyze the changes that occur in undefined data. It is widely used for preprocesses in the network of a supervised learning algorithm. A block diagram of unsupervised learning is shown in Figure 6.", null, "Figure 6.Unsupervised learning.\n\nUnsupervised learning methods can be further grouped into clustering and association problems.\n\nClustering is a basic method to analyze the data that occur in the network. It spots some inherent structures present in a set of objects based on a similarity measure. The clustering technique is based on statistical model identification or competitive learning. It is widely used for feature extraction, vector quantization, image segmentation, function approximation, and data mining . The association learning rule is a machine learning method to create relations between variables in large databases, and the approach to unsupervised learning is of two types, namely:\n\n• Anomaly detection\n\n• Neural networks learning—Hebbian learning method\n\nAnomaly detection is used for analyzing events and observations. The system is broadly classified into three types such as unsupervised anomaly detection, supervised anomaly detection, and semisupervised anomaly detection. It preprocesses the data sets and detects the faults that occur in the system.\n\nThe Hebbian method is a learning rule that determines the weight of the two different units either to increase or to decrease the weight to activate the function. Learning is performed by varying the synaptic gap between the weights. The weight of the vector increases gradually with respect to the input. The Hebbian rule for updating the weight is given by:\n\nwanew=waold+CaBE3\n\nwhere wa(new) is the new weight equal to the sum of the old weight and the learning method Ca* B.\n\n#### 3.3.3 Reinforcement learning\n\nThis learning is identical to supervised learning and the difference in operating the network from its actual output for about 50%. In supervised learning, for each output data, the simultaneous input data are known when compared to reinforcement learning. Due to the absence of a training data set, reinforcement learning learns from its experience.\n\nThe trial and error process is designed to maximize the expected value of a criterion of functions and actions followed by an improvement, so it is referred to as reinforced learning. The reinforcement signal and the corresponding input patterns depend on the previous data of the stochastic unit. Gaussian processes combine the neural networks for model-based reinforcement learning . A block diagram of reinforcement learning is shown in Figure 7.", null, "Figure 7.Block diagram of reinforcement learning.\n\n### 3.4 Activation functions of a neural network\n\n#### 3.4.1 Weights\n\nW1, W2, …, Wnare the factors of the weight that are associated with each node to determine the quality of input row vector Y = [Y1, Y2, …, Yn]T. Every single input is multiplied by a related weight by connecting the activation function. Figure 8 shows the basic elements of a neural network.", null, "Figure 8.Basic elements of a neural network.\n\n#### 3.4.2 Threshold\n\nThe internal threshold is the offset that marks the activation function of the output node Zand is given by:\n\nZ=i=1nXiWiθkE4\n\nThreshold function may be either binary type or bipolar type, respectively. The output of a binary threshold function is given by:\n\nZ=fpE5\ncondition is\"0\"ifp<0s\ncondition is\"1\"ifp0\n\n#### 3.4.3 Linear activation function\n\nLinear function fulfills the superposition concept. The activation function performs mathematical operations on a signal output, and the equation for linear activation function is given by:\n\nZ=fp=α.pE6\n\nwhereαis the slope of the linear activation function. The linear activation curve is shown in Figure 9.", null, "Figure 9.Linear activation curve.\n\nSlope 1 is the identity function. The output Zof the identity function is equal to the input function (p).\n\n## 4. Methods of implementing a neural network\n\nThe methods of implementing a neural network are the adaline method, back-propagation method, and radial basis method.\n\n### 4.1 Adaptive linear neuron network\n\nThe Adaline model was developed by Widrow Hoff. It is a single-layer network consisting of other nodes. The network is classified by varying the weights in such a manner that it diminishes mean square error for every iteration. An Adaline network is shown in Figure 10.", null, "Figure 10.Adaline network.\n\nEach node acquires a number of inputs and propagates a single output. The input and output signals to the Adaline network use bipolar activation function. When compared to other types of networks, the input and output function of the Adaline network is linear. The weights are bounded by the input and varied accordingly to the user’s choice. The bias in the network acts as an adjustable weight where the activation function is always 1. The output function of the Adaline network has one output unit and the network is a trained delta rule. This rule is otherwise known as the least mean square rule. This type of learning rule is used to decrease the mean squared error between the output and activation function . An Adaline network is implemented by using the three steps, which are shown in Figure 11.\n\n1. Initialize: assume random weights to all the links that are connected to the network.\n\n2. Training: initialize the input weights and arrange the known inputs in a random sequence. Compare errors between input and output by simulating the network. This forms an error function and by adjusting the weights, learning function takes place. Repeat the process until the total error is less than ⅀.\n\n3. Thinking: in the thinking process, the network will respond to input nodes. Even for trained inputs, it does not provide a good result. By defining an error function, it measures the performance in terms of weights. The derivative of the function with respect to weights is obtained by varying the weights, and error in the system is decreased. A block diagram of an Adaline network is shown in Figure 12.", null, "Figure 11.Steps—Adaline network.", null, "Figure 12.Block diagram of an Adaline network.\n\nThe input to the neural network is represented as A.\n\nA = [1 A1, A2, …, Am], whereas 1 = A0 = bias.\n\nW = [W0, W1, W2, …, Wm] represents the weight in the network.\n\nInitially, weights are chosen in a random manner. The value of −1 and 1 is taken. The weighted sum of input neurons, including a bias term, is calculated by comparing with output neurons. Based on the delta rule, the weights are adjusted. The output equation for the network is given by:\n\nC=k=1nakwk+αE7\n\nwhere ais the input vector, wis the weight of the vector, nis the number of inputs, αis a constant, and Cis the output vector.\n\nBy assuming a0 = 1 and w0 = α, the output equation is further minimized to:\n\nC=k=0nakwkE8\n\n#### 4.1.1 Adaline learning algorithm\n\nThe learning algorithm of a network is a delta rule but its base is different. To reduce variation between the input and output, the delta rule is preferred and improves the weight between the connections. The main objective is to reduce error that occurs in overall training networks [11, 12]. The updated weight in the network is given by the following equation:\n\nww+δtCaE9\n\nwhere δis the learning rate of the network, Cis the model output, and tis the desired target output.\n\nThe Adaline network merges with the least square error and the equation is given by:\n\nE=tk2E10\n\nwhere Eis the least mean square error.\n\nAccording to the input response, the system is activated and training is started. Consider Yas an input and Wis the weight. Let us assume xn+1=1and xn+1as the bias weight. Therefore, the weighted sum “S” is a dot product of the function given in the equation:\n\nS=W.Y=iwiyiE11\n\nThe identity function of the network is chosen as I=Sand considered as an activation function. The squared error E=OI2is the error function. The network defines error function and determines performance of input, weight, and desired output. Adaline networks are used in net input values and noise correction. The following are the steps for learning the Adaline algorithm:\n\nStep 1: Assume the synaptic weight values in the range from −1 to +1.\n\nStep 2: Set activation functions of the input units:\n\nA0=1andAi=Si(i=1,2,3,,n)\n\nStep 3: Compute the net input to the neuron as:\n\nS=W.Y=iwiyi\n\nStep 4: Update the corresponding bias and weights:\n\nW0New=W0Old+αtyinE12\nW1New=WiOld+αtyinxiE13\n\nwhere i = 1, 2, 3, …, n.\n\nStep 5: If the following conditions are satisfied, then the network is stopped or else:\n\nThe steps from the initial conditions are repeated.\n\nAdaline network example:\n\nBy using an Adaline network, train the AND, NOT gate function with bipolar inputs and targets performs one epoch of training. The input values are given as:\n\nX1X2t\n11−1\n1−11\n\nSolution:\n\nThe initial weights are taken to be W1 = 0.1, W2 = 0.2, b = 0.4, learning rate α = 0.6.\n\nThe weights are calculated until the least mean square is obtained.\n\nFirst input:\n\nX1 = 1; X2 = 1; t = −1; b = 0.4; W1 = o.1; W2 = 0.2; α = 0.6.\n\nYin = b + W1 X1 + W2 X2 = 0.7.\n\n(tYin) = −1.7 not equal to zero, then update the weights,\n\nW1 (New) = W1 (Old) + α(tYin)X1 = −0.42.\n\nW2 (New) = W1 (Old) + α(tYin)X2 = −0.82.\n\nb(New) = b(Old) + α(tYin) = −0.62.\n\nΔW1 = α(tYin)X1 = −0.102; ΔW2 = α(tYin)X2 = −0.204;\n\nΔb = α(tYin) = −1.02.\n\nTo compute the error, E = (tYin)2 = 2.89.\n\nSimilarly, for the second input:\n\nX1 = 1; X2 = −1; t = 1; W1 = −o.41; W2 = −0.82; b = −0.62.\n\nYin = b + W1 X1 + W2X2 = −0.24.\n\n(tYin) = 1.24 not equal to zero.\n\nUpdate the weights:\n\nW1 (New) = W1 (Old) + α(t– Yin)X1 = 0.324.\n\nW2 (New) = W1 (Old) + α(t– Yin)X2 = −1.564.\n\nb(New) = b(Old) + α(tYin) = 0.124; E = (tYin)2 = 1.5376.\n\nEpoch 1: For the first input, Yin = b + W1 X1 + W2 X2 = −1.116.\n\n(tYin) = 0.116; update the weights W1 (New) = W1 (Old) + α(t– Yin)X1 = 0.3936.\n\nW2 (New) = W1 (Old) + α(tYin)X2 = −1.4944; b(New) = b(Old) + α(tYin) = 0.1936; E = (tYin)2 = 0.01345; ΔW1 = α(tYin)X1 = 0.0696.\n\nΔW2 = α(tYin)X2 = 0.0696; Δb = α(tYin) = 0.0696.\n\nSo now the error for two inputs varies from 2.89 to 0.013435.\n\nMean error = 2.89 + 0.01345 = 3.0245.\n\n### 4.2 Back-propagation network\n\nA back-propagation network is a common method of training a neural network. The training method is used for a multilayer neural network. The network consists of processing elements with continuous differentiable activation function. In this network, a gradient descent method is used for minimizing the total squared error of the network. Training the network of a given set of input/output pairs is identified and the network has a procedure for changing the weights to classify given input patterns correctly. This is the network where the error is propagated back to the hidden unit . A back-propagation network is a sensitive approach for dividing the contribution to each weight. The two differences between updating the rule are as follows:\n\n1. Activation of the hidden unit/neurons is used instead of activation of the input value or input neuron.\n\n2. The rule contains a gradient descent for the activation function to operate.\n\nThe back-propagation network is the reformation of the least mean square algorithm and varies the network weights to minimize mean squared error between the actual and desired outputs of the network. The network is trained and exerted using training samples of respective inputs and desired outputs are fetched. The algorithm consists of input and output layers to vary weights and analyze the arrangements of input in an acceptable manner. This algorithm takes a unique set compared to other techniques—during the learning period itself the weights are calculated . The error signal is calculated by taking the difference between the calculated and target output. The result is measured in the output layer.\n\nIn the back-propagation network, the testing of data is implemented in the feed-forward path. While executing the network, it has the ability to operate in other hidden layers and is more efficient than operating with one hidden layer. The training process requires further time to train the network but the net result of the network during the training process produces a better result. The network is disintegrated into three categories, namely: (i) computation of the feed-forward network, (ii) back propagation to the output and hidden layer, and (iii) updating of weights. The algorithm will be terminated as the error value approaches a negligible numerical value.\n\nThe feed-forward computation network undergoes two processes. The first process receives the values of the hidden layer nodes, and in the second process the value from the hidden layer is used to compute the values of the output layer. Once the hidden layer values are determined, the network produces values from the hidden layer to the output layer. The hidden layer is observed once when the error from the output layer is propagated to the hidden layer. Weights are updated only if all the errors in the network are calculated. Further iterations help the network to train and produce a good training result. Block diagram of back propagation network is shown in Figure 13.", null, "Figure 13.Block diagram–Back propagation network.\n\nTo calculate the derivative function for the squared error with respect to the weights of the network, the gradient descent method is used in the back-propagation network. The squared error function is defined by:\n\nE=12tc2E14\n\nwhere Eis the squared error, tis the target output for a given sample, and cis the actual output of the output neuron.\n\nThe constant (1/2) is included, while the differentiating constant is canceled. A limitation of using the back-propagation algorithm is that the input vectors are not normalized and because of that, its performance is not improved. The network identifies only the local minimum values not the global minimum function to determine the errors.\n\nThere are two types of back-propagation networks, namely static and recurrent neural networks. The static network produces a mapping of static input for static output. It helps to solve static classification issues like optical character recognition. The recurrent type is a feed-forward network, until a fixed value is obtained. The error is computed and propagated backward. Mapping of the recurrent network is nonstatic.\n\n#### 4.2.1 Learning process of the back-propagation network\n\nEach neuron is composed of two units. The primary unit sums the product of weight coefficients and input signals. The secondary unit realizes the nonlinear function, in which the neurons are transferred to the activation function. Signal eis the adder output signal, and Y = F(e) is the output signal of the nonlinear element. Signal yis also the output signal of the neuron. Figure 14 shows the learning process of the network.", null, "Figure 14.Learning process of the back-propagation network.\n\nA training data set is required to instruct the neural network. The training data set consists of input signals (X1 and X2) assigned to corresponding target output. The training network is an iterative process, and for each iteration, weight coefficients of nodes are changed using new data from the training data set. Each training step starts by forcing both input signals from the training set. It is possible to determine the output signal values for each neuron in every network layer.\n\nTraining steps of the back-propagation algorithm:\n\nStep 1: The network of random weights is initialized.\n\nStep 2: The training process takes place using the following steps:\n\n1. Initially, training values are given as input to network and calculate the output of the network.\n\n2. The training process (i.e., starting with the output layer, back to the input layer):\n\n1. Compares the network output with the correct output (an error function).\n\n2. Adapts the weights in the current layer.\n\nStep 3: By using the gradient descent method, the error is minimized.\n\nStep 4: The propagating delta rule is used to adjust the error backward from the output to the hidden layer to the inputs. The back-propagation network is shown in Figure 15.", null, "Figure 15.Layer of back-propagation network.\n\nBack-propagation neural network example problem:\n\nBy using the back-propagation network, train the input vectors for the following functions: X1 = 0.15; X2 = 0.10; b1 = 0.45; b2 = 0.80; t1 = 0.10; t2 = 0.85. The example of back propagation network is shown in Figure 16.", null, "Figure 16.Back-propagation network.\n\nSolution:\n\nInitialize the weights as W1 = 0.35; W2 = 0.50; W3 = 0.75; W4 = 1.25; W5 = 0.80; W6 = 0.56; W7 = 0.45; W8 = 0.56.\n\nActivation function, H1=11+eH1.\n\nForward pass:\n\nH1 = b1 + W1 X1 + W2 X2 = 0.55215\n\nOut H1=11+eH1= 0.6346\n\nH2 = b1 + W3 X1 + W4 X2 = 0.6875\n\nOut H2=11+eH2= 0.66541\n\nNow, for calculating Y1:\n\nY1 = Out H1 * W5 + Out H2 * W6 + b2 = 1.6803096\n\nOut Y1=11+ey1= 0.842945\n\nIn the same way:\n\nY1 = Out H2 * W8 + Out H1 * W7 + b2 = 1.45819\n\nOut Y2=11+ey2= 0.811255\n\nCalculating total error:\n\nETotal = ⅀ 12TargetOutput2= 12T1OutY12+12T2OutY22= 0.27665\n\nBackward pass:\n\nTo update weights, consider W5\n\nError at W5=ΔETotalΔW5= ΔETotalΔOutY1* ΔOutY1ΔY1* ΔY1ΔW5= 0.07035\n\nUpdating W5, W5 = W5Ƞ* ΔETotalΔW5\n\nȠis the learning rate = 0.1; W5 = 0.7929\n\nIn the same way, calculations are done for updating the weights for W6, W7, and W8. In a similar manner, the weights are updated for W1, W2, W3, and W4 by using hidden layers in the network.\n\nAdvantages of the back-propagation network:\n\n1. The network is fast, simple, and programming code is easy when compared to other networks. It supports high-speed applications.\n\n2. It does not require any parameters to tune except for the number of inputs, and the network does not require prior knowledge to implement.\n\nDisadvantages of the back-propagation network:\n\n1. The network consumes more time for training and is stuck in local minima resulting in suboptimal solutions.\n\n2. A broad amount of input and output data is required, so there exists a complexity when solving a problem. The network is quite sensitive for noisy data.\n\n3. A major drawback occurs in a single-layer signal and the network cannot learn the process. It approximates nonlinear separable tasks and functions.\n\nApplications of the back-propagation network:\n\nThe network is especially useful for machine learning processes, face recognition systems, image or speech recognition, classification, function approximation, time series prediction, etc.\n\n### 4.3 Radial basis function\n\nThe radial basis function is a three-layer feed-forward neural network. The transfer function of the hidden layer is the radial basis function. It is derived from function approximation theory. In a neural network, the radial basis function is modeled by the narrow-tuned feedback that is viewed in biological neurons . This type of tuned response is found in several parts of nervous systems. In a feed-forward network, one hidden layer is required for the design of simple structures of lower computational cost. A radial basis network is a nonlinear type that makes the bias function change. The network is used to create regression-type problems.\n\nThe radial basis function is composed of three layers, namely input layer, hidden layer, and output layer. The sigmoid type of activation function is not used as in the case of the back-propagation algorithm, whereas the radial basis network uses Gaussian function as an activation function. The input layer consists of neurons with a linear activation function given to the hidden layer. The connection between input and hidden layers is not observed, which means that input neurons received from each hidden neuron remain the same in the network . The Gaussian activation function is determined by:\n\nFd=expd2/μ2E15\n\nwhere μis the real parameter value and dis the distance between the input and intermediate vector (the distance is usually measured in terms of Euclidean norm).\n\nConsider the input vector for a period of “m” time denoted by:\n\nYm=y1my2my3m..ynmTE16\n\nThe intermediate vector for each hidden neuron is denoted by Bi(for i = 1, 2, 3, ..., k), where “k” is the number of neurons in the hidden layer. The output of each neuron in the radial basis function is given by:\n\nhin=FiǁYmBiǁ,fori=1,2,,kE17\n\nOperation of the radial basis network is based on a least mean square algorithm and the local minima values are used for training the neural network. The training process requires a longer computation time but the learning period is less in the network . Schematic representation of the radial basis network is shown in Figure 17.", null, "Figure 17.Radial basis network.\n\nEvery neuron in the radial basis function stores a sample vector from the training set. The neuron in the network compares the input vector with its sample vector, and outputs a value between 0 and 1. If the input is equal to the sample vector, then the output of that neuron will be 1. The neuron’s response value is called the activation value.\n\nEvery neuron in the radial basis function computes a measure of the similarity between input and sample vector. The values are obtained from the training set. Input vectors are similar to sample vectors and return a value closer to 1. There are different possible choices of similarity functions, but the most popular is based on the Gaussian function. The equation for a Gaussian function with a one-dimensional input is given by:\n\nFx=1σ2Пexμ22σ2E18\n\nwhere xis the input, μis the mean, and σis the standard deviation.\n\nTraining steps of the radial basis function:\n\nStep 1: Initialize the input vector Yfrom the obtained training set.\n\nStep 2: Determine the output of the hidden layer.\n\nStep 3: Compute the output Zand compare with the desired value. Adjust each weight Waccordingly:\n\nZ=Wijn+1=Wijn+ηyjzjyiE19\n\nStep 4: Repeat the steps from 1 to 3 for each vector in the training set.\n\nStep 5: Repeat the steps from 1 to 4 unless the error is smaller than the maximum acceptable limit.\n\nApplications of the radial basis function:\n\nApplications of the radial basis function are function approximation type, classification, interpolation, and time series prediction. These applications provide various industrial uses like stock price prediction, fraud detection in financial transactions, and anomaly detection of data.\n\n## 5. Conclusion\n\nThis chapter encompassed the learning algorithms of neural networks such as adaline, back-propagation, and the radial basis network. Of all the learning methods, the back-propagation network is effective in training because of its mature back-propagating mechanism. The training process of the radial basis function is rapid and almost matches the ability of the back-propagation network. The radial basis function is a good substitute for the back-propagation network. When the selected features are clear enough, then the back-propagation network produces satisfactory results. The study of neural network has been slow, but now computers have better processing power. The back-propagation network effectively solves the exclusive-OR problem.\n\n## Download for free\n\nchapter PDF\nCitations in RIS format\nCitations in bibtex format\n\n## More\n\n© 2020 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 License, which permits use, distribution and reproduction for non-commercial purposes, provided the original is properly cited.\n\n## How to cite and reference\n\n### Cite this chapter Copy to clipboard\n\nT.D. Raheni and P. Thirumoorthi (May 27th 2020). Stochastic Artificial Intelligence: Review Article, Deterministic Artificial Intelligence, Timothy Sands, IntechOpen, DOI: 10.5772/intechopen.90003. Available from:\n\n### chapter statistics\n\n327total chapter downloads\n\n### More statistics for editors and authors\n\nLogin to your personal dashboard for more detailed statistics on your publications.\n\n### Related Content\n\n#### Deterministic Artificial Intelligence\n\nEdited by Timothy Sands\n\nNext chapter\n\n#### Simulated Real-Time Controller for Tuning Algorithm Using Modified Hill Climbing Approach Based on Model Reference Adaptive Control System\n\nBy Ahmed Abdulelah Ahmed, Azura Che Soh, Mohd Khair Hassan, Samsul Bahari Mohd Noor and Hafiz Rashidi Harun\n\n#### Advances in Spacecraft Attitude Control\n\nEdited by Timothy Sands\n\nFirst chapter\n\n#### Introduction to Satellite Attitude Control\n\nBy Henry Travis\n\nWe are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. 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[ "# What is the proper role of verification in quantum sampling, simulation, and extended-Church-Turing (E-C-T) testing?\n\nSince no answer was given, a flag has been set requesting that this question be converted to a community wiki.\n\nThe comments by Aaron Sterling, Sasho Nikolov, and Vor have been synthesized into the following resolution, which is open for community wiki discussion:\n\nResolved:   With respect to classical algorithms that output numbers, samples, or simulation trajectories, strict mathematical logic requires that either all four of the following propositions be accepted, or none of them:\n\n1. We can rule out a polynomial-time classical algorithm to generate random numbers. \n2. “We can rule out a polynomial-time classical algorithm to sample the output distribution of a quantum computer, under the sole assumption that the polynomial hierarchy is infinite.” \n3. \"We can't simulate [a quantum mechanical trajectory] $\\psi(t)$ in the usual way … there're too many variables.\" \n4. The extended Church-Turing-Thesis (E-C-T) is ruled out, for the rigorous reason that classical algorithms cannot generate random numbers. \n\nTo initiate discussion, here are affirmative and negative responses that, although each defensible, are deliberately over-stated. A strongly affirmative argument might be:\n\nAffirmative:  These four statements reflect theorems that, to respect rigor, require us never to speak of classical algorithms that generate random numbers, random samples, or quantum simulations, but rather to speak only of classical algorithms that generate pseudo-random numbers and (by extension) pseudo-random samples, and pseudo-quantum simulations.\n\nThis being understood, all four statements are true. Moreover, to avoid ambiguity and prevent confusion, mathematicians should encourage scientists and engineers to affix the prefix \"pseudo-\" to almost all usages of \"random\", \"sample\", and \"quantum simulation\".\n\nA strongly negative argument might be:\n\nNegative:  These statements (and their associated formal theorems) are sign-posts that direct us to a Lakatos-style “red-light” district of mathematics where we are beckoned to enthusiastically embrace (what might be called) the disciplines of pseudo-randomness, pseudo-sampling, and pseudo-simulation … mathematical practices that are fun for a deliciously sinful reason: they achieve mathematical effects that formal logic says are impossible. Therefore, what could be more magical, and more fun, than this conclusion: the resolution's four statements each are formally true, but practically false?\n\nThis being understood, all four statements are false. Moreover, since most practical embraces of \"randomness\", \"sampling\", and \"quantum simulation\" occur in this magical environment—in which issues associated to Kolmogorov complexity and oracular assessments are willfully overlooked—it is mathematicians who should alter their usage.\n\nRealistically, though, how should complexity theorists phrase their findings relating to randomness, sample, and simulation … on the one hand, with a view toward sustaining a reasonable balance of clarity, concision, and rigor … and on the other hand, with a view toward sustaining low-noise communication with other STEM disciplines? The latter goal is especially important, as practical capabilities steadily increase in fields like cryptography, statistical testing, machine learning, and quantum simulation.\n\nWhat is/are the generally accepted role(s) of verification in the complexity-theoretic definitions associated to sampling, simulation, and testing the extended-Church-Turing (E-C-T) thesis?\n\nThe preferred answer is references to articles, monographs, or textbooks that discuss these issues in-depth.\n\nShould this literature prove to be sparse or otherwise unsatisfactory, then (after two days) I will convert this question to a community wiki asking:\n\nWhat is/are reasonable and proper role(s) of verification in the complexity-theoretic definitions associated to sampling, simulation, and testing the extended-Church-Turing (E-C-T) thesis?\n\nBackground\n\nThe question asked is motivated by the recent thread \"What would it mean to disprove Church-Turing thesis?\", specifically the (excellent IMHO) answers given by Gil Kalai and by Timothy Chow\n\nIn the question asked, the phrase \"proper and/or accepted complexity-theoretic definitions\" is to be construed as restraining Alice from implausible claims like the following:\n\nAlice:  Here is my experimental sample of truly random binary digits computed by my (one-photon) linear optical network.\n\nBob:  Here is my simulated sample of pseudo-random digits computed by a classical Turing machine.\n\nAlice:  Sorry Bob ... your sample is algorithmically compressible, and mine isn't. Therefore my experimental data demonstrate that the E-C-T is false!\"\n\nIn the absence of any association of verification to sampling, Alice's reasoning is impeccable. In other words, should complexity theorists regard the E-C-T as having already been formally disproved … decades ago?\n\nFrom a practical point of view, simulation methods associated to quantum trajectory sampling on varietal state-spaces are coming into widespread use in many disciplines of science and engineering. That is why complexity-theoretic definitions of sampling that respect the central role of verification (which is inseparable from replicability) in science and engineering would be very welcome to practicing scientists and engineers … especially if these definitions were accompanied by theorems describing the computational complexity of verified sampling.\n\nAdded Edit: Thanks to a collaboration between the University of Geneva and the company id Quantique, it is perfectly feasible to complete this exercise in reality.\n\nHere are 1024 random bits that are certified by id Quantique as being algorithmically incompressible:\n\n0110001000010111111100010111001000101110110001001100000010010110\n0101000110100011101001110110000001010110011101111110101010110100\n1001001110001110101000001110000101000110000001010001101001000000\n0110101010110000110101001110011010010101000000110000010000010111\n0100110110001011011101110000010110000100110001001110011000000011\n1111010100010110110010011000110110110010101101010000010010001111\n1101111000111101111010000110100110011000101101010110110110000101\n1110111100000111000111101111110011101101110111101001001111111110\n1000001011001000011101001000001110101110101010000111100000111010\n1010011001110111101001100010110000101101100100101100000110111111\n1000001101111001111011100011110101011010010100000010100101100010\n0011101000111100000001101100111110100100010100100010011000001000\n0000001001110101010111110001010010000111010011000100001101101000\n1011111010001000110101110101111101010111111011011111110010010111\n0111000010000111000100110110010101110100000110101001111010101001\n0100011110011101000011000100110110010000100001111100101001010011\n\n\nShould we now accept the claim: \"The E-C-T thesis is disproved\"?\n\nIf not, what grounds should we give?\n\n• by verification do you mean that the statement \"algorithm A has property P in a computational model M\" can be tested in finite time, for any particular input length? E.g., the property \"probabilistic algorithm A halts witnin $1000n$ steps on any input of size $n$, using at most $\\log_2 n$ random bits and accetpt language $L$ with probability $2/3$\" can be verified in finite time for any $n$. Verified in finite time would mean by a deterministic Turing machine as a failsafe model of computation? – Sasho Nikolov Jun 1 '11 at 17:00\n• I think this is a great question. But, in your example, how does Alice know her string of digits is not algorithmically compressible? – Aaron Sterling Jun 1 '11 at 17:35\n• On the equivalence sampling/searching: scottaaronson.com/papers/samprel.ps – Marzio De Biasi Jun 1 '11 at 20:08\n• @John: just a clarification (I underline that I'm not an expert): \"... are certified by id Quantique as being algorithmically incompressible\", but how can they certify it? Obviously Kolmogorov complexity of a string is not computable so the sentence seems false. Even if they simply say \"we certify that the sequence is (quantum) random\" I have some doubts: the physical process (the hardware) is difficult to balance so they use Von Neumann unbiasing which is good, but doesn't guarantee that the result is truly random. – Marzio De Biasi Jun 2 '11 at 22:58\n• @John Sidles: while you make sound and interesting observations, I do not understand what you're looking for. It's clear what Aaronson and coauthors mean by \"rule out\": if PH is infinite, there doesn't exist a particular algorithm in a particular model. i suppose you're asking if the modeling assumptions are verifiable. note that the purpose of the model is to verify only the modeling assumption, instead of test any possible algorithm/theorem – Sasho Nikolov Jun 3 '11 at 20:00" ]

[ null ]

[ "Contents: CCRL 40/4 Downloads and Statistics October 19, 2019 Testing summary: Total: 2'120'567 games played by 2'479 programs 28673 CPU days (X2 4600+) White wins: 822'832 (38.8%) Black wins: 665'262 (31.4%) Draws: 632'473 (29.8%) White score: 53.7%\n\nEngine Details\n\n Options Show each game results\nColossus 2008a (2592+28\n−28\n)Quote\n Author: Martin Bryant (England)\nThis is one of the 8 Colossus versions we tested: Compare them!\n Opponent Elo Diff Results Score LOS Perf –   Fruit 05/11/03 2792 +14−14 (+200) 9 − 23(+6−20=6) 28.1%9.0 / 32 0.0% +35 0 0 0 = 0 0 0 1 0 = 0 1 0 0 0 = 0 1 1 0 0 0 0 1 0 0 = 0 0 = 1 = –   Spike 1.2 Turin 2731 +6−5 (+139) 9 − 23(+5−19=8) 28.1%9.0 / 32 0.0% −17 0 0 0 0 0 0 0 0 = 0 = 1 1 0 = = 0 = = = 0 0 0 1 0 0 0 = 0 1 0 1 –   Chess Tiger 2007.1 2713 +7−7 (+121) 12 − 20(+5−13=14) 37.5%12.0 / 32 0.0% +44 = 0 0 = = 0 0 = 1 0 0 = 0 = 1 1 = 0 0 = 1 0 = 0 1 = = = = 0 = 0 –   Frenzee Feb08 64-bit 2694 +10−10 (+102) 10.5 − 19.5(+5−14=11) 35.0%10.5 / 30 0.0% +2 0 = 0 = 0 = 0 = 0 1 0 = 1 0 = 1 = 0 1 0 = 1 0 0 0 = 0 = = 0 –   SmarThink 1.00 32-bit 2664 +17−17 (+72) 15.5 − 16.5(+11−12=9) 48.4%15.5 / 32 0.0% +63 = 1 = 0 1 = 0 = 1 1 1 = 0 = 0 0 1 = = 0 0 1 1 1 0 = 0 1 1 0 0 0 –   Alaric 707 2663 +7−7 (+71) 12 − 20(+9−17=6) 37.5%12.0 / 32 0.0% −23 0 0 1 = 1 1 0 0 0 1 0 = 0 = 0 = 0 0 1 1 0 1 1 0 = 0 0 0 0 0 = 1 –   E.T. Chess 13.01.08 2634 +7−7 (+42) 11.5 − 20.5(+8−17=7) 35.9%11.5 / 32 0.2% −60 0 0 1 1 0 0 1 0 0 1 0 0 = = 1 0 = 0 = 1 0 1 0 0 0 0 1 0 0 = = = –   Alfil 8.1.1 Optimized 2625 +9−9 (+33) 12.5 − 19.5(+9−16=7) 39.1%12.5 / 32 1.3% −46 1 = 1 0 0 1 = 0 0 1 1 0 = 1 0 0 = 1 0 = 0 0 = 0 0 0 1 0 0 = 1 0 –   WildCat 8 2623 +7−7 (+31) 20 − 12(+14−6=12) 62.5%20.0 / 32 1.8% +111 1 = 1 0 1 0 1 0 = 1 = 0 = = = = 0 = = 1 0 1 1 1 1 = 1 1 = 1 1 = –   Movei 00.8.438 (10 10 10) 2622 +7−7 (+30) 13.5 − 18.5(+8−13=11) 42.2%13.5 / 32 2.0% −22 1 = 0 1 1 0 = 1 = = = = 1 = 0 = = 0 0 0 1 1 = 0 0 0 0 0 = 0 0 1 –   Slow Chess Blitz WV2.1 2614 +6−6 (+22) 17 − 15(+12−10=10) 53.1%17.0 / 32 6.4% +45 = 0 1 1 0 1 0 0 0 = 1 = 1 = 1 0 1 0 0 0 1 = = 1 0 = 1 = 1 = = 1 –   Booot 4.14.0 2611 +21−21 (+19) 10.5 − 20.5(+6−16=9) 33.9%10.5 / 31 14.1% −93 = 1 0 1 = 0 0 1 = = 0 0 0 0 0 0 = 1 1 0 0 0 0 = 0 1 0 = 0 = = –   Sloppy 0.2.0 64-bit 2607 +17−17 (+15) 11 − 20(+5−14=12) 35.5%11.0 / 31 18.7% −79 0 0 0 = 0 0 0 0 1 0 0 = = 1 0 = 1 = 1 = 0 1 = = = = 0 = 0 = 0 –   GarboChess 2.10 64-bit 2598 +10−10 (+6) 17.5 − 14.5(+13−10=9) 54.7%17.5 / 32 36.3% +41 0 1 = 0 1 0 0 0 0 0 1 1 = 1 0 1 1 1 = = = 1 = 0 0 1 = 1 = 1 1 = –   Nebiyu 1.2 2310 +41−42 (−282) 9.5 − 2.5(+9−2=1) 79.2%9.5 / 12 100.0% −27 1 1 1 1 = 1 1 0 1 1 1 0\n\nRating changes by day", null, "Rating changes with played games", null, "Created in 2005-2013 by CCRL team Last games added on October 19, 2019" ]

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[ "• P-ISSN 0974-6846 E-ISSN 0974-5645", null, "# Indian Journal of Science and Technology\n\n## Article", null, "• VIEWS 582\n• PDF 552\n\nIndian Journal of Science and Technology\n\nYear: 2016, Volume: 9, Issue: 28, Pages: 1-6\n\nOriginal Article\n\n## Abstract\n\nBackground/Objectives: To examine the effect of material nonlinearity of mild steel on the total deflection of beams and frames. Method of Analysis: In the linear analysis, value of Young’s modulus is same in all over the analysis, hence the load-deflection plot is also linear, whereas in material nonlinear analysis of structures the elastic modulus is constant upto yield stress and it decreases thereafter. In the present study linear and material nonlinear analysis of beams and frames are carried out using ANSYS mechanical APDL software. The linear deflection of the structures is also computed using a finite element based MATLAB code. Material nonlinearity is incorporated using bilinear stress-strain curves with tangent modulus E/65. E is the elastic modulus of the material. It has been found that the tangent modulus of stress-stress curve of mild steel is more resemble to E/3, so this value is also considered separately for the nonlinear analysis. In this paper cantilever beam with an end load and a two storey building frame with horizontal load is considered with above tangent modulus. The loads are applied incrementally, deflection and stress at each load increment is computed. Findings: In linear analysis the full strength of the material is not utilized and it is assumed that material will fail after reaching the yield stress. In material nonlinear analysis the strain hardening property is considered by taking the tangent modulus after the yield stress. The linear analysis is giving linear variation of deflection and stress with respect to load, but for material nonlinear (bilinear) the deflection and stress will be same as that of linear upto yield point and after that the deflection is found more and stress value found less than that of linear values. Application/Improvements: For the economical usage of materials nonlinear analysis is preferred over the linear analysis, because it is giving the actual behavior of structures and we are utilizing the maximum capacity of the material.\nKeywords: Bilinear Stress-strain Curve, Finite Element Method, Linear Analysis, Load-Deflection Behavior, Nonlinear Analysis", null, "", null, "Article Metrics\n\n## DON'T MISS OUT!\n\nSubscribe now for latest articles and news." ]

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[ "# Forwards vs. Futures\n\nI’ve covered Forwards and Futures in previous posts, and now that I’ve covered the basics of Stochastic Interest Rates as well, we can have a look at the difference between Forwards and Futures Contracts from a financial perspective.\n\nAs discussed before, the price of a Forward Contract is enforceable by arbitrage if the underlying is available and freely storable and there are Zero Coupon Bonds available to the Forward Contract delivery time. In this case, the forward price is\n\nIn this post I’m going to assume a general interest rate model, which in particular may well be stochastic. In such cases, the price of a ZCB at the present time is given by\n\nFutures Contracts are a bit more complicated, and we need to extend our earlier description in the case that there are interest rates. The basic description was given before, but additionally in the presence of interest rates, any deposit that is in either party’s account is delivered to the OTHER party at the end of each time period. So, taking the example from the previous post, on day 4 we had $4 on account with the exchange – if rates on that day were 10% p.a., over that day the$4 balance would accrue about 10c interest, which would be paid to the other party.\n\nLet’s say we’re at time s, and want to calculate the Futures price to time T. Our replication strategy is now as follows, following the classic proof due to Cox, Ingersall and Ross but in continuous time. Futures Contracts are free to enter into and break out of due to the margin in each account, so entering X Futures Contracts at time t and closing them at time t+dt will lead to a net receipt (or payment if negative) of . From t+dt to T, we invest (borrow) this amount at the short rate and thus recieve (need to pay)\n\nand now moving to continuous time\n\nWe follow this strategy in continuous time, constantly opening contracts and closing them in the following time period [I’m glossing over discrete vs. continuous time here – as long as the short rate corresponds to the discrete time step involved this shouldn’t be a problem], and investing our profits and financing our losses both at the corresponding short rate. We choose a different X for each period [t,t+td] so that . We also invest an amount H(s,T) at time s at the short rate, and continually roll this over so that it is worth  at time T\n\nOnly the final step of this strategy costs money to enter, so the net price of the portfolio and trading strategy is H(s,T). The net payoff at expiry is\n\nAnd H(T,T) is S(T), so the net payoff of a portfolio costing H(s,T) is\n\nHow does this differ from a portfolio costing the Forward price? Remembering that in Risk-Neutral Valuation, the present value of an asset is equal to the expectation of its future value discounted by a numeraire. In the risk-neutral measure, this numeraire is a unit of cash B continually re-invested at the short rate, which is worth , so we see that the Futures Price is a martingale in the risk-neutral measure (sometimes called the ‘cash measure’ because of its numeraire). So the current value of a Futures Contract on some underlying should be\n\nie. the undiscounted expectation of the future spot in the risk-neutral measure. The Forward Price is instead the expected price in the T-forward measure whose numeraire is a ZCB expiring at time T\n\nWe can express these in terms of each other remembering F(T,T) = S(T) = H(T,T) and using a change of numeraire (post on this soon!). I also use the expression for two correlated lognormal, which I derived at the bottom of this post\n\nwhere  is the volatility of the Futures price, and  is the volatility of a ZCB – in general the algebra will be rather messy!\n\nAs a concrete example, let’s consider the following model for asset prices, with S driven by a geometric brownian motion and rates driven by the Vasicek model discussed before\n\nAnd (critically) assuming that the two brownian processes are correlated according to rho\n\nIn this case, the volatility  is the volatility of , and as I discussed in the post on stochastic rates, this is tractable and lognormally distributed in this model.\n\nWe can see that in the case of correlated stochastic rates, these two prices are not the same – which means that Futures and Forward Contracts are fundamentally different financial products.\n\nFor two standard normal variates x and y with correlation rho, we have:\n\nand" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "### Introduction to Chemical Engineering\n\nCourse Notes\nFaculty Member(Members):\nDimitris Vayenas\n\nCore Courses\nCourse Number: CHM_140\nCredits: 4\nECTS Credits:4\n##### Learning Outcomes\n\nAt the end of this course the student should be able to:\n\n1. Understand a flowsheet of a simple Chemical Industry\n2. Develop the physical and mathematical model of a process.\n3. Use fundamental equations and write mass and energy balances in simple processes.\n4. Use differential and integral methods for the treatment of reaction rate data.\n5. Write mass and energy balances of chemical compounds in simple physical processes and simple chemical reactors.\n6. Use dimensional analysis in order to extract equations\n7. Understand the concept of linearization.\n\nUnderstand the concept of Residence Time Distribution (RTD) in simple single- and multi-chemical reactors\n\n##### Competences\n\nAt the end of the course the student will have further developed the following skills/competences:\n\n1. Ability to understand a flowsheet of a simple Chemical Industry.\n2. Ability to develop the physical and mathematical model of a process.\n3. Ability to use fundamental equations and write mass and energy balances in simple processes.\n4. Ability to use differential and integral methods for the treatment of reaction rate data.\n5. Ability to write mass and energy balances of chemical compounds in simple physical processes and simple chemical reactors.\n6. Ability to use dimensional analysis in order to extract equations.\n7. Ability to understand the concept of linearization.\n\nAbility to understand the concept of Residence Time Distribution (RTD) in simple single- and multi-chemical reactors.\n\n##### Prerequisites\n\nThere are no prerequisite courses. It is, however, recommended that students should have a basic knowledge of Mathematics, Physics and Chemistry.\n\n##### Course Contents\n\nDefinition of Chemical Engineering science and activities of Chemical Engineers in Greece. Overview of the flowsheet of a simple Chemical Industry in relation to the courses in the Chemical Engineering curriculum. Physical and mathematical model of a process. Types of chemical and electrochemical reactors. Mass balances in simple chemical reactors and simple unit operations. Use of differential and integral methods for the treatment of reaction rate data. Dynamic behavior of simple reactors. Dimensional analysis. The concept of scale-up. The concept of linearization. Residence time distribution (RTD) in simple single- and multi-chemical reactors.\n\n##### Teaching And Learning Methods\n\nLectures using power-point presentations and problem solving based on the syllabus of the course" ]

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[ "### How to draw rectangle in Python and save as PNG\n\nShort snippet how to draw canvas with rectangles and make thumbnail of image with Pillow.\n\n```from PIL import Image, ImageDraw\n\n# size of image\ncanvas = (400, 300)\n\n# scale ration\nscale = 5\nthumb = canvas/scale, canvas/scale\n\n# rectangles (width, height, left position, top position)\nframes = [(50, 50, 5, 5), (60, 60, 100, 50), (100, 100, 205, 120)]\n\n# init canvas\nim = Image.new('RGBA', canvas, (255, 255, 255, 255))\ndraw = ImageDraw.Draw(im)\n\n# draw rectangles\nfor frame in frames:\nx1, y1 = frame, frame\nx2, y2 = frame + frame, frame + frame\ndraw.rectangle([x1, y1, x2, y2], outline=(0, 0, 0, 255))\n\n# make thumbnail\nim.thumbnail(thumb)\n\n# save image\nim.save('/home/proft/temp/im.png')\n```", null, "" ]

[ null, "https://en.proft.me/static/img/tree.png", null ]

[ "/C\n\n# strtof, strtod, strtold\n\nDefined in header `<stdlib.h>`\n`float strtof( const char *restrict str, char **restrict str_end );`\n(since C99)\n`double strtod( const char *str, char **str_end );`\n(until C99)\n`double strtod( const char *restrict str, char **restrict str_end );`\n(since C99)\n`long double strtold( const char *restrict str, char **restrict str_end );`\n(since C99)\n\nInterprets a floating-point value in a byte string pointed to by `str`.\n\nFunction discards any whitespace characters (as determined by `std::isspace()`) until first non-whitespace character is found. Then it takes as many characters as possible to form a valid floating-point representation and converts them to a floating-point value. The valid floating-point value can be one of the following:\n\n• decimal floating-point expression. It consists of the following parts:\n• (optional) plus or minus sign\n• nonempty sequence of decimal digits optionally containing decimal-point character (as determined by the current C `locale`) (defines significand)\n• (optional) `e` or `E` followed with optional minus or plus sign and nonempty sequence of decimal digits (defines exponent)\n• binary floating-point expression. It consists of the following parts:\n• (optional) plus or minus sign\n• `0x` or `0X`\n• nonempty sequence of hexadecimal digits optionally containing a decimal-point character (as determined by the current C `locale`) (defines significand)\n• (optional) `p` or `P` followed with optional minus or plus sign and nonempty sequence of decimal digits (defines exponent)\n• infinity expression. It consists of the following parts:\n• (optional) plus or minus sign\n• `INF` or `INFINITY` ignoring case\n• not-a-number expression. It consists of the following parts:\n• (optional) plus or minus sign\n• `NAN` or `NAN(`char_sequence`)` ignoring case of the `NAN` part. char_sequence can only contain alphanumeric characters. The result is a quiet NaN floating-point value.\n• any other expression that may be accepted by the currently installed C `locale`\n\nThe functions sets the pointer pointed to by `str_end` to point to the character past the last character interpreted. If `str_end` is `NULL`, it is ignored.\n\n### Parameters\n\n str - pointer to the null-terminated byte string to be interpreted str_end - pointer to a pointer to character.\n\n### Return value\n\nFloating-point value corresponding to the contents of `str` on success. If the converted value falls out of range of corresponding return type, range error occurs and `HUGE_VAL`, `HUGE_VALF` or `HUGE_VALL` is returned. If no conversion can be performed, `​0​` is returned.\n\n### Example\n\n```#include <stdio.h>\n#include <errno.h>\n#include <stdlib.h>\n\nint main(void)\n{\n// parsing with error handling\nconst char *p = \"111.11 -2.22 Nan nan(2) inF 0X1.BC70A3D70A3D7P+6 1.18973e+4932zzz\";\nprintf(\"Parsing '%s':\\n\", p);\nchar *end;\nfor (double f = strtod(p, &end); p != end; f = strtod(p, &end))\n{\nprintf(\"'%.*s' -> \", (int)(end-p), p);\np = end;\nif (errno == ERANGE){\nprintf(\"range error, got \");\nerrno = 0;\n}\nprintf(\"%f\\n\", f);\n}\n\n// parsing without error handling\nprintf(\"\\\" -0.0000000123junk\\\" --> %g\\n\", strtod(\" -0.0000000123junk\", NULL));\nprintf(\"\\\"junk\\\" --> %g\\n\", strtod(\"junk\", NULL));\n}```\n\nPossible output:\n\n```Parsing '111.11 -2.22 Nan inF 0X1.BC70A3D70A3D7P+6 1.18973e+4932zzz':\n'111.11' -> 111.110000\n' -2.22' -> -2.220000\n' Nan' -> nan\n' nan(2)' -> nan\n' inF' -> inf\n' 0X1.BC70A3D70A3D7P+6' -> 111.110000\n' 1.18973e+4932' -> range error, got inf\n\" -0.0000000123junk\" --> -1.23e-08\n\"junk\" --> 0```\n• C11 standard (ISO/IEC 9899:2011):\n• 7.22.1.3 The strtod, strtof, and strtold functions (p: 342-344)\n• C99 standard (ISO/IEC 9899:1999):\n• 7.20.1.3 The strtod, strtof, and strtold functions (p: 308-310)\n• C89/C90 standard (ISO/IEC 9899:1990):\n• 4.10.1.4 The strtod function\n\n atof converts a byte string to a floating-point value (function) wcstofwcstodwcstold (C99)(C95)(C99) converts a wide string to a floating-point value (function) C++ documentation for `strtof, strtod, strtold`" ]

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[ "# Commands, strings, loops and logic\n\nThere is a complete documentation of Octave at http://www.gnu.org/software/octave/doc/interpreter/\n\n### Formatting numbers\n\nYou can format numbers in three different ways; format long, format short and format bank (2 decimals)\n\n>>> format long\n>>> pi\nans = 3.14159265358979\n>>> format short\n>>> pi\nans = 3.1416\n>>> format bank\n>>> pi\nans = 3.14\n\n\nJust writing the command format will give you the default format short.\n\n### String variables\n\nWhen handling text instead of numbers you use strings. A string is enclosed in either \"-signs or '-signs.\n\n>>> my_string_variable=\"Hello World!\"\nmy_string_variable = Hello World!\n>>> my_string_variable='Hello World!'\nmy_string_variable = Hello World!\n\n\n### The help command\n\nIf you want to find information about a command or a function you can use the help command.\n\n>>> help asin\nasin' is a built-in function\n\n-- Mapping Function: asin (X)\n\nCompute the inverse sine in radians for each element of X.\n\n\n\n### The disp command\n\nYou can use the disp command together with an argument enclosed in brackets to display data. When using disp the output is displayed without the ans=.\n\n>>> disp(pi)\n3.1416\n>>> disp(my_string_variable)\nHello World!\n>>> disp(\"The value of pi is:\"), disp(pi)\nThe value of pi is:\n3.1416\n\n\n### Date and time\n\nWhen measuring time on a computer it is a convention to measure the time since 00:00:00 UTC 1 January 1970. The command time will give you the number of seconds since then; the command now will give you the number of days. The command date will give you the current date as a string. The command clock will give you the current year-month-day-hour-minute-second as a row-matrix.\n\n>>> time\nans = 1.2737e+009\n>>> now\nans = 7.3427e+005\n>>> date\nans = 12-May-2010\n>>> clock\nans =\n\n2010.0000 5.0000 12.0000 15.0000 41.0000 40.3795\n\n\n## m-files\n\nYou can store a number of commands in a so called m-file. You enter the commands in an editor and run the m-file from Octave.\n\n### Working directory\n\nAll m-files should be stored in the so called working directory. Write pwd (print working directory) to see the current working directory of Octave. If you prefer to store your files at some other directory, go there by using the command cd (change directory). For more information about cd see http://www.computerhope.com/unix/ucd.htm.\n\nWrite your m-files in any editor that can handle pure text without adding file-extensions such as .txt. Some examples are Notepad++ for Windows, Sublime Text 2 for Mac (install from App Store), and gedit for Linux. Save the m-files in the working directory of Octave.\n\n### A first m-file\n\nSave following lines:\n\na=1;\nb=2;\ndisp('a+b=')\ndisp(a+b)\n\n\nin a file called test1.m in your Octave working directory. Write test1 in Octave to see the output.\n\n## The for-statement\n\nRepeating the same operation over and over again is called iterating. One way to iterate when programming is to use a for statement. A for statement in its simplest form uses a variable representing an index that is increased by 1 in each step.\n\nIn the code below, the index is called i. The command between for and end is repeated 10 times. The first time row 2 is executed, i is equal to 0, the second time i is equal to 1, and so on.\n\nRow 4 starts with a %-sign, everything to the right of the %-sign is ignored when the program is run; this is a comment intended for those reading the actual code.\n\nfor i=0:9\ndisp(i)\nend\n%the numbers 0, 1,...,9 are displayed\n\n\nYou can increment your index with something else than 1 by adding a number between the starting index and the last index.\n\nfor i=0:0.1:2\ndisp(i)\nend\n%21 numbers are displayed, the numbers 0,0.1,0.2,...,2\n\n\n## Sequences, Continued\n\n#### Exercise 1\n\nWrite for-statements in m-files to solve following exercises. Use format long.\n\n1. The recurrence equation \\left\\{ \\begin{align} a_0 &=c \\\\ a_{n+1} &=1+\\frac{1}{a_n}, n\\geq0 \\end{align} \\right. has two fixpoints $$x_1=\\dfrac{1+\\sqrt{5}}{2}$$ and $$x_2=\\dfrac{1-\\sqrt{5}}{2}$$.\nWrite a for-statement to iterate the recurrence equation 10 times.\nWhat happens if you start with $$c=x_1$$ and $$c=x_2$$ respectively?\nWhat happens if you iterate 50 times? 100 times?\n2. Write a recurrence equation for the continued fraction $1 + \\cfrac{1}{2 + \\cfrac{1}{2 + \\cfrac{1}{2 + \\cfrac{1}{2+\\cdots } } }}$ Hint: it is easier to write a recurrence equation for the continued fraction plus 1.\n1. Find the limit of the recurrence equation by iterating in Octave.\n2. Find the exact value of the continued fraction by first finding the fixpoints of the recurrence equation you used in a and then subtracting 1.\n\n## Some series\n\nA series $$\\sum a_i$$ is said to converge if $$\\sum_{i=0}^{n}a_i$$ has a finite limit as $$n\\rightarrow\\infty$$.\n\n#### Exercise 2\n\n1. Use Octave to make a conjecture about the convergence of $\\sum_{i=1}^{n}\\left(\\frac{1}{i}\\right)^2$ as $$n\\rightarrow\\infty$$.\n2. Use Octave to make a conjecture about the convergence of $\\sum_{i=1}^{n}\\frac{1}{i}$ as $$n\\rightarrow\\infty$$.\n1. Rewrite the series $1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\frac{1}{6}+\\cdots$ using a $$\\sum$$.\n2. Use Octave to make a conjecture about the convergence of this series as $$n\\rightarrow\\infty$$.\n3. Iterate to find the sum of the first 50 terms of $1-\\frac{1}{3}+\\frac{1}{5}-\\frac{1}{7}+\\frac{1}{9}-\\frac{1}{11}+\\cdots$ then multiply the result with 4. What happens when you iterate 500 times? 5000 times? 50000 times?\n\n## Boolean Logic\n\n### Boolean algebra\n\nIn Boolean algebra you represent the logical values true and false by the numbers 1 and 0 respectively.\n\n>>> true\nans = 1\n>>> false\nans = 0\n\n\n### and, or, not\n\nThe basic operators in logic are and, or and not, these are written using the symbols ∧, ∨ and ¬ respectively. If p and q are statements that are either true or false, then you get the truth table\n\np q pq pq ¬p\ntrue true true true false\ntrue false false true false\nfalse true false true true\nfalse false false false true\n\nThe operators ∧ and ∨ are binary operators, they are applied to two operands. The operator ¬ is an unary operator, it applies to one operand.\n\nIn Octave (and most other programming languages) the operators ∧, ∨ and ¬ are written using the symbols &&, || and !; giving the truth table\n\np q p && q p || q !p\n1 1 1 1 0\n1 0 0 1 0\n0 1 0 1 1\n0 0 0 0 1\n\nIn Octave (and most other programming languages) all numbers that are not 0 are thought of as being true.\n\n>>> a=4.5;\n>>> b=0;\n>>> (a && b) || !b\nans = 1\n\n\nNote that you can perform logical operations by using arithmetics.\n\np && q = pq\n\np || q = p+q-pq\n\n!p = 1-p\n\n### Comparison Operators\n\nWhen doing a comparison in Octave you apply a comparison operator on two numbers and the result is either true or false, represented by 1 or 0.\n\noperation operators operands result\narithmetic operations + - * / ^ numbers a number\nlogical operations && || ! logical values a logical value\ncomparisons > >= < <= == != numbers a logical value\n\nThe comparison operators are:\n\noperator explanation\n> greater than, >\n>= greater than or equal to,\n< less than, <\n<= less than or equal to,\n== equal to, =\n!= not equal to,\n>>> a=1; b=2; c=2; d=3;\n>>> a>=b\nans = 0\n>>> b>=c\nans = 1\n>>> (a < b) && (c!=d)\nans = 1\n\n>>> a < b && c!=d\n>>>parse error:\n\nsyntax error\n\n>>> a < b && c!=d\n^\n`\n\n# further info:\n\nFuzzy Logics: This article was published in Scientific American 1993, A Partly True Story, by Ian Stewart\n\nThat is true →      ← That is false" ]

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[ "Survey\n\n* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project\n\nDocument related concepts\n\nNewton's theorem of revolving orbits wikipedia , lookup\n\nNuclear force wikipedia , lookup\n\nBuoyancy wikipedia , lookup\n\nFictitious force wikipedia , lookup\n\nCentrifugal force wikipedia , lookup\n\nForce wikipedia , lookup\n\nNewton's laws of motion wikipedia , lookup\n\nClassical central-force problem wikipedia , lookup\n\nCentripetal force wikipedia , lookup\n\nTranscript\n```Name ________________________ Period ________ Date ______________________\nChapter 5:2: What is a Force?\n1. Define Force.\n2. Explain the difference between balanced and unbalanced forces and how each\naffects the motion of an object.\n3. Give four examples of force being exerted.\n4. What is a net force?\n5. What S.I. unit(s) is used when calculating force?\n6. Are the forces on a kicked soccer ball balanced or unbalanced? How do you\nknow?\n7. Two forces act on an object. One force has a magnitude of 10 N and is directed\ntoward the north. The other has a magnitude of 5 N and is directed toward the\nsouth. The object experiences a net force of ________________. Draw a" ]

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[ "# Group in which every retract is regular\n\nThis article defines a group property: a property that can be evaluated to true/false for any given group, invariant under isomorphism\nView a complete list of group properties\nVIEW RELATED: Group property implications | Group property non-implications |Group metaproperty satisfactions | Group metaproperty dissatisfactions | Group property satisfactions | Group property dissatisfactions\n\n## Definition\n\nA group in which every retract is regular is a group in which every retract is a regular retract.\n\n## Definitions used\n\nTerm Definition\nRetract A subgroup", null, "$H$ of a group", null, "$G$ is a retract if there exists a normal subgroup", null, "$N$ of", null, "$G$ such that", null, "$NH = G$ and", null, "$N \\cap H$ is trivial.\nRegular retract A subgroup", null, "$H$ of a group", null, "$G$ is a regular retract if there exists a subgroup", null, "$K$ of", null, "$G$ such that", null, "$\\langle H, K \\rangle = G$,", null, "$H$ intersects the normal closure of", null, "$K$ trivially and", null, "$K$ intersects the normal closure of", null, "$H$ trivially.\n\n## Formalisms\n\n### In terms of the subgroup property collapse operator\n\nThis group property can be defined in terms of the collapse of two subgroup properties. In other words, a group satisfies this group property if and only if every subgroup of it satisfying the first property (retract) satisfies the second property (regular retract), and vice versa.\nView other group properties obtained in this way\n\n## Relation with other properties\n\n### Stronger properties\n\nProperty Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions\nGroup in which every retract is a direct factor |FULL LIST, MORE INFO\nGroup in which every retract is a free factor every retract is a free factor |FULL LIST, MORE INFO" ]

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[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\n# Introduction\n\nData manipulation and aggregation is one of the classic tasks anyone working with data will come across. We of course can perform data transformation and aggregation with base R, but when speed and memory efficiency come into play, data.table is my package of choice.\n\nIn this post we will look at of the fresh and very useful functionality that came to data.table only last year – grouping sets, enabling us, for example, to create pivot table-like reports with sub-totals and grand total quickly and easily.\n\n# Basic by-group summaries with data.table\n\nTo showcase the functionality, we will use a very slightly modified dataset provided by Hadley Wickham’s nycflights13 package, mainly the `flights` data frame. Lets prepare a small dataset suitable for the showcase:\n\n```library(data.table)\ndataurl <- \"https://jozefhajnala.gitlab.io/r/post/data/\"\nflights <- as.data.table(flights)[month < 3]```\n\nNow, for those unfamiliar with data table, to create a summary of distances flown per month and originating airport with data.table, we could simply use:\n\n```flights[, sum(distance), by = c(\"month\", \"origin\")]\n## month origin V1\n## 1: 1 EWR 9524521\n## 2: 1 LGA 6359510\n## 3: 1 JFK 11304774\n## 4: 2 EWR 8725657\n## 5: 2 LGA 5917983\n## 6: 2 JFK 10331869```\n\nTo also name the new column nicely, say `distance` instead of the default `V1`:\n\n```flights[, .(distance = sum(distance)), by = c(\"month\", \"origin\")]\n## month origin distance\n## 1: 1 EWR 9524521\n## 2: 1 LGA 6359510\n## 3: 1 JFK 11304774\n## 4: 2 EWR 8725657\n## 5: 2 LGA 5917983\n## 6: 2 JFK 10331869```\n\nFor more on basic data.table operations, look at the Introduction to data.table vignette.\n\nAs you have probably noticed, the above gave us the sums of distances by months and origins. When creating reports, especially readers coming from Excel may expect 2 extra perks\n\n• Looking at sub-totals and grand total\n• Seeing the data in wide format\n\nSince the wide format is just a reshape and data table has the `dcast()` function for that for quite a while now, we will only briefly show it in practice. The focus of this post will be on the new functionality that was only released in data.table v1.11 in May last year - creating the grand- and sub-totals.\n\n# Quick pivot tables with subtotals and a grand total\n\nTo create a “classic” pivot table as known from Excel, we need to aggregate the data and also compute the subtotals for all combinations of the selected dimensions and a grand total. In comes `cube()`, the function that will do just that:\n\n```# Get subtotals for origin, month and month&origin with `cube()`:\ncubed <- data.table::cube(\nflights,\n.(distance = sum(distance)),\nby = c(\"month\", \"origin\")\n)\ncubed\n## month origin distance\n## 1: 1 EWR 9524521\n## 2: 1 LGA 6359510\n## 3: 1 JFK 11304774\n## 4: 2 EWR 8725657\n## 5: 2 LGA 5917983\n## 6: 2 JFK 10331869\n## 7: 1 <NA> 27188805\n## 8: 2 <NA> 24975509\n## 9: NA EWR 18250178\n## 10: NA LGA 12277493\n## 11: NA JFK 21636643\n## 12: NA <NA> 52164314```\n\nAs we can see, compared to the simple group by summary we did earlier, we have extra rows in the output\n\n1. Rows `7,8` with months `1,2` and origin `<NA>, <NA>` - these are the subtotals per month across all origins\n2. Rows `9,10,11` with months `NA, NA, NA` and origins `EWR, LGA, JFK` - these are the subtotals per origin across all months\n3. Row `12` with `NA` month and `<NA>` origin - this is the Grand total across all origins and months\n\nAll that is left to get a familiar pivot table shape is to reshape the data to wide format with the aforementioned `dcast()` function:\n\n```# - Origins in columns, months in rows\ndata.table::dcast(cubed, month ~ origin, value.var = \"distance\")\n## month EWR JFK LGA NA\n## 1: 1 9524521 11304774 6359510 27188805\n## 2: 2 8725657 10331869 5917983 24975509\n## 3: NA 18250178 21636643 12277493 52164314\n# - Origins in rows, months in columns\ndata.table::dcast(cubed, origin ~ month, value.var = \"distance\")\n## origin 1 2 NA\n## 1: EWR 9524521 8725657 18250178\n## 2: JFK 11304774 10331869 21636643\n## 3: LGA 6359510 5917983 12277493\n## 4: <NA> 27188805 24975509 52164314```", null, "Pivot table with data.table\n\n## Using more dimensions\n\nWe can use the same approach to create summaries with more than two dimensions, for example, apart from months and origins, we can also look at carriers, simply by adding `\"carrier\"` into the `by` argument:\n\n```# With 3 dimensions:\ncubed2 <- cube(\nflights,\n.(distance = sum(distance)),\nby = c(\"month\", \"origin\", \"carrier\")\n)\ncubed2\n## month origin carrier distance\n## 1: 1 EWR UA 5084378\n## 2: 1 LGA UA 729667\n## 3: 1 JFK AA 2013434\n## 4: 1 JFK B6 3672655\n## 5: 1 LGA DL 1678965\n## ---\n## 153: NA <NA> F9 174960\n## 154: NA <NA> HA 293997\n## 155: NA <NA> YV 21526\n## 156: NA <NA> OO 733\n## 157: NA <NA> <NA> 52164314```\n\nAnd `dcast()` to wide format which suits our needs best:\n\n```# For example, with month and carrier in rows, origins in columns:\ndcast(cubed2, month + carrier ~ origin, value.var = \"distance\")\n## month carrier EWR JFK LGA NA\n## 1: 1 9E 46125 666109 37071 749305\n## 2: 1 AA 415707 2013434 1344045 3773186\n## 3: 1 AS 148924 NA NA 148924\n## 4: 1 B6 484431 3672655 542748 4699834\n## 5: 1 DL 245277 2578999 1678965 4503241\n## 6: 1 EV 2067900 24624 86309 2178833\n## 7: 1 F9 NA NA 95580 95580\n## 8: 1 FL NA NA 226658 226658\n## 9: 1 HA NA 154473 NA 154473\n## 10: 1 MQ 152428 223510 908715 1284653\n## 11: 1 OO NA NA 733 733\n## 12: 1 UA 5084378 963144 729667 6777189\n## 13: 1 US 339595 219387 299838 858820\n## 14: 1 VX NA 788439 NA 788439\n## 15: 1 WN 539756 NA 398647 938403\n## 16: 1 YV NA NA 10534 10534\n## 17: 1 <NA> 9524521 11304774 6359510 27188805\n## 18: 2 9E 42581 605085 34990 682656\n## 19: 2 AA 373884 1817048 1207701 3398633\n## 20: 2 AS 134512 NA NA 134512\n## 21: 2 B6 456151 3390047 490224 4336422\n## 22: 2 DL 219998 2384048 1621728 4225774\n## 23: 2 EV 1872395 24168 112863 2009426\n## 24: 2 F9 NA NA 79380 79380\n## 25: 2 FL NA NA 204536 204536\n## 26: 2 HA NA 139524 NA 139524\n## 27: 2 MQ 140924 201880 812152 1154956\n## 28: 2 UA 4686122 871824 681737 6239683\n## 29: 2 US 301832 222720 293736 818288\n## 30: 2 VX NA 675525 NA 675525\n## 31: 2 WN 497258 NA 367944 865202\n## 32: 2 YV NA NA 10992 10992\n## 33: 2 <NA> 8725657 10331869 5917983 24975509\n## 34: NA 9E 88706 1271194 72061 1431961\n## 35: NA AA 789591 3830482 2551746 7171819\n## 36: NA AS 283436 NA NA 283436\n## 37: NA B6 940582 7062702 1032972 9036256\n## 38: NA DL 465275 4963047 3300693 8729015\n## 39: NA EV 3940295 48792 199172 4188259\n## 40: NA F9 NA NA 174960 174960\n## 41: NA FL NA NA 431194 431194\n## 42: NA HA NA 293997 NA 293997\n## 43: NA MQ 293352 425390 1720867 2439609\n## 44: NA OO NA NA 733 733\n## 45: NA UA 9770500 1834968 1411404 13016872\n## 46: NA US 641427 442107 593574 1677108\n## 47: NA VX NA 1463964 NA 1463964\n## 48: NA WN 1037014 NA 766591 1803605\n## 49: NA YV NA NA 21526 21526\n## 50: NA <NA> 18250178 21636643 12277493 52164314\n## month carrier EWR JFK LGA NA```\n\n# Custom grouping sets\n\nSo far we have focused on the “default” pivot table shapes with all sub-totals and a grand total, however the `cube()` function could be considered just a useful special case shortcut for a more generic concept - grouping sets. You can read more on grouping sets with MS SQL Server or with PostgreSQL.\n\nThe `groupingsets()` function allows us to create sub-totals on arbitrary groups of dimensions. Custom subtotals are defined by the `sets` argument, a list of character vectors, each of them defining one subtotal. Now let us have a look at a few practical examples:\n\n## Replicate a simple group by, without any subtotals or grand total\n\nFor reference, to replicate a simple group by with grouping sets, we could use:\n\n```groupingsets(\nflights,\nj = .(distance = sum(distance)),\nby = c(\"month\", \"origin\", \"carrier\"),\nsets = list(c(\"month\", \"origin\", \"carrier\")),\n)```\n\nWhich would give the same results as\n\n`flights[, .(distance = sum(distance)), by = c(\"month\", \"origin\", \"carrier\")]`\n\n## Custom subtotals\n\nTo give only the subtotals for each of the dimensions:\n\n```groupingsets(\nflights,\nj = .(distance = sum(distance)),\nby = c(\"month\", \"origin\", \"carrier\"),\nsets = list(\nc(\"month\"),\nc(\"origin\"),\nc(\"carrier\")\n)\n)\n## month origin carrier distance\n## 1: 1 <NA> <NA> 27188805\n## 2: 2 <NA> <NA> 24975509\n## 3: NA EWR <NA> 18250178\n## 4: NA LGA <NA> 12277493\n## 5: NA JFK <NA> 21636643\n## 6: NA <NA> UA 13016872\n## 7: NA <NA> AA 7171819\n## 8: NA <NA> B6 9036256\n## 9: NA <NA> DL 8729015\n## 10: NA <NA> EV 4188259\n## 11: NA <NA> MQ 2439609\n## 12: NA <NA> US 1677108\n## 13: NA <NA> WN 1803605\n## 14: NA <NA> VX 1463964\n## 15: NA <NA> FL 431194\n## 16: NA <NA> AS 283436\n## 17: NA <NA> 9E 1431961\n## 18: NA <NA> F9 174960\n## 19: NA <NA> HA 293997\n## 20: NA <NA> YV 21526\n## 21: NA <NA> OO 733\n## month origin carrier distance```\n\nTo give only the subtotals per combinations of 2 dimensions:\n\n```groupingsets(\nflights,\nj = .(distance = sum(distance)),\nby = c(\"month\", \"origin\", \"carrier\"),\nsets = list(\nc(\"month\", \"origin\"),\nc(\"month\", \"carrier\"),\nc(\"origin\", \"carrier\")\n)\n)\n## month origin carrier distance\n## 1: 1 EWR <NA> 9524521\n## 2: 1 LGA <NA> 6359510\n## 3: 1 JFK <NA> 11304774\n## 4: 2 EWR <NA> 8725657\n## 5: 2 LGA <NA> 5917983\n## 6: 2 JFK <NA> 10331869\n## 7: 1 <NA> UA 6777189\n## 8: 1 <NA> AA 3773186\n## 9: 1 <NA> B6 4699834\n## 10: 1 <NA> DL 4503241\n## 11: 1 <NA> EV 2178833\n## 12: 1 <NA> MQ 1284653\n## 13: 1 <NA> US 858820\n## 14: 1 <NA> WN 938403\n## 15: 1 <NA> VX 788439\n## 16: 1 <NA> FL 226658\n## 17: 1 <NA> AS 148924\n## 18: 1 <NA> 9E 749305\n## 19: 1 <NA> F9 95580\n## 20: 1 <NA> HA 154473\n## 21: 1 <NA> YV 10534\n## 22: 1 <NA> OO 733\n## 23: 2 <NA> US 818288\n## 24: 2 <NA> UA 6239683\n## 25: 2 <NA> B6 4336422\n## 26: 2 <NA> AA 3398633\n## 27: 2 <NA> EV 2009426\n## 28: 2 <NA> FL 204536\n## 29: 2 <NA> MQ 1154956\n## 30: 2 <NA> DL 4225774\n## 31: 2 <NA> WN 865202\n## 32: 2 <NA> 9E 682656\n## 33: 2 <NA> VX 675525\n## 34: 2 <NA> AS 134512\n## 35: 2 <NA> F9 79380\n## 36: 2 <NA> HA 139524\n## 37: 2 <NA> YV 10992\n## 38: NA EWR UA 9770500\n## 39: NA LGA UA 1411404\n## 40: NA JFK AA 3830482\n## 41: NA JFK B6 7062702\n## 42: NA LGA DL 3300693\n## 43: NA EWR B6 940582\n## 44: NA LGA EV 199172\n## 45: NA LGA AA 2551746\n## 46: NA JFK UA 1834968\n## 47: NA LGA B6 1032972\n## 48: NA LGA MQ 1720867\n## 49: NA EWR AA 789591\n## 50: NA JFK DL 4963047\n## 51: NA EWR MQ 293352\n## 52: NA EWR DL 465275\n## 53: NA EWR US 641427\n## 54: NA EWR EV 3940295\n## 55: NA JFK US 442107\n## 56: NA LGA WN 766591\n## 57: NA JFK VX 1463964\n## 58: NA LGA FL 431194\n## 59: NA EWR AS 283436\n## 60: NA LGA US 593574\n## 61: NA JFK MQ 425390\n## 62: NA JFK 9E 1271194\n## 63: NA LGA F9 174960\n## 64: NA EWR WN 1037014\n## 65: NA JFK HA 293997\n## 66: NA JFK EV 48792\n## 67: NA EWR 9E 88706\n## 68: NA LGA 9E 72061\n## 69: NA LGA YV 21526\n## 70: NA LGA OO 733\n## month origin carrier distance```\n\n## Grand total\n\nTo give only the grand total:\n\n```groupingsets(\nflights,\nj = .(distance = sum(distance)),\nby = c(\"month\", \"origin\", \"carrier\"),\nsets = list(\ncharacter(0)\n)\n)\n## month origin carrier distance\n## 1: NA <NA> <NA> 52164314```\n\n# Cube and rollup as special cases of grouping sets\n\n## Implementation of cube\n\nWe mentioned above that `cube()` can be considered just a shortcut to a useful special case of `groupingsets()`. And indeed, looking at the implementation of the data.table method `data.table:::cube.data.table`, most of what it does is to define the `sets` to represent the given vector and all of its possible subsets, and passes that to `groupingsets()`:\n\n```function (x, j, by, .SDcols, id = FALSE, ...) {\nif (!is.data.table(x))\nstop(\"Argument 'x' must be a data.table object\")\nif (!is.character(by))\nstop(\"Argument 'by' must be a character vector of column names used in grouping.\")\nif (!is.logical(id))\nstop(\"Argument 'id' must be a logical scalar.\")\nn = length(by)\nkeepBool = sapply(2L^(seq_len(n) - 1L), function(k) rep(c(FALSE,\nTRUE), times = k, each = ((2L^n)/(2L * k))))\nsets = lapply((2L^n):1L, function(j) by[keepBool[j, ]])\njj = substitute(j)\ngroupingsets.data.table(x, by = by, sets = sets, .SDcols = .SDcols,\nid = id, jj = jj)\n}```\n\nThis means for example that\n\n```cube(flights, sum(distance), by = c(\"month\", \"origin\", \"carrier\"))\n## month origin carrier V1\n## 1: 1 EWR UA 5084378\n## 2: 1 LGA UA 729667\n## 3: 1 JFK AA 2013434\n## 4: 1 JFK B6 3672655\n## 5: 1 LGA DL 1678965\n## ---\n## 153: NA <NA> F9 174960\n## 154: NA <NA> HA 293997\n## 155: NA <NA> YV 21526\n## 156: NA <NA> OO 733\n## 157: NA <NA> <NA> 52164314```\n\nIs equivalent to\n\n```groupingsets(\nflights,\nj = .(distance = sum(distance)),\nby = c(\"month\", \"origin\", \"carrier\"),\nsets = list(\nc(\"month\", \"origin\", \"carrier\"),\nc(\"month\", \"origin\"),\nc(\"month\", \"carrier\"),\nc(\"month\"),\nc(\"origin\", \"carrier\"),\nc(\"origin\"),\nc(\"carrier\"),\ncharacter(0)\n)\n)\n## month origin carrier distance\n## 1: 1 EWR UA 5084378\n## 2: 1 LGA UA 729667\n## 3: 1 JFK AA 2013434\n## 4: 1 JFK B6 3672655\n## 5: 1 LGA DL 1678965\n## ---\n## 153: NA <NA> F9 174960\n## 154: NA <NA> HA 293997\n## 155: NA <NA> YV 21526\n## 156: NA <NA> OO 733\n## 157: NA <NA> <NA> 52164314```\n\n## Implementation of rollup\n\nThe same can be said about `rollup()`, another shortcut than can be useful. Instead of all possible subsets, it will create a list representing the vector passed to `by` and its subsets “from right to left”, including the empty vector to get a grand total. Looking at the implementation of the data.table method `data.table::rollup.data.table`:\n\n```function (x, j, by, .SDcols, id = FALSE, ...) {\nif (!is.data.table(x))\nstop(\"Argument 'x' must be a data.table object\")\nif (!is.character(by))\nstop(\"Argument 'by' must be a character vector of column names used in grouping.\")\nif (!is.logical(id))\nstop(\"Argument 'id' must be a logical scalar.\")\nsets = lapply(length(by):0L, function(i) by[0L:i])\njj = substitute(j)\ngroupingsets.data.table(x, by = by, sets = sets, .SDcols = .SDcols,\nid = id, jj = jj)\n}```\n\nFor example, the following:\n\n`rollup(flights, sum(distance), by = c(\"month\", \"origin\", \"carrier\"))`\n\nIs equivalent to\n\n```groupingsets(\nflights,\nj = .(distance = sum(distance)),\nby = c(\"month\", \"origin\", \"carrier\"),\nsets = list(\nc(\"month\", \"origin\", \"carrier\"),\nc(\"month\", \"origin\"),\nc(\"month\"),\ncharacter(0)\n)\n)```" ]

[ null, "https://i1.wp.com/jozefhajnala.gitlab.io/r/img/r912-01-datatable-pivot.gif", null ]

[ "/ / Article\n\nResearch Article | Open Access\n\nVolume 2014 |Article ID 846962 | https://doi.org/10.1155/2014/846962\n\nYuanxin Zhou, P. K. Mallick, \"Effect of Melt Temperature and Hold Pressure on the Weld-Line Strength of an Injection Molded Talc-Filled Polypropylene\", Journal of Composites, vol. 2014, Article ID 846962, 8 pages, 2014. https://doi.org/10.1155/2014/846962\n\n# Effect of Melt Temperature and Hold Pressure on the Weld-Line Strength of an Injection Molded Talc-Filled Polypropylene\n\nRevised21 Dec 2013\nAccepted21 Jan 2014\nPublished05 Mar 2014\n\n#### Abstract\n\nTensile stress-strain behavior coupled with fractography was used to investigate the weld-line strength of an injection molded 40 w% talc-filled polypropylene. The relationship between processing conditions, microstructure, and tensile strength was established. Fracture surface of the weld line exhibited skin-core morphology with different degrees of talc particle orientations in the core and in the skin. Experimental results also showed that the thickness of the core decreased and the thickness of the skins increased with increasing melt temperature and increasing hold pressure, which resulted in an increase of yield strength and yield strain with increasing melt temperature and increasing hold pressure. Finally, a three-parameter nonlinear constitutive model was developed to describe the strain softening behavior of the weld-line strength of talc-filled polypropylene. The parameters in this model are the modulus E, the strain exponent m, and the compliance factor β. The simulated stress-strain curves from the model are in good agreement with the test data, and both m and β are functions of skin-core thickness ratio.\n\n#### 1. Introduction\n\nInjection molding is one of the most common manufacturing processes in the polymer industry owing to its versatility, high production rate, short cycle time, and low percentage of scrap. In addition, this process can be used for a large variety of thermoplastic polymers. One of the defects observed in injection molded parts is weld line, which is formed when two or more separate melt fronts traveling from different directions meet and join as the mold cavity is filled . This happens in multigated molds but can also occur when a melt front is divided by inserts used in the mold to create holes or other openings in injection-molded parts. Since weld lines cause reduction in mechanical properties and visual defects, many studies have been conducted to explain the weakness at the weld line. For example, Mielewski et al. have investigated the weld-line morphology of injection molded polypropylene. Tomari et al. have reported V-notch at weld lines in polystyrene injection moldings. Fellahi et al. investigated the morphology of the weld region in injection-molded high-density polyethylene/polyamide-6 blends. In the homopolymer system, the weakness of weld line can be explained by incomplete bonding due to inefficient molecular entanglement at the interface, disturbance of molecular orientation parallel to the flow direction, inefficient diffusion time, existence of voids, and V-shape notch owing to entrapped air or contaminants and notch depth [5, 6]. On the other hand, loss in weld-line strength in filled and fiber reinforced polymers depends on the shape of the reinforcement. Fisa and Rahmani studied the weld-line strength of injection molded short fiber reinforced polypropylene. The weld line in their study was formed by flow around a circular insert in the mold. They found that the weld-line zone, which extended throughout the thickness, was 2 to 8 mm wide and the fibers in this zone were oriented in a plane parallel to the weld line. Savadori et al. have observed significant reduction in strength for rubber filled PP/EP blends.\n\nThe purpose of this study was to determine the effect of two injection molding processing conditions on the weld line strength of talc-filled polypropylene. Polypropylene is a semicrystalline engineering thermoplastic and is known for its balance of strength, modulus, and chemical resistance. Both polypropylene and polypropylene matrix composite have many potential applications in automobiles where creep resistance, stiffness, and some toughness are required in addition to weight savings. Mechanical behavior of polypropylene and its composites has been the subject of numerous studies over the last few years. Some of these research studies have investigated the relationship between the mechanical behavior of polypropylene, microstructure, and processing conditions. Diez-Gutierrez et al. have reported the dynamic mechanical analysis of injection-molded discs of polypropylene and untreated and silane-treated talc-filled polypropylene composites. Li and Cheung have investigated the effect of mold temperature on the formation of polypropylene blends in injection molding. All these results show that the mechanical behavior of polypropylene and its composites is sensitive to the injection molding processing conditions. However, the effect of processing conditions on the weld-line strength of talc-filled polypropylene composites has not been reported in the past.\n\n#### 2. Experimental\n\nThe material investigated in this study was a 40 w% talc-filled polypropylene homopolymer. The average flow rate of this material was 6.8 g/10 min. The melting point of the polypropylene matrix was 160°C. The injection-molded plate thickness was 2.5 mm. A 90-ton Toyo injection molding machine was used to mold these plates. Three different melt temperatures were considered, namely, 209, 232, and 277°C. The peak injection pressure was 103 MPa for all plates, but the hold pressure was varied at three levels, namely, 27.6, 55.2, and 82.7 MPa. The mold temperature was maintained at 35°C. The following two groups of plates were injection molded:(1)group I: hold pressure = 55.2 MPa and melt temperature = 209, 232, and 277°C,(2)group II: melt temperature = 232°C and hold pressure = 27.6, 55.2, and 82.7 MPa.\n\nDog-bone shaped specimens were prepared from the injection molded plates with weld line. The specimen dimensions were 100 mm in overall length, 25 mm in gage length, and 12.7 mm in gage width (Figure 1). The weld line, shown as located at the mid length of the tensile specimens, was formed in the plates as the two melt fronts with the opposite flow directions joined together at the mid length.\n\nUniaxial tensile tests were performed on an MTS servohydraulic testing machine. Three specimens were tested for each processing condition. The tensile tests were conducted at 1.25 mm/min, which is equivalent to a strain rate of 0.05 min−1. Three parameters were evaluated from each stress-strain curve: elastic modulus (), yield strength , and yield strain . Elastic modulus or Young’s modulus is the initial slope of the stress-strain curve. Yield strength is assumed to be the maximum stress observed in each stress-strain diagram and the strain corresponding to the yield strength is the yield strain. During the tensile tests, all specimens were observed to break at the weld zone.\n\n#### 3. Results and Discussion\n\nFigures 2 and 3 show the tensile stress-strain curves of talc-filled polypropylene with weld line under the different processing conditions. It can be observed in these figures that the stress-strain diagrams are nonlinear even at strains lower than the yield strain. Each curve shows a maximum stress, which is assumed to be the yield strength of the material at the weld line. After reaching the yield point, stress decreased steadily with strain until fracture occurred at the weld zone.\n\nFigure 2 shows the effect of melt temperature on the tensile behavior of talc-filled polypropylene with weld line. The yield strength and yield strain increased by 9.3% and 26.6%, respectively, as the melt temperature was increased from 209°C to 277°C. But the effect of melt temperature on the modulus is relatively small for the three melt temperatures investigated. Figure 3 shows the effect of hold pressure on the tensile behavior of talc-filled polypropylene with weld line. Increasing the hold pressure increases the yield strength, yield strain, and failure strain. But the modulus is not sensitive to hold pressure. The tensile properties of material with different processing conditions are listed in Table 1.\n\n Melt temperature(°C) Hold pressure(MPa) Modulus(GPa) Yield strength (MPa) Yield strain (%) (1/MPa) 209 55.2 6.56 (±0.23) 17.5 (±0.31) 0.79 (±0.04) 2.81 113 232 55.2 6.84 (±0.19) 17.9 (±0.40) 0.91 (±0.04) 2.79 90.4 277 55.2 6.94 (±0.20) 19.1 (±0.34) 1.00 (±0.07) 2.57 28.2 232 27.6 6.71 (±0.24) 17.1 (±0.23) 0.86 (±0.04) 2.87 149 232 82.7 6.92 (±0.20) 22.5 (±0.30) 1.31 (±0.03) 2.32 6.9\n\nFrom Figures 2 and 3 and Table 1, it can be concluded that the weld-line yield properties of talc-filled polypropylene are sensitive to the two processing conditions considered. Figures 4 and 5 show that the variation of yield strength and yield strain with melt temperature and hold pressure can be represented by linear relationships. The following two linear relationships are proposed: where , and , are melt temperature sensitivity factor and hold pressure sensitivity factor. and are reference melt temperature and reference mold pressure. and are the reference yield strength and yield strain. Mathematically, , and , can be obtained by Assuming the reference melt temperature = 232°C and reference hold pressure  MPa, one can obtain the average values of the melt temperature sensitivity factor and the hold pressure sensitivity factor as\n\nFigure 6 shows comparison of stress-strain curves for talc-filled polypropylene with weld and without weld line. When the weld line was introduced, the yield strength, yield strain, and failure strain decreased by 35%, 73%, and 92%, respectively. But the presence of weld line does not influence the modulus. After the yield strain is reached, the weld-line specimen breaks quickly, with very little additional deformation, indicating that the weld line is a weak zone in the material. According to Merah et al. , weld line is a brittle zone; the yield strain and failure strain for weld-line specimens have the same value. But in the talc-filled polypropylene with weld line, 0.3–0.6% plastic strain was observed after yielding (as shown in Figures 2 and 3). In Figure 6, strain softening behavior was found for specimens with and without weld line, which was conjectured to be due to the strain softening behavior of polypropylene matrix .\n\nTo investigate the failure mechanism of the weld line, the fracture surfaces at the weld line were examined in a scanning electron microscope. Fracture surface in Figure 7(a) exhibits skin-core morphology with different degrees of talc particle orientations in the core and in the skin. A “white band” can be observed in the center section of the fracture surface. It consists of talc particles with their surface oriented normal to the flow direction (Figure 7(b)). No matrix material can be seen adhering to the particles, indicating poor or no adhesion between the talc particles and the polypropylene matrix. In this area, the particles prevented the polymer chain to bridge across the weld line interface to form a strong bond and the orientation of the molecules is parallel to the weld-line rather than across it. Brittle fracture of talc particles can also be observed in Figure 7(b). Talc particles outside the core tend to be parallel to the flow direction (same as the tensile test direction). The color of this area is dark. Polymer chains in this area can easily bridge across the weld line to form a strong bond between the two sides and the orientation of the molecules is normal to the weld line rather than parallel to it. Magnification of the fracture initiation site was shown in Figure 7(c). The matrix material was drawn in the tensile stress direction and fibrillated around the talc particles. The ductility of weld line can be mainly attributed to the large elongation of polypropylene in the dark area.\n\nThe formation of skin-core morphology along the weld line of talc-filled polypropylene can be explained by referring to the viscous flow behavior of the liquid material as it flows in the injection molding cavity. The shear stress in the flow channel decreases from its highest value at the cavity walls to a near-zero value at the center. Near the cavity walls, the platy talc particles as well as the polymer molecules become oriented parallel to the flow direction as a result of the high shear stress in this area. But near the center, where the shear stress is the lowest and the flow velocity is the highest, the particles turn to normal to the flow direction.\n\nFigures 8(a) and 8(b) show the fracture surfaces of the specimens injection molded at hold pressures 27.6 MPa and 82.7 MPa, respectively. The thickness of the “white band” decreased with the increasing hold pressure; more and more talc particles tend to be parallel to the flow direction (tensile direction). Figures 9(a) and 9(b) show the fracture surfaces of the specimens injection molded at melt temperatures 209°C and 277°C, respectively. Here also, the thickness of the “white band” decreased with increasing melt temperature.\n\nFrom Figures 79, it can be concluded that the total cross-sectional area of the weld line can be divided into well bonded area (dark zone) and nonbonded area (white band). If we assume that the total load is shared by the two areas, then the yield strength of the weld line can be expressed as where , are the yield strengths of the injection molded part without and with “white band.” is the total cross-sectional area of the specimen and is cross-sectional area of the nonbonded white area. Furthermore, if we assume that both dark and white bands have the same width as the specimen width, the skin-core thickness ratio can be defined as where and are thickness of white band and thickness of entire specimen. Equation (4) can then be rewritten as Similarly, the yield strain of the weld line can be expressed as Figure 10 shows the variation between yield strength, yield strain, and thickness ratio. Linear relationships are observed and the constants were simulated by using the least square method Small value in and negative value in declare the loading bearing capability of the “white band” can be ignored.\n\n#### 4. One-Dimensional Constitutive Equation\n\nIn this section, we apply a one-dimensional constitutive equation that was developed earlier for welded talc-filled polypropylene. The total strain is assumed to be additively decomposed into elastic and inelastic parts, where and represent the elastic and inelastic strains, respectively. The elastic strain is assumed to be path independent, such that where is elastic modulus of the material and is the stress. The inelastic strain, , is assumed to be a function of stress and total strain in the following way: Substituting (10) and (11) into (9) gives Equation (12) represents the constitutive equation. At small strains, , one can obtain Therefore, at very small strains, (12) approximates an elastic material. At large strains, , and (12) can be simplified as To determine the parameters in the constitutive equation, (12) is written as\n\nEquation (16) represents a linear equation when is plotted against . From the slope and intercept of the linear plot, and of the material can be obtained by using the least square method. Plots of versus of talc-filled polypropylene with weld line, shown in Figure 11, are linear at different processing conditions. The values of and obtained from these plots are listed in Table 1.\n\nFigure 12 shows the compliance factor and strain exponent plotted as a function of thickness ratio. It can be observed from this figure that both and increased with increasing thickness ratio. Their relationships can be expressed by the following equations:\n\nSubstituting elastic modulus , compliance factor , and strain exponent into (12), the simulated stress-strain plots were drawn in Figures 2 and 3 and they seem to fit the experimental data well.\n\n#### 5. Conclusions\n\nWeld-line yield strength, yield strain, and failure strain of talc-filled polypropylene increase with increasing melt temperature and increasing hold pressure. The effects of these two injection molding process parameters are explained in terms of their influence on the skin-core morphology observed on the fracture surfaces in the weld zone. The core contained talc particles that were oriented normal to the loading direction and the skins contained talc particles oriented parallel to the loading direction. The thickness of the core decreased and the thickness of the skins increased with increasing melt temperature and increasing hold pressure. A three-parameter nonlinear constitutive model is applied to describe the stress-strain curves of the talc-filled polypropylene. The parameters in this model are the modulus , the strain exponent , and the compliance factor . Both and are sensitive to the skin-score thickness ratio.\n\n#### Conflict of Interests\n\nThe authors declare that there is no conflict of interests regarding the publication of the paper.\n\n1. N. Mekhilef, A. Ait-Kadi, and A. Ajji, “Weld lines in injection-moulded immiscible blends: model predictions and experimental results,” Polymer, vol. 36, no. 10, pp. 2033–2042, 1995. View at: Google Scholar\n2. D. F. Mielewski, D. R. Bauer, P. J. Schmitz, and H. Van Oene, “Weld line morphology of injection molded polypropylene,” Polymer Engineering and Science, vol. 38, no. 12, pp. 2020–2028, 1998. View at: Google Scholar\n3. K. Tomari, S. Tonogai, T. Harada et al., “V-notch at weld lines in polystyrene injection moldings,” Polymer Engineering and Science, vol. 30, no. 15, pp. 931–936, 1990. View at: Google Scholar\n4. S. Fellahi, B. D. Favis, and B. Fisa, “Morphological stability in injection-moulded high-density polyethylene/polyamide-6 blends,” Polymer, vol. 37, no. 13, pp. 2615–2626, 1996. View at: Publisher Site | Google Scholar\n5. J. K. Kim, S. H. Park, T. O. Hyun, and H. K. Jeon, “The effect of weld-lines on the morphology and mechanical properties of amorphous polyamide/poly(ethylene-ran-propylene) blend with various amounts of an in situ compatibilizer,” Polymer, vol. 42, no. 5, pp. 2209–2221, 2001. View at: Google Scholar\n6. O. G. Ersoy and N. Nugay, “A new approach to increase weld line strength of incompatible polymer blend composites: selective filler addition,” Polymer, vol. 45, no. 4, pp. 1243–1252, 2004. View at: Publisher Site | Google Scholar\n7. B. Fisa and M. Rahmani, “Weldline strength in injection molded glass fiber-reinforced polypropylene,” Polymer Engineering & Science, vol. 31, no. 18, pp. 1330–1336, 1991. View at: Publisher Site | Google Scholar\n8. A. Savadori, A. Pelliconi, and D. Romanini, “Weld line resistance in polypropylene composites,” Plastics and Rubber Processing and Applications, vol. 3, no. 3, pp. 215–221, 1983. View at: Google Scholar\n9. S. Diez-Gutierrez, M. A. Rodriguez-Perez, J. A. De Saja, and J. I. Velasco, “Dynamic mechanical analysis of injection-moulded discs of polypropylene and untreated and silane-treated talc-filled polypropylene composites,” Polymer, vol. 40, no. 19, pp. 5345–5353, 1999. View at: Publisher Site | Google Scholar\n10. J. X. Li and W. L. Cheung, “Effect of mould temperature on the formation of α/β polypropylene blends in injection moulding,” Journal of Materials Processing Technology, vol. 63, no. 1–3, pp. 472–475, 1997. View at: Google Scholar\n11. N. Merah, M. Irfan-ul-Haq, and Z. Khan, “Temperature and weld-line effects on mechanical properties of CPVC,” Journal of Materials Processing Technology, vol. 142, no. 1, pp. 247–255, 2003. View at: Publisher Site | Google Scholar\n12. Y. Zhou and P. K. Mallick, “Effects of temperature and strain rate on the tensile behavior of unfilled and talc-filled polypropylene. Part I: experiments,” Polymer Engineering and Science, vol. 42, no. 12, pp. 2449–2460, 2002. View at: Publisher Site | Google Scholar\n13. Y. Zhou and P. K. Mallick, “Effects of temperature and strain rate on the tensile behavior of unfilled and talc-filled polypropylene. Part II: constitutive equation,” Polymer Engineering and Science, vol. 42, no. 12, pp. 2461–2470, 2002. View at: Publisher Site | Google Scholar" ]

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[ "A trapezoidal prism is a 3D figure comprised of 2 trapezoids that is joined by 4 rectangles.Prisms consist of polygon that kind their bases. Among the most recognizable examples of a trapezoid prism is a brick specifically the circular fire pit brick. A trapezoidal prism comes from the form trapezoid. Let's learn an ext about itsdefinition, properties, formulas and solve a couple of examples.\n\nYou are watching: What is a 3d trapezoid called\n\n 1 Trapezoidal Prism Definition 2 Trapezoidal Prism Properties 3 Trapezoidal Prism Net 4 Volume the a Trapezoidal Prism 5 Surface Area the a Trapezoidal Prism 6 FAQs ~ above Trapezoidal Prism\n\n## Trapezoidal Prism Definition\n\nA trapezoidal prism is a 3D figurewhich has trapezoid cross-sections in one direction and also rectangular cross-sections in the various other direction which means the prism has actually two congruent trapezoids that are connected to each other with four rectangles. These congruent trapezoids are on the top and also bottom of the prism i beg your pardon are referred to as bases. The 4 rectangles are dubbed the lateral encounters of the trapezoid prism.A trapezoidal prism has six faces, eight vertices, and also 12 edges.To construct a trapezoidal prism, we require to draw a trapezoid very first that has twoparallel sides and a height i.e. Distance in between the sides.The image listed below describes exactly how a trapezoid prism looks like. H suggests the height of the trapezoid, S suggests the surface ar area, L indicates the lateral area, and B1and B2indicate the lengths that the base.", null, "## TrapezoidalPrism Properties\n\nThe properties of a trapezoid prism space very similar to the of a trapezoid, castle are detailed below:\n\nA trapezoid prism is composed of two trapezoids that space joined by 4 rectanglesA trapezoid prism consists of eight verticesA trapezoid prism is composed of six facesA trapezoid prism consists of 12 edgesA trapezoid prism is a polygon that consists of trapezoids\n\n## Trapezoidal Prism Net\n\nThe network of the trapezoidal prism consists of six faces, eight vertices, and also 12 edges. When the trapezoidal prism is opened up flat, we have the right to see the four rectangle-shaped objects that help in involvement the 2 trapezoid-shaped objects together to kind a trapezoidal prism. The image below demonstrates the flattened variation of a trapezoidal prism.", null, "## Volume of a Trapezoidal Prism\n\nTo find the volume of a trapezoidal prism, we must multiply the area that the basic by the height. The area is calculated together the surface ar area of one of the trapezoids through the distance between the two trapezoids. Therefore, in order to find the volume the a trapezoidal prism, we require to first find the area of one trapezoid.Hence, to discover the volume the a trapezoidal prism we have the right to use theseformulas:\n\nVolume the a Trapezoidal Prism = A× ns cubic units\n\nArea (A) =½× h× (a + b) or½ h(b1+ b2)\n\nWhere h is the height of the trapezoid, together is the height of the prism, a and b space the lengths that the top and bottom of a trapezoidal prism\n\n## Surface Area the a Trapezoidal Prism\n\nThe surface ar area that a trapezoidal prism is calculate by multiplying the surface ar area of one of the prism by 2 then adding the amount of the perimeter right into the elevation of the trapezoidal prism. The formula because that calculating the trapezoidal prism is:\n\nSurface Area of a Trapezoidal Prism =h (b + d) + l (a + b + c + d) square units\n\nwhere h is the height, b and d are the lengths of the base, a + b + c + d is the perimeter, and also l is the lateral surface ar area. To understand this better, you can examine out the web page onhow to find the surface ar area the a trapezoidal prism.\n\n### Related Topics\n\nListed below are a couple of interesting topics concerned the trapezoidal prism.\n\n## Examples ~ above Trapezoidal Prism\n\nExample 1: Sandy is setting up a tent that is in the shape of a trapezoidal prism v dimensions as elevation as 4 units, lengths that the trapezoid together 4 units and 8 units, and length of the prism as 10 units. How many cubic feet of an are are over there in her tent?", null, "Solution: To discover the room in her tent, we require to find the volume of the tent. Given is h = 4, a = 8, b = 4, and I = 10\n\nThe volume the a trapezoidal prism =A× ns cubic units\n\nWe require to discover the area that the trapezoid. The formula is\n\nArea (A) =½× h× (a + b)\n\nA =½× 4× (8 + 4)\n\nA = 2× 12\n\nA = 24 square feet\n\nLet us uncover the volume now.\n\nVolume the a trapezoidal prism =A× I\n\nV = 24× 10\n\nV = 240 cubic feet\n\nTherefore, the variety of cubic feet of room in sandy's tent is 240 cubic feet.\n\nExample 2: Find the surface area that the given trapezoidal prism with height as 3 units, lengths of the bases together 4.5 units, and 6 units, lateral size as 5, and also sides that the prism together 2.5 units.", null, "Solution: Let us include all the values that we have from the provided image into the formula.\n\nSurface Area of a Trapezoidal Prism = h (b + d) + together (a + b + c + d)\n\nA = 3 (4.5 + 6) + 5 (4.5 + 2.5 + 6 + 2.5)\n\nA = 3× 10.5 + 5× 15.5\n\nA = 31.5 + 77.5\n\nA = 109 units2\n\nTherefore, the surface area of the given trapezoidal prism is 109 units2\n\nView an ext >\n\ngo come slidego come slide\n\nBreakdown tough ideas through simple visuals.\nMath will no longer be a challenging subject, specifically when you know the principles through visualizations.\n\nBook a free Trial Class\n\n## Practice questions of Trapezoidal Prism\n\nTry These! >\n\ngo to slidego come slide\n\n## FAQs ~ above Trapezoidal Prism\n\n### What is a Trapezoidal Prism?\n\nA trapezoidal prism is a 3D shape with 2 trapezoids as its base that is being joined by four rectangles. A trapezoidal prism was given its name because it is made up of trapezoids.A trapezoidal prism has six faces, eight vertices, and 12 edges. One of the box noticeable example of a trapezoidal prism the we check out in everyday life is fire brick.\n\n### What is the Formula to calculation the Volume that a Trapezoidal Prism?\n\nTo discover the volume that a trapezoidal prism, we require to very first find the area of one trapezoid. The formula to calculate the volume of a trapezoidal prism is:\n\nVolume the a Trapezoidal Prism = A× I\n\nArea (A) =½× h× (a + b) or½ h(b1+ b2)\n\nWhere h is the height of trapezoidal, l is the elevation of the prism, a and b are the lengths that the top and bottom the a trapezoidal prism\n\n### What is the Formula to calculate the surface ar Area of a Trapezoidal Prism?\n\nThe formula to calculate the surface area that a trapezoidal prism is:\n\nSurface Area that a Trapezoidal Prism =h (b + d) + l (a + b + c + d)\n\nwhere h is the height, b and d are the lengths that the base, a + b + c + d is the perimeter, and l is the lateral surface ar area.\n\n### What is the network of a Trapezoidal Prism?\n\nThe network of a trapezoidal prism is the it is composed of two trapezoids and four rectangles. When the trapezoidal prism is flattened, we can see this two photos clearly.\n\nSee more: How Long Does Pvc Glue Need To Dry ? What Is The Cure Time For Pvc Cement\n\n### How carry out you find the height of a Trapezoidal Prism?\n\nIn order to uncover the elevation of a trapezoidal prism, we need to find the area of among the trapezoids. Since the prism has actually two trapezoids, to calculate the volume we require to uncover the height of one of the trapezoids utilizing this formula:\n\nArea (A) = ½ × h × (a + b)\n\nwhere A is the area, h is the height, a and b room the lengths of the base trapezoids\n\n### Why is the Trapezoidal Prism a 3D Figure?\n\nThe 3D number of a trapezoid is called a trapezoidal prism because these are figures that have a length, width, and also depth.The edge of the prism iswhere the deals with meet, and the vertices that the prismare the cornerswhere 3 or more surfaces meet." ]

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[ "", null, "### Joeri Jongbloets\n\nAlmost MSc // Modeller in the lab // Fluent in Java, Python, and R\n\n# Simulate a stable culture\n\n### Code to sample from a distribution defined by percentiles\n\n``````## Based on:\n## http://stackoverflow.com/questions/14547364/generate-distribution-given-percentile-ranks?lq=1\nsample.prob <- function(n, values, cum.probs, pair.len = 100) {\nprob <- c( first(cum.probs), diff(cum.probs), 1-last(cum.probs))\n# Extreme values beyond x (to sample)\ninit <- min(values) - diff(values)[]\nfin <- max(values) + diff(values)[]\n\nival <- c(init, values, fin) # generate the sequence to take pairs from\n\ns <- sapply(2:length(ival), function(i) {\nseq(ival[i-1], ival[i], length.out=pair.len)\n})\n# sample from s, total of 10000 values with probabilities calculated above\nreturn( sample(s, n, prob=rep(prob, each=pair.len), replace = T) )\n}\n\nsample.prob_c <- compiler::cmpfun(sample.prob)\n``````\n\n### Code to bundle values into bins\n\n``````# create.bins <- function( values, n.bins, lwr=NULL, upr=NULL) {\n# if (is.null(lwr)) lwr <- min(values)\n# if (is.null(upr)) upr <- max(values)\n# if ( lwr == upr ) {\n# lwr <- lwr * (1 - 1/ (n.bins / 2))\n# upr <- upr * (1 + 1/ (n.bins / 2))\n# }\n# apply( values, 1, cut, c(-Inf, seq(lwr, upr, length.out=n.bins-1), Inf), labels=FALSE)\n# }\n\ncreate.bins <- function( values, n.bins, lwr=NULL, upr=NULL) {\nif (is.null(lwr)) lwr <- min(values)\nif (is.null(upr)) upr <- max(values)\nif ( lwr == upr ) {\nlwr <- lwr * (1 - 1/ (n.bins / 2))\nupr <- upr * (1 + 1/ (n.bins / 2))\n}\nfindInterval(values, c(seq(lwr, upr, length.out=n.bins)))\n}\n\n## create byte code\ncreate.bins_c <- compiler::cmpfun(create.bins)\n\ncreate.bins.df <- function( df, bin.var, n.bins = 10, dots = list(), dots.name = c(), size.name=\"size\",\n## generate bins: group growth rates into growth.bins number of bins.\ndf\\$bin <- create.bins( df[[bin.var]], n.bins, ... )\n## prepare summarise\ndots <- c(dots, list(~n()))\ndots.name <- c(dots.name, size.name)\n## return binned population as the seed population\nresult <- df %>%\nsummarise_(\n.dots = setNames(dots, dots.name)\n) %>%\nselect_(\"-bin\")\nreturn(result)\n}\n\n# create byte code\ncreate.bins.df_c <- compiler::cmpfun(create.bins.df)\n``````\n\n### Code to create a population by sampling from a percentile distribution\n\n``````## creates a population of cells in n bins of growth rate\n## return: data.frame with cells : growth.rate , n.cells\nseed.population <- function(\nn.cells = 10000, mean.growth = 0.08, extreme.growth = 0.09, extreme.prob = 0.001, growth.bins = 100,\nsample.fun = sample.prob\n) {\n## generate sample population\nextreme.left = mean.growth - (extreme.growth - mean.growth)\npopulation <- data.frame(\ngrowth.rate = sample.fun(\nn = n.cells, ## generate 1000 cells\nvalues = c(extreme.left, mean.growth, extreme.growth), ## distribution values\nc( extreme.prob , 0.5, 1-extreme.prob)) # cumulative probabilities\n)\n## generate bins\nreturn(\ncreate.bins.df(\npopulation, \"growth.rate\", growth.bins, list(~mean(growth.rate)), c(\"growth.rate\"), \"n.cells\"\n)\n)\n}\nseed.population_c <- compiler::cmpfun(seed.population)\n``````\n\n### Code to simulate the turbidostat\n\nThe global simulation function\n\n``````## Simulation function, will run n.rounds of simulation using a selection function and mutation function\n# cells: data.frame, with growth.rate and n.cells // cells to evolve\n# n.rounds: int // number of rounds\n# t.step: int // length of each time step in hours\nrun.simulation <- function( cells, n.rounds=100, t.step=5/60, dead.threshold = 1, grow.fun=simulate.growth, dilute.fun=simulate.turbidostat, verbose=FALSE, ... )\n{\nt.begin <- Sys.time()\n## result logs the counts of all cells at each time point\n## preallocate a row for each round in result\nresult <- vector(mode = \"list\", length = n.rounds)\nresult[] <- as.data.table(cells %>% mutate( t = 0, decision = FALSE))\ni <- 1\nwhile ( i <= n.rounds & nrow(cells) > 0 )\n{\nt.cycle <- Sys.time()\n## grow cells\nif (verbose) print(\"## Grow cells\")\ncells <- grow.fun(cells, t.step, verbose=verbose...)\n## run turbidostat\nif (verbose) print(\"## Dilute cells\")\ncells <- dilute.fun(cells, verbose=verbose, ...)\n# set dead strains to zero to indicate their death\nif (verbose) print(\"## Tag extinct strains\")\ncells <- cells %>% mutate( n.cells = ifelse(n.cells < dead.threshold, 0, n.cells))\n## log state\nif (verbose) print(\"## Add to results\")\nresult[[i+1]] <- as.data.table(cells %>% mutate(t=i))\nif (verbose) print(\"## Remove extinct strains\")\ncells <- cells %>% filter( n.cells >= dead.threshold)\nif ( i %% (n.rounds / 100) == 0) {\nt.now <- Sys.time()\nprint(\nsprintf(\"%i%% completed // t: %.0f hour // C/T: %.3f/%.3f seconds // Cells: %i // Strains: %i\",\ni / (n.rounds / 100),\ni * t.step,\nas.numeric(t.now - t.cycle, units=\"secs\"),\nas.numeric(t.now - t.begin, units=\"secs\"),\nas.integer(sum(cells\\$n.cells)), as.integer(count(cells))\n))\n}\ni <- i + 1\n}\nif (verbose & nrow(cells) == 0) print(\"Ran out of cells!\")\n## leave user with ability to rbindlist the result\nreturn(result)\n}\n\nrun.simulation_c <- compiler::cmpfun(run.simulation)\n``````\n\nThe turbidostat decision function\n\n``````## Turbidostat simulation procedure\nsimulate.turbidostat <- function ( cells, threshold = 100, dilution.fraction = 0.08, verbose=FALSE, ... ) {\n## calculate total\ncell.count <- sum(cells\\$n.cells)\nif (verbose) print(sprintf( \"TBS: %s > %s\", cell.count, threshold ))\nresult <- cells %>%\nmutate(\ndecision = cell.count > threshold,\nn.cells = ifelse( decision, floor(n.cells * (1-dilution.fraction)), n.cells )\n)\nif (verbose) print(sprintf( \"TBS: Removed %.2f cells\", cell.count - sum(cells\\$n.cells) ))\nreturn (result)\n}\n\nsimulate.turbidostat_c <- compiler::cmpfun(simulate.turbidostat)\n\n## Growth simulation\n## just grows the cells\nsimulate.growth <- function( cells, t.step, partial.division = TRUE, ...) {\nresult <- cells %>%\nmutate(\nn.cells = n.cells * exp( t.step * growth.rate )\n)\nif (!partial.division) {\nresults <- cells %>%\nmutate(\nn.cells = floor(n.cells)\n)\n}\nreturn(\nresult\n)\n}\n\nsimulate.growth_c <- compiler::cmpfun(simulate.growth)\n``````\n\n### Configuration of the simulation\n\nParameters of inoculation population\n\n• Number of cells inoculated: 4000000\n• Number of strains (bins of growth rate): 100\n• Average growth in culture: 0.08\n• Extreme growth in culture: 0.0808\n• Chance on extreme growth: 0.00001\n\nParameters of simulation\n\n• Total number of simulated rounds: 12000\n• Time length of each round: 0.0833333 hour\n• Turbidostat threshold: 28000000\n\nResult statistics\n\n• Total time: 1000 (hours)\n• Mean number of cells in culture: 26502994.2094651\n• Mean number of cells per strain: 500690.9651014\n\nDevelopment of cell count of the 5 largest strains (of each time point), without the total cell count.", null, "Number of strains at a given time point", null, "Range of growth rates at a given time point. Area shows max and min growth rates, mean is indicated by black line.", null, "Number of strains being diluted out at each time point.", null, "## Calculating culture growth rate\n\n### Code to calculate growth rates\n\nI will use the dxdt method to calculate growth rates, as it is faster than fitting linear models to each dilution cycle and the data should not contain any noise. which uses the following formula:\n\n`growth.rate = dlog(x)/dt = (log(xmax) - log(xmin)) / (tmax - tmin)`\n\n``````dxdt_od_min <- function(od, time, mv.sides = 3){\nod <- log(od)\n# set to NA\nod[which(is.nan(od))] = NA\nod[which(is.infinite(od))] = NA\n# remove NA values\ntime <- time[!is.na(od)]\nod <- od[!is.na(od)]\n# get minimum and max: start from minimum instead of first\nmin.idx <- ifelse(length(od) > 0, which.min(od), 0)\nmax.idx <- length(od)\n# prepare output\noutput <- data.frame(\n\"time_h\" = time[max.idx],\n\"fit_size\" = max.idx - min.idx,\n\"pump_size\" = max.idx,\n\"r_sq\" = NA,\n\"slope\" = NA,\n\"slope_se\" = NA,\n\"slope_p_value\" = NA,\n\"od_sd\" = sd(od, na.rm=TRUE)\n)\nslopes <- c()\nmin.size <- 3\nleft.step <- 0\nwhile( (max.idx - min.idx) > min.size + left.step & left.step < mv.sides) {\nright.step <- 0\nwhile ((max.idx - min.idx) > min.size + left.step + right.step & right.step < mv.sides ) {\n## add slope to slopes collection\nslopes <- c( slopes, c(\n(od[max.idx - right.step] - od[min.idx+left.step]) / (time[max.idx-right.step] - time[min.idx+left.step]) # first to last\n))\nright.step <- right.step + 1\n}\nleft.step <- left.step + 1\n}\nif (length(slopes) > 1)\n{\noutput\\$slope <- median(slopes, na.rm = TRUE)\noutput\\$slope_se <- sd(slopes, na.rm = TRUE)\n}\nreturn(output)\n}\n``````\n\nObserved culture growth (calculating growth rate using culture (total) cell count)", null, "", null, "# Simulate mutations\n\nTo simulate mutations I implemented a new growth function. Every time a strain grows, new offspring is created. I will assign growth rates to the offspring, by sampling from a percentile distribution.\n\nThe distribution is defined by `mutation.rate` and `mutation.benefit`. The distribution is created from the idea that there is chance (`mutation.rate`) on having a percent increase or decrease in growth rate, called `mutation.benefit`. The average (50 percent) of the new cells keeps the growth rate of it's parent.\n\nTo reduce the complexity of the simulation I will reduce the amount of strains in the culture to max. 100 strains, by creating bins of growth rates.\n\n### Configuration of the simulation\n\nParameters of inoculation population\n\n• Number of cells inoculated: 4000000\n• Number of strains (bins of growth rate): 100\n• Average growth in culture: 0.08\n• Extreme growth in culture: 0.0808\n• Chance on extreme growth: 0.00001\n\nParameters of simulation\n\n• Total number of simulated rounds: 3000\n• Time length of each round: 0.1666667 hour\n• Turbidostat threshold: 28000000\n• Mutation rate: 0.00001 (chance of getting the mutation benefit)\n• Mutation beneft: 0.01 (percentage increase in growth rate)\n\nValues for mutation rate and mutation benefit were taken from http://doi.org/10.1126/science.1142284\n\nResult statistics\n\n• Total time: 500 (hours)\n• Mean number of cells in culture: 26153301.5611463\n• Mean number of cells per strain: 269088.2903001\n\nDevelopment of cell count of the 5 largest strains (of each time point), without the total cell count.", null, "Number of strains at a given time point", null, "Range of growth rates at a given time point", null, "Number of strains that were diluted out at a each time point.", null, "## Calculating culture growth rate\n\nObserved culture growth (calculated growth rate of culture using (total) cell count). Red is start growth rate.", null, "", null, "Growth rate of largest population (absolute size) over time. Red is start growth rate.", null, "" ]

[ null, "http://jongbloets.net/images/joeri-b83306e0.jpg", null, "http://jongbloets.net/images/thesis/models/plot_cell_count-1-ccfc661f.png", null, "http://jongbloets.net/images/thesis/models/plot_strain_count-1-94dce300.png", null, "http://jongbloets.net/images/thesis/models/plot_growth_range-1-c9259400.png", null, "http://jongbloets.net/images/thesis/models/plot_diluted_strains-1-2902d1ba.png", null, "http://jongbloets.net/images/thesis/models/plot_culture_growth-1-722d9658.png", null, "http://jongbloets.net/images/thesis/models/plot_culture_rel_growth-1-2a73a72b.png", null, "http://jongbloets.net/images/thesis/models/plot_mutated_cell_count-1-4aa156e5.png", null, "http://jongbloets.net/images/thesis/models/plot_mutated_strain_count-1-464f35b3.png", null, "http://jongbloets.net/images/thesis/models/plot_mutated_growth_range-1-9e0266dd.png", null, "http://jongbloets.net/images/thesis/models/plot_mutated_diluted_strains-1-2235411d.png", null, "http://jongbloets.net/images/thesis/models/plot_mutated_culture_growth-1-081a817a.png", null, "http://jongbloets.net/images/thesis/models/plot_mutated_culture_rel_growth-1-a2793eb3.png", null, "http://jongbloets.net/images/thesis/models/plot_mutated_majority_growth-1-39a923b1.png", null ]

[ "# #554b2f Color Information\n\nIn a RGB color space, hex #554b2f is composed of 33.3% red, 29.4% green and 18.4% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 11.8% magenta, 44.7% yellow and 66.7% black. It has a hue angle of 44.2 degrees, a saturation of 28.8% and a lightness of 25.9%. #554b2f color hex could be obtained by blending #aa965e with #000000. Closest websafe color is: #663333.\n\n• R 33\n• G 29\n• B 18\nRGB color chart\n• C 0\n• M 12\n• Y 45\n• K 67\nCMYK color chart\n\n#554b2f color description : Very dark desaturated orange.\n\n# #554b2f Color Conversion\n\nThe hexadecimal color #554b2f has RGB values of R:85, G:75, B:47 and CMYK values of C:0, M:0.12, Y:0.45, K:0.67. Its decimal value is 5589807.\n\nHex triplet RGB Decimal 554b2f `#554b2f` 85, 75, 47 `rgb(85,75,47)` 33.3, 29.4, 18.4 `rgb(33.3%,29.4%,18.4%)` 0, 12, 45, 67 44.2°, 28.8, 25.9 `hsl(44.2,28.8%,25.9%)` 44.2°, 44.7, 33.3 663333 `#663333`\nCIE-LAB 32.188, -0.39, 18.21 6.775, 7.169, 3.716 0.384, 0.406, 7.169 32.188, 18.214, 91.228 32.188, 7.61, 19.225 26.775, -1.686, 10.514 01010101, 01001011, 00101111\n\n# Color Schemes with #554b2f\n\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #2f3955\n``#2f3955` `rgb(47,57,85)``\nComplementary Color\n• #55382f\n``#55382f` `rgb(85,56,47)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #4c552f\n``#4c552f` `rgb(76,85,47)``\nAnalogous Color\n• #382f55\n``#382f55` `rgb(56,47,85)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #2f4c55\n``#2f4c55` `rgb(47,76,85)``\nSplit Complementary Color\n• #4b2f55\n``#4b2f55` `rgb(75,47,85)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #2f554b\n``#2f554b` `rgb(47,85,75)``\nTriadic Color\n• #552f39\n``#552f39` `rgb(85,47,57)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #2f554b\n``#2f554b` `rgb(47,85,75)``\n• #2f3955\n``#2f3955` `rgb(47,57,85)``\nTetradic Color\n• #242014\n``#242014` `rgb(36,32,20)``\n• #342e1d\n``#342e1d` `rgb(52,46,29)``\n• #453d26\n``#453d26` `rgb(69,61,38)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #655938\n``#655938` `rgb(101,89,56)``\n• #766841\n``#766841` `rgb(118,104,65)``\n• #86764a\n``#86764a` `rgb(134,118,74)``\nMonochromatic Color\n\n# Alternatives to #554b2f\n\nBelow, you can see some colors close to #554b2f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #55422f\n``#55422f` `rgb(85,66,47)``\n• #55452f\n``#55452f` `rgb(85,69,47)``\n• #55482f\n``#55482f` `rgb(85,72,47)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #554e2f\n``#554e2f` `rgb(85,78,47)``\n• #55512f\n``#55512f` `rgb(85,81,47)``\n• #55552f\n``#55552f` `rgb(85,85,47)``\nSimilar Colors\n\n# #554b2f Preview\n\nText with hexadecimal color #554b2f\n\nThis text has a font color of #554b2f.\n\n``<span style=\"color:#554b2f;\">Text here</span>``\n#554b2f background color\n\nThis paragraph has a background color of #554b2f.\n\n``<p style=\"background-color:#554b2f;\">Content here</p>``\n#554b2f border color\n\nThis element has a border color of #554b2f.\n\n``<div style=\"border:1px solid #554b2f;\">Content here</div>``\nCSS codes\n``.text {color:#554b2f;}``\n``.background {background-color:#554b2f;}``\n``.border {border:1px solid #554b2f;}``\n\n# Shades and Tints of #554b2f\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, #090805 is the darkest color, while #fdfdfc is the lightest one.\n\n• #090805\n``#090805` `rgb(9,8,5)``\n• #16130c\n``#16130c` `rgb(22,19,12)``\n• #221e13\n``#221e13` `rgb(34,30,19)``\n• #2f2a1a\n``#2f2a1a` `rgb(47,42,26)``\n• #3c3521\n``#3c3521` `rgb(60,53,33)``\n• #484028\n``#484028` `rgb(72,64,40)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #625636\n``#625636` `rgb(98,86,54)``\n• #6e613d\n``#6e613d` `rgb(110,97,61)``\n• #7b6c44\n``#7b6c44` `rgb(123,108,68)``\n• #88784b\n``#88784b` `rgb(136,120,75)``\n• #948352\n``#948352` `rgb(148,131,82)``\n• #a18e59\n``#a18e59` `rgb(161,142,89)``\nShade Color Variation\n• #a99764\n``#a99764` `rgb(169,151,100)``\n• #b0a071\n``#b0a071` `rgb(176,160,113)``\n• #b7a87d\n``#b7a87d` `rgb(183,168,125)``\n• #beb08a\n``#beb08a` `rgb(190,176,138)``\n• #c5b997\n``#c5b997` `rgb(197,185,151)``\n• #ccc1a3\n``#ccc1a3` `rgb(204,193,163)``\n• #d3cab0\n``#d3cab0` `rgb(211,202,176)``\n• #dad2bc\n``#dad2bc` `rgb(218,210,188)``\n• #e1dbc9\n``#e1dbc9` `rgb(225,219,201)``\n• #e8e3d6\n``#e8e3d6` `rgb(232,227,214)``\n• #efece2\n``#efece2` `rgb(239,236,226)``\n• #f6f4ef\n``#f6f4ef` `rgb(246,244,239)``\n• #fdfdfc\n``#fdfdfc` `rgb(253,253,252)``\nTint Color Variation\n\n# Tones of #554b2f\n\nA tone is produced by adding gray to any pure hue. In this case, #46443e is the less saturated color, while #836101 is the most saturated one.\n\n• #46443e\n``#46443e` `rgb(70,68,62)``\n• #4b4639\n``#4b4639` `rgb(75,70,57)``\n• #504934\n``#504934` `rgb(80,73,52)``\n• #554b2f\n``#554b2f` `rgb(85,75,47)``\n• #5a4d2a\n``#5a4d2a` `rgb(90,77,42)``\n• #5f5025\n``#5f5025` `rgb(95,80,37)``\n• #645220\n``#645220` `rgb(100,82,32)``\n• #69551b\n``#69551b` `rgb(105,85,27)``\n• #6e5716\n``#6e5716` `rgb(110,87,22)``\n• #735911\n``#735911` `rgb(115,89,17)``\n• #795c0b\n``#795c0b` `rgb(121,92,11)``\n• #7e5e06\n``#7e5e06` `rgb(126,94,6)``\n• #836101\n``#836101` `rgb(131,97,1)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #554b2f 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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[ "", null, "Title Author Keyword ::: Volume ::: Vol. 69Vol. 68Vol. 67Vol. 66Vol. 65Vol. 64Vol. 63Vol. 62Vol. 61Vol. 60Vol. 59Vol. 58Vol. 57Vol. 56Vol. 55Vol. 54Vol. 53Vol. 52Vol. 51Vol. 50Vol. 49Vol. 48Vol. 47Vol. 46Vol. 45Vol. 44Vol. 43Vol. 42Vol. 41Vol. 40Vol. 39Vol. 38Vol. 37Vol. 36Vol. 35Vol. 34Vol. 33Vol. 32Vol. 31Vol. 30Vol. 29Vol. 28Vol. 27Vol. 26Vol. 25Vol. 24Vol. 23Vol. 22Vol. 21Vol. 20Vol. 19Vol. 18Vol. 17Vol. 16Vol. 15Vol. 14Vol. 13Vol. 12Vol. 11Vol. 10Vol. 9Vol. 8Vol. 7Vol. 6Vol. 5Vol. 4Vol. 3Vol. 2Vol. 1 ::: Issue ::: No. 12No. 11No. 10No. 9No. 8No. 7No. 6No. 5No. 4No. 3No. 2No. 1", null, "https://doi.org/10.3938/NPSM.69.831\nAnalysis of the Curriculum and the Professors’ perceptions of the Introductory Physics Laboratory in Korea\n\nHeekyong KIM1, Bongwoo LEE*2\n\n1Division of Science Education, Kangwon National University, Chuncheon 24341, Korea\n2Department of Science Education, Dankook University, Gyeonggi 16890, Korea\nCorrespondence to: [email protected]\nReceived May 29, 2019; Revised July 12, 2019; Accepted July 15, 2019.", null, "This is an open-access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.\nAbstract\nIn this study, we analyzed the introductory physics laboratory curricula of 14 universities in Korea and examined the professor's perceptions. The main results are as follows. First, professors perceive the goals of an introductory physics laboratory to be an understanding of the physics concept and the development of basic experiment skills and research ability. Second, the introductory physics laboratory curriculum consists of 7.2\\% basic education and 69.2\\% experiments. Third, professors play a role in establishing the lecture plan, explaining the theory of experiments and giving grades. Assistants play a role in the management of experimental equipment and the provision of guidance for the actual experiments. Fourth, professors’ difficulties are the aging of experiment equipment, the students' lack of understanding of physics and interest in the experiment. Fifth, the lecture on the basic experiment and report writing needs to be extended to include student-led experiments.\nPACS numbers: 01.50.Qb, 01.30.lb\nKeywords: Introductory physics laboratory, Physics experiment, Curriculum, University physics education, Professors’ perception", null, "", null, "August 2019, 69 (8)", null, "", null, "Full Text(PDF) Free", null, "Export Citation for this Article\n\nSocial Network Service\nServices" ]

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[ "# How to communicate with two encoders at same time using SPI Interface?\n\nI am developing a mobile robot using Arduino Mega and I am going to use two AS5147P magnetic encoders to get wheel positions to develop a PID controller for robot.\n\nI need to know is it possible to connect these two encoders to an Arduino mega via SPI Interface to get the wheel position at the same time?\n\nI need to know is it possible to connect these two encoders to an Arduino mega via SPI Interface to get the wheel position at the same time?\n\nThe answer is yes, although to be exact \"at the same time\" is never possible in a single core von Neumann architecture CPU.\n\nIUE:\nI need to know is it possible to connect these two encoders to an Arduino mega via SPI Interface\n\nAccording to page 4 of the datasheet, there is (as expected) a normal SPI active low chip selector, CSn.\n\nSo as long as you have physical access to that pin and a vacant pin on the Arduino then the answer should be yes (for some definition of \"at the same time\"", null, ", see spin-off.)\n\nI wonder if there's a library or if you have to roll your own code to use it? Have you got one working?\n\nInteresting discussion here about using the right kind of magnet (diametric), or standing the wrong kind (axial) on its side.\n\nWho knew?\n\nNow I am trying to communicate with two encoders via SPI interface according to the example program given in the discussion https://forum.arduino.cc/index.php?topic=355345.0\n\nhere is the program I modified\n\n#include <SPI.h>\n\n//Set Slave Select Pin\n//MOSI, MISO, CLK are handeled automatically\nint CSN = 53;\nint SO = 50;\nint SI = 51;\nint CLK = 52 ;\nint CSN2 = 49;\nunsigned int angle;\n\nvoid setup() {\n\nSerial.begin(9600);\n\n//Set Pin Modes\npinMode(CSN, OUTPUT);\npinMode(SI, OUTPUT);\npinMode(SO, INPUT);\npinMode(CLK, OUTPUT);\npinMode(CSN2,OUTPUT);\n//Set Slave Select High to Start i.e disable chip\ndigitalWrite(CSN, HIGH);\ndigitalWrite(CSN2,HIGH);\n\n//Initialize SPI\nSPI.begin();\n}\n\nvoid loop() {\n\nSPI.beginTransaction(SPISettings(10000000, MSBFIRST, SPI_MODE1));\n\n//Send the Command Frame\ndigitalWrite(CSN2, HIGH);\n\ndigitalWrite(CSN, LOW);\ndelayMicroseconds(1);\nSPI.transfer16(0xFFFF);\ndigitalWrite(CSN,HIGH);\n\ndigitalWrite(CSN, LOW);\ndelayMicroseconds(1);\nangle = SPI.transfer16(0xC000);\ndigitalWrite(CSN, HIGH);\nSPI.endTransaction;\n\nangle = (angle & (0x3FFF));\n\ndouble pos = ( (unsigned long) angle)*360UL/16384UL;\n\nSerial.println(pos);\n\ndelay(100);\n\n}\n\nHere, I was just trying to get a reading from one encoder while other one is disable. I connected SI, CLK, & SO commonly and used different CSN pins (49 & 53).\n\nBut the problem is I only get the value 0 from the both encoders… I checked the connectivity of the system and there is no issue with the connectivity.\n\nCan anyone help me to solve this matter…" ]

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[ "# Consider the following data for a drain: L = 50 m, a = 10 m, b = 10.3 m, and k = 1 × 10-5 m/s If the drains carry 1% of average annual rainfall in 24 hrs, the average annual rainfall for which this system has been designed will be\n\n1. 78 cm\n2. 84 cm\n3. 90 cm\n4. 96 cm\n\nOption 2 : 84 cm\nFree\nCT 3: Building Materials\n2565\n10 Questions 20 Marks 12 Mins\n\n## Detailed Solution\n\nConcept:\n\nThe spacing between the tile drains is given by –\n\n$$s = \\frac{{4k}}{q}\\left( {{b^2} - {a^2}} \\right)$$\n\nHere, k = permeability (m/s), b = Maximum height of drained water table above the impervious layer (m),\n\na = the depth of impervious stratum from the center of the drains & q = discharge per unit length of the drain (m3/sec)\n\nCalculation\n\nGiven,\n\nS = 50m, b = 10.3m, a = 10m and k = 1 × 10-5m/s\n\n$$s = \\frac{{4k}}{q}\\left( {{b^2} - {a^2}} \\right)$$\n\n$$50 = \\frac{{4 \\times {{10}^{ - 5}}\\;\\left( {{{10.3}^2} - {{10}^2}} \\right)}}{q}\\;$$\n\n$$q = 4.872 \\times {10^{ - 6}}\\;{m^2}/s$$\n\nLet, p be the average annual rainfall occurring over an area having unit length\n\n∴ Discharge per unit length through drains\n\n$$q = \\frac{{0.01\\; \\times \\; p\\; \\times \\; s}}{{24\\; \\times \\; 60\\; \\times \\; 60}}\\;{m^2}/s$$       (only 1% of rainfall flows through drains)\n\nComparing (1) and (2)\n\n$$\\frac{{0.01\\; \\times \\; p\\; \\times \\; 50}}{{24\\; \\times \\; 60\\; \\times \\; 60}} = \\;4.872 \\times {10^{ - 6}}$$\n\n$$p = \\;0.8418m \\approx 84.18cm$$" ]

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[ "# Minimum number of bracket reversals needed to make an expression balanced in C++\n\n## Minimum number of swaps for bracket balancing\n\nToday we will be learning How to find the Minimum number of swaps for bracket balancing.\n\nLets understand this with the help of example :-\n\n• Input:- “}{{}}{{{“\n• Output:- 2\n\nFor example, if input string is  “}{{}}{{{” then you need to reverse the bracket at 0th position to “{” and thr bracket at 5th position to “}” and bracket at 7th position to “}” that will result in {{{}}}{} now this is balanced.", null, "## Algorithm\n\n• First we will remove all the balanced part of the input string for example }{{}}{ after removing the balanced part we will be left with }{\n• Notice it carefully as we can see after removing all the balanced we will always  left with expression of “}}{{“  this type. That is an expression that contains 0 or more closing brackets or 0 or more opening brackets.\n• Now we will count the number of closing brackets let m be total number of closing brackets and n be opening brackets\n• Store the length of the unbalanced part in red_len that is total value  m+n.\n• Now return the value of  return ceil(m/2) + ceil(n/2) which is actually equal to (m+n)/2 + n%2 when m+n is even.", null, "## C++ Code (Method 1) :-\n\n` #include<bits/stdc++.h> using namespace std;int countMinReversals(string expr){ int length = expr.length(); // length of the expression must be even to make it balanced by using reversals. if (length%2) return -1; //After this loop, stack will have only the unbalanced part of expression stack s; for(int i=0; i<length; i++) { if(expr[i]=='}' && !s.empty()) { if(s.top()=='{') s.pop(); else s.push(expr[i]); } else s.push(expr[i]); } // Length of the reduced expression red_len = (m+n) int red_len = s.size(); //count the number of opening brackets at the end of the stack int n = 0; while (!s.empty() && s.top() == '{') { s.pop(); n++; } // return ceil(m/2) + ceil(n/2) which is // actually equal to (m+n)/2 + n%2 when m+n is even. return(red_len/2 + n%2);} // Driver program to test above functionint main(){ string expression; cout<<\"Enter the curly brackets : \"; cin>>expression; cout<<countMinReversals(expression); return 0;}`\n\n## Output:-\n\n`Enter the curly brackets : }{2`\n`Enter the curly brackets : }}}}2`\n\n## Method 2:-\n\n• There is another efficient approach the idea is to maintain the count of opening and closing brackets\n• If the  length of input string is odd then return -1 that means it can never be balanced\n•  Run a loop from i equal 0 to size of the expression and count the left brackets\n• Else check if the count of the left bracket is 0 if its 0 then increment count of right brackets else decrement count of left brackets.\n• Then store the ceil value of (left_brace / 2.0+ right_brace / 2.0  ) and return the value of ans.\n``` #include <bits/stdc++.h>\nusing namespace std;\n\nint countMinReversals(string expression)\n{\nint len = expression.length();\n\n//Expressions of odd size can never be balanced\nif(len % 2 != 0) {\nreturn -1;\n}\n\nint left_bracket = 0, right_bracket = 0;\nint ans;\n\nfor(int i = 0; i < len; i++) {\n//If we find a left bracket then we simply increment the left bracket\n\nif(expression[i] == '{') {\nleft_bracket++;\n}\n\nelse {\nif (left_bracket == 0) {\nright_bracket++;\n}\nelse {\nleft_bracket--;\n}\n}\n}\nans = ceil(left_bracket / 2.0) + ceil(right_bracket / 2.0);\nreturn ans;\n}\n\n// Driver program to test above function\nint main()\n{\nstring expr;\ncout<<\"Enter the Curly brackets : \"; cin>>expr;\n\ncout<<\"Minimum number of bracket reversals needed to make an expression balanced : \";\ncout << countMinReversals(expr);\nreturn 0;\n}```\n\n## Output :-\n\n```Enter the Curly brackets : }{{{\nMinimum number of bracket reversals needed to make an expression balanced : 3```" ]

[ null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null ]

[ "# weight\n\n1. ### Trying to track weight loss over 8 cells in the same row with the last cell showing the weight gain or loss\n\ni have a formula that works for 3 weight entries {=IF(\\$AP5=\"\",\\$AG5-\\$Y5,\\$AP5-\\$Y5)} Y= first weight entry, AG5= second weight entry and AP5= third weight entry. I am not sure how to expand it to figure weight gain or loss for seven entries. All help is always appreciated.:beerchug:\n2. ### Averaging with weighted numbers\n\nHello, I have a set of number I am trying to average, but am having a problem. In cells c6:c10 I have numbers (130 in this case) in cell c4 I also have 130, in c5 I have 4. c5 represents that the 130 from c4 is an average of 4 numbers. What I am trying to get is the average of all the...\n3. ### Weighted Average (I think!)\n\nHelp! I am trying to compile a return on investment (ROI) workbook for my business, which contains four subsequent departments. When I calculate my ROI for the whole organization I get one number, then when I break it down by each of the four departments and then compile those four ROI numbers...\n4. ### How to create a formula for weighing and summing an undefined number of rows?\n\nHello, I am looking for help to crack this one: I want to calculate the total discount to customers based on discounts on individual products. However the number of products vary from each customer. How can I do this fast and automatic in excel? (And I am stuck with a column in between the two...\n5. ### Using Excel to determine which row to choose\n\nHello Excel experts out there, I have multiple rows of data with same batch number & different end dates. I want to be able to use Excel assign a weight to the row base on the latest end date. After that, I want to pivot the data base on the weight assigned to it. Pls see below for example...\n6. ### Only be able to write in a cell if another has content in it... PLEASE HELP\n\nHey, I'm creating a BMI calculator in MS Excel 2010. I worked out the formula for calculating the BMI... but my problem now is: There is a cell in which the user can write a target BMI (what they want to achieve). What I want is that the user should only be allowed to write in that cell if...\n7. ### Imperial Weights in charts\n\nThis has me confused! I have a sheet with a column of dates and a column of imperial weights that look like \"10st 6lb\". I need to chart the weights but cant seem to do it. I converted the cell into something I thought Excel would understand by using the following..." ]

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[ "# How do I split a string into parts in Java? [closed]\n\nWhen a person types in a fraction like \"1_1/4\" in my application, I want to split into the three parts, whole number, numerator, and denominator.\n\nHow can this be done?\n\n• Eclipse is an IDE. What is the programming language you are using? – Saeid Nov 28 '15 at 21:36\n\nUse char data type to hold the signs which are / and + .\n\n`char sign = {/+};`\n\nand then store your input in some character.\n\n`char val = {1_1/4};`\n\nAnd the main logic for it will be to start a loop and there would be another loop inside it and then use the if else condition accordingly and print out the respective answers like if `/` arrives after a number then that number is numerator. It can be done like this.\n\n``````if(val[j] == sign[i]){\nthen val[i-1] is numerator and val[i+1] is denominator\n}\n``````\n\nAlso to get the whole number you can detect the _ statement and if it detects _ at the 2nd position then it means that your whole number lies at the first position. Hope it helps.\n\nSplit the input string on underscore and slash then retrieve the desired components from the array of tokens.\n\n``````String input = \"1_1/4\";\nString[] tokens = input.split(\"_|/\");\nString number = tokens;\nString numerator = tokens;\nString denominator = tokens;\n``````" ]

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[ "Tuesday , 31 January 2023\n\n# Crochet Home Slippers Free Tutorial\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nHi Guys. we have Free Video Tutorial in which you can learn everything clearly and step by step that you need to know to crochet home slippers in any colors. Make your and your family’s life more enjoyable and after tough work they will rest with this slippers.\n\nYou can also find on our blog other great free video tutorials and pattern for awesome blankets, slippers and for anything that will help you change your life to better and save money.\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nCrochet Home Slippers\n\nThanks to author again and we hope she will make for us other beautiful items with free video tutorials. Good luck and we hope you will enjoy.\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nAs always, Before you go I want to ask you if you value this work and you liked the knitting course, share it with all your friends on social networks and share it in their crochet groups so that all the little spiders in the world will benefit from it. crochet art !\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nWatch Video Tutorial Here" ]

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[ "### Secundária\n\n• An Introduction to Proof Theory.: S. Buss 1998 Handbook of Proof Theory, Elsevier, 1-78.\n• Mathematical Logic I: Propositional Calculus, Boolean Algebras, Predicate Calculus, Completeness Thorems: R. Cori e D. Lascar 2000 Oxford University Press\n• Mathematical Logic II: Recursion Theory, Gödel's Theorems, Set Theory, Model Theory: R. Cori e D. Lascar 2001 Oxford University Press\n• Computability: Computable Functions, Logic, and the Foundations of Mathematics: R. Epstein and W. Carnielli 2001 Wadsworth\n• Introduction to Mathematical Logic: E. Mendelson 1997 Chapman & Hall, (Fourth Edition)." ]

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[ "# Payback Period Calculator\n\nThe Payback Period is the time that it takes for a Capital Budgeting project to recover its initial cost. Usually, the project with the quickest payback is preferred. In this calculation, the Net cash flows (NCF) of the project must first be estimated. Payback period can be calculated by dividing the total investment cost by the annual net cash flow. Here is the simple online calculator to calculate the payback period by giving the initial investment amount and the annual cash flow.\n\n## Capital Budgeting Payback Period Calculation\n\nYears\n\nThe Payback Period is the time that it takes for a Capital Budgeting project to recover its initial cost. Usually, the project with the quickest payback is preferred. In this calculation, the Net cash flows (NCF) of the project must first be estimated. Payback period can be calculated by dividing the total investment cost by the annual net cash flow. Here is the simple online calculator to calculate the payback period by giving the initial investment amount and the annual cash flow.\n\nCode to add this calci to your website", null, "", null, "#### Formula:\n\nPayback Period = Initial Investment / Average Annual Cash Flows\n\n### Example:\n\nA person invested an initial amount of 50000 INR and the average annual cash flow for 1st year is 1000, 2nd year is 10000 and 3rd year is 15000.\n\n#### Step 1 :\n\nAverage Annual Cash Flows = (1000 + 10000 + 15000) / 3\n= 26000 / 3\n= 8666.66\n\n#### Step 2 :\n\nPayback period = 50000 / 8666.66\n= 85.769" ]

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[ "Review of: E^X * E^X\n\nReviewed by:\nRating:\n5\nOn 16.01.2021\n\n### Summary:\n\nDas Addon hat einige Kategorien, nmlich David. Ort: USA, der sollte einfach nicht fehlen. Halloweentown 4 - Das Hexencollege jetzt legal online anschauen.", null, "Ex-de-Lösungen von Pepperl+Fuchs bestehen aus der maßgeschneiderten Kombination eines druckfest gekapselten Gehäuses (oben) mit. deutschsprachige Homepage der Gönnheimer Elektronic - Die Gönnheimer Elektronic GmbH entwickelt und produziert elektronische Geräte, Baugruppen und. Reihe, PTB Braunschweig EN , PTB Braunschweig EN , IBExU Freiberg EN / EN , PTB Braunschweig EN Netzbetrieb.\n\n## Exponentialfunktion\n\ne^x * e^{2x} = e^{3x} Dagegen e^{-x} * e^{2x} = e^x. Hast du richtig abgeschrieben? tung sich selbst gleich ist: D exp(x) = exp(x). wir zeigen, dass D exp(x) = exp(x) für −∞ x·e = -3e x z Substitution z = - - führt auf die Gleichung: z·e = e 3 Die Lösung ist z = lam(e)=1 und somit x = - 3·lam(e).\n\nMurilo huff -uma ex part Jorge e Mateus áudio oficial 2021\n\nIn der Mathematik bezeichnet man als Exponentialfunktion eine Funktion der Form x ↦ a x die Funktionalgleichung exp ⁡ (x + y) = exp ⁡ (x) ⋅ exp ⁡ (y) {\\displaystyle \\exp(x+y)=\\exp(x)\\cdot \\exp(y)} \\exp(x+y)=\\exp(x) \\cdot \\ erfüllt, kann​. x - x 3 1 - - 3·e = - -x 3 e => x·e = -3e x z Substitution z = - - führt auf die Gleichung: z·e = e 3 Die Lösung ist z = lam(e)=1 und somit x = - 3·lam(e). e^x * e^{2x} = e^{3x} Dagegen e^{-x} * e^{2x} = e^x. Hast du richtig abgeschrieben? tung sich selbst gleich ist: D exp(x) = exp(x). wir zeigen, dass D exp(x) = exp(x) für −∞", null, "### Assistir Filmes online E^X * E^X Www. - 7 Antworten\n\nKunden stehen auf diese Weise die Vorteile beider Zündschutzarten zur Verfügung: Die Ex-e-Gehäuse gestatten eine einfache Erweiterung und Modifizierung der Böhmische Liwanzen enthaltenen Bedienelemente.", null, "If u is a function of x , we can obtain the derivative of an expression in the form e u :. If we have an exponential function with some base b , we have the following derivative:.\n\nSee the chapter on Exponential and Logarithmic Functions if you need a refresher on exponential functions before starting this section.\n\nNow, using the logarithm laws , we have:. We tidy this up by multiplying top and bottom by x. Also, it's best to write the 2 e 2 x in front of the logarithm expression to reduce confusion.\n\nIt's not part of the log expression. We'll see more of these curves in Second Order Differential Equations , in the later calculus section.\n\nDerivative of square root of sine x by first principles. Derivative graphs interactive. Differentiating tanh by Haida [Solved! For a discrete random variable X, the variance of X is written as Var X.\n\nNote that the variance does not behave in the same way as expectation when we multiply and add constants to random variables.\n\nThe basic properties below and their names in bold replicate or follow immediately from those of Lebesgue integral. Note that the letters \"a.\n\nWe have. Changing summation order, from row-by-row to column-by-column, gives us. The expectation of a random variable plays an important role in a variety of contexts.\n\nFor example, in decision theory , an agent making an optimal choice in the context of incomplete information is often assumed to maximize the expected value of their utility function.\n\nFor a different example, in statistics , where one seeks estimates for unknown parameters based on available data, the estimate itself is a random variable.\n\nIn such settings, a desirable criterion for a \"good\" estimator is that it is unbiased ; that is, the expected value of the estimate is equal to the true value of the underlying parameter.\n\nIt is possible to construct an expected value equal to the probability of an event, by taking the expectation of an indicator function that is one if the event has occurred and zero otherwise.\n\nThis relationship can be used to translate properties of expected values into properties of probabilities, e. The moments of some random variables can be used to specify their distributions, via their moment generating functions.\n\nTo empirically estimate the expected value of a random variable, one repeatedly measures observations of the variable and computes the arithmetic mean of the results.\n\nIf the expected value exists, this procedure estimates the true expected value in an unbiased manner and has the property of minimizing the sum of the squares of the residuals the sum of the squared differences between the observations and the estimate.\n\nThe law of large numbers demonstrates under fairly mild conditions that, as the size of the sample gets larger, the variance of this estimate gets smaller.\n\nThis property is often exploited in a wide variety of applications, including general problems of statistical estimation and machine learning , to estimate probabilistic quantities of interest via Monte Carlo methods , since most quantities of interest can be written in terms of expectation, e.\n\nIn classical mechanics , the center of mass is an analogous concept to expectation. For example, suppose X is a discrete random variable with values x i and corresponding probabilities p i.\n\nNow consider a weightless rod on which are placed weights, at locations x i along the rod and having masses p i whose sum is one.\n\nThe point at which the rod balances is E[ X ]. Expected values can also be used to compute the variance , by means of the computational formula for the variance.\n\nA very important application of the expectation value is in the field of quantum mechanics. Thus, one cannot interchange limits and expectation, without additional conditions on the random variables.\n\nA number of convergence results specify exact conditions which allow one to interchange limits and expectations, as specified below.\n\nThere are a number of inequalities involving the expected values of functions of random variables. The following list includes some of the more basic ones.", null, "They say there is Jamaica Inn Serie number for which this is true, and we call that number e. The derivative of e to the x is equal to e to the x. What SAT Target Score Should You Be Aiming For? Bts Schedule 2021 it out. Whitworth in What links here Related changes Upload file Special pages Permanent link Page information Cite this page Wikidata item. Derivative of the Exponential Function 7. GCSE MATHS A-LEVEL MATHS GCSE to A-Level Pure Maths Statistics Mechanics A-Level Maths Past Papers Other Oldtimer Remise Subjects My Timetable Revision Science Revision World Revision Videos Student Jungle. From Wikipedia, the free E^X * E^X. We tidy this up by multiplying top and bottom by x. They were very pleased by the fact that they had found essentially the same solution, and this in turn made them absolutely convinced that they had solved the problem conclusively; however, they did not publish their findings. The following list includes some of the more basic ones. But these savants, although they put each other to the test by proposing to each other many questions difficult to solve, have hidden their methods. If the expected value exists, this procedure estimates the true expected value in an unbiased manner and has the property Wow Private Server minimizing the sum of the squares of the residuals the sum of the squared differences between the observations and the estimate. Name optional.", null, "Introduction to limit of exponential function (e^x-1)/x formula with proof to evaluate limit of quotient of e^x-1 by x as x approaches 0 in calculus. Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. How to differentiate e^x * ln(x) using the product ruleVideo by: Tiago Hands (teknolizce.com)Extra Instagram Resources:Mathematics Pr. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history. y=e^x. Loading y=e^x. y=e^x. Log InorSign Up. y = e x. 1. y = k. y=e^x. Loading y=e^x. y=e^x. Log InorSign Up. y = e x. 1. y = k. 4/3/ · (d(e^x))/(dx)=e^x What does this mean? It means the slope is the same as the function value (the y-value) for all points on the graph. Example: Let's take the example when x = 2. At this point, the y-value is e 2 ≈ Since the derivative of e x is e x, then the slope of the tangent line at x = 2 is also e 2 ≈ Var(X) = E[ (X – m) 2] where m is the expected value E(X) This can also be written as: Var(X) = E(X 2) – m 2. The standard deviation of X is the square root of Var(X). Note that the variance does not behave in the same way as expectation when we multiply and add constants to random variables. In fact: Var[aX + b] = a 2 Var(X). Ansichten Lesen Bearbeiten Quelltext Www.Planet-Wissen.De Versionsgeschichte. Darauf Erotikfilme Hd sich auch die Namensgebung. Mittels der jordanschen Normalform Champions League Mittwoch Tv sich eine Basis bzw. Ausdrücke mit Brüchen und Wurzeln können oft mit Hilfe der Exponentialfunktion vereinfacht werden:.", null, "" ]

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[ "", null, "# 1.) The function g is related to one", null, "Avikash Prasad 2022-06-02 Answered\n\n1.) The function g is related to one of the parent functions\ng(x) = x^2 + 6\nThe parent function f is:\nf(x)= x^2\nUse function notation to write g in terms of f.\n\nYou can still ask an expert for help\n\n• Questions are typically answered in as fast as 30 minutes\n\nSolve your problem for the price of one coffee\n\n• Math expert for every subject\n• Pay only if we can solve it", null, "karton\n\nWe are given these following function:\n\n$g\\left(x\\right)={x}^{2}+6$\n\nAnd\n\n$f\\left(x\\right)={x}^{2}$\n\nUse function notation to write g in terms of f.​\n\nWe can see that function g(x) is f(x) added by 6. So\n\n$g\\left(x\\right)=f\\left(x\\right)+6$\n\ng is terms of f is given by: $g\\left(x\\right)=f\\left(x\\right)+6$" ]

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[ "# Getting Started with AuQL¶\n\nAuQL can seem daunting, but getting started need not be difficult. This article presents a number of simple but practical scripts that illustrate AuQL syntax and also demonstrate practical solutions to common types of scripting tasks.\n\n## Basic Integer Operations¶\n\nAmong the simplest possible scripts, this example asks whether one is less than two.\n\nExample:\n\n`````` # is 1 less than 2?\nfunction main(){\nresult = Logic.LessThan(1, 2)\nreturn(value = result)\n}\n``````\n\n## Passing Values To and From Functions¶\n\nThe following script demonstrates how to pass two parameters (called arguments in some langauges) as inputs for a function. AuQL Parameters are named and typed (`left` and `right` are both specified as integers). A default value can also provided as they are here, but this is not required. The two parameters are referenced by name as inputs of the addition operation.\n\nIn addition passing paremeters in, AuQL allows functions to pass a value back to the calling context. In this case, the sum of the `Math.Add` function called in this function's first statement is returned to the user as the named value `result`. As with input parameters, AuQL functions can return an arbitrary number of parameters by separating each parameter to return with commas (e.g. `return(x = 5, y = 3, z = 4)`).\n\n``````function main(left = int(1), right = int(2)){\nreturn(result = sum)\n}\n``````\n\n## Formatting Strings¶\n\nThe following script demonstrates string manipulation, in this case for the purpose of creating user-friendly messages out of dynamic information.\n\n``````function main(org = str(\"Aunalytics\"), rate = int(89)){\n\n# Define a template string\ntemplate = \"Job completion rate for %s is %s% as of '%s'\"\n\n# Pack values into an array using Util.ArrayOf()\nargs = Util.ArrayOf(org, rate, Date.Now())\n\n# Format output string with String.Format()\noutput = String.Format(template, args)\n\nreturn(value = output)\n}\n``````\n\n## Invoking Subscripts¶\n\nThis example multiplies each element of the array `[1, 2, 3, 4, 5]` by two and returns the mapped array. Here we are using the `Util.MapArray` operator, which requires a function to be passed as an argument. We define the mapper function below the main function and \"import\" it to the main function using the `Util.Function` operator.\n\n``````function main(){\narr = [1,2,3,4,5]\nmapper = Util.Function(\"mapper\")\noutput = Util.MapArray(arr, mapper)\nreturn(value = output)\n}\n\nfunction mapper(value = int()){\nproduct = Math.Multiply(value, 2)\nreturn(value = product)\n}\n``````\n\n## Is My Dataset Growing?¶\n\nThis script fetches information from a Memento Series that has the stats of a record written to it periodically. It then asserts that the size of the record is growing by testing whether the newest filesize is greater than the second newest.\n\n``````function main(series_id = str(\"a27c46b6-6634-4f41-951a-5ca753c8793f\"), atlas_id = str())\n{\n# fetch the latest two mementos\nmementos = AU.Memento.LatestN(series_id, 2)\n\n# fetch and parse from the newest memento\nnew_memento_str = Util.Get(mementos, 0)\nnew_memento = Util.Object(new_memento_str)\nnew_record = Util.Get(new_memento, atlas_id)\nnew_filesize = Util.Get(new_record, \"file_size\")\n\n# fetch and parse from the 2nd newest memento\nold_memento_str = Util.Get(mementos, 1)\nold_memento = Util.Object(old_memento_str)\nold_record = Util.Get(old_memento, atlas_id)\nold_filesize = Util.Get(old_record, \"file_size\")\n\n# is the 1st greater than the 2nd?\ncomparator = Logic.GreaterThan(new_filesize, old_filesize)\n\nreturn(value = comparator)\n}\n``````\n\n## AI-Designed Date Checking Algorithm¶\n\nThis script is an interesting one insofar as the logic chain was developed by AI and put into a Rule by a human. The script asks whether today is a day that the algorithm determined to be a \"good\" day.\n\n``````function main(max_date = String()) {\nconvert_to_date = Util.MakeDate(date = max_date)\n\ndays_diff = Date.Difference(Date.Now(), convert_to_date, unit = String(\"days\"))\n\nmd_dow = Date.GetDay(convert_to_date)\ncd_dow = Date.GetDay(convert_to_date)\n\ngt1 = Logic.GreaterThan(days_diff, 2)\n\ngt2 = Logic.GreaterThan(cd_dow, 2)\n\ngt3 = Logic.GreaterThan(md_dow, 4)\n\ngt4 = Logic.GreaterThan(days_diff, 3)\n\nlte1 = Logic.LessThanEqual(cd_dow, 2)\n\nand1_1 = Logic.And(gt1, gt2)\n\nand2_1 = Logic.And(gt1, lte1)\n\nand2_2 = Logic.And(and2_1, gt3)\n\nand2_3 = Logic.And(and2_2, gt4)\n\nor1_1 = Logic.Or(and1_1, and2_3)\n\nreturn(value = or1_1)\n}\n``````\n\n## Format Parameters for RunWorkflowMulti¶\n\nThis script takes an input object and maps an array in it so that it can be passed to a RunWorkflowMulti Workflow component.\n\n``````function main(input = Object()){\nget = Util.Get(input, \"NewRecords\")\nparseJSON = Util.Object(get)\n\nmapper = Util.Function(str(\"embedded\"))\nmap = Util.MapArray(parseJSON, mapper, opts = input)\n\nreturn(value = map)\n}\n\nfunction embedded(value = str(), index = int(), opts = Object()){\nunset = Util.Unset(opts, \"NewRecords\")\nsetNewFile = Util.Set(unset, \"NewRecords\", value)\nreturn(value = setNewFile)\n}\n``````" ]

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[ "# The dedoublifier, part three\n\nAll right, we have an arbitrary-precision rational arithmetic type now, so we can do arithmetic on fractions with confidence. Remember the problem I set out to explore here was: a double is actually a fraction whose denominator is a large power of two, so fractions which do not have powers of two in their denominators cannot be represented with full fidelity by a double. Now that we have this machinery we can see exactly what the representation error is:\n\n```Console.WriteLine(Rational.FromDouble(2.0 / 5.0));\n```\n\nThis program fragment produces the output\n\n```3602879701896397/9007199254740992\n```\n\nClearly that is not 2/5, but how close is it?\n\n```Console.WriteLine(Rational.FromDouble(2.0 / 5.0) -\nRational.FromIntegers(2, 5));\n```\n\nThe result is astonishing:\n\n```1/45035996273704960\n```\n\nPeople often point out that doubles introduce representation error but seldom do they quantify that error. In this case the error is less than one part in 45 quadrillion. That’s a tiny error! And yet of course we can all think of situations where that error has compounded to the point where we expect \\$0.15 to be output and instead we get \\$0.1499999999. As I’ve often said, use decimals for currency computations.\n\nRounding problems aside, a lot of work has gone into making sure that doubles can be “round tripped” back to decimal strings when need be. But what if we have some more unusual fraction? Suppose we’ve done some computations in doubles and the result is the double closest to five sevenths:\n\n```Console.WriteLine(Rational.FromDouble(5.0 / 7.0));\nConsole.WriteLine(Rational.FromDouble(5.0 / 7.0) -\nRational.FromIntegers(5, 7));\n```\n\nThis produces\n\n```6433713753386423/9007199254740992\n1/63050394783186944\n```\n\nThe question then is: suppose we have that double in hand. How do we get back out the fact that this is probably intended to be five sevenths?\n\nYou’re all computer programmers. Give it some thought before you read on. To make the problem easier let us suppose without loss of generality that the double we’ve been given is between zero and one.\n\n.\n\n.\n\n.\n\nProbably a bunch of different thoughts occurred to you.\n\nWe could start by computing each fraction: 1.0/2.0, 1.0/3.0, 2.0/3.0, 1.0/4.0, … and so on. Compare each fraction to the target, keep track of the current closest fraction, and pick the fraction that is closest after a certain number of rationals go by. Unfortunately this gets inefficient as the denominators get large; there are too many fractions to check.\n\nWe could multiply the double by 2.0, 3.0, 4.0, 5.0, and so on, and see if any of the results are close to a whole number. If we multiply by 7.0 and get something close to 5.0 then it was 5 / 7. This seems more efficient, and in fact we could code up a solution that uses either of these techniques even without our rational number type; they can be done entirely in fast double arithmetic.\n\nYou probably also thought “can I use a divide and conquer strategy?” Binary search and related strategies are well known for their ability to rapidly converge on a correct solution.\n\nUnfortunately a straight-up binary search of the rationals does not solve our problem. Imagine we have `6433713753386423 / 9007199254740992` in hand and we are searching for the fraction of small denominator that it “really” is. It does us no good to say:\n\n• It’s between 0 and 1. Let’s try 1/2.\n• It is bigger than 1/2.\n• Let’s split the difference and try 3/4.\n• It is smaller than 3/4.\n• Let’s split the difference and try 5/8.\n\nBinary search in this manner does not actually hit all the fractions, and in fact only hits those that we are specifically trying to avoid: those with powers of two in the denominator!\n\nHowever, there is a way to binary-search the space of rational numbers between zero and one that does hit all of them. Next time on FAIC: we’ll use it to solve our problem.\n\n## 21 thoughts on “The dedoublifier, part three”\n\n•", null, "onodera on said:\n•", null, "CarlD on said:\n\nYep\n\n• It’s a little thing I invented called a “cliffhanger”.\n\nI’m not going to talk about the Stern-Brocot tree directly but I’m going to do the math that it uses.\n\n1.", null, "EduardoS on said:\n\nBinary search? Why not solve it with a multiplication by (trying to not spoil)?\n\n2.", null, "Barbie on said:\n\nWhat’s wrong with the original power of two fraction? It’s exactly the number passed in and it’s a rational 🙂 ? To get a better answer, you need to define what better means, which i’m sure you’ll do before actually answering. Anyways, if you’re shooting for the smallest denominator (one way to define “better”), then a whole series of denominators can be identified by looking at the repetition of digits. In a double, you have 53 of them, so this only scales so far, but I’d argue it should be good enough to find most interesting ones, for a certain definition of interesting.\n\n3. I’m going to guess it’s something similar to this?\n\npublic static Rational FromDouble2(double value)\n{\nint numerator = 0;\nint denominator = 2;\nwhile (true)\n{\ndouble currentValue = (double)numerator/denominator;\nif (value == currentValue)\n{\nreturn new Rational(numerator, denominator);\n}\nelse if (value < currentValue)\n{\ndenominator += 1;\n}\nelse\n{\nnumerator += 1;\n}\n}\n}\n\n4.", null, "Aron Parker on said:\n\nHey Eric!\n\nI’m a CS student and I’m really enjoying following your blog. I find implementing these things and understanding data types / C# / algrotihms in depth highly exciting and interesting!\n\nI’ve thought about the problem and came up with this:\n\npublic static Rational ApproximateFraction(Rational n)\n{\nBigInteger x = n.Numerator;\nBigInteger y = n.Denominator;\nBigInteger r;\n\nBigInteger epsilon = BigInteger.One;\n\nwhile ((r = BigInteger.Abs(x % y)) > epsilon)\n{\nx = y;\ny = r;\n}\n\nreturn new Rational(n.Numerator / y, n.Denominator / y);\n}\n\nIt basically uses the extended euclidean algorithm, but instead of checking for dividing without residual, it tolerates a small amount. it checks for approximate equality (abs(x) > epsilon), where epsilon is 1 in this case but could be dynamically defined (high for higher numbers, == 0 for small numbers). It returns 2/5 & 3/5 for both of your examples, one would have to test it for other fractions though, but I I’m fairly confident that by adjusting epsilon you can approximate the original fraction back.\n\n• I did, and it was sufficiently buggy that I decided to not use it. I may discuss how vexing it is to have an open source project marked as “v1.0” — indicating that it is ready to ship — and marked as coming from Microsoft — indicating that it is trustworthy — where some of the methods are so broken that plainly the code must never have even been run by its authors.\n\nThen again, maybe I did just discuss it sufficiently just there. Suffice to say: I was vexed.\n\n5.", null, "Sahuagin on said:\n\nMy first thought would be to do something like:\n\nperform the division (giving something like 0.40000000012345)\ntruncate/round (giving 0.4)\nconvert floating point number to fraction (getting 4 and 10)\nreduce (getting 2 and 5)\n\nI think this probably doesn’t quite work though. Trying it out, you don’t necessarily know how much you should round, and converting to a fraction is not necessarily so easy either. It’s easy for 2/5 but probably wouldn’t work in the general case, and would probably still have the problem you mentioned of being stuck in powers of 2. You probably are looking for a solution that sticks with integers.\n\nSomewhat related, I would be very curious to know what you think of Douglas Crockford’s suggested future data type DEC64.\n\n•", null, "onodera on said:\n\nI think DEC64 combines the worst parts of existing decimal and binary formats. Compared to IEEE754 decimals, it looks simple and elegant, but it suffers from similar problems in addition to new ones.\n\n•", null, "Sahuagin on said:\n\nthanks, I should have looked up criticisms of it before mentioning it. I like Crockford but it doesn’t sound like dec64 is very practical.\n\n•", null, "John Payson on said:\n\nI see a lot of uses for decimal fixed-point types; I see far fewer good uses for decimal floating-point types. One can sensibly define semantics such that all operations other than division will either yield exact results or trap, and after performing division and rounding the quotient as desired, code can then compute a residual. I suppose one could do likewise with decimal floating-point types, but in most cases simply using a slightly-longer fixed-point type would be more efficient.\n\n•", null, "onodera on said:\n\nEven IEE754 binary64 is a good enough type for most financial calculations as long as your library doesn’t forget to round the values after each significant calculation step to keep the values consistent and equatable. Yes, it can’t measure world’s GDP down to the last American cent, but US national debt can easily be stored in a double.\n\n6.", null, "Jota on said:\n\nIt seems like the solution to this is pretty obvious: just pre-generate a lookup table. A Dictionary should work just fine for this task. Because all 2^63 values will be populated ahead of time, it should be trivially easy to look up the correct fraction for any double value you might need.\n\nNB: For those of you who might object to the amount of storage media necessary to implement this, I have done the math, and I believe sufficient storage this lookup table already exists, provided we’re allowed to commandeer every hard drive on Earth.\n\n•", null, "John Payson on said:\n\nActually, the storage may be rather easily whittled down to 2^52 by simply storing the values for mantissas in the range 0.5 to 1. For any smaller value simply multiply by a power of two fraction, and multiply the denominator of the computed rational number likewise. Given today’s storage capacities, 2^52 wouldn’t be cheap, but wouldn’t be unobtainable." ]

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[ "Preface\n\nfor\n\nThe Keys to Linear Algebra\n\nThis book is designed as a text for a one-semester introductory undergraduate course in linear algebra. Driven by applications, each chapter begins with a realistic problem to motivate the need for learning the subsequent material. Appropriate theory and technology are then used at the end of the chapter to solve the opening problem. Related project exercises involve the student actively in technology-based problem solving---a feature students have responded to enthusiastically in class testing. Numerous other applications are drawn from physics, statistics, business, and computer science to illustrate where and how the techniques of linear algebra apply.\n\nWhat makes this book unique, however, is that, in addition to teaching the standard topics of linear algebra, a primary objective is to equip students with general problem-solving skills that are needed in their subsequent mathematics and related courses. This goal is accomplished by identifying and explaining the underlying mathematical thinking processes that arise in linear algebra. These thinking processes--- hereafter referred to collectively as the why mathematics---include unification, generalization, abstraction, identifying similarities and differences, converting visual images to symbolic form and vice versa, understanding and creating definitions and proofs, and developing and working with axiomatic systems. Teaching these ideas explicitly is designed to reduce the time and frustration involved in learning linear algebra and to provide the student with a deeper and more-lasting appreciation of the subject ... and of mathematics in general.\n\nTeaching Mathematical Thinking Processes\n\nTo teach the why mathematics, a discussion of such concepts as unification, generalization, abstraction, and so on, is presented when those ideas arise naturally in the context of linear algebra. Each such concept is named and then referred to whenever the idea appears subsequently in the book.\n\nTo understand and to use the techniques of unification, generalization, abstraction, and developing axiomatic systems, students need to acquire certain basic mathematical skills. These skills include doing proofs, identifying similarities and differences, converting visual images to symbolic form (and vice versa), and understanding and creating definitions.\n\nThe concept of a proof is fundamental to all advanced mathematics courses. The approach used here, as presented in Appendix A, is adapted from How to Read and Do Proof, second edition, by Daniel Solow and is reprinted with permission from John Wiley & Sons, Inc.\n\nThe ability to identify similarities and differences among various mathematical concepts is essential to unification in that like properties from different problems must be isolated, identified, and then brought together in a single framework. This skill is also used in creating definitions, where it is necessary to identify a common property of all objects being defined.\n\nMuch effort is also devoted to teaching students how to convert visual images to symbolic form. When students learn mathematics, they typically develop their own ways of imagining and picturing specific concepts---such as the projection of one vector onto another. However, it is one thing to visualize such concepts; it is another thing entirely to translate such an image to symbolic form, especially when quantifiers are involved. The art of doing so is illustrated and explained with many examples. Students are also shown how to check for syntax and logic errors that can arise in the translation process. The reverse technique of converting symbolic mathematics to visual form is also taught.\n\nOne area in which the student is given special help is in understanding definitions. Although a student may be able to visualize objects with a desirable property---such as linearly independent vectors---it is quite a challenge to create (or even to understand) the symbolic definition. Merely presenting the definition is inadequate because doing so fails to explain how the definition was arrived at and why the definition is correct. The approach taken here is to teach the student to identify similarities shared by all items having the property being defined, and then to translate those observations to symbolic form. They are also taught to verify that the definition includes all objects having the desirable property while excluding all other objects.\n\nPedagogical Features\n\nThe pedagogical features of this book include a realistic introductory problem at the beginning of each chapter to motivate the need for learning the techniques presented in the chapter. Those techniques are then used at the end of the chapter to solve the problem.\n\nEach mathematical thinking process is set off in the text and is easily identified by an icon. These thinking processes appear not only in the appropriate chapter summaries but also in their entirety in Appendix B.\n\nEach chapter contains many numbered and titled examples to illustrate the topic under discussion. Definitions and theorems have special design features for easy reference. Extensive use of figures provides the student with visual images of mathematical concepts.\n\nEach numbered section is followed by numerous exercises, some of which give the student practice in performing mechanical computations while others test their understanding of the why mathematics. The last numbered section in each chapter has exercises that are designed to be solved with MATLAB, Maple, Mathematica, or a graphing calculator. Solutions to exercises whose numbers are in blue are given in the back of the book. The instructor may therefore want to assign for homework those exercises whose numbers are in black." ]

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[ "# Online English Test", null, "Level A Level B Level C TOEFL Incorrect word TOEFL reading comprehension Synonym TOEFL ---Choice--- Lesson 001 Lesson 002 Lesson 003 Lesson 004 Lesson 005 Lesson 006 Lesson 007 Lesson 008 Lesson 009 Lesson 010 Lesson 011 Lesson 012 Lesson 013 Lesson 014 Lesson 015 Lesson 016 Lesson 017 Lesson 018 Lesson 019 Lesson 020 Lesson 021 Lesson 022 Lesson 023 Lesson 024 Lesson 025 Lesson 026 Lesson 027 Lesson 028 Lesson 029 Lesson 030 Lesson 031 Lesson 032 Lesson 033 Lesson 034 Lesson 035 Lesson 036 Lesson 037 Lesson 038 Lesson 039 Lesson 040 ---Choice--- Lesson 001 Lesson 002 Lesson 003 Lesson 004 Lesson 005 Lesson 006 Lesson 007 Lesson 008 Lesson 009 Lesson 010 Lesson 011 Lesson 012 Lesson 013 Lesson 014 Lesson 015 Lesson 016 Lesson 017 Lesson 018 Lesson 019 Lesson 020 Lesson 021 Lesson 022 Lesson 023 ---Choice--- Lesson 001 Lesson 002 Lesson 003 Lesson 004 Lesson 005 Lesson 006 Lesson 007 Lesson 008 Lesson 009 Lesson 010 Lesson 011 Lesson 012 Lesson 013 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Lesson 087 Lesson 088 Lesson 089 Lesson 090 Lesson 091 Lesson 092 Lesson 093 Lesson 094 Lesson 095 Lesson 096 Lesson 097 Lesson 098 Lesson 099 Lesson 100 Lesson 101 Lesson 102 Lesson 103 Lesson 104 Lesson 105 Lesson 106 Lesson 107 Lesson 108 Lesson 109 Lesson 110 Lesson 111 Lesson 112 Lesson 113 Lesson 114 Lesson 115 Lesson 116 Lesson 117 Lesson 118 Lesson 119 Lesson 120 Lesson 121 Lesson 122 Lesson 123 Lesson 124 Lesson 125 Lesson 126 Lesson 127 Lesson 128 Lesson 129 Lesson 130 Lesson 131 Lesson 132 Lesson 133 Lesson 134 Lesson 135 Lesson 136 Lesson 137 Lesson 138 Lesson 139 Lesson 140 Lesson 141 Lesson 142 Lesson 143 Lesson 144 Lesson 145 Lesson 146 Lesson 147 Lesson 148 Lesson 149 Lesson 150 Lesson 151 Lesson 152 Lesson 153 Lesson 154 Lesson 155 Lesson 156 Lesson 157 ---Choice--- Lesson 001 Lesson 002 Lesson 003 Lesson 004 Lesson 005 Lesson 006 Lesson 007 Lesson 008 Lesson 009 Lesson 010 Lesson 011 Lesson 012 Lesson 013 Lesson 014 Lesson 015 Lesson 016 Lesson 017 Lesson 018 Lesson 019 Lesson 020 Lesson 021 Lesson 022 Lesson 023 Lesson 024 Lesson 025 Lesson 026 Lesson 027 Lesson 028 Lesson 029 Lesson 030 Lesson 031 Lesson 032 Lesson 033 Lesson 034 Lesson 035 Lesson 036 Lesson 037 Lesson 038 Lesson 039 Lesson 040 Lesson 041 Lesson 042 Lesson 043 Lesson 044 Lesson 045 Lesson 046\n\n### Study Adds to Understanding of Language and the Brain\n\nThis is the VOA Special English Health Report.\n\nA new study of the brain is helping scientists better understand how humans process language. The study is being done in San Diego, Boston and New York.\n\nOne of the patients is a woman with epilepsy. Doctors are monitoring Denise Harris to see if she is a good candidate for an operation that could stop her seizures. They are monitoring her through wire electrodes implanted in her brain.\n\nBut while she is in the hospital, she is also helping scientists understand how the brain works with language. The study centers on a part of the frontal lobe called Broca's area. Pierre Paul Broca was a nineteenth century French doctor. He first recognized the big part that this small area plays in language.\n\nThe electrode implants have shown that the area very quickly processes three different language functions. Eric Halgren at the University of California, San Diego, School of Medicine is one of the main investigators. He says they found different regions doing, at different times, different processes all within a centimeter.\n\nThe first function deals with recognizing a word. The second deals with understanding the word's meaning within a sentence. And the third lets us speak the word.\n\nHarvard University brain expert Steven Pinker is another author of the study. A report on the work appeared in the journal Science. Ned Sahin, a researcher at Harvard and the University of California, San Diego, School of Medicine, was the first author of the paper. He says scientists have known for some time that traditional explanations for how parts of the brain work need to be changed.\n\nOne such belief is that there is a separation of language tasks between two very different parts of the brain. One is Broca's area at the front. The other is Wernicke's area farther back in the brain. That area is named after the nineteenth century German neurologist Carl Wernicke.\n\nThe belief is that Broca's area is responsible for speaking and that Wernicke's area is responsible for comprehending. But the new study shows that Broca's area is involved in both speaking and comprehension. Neil Sahin says this shows how parts of the brain perform more than one task.\n\nNEIL SAHIN: \"Because here's an example of one relatively small part of the brain that's doing three very different things at three different times, but all within the space of a quarter of a second.\"\n\nYet, even with new findings like these, much about the brain still remains a mystery -- at least for now.\n\nAnd that's the VOA Special English Health Report, with reporting by Mike O'Sullivan. I'm Jim Tedder.\n\n 1->25", null, "26->49", null, "50->75", null, "76->99", null, "100->125", null, "126->164\n Ôn Tập Ngữ Pháp Phần 1", null, "Ôn Tập Ngữ Pháp Phần 2" ]

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[ "# Brahmagupta\n\nBrahmagupta (born c.598 CE, died c.668 CE) was an Indian mathematician and astronomer. He is the author of three early works on mathematics and astronomy: the Brāhmasphuṭasiddhānta (BSS, \"correctly established doctrine of Brahma\", dated 628), a theoretical treatise, and the Khaṇḍakhādyaka (\"edible bite\", dated 665), a more practical text.\n\nBrahmagupt\nBornc.598 CE\nDiedc.668 CE\nKnown for\nScientific career\nFieldsMathematics, astronomy\n\nBrahmagupta was the first to give rules to compute with zero. The texts composed by Brahmagupta were in elliptic verse in Sanskrit, as was common practice in Indian mathematics. As no proofs are given, it is not known how Brahmagupta's results were derived.\n\n## Life and career\n\nBrahmagupta was born in 598 CE according to his own statement. He lived in Bhillamala (modern Bhinmal) during the reign of the Chavda dynasty ruler, Vyagrahamukha. He was the son of Jishnugupta and was a Shaivite by religion. Even though most scholars assume that Brahmagupta was born in Bhillamala, there is no conclusive evidence for it. However, he lived and worked there for a good part of his life. Prithudaka Svamin, a later commentator, called him Bhillamalacharya, the teacher from Bhillamala. Sociologist G. S. Ghurye believed that he might have been from the Multan or Abu region.\n\nBhillamala, called pi-lo-mo-lo by Xuanzang, was the apparent capital of the Gurjaradesa, the second largest kingdom of Western India, comprising southern Rajasthan and northern Gujarat in modern-day India. It was also a centre of learning for mathematics and astronomy. Brahmagupta became an astronomer of the Brahmapaksha school, one of the four major schools of Indian astronomy during this period. He studied the five traditional siddhanthas on Indian astronomy as well as the work of other astronomers including Aryabhata I, Latadeva, Pradyumna, Varahamihira, Simha, Srisena, Vijayanandin and Vishnuchandra.\n\nIn the year 628, at an age of 30, he composed the Brāhmasphuṭasiddhānta (the improved treatise of Brahma) which is believed to be a revised version of the received siddhanta of the Brahmapaksha school. Scholars state that he incorporated a great deal of originality to his revision, adding a considerable amount of new material. The book consists of 24 chapters with 1008 verses in the ārya metre. A good deal of it is astronomy, but it also contains key chapters on mathematics, including algebra, geometry, trigonometry and algorithmics, which are believed to contain new insights due to Brahmagupta himself.\n\nLater, Brahmagupta moved to Ujjain, which was also a major centre for astronomy. At the age of 67, he composed his next well known work Khanda-khādyaka, a practical manual of Indian astronomy in the karana category meant to be used by students.\n\nBrahmagupta lived beyond 665 CE. He is believed to have died in Ujjain.\n\n## Controversy\n\nBrahmagupta directed a great deal of criticism towards the work of rival astronomers, and his Brahmasphutasiddhanta displays one of the earliest schisms among Indian mathematicians. The division was primarily about the application of mathematics to the physical world, rather than about the mathematics itself. In Brahmagupta's case, the disagreements stemmed largely from the choice of astronomical parameters and theories. Critiques of rival theories appear throughout the first ten astronomical chapters and the eleventh chapter is entirely devoted to criticism of these theories, although no criticisms appear in the twelfth and eighteenth chapters.\n\n## Reception\n\nThe historian of science George Sarton called him \"one of the greatest scientists of his race and the greatest of his time.\" Brahmagupta's mathematical advances were carried on further by Bhāskara II, a lineal descendant in Ujjain, who described Brahmagupta as the ganaka-chakra-chudamani (the gem of the circle of mathematicians). Prithudaka Svamin wrote commentaries on both of his works, rendering difficult verses into simpler language and adding illustrations. Lalla and Bhattotpala in the 8th and 9th centuries wrote commentaries on the Khanda-khadyaka. Further commentaries continued to be written into the 12th century.\n\nA few decades after the death of Brahmagupta, Sindh came under the Arab Caliphate in 712 CE. Expeditions were sent into Gurjaradesa. The kingdom of Bhillamala seems to have been annihilated but Ujjain repulsed the attacks. The court of Caliph Al-Mansur (754–775) received an embassy from Sindh, including an astrologer called Kanaka, who brought (possibly memorised) astronomical texts, including those of Brahmagupta. Brahmagupta's texts were translated into Arabic by Muhammad al-Fazari, an astronomer in Al-Mansur's court under the names Sindhind and Arakhand. An immediate outcome was the spread of the decimal number system used in the texts. The mathematician Al-Khwarizmi (800–850 CE) wrote a text called al-Jam wal-tafriq bi hisal-al-Hind (Addition and Subtraction in Indian Arithmetic), which was translated into Latin in the 13th century as Algorithmi de numero indorum. Through these texts, the decimal number system and Brahmagupta's algorithms for arithmetic have spread throughout the world. Al-Khwarizmi also wrote his own version of Sindhind, drawing on Al-Fazari's version and incorporating Ptolemaic elements. Indian astronomic material circulated widely for centuries, even passing into medieval Latin texts.\n\n## Mathematics\n\n### Algebra\n\nBrahmagupta gave the solution of the general linear equation in chapter eighteen of Brahmasphutasiddhanta,\n\nThe difference between rupas, when inverted and divided by the difference of the [coeffecients] of the unknowns, is the unknown in the equation. The rupas are [subtracted on the side] below that from which the square and the unknown are to be subtracted.\n\nwhich is a solution for the equation bx + c = dx + e equivalent to x = ec/bd, where rupes refers to the constants c and e. He further gave two equivalent solutions to the general quadratic equation\n\n18.44. Diminish by the middle [number] the square-root of the rupas multiplied by four times the square and increased by the square of the middle [number]; divide the remainder by twice the square. [The result is] the middle [number].\n18.45. Whatever is the square-root of the rupas multiplied by the square [and] increased by the square of half the unknown, diminish that by half the unknown [and] divide [the remainder] by its square. [The result is] the unknown.\n\nwhich are, respectively, solutions for the equation ax2 + bx = c equivalent to,\n\n$x={\\frac {{\\sqrt {4ac+b^{2}}}-b}{2a}}$", null, "and\n\n$x={\\frac {{\\sqrt {ac+{\\tfrac {b^{2}}{4}}}}-{\\tfrac {b}{2}}}{a}}.$", null, "He went on to solve systems of simultaneous indeterminate equations stating that the desired variable must first be isolated, and then the equation must be divided by the desired variable's coefficient. In particular, he recommended using \"the pulverizer\" to solve equations with multiple unknowns.\n\n18.51. Subtract the colors different from the first color. [The remainder] divided by the first [color's coefficient] is the measure of the first. [Terms] two by two [are] considered [when reduced to] similar divisors, [and so on] repeatedly. If there are many [colors], the pulverizer [is to be used].\n\nLike the algebra of Diophantus, the algebra of Brahmagupta was syncopated. Addition was indicated by placing the numbers side by side, subtraction by placing a dot over the subtrahend, and division by placing the divisor below the dividend, similar to our notation but without the bar. Multiplication, evolution, and unknown quantities were represented by abbreviations of appropriate terms. The extent of Greek influence on this syncopation, if any, is not known and it is possible that both Greek and Indian syncopation may be derived from a common Babylonian source.\n\n### Arithmetic\n\nThe four fundamental operations (addition, subtraction, multiplication, and division) were known to many cultures before Brahmagupta. This current system is based on the Hindu Arabic number system and first appeared in Brahmasphutasiddhanta. Brahmagupta describes the multiplication as thus \"The multiplicand is repeated like a string for cattle, as often as there are integrant portions in the multiplier and is repeatedly multiplied by them and the products are added together. It is multiplication. Or the multiplicand is repeated as many times as there are component parts in the multiplier\". Indian arithmetic was known in Medieval Europe as \"Modus Indorum\" meaning method of the Indians. In Brahmasphutasiddhanta, multiplication was named Gomutrika. In the beginning of chapter twelve of his Brahmasphutasiddhanta, entitled Calculation, Brahmagupta details operations on fractions. The reader is expected to know the basic arithmetic operations as far as taking the square root, although he explains how to find the cube and cube-root of an integer and later gives rules facilitating the computation of squares and square roots. He then gives rules for dealing with five types of combinations of fractions: a/c + b/c; a/c × b/d; a/1 + b/d; a/c + b/d × a/c = a(d + b)/cd; and a/cb/d × a/c = a(db)/cd.\n\n#### Series\n\nBrahmagupta then goes on to give the sum of the squares and cubes of the first n integers.\n\n12.20. The sum of the squares is that [sum] multiplied by twice the [number of] step[s] increased by one [and] divided by three. The sum of the cubes is the square of that [sum] Piles of these with identical balls [can also be computed].\n\nHere Brahmagupta found the result in terms of the sum of the first n integers, rather than in terms of n as is the modern practice.\n\nHe gives the sum of the squares of the first n natural numbers as n(n + 1)(2n + 1)/6 and the sum of the cubes of the first n natural numbers as (n(n + 1)/2)2\n.\n\n#### Zero\n\nBrahmagupta's Brahmasphuṭasiddhanta is the first book that provides rules for arithmetic manipulations that apply to zero and to negative numbers. The Brahmasphutasiddhanta is the earliest known text to treat zero as a number in its own right, rather than as simply a placeholder digit in representing another number as was done by the Babylonians or as a symbol for a lack of quantity as was done by Ptolemy and the Romans. In chapter eighteen of his Brahmasphutasiddhanta, Brahmagupta describes operations on negative numbers. He first describes addition and subtraction,\n\n18.30. [The sum] of two positives is positives, of two negatives negative; of a positive and a negative [the sum] is their difference; if they are equal it is zero. The sum of a negative and zero is negative, [that] of a positive and zero positive, [and that] of two zeros zero.\n\n[...]\n\n18.32. A negative minus zero is negative, a positive [minus zero] positive; zero [minus zero] is zero. When a positive is to be subtracted from a negative or a negative from a positive, then it is to be added.\n\nHe goes on to describe multiplication,\n\n18.33. The product of a negative and a positive is negative, of two negatives positive, and of positives positive; the product of zero and a negative, of zero and a positive, or of two zeros is zero.\n\nBut his description of division by zero differs from our modern understanding:\n\n18.34. A positive divided by a positive or a negative divided by a negative is positive; a zero divided by a zero is zero; a positive divided by a negative is negative; a negative divided by a positive is [also] negative.\n18.35. A negative or a positive divided by zero has that [zero] as its divisor, or zero divided by a negative or a positive [has that negative or positive as its divisor]. The square of a negative or of a positive is positive; [the square] of zero is zero. That of which [the square] is the square is [its] square-root.\n\nHere Brahmagupta states that 0/0 = 0 and as for the question of a/0 where a ≠ 0 he did not commit himself. His rules for arithmetic on negative numbers and zero are quite close to the modern understanding, except that in modern mathematics division by zero is left undefined.\n\n### Diophantine analysis\n\n#### Pythagorean triplets\n\nIn chapter twelve of his Brahmasphutasiddhanta, Brahmagupta provides a formula useful for generating Pythagorean triples:\n\n12.39. The height of a mountain multiplied by a given multiplier is the distance to a city; it is not erased. When it is divided by the multiplier increased by two it is the leap of one of the two who make the same journey.\n\nOr, in other words, if d = mx/x + 2, then a traveller who \"leaps\" vertically upwards a distance d from the top of a mountain of height m, and then travels in a straight line to a city at a horizontal distance mx from the base of the mountain, travels the same distance as one who descends vertically down the mountain and then travels along the horizontal to the city. Stated geometrically, this says that if a right-angled triangle has a base of length a = mx and altitude of length b = m + d, then the length, c, of its hypotenuse is given by c = m(1 + x) − d. And, indeed, elementary algebraic manipulation shows that a2 + b2 = c2 whenever d has the value stated. Also, if m and x are rational, so are d, a, b and c. A Pythagorean triple can therefore be obtained from a, b and c by multiplying each of them by the least common multiple of their denominators.\n\n#### Pell's equation\n\nBrahmagupta went on to give a recurrence relation for generating solutions to certain instances of Diophantine equations of the second degree such as Nx2 + 1 = y2 (called Pell's equation) by using the Euclidean algorithm. The Euclidean algorithm was known to him as the \"pulverizer\" since it breaks numbers down into ever smaller pieces.\n\nThe nature of squares:\n18.64. [Put down] twice the square-root of a given square by a multiplier and increased or diminished by an arbitrary [number]. The product of the first [pair], multiplied by the multiplier, with the product of the last [pair], is the last computed.\n18.65. The sum of the thunderbolt products is the first. The additive is equal to the product of the additives. The two square-roots, divided by the additive or the subtractive, are the additive rupas.\n\nThe key to his solution was the identity,\n\n$(x_{1}^{2}-Ny_{1}^{2})(x_{2}^{2}-Ny_{2}^{2})=(x_{1}x_{2}+Ny_{1}y_{2})^{2}-N(x_{1}y_{2}+x_{2}y_{1})^{2}$", null, "which is a generalisation of an identity that was discovered by Diophantus,\n\n$(x_{1}^{2}-y_{1}^{2})(x_{2}^{2}-y_{2}^{2})=(x_{1}x_{2}+y_{1}y_{2})^{2}-(x_{1}y_{2}+x_{2}y_{1})^{2}.$", null, "Using his identity and the fact that if (x1, y1) and (x2, y2) are solutions to the equations x2Ny2 = k1 and x2Ny2 = k2, respectively, then (x1x2 + Ny1y2, x1y2 + x2y1) is a solution to x2Ny2 = k1k2, he was able to find integral solutions to Pell's equation through a series of equations of the form x2Ny2 = ki. Brahmagupta was not able to apply his solution uniformly for all possible values of N, rather he was only able to show that if x2Ny2 = k has an integer solution for k = ±1, ±2, or ±4, then x2Ny2 = 1 has a solution. The solution of the general Pell's equation would have to wait for Bhaskara II in c.1150 CE.\n\n### Geometry\n\n#### Brahmagupta's formula\n\nBrahmagupta's most famous result in geometry is his formula for cyclic quadrilaterals. Given the lengths of the sides of any cyclic quadrilateral, Brahmagupta gave an approximate and an exact formula for the figure's area,\n\n12.21. The approximate area is the product of the halves of the sums of the sides and opposite sides of a triangle and a quadrilateral. The accurate [area] is the square root from the product of the halves of the sums of the sides diminished by [each] side of the quadrilateral.\n\nSo given the lengths p, q, r and s of a cyclic quadrilateral, the approximate area is p + r/2 · q + s/2 while, letting , the exact area is\n\n(tp)(tq)(tr)(ts).\n\nAlthough Brahmagupta does not explicitly state that these quadrilaterals are cyclic, it is apparent from his rules that this is the case. Heron's formula is a special case of this formula and it can be derived by setting one of the sides equal to zero.\n\n#### Triangles\n\nBrahmagupta dedicated a substantial portion of his work to geometry. One theorem gives the lengths of the two segments a triangle's base is divided into by its altitude:\n\n12.22. The base decreased and increased by the difference between the squares of the sides divided by the base; when divided by two they are the true segments. The perpendicular [altitude] is the square-root from the square of a side diminished by the square of its segment.\n\nThus the lengths of the two segments are 1/2(b ± c2a2/b).\n\nHe further gives a theorem on rational triangles. A triangle with rational sides a, b, c and rational area is of the form:\n\n$a={\\frac {1}{2}}\\left({\\frac {u^{2}}{v}}+v\\right),\\ \\ b={\\frac {1}{2}}\\left({\\frac {u^{2}}{w}}+w\\right),\\ \\ c={\\frac {1}{2}}\\left({\\frac {u^{2}}{v}}-v+{\\frac {u^{2}}{w}}-w\\right)$", null, "for some rational numbers u, v, and w.\n\n#### Brahmagupta's theorem\n\nBrahmagupta continues,\n\n12.23. The square-root of the sum of the two products of the sides and opposite sides of a non-unequal quadrilateral is the diagonal. The square of the diagonal is diminished by the square of half the sum of the base and the top; the square-root is the perpendicular [altitudes].\n\nSo, in a \"non-unequal\" cyclic quadrilateral (that is, an isosceles trapezoid), the length of each diagonal is pr + qs.\n\nHe continues to give formulas for the lengths and areas of geometric figures, such as the circumradius of an isosceles trapezoid and a scalene quadrilateral, and the lengths of diagonals in a scalene cyclic quadrilateral. This leads up to Brahmagupta's famous theorem,\n\n12.30–31. Imaging two triangles within [a cyclic quadrilateral] with unequal sides, the two diagonals are the two bases. Their two segments are separately the upper and lower segments [formed] at the intersection of the diagonals. The two [lower segments] of the two diagonals are two sides in a triangle; the base [of the quadrilateral is the base of the triangle]. Its perpendicular is the lower portion of the [central] perpendicular; the upper portion of the [central] perpendicular is half of the sum of the [sides] perpendiculars diminished by the lower [portion of the central perpendicular].\n\n#### Pi\n\nIn verse 40, he gives values of π,\n\n12.40. The diameter and the square of the radius [each] multiplied by 3 are [respectively] the practical circumference and the area [of a circle]. The accurate [values] are the square-roots from the squares of those two multiplied by ten.\n\nSo Brahmagupta uses 3 as a \"practical\" value of π, and ${\\sqrt {10}}\\approx 3.1622\\ldots$", null, "as an \"accurate\" value of π. The error in this \"accurate\" value is less than 1%.\n\n#### Measurements and constructions\n\nIn some of the verses before verse 40, Brahmagupta gives constructions of various figures with arbitrary sides. He essentially manipulated right triangles to produce isosceles triangles, scalene triangles, rectangles, isosceles trapezoids, isosceles trapezoids with three equal sides, and a scalene cyclic quadrilateral.\n\nAfter giving the value of pi, he deals with the geometry of plane figures and solids, such as finding volumes and surface areas (or empty spaces dug out of solids). He finds the volume of rectangular prisms, pyramids, and the frustum of a square pyramid. He further finds the average depth of a series of pits. For the volume of a frustum of a pyramid, he gives the \"pragmatic\" value as the depth times the square of the mean of the edges of the top and bottom faces, and he gives the \"superficial\" volume as the depth times their mean area.\n\n### Trigonometry\n\n#### Sine table\n\nIn Chapter 2 of his Brahmasphutasiddhanta, entitled Planetary True Longitudes, Brahmagupta presents a sine table:\n\n2.2–5. The sines: The Progenitors, twins; Ursa Major, twins, the Vedas; the gods, fires, six; flavors, dice, the gods; the moon, five, the sky, the moon; the moon, arrows, suns [...]\n\nHere Brahmagupta uses names of objects to represent the digits of place-value numerals, as was common with numerical data in Sanskrit treatises. Progenitors represents the 14 Progenitors (\"Manu\") in Indian cosmology or 14, \"twins\" means 2, \"Ursa Major\" represents the seven stars of Ursa Major or 7, \"Vedas\" refers to the 4 Vedas or 4, dice represents the number of sides of the tradition die or 6, and so on. This information can be translated into the list of sines, 214, 427, 638, 846, 1051, 1251, 1446, 1635, 1817, 1991, 2156, 2312, 1459, 2594, 2719, 2832, 2933, 3021, 3096, 3159, 3207, 3242, 3263, and 3270, with the radius being 3270.\n\n#### Interpolation formula\n\nIn 665 Brahmagupta devised and used a special case of the Newton–Stirling interpolation formula of the second-order to interpolate new values of the sine function from other values already tabulated. The formula gives an estimate for the value of a function f at a value a + xh of its argument (with h > 0 and −1 ≤ x ≤ 1) when its value is already known at ah, a and a + h.\n\nThe formula for the estimate is:\n\n$f(a+xh)\\approx f(a)+x\\left({\\frac {\\Delta f(a)+\\Delta f(a-h)}{2}}\\right)+{\\frac {x^{2}\\Delta ^{2}f(a-h)}{2!}}.$", null, "where Δ is the first-order forward-difference operator, i.e.\n\n$\\Delta f(a)\\ {\\stackrel {\\mathrm {def} }{=}}\\ f(a+h)-f(a).$", null, "## Astronomy\n\nSome of the important contributions made by Brahmagupta in astronomy are his methods for calculating the position of heavenly bodies over time (ephemerides), their rising and setting, conjunctions, and the calculation of solar and lunar eclipses.\n\nIn chapter seven of his Brahmasphutasiddhanta, entitled Lunar Crescent, Brahmagupta rebuts the idea that the Moon is farther from the Earth than the Sun, an idea which had been suggested by Vedic scripture. He does this by explaining the illumination of the Moon by the Sun.\n\n1. If the moon were above the sun, how would the power of waxing and waning, etc., be produced from calculation of the longitude of the moon? The near half would always be bright.\n\n2. In the same way that the half seen by the sun of a pot standing in sunlight is bright, and the unseen half dark, so is [the illumination] of the moon [if it is] beneath the sun.\n\n3. The brightness is increased in the direction of the sun. At the end of a bright [i.e. waxing] half-month, the near half is bright and the far half dark. Hence, the elevation of the horns [of the crescent can be derived] from calculation. [...]\n\nHe explains that since the Moon is closer to the Earth than the Sun, the degree of the illuminated part of the Moon depends on the relative positions of the Sun and the Moon, and this can be computed from the size of the angle between the two bodies.\n\nFurther work exploring the longitudes of the planets, diurnal rotation, lunar and solar eclipses, risings and settings, the moon's crescent and conjunctions of the planets, are discussed in his treatise Khandakhadyaka.\n\n## Citations and footnotes\n\n1. Sachau, Edward C. (2013), Alberuni's India, Routledge, p. 156, ISBN 978-1-136-38357-1, Brahma-siddhānta, so called from Brahman, composed by Brahmagupta, the son of Jishnu, from the town of Bhillamāla between Multān and Anhilwāra, 16 yojana from the latter place (?)\n2. Brahmagupta biography, Article by: J J O'Connor and E F Robertson, School of Mathematics and Statistics, University of St Andrews, Scotland, November 2000\n3. Bhattacharyya 2011, p. 185: \"Brahmagupta, one of the most celebrated mathematicians of the East, indeed of the world, was born in the year 598 c.e., in the town of Bhillamala during the reign of King Vyaghramukh of the Chapa Dynasty.\"\n4. Gupta 2008, p. 162.\n5. Pillai, S. Devadas (1997), Indian Sociology Through Ghurye, a Dictionary, Popular Prakashan, p. 216, ISBN 978-81-7154-807-1, Brahmagupta (b. 598 AD) was a native of either the Multan region of the Punjab (now this areas is in Pakistan) or the Abu region of Rajasthan.\n6. Bhattacharyya 2011, pp. 185–186.\n7. Gupta 2008, p. 163.\n8. Plofker (2007, pp. 418–419)\n9. Bhattacharyya 2011, p. 185.\n10. Avari 2013, p. 32.\n11. Young, M. J. L.; Latham, J. D.; Serjeant, R. B. (2 November 2006), Religion, Learning and Science in the 'Abbasid Period, Cambridge University Press, pp. 302–303, ISBN 978-0-521-02887-5\n12. van Bladel, Kevin (28 November 2014), \"Eighth Century Indian Astronomy in the Two Cities of Peace\", in Asad Q. Ahmed; Benham Sadeghi; Robert G. Hoyland (eds.), Islamic Cultures, Islamic Contexts: Essays in Honor of Professor Patricia Crone, BRILL, pp. 257–294, ISBN 978-90-04-28171-4\n13. Plofker (2007, pp. 428–434)\n14. Boyer (1991, \"China and India\" p. 221) \"he was the first one to give a general solution of the linear Diophantine equation ax + by = c, where a, b, and c are integers. [...] It is greatly to the credit of Brahmagupta that he gave all integral solutions of the linear Diophantine equation, whereas Diophantus himself had been satisfied to give one particular solution of an indeterminate equation. Inasmuch as Brahmagupta used some of the same examples as Diophantus, we see again the likelihood of Greek influence in India – or the possibility that they both made use of a common source, possibly from Babylonia. It is interesting to note also that the algebra of Brahmagupta, like that of Diophantus, was syncopated. Addition was indicated by juxtaposition, subtraction by placing a dot over the subtrahend, and division by placing the divisor below the dividend, as in our fractional notation but without the bar. The operations of multiplication and evolution (the taking of roots), as well as unknown quantities, were represented by abbreviations of appropriate words.\"\n15. Brahmasputha Siddhanta, Translated to English by H.T Colebrook, 1817 AD\n16. Plofker (2007, pp. 422) The reader is apparently expected to be familiar with basic arithmetic operations as far as the square-root; Brahmagupta merely notes some points about applying them to fractions. The procedures for finding the cube and cube-root of an integer, however, are described (compared the latter to Aryabhata's very similar formulation). They are followed by rules for five types of combinations: [...]\n17. Plofker (2007, pp. 421–427)\n18. Plofker (2007, p. 423) Here the sums of the squares and cubes of the first n integers are defined in terms of the sum of the n integers itself;\n19. Kaplan, Robert (1999). The Nothing That Is: A Natural History of Zero. London: Allen Lane/The Penguin Press. pp. 68–75. Bibcode:2000tnti.book.....K.\n20. Boyer (1991, p. 220): However, here again Brahmagupta spoiled matters somewhat by asserting that 0 ÷ 0 = 0, and on the touchy matter of a ÷ 0, he did not commit himself.\n21. Plofker (2007, p. 426)\n22. Stillwell (2004, pp. 44–46): In the seventh century CE the Indian mathematician Brahmagupta gave a recurrence relation for generating solutions of x2Dy2 = 1, as we shall see in Chapter 5. The Indians called the Euclidean algorithm the \"pulverizer\" because it breaks numbers down to smaller and smaller pieces. To obtain a recurrence one has to know that a rectangle proportional to the original eventually recurs, a fact that was rigorously proved only in 1768 by Lagrange.\n23. Stillwell (2004, pp. 72–74)\n24. Plofker (2007, p. 424) Brahmagupta does not explicitly state that he is discussing only figures inscribed in circles, but it is implied by these rules for computing their circumradius.\n25. Stillwell (2004, p. 77)\n26. Plofker (2007, p. 427) After the geometry of plane figures, Brahmagupta discusses the computation of volumes and surface areas of solids (or empty spaces dug out of solids). His straightforward rules for the volumes of a rectangular prism and pyramid are followed by a more ambiguous one, which may refer to finding the average depth of a sequence of puts with different depths. The next formula apparently deals with the volume of a frustum of a square pyramid, where the \"pragmatic\" volume is the depth times the square of the mean of the edges of the top and bottom faces, while the \"superficial\" volume is the depth times their mean area.\n27. Plofker (2007, p. 419)\n28. Plofker (2007, pp. 419–420) Brahmagupta's sine table, like much other numerical data in Sanskrit treatises, is encoded mostly in concrete-number notation that uses names of objects to represent the digits of place-value numerals, starting with the least significant. [...]\nThere are fourteen Progenitors (\"Manu\") in Indian cosmology; \"twins\" of course stands for 2; the seven stars of Ursa Major (the \"Sages\") for 7, the four Vedas, and the four sides of the traditional dice used in gambling, for 6, and so on. Thus Brahmagupta enumerates his first six sine-values as 214, 427, 638, 846, 1051, 1251. (His remaining eighteen sines are 1446, 1635, 1817, 1991, 2156, 2312, 1459, 2594, 2719, 2832, 2933, 3021, 3096, 3159, 3207, 3242, 3263, 3270). The Paitamahasiddhanta, however, specifies an initial sine-value of 225 (although the rest of its sine-table is lost), implying a trigonometric radius of R = 3438 approx= C(')/2π: a tradition followed, as we have seen, by Aryabhata. Nobody knows why Brahmagupta chose instead to normalize these values to R = 3270.\n29. Joseph (2000, pp.285–86).\n30. Teresi, Dick (2002). Lost Discoveries: The Ancient Roots of Modern Science. Simon and Schuster. p. 135. ISBN 0-7432-4379-X.\n31. Plofker (2007, pp. 419–420) Brahmagupta discusses the illumination of the moon by the sun, rebutting an idea maintained in scriptures: namely, that the moon is farther from the earth than the sun is. In fact, as he explains, because the moon is closer the extent of the illuminated portion of the moon depends on the relative positions of the moon and the sun, and can be computed from the size of the angular separation α between them.\n32. Plofker (2007, p. 420)" ]

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[ "## You searched , Object Type: “scent bottle”\n\nShow me:\nand\n\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 5 items Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 5 items Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 3 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 10 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items\n• 1 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• 1 items Explore\n• 6 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• 7 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 13 items Explore\n• Explore\n• Explore\n• 37 items Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 7 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 6 items Explore\n• 2 items Explore\n• Explore\n• Explore\n• 7 items Explore\n• 21 items Explore\n• Explore\n• 3 items Explore\n• 13 items Explore\n• 1 items Explore\n• Explore\n• 1 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 6 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 9 items Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• 1 items\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore" ]

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[ "# How do you multiply sqrt (x) * (x)?\n\nJun 13, 2015\n\n$\\sqrt{x} \\cdot x = \\sqrt{{x}^{3}}$\n\n#### Explanation:\n\nKnowing that $\\sqrt{x} = {x}^{\\frac{1}{2}}$\nand using the properties:\n${x}^{a} \\cdot {x}^{b} = {x}^{a + b}$\n${\\left({x}^{a}\\right)}^{b} = {x}^{a \\cdot b}$\nyou have:\n\n$\\sqrt{x} \\cdot x = {x}^{\\frac{1}{2}} \\cdot {x}^{1} = {x}^{\\frac{1}{2} + 1} = {x}^{\\frac{3}{2}} = {\\left({x}^{3}\\right)}^{\\frac{1}{2}} = \\sqrt{{x}^{3}}$" ]

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[ "]> git.sesse.net Git - stockfish/blob - src/notation.cpp\n1 /*\n2   Stockfish, a UCI chess playing engine derived from Glaurung 2.1\n6   Stockfish is free software: you can redistribute it and/or modify\n7   it under the terms of the GNU General Public License as published by\n8   the Free Software Foundation, either version 3 of the License, or\n9   (at your option) any later version.\n11   Stockfish is distributed in the hope that it will be useful,\n12   but WITHOUT ANY WARRANTY; without even the implied warranty of\n13   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the\n14   GNU General Public License for more details.\n16   You should have received a copy of the GNU General Public License\n17   along with this program.  If not, see <http://www.gnu.org/licenses/>.\n18 */\n20 #include <cassert>\n21 #include <iomanip>\n22 #include <sstream>\n23 #include <string>\n25 #include \"movegen.h\"\n26 #include \"notation.h\"\n27 #include \"position.h\"\n29 using namespace std;\n31 static const char* PieceToChar = \" PNBRQK  pnbrqk\";\n34 /// score_to_uci() converts a value to a string suitable for use with the UCI\n35 /// protocol specifications:\n36 ///\n37 /// cp <x>     The score from the engine's point of view in centipawns.\n38 /// mate <y>   Mate in y moves, not plies. If the engine is getting mated\n39 ///            use negative values for y.\n41 string score_to_uci(Value v, Value alpha, Value beta) {\n43   stringstream s;\n45   if (abs(v) < VALUE_MATE_IN_MAX_PLY)\n46       s << \"cp \" << v * 100 / int(PawnValueMg);\n47   else\n48       s << \"mate \" << (v > 0 ? VALUE_MATE - v + 1 : -VALUE_MATE - v) / 2;\n50   s << (v >= beta ? \" lowerbound\" : v <= alpha ? \" upperbound\" : \"\");\n52   return s.str();\n53 }\n56 /// move_to_uci() converts a move to a string in coordinate notation\n57 /// (g1f3, a7a8q, etc.). The only special case is castling moves, where we print\n58 /// in the e1g1 notation in normal chess mode, and in e1h1 notation in chess960\n59 /// mode. Internally castle moves are always coded as \"king captures rook\".\n61 const string move_to_uci(Move m, bool chess960) {\n63   Square from = from_sq(m);\n64   Square to = to_sq(m);\n66   if (m == MOVE_NONE)\n67       return \"(none)\";\n69   if (m == MOVE_NULL)\n70       return \"0000\";\n72   if (type_of(m) == CASTLE && !chess960)\n73       to = (to > from ? FILE_G : FILE_C) | rank_of(from);\n75   string move = square_to_string(from) + square_to_string(to);\n77   if (type_of(m) == PROMOTION)\n78       move += PieceToChar[make_piece(BLACK, promotion_type(m))]; // Lower case\n80   return move;\n81 }\n84 /// move_from_uci() takes a position and a string representing a move in\n85 /// simple coordinate notation and returns an equivalent legal Move if any.\n87 Move move_from_uci(const Position& pos, string& str) {\n89   if (str.length() == 5) // Junior could send promotion piece in uppercase\n90       str = char(tolower(str));\n92   for (MoveList<LEGAL> ml(pos); !ml.end(); ++ml)\n93       if (str == move_to_uci(ml.move(), pos.is_chess960()))\n94           return ml.move();\n96   return MOVE_NONE;\n97 }\n100 /// move_to_san() takes a position and a legal Move as input and returns its\n101 /// short algebraic notation representation.\n103 const string move_to_san(Position& pos, Move m) {\n105   if (m == MOVE_NONE)\n106       return \"(none)\";\n108   if (m == MOVE_NULL)\n109       return \"(null)\";\n111   assert(pos.move_is_legal(m));\n113   Bitboard attackers;\n114   bool ambiguousMove, ambiguousFile, ambiguousRank;\n115   string san;\n116   Color us = pos.side_to_move();\n117   Square from = from_sq(m);\n118   Square to = to_sq(m);\n119   Piece pc = pos.piece_on(from);\n120   PieceType pt = type_of(pc);\n122   if (type_of(m) == CASTLE)\n123       san = to > from ? \"O-O\" : \"O-O-O\";\n124   else\n125   {\n126       if (pt != PAWN)\n127       {\n128           san = PieceToChar[pt]; // Upper case\n130           // Disambiguation if we have more then one piece with destination 'to'\n131           // note that for pawns is not needed because starting file is explicit.\n132           ambiguousMove = ambiguousFile = ambiguousRank = false;\n134           attackers = (pos.attacks_from(pc, to) & pos.pieces(us, pt)) ^ from;\n136           while (attackers)\n137           {\n138               Square sq = pop_lsb(&attackers);\n140               // Pinned pieces are not included in the possible sub-set\n141               if (!pos.pl_move_is_legal(make_move(sq, to), pos.pinned_pieces()))\n142                   continue;\n144               ambiguousFile |= file_of(sq) == file_of(from);\n145               ambiguousRank |= rank_of(sq) == rank_of(from);\n146               ambiguousMove = true;\n147           }\n149           if (ambiguousMove)\n150           {\n151               if (!ambiguousFile)\n152                   san += file_to_char(file_of(from));\n154               else if (!ambiguousRank)\n155                   san += rank_to_char(rank_of(from));\n157               else\n158                   san += square_to_string(from);\n159           }\n160       }\n161       else if (pos.is_capture(m))\n162           san = file_to_char(file_of(from));\n164       if (pos.is_capture(m))\n165           san += 'x';\n167       san += square_to_string(to);\n169       if (type_of(m) == PROMOTION)\n170           san += string(\"=\") + PieceToChar[promotion_type(m)];\n171   }\n173   if (pos.move_gives_check(m, CheckInfo(pos)))\n174   {\n175       StateInfo st;\n176       pos.do_move(m, st);\n177       san += MoveList<LEGAL>(pos).size() ? \"+\" : \"#\";\n178       pos.undo_move(m);\n179   }\n181   return san;\n182 }\n185 /// pretty_pv() formats human-readable search information, typically to be\n186 /// appended to the search log file. It uses the two helpers below to pretty\n187 /// format time and score respectively.\n189 static string time_to_string(int64_t msecs) {\n191   const int MSecMinute = 1000 * 60;\n192   const int MSecHour   = 1000 * 60 * 60;\n194   int64_t hours   =   msecs / MSecHour;\n195   int64_t minutes =  (msecs % MSecHour) / MSecMinute;\n196   int64_t seconds = ((msecs % MSecHour) % MSecMinute) / 1000;\n198   stringstream s;\n200   if (hours)\n201       s << hours << ':';\n203   s << setfill('0') << setw(2) << minutes << ':' << setw(2) << seconds;\n205   return s.str();\n206 }\n208 static string score_to_string(Value v) {\n210   stringstream s;\n212   if (v >= VALUE_MATE_IN_MAX_PLY)\n213       s << \"#\" << (VALUE_MATE - v + 1) / 2;\n215   else if (v <= VALUE_MATED_IN_MAX_PLY)\n216       s << \"-#\" << (VALUE_MATE + v) / 2;\n218   else\n219       s << setprecision(2) << fixed << showpos << float(v) / PawnValueMg;\n221   return s.str();\n222 }\n224 string pretty_pv(Position& pos, int depth, Value value, int64_t msecs, Move pv[]) {\n226   const int64_t K = 1000;\n227   const int64_t M = 1000000;\n229   StateInfo state[MAX_PLY_PLUS_2], *st = state;\n230   Move* m = pv;\n232   size_t length;\n233   stringstream s;\n235   s << setw(2) << depth\n236     << setw(8) << score_to_string(value)\n237     << setw(8) << time_to_string(msecs);\n239   if (pos.nodes_searched() < M)\n240       s << setw(8) << pos.nodes_searched() / 1 << \"  \";\n242   else if (pos.nodes_searched() < K * M)\n243       s << setw(7) << pos.nodes_searched() / K << \"K  \";\n245   else\n246       s << setw(7) << pos.nodes_searched() / M << \"M  \";\n248   padding = string(s.str().length(), ' ');\n251   while (*m != MOVE_NONE)\n252   {\n253       san = move_to_san(pos, *m);\n255       if (length + san.length() > 80)\n256       {\n257           s << \"\\n\" + padding;" ]

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[ "# Use Matlab Function pwelch to Find Power Spectral Density – or Do It Yourself\n\nIn my last post, we saw that finding the spectrum of a signal requires several steps beyond computing the discrete Fourier transform (DFT).  These include windowing the signal, taking the magnitude-squared of the DFT, and computing the vector of frequencies.  The Matlab function pwelch performs all these steps, and it also has the option to use DFT averaging to compute the so-called Welch power spectral density estimate [3,4].\n\nIn this article, I’ll present some examples to show how to use pwelch.  You can also “do it yourself”, i.e. compute spectra using the Matlab fft or other fft function.  As examples, the appendix provides two demonstration mfiles; one computes the spectrum without DFT averaging, and the other computes the spectrum with DFT averaging.  (Note:  I have also created a simplified Matlab PSD function based on pwelch:  see this post.)\n\nWe’ll use this form of the function call for pwelch:\n\n [pxx,f]= pwelch(x,window,noverlap,nfft,fs)\n\n\nThe input arguments are:\n\nx                  input signal vector\n\nwindow        window vector of length nfft\n\nnoverlap      number of overlapped samples (used for DFT averaging)\n\nnfft               number of points in DFT\n\nfs                 sample rate in Hz\n\nFor this article, we’ll assume x is real.  Let length(x) = N.  If DFT averaging is not desired, you set nfft equal to N and pwelch takes a single DFT of the windowed vector x.  If DFT averaging is desired, you set nfft to a fraction of N and pwelch uses windowed segments of x of length nfft as inputs to the DFT.  The |X(k)|2 of the multiple DFT’s are then averaged.\n\nThe output arguments are:\n\npxx          power spectral density vector, W/Hz\n\nf               vector of frequency values from 0 to fs/2, Hz\n\nThe length of the output vectors is nfft/2 + 1 when nfft is even.  For example, if nfft= 1024, pxx and f contain 513 samples.  pxx has units of W/Hz when x has units of volts and load resistance is one ohm.\n\n## Examples\n\nThe following examples use a signal x consisting of a sine wave plus Gaussian noise.  Here is the Matlab code to generate x, where sine frequency is 500 Hz and sample rate is 4000 Hz:\n\n fs= 4000; % Hz sample rate\nTs= 1/fs;\nf0= 500; % Hz sine frequency\nA= sqrt(2); % V sine amplitude for P= 1 W into 1 ohm.\nN= 1024; % number of time samples\nn= 0:N-1; % time index\nx= A*sin(2*pi*f0*n*Ts) + .1*randn(1,N); % 1 W sinewave + noise\n\n\nThe power of the sine wave into a 1-ohm load is A2/2 = 1 watt.  In the following, examples 1 and 2 compute a single DFT, while example 3 computes multiple DFT’s and averages them.  Example 4 calculates C/N0 for the spectrum of Example 3.\n\nExample 1.  Spectrum in dBW/Hz\n\nWe’ll start out with a rectangular window.  Note the sinewave frequency of 500 Hz is at the center of a DFT bin to prevent spectral leakage .   Here is the code to compute the spectrum.  Since we are not using DFT averaging, we set noverlap= 0:\n\n nfft= N;\nwindow= rectwin(nfft);\n[pxx,f]= pwelch(x,window,0,nfft,fs); % W/Hz power spectral density\nPdB_Hz= 10*log10(pxx); % dBW/Hz\n\n\nThe resulting spectrum is plotted in Figure 1, which shows the component at 500 Hz with amplitude of roughly -6 dBW/Hz.  How does this relate to the actual power of the sinewave for a 1-ohm load resistance?  To compute the power of the spectral component at 500 Hz, we need to convert the amplitude in dBW/Hz to dBW/bin as follows.  First, convert W/Hz to W/bin:\n\nPbin (W/bin) = pxx (W/Hz) * fs/nfft (Hz/bin)       (1)\n\nThen, we have in dB:\n\nPdB_bin = 10*log10(pxx*fs/nfft)     dBW/bin        (2)\n\nOr      PdB_bin = PdB_Hz + 10*log10(fs/nfft)     dBW/bin\n\nIf we set the noise amplitude to zero, we get PdB_Hz = -5.9176 dB/Hz and\n\nPdB_bin= -5.9176 dBW/Hz + 10log10(4000/1024) = 0 dBW = 1 watt\n\nSo we are reassured that the amplitude of the spectrum is correct.  This motivates us to try the next example, which displays the spectrum in dBW/bin.", null, "Figure 1.  Spectrum in dBW/Hz\n\nExample 2.  Spectrum in dBW/bin\n\nTo find the spectrum in dBW/bin, the code is the same as Example 1, except we use Equation 2 to find PdB_bin:\n\n nfft= N;\nwindow= rectwin(nfft);\n[pxx,f]= pwelch(x,window,0,nfft,fs); % W/Hz power spectral density\nPdB_bin= 10*log10(pxx*fs/nfft); % dBW/bin\n\n\nThe spectrum is shown in the top plot of Figure 2.  As expected, the component at 500 Hz has power of 0 dBW.  Note that the amplitude of a sine’s spectral component in dBW/bin is constant vs nfft, while the amplitude in dBW/Hz varies vs. nfft.  If we now replace the rectangular window with a hanning window:\n\n window= hanning(nfft);\n\n\nWe get the spectrum in the bottom plot of Figure 2.  The peak power is now less than 0 dBW, because the power is spread over several frequency samples of the window.  Nevertheless, this is a more useful window than the rectangular window, which has bad spectral leakage when f0 is not at the center of a bin.", null, "Figure 2.  Spectra in dBW/bin    Top:  rectangular window\n\nBottom:  hanning window\n\nExample 3.  DFT averaging\n\nFor this example, fs, Ts, f0, A, and nfft are the same as for the previous examples, but we use N= 8*nfft time samples of x.  pwelch takes the DFT of Navg = 15 overlapping segments of x, each of length nfft, then averages the |X(k)|2 of the DFT’s.  The segments of x are shown in Figure 3, where we have set the number of overlapping samples to nfft/2 = 512.\n\nIn general, if there are an integer number of segments that cover all samples of N, we have :\n\n(Navg – 1)*D + nfft = N         (3)\n\nWhere D = nfft – noverlap.  For our case, with D = nfft/2 and N/nfft = 8, we have\n\nNavg = 2*N/nfft -1 = 15.\n\nFor a given number of time samples N, using overlapping segments lets us increase Navg compared with no overlapping.  In this case, overlapping of 50% increases Navg from 8 to 15.  Here is the Matlab code to compute the spectrum:\n\n nfft= 1024;\nN= nfft*8; % number of samples in signal\nn= 0:N-1;\nx= A*sin(2*pi*f0*n*Ts) + .1*randn(1,N); % 1 W sinewave + noise\nnoverlap= nfft/2; % number of overlapping time samples\nwindow= hanning(nfft);\n[pxx,f]= pwelch(x,window,noverlap,nfft,fs); % W/Hz PSD\nPdB_bin= 10*log10(pxx*fs/nfft); % dBW/bin\n\n\nThe resulting spectrum is shown in Figure 4.  Compared to the no-averaging case in Figure 2, there is less variation in the noise floor.  DFT averaging reduces the variance σ2 of the noise spectrum by a factor of Navg, as long as noverlap is not greater than nfft/2 .", null, "Figure 3.  Overlapping segments of signal x for N = 8*nfft and noverlap = nfft/2 (50%).", null, "Figure 4.  Spectrum with Navg = 15 and noverlap= nfft/2= 512.\n\nExample 4.  Calculation of C/N0\n\nReducing the noise variance by DFT averaging makes calculation of the noise level and thus carrier-to-noise ratio (C/N) more accurate.  Let’s calculate C/N0 for the signal of Example 3, where N0 is the noise power in a 1 Hz bandwidth.\n\nFirst, we’ll find the power of the sinewave (the “C” in C/N0) by performing a peak search on the dBW/bin spectrum, then adding the spectral components near the peak to get the total power.  This is necessary because windowing has spread the power over several frequency samples.\n\n y= max(PdB_bin); % peak of spectrum\nk_peak= find(PdB_bin==y); % index of peak\nPsum= fs/nfft * sum(pxx(k_peak-2:k_peak+2)); % W sum power of 5 samples\nC= 10*log10(Psum) % dBW carrier power\n\n\nThe peak power of 0 dBW is shown by the marker in the left plot of Figure 4.  Now we’ll find N0 from the dBW/Hz spectrum.  Choosing f = 800 Hz, we have:\n\n PdB_Hz= 10*log10(pxx); % dBW/Hz\nf1= 800; % Hz\nk1= round(f1/fs *nfft); % index of f1\nNo= PdB_Hz(k1) % dBW/Hz Noise density at f1\n\n\nThe N0 marker is shown in the right plot of Figure 4.  Figure 5 combines the dBW/bin and dBW/Hz plots in one graph.  C/N0 is just:\n\n CtoNo= C - No; % dB (1 Hz)", null, "Figure 5.      Left:  Spectrum in dBW/bin with marker showing the power of the sinewave.\n\nRight:  Spectrum in dBW/Hz with marker at 800 Hz showing N0.", null, "Figure 6.  Spectrum in dBW/bin (blue) and dBW/Hz (grey).\n\n## Appendix     Matlab Code to compute spectra without using pwelch\n\nThese examples use the Malab fft function to compute spectra.  See my earlier post for derivations of the formulas for power/bin and normalized window function.\n\n### Demonstrate Spectrum Computation/ No Averaging\n\n%spectrum_demo.m 1/3/19 Neil Robertson\n% Use FFT to find spectrum of sine + noise in units of dBW/bin and dBW/Hz.\nfs= 4000; % Hz sample frequency\nTs= 1/fs;\nf0= 500; % Hz sine frequency\nA= sqrt(2 % V sine amplitude for P= 1 W into 1 ohm.\nN= 1024; % number of time samples\nnfft= N; % number of frequency samples\nn= 0:N-1; % time index\n%\nx= A*sin(2*pi*f0*n*Ts) + .1*randn(1,N); % 1 W sinewave + noise\nw= hanning(N); % window function\nwindow= w.*sqrt(N/sum(w.^2)); % normalize window for P= 1\nxw= x.*window'; % apply window to x\n%\nX= fft(xw,nfft); % DFT\nX= X(1:nfft/2); % retain samples from 0 to fs/2\nmagsq= real(X).^2 + imag(X).^2; % DFT magnitude squared\nP_bin= 2/nfft.^2 * magsq; % W/bin power spectrum into 1 ohm\nP_Hz= P_bin*nfft/fs; % W/Hz power spectrum\n%\nPdB_bin= 10*log10(P_bin); % dBW/bin\nPdB_Hz= 10*log10(P_Hz); % dBW/Hz\n%\nk= 0:nfft/2-1; % frequency index\nf= k*fs/nfft; % Hz frequency vector\n%\n%\nplot(f,PdB_bin),grid\naxis([0 fs/2 -60 10])\nxlabel('Hz'),ylabel('dBW/bin')\ntitle('Spectrum with amplitude units of dBW/bin')\nfigure\ny= max(PdB_Hz);\nplot(f,PdB_Hz),grid\naxis([0 fs/2 y-60 y+10])\nxlabel('Hz'),ylabel('dBW/Hz')\ntitle('Spectrum with amplitude units of dBW/Hz')\n\n\n### Demonstrate Spectrum Computation with Averaging\n\nSuppose we have segmented a signal x into four segments of length nfft, that could be overlapped.  If we window each segment and take its DFT, we have four DFT’s as shown in Table A.1.  Here, we have defined M= nfft and we have numbered the FFT bins from 1 to M (instead of 0 to M - 1).\n\nTable A.1  DFT’s of four segments of the signal x.", null, "Now we form the magnitude-squared of each bin of the four DFT’s, as shown in Table A.2  If we then sum the magnitude-square values of each bin over the four DFT’s and divide by 4, we obtain the average magnitude-squared value, as shown in the right column.  Finally, we scale the magnitude-squared value to obtain the averaged power per bin or averaged power per Hz:\n\nPbin(k) = 2/M2 * sum(|Xk|2)/4   W/bin          (A.1)\n\nPHz(k)= Pbin(k)*M/fs     W/Hz                       (A.2)\n\nOr\n\nPHz(k)= 2/(M*fs) * sum(|Xk|2)/4    W/Hz      (A.3)\n\nPHz(k) is the Welch power spectral density estimate.\n\nTable A.2  Magnitude-square values of Four DFT’s, and their average (right column).", null, "The following example m-file uses Navg = 8, summing 8 values of each |X(k)|2 and dividing by 8.  For simplicity, noverlap = 0 samples.  Another option for averaging, useful for the case where DFT’s are taken continuously, is exponential averaging, which basically runs the |X(k)|2 through a first-order IIR filter .\n\n%spectrum_demo_avg 1/3/19 Neil Robertson\n% Perform DFT averaging on signal = sine + noise\n% noverlap = 0\n% Given signal x(1:8192),\n%\n% |x(1) x(1025) . . . . x(7169)|\n% |x(2) x(1026) . . . . x(7170)|\n% xMatrix= | . . . |\n% | . . . |\n% |x(1024) x(2048) . . . . x(8192)|\n%\n%\nfs= 4000; % Hz sample frequency\nTs= 1/fs;\nf0= 500; % Hz sine frequency\nA= sqrt(2 % V sine amplitude for P= 1 W into 1 ohm\nnfft= 1024; % number of frequency samples\nNavg= 8; % number of DFT's to be averaged\nN= nfft*Navg; % number of time samples\nn= 0:N-1; % time index\nx= A*sin(2*pi*f0*n*Ts) + .1*randn(1,N); % 1 W sinewave + noise\nw= hanning(nfft); % window function\nwindow= w.*sqrt(nfft/sum(w.^2)); % normalize window for P= 1 W\n%\n% Create matrix xMatrix with Navg columns,\n% each column a segment of x of length nfft\nxMatrix= reshape(x,nfft,Navg);\nmagsq_sum= zeros(nfft/2);\nfor i= 1:Navg\nx_column= xMatrix(:,i);\nxw= x_column.*window; % apply window of length nfft\nX= fft(xw); % DFT\nX= X(1:nfft/2); % retain samples from 0 to fs/2\nmagsq= real(X).^2 + imag(X).^2; % DFT magnitude squared\nmagsq_sum= magsq_sum + magsq; % sum of DFT mag squared\nend\nmag_sq_avg= magsq_sum/Navg; % average of DFT mag squared\nP_bin= 2/nfft.^2 *mag_sq_avg; % W/bin power spectrum\nP_Hz= P_bin*nfft/fs; % W/Hz power spectrum\nPdB_bin= 10*log10(P_bin); % dBW/bin\nPdB_Hz= 10*log10(P_Hz); % dBW/Hz\nk= 0:nfft/2 -1; % frequency index\nf= k*fs/nfft; % Hz frequency vector\n%\nplot(f,PdB_bin),grid\naxis([0 fs/2 -60 10])\nxlabel('Hz'),ylabel('dBW/bin')\ntitle('Averaged Spectrum with amplitude units of dBW/bin')\nfigure\nplot(f,PdB_Hz),grid\naxis([0 fs/2 -60 10])\nxlabel('Hz'),ylabel('dBW/Hz')\ntitle('Averaged Spectrum with amplitude units of dBW/Hz')\n\n\n## References\n\n1.  Robertson, Neil, “Evaluate Window Functions for the Discrete Fourier Transform”  https://www.dsprelated.com/showarticle/1211.php\n2. Mathworks website   https://www.mathworks.com/help/signal/ref/pwelch.html\n3. Oppenheim, Alan V. and Shafer, Ronald W., Discrete-Time Signal Processing, Prentice Hall, 1989, pp. 737—742.\n4. Welch, Peter D., “The Use of Fast Fourier Transform for the Estimation of Power Spectra:  A Method Based on Time Averaging Over Short, Modified Periodograms”, IEEE Trans. Audio and Electroacoustics, vol AU-15, pp. 70 – 73, June 1967.  https://www.utd.edu/~cpb021000/EE%204361/Great%20DSP%20Papers/Welchs%20Periodogram.pdf\n5. Lyons, Richard G., Understanding Digital Signal Processing, 2nd Ed., Prentice Hall, 2004, section 3.8.\n6. Oppenheim and Shafer, p. 738.\n7. Robertson, Neil, “The Power Spectrum”, https://www.dsprelated.com/showarticle/1004.php\n8. Lyons, section 11.5.\n\nNeil Robertson     January, 2019\n\nPrevious post by Neil Robertson:\nEvaluate Window Functions for the Discrete Fourier Transform\nNext post by Neil Robertson:\nCompute the Frequency Response of a Multistage Decimator\n\n[ - ]\nComment by April 2, 2019", null, "Thanks Neil for the very informative article. I'm just wondering at the end where you calculate the SNR, should you have taken the processing gain into account. The noise floor as shown in figure 6 is lowered by the processing gain?\n\n[ - ]\nComment by April 2, 2019", null, "Hi gartlant,\n\nIn figure 6, the left plot has amplitude units of dBW/bin, and the right plot has amplitude units of dBW/Hz.  This is the reason for the difference in noise floor (and difference in carrier level as well).\n\nRegarding processing gain due to DFT averaging:  If you look at the un-averaged spectrum in Figure 2, and you try to put a marker on the noise, you can get a 20 dB variation of the noise reading depending on exactly where you place the marker.  In the averaged spectrum of figure 5 and 6, we have less variation in the noise floor, so maybe only 6 dB variation.  You can say processing gain is inherent in the plotted spectrum.\n\nregards,\n\nNeil\n\n[ - ]\nComment by July 25, 2019", null, "Hi Neil, thank you for the brilliant article. I am new to DSP and Matlab and will be using both to analyse long term acoustic data. Can you explain/point me to any resources for why I would/wouldn't use DFT averaging? I want to apply pwelch using a Hanning window with 50% overlap, in 1Hz frequency bins and ultimately transform the results (using 10log10?) to get values in db re 1 micro Pascal. How does this impact the nfft if my sampling rate is 144kHz? Do the two even relate? Thanks for any help you can offer,\n\n[ - ]\nComment by July 25, 2019", null, "Hi,\n\nWhat is the maximum frequency of interest of your signal?\n\nregards,\n\nNeil\n\n[ - ]\nComment by October 15, 2019", null, "Hallo Neil,\n\nThank you for sharing knowledge in DSP, it is very helpful. I'm learning a lot, by I still need your help.\n\nI'm using a GNSS receiver for some measurements and some times the results are\nnot what I inspected. Researches in the Internet tell me that by plotting the power spetrum (density) of the signal I can see if there are some interferences or any others signals that can disturb the receiver.\n\nFor GNSS receivers, a Interference threshold mask is defined by ICAO in the GNSS SARPs (Section 3.7 of appendix B of\nICAO Annex 10 to the Chicago Convention, amendment 76, Nov. 2001), and helps to identify interferences, or others unwanted signals.\n\nTo fullfill this requirements I assume that I need to plot the power spectrum density from noise floor of the signal that is  approx. -164.5 dBW.\n\nData:\nI/Q samples\nNumber of Samples 2000\nSampling frequency 60 MHz\nFrequency(Center) 1584 MHz\n\nMy questions:\n1) Is my thinking correct?\n2) can I plot this power spectrum density with I/Q Samples?\n3) How or references?\n\nIf more data are needed, please let me know, I will try provide these data.\n\nThank you against for your help.\n[ - ]\nComment by October 15, 2019", null, "Hi,\n\nYou can look at the spectrum of the I or Q channel using pwelch.  You should use 2^N samples, for example 2048.  With 2048 samples, the frequency resolution is 60E6/2048 = 29.3 kHz.\n\nThe receiver converts the signal centered at 1584 MHz to baseband I/Q.  The baseband frequency range is 0 to fs/2 = 0 to 30 MHz.\n\nYou can also find the spectrum of the complex baseband signal I + jQ, although I did not discuss that in the post.\n\nregards,\n\nNeil\n\n[ - ]\nComment by October 16, 2019", null, "Hi Neil,\n\nDank you.\n\nI was not clear enough.\n\nWhat are the steps to estimate self noise (noise floor), if there is one?\n\nBest regards\n\nLevi\n\n[ - ]\nComment by October 16, 2019", null, "Levi,\n\nThere is not a short answer to your question.  Here are a couple of references that may apply:\n\nregards,\n\nNeil\n\n[ - ]\nComment by February 17, 2020", null, "Hi, thanks for the great article. I'm trying to calculate the SNR of my ECG signal. But I do not have a characteristic peak that I'm looking for. Instead, I'm using a frequency range (everything less than 40 Hz) and summing over that frequency range, just as you did with the additional 5 samples around your peak. My sampling rate is 2000.\n\nI'm having an issue in determining how I should calculate the noise. I thought that I should use a range of of frequencies from the dB/Hz spectrum, instead of choosing a single point, as you showed. But when I do use that range of frequencies (800-900 Hz), the SNR values are unrealistically high. However, if I use the same range from the dB/bin spectrum (which is not what you mentioned in the article), then the SNR value is reasonable. Below is my code:\n\nnpoints = 2000;\n\nnfft = npoints;\n\nwindow = hanning(nfft);\n\nnoverlap = floor(npoints*0.50);\n\ncrit_f = 40;\n\nnoise_f1 = 800;\n\nnoise_f2 = 900;\n\n[pxx,f] = pwelch(ecgnostretch,window,noverlap,nfft,Fs);\n\nPdB_bin = 10*log10(pxx*Fs/nfft);\n\nfigure\n\nplot(f,PdB_bin)\n\nind = find(f==crit_f);\n\nPsum = Fs/nfft * sum(pxx(1:ind));\n\nS = 10*log10(Psum);\n\nPdB_Hz = 10*log10(pxx);\n\n% k1 = round(noise_f/Fs*nfft);\n\nind1 = find(f==noise_f1);\n\nind2 = find(f==noise_f2);\n\nNo = Fs/nfft * mean(pxx(ind1:ind2)); %for noise, we should average, that's why use mean instead of sum\n\nSNR_ecgnostretch = S-No;\n\nCould you offer any suggestions as to if this is an accurate way to calculate the SNR?\n\n[ - ]\nComment by February 17, 2020", null, "Hi fazeem,\n\nI think you have a mistake in computing No: it would need to be a dB quantity.  I am not familiar with ECG; however, here is an example of SNR calculation.  Note that SNR and S/No are two different things.  Since you say you are interested in SNR, that's what I calculate here.  The result is SNR_dB = 40.\n\nregards,\n\nNeil\n\n%snr1.m   2/17/20 nr\n\n% compute SNR of sine + gaussian noise\n\n% generate signal\n\nfs= 4000;\n\nTs= 1/fs;\n\nf0= 500;\n\nA= sqrt(2);\n\nN= 8192;\n\nn= 0:N-1;\n\n% signal with SNR of 40 dB  (20 log10(.01) = -40 dBW)\n\nx= A*sin(2*pi*f0*n*Ts) + .01*randn(1,N);\n\nnfft= 2048;\n\nnoverlap= nfft/2;\n\nwindow= hanning(nfft);\n\n[p,f]= pwelch(x,window,noverlap,nfft,fs);\n\ndBW_bin= 10*log10(p*fs/nfft);\n\n% sum noise, excluding carrier neighborhood\n\nsump= sum(p(1:252)) + sum(p(262:end));\n\npnoise_bin= sump*fs/nfft;   % total noise power in W\n\nN= 10*log10(pnoise_bin);    % dBW\n\nSNR_dB= 0 - N           %  note: signal power is 0 dBW\n\nplot(f,dBW_bin),grid\n\naxis([0 fs/2 -90 10])\n\nxlabel('Hz'),ylabel('dBW/bin')", null, "[ - ]\nComment by February 17, 2020", null, "Thank you very much for your reply. For ECG, I don't have any characteristic peak, so I'm just using the frequency range between 0-40 Hz as that is the relevant frequency range for this signal.\n\nIt seems that, for the noise, I need to select my entire noise range as that outside my frequency range of interest, i.e. everything above 40 Hz. If I do that, I still get a negative SNR, which is probably due to some of the noise values being above 0 dB.  I think this is due to my data acquisition system as all my PSD plots show similar behavior. If I pick a frequency range which is for sure in the noise floor (i.e. 800 to 1000 Hz), I can get an SNR of about 1, which doesn't make sense just by looking at the raw signal.\n\nAttached is an example of the PSD plot (dbW/bin on the y axis, frequency on x) I'm getting along with the code. Below that is a picture of one of the signals.", null, "", null, "Also, why is 0 used as the power of the signal?\n\n[ - ]\nComment by February 18, 2020", null, "Fazeem,\n\nThe power of the signal in my example is 1 watt = 0 dBW.\n\nLooking at your signal spectrum, I guess it makes sense to sum the power/bin from 0 to 40 Hz.  One way to find the noise power would be to use a signal with the transducer not attached to a person (or attached to a dead person), then just sum the noise spectrum.\n\nIf that is not possible, you could assume that the noise floor is flat.  Then you could find No near 800 Hz or so, where No = 10*log10(pxx) at a single point (note pxx is in W/Hz).  You could average No over a few nearby points if desired.  Then total noise would be No + 10*log10(fs/2) dBW.\n\nAgain, I don't know anything about ECG. I am not responsible for the use or misuse of this advice.\n\nregards,\n\nNeil\n\n[ - ]\nComment by February 18, 2020", null, "Thank you again for the response. I think what you've mentioned is definitely logical and applies to my case actually. I've reviewed some literature and this seems to be a plausible strategy. Thank you for the comments.\n\n[ - ]\nComment by February 19, 2020", null, "Hello ,\n\nI would like to ask you what are the best way to calculate the parameters used in pwelch function.\n\nI have a signal length of N= 5678000 and a sampling rate of fs= 1000.\n\nThen I divide this signal in 3 depending on the onset and offset of a stimulus. For example N1= 1126535, N2=951465, N3=3600000.\n\nI'm using the pwelch function in matlab but I don know what are the best parameters to choose for the window, the noverlap and the nfft.\n\nI'm using now :\n\n[pxx1,f1] = pwelch(BigMatPre(chs(e),:),2000,1000,2000,fs);\n\n(BigMatPre is my signal)\n\nAll my best,\n\nsarah\n\n[ - ]\nComment by February 19, 2020", null, "Sarah,\n\nIn my post, I use 5 inputs to pwelch, but you are showing 6 inputs, so I'm a little confused.\n\nFirst of all, the computation will be faster if you choose nfft as a power of 2:  you could use 2048 instead of 2000, although then the frequency step would not be a simple fraction.\n\nFrequency resolution depends on nfft:  For fs= 1000 and nfft = 2000, you have a resolution of fs/nfft = 0.5 Hz per bin.  Higher values of nfft give better resolution at the expense of less spectrum averaging.  So you may want to try different values of nfft to get a feeling for the trade-off.  For example, you could try nfft = 4000 (or 4096).\n\nThe choice of noverlap = nfft/2 is probably OK.\n\nFor a window, you could start with a Hanning window.  See my post discussing window functions:\n\nhttps://www.dsprelated.com/showarticle/1211.php\n\nregards,\n\nNeil\n\n[ - ]\nComment by February 19, 2020", null, "BigMatPre(chs(e),:) is because is a loop analysing the pwelch for e row.\n\nI have an other question if you can help me.\n\nI then want to compare the PSD of different groups, for example between the frequency range [40-80] but I m confuised. Is it good to compare the mean of a normalize PSD (mean of PSD value between 40-80 / mean of the PSD along [0.5-80] or the area under the curve?\n\nthak you\n\n[ - ]\nComment by February 19, 2020", null, "Sorry, I don't understand the question.\n\nNeil\n\n[ - ]\nComment by February 20, 2020", null, "Oh sorry,\n\nI mean how can you compare the PSD of two groups? By a numeric way not only the shape of the PSD.\n\nI was thinking to take the PSD values for a particular frequency range and compare statistically. Ex: the mean PSD power of 3 individual from a group 1 and the mean PSD power of 3 individual from group 2\n\n[ - ]\nComment by February 20, 2020", null, "Sarah,\n\nNow you are getting beyond my knowledge base!  This link might conceivably be of some help:\n\nregards,\n\nNeil\n\n[ - ]\nComment by February 19, 2020", null, "Sarah,\n\nIn your code, I think there may be a mistake in your call to pwelch.  The 2nd input is \"2000\".  However, the 2nd input should be a window function, for example hanning(2000).\n\nregards,\n\nNeil\n\n[ - ]\nComment by February 20, 2020", null, "Yes I change the values to a choose the nfft as a power of 2 and I tried hanning and hamming windows\n\nThank you !!!\n\n[ - ]\nComment by June 29, 2020", null, "Hello,\n\none of the things is not that clear:\n\npxx,f]= pwelch(x,window,noverlap,nfft,fs);  % W/Hz  PSD\n\nPdB_bin= 10*log10(pxx*fs/nfft);              % dBW/bin\n\nBut pwelch return a RMS value. To go to the relative peak power we should add 3dB. Is that right?\n\nthanks\n\n[ - ]\nComment by June 29, 2020", null, "Deerfield,\n\nYes, pxx is average power density.\n\nIf the signal is a sinewave, you are correct.  You would add 3 dB (actually 10*log10(2) = 3.01 dB) to PdB_bin to get the peak power.\n\nregards,\n\nNeil\n\n[ - ]\nComment by June 30, 2020", null, "Thanks Neirober, understood. What about th esignal is a complex modulated signal (QPSK) or even a dual tone?\n\nthanks,\n\nchristian\n\n[ - ]\nComment by June 30, 2020", null, "Christian,\n\nI assume by \"complex modulated signal (QPSK)\" you mean the real part of\n\n(I + jQ)e^(jwt).\n\nIn general, the average power of a spectrum is\n\n$$P_{av}= 1/N^2 \\sum|X(k)|^2$$\n\nThis applies to any real signal.  See my post on the power spectrum at https://www.dsprelated.com/showarticle/1004.php\n\nregards,\n\nNeil\n\n[ - ]\nComment by July 1, 2020", null, "Hello Neil,\n\nthank you very much for this clarification. Coming back to the 3dB that i need to add to move from power average to peak power for a sine, does this apply also in this case of QPSK or dual tone? Should i add 3dB in any case when i use pwelch? This is still not really clear to me.\n\nthanks a lot\n\n[ - ]\nComment by July 1, 2020", null, "Christian,\n\nThe reason you add 3 dB for a sine is (of course) that the peak power of a sinewave is 3 dB above the average power.  This is not the case for the other waveforms.\n\nThe easiest way to find peak power of the other waveforms would be to find the peak voltage of the time signal.  Then instantaneous peak power is just Vpeak^2/R = Vpeak^2 for R = 1 ohm.\n\nIn most cases in engineering, we are interested in average power and not peak power.  For example, SNR is based on average power.\n\nFinally, note that there is a difference between instantaneous peak power and the peak envelope power (PEP) of a modulated signal.  Peak envelope power is the average power of the modulating sinewave at the peak of the modulation envelope.\n\nregards,\n\nNeil\n\n[ - ]\nComment by July 5, 2020", null, "Hi Neil,\n\nthank you very much. to summarize, for singletone/dual tone i use the pwelch function with this function to convert rms power to dBFS peak power:\n\nfunction power_dBFS = rmsPower2peakPower_dBFS(varargin)\n\ninput_parameter = inputParser;\n\nparse(input_parameter, varargin{:});\n\npower_dBFS = 10*log10(input_parameter.Results.rms_power_W_Hz * input_parameter.Results.RBW_Hz);\n\nif(strcmp(input_parameter.Results.rms_peak, 'peak'))\n\npower_dBFS = power_dBFS + 3;\n\nend\n\nend\n\nFor more complex modulation, i cannot use this function.\n\nthanks,\n\nchristian\n\nTo post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments." ]

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[ "# Queries to count subarrays consisting of given integer as the last element\n\n• Last Updated : 19 Jan, 2022\n\nGiven an array arr[] and an array query[] consisting of Q queries, the task for every ith query is to count the number of subarrays having query[i] as the last element.\nNote: Subarrays will be considered to be different for different occurrences of X.\n\nExamples:\n\nInput: arr[] = {1, 5, 4, 5, 6}, Q=3, query[]={1, 4, 5}\nOutput: 1 3 6\nExplanation:\nQuery 1: Subarrays having 1 as the last element is {1}\nQuery 2: Subarrays having 4 as the last element are {4}, {5, 4}, {1, 5, 4}. Therefore, the count is 3.\nQuery 3: Subarrays having 5 as the last element are {1, 5}, {5}, {1, 5, 4, 5}, {5}, {4, 5}, {5, 4, 5}. Therefore, the count is 6.\n.\nInput: arr[] = {1, 2, 3, 3}, Q = 3, query[]={3, 1, 2}\nOutput: 7 1 2\n\nNaive Approach: The simplest approach to solve the problem is to generate all the subarrays for each query and for each subarray, check if it consists of X as the last element.\nTime Complexity: O(Q×N3)\nAuxiliary Space: O(1)\n\nEfficient Approach: To optimize the above approach, the idea is to use Hashing. Traverse the array and for every array element arr[i], search for its occurrence in the array. For every index, say idx, in which arr[i] is found, add (idx+1) to the count of subarrays having arr[i] as the last element.\n\nFollow the steps below to solve the problem:\n\n• Initialize an unordered map, say mp, to store the number of subarrays having X as the last element.\n• Traverse the array and for every integer arr[i], increase mp[arr[i]] by (i+1).\n• Traverse the array query[] and for each query query[i], print mp[query[i]].\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to perform queries to count the``// number of subarrays having given numbers``// as the last integer``int` `subarraysEndingWithX(``    ``int` `arr[], ``int` `N,``    ``int` `query[], ``int` `Q)``{``    ``// Stores the number of subarrays having``    ``// x as the last element``    ``unordered_map<``int``, ``int``> mp;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Stores current array element``        ``int` `val = arr[i];` `        ``// Add contribution of subarrays``        ``// having arr[i] as last element``        ``mp[val] += (i + 1);``    ``}` `    ``// Traverse the array of queries``    ``for` `(``int` `i = 0; i < Q; i++) {` `        ``int` `q = query[i];` `        ``// Print the count of subarrays``        ``cout << mp[q] << ``\" \"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 1, 5, 4, 5, 6 };` `    ``// Number of queries``    ``int` `Q = 3;` `    ``// Array of queries``    ``int` `query[] = { 1, 4, 5 };` `    ``// Size of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``subarraysEndingWithX(arr, N, query, Q);` `    ``return` `0;``}`\n\n## Java\n\n `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to perform queries to count the``// number of subarrays having given numbers``// as the last integer``static` `void` `subarraysEndingWithX(``    ``int` `arr[], ``int` `N,``    ``int` `query[], ``int` `Q)``{``    ``// Stores the number of subarrays having``    ``// x as the last element``    ``HashMap mp = ``new` `HashMap();` `    ``// Traverse the array``    ``for` `(``int` `i = ``0``; i < N; i++) {` `        ``// Stores current array element``        ``int` `val = arr[i];` `        ``// Add contribution of subarrays``        ``// having arr[i] as last element``        ``if``(mp.containsKey(val))``            ``mp.put(val, mp.get(val)+(i+``1``));``        ``else``            ``mp.put(val, i+``1``);``    ``}` `    ``// Traverse the array of queries``    ``for` `(``int` `i = ``0``; i < Q; i++) {` `        ``int` `q = query[i];` `        ``// Print the count of subarrays``        ``System.out.print(mp.get(q)+ ``\" \"``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// Given array``    ``int` `arr[] = { ``1``, ``5``, ``4``, ``5``, ``6` `};` `    ``// Number of queries``    ``int` `Q = ``3``;` `    ``// Array of queries``    ``int` `query[] = { ``1``, ``4``, ``5` `};` `    ``// Size of the array``    ``int` `N = arr.length;` `    ``subarraysEndingWithX(arr, N, query, Q);` `}``}` `// This code is contributed by 29AjayKumar`\n\n## Python3\n\n `# Python 3 program for the above approach` `# Function to perform queries to count the``# number of subarrays having given numbers``# as the last integer``def` `subarraysEndingWithX(arr, N, query, Q):``  ` `    ``# Stores the number of subarrays having``    ``# x as the last element``    ``mp ``=` `{}` `    ``# Traverse the array``    ``for` `i ``in` `range``(N):``      ` `        ``# Stores current array element``        ``val ``=` `arr[i]` `        ``# Add contribution of subarrays``        ``# having arr[i] as last element``        ``if` `val ``in` `mp:``            ``mp[val] ``+``=` `(i ``+` `1``)``        ``else``:``            ``mp[val] ``=` `mp.get(val, ``0``) ``+` `(i ``+` `1``);` `    ``# Traverse the array of queries``    ``for` `i ``in` `range``(Q):``        ``q ``=` `query[i]` `        ``# Print the count of subarrays``        ``print``(mp[q],end ``=` `\" \"``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Given array``    ``arr ``=`  `[``1``, ``5``, ``4``, ``5``, ``6``]``    ` `    ``# Number of queries``    ``Q ``=` `3``    ` `    ``# Array of queries``    ``query  ``=` `[``1``, ``4``, ``5``]``    ` `    ``# Size of the array``    ``N ``=` `len``(arr)``    ``subarraysEndingWithX(arr, N, query, Q)``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`\n\n## C#\n\n `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG``{` `    ``// Function to perform queries to count the``    ``// number of subarrays having given numbers``    ``// as the last integer``    ``static` `void` `subarraysEndingWithX(``int``[] arr, ``int` `N,``                                     ``int``[] query, ``int` `Q)``    ``{``      ` `        ``// Stores the number of subarrays having``        ``// x as the last element``        ``Dictionary<``int``, ``int``> mp``            ``= ``new` `Dictionary<``int``, ``int``>();` `        ``// Traverse the array``        ``for` `(``int` `i = 0; i < N; i++) {` `            ``// Stores current array element``            ``int` `val = arr[i];` `            ``// Add contribution of subarrays``            ``// having arr[i] as last element``            ``if` `(mp.ContainsKey(val))``                ``mp[val] = mp[val] + (i + 1);``            ``else``                ``mp[val] = i + 1;``        ``}` `        ``// Traverse the array of queries``        ``for` `(``int` `i = 0; i < Q; i++)``        ``{``            ``int` `q = query[i];` `            ``// Print the count of subarrays``            ``Console.Write(mp[q] + ``\" \"``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``      ` `        ``// Given array``        ``int``[] arr = { 1, 5, 4, 5, 6 };` `        ``// Number of queries``        ``int` `Q = 3;` `        ``// Array of queries``        ``int``[] query = { 1, 4, 5 };` `        ``// Size of the array``        ``int` `N = arr.Length;``        ``subarraysEndingWithX(arr, N, query, Q);``    ``}``}` `// This code is contributed by chitranayal.`\n\n## Javascript\n\n `         `\n\nOutput:\n\n`1 3 6`\n\nTime Complexity: O(Q + N)\nAuxiliary Space:O(N)\n\nMy Personal Notes arrow_drop_up" ]

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[ "# es.para.d: es.para.d In es.dif: Compute Effect Sizes of the Difference\n\n## Description\n\nCalculates the effect size of the difference, its variance and confidence interval from statistics of two datasets under the variance equality. Default function returns Hedges' d (unbiased=TRUE). When unbiased=FALSE, it returns Cohen's d. The confidence interval is given by binary search. Therefore, the confidence interval has rounding error, and alpha with many digits will need very long computation time. In some situations, the confidence interval may be inaccurate, and the warning message appears because of the limitation of the function pt(), which is used for the calculation.\n\n## Usage\n\n `1` ```es.para.d(mean1,mean2,var1,var2,n1,n2,alpha=0.05,unbiased=TRUE,vector_out=FALSE) ```\n\n## Arguments\n\n `mean1` Mean of dataset 1. `mean2` Mean of dataset 2. `var1` Unbiased (divided by n1-1) variance of dataset 1. `var2` Unbiased (divided by n2-1) variance of dataset 2. `n1` Sample size of dataset 1. `n2` Sample size of dataset 2. `alpha` Confidence level, or type I error rate for the confidence interval. The default value gives 95% CI. `unbiased` When true (default), the bias corrected value is returned. When false, the value based on the whole population sample is returned. `vector_out` Whether output is in vector or not. When in vector, it is c(effect size, the variance, lower CI bound, higher CI bound).\n\n## References\n\nAoki S. (2020) Effect sizes of the differences between means without assuming the variance equality and between a mean and a constant. Heliyon 6(1): e03306.\n\nAoki S., Ito M. & Shimada M. (2019) Effect sizes of the differences between means without assuming the variance equality and between a mean and a constant. arXiv:1901.09581.\n\nCohen, J. (1988). Statistical Power Analysis for the Behavioral Sciences, 2nd edition. New York: Academic Press.\n\nHedges, L. V. (1981). Distribution theory for Glass's estimator of effect size and related estimators. Journal of Educational Statistics 6, 107–128.\n\nHedges, L. V. and Olkin I. (1985). Statistical Methods for Meta-analysis. Orlando: Academic Press.\n\n## Examples\n\n `1` ```es.para.d(2,3,1,2,4,3) ```\n\nes.dif documentation built on May 2, 2020, 9:05 a.m." ]

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[ "float has 23 mantissa bits + 1 hidden bit: log(2 24)÷log(10) = 7.22 digits. Software Development Forum . Here, we will learn about float and double data types in java? Difference between Slack and Float. Float. Home. Reported In shows products that are verified to work for the solution described in this article. Software. Float is a datatype which is used to represent the floating point numbers. Transact-SQL Syntax Conventions. Noun A mason’s tool, used in spreading and dressing mortar, and breaking bricks to shape them. C# language specification. The reason is that floating-point values and integers are handled differently inside the computer. According to IEEE, it has a 64-bit floating point precision. CONTENTS. The main difference between Float and Double is that the former is the single precision (32-bit) floating point data, while the latter is double precision (64-bit) floating point data type. Creating Double-Precision Data long double: Real floating-point type, usually mapped to an extended precision floating-point number format. English . There are eight primitive datatypes supported by Java. Float floatObject = new Float(f); Initially primitives were … Ok, so when we talk about Slack, we talk about inactivity while Float is about activity. If you have good sky conditions, you only have to wait for the algorithm to resolve, which usually takes between 30-90 seconds. These terms are often used interchangeable mainly because they are super similar. Decimal’s declaration and functioning is similar to Double. But there is one big difference between floating point values and decimal (numeric) values. Summary. As verbs the difference between trowel and float is that trowel is to apply a subtsance with a trowel while float is of an object or substance, to be supported by a liquid of greater density than the object so as that part of the object or substance remains above the surface. The Difference between Float Glass and Sheet Glass. The widening occurs in a byte, short, int, long, float, double order. The terms “floatplane” and “seaplane” are used interchangeably in some countries, but technically have different meanings. The key difference between int and long is that int is 32 bits in width while long is 64 bits in width. The key difference between float and double is that float is a single precision 32 bit IEEE 754 floating point data type while double is a ... a smaller data type to a larger data type, there is no casting required. For example, there are approximately 8,388,607 single-precision numbers between 1.0 and 2.0, while there are only about 8191 between 1023.0 and 1024.0. Both a floatplane and a seaplane can take-off from, and land on, water such as oceans, seas, rivers, and gulfs. 8 bytes. For example: part of univar output: Mean S.D. The default value of n is 53. For representing floating point numbers, we use float, double and long double.. What’s the difference ? It is a 32-bit IEEE 754 single precision floating point number ( 1-bit for the sign, 8-bit for exponent, 23*-bit for the value. This article discusses the difference between int and long. Similarities Between int and long 5. Float is an object; float is a primitive. float f=1.0f; Float floatObject = f; or explicitly. There are significant differences between float and double, although both are floating point numbers ! LabWindows/CVI 6.0 Full LabVIEW Base Issue Details I am considering using either float datatype or … The Real Difference between Integers and Floating-Point Values. Variable is the name given to a location that stores data. Format float variable to long 10 Jul 2018, 09:42. In Java, the float and double data types store same value for x.0, x.5 but it stores different values for x.1, x.2, x.3, x.4, x.6, x.7, x.8, and x.9 where x is any integer number. By Dan Gookin . It is melted at high temperature in a melting furnace. The basic difference between the type int and long is of their width where int is 32 bit, and long is 64 bits. 2008macedonkon3 0 Newbie Poster . Float glass: It is made of sea sand, quartz sandstone powder, soda, dolomite and other raw materials in a certain proportion. Both can transport people … double has 2x more precision then float.. float is a 32 bit IEEE 754 single precision Floating Point Number1 bit for the sign, (8 bits for the exponent, and 23* for the value), i.e. The ISO synonym for real is float(24). Overview and Key Difference 2. Submitted by Preeti Jain, on January 31, 2018 1) float data type in java. Difference between float and double in C/C++ C C++ Server Side Programming Programming As we know that in C/C++ we require float and double data type for the representation of Floating point numbers i.e the numbers which have decimal part with them.Now on the basis of precision provided by both of these data types we can differentiate between both of them. It has 6 decimal digits of precision. Side by Side Comparison – int vs long in Tabular Form 6. Different behaviors of storing the same value using float and double data type. Float takes 4 bytes for storage. What is long 4. float [ (n)] Where n is the number of bits that are used to store the mantissa of the float number in scientific notation and, therefore, dictates the precision and storage size. What is the Precision Difference Between Float and Double Datatypes? This solution might also apply to other similar products or applications. Double is called “double” because it’s basically a double precision version of Float. int * Int data type is a 32-bit signed two's complement integer. What is int? We use DECIMAL data type to store exact numeric values, where we do not want precision but exact and accurate values. float double; Size: 4 bytes: Size: 8 bytes: Precision: In general, 7 decimal digits precision: Precision: In general, 15 decimal digits precision: Example: 3.56f, 3e5f etc. Same relationship as Integer and int, Double and double, Long and long. On any computer, mathematically equivalent expressions can produce different values using floating-point arithmetic. Primitive datatypes are predefined by the language and named by a keyword. However, you can convert any floating-point type to any other floating-point type with the explicit cast. 1) float is a data type (or in another words it is a keyword which has special meaning) in java. Actual properties unspecified. Creating Floating-Point Data. Difference between Decimal, Float and Double. For more information, see Built-in numeric conversions. Float stands for Floating constant, meaning, it is any signed or unsigned numbers that is represented having decimal point. Here's how the number of digits are calculated: double has 52 mantissa bits + 1 hidden bit: log(2 53)÷log(10) = 15.95 digits. The n value will determine if the value is kept as float or changed to double. Compare float and double in java. According to IEEE, it has a 32-bit floating point precision. In terms of data types, an integer belongs to a set of mathematical integers whose value is the same as a corresponding mathematical integer. Difference between float and double . When assigning a larger data type to a small data type, it is necessary to do the casting. So Decimals have much higher precision and are usually used within monetary (financial) applications that require a high degree of accuracy. Reported In. The float value is a 32-bit number with floating point , the variable double is 64-bit or 80-bit long and therefore more precise. Languages such as C++ use variables in the program. Qingdao Creation Classic Glass Co.,Ltd | Updated: Feb 21, 2019. trowel . That is how long an activity can be delayed without having some kind of impact. Double takes 8 bytes for storage. Programming Forum . A variable is a name given to a memory location that stores data. The numbers are very large, up to the hundreds of millions, but when I use the univar or sum command, i'm losing precision on my 5 number summary. Syntax. The main difference between double and long double is that double is used to represent a double precision floating point while long precision is used to represent extended precision floating point value.. 11 Years Ago. It has the double precision or you can say two times more precision than float. In case the UNI-GR1 falls back to a Float status during measurements, it usually indicates sky conditions get more challenging or have simply changed, because you walked around a corner. Huge difference. Floating slabs are concrete slabs that are laying over the ground, without any kind of anchoring, as if it simply sits on it and floats.The main application of floating slabs is to use as a base foundation for sheds, manufacturing workshops, home additional room, or garages.. or. A Decimal type can store a Maximum of 65 … As the name implies, a double has 2x the precision of float .In general a double has 15 decimal digits of precision, while float has 7.. The liquid glass flows continuously from the kiln to the liquid metal surface and floats on it. 1. If n is specified, it must be a value between 1 and 53. Example: +11.735 , -7.5 , +6.628 In current versions, float is expressed as float(n). FLOAT is accurate to approximately 7 decimal places, and DOUBLE upto 14. Hi y'all, I'm working with a variable that is a numerical but is formatted as a float. Precision differences. Updated Nov 10, 2017. The main difference is Floats and Doubles are binary floating point types and a Decimal will store the value as a floating decimal point type. long int (long) Long Integer. An integer exists inside the computer as a true binary value. There is only one implicit conversion between floating-point numeric types: from float to double. What is int 3. When writing programs, it is necessary to store data. Difference Between float and double. What Is a Floating Slab? Key Difference: Generally, Integers can be described as whole numbers meaning that they do not have any fractional parts, whereas float describes a number that can be only written in a decimal number system. 4bytes in size, with number range of, signed: -2147483648 to 2147483647 and unsigned: 0 to 4294967295. The types int and long when counted in bytes instead of bits the type int is 4 bytes and the type long is just twice if type int i.e. Use double-precision to store values greater than approximately 3.4 x 10 38 or less than approximately -3.4 x 10 38. Learn: Difference between float and double in java? For numbers that lie between these two limits, you can use either double- or single-precision, but single requires less memory. In the case of … 2) float takes 4 bytes(i.e. In programming, it is necessary to store data. Example: 3.56, 3e5 etc. The 7.475 billion floating shares are the shares considered for the free float, market capitalization index weightings, such as in the S&P 500. Discussion / Question . The main difference between long and double in Java is that long is a data type that stores 64 bit two’s complement integer while double is a data type that stores double prevision 64 bit IEEE 754 floating point. If a floating-point value can also be a whole number, why bother using integers in your programs at all? float can be converted to Float by autoboxing, e.g. Basically all of them represent the decimal values such as 3.14 The main difference between them is that in float we can store values upto 4 bytes (6 places after decimal point) Double upto 8 bytes And long double even more than float and double. Meaning ) in java meaning ) in java float value is a datatype which is used to represent floating... The widening occurs in a byte, short, int, double and long is that floating-point values integers... The kiln to the liquid metal surface and floats on difference between float and long value can also be a value 1...: Mean S.D relationship as integer and int, long, float double... Similar to double decimal places, and breaking bricks to shape them without having some kind of impact floatObject. Comparison – int vs long in Tabular Form 6 also be a value 1. Products or applications Preeti Jain, on January 31, 2018 1 ) float is expressed float. 1 and 53 a variable that is represented having decimal point degree of accuracy 8191. Seaplane ” are used interchangeably in some countries, but technically have meanings. Solution might also apply to other similar products or applications, used spreading. Byte, short, int, double order float is a numerical but is formatted as a float,! The ISO synonym for Real is float ( 24 ) type, it is necessary to the. Float stands for floating constant, meaning, it has a 32-bit floating numbers. Although both are floating point numbers, float is expressed as float changed... The program: 0 to 4294967295 double data types in java without having some kind of.. About activity x 10 38 or less than approximately 3.4 x 10.... C++ use variables in the case of … What is the name given to small! Usually used within monetary ( financial ) applications that require a high degree of accuracy require a high degree accuracy. While there are only about 8191 between 1023.0 and 1024.0 difference between float and long value between 1 and 53 number. Mason ’ s the difference between int and long is that int is 32 bits in width as!: difference between float and double data types in java do not want precision but exact accurate. About Slack, we use float, double order limits, you convert! To shape them floating-point type, usually mapped to an extended precision floating-point number.... Bits + 1 hidden bit: log ( 2 24 ) 32-bit number with floating point numbers representing!, so when we talk about inactivity while float is expressed as float ( 24 ) ÷log ( )... 1 ) float data type is a 32-bit number with floating point precision ” because it ’ s the?! Are only about 8191 between 1023.0 and 1024.0 ) = 7.22 digits ; float =... Melted at high temperature in a melting furnace the case of … What is the precision difference between float double!, you can use either double- or single-precision, but single requires less memory one implicit conversion floating-point! Determine if the value is a keyword it is any signed or unsigned that. Is represented having decimal point as integer and int, double order activity! Shape them y'all, I 'm working with a variable is a data type to memory... Has a 32-bit number with floating point values and decimal ( numeric ) values between and! One implicit conversion between floating-point numeric types: from float to double versions, float is a type... Has 23 mantissa bits + 1 hidden bit: log ( 2 24 ) for floating,... More precise the kiln to the liquid metal surface and floats on it high degree accuracy! Decimal ( numeric ) values type ( or in another words it is a keyword in this article the. Floating-Point number format handled differently inside the computer limits, you can convert any floating-point type to a location! Of impact by Preeti Jain, on January 31, 2018 1 ) float is a signed! Decimal ’ s tool, used in spreading and dressing mortar, long! Expressions can produce different values using floating-point arithmetic melted at high temperature in a melting furnace integers in programs... 1 ) float data type ( or in another words it is necessary to values... To IEEE, it has a 32-bit floating point numbers data type in java ( 2 24.. A high degree of accuracy variable is a keyword to an extended floating-point! Jul 2018, 09:42 long and long is that floating-point values and integers are differently... The same value using float and double upto 14 approximately 8,388,607 single-precision between... Declaration and functioning is similar to double products or difference between float and long datatype which is used to represent the floating numbers... Byte, short, int, long, float is accurate to 7. And “ seaplane ” are used interchangeably in some countries, but technically have different meanings 64-bit. Float f=1.0f ; float floatObject = f ; or explicitly to 4294967295 higher precision are. 0 to 4294967295 and dressing mortar, and double in java Jain, on January,... Memory location that stores data double precision version of float: 0 4294967295. ( 2 24 ) ÷log ( 10 ) = 7.22 digits is expressed as float or changed to double to! Double in java and long the casting temperature in a byte, short,,... ; or explicitly for numbers that lie between these two limits, you can convert any floating-point,! When assigning a larger data type to any other floating-point type to any floating-point. The type int and long double: Real floating-point type to any other floating-point type with the explicit cast S.D. Variable to long 10 Jul 2018, 09:42 in java Co., Ltd | Updated: Feb 21,.... Flows continuously from the kiln to the liquid metal surface and floats it. To a small data type ( or in another words it is signed... Functioning is similar to double different meanings on any computer, mathematically equivalent expressions can different! Has special meaning ) in java to 4294967295 and floats on it declaration and functioning is to! Tabular Form 6, signed: -2147483648 to 2147483647 and unsigned: to. Such as C++ use variables in the case of … What is the precision difference the! With number range of, signed: -2147483648 to 2147483647 and unsigned: 0 to 4294967295 bit, long!, 2019 it has a 32-bit number with floating point values and decimal ( numeric ) values a given. Precision floating-point number format output: Mean S.D a name given to a location that stores data is... Differences between float and double data types in java the case of … What the. ( financial ) applications that require a high degree of accuracy same using... Where int is 32 bits in width while long is that floating-point values and integers are handled differently the... Between these two limits, you can convert any floating-point type to a small data type to a memory that.: part of univar output: Mean S.D only about 8191 between 1023.0 and.... Article discusses the difference between int and long double: Real floating-point type to store exact numeric values, we! The liquid metal surface and floats on it int data type ( or in another words it is necessary store... And floats on it will learn about float and double data types in java are handled differently inside computer... Decimal data type, it has a 64-bit floating point numbers 24 ÷log! In a byte, short, int, long and therefore more.. Number, why bother using integers in your programs at all will determine if the is... 7.22 digits activity can be delayed without having some kind of impact ok, so we! Which is used to represent the floating point precision keyword which has special meaning ) in java the program having! Than approximately -3.4 x 10 38 functioning is similar to double formatted as a true binary value equivalent... Jul 2018, 09:42 Slack, we talk about inactivity while float is accurate to 7., short, difference between float and long, long, float, double and long is of their width where int 32! Be a value between 1 difference between float and long 53 precision but exact and accurate values 10 2018. Used to represent the floating point numbers, we will learn about float and double, although are. Upto 14 between int and long is of their width where int 32... Used within monetary ( financial ) applications that require a high degree of accuracy is similar double... ( numeric ) values where int is 32 bit, and double data type a. Double upto 14 some countries, but technically have different meanings float variable long! A melting furnace: log ( 2 24 ) ÷log ( 10 =. Value is kept as float difference between float and long changed to double or applications continuously from the kiln to liquid. 7 decimal places, and long is that floating-point values and integers are handled differently inside the computer as float. Type in java in the program memory location that stores data ok, so when we talk about Slack we. Exact and accurate values ok, so when we talk about inactivity float. However, you can use either double- or single-precision, but technically have different meanings x 10 38 furnace! Is 32 bit, and breaking bricks to difference between float and long them a double precision version float. Part of univar output: Mean S.D Comparison – int vs long in Form! Is formatted as a true binary value a name given to a location that stores data therefore more.... Output: Mean S.D either double- or single-precision, but single requires less.... Float, double order the value is kept as float or changed to...." ]

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[ "Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nScilab 5.3.0\nChange language to: English - Português - 日本語\n\nManuel Scilab >> xcos > Batch functions > steadycos\n\nFinds an equilibrium state of a general dynamical system described by a xcos diagram\n\n### Calling Sequence\n\n`[X,U,Y,XP]=steadycos(scs_m,X,U,Y,Indx,Indu,Indy [,Indxp [,param ] ])`\n\n### Description\n\nThis function finds the steady state for a given system described by a xcos diagram. The diagram consists in general of a Super block with input and output port blocks. The steady state concern only the continuous-time dynamics.\n\n`[X,U,Y,XP]=steadycos(scs_m,X,U,Y,Indx,Indu,Indy [,Indxp [,param ] ])`\n\n### Arguments\n\n• scs_m : a xcos data structure\n\n• X : column vector. Continuous state. Can be set to [] if zero.\n\n• U : column vector. Input. Can be set to [] if zero.\n\n• Y : column vector. Output. Can be set to [] if zero.\n\n• Indx : index of entries of X that are not fixed. If all can vary, set to 1:\\$\n\n• Indu : index of entries of U that are not fixed. If all can vary, set to 1:\\$\n\n• Indy : index of entries of Y that are not fixed. If all can vary, set to 1:\\$\n\n• Indxp : index of entries of XP (derivative of x) that need not be zero. If all can vary, set to 1:\\$. Default [].\n\n• param : list with two elements (default list(1.d-6,0)). param(1): scalar. Perturbation level for linearization; the following variation is used del([x;u])_i = param(1)+param(1)*1d-4*abs([x;u])_i. param(2): scalar. Time t.\n\n• X : steady state X\n\n• U : stationary input U\n\n• Y : output corresponding to steady state found\n\n• XP : derivative of the state corresponding to steady state found\n\n### Authors\n\nRamine Nikoukhah - INRIA" ]

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[ "# Get the Eye to Explore True Math\n\n### Beware Of Fraud Math\n\nRaghavendra'sAnalysis\n\nCurriculum Developers\n\nFastest Division (NEW Methods)\n\nMore About 911\n\nImportant Note\n\nRaghavendra's Analysis\n\nApplications & Advantage\n\nMany Difficult Tricks Vs One Easy Math\n\nFastest Conversions (All New)\n\nGames\n\nUpgrade Calculator Free\n\nFastest Conversions (Easy and Best)\n\nHexadecimal To Decimal (High Speed)\n\nDecimal To Binary (Without division)\n\nDecimal To Octal, Hexadecimal\n\nBinary To Decimal\n\nEasy formulas for Binary Octal Arithmetic operations(As Easy as Decimal Arithmetic)\n\nBlog\n\nGuidelines For Readers\n\nFAQs\n\nContact Author\n\nBook In Other Languages\n\n Most Surprising Division Now rapidly Divide any number by 9,99, 999, etc. just doing only addition calculations. Experience the joy of doing division without doing any divisions from Number System book. The book also contains amazing derivations to automatically get formulas.\n\n Sum Of Natural Numbers    Raghavendra's analysis is the one and only TRUE algebraic method to derive formulas for finding sums of: a + ar + ar2 + ar3 + ...+ arn-1 (GP) 1 + 2 + 3 + . . . + n 12 + 22 + 32 + . . . + n2   13 + 23 + 33 + . . . + n3  etc. So what method you want Learn?\n\n Can You Imagine This?     Can you imagine to do the following calculations rapidly just learning one method?    Derive formulas to find sums Geometric progression, sums of natural numbers, first n squares, cubes, sum of odd numbers, breaking one series into sums of other series, converting one series to another, deriving formulas for finding sums of different types of series, derive 5 more new methods for rapid converting decimal number to binary, octal, hexadecimal etc. and vice versa., developing new and easy methods for binary, octal, etc. arithmetic operations, develop new number system, arithmetic operations, developing methods to convert decimal number to binary, binary arithmetic operations in normal calculator, etc.    Difficult to believe that different problems of different chapters possible to be solved in one method? 'Raghavendra's Analysis' the new invention made possible all these very easily using only kids level mathematics calculations like addition, subtraction, multiplication and division of some small numbers. Easy Intelligent Math Or Difficult poor Math? Only Two choices! Change To New Or Follow the Failed Ways.\n\n Raghavendra's Analysis    List of Series Sums formulas Derived Using Raghavend's Analysis.Click Here.    List of new and fastest number system conversion methods invented (derived) using Raghavendra's analysis. Click Here.    Name of New Number System and Arithmetic operations invented using Raghavendra's analysis.\n\n Our Challenge    Presently students are learning answers getting neat tricks as mathematics. Because their teachers teaches that lessons without telling the them the truth.    Learning Raghavendra's analysis is must needed to study or teach derivations for formulas to find sums of Geometric progression,  sums of first n natural numbers, squares, cubes, etc.    So for those who wants to learn or teach true math, there is only one choice; that is Raghavendra's Analysis. Its Time for students to demand or learn Real and Quality math education. Aware Now\n\n How to write a program to convert any decimal number (+ve or -ve) to binary, octal and vice versa using only one formula?    Using 'Quotient method' a new formula invented using Raghavendra's analysis gives the algorithm to write the said program easily. The research book also contains extremely powerful formulas to easily develop several verities of useful software.\n\nMany Fraud Vs  One Real Mathematics\n\nLearning one real algebraic method is more useful than learning several answers getting tricks. Raghavendra's Analysis is the best proof for this.\n\nExcept Raghavendra's Analysis, There is no real algebraic derivations to derive formulas to find sums of several types of series like geometric series, sums of first n natural numbers, squares, cubes, etc.\n\nSo other than Raghavendra's Analysis it requires to use different and many answer getting tricks to solve different problems related to series. But if a student learn one derivation using Raghavendra's, then he can automatically derive 'n' number of formulas. Below are the advantages of learning real mathematics, i.e, Raghavendra's analysis.\n\nRAGHAVENDRA’S ANALYSIS\n\nVs MANY OTHER METHODS\n\n### Right Decision To Reach Right Destination In Right Time", null, "Higher Education\n\nSecondary Education\n\nHigh School\n\nElementary Education\n\nCompetitive Exams\n\nTeachers & Professors\n\nParents & Guardians\n\nResearch & Development\n\nComputer Education\n\nSoftware Development\n\nFuture Technology\n\nGovernment\n\nGift This Book\n\nBulk Orders\n\n Question? Which is the one and only Genuine method to derive formulas for finding sums of Geometric progression, first n natural numbers, squares, cubes, etc.?\n\n What a surprise? If many numbers are added to one number, the total not changes! How? Learn DCB(New) number system!\n\n Is it Possible to do binary arithmetic operations including fractions in a normal calculator?    Yes, using \"Add method\" even children can do these just like playing games.    Note 1 : It is not possible to do fractional parts calculation of binary arithmetic operations top scientific calculator; but \"Add Method\" makes this possible even in normal calculator.   Note 2 : \"Add Method\" is the new method introduced to math  from \"Number System\" text.\n\n Headache, Tension from Math Virus?    Get Freedom from lots of terror solutions, complex formulas and calculations just by learning one method Raghavendra's Analysis.\n\n Kids Vs College Students    Kids easily learned and solved Higher education level complex problems faster than University students. \"Raghavendra's analysis\" is the secret behind this shocking news.\n\n Kick Terror Math Before It Harms    Learning or By hearting the poor and  terror answer getting solutions makes heavy damages in the students mind resulting them to accept foundationless thoughts without thinking. Only Raghavendra's Analysis opens students eyes to explore the useful truth and secrets of the number system.\n\n This book consists of ample of invaluable information's which help the students to a great extent. This book aims primarily to creat intrest in students. Read More\n\n Preface     Number System: Automatic Algebraic Solutions is an innovative text that introduces high quality methods for easily learning various concepts of numner system.\n\nNumber System: Automatic Algebraic Solutions: is the must book for true and high quality education. No other book can provide better and fast solutions than this.\n\n### Are You Ready For The Next Generation Mathematics?", null, "Copyright © 2003 – 2010" ]

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[ "## Convex functions, second derivatives and Hessian matrices\n\nIn single variable calculus, a twice differentiable function $$f:(a,b)\\to\\mathbb{R}$$ is convex if and only if $$f^{\\prime\\prime}(x)\\ge 0$$ for all $$x\\in(a,b)$$. It is not too hard to extend this result to functions defined on more general spaces:\n\n## Differentiation done correctly: 5. Maxima and minima\n\nNavigation: 1. The derivative | 2. Higher derivatives | 3. Partial derivatives | 4. Inverse and implicit functions | 5. Maxima and minima\n\nIn this final post, we are going to look at some applications of differentiation to locating maxima and minima of real valued functions. In order to do this, we will be using Taylor’s theorem (covered in part 2) to prove the higher derivative test for functions on Banach spaces, and the implicit function theorem (covered in part 4) to prove a special case of the method of Lagrange multipliers.\n\n## Differentiation done correctly: 3. Partial derivatives\n\nNavigation: 1. The derivative | 2. Higher derivatives | 3. Partial derivatives | 4. Inverse and implicit functions | 5. Maxima and minima\n\nWhile we saw that differentiable maps may be naturally split into component functions when the codomain is a product of Banach spaces, the situation for the domain is more complicated. (This is partly due to the fact that as a topological space, there is no natural injection into a product of Banach spaces.) In this post, we will look at how the existence of partial derivatives relates to differentiability, how the symmetry of higher derivatives (covered in part 2) affects mixed partial derivatives, and finally a short proof of differentiation under the integral sign.\n\n## Differentiation done correctly: 2. Higher derivatives\n\nNavigation: 1. The derivative | 2. Higher derivatives | 3. Partial derivatives | 4. Inverse and implicit functions | 5. Maxima and minima\n\nLast time, we covered the definition of the derivative and its basic properties, which all turn out to be quite similar to their single variable counterparts. Now we are going to explore higher derivatives. In traditional multivariable calculus, true higher derivatives do not exist (except in a specific situation which will be discussed in part 5). Of course, we have so-called “mixed/higher partial derivatives”, which are coordinate-dependent and notationally tricky to work with. As a consequence, the usual statement of Taylor’s theorem in $$\\mathbb{R}^n$$ ends up being ugly and hard to remember. In reality, Taylor’s theorem for Banach spaces looks almost exactly the same as the single variable Taylor’s theorem!\n\n## Some series convergence problems\n\nHere are some series convergence problems that I gathered quite a while ago. A few of them are a bit tricky.\n\n## Every continuous open mapping of R into R is monotonic\n\nI’m willing to bet that most students who have used Rudin’s Principles of Mathematical Analysis have encountered this problem:\n\n15. Call a mapping of $$X$$ into $$Y$$ open if $$f(V)$$ is an open set in $$Y$$ whenever $$V$$ is an open set in $$X$$.\n\nProve that every continuous open mapping of $$\\mathbb{R}$$ into $$\\mathbb{R}$$ is monotonic.\n\n## Obvious?\n\nHere is one “solution” that is fairly intuitive. It relies on finding a minimum or maximum and considering the image of a small neighborhood around that min/max:\n\n## PAE patch updated for Windows 8\n\nNote: An updated version for Windows 8.1 is available.\n\nThis patch allows you to use more than 3/4GB of RAM on an x86 Windows system. Works on Vista, 7, 8, has been tested on Windows Vista SP2, Windows 7 SP0, Windows 7 SP1 and Windows 8 SP0. Instructions and source code included.\n\nBefore using this patch, make sure you have fully removed any other “RAM patches” you may have used. This patch does NOT enable test signing mode and does NOT add any watermarks.\n\nNote: I do not offer any support for this. If this did not work for you, either:\n\n• You cannot follow instructions correctly, or\n\n## Free product of free groups and group presentations\n\nHere is Problem 9-4(b) from Introduction to Topological Manifolds by John M. Lee:\n\nLet $$S_1$$ and $$S_2$$ be disjoint sets, and let $$R_i$$ be a subset of the free group $$F(S_i)$$ for $$i=1,2$$. Prove that $$\\langle S_1 \\cup S_2 \\mid R_1 \\cup R_2 \\rangle$$ is a presentation of the free product group $$\\langle S_1 \\mid R_1 \\rangle * \\langle S_2 \\mid R_2 \\rangle$$.\n\nThe proof is fairly straightforward if we stick to the universal properties of the free group, free product and quotient group. Here I will write out the full details of the proof, which is more fun to do than to read. There are three basic steps to showing that two things are isomorphic:\n\n1. Set up the canonical maps and choose names for them. These maps are given to you by the definition of the free group, free product, etc.\n2. Repeatedly use the existence part of the universal properties to derive two maps.\n3. Use the uniqueness part of the universal properties to show that the maps are inverses to each other." ]

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[ "", null, "## Standardize in Excel – Let’s compare apples with oranges\n\nCan you compare apples with oranges? Generally speaking, the answer is no.  However, if we use standard deviation as a measure, then we can.  Using standard deviation to compare data with different units is called standardizing.  By standardizing data, you are calculating the Z score.  Standardized values have no units and hence allow you to compare apples with oranges. In Excel, this is easily done using Standardize.\n\n### What is Z Score?\n\nZ scores allow us to compare datasets measured in different units.  Z score is how many standard deviations above or below average a data point is.\n\nFor any dataset the average Z score will be = 0 and the standard deviation of Z Scores =1.  Any Z score above 2 or below -2 is an outlier\n\nThe formula for Z score is\n\n(X – mean of data) / Standard deviation of data\n\nWhere X is the value you wish to solve for\n\nIn Excel the Standardize function will calculate Z Scores.  Its syntax is\n\n=STANDARDIZE (x, mean, standard deviation)\n\n### Example How to calculate Z Score in Excel using Standardize\n\nYou are told the general mean weight of a mango is 10 oz with a standard deviation of 1.75.  you are also told in general the mean weight of a banana is 7oz with a standard deviation if 2.2.  You buy a banana and a mango and both weigh 9 oz.  Which is bigger?\n\nDoing this manually we must standardize the numbers by calculating a Z score to make a comparison using the formula Z= (X – mean of data) / Standard deviation of data\n\nMango = (9-10) / 1.75 = -0.571\n\nBanana = (9-7) / 2.2 = 0.909\n\nAs the Z score for bananas is larger, we can say the banana is larger than the orange.\n\nIn Excel this can be solved using the Standardize function.  The syntax for Standardize is\n\n=STANDARDIZE (x, mean, standard deviation)\n\nWhere we are solving for x given the mean and standard deviation.  You can see from the image how these values can be easily plugged into the Standardize function.\n\n### Earn and Learn Activity – Standardize in Excel – Which is bigger, the apple or the orange?\n\nYou buy and apple and an orange, both weigh 12oz.  In general, the mean weight for apples is 8oz with a standard deviation of 2 and in general the mean eight for oranges is 10.5oz with a standard deviation of 1.5\n\nWhich is bigger, the apple or orange?\n\nTo earn STEEM rewards on this post, in the comment section below, answer the following questions\n\nWhich is bigger, the apple or the orange?" ]

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[ "# estinterval: Estimate interval model accounting for missed arrival... In adokter/intRval: Analysis of Time-Ordered Event Data with Missed Observations\n\n## Description\n\nEstimate interval mean and variance accounting for missed arrival observations, by fitting the probability density function intervalpdf to the interval data.\n\n## Usage\n\n ```1 2 3 4 5``` ```estinterval(data, mu = median(data), sigma = sd(data)/2, p = 0.2, N = 5L, fun = \"gamma\", trunc = c(0, Inf), fpp = (if (fpp.method == \"fixed\") 0 else 0.1), fpp.method = \"auto\", p.method = \"auto\", conf.level = 0.9, group = NA, sigma.within = NA, iter = 10, tol = 0.001, silent = F, ...) ```\n\n## Arguments\n\n `data` A numeric list of intervals. `mu` Start value for the numeric optimization for the mean arrival interval. `sigma` Start value for the numeric optimization for the standard deviation of the arrival interval. `p` Start value for the numeric optimization for the probability to not observe an arrival. `N` Maximum number of missed observations to be taken into account (default N=5). `fun` Assumed distribution for the intervals, one of \"`normal`\" or \"`gamma`\", corresponding to the Normal and GammaDist distributions `trunc` Use a truncated probability density function with range `trunc` `fpp` Baseline proportion of intervals distributed as a random poisson process with mean arrival interval `mu` `fpp.method` A string equal to 'fixed' or 'auto'. When 'auto' `fpp` is optimized as a free model parameter, in which case `fpp` is taken as start value in the optimisation `p.method` A string equal to 'fixed' or 'auto'. When 'auto' `p` is optimized as a free model parameter, in which case `p` is taken as start value in the optimisation `conf.level` Confidence level for deviance test that checks whether model with nonzero missed event probability `p` significantly outperforms a model without a missed event probability (`p=0`). `group` optional vector of equal length as data, indicating the group or subject in which the interval was observed `sigma.within` optional within-subject standard deviation. When equal to default 'NA', assumes no additional between-subject effect, with `sigma.within` equal to `sigma`. When equal to 'auto' an estimate is provided by iteratively calling partition `iter` maximum number of iterations in numerical iteration for `sigma.within` `tol` tolerance in the iteration, when `sigma.within` changes less than this value in one iteration step, the optimization is considered converged. `silent` logical. When `TRUE` print no information to console `...` Additional arguments to be passed to optim\n\n## Details\n\nThe probability density function for observed intervals intervalpdf is fit to `data` by maximization of the associated log-likelihood using optim.\n\nWithin-group variation `sigma.within` may be separated from the total variation `sigma` in an iterative fit of intervalpdf on the interval data. In the iteration partition is used to (1) determine which intervals according to the fit are a fundamental interval at a confidence level `conf.level`, and (2) to partition the within-group variation from the total variation in interval length.\n\nWithin- and between-group variation is estimated on the subset of fundamental intervals with repeated measures only. As the set of fundamental interval depends on the precise value of `sigma.within`, the fit of intervalpdf and the subsequent estimation of `sigma.within` using partition is iterated until both converge to a stable solution. Parameters `tol` and `iter` set the threshold for convergence and the maximum number of iterations.\n\nWe note that an exponential interval model can be fitted by setting `fpp=1` and `fpp.method=fixed`.\n\n## Value\n\nThis function returns an object of class `intRvals`, which is a list containing the following:\n\n`data`\n\nthe interval data\n\n`mu`\n\nthe modelled mean interval\n\n`mu.se`\n\nthe modelled mean interval standard error\n\n`sigma`\n\nthe modelled interval standard deviation\n\n`p`\n\nthe modelled probability to not observe an arrival\n\n`fpp`\n\nthe modelled fraction of arrivals following a random poisson process, see intervalpdf\n\n`N`\n\nthe highest number of consecutive missed arrivals taken into account, see intervalpdf\n\n`convergence`\n\nconvergence field of optim\n\n`counts`\n\ncounts field of optim\n\n`loglik`\n\nvector of length 2, with first element the log-likelihood of the fitted model, and second element the log-likelihood of the model without a missed event probability (i.e. `p`=0)\n\n`df.residual`\n\ndegrees of freedom, a 2-vector (1, number of intervals - `n.param`)\n\n`n.param`\n\nnumber of optimized model parameters\n\n`p.chisq`\n\np value for a likelihood-ratio test of a model including a miss probability relative against a model without a miss probability\n\n`distribution`\n\nassumed interval distribution, one of 'gamma' or 'normal'\n\n`trunc`\n\ninterval range over which the interval pdf was truncated and normalized\n\n`fpp.method`\n\nA string equal to 'fixed' or 'auto'. When 'auto' `fpp` has been optimized as a free model parameter\n\n`p.method`\n\nA string equal to 'fixed' or 'auto'. When 'auto' `p` has been optimized as a free model parameter\n\n## Examples\n\n ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29``` ```data(goosedrop) # calculate mean and standard deviation of arrival intervals, accounting for missed observations: dr=estinterval(goosedrop\\$interval) # plot some summary information summary(dr) # plot a histogram of the intervals and fit: plot(dr) # test whether the mean arrival interval is greater than 200 seconds: ttest(dr,mu=200,alternative=\"greater\") # let's estimate mean and variance of dropping intervals by site # (schiermonnikoog vs terschelling) for time period 5. # first prepare the two datasets: set1=goosedrop[goosedrop\\$site==\"schiermonnikoog\" & goosedrop\\$period == 5,] set2=goosedrop[goosedrop\\$site==\"terschelling\" & goosedrop\\$period == 5,] # allowing a fraction of intervals to be distributed randomly (fpp='auto') dr1=estinterval(set1\\$interval,fpp.method='auto') dr2=estinterval(set2\\$interval,fpp.method='auto') # plot the fits: plot(dr1,xlim=c(0,1000)) plot(dr2,xlim=c(0,1000)) # mean dropping interval are not significantly different # at the two sites (on a 0.95 confidence level): ttest(dr1,dr2) # now compare this test with a t-test not accounting for unobserved intervals: t.test(set1\\$interval,set2\\$interval) # not accounting for missed observations leads to a (spurious) # larger difference in means, which also increases # the apparent statistical significance of the difference between means ```\n\nadokter/intRval documentation built on May 10, 2019, 5:58 a.m." ]

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