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Matrix A is decomposed in three matrices (L, U, P) where P * A = L * U",examples:["lup([[2, 1], [1, 4]])","lup(matrix([[2, 1], [1, 4]]))","lup(sparse([[2, 1], [1, 4]]))"],seealso:["lusolve","lsolve","usolve","matrix","sparse","slu","qr"]},XA={name:"lusolve",category:"Algebra",syntax:["x=lusolve(A, b)","x=lusolve(lu, b)"],description:"Solves the linear system A * x = b where A is an [n x n] matrix and b is a [n] column vector.",examples:["a = [-2, 3; 2, 1]","b = [11, 9]","x = lusolve(a, b)"],seealso:["lup","slu","lsolve","usolve","matrix","sparse"]},QA={name:"polynomialRoot",category:"Algebra",syntax:["x=polynomialRoot(-6, 3)","x=polynomialRoot(4, -4, 1)","x=polynomialRoot(-8, 12, -6, 1)"],description:"Finds the roots of a univariate polynomial given by its coefficients starting from constant, linear, and so on, increasing in degree.",examples:["a = polynomialRoot(-6, 11, -6, 1)"],seealso:["cbrt","sqrt"]},jA={name:"qr",category:"Algebra",syntax:["qr(A)"],description:"Calculates the Matrix QR decomposition. Matrix `A` is decomposed in two matrices (`Q`, `R`) where `Q` is an orthogonal matrix and `R` is an upper triangular matrix.",examples:["qr([[1, -1, 4], [1, 4, -2], [1, 4, 2], [1, -1, 0]])"],seealso:["lup","slu","matrix"]},KA={name:"rationalize",category:"Algebra",syntax:["rationalize(expr)","rationalize(expr, scope)","rationalize(expr, scope, detailed)"],description:"Transform a rationalizable expression in a rational fraction. If rational fraction is one variable polynomial then converts the numerator and denominator in canonical form, with decreasing exponents, returning the coefficients of numerator.",examples:['rationalize("2x/y - y/(x+1)")','rationalize("2x/y - y/(x+1)", true)'],seealso:["simplify"]},e8={name:"resolve",category:"Algebra",syntax:["resolve(node, scope)"],description:"Recursively substitute variables in an expression tree.",examples:['resolve(parse("1 + x"), { x: 7 })','resolve(parse("size(text)"), { text: "Hello World" })','resolve(parse("x + y"), { x: parse("3z") })','resolve(parse("3x"), { x: parse("y+z"), z: parse("w^y") })'],seealso:["simplify","evaluate"],mayThrow:["ReferenceError"]},t8={name:"simplify",category:"Algebra",syntax:["simplify(expr)","simplify(expr, rules)"],description:"Simplify an expression tree.",examples:['simplify("3 + 2 / 4")','simplify("2x + x")','f = parse("x * (x + 2 + x)")',"simplified = simplify(f)","simplified.evaluate({x: 2})"],seealso:["simplifyCore","derivative","evaluate","parse","rationalize","resolve"]},a8={name:"simplifyConstant",category:"Algebra",syntax:["simplifyConstant(expr)","simplifyConstant(expr, options)"],description:"Replace constant subexpressions of node with their values.",examples:['simplifyConstant("(3-3)*x")','simplifyConstant(parse("z-cos(tau/8)"))'],seealso:["simplify","simplifyCore","evaluate"]},r8={name:"simplifyCore",category:"Algebra",syntax:["simplifyCore(node)"],description:"Perform simple one-pass simplifications on an expression tree.",examples:['simplifyCore(parse("0*x"))','simplifyCore(parse("(x+0)*2"))'],seealso:["simplify","simplifyConstant","evaluate"]},n8={name:"slu",category:"Algebra",syntax:["slu(A, order, threshold)"],description:"Calculate the Matrix LU decomposition with full pivoting. Matrix A is decomposed in two matrices (L, U) and two permutation vectors (pinv, q) where P * A * Q = L * U",examples:["slu(sparse([4.5, 0, 3.2, 0; 3.1, 2.9, 0, 0.9; 0, 1.7, 3, 0; 3.5, 0.4, 0, 1]), 1, 0.001)"],seealso:["lusolve","lsolve","usolve","matrix","sparse","lup","qr"]},s8={name:"symbolicEqual",category:"Algebra",syntax:["symbolicEqual(expr1, expr2)","symbolicEqual(expr1, expr2, options)"],description:"Returns true if the difference of the expressions simplifies to 0",examples:['symbolicEqual("x*y","y*x")','symbolicEqual("abs(x^2)", "x^2")','symbolicEqual("abs(x)", "x", {context: {abs: {trivial: true}}})'],seealso:["simplify","evaluate"]},i8={name:"usolve",category:"Algebra",syntax:["x=usolve(U, b)"],description:"Finds one solution of the linear system U * x = b where U is an [n x n] upper triangular matrix and b is a [n] column vector.",examples:["x=usolve(sparse([1, 1, 1, 1; 0, 1, 1, 1; 0, 0, 1, 1; 0, 0, 0, 1]), [1; 2; 3; 4])"],seealso:["usolveAll","lup","lusolve","lsolve","matrix","sparse"]},o8={name:"usolveAll",category:"Algebra",syntax:["x=usolve(U, b)"],description:"Finds all solutions of the linear system U * x = b where U is an [n x n] upper triangular matrix and b is a [n] column vector.",examples:["x=usolve(sparse([1, 1, 1, 1; 0, 1, 1, 1; 0, 0, 1, 1; 0, 0, 0, 1]), [1; 2; 3; 4])"],seealso:["usolve","lup","lusolve","lsolve","matrix","sparse"]},l8={name:"abs",category:"Arithmetic",syntax:["abs(x)"],description:"Compute the absolute value.",examples:["abs(3.5)","abs(-4.2)"],seealso:["sign"]},m8={name:"add",category:"Operators",syntax:["x + y","add(x, y)"],description:"Add two values.",examples:["a = 2.1 + 3.6","a - 3.6","3 + 2i","3 cm + 2 inch",'"2.3" + "4"'],seealso:["subtract"]},u8={name:"cbrt",category:"Arithmetic",syntax:["cbrt(x)","cbrt(x, allRoots)"],description:"Compute the cubic root value. If x = y * y * y, then y is the cubic root of x. When `x` is a number or complex number, an optional second argument `allRoots` can be provided to return all three cubic roots. If not provided, the principal root is returned",examples:["cbrt(64)","cube(4)","cbrt(-8)","cbrt(2 + 3i)","cbrt(8i)","cbrt(8i, true)","cbrt(27 m^3)"],seealso:["square","sqrt","cube","multiply"]},p8={name:"ceil",category:"Arithmetic",syntax:["ceil(x)"],description:"Round a value towards plus infinity. If x is complex, both real and imaginary part are rounded towards plus infinity.",examples:["ceil(3.2)","ceil(3.8)","ceil(-4.2)"],seealso:["floor","fix","round"]},c8={name:"cube",category:"Arithmetic",syntax:["cube(x)"],description:"Compute the cube of a value. The cube of x is x * x * x.",examples:["cube(2)","2^3","2 * 2 * 2"],seealso:["multiply","square","pow"]},f8={name:"divide",category:"Operators",syntax:["x / y","divide(x, y)"],description:"Divide two values.",examples:["a = 2 / 3","a * 3","4.5 / 2","3 + 4 / 2","(3 + 4) / 2","18 km / 4.5"],seealso:["multiply"]},h8={name:"dotDivide",category:"Operators",syntax:["x ./ y","dotDivide(x, y)"],description:"Divide two values element wise.",examples:["a = [1, 2, 3; 4, 5, 6]","b = [2, 1, 1; 3, 2, 5]","a ./ b"],seealso:["multiply","dotMultiply","divide"]},v8={name:"dotMultiply",category:"Operators",syntax:["x .* y","dotMultiply(x, y)"],description:"Multiply two values element wise.",examples:["a = [1, 2, 3; 4, 5, 6]","b = [2, 1, 1; 3, 2, 5]","a .* b"],seealso:["multiply","divide","dotDivide"]},g8={name:"dotPow",category:"Operators",syntax:["x .^ y","dotPow(x, y)"],description:"Calculates the power of x to y element wise.",examples:["a = [1, 2, 3; 4, 5, 6]","a .^ 2"],seealso:["pow"]},d8={name:"exp",category:"Arithmetic",syntax:["exp(x)"],description:"Calculate the exponent of a value.",examples:["exp(1.3)","e ^ 1.3","log(exp(1.3))","x = 2.4","(exp(i*x) == cos(x) + i*sin(x)) # Euler's formula"],seealso:["expm","expm1","pow","log"]},y8={name:"expm",category:"Arithmetic",syntax:["exp(x)"],description:"Compute the matrix exponential, expm(A) = e^A. The matrix must be square. Not to be confused with exp(a), which performs element-wise exponentiation.",examples:["expm([[0,2],[0,0]])"],seealso:["exp"]},w8={name:"expm1",category:"Arithmetic",syntax:["expm1(x)"],description:"Calculate the value of subtracting 1 from the exponential value.",examples:["expm1(2)","pow(e, 2) - 1","log(expm1(2) + 1)"],seealso:["exp","pow","log"]},x8={name:"fix",category:"Arithmetic",syntax:["fix(x)"],description:"Round a value towards zero. If x is complex, both real and imaginary part are rounded towards zero.",examples:["fix(3.2)","fix(3.8)","fix(-4.2)","fix(-4.8)"],seealso:["ceil","floor","round"]},b8={name:"floor",category:"Arithmetic",syntax:["floor(x)"],description:"Round a value towards minus infinity.If x is complex, both real and imaginary part are rounded towards minus infinity.",examples:["floor(3.2)","floor(3.8)","floor(-4.2)"],seealso:["ceil","fix","round"]},N8={name:"gcd",category:"Arithmetic",syntax:["gcd(a, b)","gcd(a, b, c, ...)"],description:"Compute the greatest common divisor.",examples:["gcd(8, 12)","gcd(-4, 6)","gcd(25, 15, -10)"],seealso:["lcm","xgcd"]},E8={name:"hypot",category:"Arithmetic",syntax:["hypot(a, b, c, ...)","hypot([a, b, c, ...])"],description:"Calculate the hypotenusa of a list with values. ",examples:["hypot(3, 4)","sqrt(3^2 + 4^2)","hypot(-2)","hypot([3, 4, 5])"],seealso:["abs","norm"]},D8={name:"invmod",category:"Arithmetic",syntax:["invmod(a, b)"],description:"Calculate the (modular) multiplicative inverse of a modulo b. Solution to the equation ax ≣ 1 (mod b)",examples:["invmod(8, 12)","invmod(7, 13)","invmod(15151, 15122)"],seealso:["gcd","xgcd"]},A8={name:"lcm",category:"Arithmetic",syntax:["lcm(x, y)"],description:"Compute the least common multiple.",examples:["lcm(4, 6)","lcm(6, 21)","lcm(6, 21, 5)"],seealso:["gcd"]},M8={name:"log",category:"Arithmetic",syntax:["log(x)","log(x, base)"],description:"Compute the logarithm of a value. If no base is provided, the natural logarithm of x is calculated. If base if provided, the logarithm is calculated for the specified base. log(x, base) is defined as log(x) / log(base).",examples:["log(3.5)","a = log(2.4)","exp(a)","10 ^ 4","log(10000, 10)","log(10000) / log(10)","b = log(1024, 2)","2 ^ b"],seealso:["exp","log1p","log2","log10"]},S8={name:"log10",category:"Arithmetic",syntax:["log10(x)"],description:"Compute the 10-base logarithm of a value.",examples:["log10(0.00001)","log10(10000)","10 ^ 4","log(10000) / log(10)","log(10000, 10)"],seealso:["exp","log"]},C8={name:"log1p",category:"Arithmetic",syntax:["log1p(x)","log1p(x, base)"],description:"Calculate the logarithm of a `value+1`",examples:["log1p(2.5)","exp(log1p(1.4))","pow(10, 4)","log1p(9999, 10)","log1p(9999) / log(10)"],seealso:["exp","log","log2","log10"]},T8={name:"log2",category:"Arithmetic",syntax:["log2(x)"],description:"Calculate the 2-base of a value. This is the same as calculating `log(x, 2)`.",examples:["log2(0.03125)","log2(16)","log2(16) / log2(2)","pow(2, 4)"],seealso:["exp","log1p","log","log10"]},$8={name:"mod",category:"Operators",syntax:["x % y","x mod y","mod(x, y)"],description:"Calculates the modulus, the remainder of an integer division.",examples:["7 % 3","11 % 2","10 mod 4","isOdd(x) = x % 2","isOdd(2)","isOdd(3)"],seealso:["divide"]},F8={name:"multiply",category:"Operators",syntax:["x * y","multiply(x, y)"],description:"multiply two values.",examples:["a = 2.1 * 3.4","a / 3.4","2 * 3 + 4","2 * (3 + 4)","3 * 2.1 km"],seealso:["divide"]},_8={name:"norm",category:"Arithmetic",syntax:["norm(x)","norm(x, p)"],description:"Calculate the norm of a number, vector or matrix.",examples:["abs(-3.5)","norm(-3.5)","norm(3 - 4i)","norm([1, 2, -3], Infinity)","norm([1, 2, -3], -Infinity)","norm([3, 4], 2)","norm([[1, 2], [3, 4]], 1)",'norm([[1, 2], [3, 4]], "inf")','norm([[1, 2], [3, 4]], "fro")']},B8={name:"nthRoot",category:"Arithmetic",syntax:["nthRoot(a)","nthRoot(a, root)"],description:'Calculate the nth root of a value. The principal nth root of a positive real number A, is the positive real solution of the equation "x^root = A".',examples:["4 ^ 3","nthRoot(64, 3)","nthRoot(9, 2)","sqrt(9)"],seealso:["nthRoots","pow","sqrt"]},O8={name:"nthRoots",category:"Arithmetic",syntax:["nthRoots(A)","nthRoots(A, root)"],description:'Calculate the nth roots of a value. An nth root of a positive real number A, is a positive real solution of the equation "x^root = A". This function returns an array of complex values.',examples:["nthRoots(1)","nthRoots(1, 3)"],seealso:["sqrt","pow","nthRoot"]},P8={name:"pow",category:"Operators",syntax:["x ^ y","pow(x, y)"],description:"Calculates the power of x to y, x^y.",examples:["2^3","2*2*2","1 + e ^ (pi * i)","pow([[1, 2], [4, 3]], 2)","pow([[1, 2], [4, 3]], -1)"],seealso:["multiply","nthRoot","nthRoots","sqrt"]},z8={name:"round",category:"Arithmetic",syntax:["round(x)","round(x, n)","round(unit, valuelessUnit)","round(unit, n, valuelessUnit)"],description:"round a value towards the nearest integer.If x is complex, both real and imaginary part are rounded towards the nearest integer. When n is specified, the value is rounded to n decimals.",examples:["round(3.2)","round(3.8)","round(-4.2)","round(-4.8)","round(pi, 3)","round(123.45678, 2)","round(3.241cm, 2, cm)","round([3.2, 3.8, -4.7])"],seealso:["ceil","floor","fix"]},R8={name:"sign",category:"Arithmetic",syntax:["sign(x)"],description:"Compute the sign of a value. The sign of a value x is 1 when x>1, -1 when x<0, and 0 when x=0.",examples:["sign(3.5)","sign(-4.2)","sign(0)"],seealso:["abs"]},I8={name:"sqrt",category:"Arithmetic",syntax:["sqrt(x)"],description:"Compute the square root value. If x = y * y, then y is the square root of x.",examples:["sqrt(25)","5 * 5","sqrt(-1)"],seealso:["square","sqrtm","multiply","nthRoot","nthRoots","pow"]},q8={name:"sqrtm",category:"Arithmetic",syntax:["sqrtm(x)"],description:"Calculate the principal square root of a square matrix. The principal square root matrix `X` of another matrix `A` is such that `X * X = A`.",examples:["sqrtm([[33, 24], [48, 57]])"],seealso:["sqrt","abs","square","multiply"]},L8={name:"sylvester",category:"Algebra",syntax:["sylvester(A,B,C)"],description:"Solves the real-valued Sylvester equation AX+XB=C for X",examples:["sylvester([[-1, -2], [1, 1]], [[-2, 1], [-1, 2]], [[-3, 2], [3, 0]])","A = [[-1, -2], [1, 1]]; B = [[2, -1], [1, -2]]; C = [[-3, 2], [3, 0]]","sylvester(A, B, C)"],seealso:["schur","lyap"]},k8={name:"schur",category:"Algebra",syntax:["schur(A)"],description:"Performs a real Schur decomposition of the real matrix A = UTU'",examples:["schur([[1, 0], [-4, 3]])","A = [[1, 0], [-4, 3]]","schur(A)"],seealso:["lyap","sylvester"]},U8={name:"lyap",category:"Algebra",syntax:["lyap(A,Q)"],description:"Solves the Continuous-time Lyapunov equation AP+PA'+Q=0 for P",examples:["lyap([[-2, 0], [1, -4]], [[3, 1], [1, 3]])","A = [[-2, 0], [1, -4]]","Q = [[3, 1], [1, 3]]","lyap(A,Q)"],seealso:["schur","sylvester"]},G8={name:"square",category:"Arithmetic",syntax:["square(x)"],description:"Compute the square of a value. The square of x is x * x.",examples:["square(3)","sqrt(9)","3^2","3 * 3"],seealso:["multiply","pow","sqrt","cube"]},H8={name:"subtract",category:"Operators",syntax:["x - y","subtract(x, y)"],description:"subtract two values.",examples:["a = 5.3 - 2","a + 2","2/3 - 1/6","2 * 3 - 3","2.1 km - 500m"],seealso:["add"]},V8={name:"unaryMinus",category:"Operators",syntax:["-x","unaryMinus(x)"],description:"Inverse the sign of a value. Converts booleans and strings to numbers.",examples:["-4.5","-(-5.6)",'-"22"'],seealso:["add","subtract","unaryPlus"]},J8={name:"unaryPlus",category:"Operators",syntax:["+x","unaryPlus(x)"],description:"Converts booleans and strings to numbers.",examples:["+true",'+"2"'],seealso:["add","subtract","unaryMinus"]},Z8={name:"xgcd",category:"Arithmetic",syntax:["xgcd(a, b)"],description:"Calculate the extended greatest common divisor for two values. The result is an array [d, x, y] with 3 entries, where d is the greatest common divisor, and d = x * a + y * b.",examples:["xgcd(8, 12)","gcd(8, 12)","xgcd(36163, 21199)"],seealso:["gcd","lcm"]},W8={name:"bitAnd",category:"Bitwise",syntax:["x & y","bitAnd(x, y)"],description:"Bitwise AND operation. Performs the logical AND operation on each pair of the corresponding bits of the two given values by multiplying them. If both bits in the compared position are 1, the bit in the resulting binary representation is 1, otherwise, the result is 0",examples:["5 & 3","bitAnd(53, 131)","[1, 12, 31] & 42"],seealso:["bitNot","bitOr","bitXor","leftShift","rightArithShift","rightLogShift"]},Y8={name:"bitNot",category:"Bitwise",syntax:["~x","bitNot(x)"],description:"Bitwise NOT operation. Performs a logical negation on each bit of the given value. Bits that are 0 become 1, and those that are 1 become 0.",examples:["~1","~2","bitNot([2, -3, 4])"],seealso:["bitAnd","bitOr","bitXor","leftShift","rightArithShift","rightLogShift"]},X8={name:"bitOr",category:"Bitwise",syntax:["x | y","bitOr(x, y)"],description:"Bitwise OR operation. Performs the logical inclusive OR operation on each pair of corresponding bits of the two given values. The result in each position is 1 if the first bit is 1 or the second bit is 1 or both bits are 1, otherwise, the result is 0.",examples:["5 | 3","bitOr([1, 2, 3], 4)"],seealso:["bitAnd","bitNot","bitXor","leftShift","rightArithShift","rightLogShift"]},Q8={name:"bitXor",category:"Bitwise",syntax:["bitXor(x, y)"],description:"Bitwise XOR operation, exclusive OR. Performs the logical exclusive OR operation on each pair of corresponding bits of the two given values. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1.",examples:["bitOr(1, 2)","bitXor([2, 3, 4], 4)"],seealso:["bitAnd","bitNot","bitOr","leftShift","rightArithShift","rightLogShift"]},j8={name:"leftShift",category:"Bitwise",syntax:["x << y","leftShift(x, y)"],description:"Bitwise left logical shift of a value x by y number of bits.",examples:["4 << 1","8 >> 1"],seealso:["bitAnd","bitNot","bitOr","bitXor","rightArithShift","rightLogShift"]},K8={name:"rightArithShift",category:"Bitwise",syntax:["x >> y","rightArithShift(x, y)"],description:"Bitwise right arithmetic shift of a value x by y number of bits.",examples:["8 >> 1","4 << 1","-12 >> 2"],seealso:["bitAnd","bitNot","bitOr","bitXor","leftShift","rightLogShift"]},eM={name:"rightLogShift",category:"Bitwise",syntax:["x >>> y","rightLogShift(x, y)"],description:"Bitwise right logical shift of a value x by y number of bits.",examples:["8 >>> 1","4 << 1","-12 >>> 2"],seealso:["bitAnd","bitNot","bitOr","bitXor","leftShift","rightArithShift"]},tM={name:"bellNumbers",category:"Combinatorics",syntax:["bellNumbers(n)"],description:"The Bell Numbers count the number of partitions of a set. A partition is a pairwise disjoint subset of S whose union is S. `bellNumbers` only takes integer arguments. The following condition must be enforced: n >= 0.",examples:["bellNumbers(3)","bellNumbers(8)"],seealso:["stirlingS2"]},aM={name:"catalan",category:"Combinatorics",syntax:["catalan(n)"],description:"The Catalan Numbers enumerate combinatorial structures of many different types. catalan only takes integer arguments. The following condition must be enforced: n >= 0.",examples:["catalan(3)","catalan(8)"],seealso:["bellNumbers"]},rM={name:"composition",category:"Combinatorics",syntax:["composition(n, k)"],description:"The composition counts of n into k parts. composition only takes integer arguments. The following condition must be enforced: k <= n.",examples:["composition(5, 3)"],seealso:["combinations"]},nM={name:"stirlingS2",category:"Combinatorics",syntax:["stirlingS2(n, k)"],description:"he Stirling numbers of the second kind, counts the number of ways to partition a set of n labelled objects into k nonempty unlabelled subsets. `stirlingS2` only takes integer arguments. The following condition must be enforced: k <= n. If n = k or k = 1, then s(n,k) = 1.",examples:["stirlingS2(5, 3)"],seealso:["bellNumbers"]},sM={name:"arg",category:"Complex",syntax:["arg(x)"],description:"Compute the argument of a complex value. If x = a+bi, the argument is computed as atan2(b, a).",examples:["arg(2 + 2i)","atan2(3, 2)","arg(2 + 3i)"],seealso:["re","im","conj","abs"]},iM={name:"conj",category:"Complex",syntax:["conj(x)"],description:"Compute the complex conjugate of a complex value. If x = a+bi, the complex conjugate is a-bi.",examples:["conj(2 + 3i)","conj(2 - 3i)","conj(-5.2i)"],seealso:["re","im","abs","arg"]},oM={name:"im",category:"Complex",syntax:["im(x)"],description:"Get the imaginary part of a complex number.",examples:["im(2 + 3i)","re(2 + 3i)","im(-5.2i)","im(2.4)"],seealso:["re","conj","abs","arg"]},lM={name:"re",category:"Complex",syntax:["re(x)"],description:"Get the real part of a complex number.",examples:["re(2 + 3i)","im(2 + 3i)","re(-5.2i)","re(2.4)"],seealso:["im","conj","abs","arg"]},mM={name:"evaluate",category:"Expression",syntax:["evaluate(expression)","evaluate(expression, scope)","evaluate([expr1, expr2, expr3, ...])","evaluate([expr1, expr2, expr3, ...], scope)"],description:"Evaluate an expression or an array with expressions.",examples:['evaluate("2 + 3")','evaluate("sqrt(16)")','evaluate("2 inch to cm")','evaluate("sin(x * pi)", { "x": 1/2 })','evaluate(["width=2", "height=4","width*height"])'],seealso:[]},uM={name:"help",category:"Expression",syntax:["help(object)","help(string)"],description:"Display documentation on a function or data type.",examples:["help(sqrt)",'help("complex")'],seealso:[]},pM={name:"distance",category:"Geometry",syntax:["distance([x1, y1], [x2, y2])","distance([[x1, y1], [x2, y2]])"],description:"Calculates the Euclidean distance between two points.",examples:["distance([0,0], [4,4])","distance([[0,0], [4,4]])"],seealso:[]},cM={name:"intersect",category:"Geometry",syntax:["intersect(expr1, expr2, expr3, expr4)","intersect(expr1, expr2, expr3)"],description:"Computes the intersection point of lines and/or planes.",examples:["intersect([0, 0], [10, 10], [10, 0], [0, 10])","intersect([1, 0, 1], [4, -2, 2], [1, 1, 1, 6])"],seealso:[]},fM={name:"and",category:"Logical",syntax:["x and y","and(x, y)"],description:"Logical and. Test whether two values are both defined with a nonzero/nonempty value.",examples:["true and false","true and true","2 and 4"],seealso:["not","or","xor"]},hM={name:"not",category:"Logical",syntax:["not x","not(x)"],description:"Logical not. Flips the boolean value of given argument.",examples:["not true","not false","not 2","not 0"],seealso:["and","or","xor"]},vM={name:"or",category:"Logical",syntax:["x or y","or(x, y)"],description:"Logical or. Test if at least one value is defined with a nonzero/nonempty value.",examples:["true or false","false or false","0 or 4"],seealso:["not","and","xor"]},gM={name:"xor",category:"Logical",syntax:["x xor y","xor(x, y)"],description:"Logical exclusive or, xor. Test whether one and only one value is defined with a nonzero/nonempty value.",examples:["true xor false","false xor false","true xor true","0 xor 4"],seealso:["not","and","or"]},dM={name:"column",category:"Matrix",syntax:["column(x, index)"],description:"Return a column from a matrix or array.",examples:["A = [[1, 2], [3, 4]]","column(A, 1)","column(A, 2)"],seealso:["row","matrixFromColumns"]},yM={name:"concat",category:"Matrix",syntax:["concat(A, B, C, ...)","concat(A, B, C, ..., dim)"],description:"Concatenate matrices. By default, the matrices are concatenated by the last dimension. The dimension on which to concatenate can be provided as last argument.",examples:["A = [1, 2; 5, 6]","B = [3, 4; 7, 8]","concat(A, B)","concat(A, B, 1)","concat(A, B, 2)"],seealso:["det","diag","identity","inv","ones","range","size","squeeze","subset","trace","transpose","zeros"]},wM={name:"count",category:"Matrix",syntax:["count(x)"],description:"Count the number of elements of a matrix, array or string.",examples:["a = [1, 2; 3, 4; 5, 6]","count(a)","size(a)",'count("hello world")'],seealso:["size"]},xM={name:"cross",category:"Matrix",syntax:["cross(A, B)"],description:"Calculate the cross product for two vectors in three dimensional space.",examples:["cross([1, 1, 0], [0, 1, 1])","cross([3, -3, 1], [4, 9, 2])","cross([2, 3, 4], [5, 6, 7])"],seealso:["multiply","dot"]},bM={name:"ctranspose",category:"Matrix",syntax:["x'","ctranspose(x)"],description:"Complex Conjugate and Transpose a matrix",examples:["a = [1, 2, 3; 4, 5, 6]","a'","ctranspose(a)"],seealso:["concat","det","diag","identity","inv","ones","range","size","squeeze","subset","trace","zeros"]},NM={name:"det",category:"Matrix",syntax:["det(x)"],description:"Calculate the determinant of a matrix",examples:["det([1, 2; 3, 4])","det([-2, 2, 3; -1, 1, 3; 2, 0, -1])"],seealso:["concat","diag","identity","inv","ones","range","size","squeeze","subset","trace","transpose","zeros"]},EM={name:"diag",category:"Matrix",syntax:["diag(x)","diag(x, k)"],description:"Create a diagonal matrix or retrieve the diagonal of a matrix. When x is a vector, a matrix with the vector values on the diagonal will be returned. When x is a matrix, a vector with the diagonal values of the matrix is returned. When k is provided, the k-th diagonal will be filled in or retrieved, if k is positive, the values are placed on the super diagonal. When k is negative, the values are placed on the sub diagonal.",examples:["diag(1:3)","diag(1:3, 1)","a = [1, 2, 3; 4, 5, 6; 7, 8, 9]","diag(a)"],seealso:["concat","det","identity","inv","ones","range","size","squeeze","subset","trace","transpose","zeros"]},DM={name:"diff",category:"Matrix",syntax:["diff(arr)","diff(arr, dim)"],description:["Create a new matrix or array with the difference of the passed matrix or array.","Dim parameter is optional and used to indicant the dimension of the array/matrix to apply the difference","If no dimension parameter is passed it is assumed as dimension 0","Dimension is zero-based in javascript and one-based in the parser","Arrays must be 'rectangular' meaning arrays like [1, 2]","If something is passed as a matrix it will be returned as a matrix but other than that all matrices are converted to arrays"],examples:["A = [1, 2, 4, 7, 0]","diff(A)","diff(A, 1)","B = [[1, 2], [3, 4]]","diff(B)","diff(B, 1)","diff(B, 2)","diff(B, bignumber(2))","diff([[1, 2], matrix([3, 4])], 2)"],seealso:["subtract","partitionSelect"]},AM={name:"dot",category:"Matrix",syntax:["dot(A, B)","A * B"],description:"Calculate the dot product of two vectors. The dot product of A = [a1, a2, a3, ..., an] and B = [b1, b2, b3, ..., bn] is defined as dot(A, B) = a1 * b1 + a2 * b2 + a3 * b3 + ... + an * bn",examples:["dot([2, 4, 1], [2, 2, 3])","[2, 4, 1] * [2, 2, 3]"],seealso:["multiply","cross"]},MM={name:"eigs",category:"Matrix",syntax:["eigs(x)"],description:"Calculate the eigenvalues and optionally eigenvectors of a square matrix",examples:["eigs([[5, 2.3], [2.3, 1]])","eigs([[1, 2, 3], [4, 5, 6], [7, 8, 9]], { precision: 1e-6, eigenvectors: false })"],seealso:["inv"]},SM={name:"filter",category:"Matrix",syntax:["filter(x, test)"],description:"Filter items in a matrix.",examples:["isPositive(x) = x > 0","filter([6, -2, -1, 4, 3], isPositive)","filter([6, -2, 0, 1, 0], x != 0)"],seealso:["sort","map","forEach"]},CM={name:"flatten",category:"Matrix",syntax:["flatten(x)"],description:"Flatten a multi dimensional matrix into a single dimensional matrix.",examples:["a = [1, 2, 3; 4, 5, 6]","size(a)","b = flatten(a)","size(b)"],seealso:["concat","resize","size","squeeze"]},TM={name:"forEach",category:"Matrix",syntax:["forEach(x, callback)"],description:"Iterates over all elements of a matrix/array, and executes the given callback function.",examples:["numberOfPets = {}","addPet(n) = numberOfPets[n] = (numberOfPets[n] ? numberOfPets[n]:0 ) + 1;",'forEach(["Dog","Cat","Cat"], addPet)',"numberOfPets"],seealso:["map","sort","filter"]},$M={name:"getMatrixDataType",category:"Matrix",syntax:["getMatrixDataType(x)"],description:'Find the data type of all elements in a matrix or array, for example "number" if all items are a number and "Complex" if all values are complex numbers. If a matrix contains more than one data type, it will return "mixed".',examples:["getMatrixDataType([1, 2, 3])","getMatrixDataType([[5 cm], [2 inch]])",'getMatrixDataType([1, "text"])',"getMatrixDataType([1, bignumber(4)])"],seealso:["matrix","sparse","typeOf"]},FM={name:"identity",category:"Matrix",syntax:["identity(n)","identity(m, n)","identity([m, n])"],description:"Returns the identity matrix with size m-by-n. The matrix has ones on the diagonal and zeros elsewhere.",examples:["identity(3)","identity(3, 5)","a = [1, 2, 3; 4, 5, 6]","identity(size(a))"],seealso:["concat","det","diag","inv","ones","range","size","squeeze","subset","trace","transpose","zeros"]},_M={name:"inv",category:"Matrix",syntax:["inv(x)"],description:"Calculate the inverse of a matrix",examples:["inv([1, 2; 3, 4])","inv(4)","1 / 4"],seealso:["concat","det","diag","identity","ones","range","size","squeeze","subset","trace","transpose","zeros"]},BM={name:"pinv",category:"Matrix",syntax:["pinv(x)"],description:"Calculate the Moore–Penrose inverse of a matrix",examples:["pinv([1, 2; 3, 4])","pinv([[1, 0], [0, 1], [0, 1]])","pinv(4)"],seealso:["inv"]},OM={name:"kron",category:"Matrix",syntax:["kron(x, y)"],description:"Calculates the kronecker product of 2 matrices or vectors.",examples:["kron([[1, 0], [0, 1]], [[1, 2], [3, 4]])","kron([1,1], [2,3,4])"],seealso:["multiply","dot","cross"]},PM={name:"map",category:"Matrix",syntax:["map(x, callback)"],description:"Create a new matrix or array with the results of the callback function executed on each entry of the matrix/array.",examples:["map([1, 2, 3], square)"],seealso:["filter","forEach"]},zM={name:"matrixFromColumns",category:"Matrix",syntax:["matrixFromColumns(...arr)","matrixFromColumns(row1, row2)","matrixFromColumns(row1, row2, row3)"],description:"Create a dense matrix from vectors as individual columns.",examples:["matrixFromColumns([1, 2, 3], [[4],[5],[6]])"],seealso:["matrix","matrixFromRows","matrixFromFunction","zeros"]},RM={name:"matrixFromFunction",category:"Matrix",syntax:["matrixFromFunction(size, fn)","matrixFromFunction(size, fn, format)","matrixFromFunction(size, fn, format, datatype)","matrixFromFunction(size, format, fn)","matrixFromFunction(size, format, datatype, fn)"],description:"Create a matrix by evaluating a generating function at each index.",examples:["f(I) = I[1] - I[2]","matrixFromFunction([3,3], f)","g(I) = I[1] - I[2] == 1 ? 4 : 0",'matrixFromFunction([100, 100], "sparse", g)',"matrixFromFunction([5], random)"],seealso:["matrix","matrixFromRows","matrixFromColumns","zeros"]},IM={name:"matrixFromRows",category:"Matrix",syntax:["matrixFromRows(...arr)","matrixFromRows(row1, row2)","matrixFromRows(row1, row2, row3)"],description:"Create a dense matrix from vectors as individual rows.",examples:["matrixFromRows([1, 2, 3], [[4],[5],[6]])"],seealso:["matrix","matrixFromColumns","matrixFromFunction","zeros"]},qM={name:"ones",category:"Matrix",syntax:["ones(m)","ones(m, n)","ones(m, n, p, ...)","ones([m])","ones([m, n])","ones([m, n, p, ...])"],description:"Create a matrix containing ones.",examples:["ones(3)","ones(3, 5)","ones([2,3]) * 4.5","a = [1, 2, 3; 4, 5, 6]","ones(size(a))"],seealso:["concat","det","diag","identity","inv","range","size","squeeze","subset","trace","transpose","zeros"]},LM={name:"partitionSelect",category:"Matrix",syntax:["partitionSelect(x, k)","partitionSelect(x, k, compare)"],description:"Partition-based selection of an array or 1D matrix. Will find the kth smallest value, and mutates the input array. Uses Quickselect.",examples:["partitionSelect([5, 10, 1], 2)",'partitionSelect(["C", "B", "A", "D"], 1, compareText)',"arr = [5, 2, 1]","partitionSelect(arr, 0) # returns 1, arr is now: [1, 2, 5]","arr","partitionSelect(arr, 1, 'desc') # returns 2, arr is now: [5, 2, 1]","arr"],seealso:["sort"]},kM={name:"range",category:"Type",syntax:["start:end","start:step:end","range(start, end)","range(start, end, step)","range(string)"],description:"Create a range. Lower bound of the range is included, upper bound is excluded.",examples:["1:5","3:-1:-3","range(3, 7)","range(0, 12, 2)",'range("4:10")',"range(1m, 1m, 3m)","a = [1, 2, 3, 4; 5, 6, 7, 8]","a[1:2, 1:2]"],seealso:["concat","det","diag","identity","inv","ones","size","squeeze","subset","trace","transpose","zeros"]},UM={name:"reshape",category:"Matrix",syntax:["reshape(x, sizes)"],description:"Reshape a multi dimensional array to fit the specified dimensions.",examples:["reshape([1, 2, 3, 4, 5, 6], [2, 3])","reshape([[1, 2], [3, 4]], [1, 4])","reshape([[1, 2], [3, 4]], [4])","reshape([1, 2, 3, 4], [-1, 2])"],seealso:["size","squeeze","resize"]},GM={name:"resize",category:"Matrix",syntax:["resize(x, size)","resize(x, size, defaultValue)"],description:"Resize a matrix.",examples:["resize([1,2,3,4,5], [3])","resize([1,2,3], [5])","resize([1,2,3], [5], -1)","resize(2, [2, 3])",'resize("hello", [8], "!")'],seealso:["size","subset","squeeze","reshape"]},HM={name:"rotate",category:"Matrix",syntax:["rotate(w, theta)","rotate(w, theta, v)"],description:"Returns a 2-D rotation matrix (2x2) for a given angle (in radians). Returns a 2-D rotation matrix (3x3) of a given angle (in radians) around given axis.",examples:["rotate([1, 0], pi / 2)",'rotate(matrix([1, 0]), unit("35deg"))','rotate([1, 0, 0], unit("90deg"), [0, 0, 1])','rotate(matrix([1, 0, 0]), unit("90deg"), matrix([0, 0, 1]))'],seealso:["matrix","rotationMatrix"]},VM={name:"rotationMatrix",category:"Matrix",syntax:["rotationMatrix(theta)","rotationMatrix(theta, v)","rotationMatrix(theta, v, format)"],description:"Returns a 2-D rotation matrix (2x2) for a given angle (in radians). Returns a 2-D rotation matrix (3x3) of a given angle (in radians) around given axis.",examples:["rotationMatrix(pi / 2)",'rotationMatrix(unit("45deg"), [0, 0, 1])','rotationMatrix(1, matrix([0, 0, 1]), "sparse")'],seealso:["cos","sin"]},JM={name:"row",category:"Matrix",syntax:["row(x, index)"],description:"Return a row from a matrix or array.",examples:["A = [[1, 2], [3, 4]]","row(A, 1)","row(A, 2)"],seealso:["column","matrixFromRows"]},ZM={name:"size",category:"Matrix",syntax:["size(x)"],description:"Calculate the size of a matrix.",examples:["size(2.3)",'size("hello world")',"a = [1, 2; 3, 4; 5, 6]","size(a)","size(1:6)"],seealso:["concat","count","det","diag","identity","inv","ones","range","squeeze","subset","trace","transpose","zeros"]},WM={name:"sort",category:"Matrix",syntax:["sort(x)","sort(x, compare)"],description:'Sort the items in a matrix. Compare can be a string "asc", "desc", "natural", or a custom sort function.',examples:["sort([5, 10, 1])",'sort(["C", "B", "A", "D"], "natural")',"sortByLength(a, b) = size(a)[1] - size(b)[1]",'sort(["Langdon", "Tom", "Sara"], sortByLength)','sort(["10", "1", "2"], "natural")'],seealso:["map","filter","forEach"]},YM={name:"squeeze",category:"Matrix",syntax:["squeeze(x)"],description:"Remove inner and outer singleton dimensions from a matrix.",examples:["a = zeros(3,2,1)","size(squeeze(a))","b = zeros(1,1,3)","size(squeeze(b))"],seealso:["concat","det","diag","identity","inv","ones","range","size","subset","trace","transpose","zeros"]},XM={name:"subset",category:"Matrix",syntax:["value(index)","value(index) = replacement","subset(value, [index])","subset(value, [index], replacement)"],description:"Get or set a subset of the entries of a matrix or characters of a string. Indexes are one-based. There should be one index specification for each dimension of the target. Each specification can be a single index, a list of indices, or a range in colon notation `l:u`. In a range, both the lower bound l and upper bound u are included; and if a bound is omitted it defaults to the most extreme valid value. The cartesian product of the indices specified in each dimension determines the target of the operation.",examples:["d = [1, 2; 3, 4]","e = []","e[1, 1:2] = [5, 6]","e[2, :] = [7, 8]","f = d * e","f[2, 1]","f[:, 1]","f[[1,2], [1,3]] = [9, 10; 11, 12]","f"],seealso:["concat","det","diag","identity","inv","ones","range","size","squeeze","trace","transpose","zeros"]},QM={name:"trace",category:"Matrix",syntax:["trace(A)"],description:"Calculate the trace of a matrix: the sum of the elements on the main diagonal of a square matrix.",examples:["A = [1, 2, 3; -1, 2, 3; 2, 0, 3]","trace(A)"],seealso:["concat","det","diag","identity","inv","ones","range","size","squeeze","subset","transpose","zeros"]},jM={name:"transpose",category:"Matrix",syntax:["x'","transpose(x)"],description:"Transpose a matrix",examples:["a = [1, 2, 3; 4, 5, 6]","a'","transpose(a)"],seealso:["concat","det","diag","identity","inv","ones","range","size","squeeze","subset","trace","zeros"]},KM={name:"zeros",category:"Matrix",syntax:["zeros(m)","zeros(m, n)","zeros(m, n, p, ...)","zeros([m])","zeros([m, n])","zeros([m, n, p, ...])"],description:"Create a matrix containing zeros.",examples:["zeros(3)","zeros(3, 5)","a = [1, 2, 3; 4, 5, 6]","zeros(size(a))"],seealso:["concat","det","diag","identity","inv","ones","range","size","squeeze","subset","trace","transpose"]},e7={name:"fft",category:"Matrix",syntax:["fft(x)"],description:"Calculate N-dimensional fourier transform",examples:["fft([[1, 0], [1, 0]])"],seealso:["ifft"]},t7={name:"ifft",category:"Matrix",syntax:["ifft(x)"],description:"Calculate N-dimensional inverse fourier transform",examples:["ifft([[2, 2], [0, 0]])"],seealso:["fft"]},a7={name:"combinations",category:"Probability",syntax:["combinations(n, k)"],description:"Compute the number of combinations of n items taken k at a time",examples:["combinations(7, 5)"],seealso:["combinationsWithRep","permutations","factorial"]},r7={name:"combinationsWithRep",category:"Probability",syntax:["combinationsWithRep(n, k)"],description:"Compute the number of combinations of n items taken k at a time with replacements.",examples:["combinationsWithRep(7, 5)"],seealso:["combinations","permutations","factorial"]},n7={name:"factorial",category:"Probability",syntax:["n!","factorial(n)"],description:"Compute the factorial of a value",examples:["5!","5 * 4 * 3 * 2 * 1","3!"],seealso:["combinations","combinationsWithRep","permutations","gamma"]},s7={name:"gamma",category:"Probability",syntax:["gamma(n)"],description:"Compute the gamma function. For small values, the Lanczos approximation is used, and for large values the extended Stirling approximation.",examples:["gamma(4)","3!","gamma(1/2)","sqrt(pi)"],seealso:["factorial"]},i7={name:"lgamma",category:"Probability",syntax:["lgamma(n)"],description:"Logarithm of the gamma function for real, positive numbers and complex numbers, using Lanczos approximation for numbers and Stirling series for complex numbers.",examples:["lgamma(4)","lgamma(1/2)","lgamma(i)","lgamma(complex(1.1, 2))"],seealso:["gamma"]},o7={name:"kldivergence",category:"Probability",syntax:["kldivergence(x, y)"],description:"Calculate the Kullback-Leibler (KL) divergence between two distributions.",examples:["kldivergence([0.7,0.5,0.4], [0.2,0.9,0.5])"],seealso:[]},l7={name:"multinomial",category:"Probability",syntax:["multinomial(A)"],description:"Multinomial Coefficients compute the number of ways of picking a1, a2, ..., ai unordered outcomes from `n` possibilities. multinomial takes one array of integers as an argument. The following condition must be enforced: every ai > 0.",examples:["multinomial([1, 2, 1])"],seealso:["combinations","factorial"]},m7={name:"permutations",category:"Probability",syntax:["permutations(n)","permutations(n, k)"],description:"Compute the number of permutations of n items taken k at a time",examples:["permutations(5)","permutations(5, 3)"],seealso:["combinations","combinationsWithRep","factorial"]},u7={name:"pickRandom",category:"Probability",syntax:["pickRandom(array)","pickRandom(array, number)","pickRandom(array, weights)","pickRandom(array, number, weights)","pickRandom(array, weights, number)"],description:"Pick a random entry from a given array.",examples:["pickRandom(0:10)","pickRandom([1, 3, 1, 6])","pickRandom([1, 3, 1, 6], 2)","pickRandom([1, 3, 1, 6], [2, 3, 2, 1])","pickRandom([1, 3, 1, 6], 2, [2, 3, 2, 1])","pickRandom([1, 3, 1, 6], [2, 3, 2, 1], 2)"],seealso:["random","randomInt"]},p7={name:"random",category:"Probability",syntax:["random()","random(max)","random(min, max)","random(size)","random(size, max)","random(size, min, max)"],description:"Return a random number.",examples:["random()","random(10, 20)","random([2, 3])"],seealso:["pickRandom","randomInt"]},c7={name:"randomInt",category:"Probability",syntax:["randomInt(max)","randomInt(min, max)","randomInt(size)","randomInt(size, max)","randomInt(size, min, max)"],description:"Return a random integer number",examples:["randomInt(10, 20)","randomInt([2, 3], 10)"],seealso:["pickRandom","random"]},f7={name:"compare",category:"Relational",syntax:["compare(x, y)"],description:"Compare two values. Returns 1 when x > y, -1 when x < y, and 0 when x == y.",examples:["compare(2, 3)","compare(3, 2)","compare(2, 2)","compare(5cm, 40mm)","compare(2, [1, 2, 3])"],seealso:["equal","unequal","smaller","smallerEq","largerEq","compareNatural","compareText"]},h7={name:"compareNatural",category:"Relational",syntax:["compareNatural(x, y)"],description:"Compare two values of any type in a deterministic, natural way. Returns 1 when x > y, -1 when x < y, and 0 when x == y.",examples:["compareNatural(2, 3)","compareNatural(3, 2)","compareNatural(2, 2)","compareNatural(5cm, 40mm)",'compareNatural("2", "10")',"compareNatural(2 + 3i, 2 + 4i)","compareNatural([1, 2, 4], [1, 2, 3])","compareNatural([1, 5], [1, 2, 3])","compareNatural([1, 2], [1, 2])","compareNatural({a: 2}, {a: 4})"],seealso:["equal","unequal","smaller","smallerEq","largerEq","compare","compareText"]},v7={name:"compareText",category:"Relational",syntax:["compareText(x, y)"],description:"Compare two strings lexically. Comparison is case sensitive. Returns 1 when x > y, -1 when x < y, and 0 when x == y.",examples:['compareText("B", "A")','compareText("A", "B")','compareText("A", "A")','compareText("2", "10")','compare("2", "10")',"compare(2, 10)",'compareNatural("2", "10")','compareText("B", ["A", "B", "C"])'],seealso:["compare","compareNatural"]},g7={name:"deepEqual",category:"Relational",syntax:["deepEqual(x, y)"],description:"Check equality of two matrices element wise. Returns true if the size of both matrices is equal and when and each of the elements are equal.",examples:["deepEqual([1,3,4], [1,3,4])","deepEqual([1,3,4], [1,3])"],seealso:["equal","unequal","smaller","larger","smallerEq","largerEq","compare"]},d7={name:"equal",category:"Relational",syntax:["x == y","equal(x, y)"],description:"Check equality of two values. Returns true if the values are equal, and false if not.",examples:["2+2 == 3","2+2 == 4","a = 3.2","b = 6-2.8","a == b","50cm == 0.5m"],seealso:["unequal","smaller","larger","smallerEq","largerEq","compare","deepEqual","equalText"]},y7={name:"equalText",category:"Relational",syntax:["equalText(x, y)"],description:"Check equality of two strings. Comparison is case sensitive. Returns true if the values are equal, and false if not.",examples:['equalText("Hello", "Hello")','equalText("a", "A")','equal("2e3", "2000")','equalText("2e3", "2000")','equalText("B", ["A", "B", "C"])'],seealso:["compare","compareNatural","compareText","equal"]},w7={name:"larger",category:"Relational",syntax:["x > y","larger(x, y)"],description:"Check if value x is larger than y. Returns true if x is larger than y, and false if not.",examples:["2 > 3","5 > 2*2","a = 3.3","b = 6-2.8","(a > b)","(b < a)","5 cm > 2 inch"],seealso:["equal","unequal","smaller","smallerEq","largerEq","compare"]},x7={name:"largerEq",category:"Relational",syntax:["x >= y","largerEq(x, y)"],description:"Check if value x is larger or equal to y. Returns true if x is larger or equal to y, and false if not.",examples:["2 >= 1+1","2 > 1+1","a = 3.2","b = 6-2.8","(a >= b)"],seealso:["equal","unequal","smallerEq","smaller","compare"]},b7={name:"smaller",category:"Relational",syntax:["x < y","smaller(x, y)"],description:"Check if value x is smaller than value y. Returns true if x is smaller than y, and false if not.",examples:["2 < 3","5 < 2*2","a = 3.3","b = 6-2.8","(a < b)","5 cm < 2 inch"],seealso:["equal","unequal","larger","smallerEq","largerEq","compare"]},N7={name:"smallerEq",category:"Relational",syntax:["x <= y","smallerEq(x, y)"],description:"Check if value x is smaller or equal to value y. Returns true if x is smaller than y, and false if not.",examples:["2 <= 1+1","2 < 1+1","a = 3.2","b = 6-2.8","(a <= b)"],seealso:["equal","unequal","larger","smaller","largerEq","compare"]},E7={name:"unequal",category:"Relational",syntax:["x != y","unequal(x, y)"],description:"Check unequality of two values. Returns true if the values are unequal, and false if they are equal.",examples:["2+2 != 3","2+2 != 4","a = 3.2","b = 6-2.8","a != b","50cm != 0.5m","5 cm != 2 inch"],seealso:["equal","smaller","larger","smallerEq","largerEq","compare","deepEqual"]},D7={name:"setCartesian",category:"Set",syntax:["setCartesian(set1, set2)"],description:"Create the cartesian product of two (multi)sets. Multi-dimension arrays will be converted to single-dimension arrays and the values will be sorted in ascending order before the operation.",examples:["setCartesian([1, 2], [3, 4])"],seealso:["setUnion","setIntersect","setDifference","setPowerset"]},A7={name:"setDifference",category:"Set",syntax:["setDifference(set1, set2)"],description:"Create the difference of two (multi)sets: every element of set1, that is not the element of set2. Multi-dimension arrays will be converted to single-dimension arrays before the operation.",examples:["setDifference([1, 2, 3, 4], [3, 4, 5, 6])","setDifference([[1, 2], [3, 4]], [[3, 4], [5, 6]])"],seealso:["setUnion","setIntersect","setSymDifference"]},M7={name:"setDistinct",category:"Set",syntax:["setDistinct(set)"],description:"Collect the distinct elements of a multiset. A multi-dimension array will be converted to a single-dimension array before the operation.",examples:["setDistinct([1, 1, 1, 2, 2, 3])"],seealso:["setMultiplicity"]},S7={name:"setIntersect",category:"Set",syntax:["setIntersect(set1, set2)"],description:"Create the intersection of two (multi)sets. Multi-dimension arrays will be converted to single-dimension arrays before the operation.",examples:["setIntersect([1, 2, 3, 4], [3, 4, 5, 6])","setIntersect([[1, 2], [3, 4]], [[3, 4], [5, 6]])"],seealso:["setUnion","setDifference"]},C7={name:"setIsSubset",category:"Set",syntax:["setIsSubset(set1, set2)"],description:"Check whether a (multi)set is a subset of another (multi)set: every element of set1 is the element of set2. Multi-dimension arrays will be converted to single-dimension arrays before the operation.",examples:["setIsSubset([1, 2], [3, 4, 5, 6])","setIsSubset([3, 4], [3, 4, 5, 6])"],seealso:["setUnion","setIntersect","setDifference"]},T7={name:"setMultiplicity",category:"Set",syntax:["setMultiplicity(element, set)"],description:"Count the multiplicity of an element in a multiset. A multi-dimension array will be converted to a single-dimension array before the operation.",examples:["setMultiplicity(1, [1, 2, 2, 4])","setMultiplicity(2, [1, 2, 2, 4])"],seealso:["setDistinct","setSize"]},$7={name:"setPowerset",category:"Set",syntax:["setPowerset(set)"],description:"Create the powerset of a (multi)set: the powerset contains very possible subsets of a (multi)set. A multi-dimension array will be converted to a single-dimension array before the operation.",examples:["setPowerset([1, 2, 3])"],seealso:["setCartesian"]},F7={name:"setSize",category:"Set",syntax:["setSize(set)","setSize(set, unique)"],description:'Count the number of elements of a (multi)set. When the second parameter "unique" is true, count only the unique values. A multi-dimension array will be converted to a single-dimension array before the operation.',examples:["setSize([1, 2, 2, 4])","setSize([1, 2, 2, 4], true)"],seealso:["setUnion","setIntersect","setDifference"]},_7={name:"setSymDifference",category:"Set",syntax:["setSymDifference(set1, set2)"],description:"Create the symmetric difference of two (multi)sets. Multi-dimension arrays will be converted to single-dimension arrays before the operation.",examples:["setSymDifference([1, 2, 3, 4], [3, 4, 5, 6])","setSymDifference([[1, 2], [3, 4]], [[3, 4], [5, 6]])"],seealso:["setUnion","setIntersect","setDifference"]},B7={name:"setUnion",category:"Set",syntax:["setUnion(set1, set2)"],description:"Create the union of two (multi)sets. Multi-dimension arrays will be converted to single-dimension arrays before the operation.",examples:["setUnion([1, 2, 3, 4], [3, 4, 5, 6])","setUnion([[1, 2], [3, 4]], [[3, 4], [5, 6]])"],seealso:["setIntersect","setDifference"]},O7={name:"zpk2tf",category:"Signal",syntax:["zpk2tf(z, p, k)"],description:"Compute the transfer function of a zero-pole-gain model.",examples:["zpk2tf([1, 2], [-1, -2], 1)","zpk2tf([1, 2], [-1, -2])","zpk2tf([1 - 3i, 2 + 2i], [-1, -2])"],seealso:[]},P7={name:"freqz",category:"Signal",syntax:["freqz(b, a)","freqz(b, a, w)"],description:"Calculates the frequency response of a filter given its numerator and denominator coefficients.",examples:["freqz([1, 2], [1, 2, 3])","freqz([1, 2], [1, 2, 3], [0, 1])","freqz([1, 2], [1, 2, 3], 512)"],seealso:[]},z7={name:"erf",category:"Special",syntax:["erf(x)"],description:"Compute the erf function of a value using a rational Chebyshev approximations for different intervals of x",examples:["erf(0.2)","erf(-0.5)","erf(4)"],seealso:[]},R7={name:"zeta",category:"Special",syntax:["zeta(s)"],description:"Compute the Riemann Zeta Function using an infinite series and Riemanns Functional Equation for the entire complex plane",examples:["zeta(0.2)","zeta(-0.5)","zeta(4)"],seealso:[]},I7={name:"mad",category:"Statistics",syntax:["mad(a, b, c, ...)","mad(A)"],description:"Compute the median absolute deviation of a matrix or a list with values. The median absolute deviation is defined as the median of the absolute deviations from the median.",examples:["mad(10, 20, 30)","mad([1, 2, 3])"],seealso:["mean","median","std","abs"]},q7={name:"max",category:"Statistics",syntax:["max(a, b, c, ...)","max(A)","max(A, dimension)"],description:"Compute the maximum value of a list of values.",examples:["max(2, 3, 4, 1)","max([2, 3, 4, 1])","max([2, 5; 4, 3])","max([2, 5; 4, 3], 1)","max([2, 5; 4, 3], 2)","max(2.7, 7.1, -4.5, 2.0, 4.1)","min(2.7, 7.1, -4.5, 2.0, 4.1)"],seealso:["mean","median","min","prod","std","sum","variance"]},L7={name:"mean",category:"Statistics",syntax:["mean(a, b, c, ...)","mean(A)","mean(A, dimension)"],description:"Compute the arithmetic mean of a list of values.",examples:["mean(2, 3, 4, 1)","mean([2, 3, 4, 1])","mean([2, 5; 4, 3])","mean([2, 5; 4, 3], 1)","mean([2, 5; 4, 3], 2)","mean([1.0, 2.7, 3.2, 4.0])"],seealso:["max","median","min","prod","std","sum","variance"]},k7={name:"median",category:"Statistics",syntax:["median(a, b, c, ...)","median(A)"],description:"Compute the median of all values. The values are sorted and the middle value is returned. In case of an even number of values, the average of the two middle values is returned.",examples:["median(5, 2, 7)","median([3, -1, 5, 7])"],seealso:["max","mean","min","prod","std","sum","variance","quantileSeq"]},U7={name:"min",category:"Statistics",syntax:["min(a, b, c, ...)","min(A)","min(A, dimension)"],description:"Compute the minimum value of a list of values.",examples:["min(2, 3, 4, 1)","min([2, 3, 4, 1])","min([2, 5; 4, 3])","min([2, 5; 4, 3], 1)","min([2, 5; 4, 3], 2)","min(2.7, 7.1, -4.5, 2.0, 4.1)","max(2.7, 7.1, -4.5, 2.0, 4.1)"],seealso:["max","mean","median","prod","std","sum","variance"]},G7={name:"mode",category:"Statistics",syntax:["mode(a, b, c, ...)","mode(A)","mode(A, a, b, B, c, ...)"],description:"Computes the mode of all values as an array. In case mode being more than one, multiple values are returned in an array.",examples:["mode(2, 1, 4, 3, 1)","mode([1, 2.7, 3.2, 4, 2.7])","mode(1, 4, 6, 1, 6)"],seealso:["max","mean","min","median","prod","std","sum","variance"]},H7={name:"prod",category:"Statistics",syntax:["prod(a, b, c, ...)","prod(A)"],description:"Compute the product of all values.",examples:["prod(2, 3, 4)","prod([2, 3, 4])","prod([2, 5; 4, 3])"],seealso:["max","mean","min","median","min","std","sum","variance"]},V7={name:"quantileSeq",category:"Statistics",syntax:["quantileSeq(A, prob[, sorted])","quantileSeq(A, [prob1, prob2, ...][, sorted])","quantileSeq(A, N[, sorted])"],description:`Compute the prob order quantile of a matrix or a list with values. The sequence is sorted and the middle value is returned. Supported types of sequence values are: Number, BigNumber, Unit Supported types of probablity are: Number, BigNumber. - -In case of a (multi dimensional) array or matrix, the prob order quantile of all elements will be calculated.`,examples:["quantileSeq([3, -1, 5, 7], 0.5)","quantileSeq([3, -1, 5, 7], [1/3, 2/3])","quantileSeq([3, -1, 5, 7], 2)","quantileSeq([-1, 3, 5, 7], 0.5, true)"],seealso:["mean","median","min","max","prod","std","sum","variance"]},J7={name:"std",category:"Statistics",syntax:["std(a, b, c, ...)","std(A)","std(A, dimension)","std(A, normalization)","std(A, dimension, normalization)"],description:'Compute the standard deviation of all values, defined as std(A) = sqrt(variance(A)). Optional parameter normalization can be "unbiased" (default), "uncorrected", or "biased".',examples:["std(2, 4, 6)","std([2, 4, 6, 8])",'std([2, 4, 6, 8], "uncorrected")','std([2, 4, 6, 8], "biased")',"std([1, 2, 3; 4, 5, 6])"],seealso:["max","mean","min","median","prod","sum","variance"]},Z7={name:"cumsum",category:"Statistics",syntax:["cumsum(a, b, c, ...)","cumsum(A)"],description:"Compute the cumulative sum of all values.",examples:["cumsum(2, 3, 4, 1)","cumsum([2, 3, 4, 1])","cumsum([1, 2; 3, 4])","cumsum([1, 2; 3, 4], 1)","cumsum([1, 2; 3, 4], 2)"],seealso:["max","mean","median","min","prod","std","sum","variance"]},W7={name:"sum",category:"Statistics",syntax:["sum(a, b, c, ...)","sum(A)","sum(A, dimension)"],description:"Compute the sum of all values.",examples:["sum(2, 3, 4, 1)","sum([2, 3, 4, 1])","sum([2, 5; 4, 3])"],seealso:["max","mean","median","min","prod","std","sum","variance"]},Y7={name:"variance",category:"Statistics",syntax:["variance(a, b, c, ...)","variance(A)","variance(A, dimension)","variance(A, normalization)","variance(A, dimension, normalization)"],description:'Compute the variance of all values. Optional parameter normalization can be "unbiased" (default), "uncorrected", or "biased".',examples:["variance(2, 4, 6)","variance([2, 4, 6, 8])",'variance([2, 4, 6, 8], "uncorrected")','variance([2, 4, 6, 8], "biased")',"variance([1, 2, 3; 4, 5, 6])"],seealso:["max","mean","min","median","min","prod","std","sum"]},X7={name:"corr",category:"Statistics",syntax:["corr(A,B)"],description:"Compute the correlation coefficient of a two list with values, For matrices, the matrix correlation coefficient is calculated.",examples:["corr([2, 4, 6, 8],[1, 2, 3, 6])","corr(matrix([[1, 2.2, 3, 4.8, 5], [1, 2, 3, 4, 5]]), matrix([[4, 5.3, 6.6, 7, 8], [1, 2, 3, 4, 5]]))"],seealso:["max","mean","min","median","min","prod","std","sum"]},Q7={name:"acos",category:"Trigonometry",syntax:["acos(x)"],description:"Compute the inverse cosine of a value in radians.",examples:["acos(0.5)","acos(cos(2.3))"],seealso:["cos","atan","asin"]},j7={name:"acosh",category:"Trigonometry",syntax:["acosh(x)"],description:"Calculate the hyperbolic arccos of a value, defined as `acosh(x) = ln(sqrt(x^2 - 1) + x)`.",examples:["acosh(1.5)"],seealso:["cosh","asinh","atanh"]},K7={name:"acot",category:"Trigonometry",syntax:["acot(x)"],description:"Calculate the inverse cotangent of a value.",examples:["acot(0.5)","acot(cot(0.5))","acot(2)"],seealso:["cot","atan"]},e4={name:"acoth",category:"Trigonometry",syntax:["acoth(x)"],description:"Calculate the hyperbolic arccotangent of a value, defined as `acoth(x) = (ln((x+1)/x) + ln(x/(x-1))) / 2`.",examples:["acoth(2)","acoth(0.5)"],seealso:["acsch","asech"]},t4={name:"acsc",category:"Trigonometry",syntax:["acsc(x)"],description:"Calculate the inverse cotangent of a value.",examples:["acsc(2)","acsc(csc(0.5))","acsc(0.5)"],seealso:["csc","asin","asec"]},a4={name:"acsch",category:"Trigonometry",syntax:["acsch(x)"],description:"Calculate the hyperbolic arccosecant of a value, defined as `acsch(x) = ln(1/x + sqrt(1/x^2 + 1))`.",examples:["acsch(0.5)"],seealso:["asech","acoth"]},r4={name:"asec",category:"Trigonometry",syntax:["asec(x)"],description:"Calculate the inverse secant of a value.",examples:["asec(0.5)","asec(sec(0.5))","asec(2)"],seealso:["acos","acot","acsc"]},n4={name:"asech",category:"Trigonometry",syntax:["asech(x)"],description:"Calculate the inverse secant of a value.",examples:["asech(0.5)"],seealso:["acsch","acoth"]},s4={name:"asin",category:"Trigonometry",syntax:["asin(x)"],description:"Compute the inverse sine of a value in radians.",examples:["asin(0.5)","asin(sin(0.5))"],seealso:["sin","acos","atan"]},i4={name:"asinh",category:"Trigonometry",syntax:["asinh(x)"],description:"Calculate the hyperbolic arcsine of a value, defined as `asinh(x) = ln(x + sqrt(x^2 + 1))`.",examples:["asinh(0.5)"],seealso:["acosh","atanh"]},o4={name:"atan",category:"Trigonometry",syntax:["atan(x)"],description:"Compute the inverse tangent of a value in radians.",examples:["atan(0.5)","atan(tan(0.5))"],seealso:["tan","acos","asin"]},l4={name:"atan2",category:"Trigonometry",syntax:["atan2(y, x)"],description:"Computes the principal value of the arc tangent of y/x in radians.",examples:["atan2(2, 2) / pi","angle = 60 deg in rad","x = cos(angle)","y = sin(angle)","atan2(y, x)"],seealso:["sin","cos","tan"]},m4={name:"atanh",category:"Trigonometry",syntax:["atanh(x)"],description:"Calculate the hyperbolic arctangent of a value, defined as `atanh(x) = ln((1 + x)/(1 - x)) / 2`.",examples:["atanh(0.5)"],seealso:["acosh","asinh"]},u4={name:"cos",category:"Trigonometry",syntax:["cos(x)"],description:"Compute the cosine of x in radians.",examples:["cos(2)","cos(pi / 4) ^ 2","cos(180 deg)","cos(60 deg)","sin(0.2)^2 + cos(0.2)^2"],seealso:["acos","sin","tan"]},p4={name:"cosh",category:"Trigonometry",syntax:["cosh(x)"],description:"Compute the hyperbolic cosine of x in radians.",examples:["cosh(0.5)"],seealso:["sinh","tanh","coth"]},c4={name:"cot",category:"Trigonometry",syntax:["cot(x)"],description:"Compute the cotangent of x in radians. Defined as 1/tan(x)",examples:["cot(2)","1 / tan(2)"],seealso:["sec","csc","tan"]},f4={name:"coth",category:"Trigonometry",syntax:["coth(x)"],description:"Compute the hyperbolic cotangent of x in radians.",examples:["coth(2)","1 / tanh(2)"],seealso:["sech","csch","tanh"]},h4={name:"csc",category:"Trigonometry",syntax:["csc(x)"],description:"Compute the cosecant of x in radians. Defined as 1/sin(x)",examples:["csc(2)","1 / sin(2)"],seealso:["sec","cot","sin"]},v4={name:"csch",category:"Trigonometry",syntax:["csch(x)"],description:"Compute the hyperbolic cosecant of x in radians. Defined as 1/sinh(x)",examples:["csch(2)","1 / sinh(2)"],seealso:["sech","coth","sinh"]},g4={name:"sec",category:"Trigonometry",syntax:["sec(x)"],description:"Compute the secant of x in radians. Defined as 1/cos(x)",examples:["sec(2)","1 / cos(2)"],seealso:["cot","csc","cos"]},d4={name:"sech",category:"Trigonometry",syntax:["sech(x)"],description:"Compute the hyperbolic secant of x in radians. Defined as 1/cosh(x)",examples:["sech(2)","1 / cosh(2)"],seealso:["coth","csch","cosh"]},y4={name:"sin",category:"Trigonometry",syntax:["sin(x)"],description:"Compute the sine of x in radians.",examples:["sin(2)","sin(pi / 4) ^ 2","sin(90 deg)","sin(30 deg)","sin(0.2)^2 + cos(0.2)^2"],seealso:["asin","cos","tan"]},w4={name:"sinh",category:"Trigonometry",syntax:["sinh(x)"],description:"Compute the hyperbolic sine of x in radians.",examples:["sinh(0.5)"],seealso:["cosh","tanh"]},x4={name:"tan",category:"Trigonometry",syntax:["tan(x)"],description:"Compute the tangent of x in radians.",examples:["tan(0.5)","sin(0.5) / cos(0.5)","tan(pi / 4)","tan(45 deg)"],seealso:["atan","sin","cos"]},b4={name:"tanh",category:"Trigonometry",syntax:["tanh(x)"],description:"Compute the hyperbolic tangent of x in radians.",examples:["tanh(0.5)","sinh(0.5) / cosh(0.5)"],seealso:["sinh","cosh"]},N4={name:"to",category:"Units",syntax:["x to unit","to(x, unit)"],description:"Change the unit of a value.",examples:["5 inch to cm","3.2kg to g","16 bytes in bits"],seealso:[]},E4={name:"bin",category:"Utils",syntax:["bin(value)"],description:"Format a number as binary",examples:["bin(2)"],seealso:["oct","hex"]},D4={name:"clone",category:"Utils",syntax:["clone(x)"],description:"Clone a variable. Creates a copy of primitive variables,and a deep copy of matrices",examples:["clone(3.5)","clone(2 - 4i)","clone(45 deg)","clone([1, 2; 3, 4])",'clone("hello world")'],seealso:[]},A4={name:"format",category:"Utils",syntax:["format(value)","format(value, precision)"],description:"Format a value of any type as string.",examples:["format(2.3)","format(3 - 4i)","format([])","format(pi, 3)"],seealso:["print"]},M4={name:"hasNumericValue",category:"Utils",syntax:["hasNumericValue(x)"],description:"Test whether a value is an numeric value. In case of a string, true is returned if the string contains a numeric value.",examples:["hasNumericValue(2)",'hasNumericValue("2")','isNumeric("2")',"hasNumericValue(0)","hasNumericValue(bignumber(500))","hasNumericValue(fraction(0.125))","hasNumericValue(2 + 3i)",'hasNumericValue([2.3, "foo", false])'],seealso:["isInteger","isZero","isNegative","isPositive","isNaN","isNumeric"]},S4={name:"hex",category:"Utils",syntax:["hex(value)"],description:"Format a number as hexadecimal",examples:["hex(240)"],seealso:["bin","oct"]},C4={name:"isInteger",category:"Utils",syntax:["isInteger(x)"],description:"Test whether a value is an integer number.",examples:["isInteger(2)","isInteger(3.5)","isInteger([3, 0.5, -2])"],seealso:["isNegative","isNumeric","isPositive","isZero"]},T4={name:"isNaN",category:"Utils",syntax:["isNaN(x)"],description:"Test whether a value is NaN (not a number)",examples:["isNaN(2)","isNaN(0 / 0)","isNaN(NaN)","isNaN(Infinity)"],seealso:["isNegative","isNumeric","isPositive","isZero"]},$4={name:"isNegative",category:"Utils",syntax:["isNegative(x)"],description:"Test whether a value is negative: smaller than zero.",examples:["isNegative(2)","isNegative(0)","isNegative(-4)","isNegative([3, 0.5, -2])"],seealso:["isInteger","isNumeric","isPositive","isZero"]},F4={name:"isNumeric",category:"Utils",syntax:["isNumeric(x)"],description:"Test whether a value is a numeric value. Returns true when the input is a number, BigNumber, Fraction, or boolean.",examples:["isNumeric(2)",'isNumeric("2")','hasNumericValue("2")',"isNumeric(0)","isNumeric(bignumber(500))","isNumeric(fraction(0.125))","isNumeric(2 + 3i)",'isNumeric([2.3, "foo", false])'],seealso:["isInteger","isZero","isNegative","isPositive","isNaN","hasNumericValue"]},_4={name:"isPositive",category:"Utils",syntax:["isPositive(x)"],description:"Test whether a value is positive: larger than zero.",examples:["isPositive(2)","isPositive(0)","isPositive(-4)","isPositive([3, 0.5, -2])"],seealso:["isInteger","isNumeric","isNegative","isZero"]},B4={name:"isPrime",category:"Utils",syntax:["isPrime(x)"],description:"Test whether a value is prime: has no divisors other than itself and one.",examples:["isPrime(3)","isPrime(-2)","isPrime([2, 17, 100])"],seealso:["isInteger","isNumeric","isNegative","isZero"]},O4={name:"isZero",category:"Utils",syntax:["isZero(x)"],description:"Test whether a value is zero.",examples:["isZero(2)","isZero(0)","isZero(-4)","isZero([3, 0, -2, 0])"],seealso:["isInteger","isNumeric","isNegative","isPositive"]},P4={name:"numeric",category:"Utils",syntax:["numeric(x)"],description:"Convert a numeric input to a specific numeric type: number, BigNumber, or Fraction.",examples:['numeric("4")','numeric("4", "number")','numeric("4", "BigNumber")','numeric("4", "Fraction")','numeric(4, "Fraction")','numeric(fraction(2, 5), "number")'],seealso:["number","fraction","bignumber","string","format"]},z4={name:"oct",category:"Utils",syntax:["oct(value)"],description:"Format a number as octal",examples:["oct(56)"],seealso:["bin","hex"]},R4={name:"print",category:"Utils",syntax:["print(template, values)","print(template, values, precision)"],description:"Interpolate values into a string template.",examples:['print("Lucy is $age years old", {age: 5})','print("The value of pi is $pi", {pi: pi}, 3)','print("Hello, $user.name!", {user: {name: "John"}})','print("Values: $1, $2, $3", [6, 9, 4])'],seealso:["format"]},I4={name:"typeOf",category:"Utils",syntax:["typeOf(x)"],description:"Get the type of a variable.",examples:["typeOf(3.5)","typeOf(2 - 4i)","typeOf(45 deg)",'typeOf("hello world")'],seealso:["getMatrixDataType"]},q4={name:"solveODE",category:"Numeric",syntax:["solveODE(func, tspan, y0)","solveODE(func, tspan, y0, options)"],description:"Numerical Integration of Ordinary Differential Equations.",examples:["f(t,y) = y","tspan = [0, 4]","solveODE(f, tspan, 1)","solveODE(f, tspan, [1, 2])",'solveODE(f, tspan, 1, { method:"RK23", maxStep:0.1 })'],seealso:["derivative","simplifyCore"]},L4={bignumber:$A,boolean:FA,complex:_A,createUnit:BA,fraction:OA,index:PA,matrix:zA,number:RA,sparse:IA,splitUnit:qA,string:LA,unit:kA,e:ff,E:ff,false:vA,i:gA,Infinity:dA,LN2:wA,LN10:yA,LOG2E:bA,LOG10E:xA,NaN:NA,null:EA,pi:hf,PI:hf,phi:DA,SQRT1_2:AA,SQRT2:MA,tau:SA,true:CA,version:TA,speedOfLight:{description:"Speed of light in vacuum",examples:["speedOfLight"]},gravitationConstant:{description:"Newtonian constant of gravitation",examples:["gravitationConstant"]},planckConstant:{description:"Planck constant",examples:["planckConstant"]},reducedPlanckConstant:{description:"Reduced Planck constant",examples:["reducedPlanckConstant"]},magneticConstant:{description:"Magnetic constant (vacuum permeability)",examples:["magneticConstant"]},electricConstant:{description:"Electric constant (vacuum permeability)",examples:["electricConstant"]},vacuumImpedance:{description:"Characteristic impedance of vacuum",examples:["vacuumImpedance"]},coulomb:{description:"Coulomb's constant",examples:["coulomb"]},elementaryCharge:{description:"Elementary charge",examples:["elementaryCharge"]},bohrMagneton:{description:"Borh magneton",examples:["bohrMagneton"]},conductanceQuantum:{description:"Conductance quantum",examples:["conductanceQuantum"]},inverseConductanceQuantum:{description:"Inverse conductance quantum",examples:["inverseConductanceQuantum"]},magneticFluxQuantum:{description:"Magnetic flux quantum",examples:["magneticFluxQuantum"]},nuclearMagneton:{description:"Nuclear magneton",examples:["nuclearMagneton"]},klitzing:{description:"Von Klitzing constant",examples:["klitzing"]},bohrRadius:{description:"Borh radius",examples:["bohrRadius"]},classicalElectronRadius:{description:"Classical electron radius",examples:["classicalElectronRadius"]},electronMass:{description:"Electron mass",examples:["electronMass"]},fermiCoupling:{description:"Fermi coupling constant",examples:["fermiCoupling"]},fineStructure:{description:"Fine-structure constant",examples:["fineStructure"]},hartreeEnergy:{description:"Hartree energy",examples:["hartreeEnergy"]},protonMass:{description:"Proton mass",examples:["protonMass"]},deuteronMass:{description:"Deuteron Mass",examples:["deuteronMass"]},neutronMass:{description:"Neutron mass",examples:["neutronMass"]},quantumOfCirculation:{description:"Quantum of circulation",examples:["quantumOfCirculation"]},rydberg:{description:"Rydberg constant",examples:["rydberg"]},thomsonCrossSection:{description:"Thomson cross section",examples:["thomsonCrossSection"]},weakMixingAngle:{description:"Weak mixing angle",examples:["weakMixingAngle"]},efimovFactor:{description:"Efimov factor",examples:["efimovFactor"]},atomicMass:{description:"Atomic mass constant",examples:["atomicMass"]},avogadro:{description:"Avogadro's number",examples:["avogadro"]},boltzmann:{description:"Boltzmann constant",examples:["boltzmann"]},faraday:{description:"Faraday constant",examples:["faraday"]},firstRadiation:{description:"First radiation constant",examples:["firstRadiation"]},loschmidt:{description:"Loschmidt constant at T=273.15 K and p=101.325 kPa",examples:["loschmidt"]},gasConstant:{description:"Gas constant",examples:["gasConstant"]},molarPlanckConstant:{description:"Molar Planck constant",examples:["molarPlanckConstant"]},molarVolume:{description:"Molar volume of an ideal gas at T=273.15 K and p=101.325 kPa",examples:["molarVolume"]},sackurTetrode:{description:"Sackur-Tetrode constant at T=1 K and p=101.325 kPa",examples:["sackurTetrode"]},secondRadiation:{description:"Second radiation constant",examples:["secondRadiation"]},stefanBoltzmann:{description:"Stefan-Boltzmann constant",examples:["stefanBoltzmann"]},wienDisplacement:{description:"Wien displacement law constant",examples:["wienDisplacement"]},molarMass:{description:"Molar mass constant",examples:["molarMass"]},molarMassC12:{description:"Molar mass constant of carbon-12",examples:["molarMassC12"]},gravity:{description:"Standard acceleration of gravity (standard acceleration of free-fall on Earth)",examples:["gravity"]},planckLength:{description:"Planck length",examples:["planckLength"]},planckMass:{description:"Planck mass",examples:["planckMass"]},planckTime:{description:"Planck time",examples:["planckTime"]},planckCharge:{description:"Planck charge",examples:["planckCharge"]},planckTemperature:{description:"Planck 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Lt)if(Number.isSafeInteger(n)&&n>0)n===e.dimensionality&&(wh(a,...e.coordinates()),r=!1);else throw new TypeError(rs(a,"dimensionality",n,"positive integer"));if(r)throw new Error(so(a,e))}static validatePointArray(e,n,a="validatePointArray"){if(Array.isArray(e)){if(Za(n))try{n=e[0].dimensionality}catch{}else if(!Number.isSafeInteger(n)||n<=0)throw new TypeError(rs(a,"dimensionality",n,"positive integer"));e.forEach(r=>Lt.validatePoint(r,n,a))}else throw new TypeError(rs(a,"maybePointsArray",e,"array"))}static centroid(e,n="centroid"){if(Lt.validatePointArray(e,void 0,n),e.length===0)throw new TypeError(no(n,"pointsArray"));let a=pa(0,e[0].dimensionality).map(r=>vB(e.map(s=>s.coordinate(r))));return new Lt(...a)}static random(e,n="random"){if(!Number.isSafeInteger(e)||e<=0)throw new TypeError(rs(n,"dimensionality",e,"positive integer"));let a=pa(0,e).map(r=>(Math.random()-.5)*2*Number.MAX_SAFE_INTEGER);return new Lt(...a)}static zero(e,n="zero"){if(!Number.isSafeInteger(e)||e<=0)throw new TypeError(rs(n,"dimensionality",e,"positive integer"));let a=pa(0,e).map(r=>0);return new Lt(...a)}coordinates(){return Ie(this,Hr).slice()}coordinate(e){if(!Number.isSafeInteger(e)||e<0||e>=this.dimensionality)throw new TypeError(lB("coordinate",e,this.dimensionality));return this.coordinates()[e]}get dimensionality(){return Ie(this,pi)}clone(){return new Lt(...this.coordinates())}distanceTo(e){Lt.validatePoint(e,this.dimensionality,"distanceTo");let n=this.coordinates(),a=e.coordinates(),r=n.map((s,i)=>{let p=s-a[i];return p*p});return Math.sqrt(r.reduce((s,i)=>s+i))}maxDistance(e){if(Lt.validatePointArray(e,this.dimensionality,"maxDistance"),e.length===0)throw new TypeError(no("maxDistance","pointsArray"));return e.reduce(([n,a],r)=>{let s=this.distanceTo(r);return s>a&&(a=s,n=r),[n,a]},[null,-1])}minDistance(e){if(Lt.validatePointArray(e,this.dimensionality,"minDistance"),e.length===0)throw new TypeError(no("minDistance","pointsArray"));return e.reduce(([n,a],r)=>{let s=this.distanceTo(r);return sa+r*n[s],0)}equals(e){let n=!1;if(e instanceof Lt&&e.dimensionality===this.dimensionality){let a=this.coordinates(),r=e.coordinates();n=a.every((s,i)=>s===r[i])}return n}toString(){return`(${this.coordinates().map(e=>e.toString()).join(",")})`}toJson(){return JSON.stringify(this.coordinates())}};Hr=new WeakMap,pi=new WeakMap;let ql=Lt;class Ut extends ql{static fromJson(e){return new Ut(...JSON.parse(e))}constructor(e,n){super(e,n)}static validatePoint(e,n="validatePoint"){super.validatePoint(e,2,n)}static random({width:e,height:n}={}){return new Ut(gh(0,e),gh(0,n))}get x(){return this.coordinates()[0]}get y(){return this.coordinates()[1]}add(e){if(!(e instanceof Ut))throw new Error(so("add",e));return new Ut(this.x+e.x,this.y+e.y)}subtract(e){if(!(e instanceof Ut))throw new Error(so("subtract",e));return new Ut(this.x-e.x,this.y-e.y)}dotProduct(e){if(!(e instanceof Ut))throw new Error(so("dotProduct",e));return this.x*e.x+this.y*e.y}clone(){return new Ut(this.x,this.y)}}var gB=dB;function hs(t){return t instanceof Buffer?Buffer.from(t):new t.constructor(t.buffer.slice(),t.byteOffset,t.length)}function dB(t){if(t=t||{},t.circles)return yB(t);return t.proto?a:n;function e(r,s){for(var i=Object.keys(r),p=new Array(i.length),m=0;m]/,bB=NB;function NB(t){var e=""+t,n=xB.exec(e);if(!n)return e;var a,r="",s=0,i=0;for(s=n.index;s ${Ie(this,Cr).toString()}`}equals(e){return e instanceof Ur&&this.toJson()===e.toJson()}clone(){return new Ur(this.source.clone(),this.destination.clone(),{weight:this.weight,label:this.label})}};Sr=new WeakMap,Cr=new WeakMap,vn=new WeakMap,Jr=new WeakMap;let ca=Ur;var ds,gn;class EB{constructor(e,n){Rt(this,ds,void 0);Rt(this,gn,void 0);$t(this,ds,{...e}),$t(this,gn,{...n}),Object.freeze(Ie(this,ds)),Object.freeze(Ie(this,gn))}get distance(){return Ie(this,ds)}get predecessor(){return Ie(this,gn)}reconstructPathTo(e){let n=[];for(;At(e);)n.push(e),e=Ie(this,gn)[e];return e===null?n.reverse():[]}}ds=new WeakMap,gn=new WeakMap;var ci,dn,Zr;class DB{constructor(e,n,a){Rt(this,ci,void 0);Rt(this,dn,void 0);Rt(this,Zr,void 0);$t(this,ci,a),$t(this,dn,{...e}),$t(this,Zr,{...n}),Object.freeze(Ie(this,dn)),Object.freeze(Ie(this,Zr))}get timeDiscovered(){return Ie(this,dn)}get timeVisited(){return Ie(this,Zr)}isAcyclic(){return Ie(this,ci)}verticesByVisitOrder(){return Object.keys(Ie(this,dn)).sort((e,n)=>Ie(this,Zr)[n]-Ie(this,Zr)[e])}}ci=new WeakMap,dn=new WeakMap,Zr=new WeakMap;const cn=new WeakMap;class Ea{static fromJson(e){return Ea.fromJsonObject(JSON.parse(e))}static fromJsonObject({vertices:e,edges:n}){let a=new Ea;return e.forEach(r=>a.addVertex(os.fromJsonObject(r))),n.forEach(r=>a.addEdge(ca.fromJsonObject(r))),a}static completeGraph(e){if(!Ft(e)||e<2)throw new Error(Ue("Graph.completeGraph","n",e));let n=new Ea,a=[];const r=pa(1,e+1);return r.forEach(s=>a[s]=n.createVertex(s)),r.forEach(s=>pa(s+1,e+1).forEach(i=>{n.createEdge(a[s],a[i]),n.createEdge(a[i],a[s])})),n}static completeBipartiteGraph(e,n){if(!Ft(e)||e<1)throw new Error(Ue("Graph.completeBipartiteGraph","n",e));if(!Ft(n)||n<1)throw new Error(Ue("Graph.completeBipartiteGraph","m",n));let a=new Ea,r=[];const s=pa(1,e+1),i=pa(e+1,e+n+1);return s.forEach(p=>r[p]=a.createVertex(p)),i.forEach(p=>r[p]=a.createVertex(p)),s.forEach(p=>i.forEach(m=>{a.createEdge(r[p],r[m]),a.createEdge(r[m],r[p])})),a}constructor(){cn.set(this,new Map)}get id(){return`${this.vertices.map(e=>`{${e.id}}`).sort().join("")}|${this.edges.map(e=>e.id).sort().join("")}`}get vertices(){return[...wn(this)]}get edges(){return[...Gr(this)]}get simpleEdges(){return[...Gr(this)].filter(e=>e.source.id!==e.destination.id)}isDirected(){return!0}isEmpty(){return this.vertices.length===0}createVertex(e,{weight:n,label:a,data:r}={}){if(this.hasVertex(Ht.idFromName(e)))throw new Error(vh("Graph.createVertex",e));const s=new os(e,{weight:n,label:a,data:r});let i=cn.get(this);return i.set(s.id,s),cn.set(this,i),s.id}addVertex(e){if(!(e instanceof Ht))throw new Error(Ue("Graph.addVertex",e));let n=cn.get(this);if(this.hasVertex(e.id))throw new Error(vh("Graph.addVertex",e));const a=new os(e.name,{weight:e.weight,label:e.label??void 0,data:e.data??void 0});return n.set(e.id,a),cn.set(this,n),a.id}hasVertex(e){const n=Xt(this,e);return At(n)&&n instanceof os}getVertex(e){return Xt(this,e)}getVertexOutDegree(e){const n=Xt(this,e);return n==null?void 0:n.outDegree()}getVertexWeight(e){const n=Xt(this,e);return(n==null?void 0:n.weight)??yl("Graph.getVertexWeight",e)}setVertexWeight(e,n){if(!Ft(n))throw new TypeError(Ue("Graph.setVertexWeight","weight",n));let a=Xt(this,e);if(At(a))a.weight=n;else throw new Error(Na("Graph.setVertexWeight",e))}getVertexLabel(e){const n=Xt(this,e);return(n==null?void 0:n.label)??yl("Graph.getVertexLabel",e)}setVertexLabel(e,n){if(!Ht.isValidLabel(n))throw new TypeError(Bm("Graph.setVertexLabel",n));let a=Xt(this,e);if(At(a))a.label=n;else throw new Error(Na("Graph.setVertexLabel",e))}getVertexData(e){const n=Xt(this,e);return(n==null?void 0:n.data)??yl("Graph.getVertexData",e)}setVertexData(e,n){if(!Ht.isValidData(n))throw new TypeError(Id("Graph.setVertexData",n));let a=Xt(this,e);if(At(a))a.data=n;else throw new Error(Na("Graph.setVertexData",e))}createEdge(e,n,{weight:a,label:r}={}){if(!this.hasVertex(e))throw new Error(Na("Graph.createEdge",e));if(!this.hasVertex(n))throw new Error(Na("Graph.createEdge",n));const s=Xt(this,e),i=Xt(this,n);return s.addEdgeTo(i,{edgeWeight:a,edgeLabel:r}).id}addEdge(e){if(!(e instanceof ca))throw new Error(Ue("Graph.addEdge",e));if(!this.hasVertex(e.source))throw new Error(Na("Graph.addEdge",e.source));if(!this.hasVertex(e.destination))throw new Error(Na("Graph.addEdge",e.destination));const n=Xt(this,e.source),a=Xt(this,e.destination);return n.addEdgeTo(a,{edgeWeight:e.weight,edgeLabel:e.label}).id}hasEdge(e){return At(this.getEdge(e))}hasEdgeBetween(e,n){const a=ao(this,e,n);return At(a)&&a instanceof ca}getEdgeBetween(e,n){return ao(this,e,n)}getEdge(e){e=AB(e);for(const n of Gr(this))if(n.id===e)return n}*getEdgesFrom(e){const n=Xt(this,e);if(Za(n))throw new Error(Na("Graph.getEdgesFrom",e));yield*n.outgoingEdges()}getEdgesInPath(e){if(!Array.isArray(e))throw new TypeError(Ue("Graph.getEdgesInPath","verticesSequence",e));const n=e.length,a=[];for(let r=0;r${i}`));a.push(p)}return a}setEdgeWeight(e,n){if(!Ft(n))throw new TypeError(Ue("Graph.setEdgeWeight","weight",n));const a=this.getEdge(e);if(At(a))a.weight=n;else throw new Error(fs("Graph.setEdgeWeight",e))}getEdgeWeight(e,n){const a=ao(this,e,n);if(At(a))return a.weight;throw new Error(fs("Graph.getEdgeWeight",`${e} -> ${n}`))}getEdgeLabel(e,n){const a=ao(this,e,n);return a==null?void 0:a.label}inducedSubGraph(e){return MB(this,e)}clone(){const e=new Ea;for(const n of wn(this))e.addVertex(n.clone());for(const n of Gr(this))e.addEdge(n.clone());return e}toString(){return[...this.edges].map(e=>e.toString()).sort().join(", ")}toJson(){return ui(this.toJsonObject())}toJsonObject(){return{vertices:[...wn(this)].sort().map(e=>e.toJsonObject()),edges:[...Gr(this)].sort(ca.compareEdges).map(e=>e.toJsonObject())}}equals(e){return e instanceof Ea&&this.toJson()===e.toJson()}isConnected(){return this.symmetricClosure().isConnected()}isStronglyConnected(){return this.stronglyConnectedComponents().size===1}isAcyclic(){return this.dfs().isAcyclic()}isBipartite(){return this.symmetricClosure().isBipartite()}isComplete(){const e=this.vertices.length;return this.simpleEdges.length===e*(e-1)}isCompleteBipartite(){const[e,n,a]=this.isBipartite(),r=this.simpleEdges.length;return e&&r===2*n.size*a.size}symmetricClosure(){let e=new Ga;for(const n of this.vertices)e.addVertex(n);for(const n of this.edges)if(!e.hasEdge(n)){const a=this.getEdgeBetween(n.destination,n.source);let r=n.weight+((a==null?void 0:a.weight)??0);e.createEdge(n.source,n.destination,{weight:r})}return e}transpose(){let e=new Ea;for(const n of this.vertices)e.addVertex(n);for(const n of this.edges)e.addEdge(n.transpose());return e}transitiveClosure(){throw new Error("Unimplemented")}bfs(e){const n=this.getVertex(e);if(!At(n))throw new Error(Na("Graph.bfs",e));let a={},r={};for(const i of this.vertices)a[i.id]=1/0;a[n.id]=0,r[n.id]=null;let s=[n];for(;s.length>0;){const i=s.pop(0),p=i.outgoingEdges();for(const m of p){const o=m.destination;a[o.id]===1/0&&(r[o.id]=i.id,a[o.id]=a[i.id]+1,s.push(o))}}return new EB(a,r)}dfs(){let e={},n={},a=0,r=!0;return this.vertices.forEach(s=>{e[s.id]||(e[s.id]=++a,[a,r]=Ul(this,s,e,n,r,a))}),new DB(e,n,r)}connectedComponents(){return this.symmetricClosure().connectedComponents()}topologicalOrdering(){const e=this.dfs();return e.isAcyclic()?e.verticesByVisitOrder():null}stronglyConnectedComponents(){const e=this.transpose().dfs().verticesByVisitOrder();let n={},a=0,r=new Set;return e.forEach(s=>{if(!n[s]){let i={},p=!0;n[s]=++a,[a,p]=Ul(this,this.getVertex(s),n,i,p,a),r.add(new Set(Object.keys(i)))}}),r}}class Ga extends Ea{static completeGraph(e){if(!Ft(e)||e<2)throw new Error(Ue("UndirectedGraph.completeGraph","n",e));let n=new Ga,a=[];const r=pa(1,e+1);return r.forEach(s=>a[s]=n.createVertex(s)),r.forEach(s=>pa(s+1,e+1).forEach(i=>{n.createEdge(a[s],a[i])})),n}static completeBipartiteGraph(e,n){if(!Ft(e)||e<1)throw new Error(Ue("UndirectedGraph.completeBipartiteGraph","n",e));if(!Ft(n)||n<1)throw new Error(Ue("UndirectedGraph.completeBipartiteGraph","m",n));let a=new Ga,r=[];const s=pa(1,e+1),i=pa(e+1,e+n+1);return s.forEach(p=>r[p]=a.createVertex(p)),i.forEach(p=>r[p]=a.createVertex(p)),s.forEach(p=>i.forEach(m=>{a.createEdge(r[p],r[m])})),a}static squareGrid(e){if(!Ft(e)||e<1)throw new Error(Ue("UndirectedGraph.squareGrid","n",e));let n=new Ga;return pa(1,e+1).forEach(a=>{pa(1,e+1).forEach(r=>{const s=`<${a}><${r}>`,i=n.createVertex(s);if(a>1){const p=Ht.idFromName(`<${a-1}><${r}>`);n.createEdge(p,i)}if(r>1){const p=Ht.idFromName(`<${a}><${r-1}>`);n.createEdge(p,i)}})}),n}static triangularGrid(e){if(!Ft(e)||e<1)throw new Error(Ue("UndirectedGraph.squareGrid","n",e));let n=new Ga;return pa(1,e+1).forEach(a=>{pa(1,e-a+2).forEach(r=>{const s=`<${a}><${r}>`,i=n.createVertex(s);if(a>1){let p=Ht.idFromName(`<${a-1}><${r}>`);n.createEdge(p,i),p=Ht.idFromName(`<${a-1}><${r+1}>`),n.createEdge(p,i)}if(r>1){const p=Ht.idFromName(`<${a}><${r-1}>`);n.createEdge(p,i)}})}),n}get edges(){return[...Gr(this)].filter(e=>e.source.id<=e.destination.id)}get simpleEdges(){return[...Gr(this)].filter(e=>e.source.id{if(!e[r.id]){let s={},i=!0;e[r.id]=++n,[n,i]=Ul(this,r,e,s,i,n),a.add(new Set(Object.keys(s)))}}),a}isConnected(){return this.connectedComponents().size===1}isBipartite(){if(this.vertices.length<2||!this.isConnected())return[!1,null,null];const e=wn(this).next().value;let n={};n[e.id]=!0;let a=[e];for(;a.length>0;){const i=a.pop(0),p=n[i.id];for(const m of i.outgoingEdges()){const o=m.destination;if(Za(n[o.id]))n[o.id]=!p,a.push(o);else if(n[o.id]===p)return[!1,null,null]}}let r=[],s=[];for(const i of wn(this))n[i.id]?r.push(i.id):s.push(i.id);return[!0,new Set(r),new Set(s)]}isComplete(){const e=this.vertices.length;return this.simpleEdges.length===e*(e-1)/2}isCompleteBipartite(){const[e,n,a]=this.isBipartite(),r=this.simpleEdges.length;return e&&r===n.size*a.size}symmetricClosure(){return this.clone()}transpose(){return this.clone()}topologicalOrdering(){return null}stronglyConnectedComponents(){return this.connectedComponents()}}var gr;const ms=class ms extends Ht{constructor(n,{weight:a,label:r,data:s,outgoingEdges:i=[]}={}){super(n,{weight:a,label:r,data:s});Rt(this,gr,void 0);if(!Array.isArray(i))throw new TypeError(Ue("GVertex constructor","outgoingEdges",i));$t(this,gr,new Map),i.forEach(p=>{if(!(p instanceof ca)||this.id!==p.source.id)throw new TypeError(Ue("GVertex constructor","outgoingEdges",i));this.addEdge(p)})}*outgoingEdges(){for(const n of Ie(this,gr).values()){const a=n.length;a>0&&(yield n[a-1])}}outDegree(){let n=0;for(const a of Ie(this,gr).values())a.length>0&&++n;return n}edgeTo(n){if(!(n instanceof ms))throw new TypeError(Ue("GVertex.edgeTo","v",n));const a=Ie(this,gr).get(n.id)??[],r=a.length;return r>0?a[r-1]:void 0}addEdge(n){if(!(n instanceof ca)||this.id!==n.source.id)throw new TypeError(Ue("GVertex.addEdge","edge",n));return dl(Ie(this,gr),n.destination,n)}addEdgeTo(n,{edgeWeight:a,edgeLabel:r}={}){if(!(n instanceof ms))throw new TypeError(Ue("GVertex.addEdgeTo","v",n));const s=new ca(this,n,{weight:a,label:r});return this.addEdge(s),s}removeEdge(n){if(!(n instanceof ca)||this.id!==n.source.id)throw new TypeError(Ue("GVertex.removeEdge","edge",n));return dl(Ie(this,gr),n.destination)}removeEdgeTo(n){if(!(n instanceof ms))throw new TypeError(Ue("GVertex.removeEdgeTo","v",n));return dl(Ie(this,gr),n)}clone(){return new ms(this.name,{weight:this.weight})}toString(){return`{${this.id}}`}};gr=new WeakMap;let os=ms;function dl(t,e,n=null){let a=[];At(n)&&a.push(n),t.set(e.id,a)}function Xt(t,e){let n=cn.get(t);return n==null?void 0:n.get(kl(e))}function ao(t,e,n){let a=Xt(t,e),r=Xt(t,n);if(!Za(r))return a==null?void 0:a.edgeTo(r)}function*wn(t){yield*cn.get(t).values()}function*Gr(t){for(const e of wn(t))yield*e.outgoingEdges()}function kl(t){return t instanceof Ht?t.id:t}function AB(t){return t instanceof ca?t.id:t}function Ul(t,e,n,a,r,s){let i={},p=[e],m=[];do{const o=p.pop();if(i[o.id])a[o.id]=++s,m.pop();else{i[o.id]=!0,p.push(o),m.push(o.id);for(const l of t.getEdgesFrom(o.id)){const u=l.destination;Za(n[u.id])?(n[u.id]=++s,p.push(u)):(l.isLoop()||!a[u.id]&&(t.isDirected()||m[m.length-1]!==u.id)&&m.indexOf(u.id)>=0)&&(r=!1)}}}while(p.length>0);return[s,r]}function MB(t,e){if(!At(e)||fB(e)===0)throw new Error(Ue("Graph.inducedSubGraph","vertices",e));let n=t.isDirected()?new Ea:new Ga;return e instanceof Set||(e=new Set(e)),e.forEach(a=>{const r=t.getVertex(a);if(!At(r))throw new Error(Na("Graph.inducedSubGraph",a));n.addVertex(r)}),t.edges.forEach(a=>{e.has(a.source.id)&&e.has(a.destination.id)&&n.addEdge(a)}),n}function yl(t,e){throw new Error(Na(t,e))}const fn=class fn extends Ht{constructor(n,a,{weight:r,label:s,data:i}={}){super(n,{weight:r,label:s,data:i});js(this,"_center");if(!(a instanceof Ut)||a.dimensionality<2)throw new Error(Ue("EmbeddedVertex()","vertexPosition",a));this._center=a.clone()}static fromJson(n){return fn.fromJsonObject(JSON.parse(n))}static fromJsonObject({name:n,position:a,weight:r=null,label:s,data:i}){return new fn(n,Ut.fromJson(a),{weight:r,label:s??void 0,data:i??void 0})}get position(){return this._center.clone()}set position(n){if(!(n instanceof Ut))throw new TypeError(Ue("EmbeddedVertex.setPosition","center",n));this._center=n.clone()}radius(){return fn.DEFAULT_VERTEX_RADIUS*this.weight}clone(){return new fn(this.name,this.position,{weight:this.weight,label:this.label,data:this.data})}toJsonObject(){let n=super.toJsonObject();return n.position=this._center.toJson(),n}toString(){return`EmbeddedVertex: ${this.toJson()}`}toSvg(n=[]){let[a,r]=this.position.coordinates();return` - - - ${this.escapedName} - `}};js(fn,"DEFAULT_VERTEX_RADIUS",15);let na=fn;const SB="/";var fi,Wr;const vr=class vr extends ca{constructor(n,a,{weight:r,label:s,isDirected:i=!1,arcControlDistance:p=void 0}={}){if(!(n instanceof na))throw new Error(Ue("EmbeddedEdge()","source",n));if(!(a instanceof na))throw new Error(Ue("EmbeddedEdge()","destination",a));super(n,a,{weight:r,label:s});Rt(this,fi,void 0);Rt(this,Wr,void 0);if(!cB(i))throw new Error(Ue("EmbeddedEdge()","isDirected",i));if($t(this,fi,i),Za(p))$t(this,Wr,super.isLoop()?vr.DEFAULT_EDGE_LOOP_RADIUS:vr.DEFAULT_EDGE_BEZIER_CONTROL_DISTANCE);else if(Ft(p))$t(this,Wr,za(p));else throw new Error(Ue("EmbeddedEdge()","arcControlDistance",p))}static 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\\{s\\}',s,i,p='P‾−{s}\\overline{P} - \\{s\\}',m,o,l,u=`0=∑u∈(P−s)0=∑u∈(P−s)(∑v∈V,(u,v)∈Ef(u,v)−∑v∈V,(v,u)∈Ef(v,u))=∑u∈(P−s)(∑v∈P,(u,v)∈Ef(u,v)−∑v∈P,(v,u)∈Ef(v,u)+∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))=∑u∈(P−s)(∑v∈P,(u,v)∈Ef(u,v)−∑v∈P,(v,u)∈Ef(v,u))+∑u∈(P−s)(∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))=∑u∈P,v∈P,(u,v)∈Ef(u,v)−∑u∈P,v∈P,(v,u)∈Ef(v,u)+∑u∈(P−s)(∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))(Note: Simple graph.)=∑u∈(P−s)(∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))(Note: First 2 summations have same region.)=(∑u∈P,v∈P‾,(u,v)∈Ef(u,v)−∑u∈P,v∈P‾,(v,u)∈Ef(v,u))−(∑v∈P‾,(s,v)∈Ef(s,v)−∑v∈P‾,(v,s)∈Ef(v,s))≤(∑u∈P,v∈P‾,(u,v)∈Ef(u,v)−∑u∈P,v∈P‾,(v,u)∈Ef(v,u))−(∑v∈V,(s,v)∈Ef(s,v)−∑v∈V,(v,s)∈Ef(v,s))=∑u∈P,v∈P‾,(u,v)∈Ef(u,v)−∑u∈P,v∈P‾,(v,u)∈Ef(v,u)−∣f∣≤∑u∈P,v∈P‾,(u,v)∈Ec(u,v)−∑u∈P,v∈P‾,(v,u)∈Ec(v,u)−∣f∣=C(P,P‾)−∣f∣0\\\\ -= \\sum_{u \\in (P - {s})} 0\\\\ -= \\sum_{u \\in (P - {s})} \\left(\\sum_{v \\in V, (u, v) \\in E} f(u, v) - \\sum_{v \\in V, (v, u) \\in E} f(v, u)\\right)\\\\ -= \\sum_{u \\in (P - {s})} \\left(\\sum_{v \\in P, (u, v) \\in E} f(u, v) - \\sum_{v \\in P, (v, u) \\in E} f(v, u) + \\sum_{v \\in \\overline{P}, (u, v) \\in E} f(u, v) - \\sum_{v \\in \\overline{P}, (v, u) \\in E} f(v, u)\\right)\\\\ -= \\sum_{u \\in (P - {s})} \\left(\\sum_{v \\in P, (u, v) \\in E} f(u, v) - \\sum_{v \\in P, (v, u) \\in E} f(v, u)\\right) + \\sum_{u \\in (P - {s})}\\left(\\sum_{v \\in \\overline{P}, (u, v) \\in E} f(u, v) - \\sum_{v \\in \\overline{P}, (v, u) \\in E} f(v, u)\\right)\\\\ -= \\sum_{u \\in P, v \\in P, (u, v) \\in E} f(u, v) - \\sum_{u \\in P, v \\in P, (v, u) \\in E} f(v, u) + \\sum_{u \\in (P - {s})}\\left(\\sum_{v \\in \\overline{P}, (u, v) \\in E} f(u, v) - \\sum_{v \\in \\overline{P}, (v, u) \\in E} f(v, u)\\right) \\text{(Note: Simple graph.)}\\\\ -= \\sum_{u \\in (P - {s})}\\left(\\sum_{v \\in \\overline{P}, (u, v) \\in E} f(u, v) - \\sum_{v \\in \\overline{P}, (v, u) \\in E} f(v, u)\\right) \\text{(Note: First 2 summations have same region.)}\\\\ -= \\left(\\sum_{u \\in P, v \\in \\overline{P}, (u, v) \\in E} f(u, v) - 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f''(\\xi)',Rs,w,Sa,Z,Zn='ξ\\xi',Aa,ss,sp='xx',Na,as,ap='x+hx + h',Ia,Qs,ns,Da,Js,C,np='f′(x)=f(x+h)−f(x)h−h2f′′(ξ)≈f(x+h)−f(x)hf'(x) = \\frac{f(x+h) - f(x)}{h} - {h \\over 2} f''(\\xi) \\approx \\frac{f(x+h) - f(x)}{h}',Ks,T,Ba,ps,pp='−h2f′′(ξ)-{h \\over 2} f''(\\xi)',Oa,Us,ts,E,Va,ms,tp='O(hk)O(h^k)',$a,ls,mp='kk',qa,Xs,G,lp=`f(x+h)=f(x)+hf′(x)+h22f′′(x)+h36f′′′(ξ1)f(x−h)=f(x)−hf′(x)+h22f′′(x)−h36f′′′(ξ2)f(x + h) = f(x) + hf'(x) + {h^2 \\over 2}f''(x) + {h^3 \\over 6}f'''(\\xi_1)\\\\ -f(x - h) = f(x) - hf'(x) + {h^2 \\over 2}f''(x) - {h^3 \\over 6}f'''(\\xi_2)`,Ys,d,Ca,es,ep='ξ1\\xi_1',Ga,is,ip='xx',Fa,cs,cp='x+hx + h',Wa,rs,rp='ξ2\\xi_2',ja,os,op='x−hx - h',Ra,hs,hp='xx',Qa,Zs,gs,Ja,sa,F,gp='f′(x)=f(x+h)−f(x−h)2h−h212f′′′(ξ1)−h212f′′′(ξ2)≈f(x+h)−f(x−h)2hf'(x) = \\frac{f(x+h) - f(x - h)}{2h} - {h^2 \\over 12} f'''(\\xi_1) - {h^2 \\over 12} f'''(\\xi_2) \\approx \\frac{f(x+h) - f(x - h)}{2h}',aa,ys,Ka,na,W,yp='f′(x)=f(x+h)−f(x−h)2h−h26f′′′(ξ)≈f(x+h)−f(x−h)2hf'(x) = \\frac{f(x+h) - f(x - h)}{2h} - {h^2 \\over 6} f'''(\\xi) \\approx \\frac{f(x+h) - f(x - h)}{2h}',pa,x,Ua,ds,dp='f(x)f(x)',Xa,vs,vp='f(x+h)f(x+h)',Ya,us,up='f(x+2h)f(x+2h)',Za,ta,fs,sn,ma,j,fp=`f(x+2h)=f(x)+2hf′(x)+2h2f′′(x)+4h33f′′′(ξ1)f(x+h)=f(x)+hf′(x)+h22f′′(x)+h36f′′′(ξ2)\\begin{gather} -f(x + 2h) = f(x) + 2hf'(x) + 2h^2 f''(x) + {4h^3 \\over 3} f'''(\\xi_1)\\\\ -f(x + h) = f(x) + hf'(x) + {h^2 \\over 2} f''(x) + {h^3 \\over 6} f'''(\\xi_2) -\\end{gather}`,la,k,an,ws,wp='f′′(x)f''(x)',nn,xs,xp='44',pn,ea,R,bp=`f(x+2h)−4f(x+h)=−3f(x)−2hf′(x)+2h33f′′′(ξ1)−2h33f′′′(ξ2)f′(x)=−3f(x)+4f(x+h)−f(x+2h)2h+h23f′′′(ξ1)−h23f′′′(ξ2)≈−3f(x)+4f(x+h)−f(x+2h)2hf(x + 2h) - 4 f(x + h) = -3 f(x) - 2h f'(x) + {2h^3 \\over 3} f'''(\\xi_1) - {2h^3 \\over 3} f'''(\\xi_2)\\\\ -f'(x) = \\frac{- 3 f(x) + 4 f(x + h) - f(x + 2h) }{2h} + {h^2 \\over 3} f'''(\\xi_1) - {h^2 \\over 3} f'''(\\xi_2) \\approx \\frac{- 3 f(x) + 4 f(x + h) - f(x + 2h)}{2h}`,ia,_,tn,bs,zp='O(h2)O(h^2)',mn,zs,kp='f(x−2h)f(x - 2h)',ln,ca,ks,Q,en,_s,_p='hh',cn,ra,P,H,$s,rn,oa,S,A,qs,on,ha,Ms,hn,ga,Es,gn,ya,M,yn,Ls,Mp='f(x)f(x)',dn,Ts,Ep='xx',vn,da,N,un,Ps,Lp='xn=x+nhx_n = x + nh',fn,va,J,Tp=`(f′(x1)f′(x2)f′(x3)f′(x4)⋮f′(xn))≈12h(−34−10⋯01−210⋯001−21⋯0001−2⋯0⋮⋮⋮⋮⋱⋮0000⋯−3)(f(x1)f(x2)f(x3)f(x4)⋮f(xn))\\begin{pmatrix} -f'(x_1)\\\\ -f'(x_2)\\\\ -f'(x_3)\\\\ -f'(x_4)\\\\ -\\vdots\\\\ -f'(x_n) -\\end{pmatrix} -\\approx -\\dfrac{1}{2h} -\\begin{pmatrix} --3 & 4 & -1 & 0 & \\cdots & 0\\\\ -1 & -2 & 1 & 0 & \\cdots & 0\\\\ -0 & 1 & -2 & 1 & \\cdots & 0\\\\ -0 & 0 & 1 & -2 & \\cdots & 0\\\\ -\\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ -0 & 0 & 0 & 0 & \\cdots & -3 -\\end{pmatrix} -\\begin{pmatrix} -f(x_1)\\\\ -f(x_2)\\\\ -f(x_3)\\\\ -f(x_4)\\\\ -\\vdots\\\\ -f(x_n) -\\end{pmatrix}`,ua,Hs,wn,fa,K,Pp=`(f(k)(x1)f(k)(x2)f(k)(x3)f(k)(x4)⋮f(k)(xn))≈(12h)k(−34−10⋯01−210⋯001−21⋯0001−2⋯0⋮⋮⋮⋮⋱⋮0000⋯−3)k(f(x1)f(x2)f(x3)f(x4)⋮f(xn))\\begin{pmatrix} -f^{(k)}(x_1)\\\\ -f^{(k)}(x_2)\\\\ -f^{(k)}(x_3)\\\\ -f^{(k)}(x_4)\\\\ -\\vdots\\\\ -f^{(k)}(x_n) -\\end{pmatrix} -\\approx -(\\dfrac{1}{2h})^k -\\begin{pmatrix} --3 & 4 & -1 & 0 & \\cdots & 0\\\\ -1 & -2 & 1 & 0 & \\cdots & 0\\\\ -0 & 1 & -2 & 1 & \\cdots & 0\\\\ -0 & 0 & 1 & -2 & \\cdots & 0\\\\ -\\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ -0 & 0 & 0 & 0 & \\cdots & -3 -\\end{pmatrix}^k -\\begin{pmatrix} -f(x_1)\\\\ -f(x_2)\\\\ -f(x_3)\\\\ -f(x_4)\\\\ -\\vdots\\\\ -f(x_n) -\\end{pmatrix}`,wa,I,D,Cs,xn,xa,Ss,bn,ba,As,zn,za,Ns,kn;return{c(){g=t("h1"),u=t("a"),b=t("span"),V=c("Numeric Differentiation"),y=o(),v=t("h2"),z=t("a"),Vs=t("span"),La=c("Finite Difference"),Gs=o(),X=t("p"),Ta=c("By defination of differentiation, we have"),Fs=o(),$=t("div"),Ws=o(),L=t("p"),Pa=c("By Taylor’s Theorem, if "),Y=t("span"),Ha=c(" is twice differentiable, we have"),js=o(),q=t("div"),Rs=o(),w=t("p"),Sa=c("where "),Z=t("span"),Aa=c(" is a real number between "),ss=t("span"),Na=c(" and "),as=t("span"),Ia=c("."),Qs=o(),ns=t("p"),Da=c("Then we have"),Js=o(),C=t("div"),Ks=o(),T=t("p"),Ba=c("In this case "),ps=t("span"),Oa=c(" become error."),Us=o(),ts=t("p"),E=t("strong"),Va=c("Generally, if error is "),ms=t("span"),$a=c(", we say the method is of order "),ls=t("span"),qa=c("."),Xs=o(),G=t("div"),Ys=o(),d=t("p"),Ca=c("where "),es=t("span"),Ga=c(" is a real number between "),is=t("span"),Fa=c(" and "),cs=t("span"),Wa=c(", and "),rs=t("span"),ja=c(" is a real number between "),os=t("span"),Ra=c(" and "),hs=t("span"),Qa=c("."),Zs=o(),gs=t("p"),Ja=c("Then we have"),sa=o(),F=t("div"),aa=o(),ys=t("p"),Ka=c("So we obtain a method of order 2. And obviously this can be written as"),na=o(),W=t("div"),pa=o(),x=t("p"),Ua=c("However, we wonder if we can get the derivative from "),ds=t("span"),Xa=c(", "),vs=t("span"),Ya=c(" and "),us=t("span"),Za=c("."),ta=o(),fs=t("p"),sn=c("Consider"),ma=o(),j=t("div"),la=o(),k=t("p"),an=c("We want to remove "),ws=t("span"),nn=c(" so we subtract "),xs=t("span"),pn=c(" times the second equation from the first one and get"),ea=o(),R=t("div"),ia=o(),_=t("p"),tn=c("The error is "),bs=t("span"),mn=c(". The "),zs=t("span"),ln=c(" case is symmetric, so we just skip it."),ca=o(),ks=t("p"),Q=t("strong"),en=c("Notice that since "),_s=t("span"),cn=c(" is used as denominator, we can’t make it too small."),ra=o(),P=t("h2"),H=t("a"),$s=t("span"),rn=c("Higher Order Derivatives"),oa=o(),S=t("h3"),A=t("a"),qs=t("span"),on=c("Differentiation Matrix"),ha=o(),Ms=t("p"),hn=c("While it’s possible to get higher order derivatives by Taylor’s Theorem, it’s often hard."),ga=o(),Es=t("p"),gn=c("So we apply first order derivative many times."),ya=o(),M=t("p"),yn=c("Notice that in the previous section, the first order derivative is a linear combination of "),Ls=t("span"),dn=c(" near "),Ts=t("span"),vn=c("."),da=o(),N=t("p"),un=c("The operation can be written as a matrix multiplication. Let "),Ps=t("span"),fn=c(", then we have"),va=o(),J=t("div"),ua=o(),Hs=t("p"),wn=c("So we can have a general formula"),fa=o(),K=t("div"),wa=o(),I=t("h3"),D=t("a"),Cs=t("span"),xn=c("Interpolation and Differentiation"),xa=o(),Ss=t("p"),bn=c("It’s a good idea to interpolate the function first, then differentiate the interpolation."),ba=o(),As=t("p"),zn=c("Typically, we can use Cubic Spline Interpolation for derivative less than 3, since it’s first and second order derivative converge to the original function."),za=o(),Ns=t("p"),kn=c("For higher order derivative, we can use Chebyshev Polynomial Interpolation or Lagrange Polynomial Interpolation. They are polynomials and easy to differentiate."),this.h()},l(s){g=m(s,"H1",{id:!0});var p=l(g);u=m(p,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Pn=l(u);b=m(Pn,"SPAN",{class:!0}),l(b).forEach(a),Pn.forEach(a),V=r(p,"Numeric Differentiation"),p.forEach(a),y=h(s),v=m(s,"H2",{id:!0});var _n=l(v);z=m(_n,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Hn=l(z);Vs=m(Hn,"SPAN",{class:!0}),l(Vs).forEach(a),Hn.forEach(a),La=r(_n,"Finite Difference"),_n.forEach(a),Gs=h(s),X=m(s,"P",{});var Sn=l(X);Ta=r(Sn,"By defination of differentiation, we have"),Sn.forEach(a),Fs=h(s),$=m(s,"DIV",{class:!0});var Hp=l($);Hp.forEach(a),Ws=h(s),L=m(s,"P",{});var ka=l(L);Pa=r(ka,"By Taylor’s Theorem, if "),Y=m(ka,"SPAN",{class:!0});var Sp=l(Y);Sp.forEach(a),Ha=r(ka," is twice differentiable, we have"),ka.forEach(a),js=h(s),q=m(s,"DIV",{class:!0});var Ap=l(q);Ap.forEach(a),Rs=h(s),w=m(s,"P",{});var B=l(w);Sa=r(B,"where "),Z=m(B,"SPAN",{class:!0});var Np=l(Z);Np.forEach(a),Aa=r(B," is a real number between "),ss=m(B,"SPAN",{class:!0});var Ip=l(ss);Ip.forEach(a),Na=r(B," and "),as=m(B,"SPAN",{class:!0});var Dp=l(as);Dp.forEach(a),Ia=r(B,"."),B.forEach(a),Qs=h(s),ns=m(s,"P",{});var An=l(ns);Da=r(An,"Then we have"),An.forEach(a),Js=h(s),C=m(s,"DIV",{class:!0});var Bp=l(C);Bp.forEach(a),Ks=h(s),T=m(s,"P",{});var _a=l(T);Ba=r(_a,"In this case "),ps=m(_a,"SPAN",{class:!0});var Op=l(ps);Op.forEach(a),Oa=r(_a," become error."),_a.forEach(a),Us=h(s),ts=m(s,"P",{});var Nn=l(ts);E=m(Nn,"STRONG",{});var Is=l(E);Va=r(Is,"Generally, if error is "),ms=m(Is,"SPAN",{class:!0});var Vp=l(ms);Vp.forEach(a),$a=r(Is,", we say the method is of order "),ls=m(Is,"SPAN",{class:!0});var $p=l(ls);$p.forEach(a),qa=r(Is,"."),Is.forEach(a),Nn.forEach(a),Xs=h(s),G=m(s,"DIV",{class:!0});var qp=l(G);qp.forEach(a),Ys=h(s),d=m(s,"P",{});var f=l(d);Ca=r(f,"where "),es=m(f,"SPAN",{class:!0});var Cp=l(es);Cp.forEach(a),Ga=r(f," is a real number between "),is=m(f,"SPAN",{class:!0});var Gp=l(is);Gp.forEach(a),Fa=r(f," and "),cs=m(f,"SPAN",{class:!0});var Fp=l(cs);Fp.forEach(a),Wa=r(f,", and "),rs=m(f,"SPAN",{class:!0});var Wp=l(rs);Wp.forEach(a),ja=r(f," is a real number between "),os=m(f,"SPAN",{class:!0});var jp=l(os);jp.forEach(a),Ra=r(f," and "),hs=m(f,"SPAN",{class:!0});var Rp=l(hs);Rp.forEach(a),Qa=r(f,"."),f.forEach(a),Zs=h(s),gs=m(s,"P",{});var In=l(gs);Ja=r(In,"Then we have"),In.forEach(a),sa=h(s),F=m(s,"DIV",{class:!0});var Qp=l(F);Qp.forEach(a),aa=h(s),ys=m(s,"P",{});var Dn=l(ys);Ka=r(Dn,"So we obtain a method of order 2. And obviously this can be written as"),Dn.forEach(a),na=h(s),W=m(s,"DIV",{class:!0});var Jp=l(W);Jp.forEach(a),pa=h(s),x=m(s,"P",{});var O=l(x);Ua=r(O,"However, we wonder if we can get the derivative from "),ds=m(O,"SPAN",{class:!0});var Kp=l(ds);Kp.forEach(a),Xa=r(O,", "),vs=m(O,"SPAN",{class:!0});var Up=l(vs);Up.forEach(a),Ya=r(O," and "),us=m(O,"SPAN",{class:!0});var Xp=l(us);Xp.forEach(a),Za=r(O,"."),O.forEach(a),ta=h(s),fs=m(s,"P",{});var Bn=l(fs);sn=r(Bn,"Consider"),Bn.forEach(a),ma=h(s),j=m(s,"DIV",{class:!0});var Yp=l(j);Yp.forEach(a),la=h(s),k=m(s,"P",{});var Ds=l(k);an=r(Ds,"We want to remove "),ws=m(Ds,"SPAN",{class:!0});var Zp=l(ws);Zp.forEach(a),nn=r(Ds," so we subtract "),xs=m(Ds,"SPAN",{class:!0});var st=l(xs);st.forEach(a),pn=r(Ds," times the second equation from the first one and get"),Ds.forEach(a),ea=h(s),R=m(s,"DIV",{class:!0});var at=l(R);at.forEach(a),ia=h(s),_=m(s,"P",{});var Bs=l(_);tn=r(Bs,"The error is "),bs=m(Bs,"SPAN",{class:!0});var nt=l(bs);nt.forEach(a),mn=r(Bs,". The "),zs=m(Bs,"SPAN",{class:!0});var pt=l(zs);pt.forEach(a),ln=r(Bs," case is symmetric, so we just skip it."),Bs.forEach(a),ca=h(s),ks=m(s,"P",{});var On=l(ks);Q=m(On,"STRONG",{});var Ma=l(Q);en=r(Ma,"Notice that since "),_s=m(Ma,"SPAN",{class:!0});var tt=l(_s);tt.forEach(a),cn=r(Ma," is used as denominator, we can’t make it too small."),Ma.forEach(a),On.forEach(a),ra=h(s),P=m(s,"H2",{id:!0});var Mn=l(P);H=m(Mn,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Vn=l(H);$s=m(Vn,"SPAN",{class:!0}),l($s).forEach(a),Vn.forEach(a),rn=r(Mn,"Higher Order Derivatives"),Mn.forEach(a),oa=h(s),S=m(s,"H3",{id:!0});var En=l(S);A=m(En,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var $n=l(A);qs=m($n,"SPAN",{class:!0}),l(qs).forEach(a),$n.forEach(a),on=r(En,"Differentiation Matrix"),En.forEach(a),ha=h(s),Ms=m(s,"P",{});var qn=l(Ms);hn=r(qn,"While it’s possible to get higher order derivatives by Taylor’s Theorem, it’s often hard."),qn.forEach(a),ga=h(s),Es=m(s,"P",{});var Cn=l(Es);gn=r(Cn,"So we apply first order derivative many times."),Cn.forEach(a),ya=h(s),M=m(s,"P",{});var Os=l(M);yn=r(Os,"Notice that in the previous section, the first order derivative is a linear combination of "),Ls=m(Os,"SPAN",{class:!0});var mt=l(Ls);mt.forEach(a),dn=r(Os," near "),Ts=m(Os,"SPAN",{class:!0});var lt=l(Ts);lt.forEach(a),vn=r(Os,"."),Os.forEach(a),da=h(s),N=m(s,"P",{});var Ea=l(N);un=r(Ea,"The operation can be written as a matrix multiplication. Let "),Ps=m(Ea,"SPAN",{class:!0});var et=l(Ps);et.forEach(a),fn=r(Ea,", then we have"),Ea.forEach(a),va=h(s),J=m(s,"DIV",{class:!0});var it=l(J);it.forEach(a),ua=h(s),Hs=m(s,"P",{});var Gn=l(Hs);wn=r(Gn,"So we can have a general formula"),Gn.forEach(a),fa=h(s),K=m(s,"DIV",{class:!0});var ct=l(K);ct.forEach(a),wa=h(s),I=m(s,"H3",{id:!0});var Ln=l(I);D=m(Ln,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Fn=l(D);Cs=m(Fn,"SPAN",{class:!0}),l(Cs).forEach(a),Fn.forEach(a),xn=r(Ln,"Interpolation and Differentiation"),Ln.forEach(a),xa=h(s),Ss=m(s,"P",{});var Wn=l(Ss);bn=r(Wn,"It’s a good idea to interpolate the function first, then differentiate the interpolation."),Wn.forEach(a),ba=h(s),As=m(s,"P",{});var jn=l(As);zn=r(jn,"Typically, we can use Cubic Spline Interpolation for derivative less than 3, since it’s first and second order derivative converge to the original function."),jn.forEach(a),za=h(s),Ns=m(s,"P",{});var Rn=l(Ns);kn=r(Rn,"For higher order derivative, we can use Chebyshev Polynomial Interpolation or Lagrange Polynomial Interpolation. 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diff --git a/docs/_app/immutable/nodes/12.62331ccd.js b/docs/_app/immutable/nodes/12.62331ccd.js deleted file mode 100644 index 52bb616..0000000 --- a/docs/_app/immutable/nodes/12.62331ccd.js +++ /dev/null @@ -1,28 +0,0 @@ -import{S as Xl,i as se,s as ae,O as _m,y as js,z as Is,A as qs,P as ne,Q as Bm,g as Cs,d as Ds,B as Qs,R as Zm,k as p,q as r,a as u,l as t,m,h as s,r as c,c as v,n as e,b as l,E as n,G as Os}from"../chunks/index.488f1ee5.js";import{L as pe}from"../chunks/layout.c8702512.js";import{P as Xn}from"../chunks/Proof.26996d3c.js";import{S as yn}from"../chunks/State.8d241e02.js";function te(W){let i,b,d,C='nn',x,g,H='nn',P;return{c(){i=p("p"),b=r("The degree of precision of a numerical integration formula is the largest integer "),d=p("span"),x=r(" such that the formula is exact for all polynomials of degree "),g=p("span"),P=r(" or less."),this.h()},l(f){i=t(f,"P",{});var M=m(i);b=c(M,"The degree of precision of a numerical integration formula is the largest integer "),d=t(M,"SPAN",{class:!0});var A=m(d);A.forEach(s),x=c(M," such that the formula is exact for all polynomials of degree "),g=t(M,"SPAN",{class:!0});var _=m(g);_.forEach(s),P=c(M," or less."),M.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline")},m(f,M){l(f,i,M),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P)},p:Os,d(f){f&&s(i)}}}function me(W){let i,b,d,C='nn',x,g,H='n−1n-1',P;return{c(){i=p("p"),b=r("If we use "),d=p("span"),x=r(" points in Lagrange Interpolation, then the degree of precision is at least "),g=p("span"),P=r("."),this.h()},l(f){i=t(f,"P",{});var M=m(i);b=c(M,"If we use "),d=t(M,"SPAN",{class:!0});var A=m(d);A.forEach(s),x=c(M," points in Lagrange Interpolation, then the degree of precision is at least "),g=t(M,"SPAN",{class:!0});var _=m(g);_.forEach(s),P=c(M,"."),M.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline")},m(f,M){l(f,i,M),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P)},p:Os,d(f){f&&s(i)}}}function le(W){let i,b,d,C='f(x)f(x)',x,g,H='n−1n-1',P,f,M='f(x)f(x)',A,_,I='n−1n-1',G;return{c(){i=p("p"),b=r("For any polynomial "),d=p("span"),x=r(" of degree "),g=p("span"),P=r(", Lagrange Interpolation restores "),f=p("span"),A=r(" exactly. So the degree of precision is at least "),_=p("span"),G=r("."),this.h()},l(L){i=t(L,"P",{});var z=m(i);b=c(z,"For any polynomial "),d=t(z,"SPAN",{class:!0});var D=m(d);D.forEach(s),x=c(z," of degree "),g=t(z,"SPAN",{class:!0});var S=m(g);S.forEach(s),P=c(z,", Lagrange Interpolation restores "),f=t(z,"SPAN",{class:!0});var j=m(f);j.forEach(s),A=c(z," exactly. So the degree of precision is at least "),_=t(z,"SPAN",{class:!0});var K=m(_);K.forEach(s),G=c(z,"."),z.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(f,"class","math math-inline"),e(_,"class","math math-inline")},m(L,z){l(L,i,z),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P),n(i,f),f.innerHTML=M,n(i,A),n(i,_),_.innerHTML=I,n(i,G)},p:Os,d(L){L&&s(i)}}}function ee(W){let i,b,d,C='nn',x,g,H='nn',P,f,M='nn',A,_,I='n−1n-1',G,L,z='nn',D;return{c(){i=p("p"),b=r("If we use "),d=p("span"),x=r(" equally spaced points, then the degree of precision is at least "),g=p("span"),P=r(" when "),f=p("span"),A=r(" is odd, and "),_=p("span"),G=r(" when "),L=p("span"),D=r(" is even."),this.h()},l(S){i=t(S,"P",{});var j=m(i);b=c(j,"If we use "),d=t(j,"SPAN",{class:!0});var K=m(d);K.forEach(s),x=c(j," equally spaced points, then the degree of precision is at least "),g=t(j,"SPAN",{class:!0});var Q=m(g);Q.forEach(s),P=c(j," when "),f=t(j,"SPAN",{class:!0});var X=m(f);X.forEach(s),A=c(j," is odd, and "),_=t(j,"SPAN",{class:!0});var ns=m(_);ns.forEach(s),G=c(j," when "),L=t(j,"SPAN",{class:!0});var V=m(L);V.forEach(s),D=c(j," is even."),j.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(f,"class","math math-inline"),e(_,"class","math math-inline"),e(L,"class","math math-inline")},m(S,j){l(S,i,j),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P),n(i,f),f.innerHTML=M,n(i,A),n(i,_),_.innerHTML=I,n(i,G),n(i,L),L.innerHTML=z,n(i,D)},p:Os,d(S){S&&s(i)}}}function ie(W){let i,b,d,C='11',x;return{c(){i=p("p"),b=r("The degree of precision is "),d=p("span"),x=r("."),this.h()},l(g){i=t(g,"P",{});var H=m(i);b=c(H,"The degree of precision is "),d=t(H,"SPAN",{class:!0});var P=m(d);P.forEach(s),x=c(H,"."),H.forEach(s),this.h()},h(){e(d,"class","math math-inline")},m(g,H){l(g,i,H),n(i,b),n(i,d),d.innerHTML=C,n(i,x)},p:Os,d(g){g&&s(i)}}}function re(W){let i,b,d,C='f(x)=1f(x) = 1',x,g,H='∫abf(x)dx=b−a\\int_a^b f(x) dx = b-a',P,f,M,A,_,I='f(x)=xf(x) = x',G,L,z='∫abf(x)dx=b2−a22\\int_a^b f(x) dx = \\dfrac{b^2-a^2}{2}',D,S,j,K,Q,X='f(x)=x2f(x) = x^2',ns,V,ps='∫abf(x)dx=b3−a33\\int_a^b f(x) dx = \\dfrac{b^3-a^3}{3}',ts;return{c(){i=p("p"),b=r("When "),d=p("span"),x=r(", "),g=p("span"),P=r(". The formula is exact."),f=u(),M=p("p"),A=r("When "),_=p("span"),G=r(", "),L=p("span"),D=r(". The formula is exact."),S=u(),j=p("p"),K=r("When "),Q=p("span"),ns=r(", "),V=p("span"),ts=r(". The formula is not exact."),this.h()},l(q){i=t(q,"P",{});var T=m(i);b=c(T,"When "),d=t(T,"SPAN",{class:!0});var ss=m(d);ss.forEach(s),x=c(T,", "),g=t(T,"SPAN",{class:!0});var U=m(g);U.forEach(s),P=c(T,". The formula is exact."),T.forEach(s),f=v(q),M=t(q,"P",{});var N=m(M);A=c(N,"When "),_=t(N,"SPAN",{class:!0});var fs=m(_);fs.forEach(s),G=c(N,", "),L=t(N,"SPAN",{class:!0});var O=m(L);O.forEach(s),D=c(N,". The formula is exact."),N.forEach(s),S=v(q),j=t(q,"P",{});var R=m(j);K=c(R,"When "),Q=t(R,"SPAN",{class:!0});var xs=m(Q);xs.forEach(s),ns=c(R,", "),V=t(R,"SPAN",{class:!0});var F=m(V);F.forEach(s),ts=c(R,". The formula is not exact."),R.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(_,"class","math math-inline"),e(L,"class","math math-inline"),e(Q,"class","math math-inline"),e(V,"class","math math-inline")},m(q,T){l(q,i,T),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P),l(q,f,T),l(q,M,T),n(M,A),n(M,_),_.innerHTML=I,n(M,G),n(M,L),L.innerHTML=z,n(M,D),l(q,S,T),l(q,j,T),n(j,K),n(j,Q),Q.innerHTML=X,n(j,ns),n(j,V),V.innerHTML=ps,n(j,ts)},p:Os,d(q){q&&s(i),q&&s(f),q&&s(M),q&&s(S),q&&s(j)}}}function ce(W){let i,b,d,C='nn',x,g,H='2n−12n-1',P;return{c(){i=p("p"),b=r("If we use "),d=p("span"),x=r(" points in Newton-Cotes formula, then the degree of precision is at most "),g=p("span"),P=r("."),this.h()},l(f){i=t(f,"P",{});var M=m(i);b=c(M,"If we use "),d=t(M,"SPAN",{class:!0});var A=m(d);A.forEach(s),x=c(M," points in Newton-Cotes formula, then the degree of precision is at most "),g=t(M,"SPAN",{class:!0});var _=m(g);_.forEach(s),P=c(M,"."),M.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline")},m(f,M){l(f,i,M),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P)},p:Os,d(f){f&&s(i)}}}function oe(W){let i,b,d,C='xix_i',x,g,H='wiw_i',P,f,M='2n2n',A,_,I,G,L,z='f(x)=∏i=1n(x−xi)2f(x) = \\prod_{i = 1}^n (x - x_i)^2',D,S,j='2n2n',K,Q,X='f(x)f(x)',ns,V,ps,ts,q,T,ss='∫abf(x)dx=∑i=1nwif(xi)=∑i=1nwi⋅0=0\\int_a^b f(x) dx = \\sum_{i=1}^n w_i f(x_i) = \\sum_{i = 1}^n w_i \\cdot 0 = 0',U,N,fs,O,R='f(x)=0f(x) = 0',xs,F,rs='x=xix = x_i',as,B,cs='ii',bs,Z,hs='f(x)>0f(x) > 0',Ps;return{c(){i=p("p"),b=r("Proof by contradiction. Suppose that some one choose points "),d=p("span"),x=r(" and weights "),g=p("span"),P=r(" such that the degree of precision is "),f=p("span"),A=r("."),_=u(),I=p("p"),G=r("Then let "),L=p("span"),D=r(", which is a polynomial of degree "),S=p("span"),K=r(". Then the formula is exact for "),Q=p("span"),ns=r("."),V=u(),ps=p("p"),ts=r("Hence"),q=u(),T=p("div"),U=u(),N=p("p"),fs=r("However, "),O=p("span"),xs=r(" holds only when "),F=p("span"),as=r(" for all "),B=p("span"),bs=r(", and "),Z=p("span"),Ps=r(" for all other places. So the integral is positive. Contradiction."),this.h()},l($){i=t($,"P",{});var k=m(i);b=c(k,"Proof by contradiction. Suppose that some one choose points "),d=t(k,"SPAN",{class:!0});var ds=m(d);ds.forEach(s),x=c(k," and weights "),g=t(k,"SPAN",{class:!0});var ms=m(g);ms.forEach(s),P=c(k," such that the degree of precision is "),f=t(k,"SPAN",{class:!0});var ws=m(f);ws.forEach(s),A=c(k,"."),k.forEach(s),_=v($),I=t($,"P",{});var Y=m(I);G=c(Y,"Then let "),L=t(Y,"SPAN",{class:!0});var J=m(L);J.forEach(s),D=c(Y,", which is a polynomial of degree "),S=t(Y,"SPAN",{class:!0});var gs=m(S);gs.forEach(s),K=c(Y,". Then the formula is exact for "),Q=t(Y,"SPAN",{class:!0});var Ls=m(Q);Ls.forEach(s),ns=c(Y,"."),Y.forEach(s),V=v($),ps=t($,"P",{});var y=m(ps);ts=c(y,"Hence"),y.forEach(s),q=v($),T=t($,"DIV",{class:!0});var w=m(T);w.forEach(s),U=v($),N=t($,"P",{});var ls=m(N);fs=c(ls,"However, "),O=t(ls,"SPAN",{class:!0});var es=m(O);es.forEach(s),xs=c(ls," holds only when "),F=t(ls,"SPAN",{class:!0});var zs=m(F);zs.forEach(s),as=c(ls," for all "),B=t(ls,"SPAN",{class:!0});var Ss=m(B);Ss.forEach(s),bs=c(ls,", and "),Z=t(ls,"SPAN",{class:!0});var _s=m(Z);_s.forEach(s),Ps=c(ls," for all other places. So the integral is positive. Contradiction."),ls.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(f,"class","math math-inline"),e(L,"class","math math-inline"),e(S,"class","math math-inline"),e(Q,"class","math math-inline"),e(T,"class","math math-display"),e(O,"class","math math-inline"),e(F,"class","math math-inline"),e(B,"class","math math-inline"),e(Z,"class","math math-inline")},m($,k){l($,i,k),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P),n(i,f),f.innerHTML=M,n(i,A),l($,_,k),l($,I,k),n(I,G),n(I,L),L.innerHTML=z,n(I,D),n(I,S),S.innerHTML=j,n(I,K),n(I,Q),Q.innerHTML=X,n(I,ns),l($,V,k),l($,ps,k),n(ps,ts),l($,q,k),l($,T,k),T.innerHTML=ss,l($,U,k),l($,N,k),n(N,fs),n(N,O),O.innerHTML=R,n(N,xs),n(N,F),F.innerHTML=rs,n(N,as),n(N,B),B.innerHTML=cs,n(N,bs),n(N,Z),Z.innerHTML=hs,n(N,Ps)},p:Os,d($){$&&s(i),$&&s(_),$&&s(I),$&&s(V),$&&s(ps),$&&s(q),$&&s(T),$&&s(U),$&&s(N)}}}function he(W){let i,b,d,C='nn',x,g,H='xix_i',P,f,M='wiw_i',A,_,I='2n−12n-1',G;return{c(){i=p("p"),b=r("If "),d=p("span"),x=r(" points "),g=p("span"),P=r(" and weights "),f=p("span"),A=r(" satisfy the degree of precision is "),_=p("span"),G=r(", then the intergation formula is called Gaussian Quadrature."),this.h()},l(L){i=t(L,"P",{});var z=m(i);b=c(z,"If "),d=t(z,"SPAN",{class:!0});var D=m(d);D.forEach(s),x=c(z," points "),g=t(z,"SPAN",{class:!0});var S=m(g);S.forEach(s),P=c(z," and weights "),f=t(z,"SPAN",{class:!0});var j=m(f);j.forEach(s),A=c(z," satisfy the degree of precision is "),_=t(z,"SPAN",{class:!0});var K=m(_);K.forEach(s),G=c(z,", then the intergation formula is called Gaussian Quadrature."),z.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(f,"class","math math-inline"),e(_,"class","math math-inline")},m(L,z){l(L,i,z),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P),n(i,f),f.innerHTML=M,n(i,A),n(i,_),_.innerHTML=I,n(i,G)},p:Os,d(L){L&&s(i)}}}function ge(W){let i,b,d,C='f(x)f(x)',x,g,H='2n−12n-1',P,f,M='f(x)=Pn(x)g(x)+r(x)f(x) = P_n(x)g(x) + r(x)',A,_,I='g(x)g(x)',G,L,z='r(x)r(x)',D,S,j='nn',K,Q,X,ns,V,ps='g(x)g(x)',ts,q,T='P0(x),P1(x),⋯ ,Pn−1(x)P_0(x), P_1(x), \\cdots, P_{n-1}(x)',ss,U,N,fs,O,R,xs='∫−11f(x)dx=∫−11Pn(x)g(x)dx+∫−11r(x)dx=∫−11r(x)dx\\int_{-1}^1 f(x) dx = \\int_{-1}^1 P_n(x)g(x) dx + \\int_{-1}^1 r(x) dx = \\int_{-1}^1 r(x) dx',F,rs,as,B,cs='r(x)r(x)',bs,Z,hs='2n−12n-1',Ps,$,k,ds,ms,ws='2n−12n-1',Y,J,gs='2n−12n-1',Ls;return{c(){i=p("p"),b=r("Let "),d=p("span"),x=r(" be a polynomial of degree "),g=p("span"),P=r(". Then "),f=p("span"),A=r(", where "),_=p("span"),G=r(", "),L=p("span"),D=r(" are two polynomial of degree less than "),S=p("span"),K=r("."),Q=u(),X=p("p"),ns=r("It’s trival that "),V=p("span"),ts=r(" can be express as a linear combination of "),q=p("span"),ss=r("."),U=u(),N=p("p"),fs=r("By the othogonality of Legendre polynomial and linearity of integration, we have"),O=u(),R=p("div"),F=u(),rs=p("p"),as=r("By the previous theorem, we know that the the integration formula is at least for "),B=p("span"),bs=r(", so the degree of precision is at least "),Z=p("span"),Ps=r("."),$=u(),k=p("p"),ds=r("Since the degree of precision is at most "),ms=p("span"),Y=r(", the degree of precision is exactly "),J=p("span"),Ls=r("."),this.h()},l(y){i=t(y,"P",{});var w=m(i);b=c(w,"Let "),d=t(w,"SPAN",{class:!0});var ls=m(d);ls.forEach(s),x=c(w," be a polynomial of degree "),g=t(w,"SPAN",{class:!0});var es=m(g);es.forEach(s),P=c(w,". Then "),f=t(w,"SPAN",{class:!0});var zs=m(f);zs.forEach(s),A=c(w,", where "),_=t(w,"SPAN",{class:!0});var Ss=m(_);Ss.forEach(s),G=c(w,", "),L=t(w,"SPAN",{class:!0});var _s=m(L);_s.forEach(s),D=c(w," are two polynomial of degree less than "),S=t(w,"SPAN",{class:!0});var Ns=m(S);Ns.forEach(s),K=c(w,"."),w.forEach(s),Q=v(y),X=t(y,"P",{});var ys=m(X);ns=c(ys,"It’s trival that "),V=t(ys,"SPAN",{class:!0});var na=m(V);na.forEach(s),ts=c(ys," can be express as a linear combination of "),q=t(ys,"SPAN",{class:!0});var is=m(q);is.forEach(s),ss=c(ys,"."),ys.forEach(s),U=v(y),N=t(y,"P",{});var Vs=m(N);fs=c(Vs,"By the othogonality of Legendre polynomial and linearity of integration, we have"),Vs.forEach(s),O=v(y),R=t(y,"DIV",{class:!0});var pa=m(R);pa.forEach(s),F=v(y),rs=t(y,"P",{});var us=m(rs);as=c(us,"By the previous theorem, we know that the the integration formula is at least for "),B=t(us,"SPAN",{class:!0});var Es=m(B);Es.forEach(s),bs=c(us,", so the degree of precision is at least "),Z=t(us,"SPAN",{class:!0});var Ms=m(Z);Ms.forEach(s),Ps=c(us,"."),us.forEach(s),$=v(y),k=t(y,"P",{});var os=m(k);ds=c(os,"Since the degree of precision is at most "),ms=t(os,"SPAN",{class:!0});var $s=m(ms);$s.forEach(s),Y=c(os,", the degree of precision is exactly "),J=t(os,"SPAN",{class:!0});var Rs=m(J);Rs.forEach(s),Ls=c(os,"."),os.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(f,"class","math math-inline"),e(_,"class","math math-inline"),e(L,"class","math math-inline"),e(S,"class","math math-inline"),e(V,"class","math math-inline"),e(q,"class","math math-inline"),e(R,"class","math math-display"),e(B,"class","math math-inline"),e(Z,"class","math math-inline"),e(ms,"class","math math-inline"),e(J,"class","math math-inline")},m(y,w){l(y,i,w),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P),n(i,f),f.innerHTML=M,n(i,A),n(i,_),_.innerHTML=I,n(i,G),n(i,L),L.innerHTML=z,n(i,D),n(i,S),S.innerHTML=j,n(i,K),l(y,Q,w),l(y,X,w),n(X,ns),n(X,V),V.innerHTML=ps,n(X,ts),n(X,q),q.innerHTML=T,n(X,ss),l(y,U,w),l(y,N,w),n(N,fs),l(y,O,w),l(y,R,w),R.innerHTML=xs,l(y,F,w),l(y,rs,w),n(rs,as),n(rs,B),B.innerHTML=cs,n(rs,bs),n(rs,Z),Z.innerHTML=hs,n(rs,Ps),l(y,$,w),l(y,k,w),n(k,ds),n(k,ms),ms.innerHTML=ws,n(k,Y),n(k,J),J.innerHTML=gs,n(k,Ls)},p:Os,d(y){y&&s(i),y&&s(Q),y&&s(X),y&&s(U),y&&s(N),y&&s(O),y&&s(R),y&&s(F),y&&s(rs),y&&s($),y&&s(k)}}}function ye(W){let i,b,d,C='H(x)H(x)',x,g,H='f(x)f(x)',P,f,M='xix_i',A,_,I,G=`H(xi)=f(xi)H′(xi)=f′(xi)H(x_i) = f(x_i)\\\\ -H'(x_i) = f'(x_i)`,L,z,D,S,j='H(x)H(x)',K,Q,X='2n−12n-1',ns,V,ps,ts=`f(x)=H(x)+f(2n)(ξ)(2n)!(π(x))2∫−11f(x)dx=∫−11H(x)dx+f(2n)(ξ)(2n)!∫−11(π(x))2dxf(x) = H(x) + \\dfrac{f^{(2n)}(\\xi)}{(2n)!}(\\pi(x))^2 \\\\ -\\int_{-1}^1 f(x) dx = \\int_{-1}^1 H(x) dx + \\dfrac{f^{(2n)}(\\xi)}{(2n)!}\\int_{-1}^1 (\\pi(x))^2 dx`,q,T,ss,U,N='H(x)H(x)',fs,O,R='2n−12n-1',xs,F,rs='H(x)H(x)',as,B,cs,bs,Z,hs='H(xi)=f(xi)H(x_i) = f(x_i)',Ps,$,k,ds='∫−11H(x)dx=∑i=1nwiH(xi)=∑i=1nwif(xi)\\int_{-1}^1 H(x) dx = \\sum_{i=1}^n w_i H(x_i) = \\sum_{i=1}^n w_i f(x_i)',ms,ws,Y,J,gs,Ls=`f(x)=H(x)+f(2n)(ξ)(2n)!(π(x))2∫−11f(x)dx=∑i=1nwif(xi)+f(2n)(ξ)(2n)!∫−11(π(x))2dxf(x) = H(x) + \\dfrac{f^{(2n)}(\\xi)}{(2n)!}(\\pi(x))^2 \\\\ -\\int_{-1}^1 f(x) dx = \\sum_{i=1}^n w_i f(x_i) + \\dfrac{f^{(2n)}(\\xi)}{(2n)!}\\int_{-1}^1 (\\pi(x))^2 dx`;return{c(){i=p("p"),b=r("Consider Hermite Interpolation Polynomial "),d=p("span"),x=r(" of "),g=p("span"),P=r(" at "),f=p("span"),A=r("."),_=u(),I=p("div"),L=u(),z=p("p"),D=r("Then "),S=p("span"),K=r(" is a polynomial of degree at most "),Q=p("span"),ns=r("."),V=u(),ps=p("div"),q=u(),T=p("p"),ss=r("Since "),U=p("span"),fs=r(" is a polynomial of degree at most "),O=p("span"),xs=r(", the integration formula is exact for "),F=p("span"),as=r("."),B=u(),cs=p("p"),bs=r("By the condition of Hermite Interpolation Polynomial, we know that "),Z=p("span"),Ps=r("."),$=u(),k=p("div"),ms=u(),ws=p("p"),Y=r("Substitute it into the previous equation, we get"),J=u(),gs=p("div"),this.h()},l(y){i=t(y,"P",{});var w=m(i);b=c(w,"Consider Hermite Interpolation Polynomial "),d=t(w,"SPAN",{class:!0});var ls=m(d);ls.forEach(s),x=c(w," of "),g=t(w,"SPAN",{class:!0});var es=m(g);es.forEach(s),P=c(w," at "),f=t(w,"SPAN",{class:!0});var zs=m(f);zs.forEach(s),A=c(w,"."),w.forEach(s),_=v(y),I=t(y,"DIV",{class:!0});var Ss=m(I);Ss.forEach(s),L=v(y),z=t(y,"P",{});var _s=m(z);D=c(_s,"Then "),S=t(_s,"SPAN",{class:!0});var Ns=m(S);Ns.forEach(s),K=c(_s," is a polynomial of degree at most "),Q=t(_s,"SPAN",{class:!0});var ys=m(Q);ys.forEach(s),ns=c(_s,"."),_s.forEach(s),V=v(y),ps=t(y,"DIV",{class:!0});var na=m(ps);na.forEach(s),q=v(y),T=t(y,"P",{});var is=m(T);ss=c(is,"Since "),U=t(is,"SPAN",{class:!0});var Vs=m(U);Vs.forEach(s),fs=c(is," is a polynomial of degree at most "),O=t(is,"SPAN",{class:!0});var pa=m(O);pa.forEach(s),xs=c(is,", the integration formula is exact for "),F=t(is,"SPAN",{class:!0});var us=m(F);us.forEach(s),as=c(is,"."),is.forEach(s),B=v(y),cs=t(y,"P",{});var Es=m(cs);bs=c(Es,"By the condition of Hermite Interpolation Polynomial, we know that "),Z=t(Es,"SPAN",{class:!0});var Ms=m(Z);Ms.forEach(s),Ps=c(Es,"."),Es.forEach(s),$=v(y),k=t(y,"DIV",{class:!0});var os=m(k);os.forEach(s),ms=v(y),ws=t(y,"P",{});var $s=m(ws);Y=c($s,"Substitute it into the previous equation, we get"),$s.forEach(s),J=v(y),gs=t(y,"DIV",{class:!0});var Rs=m(gs);Rs.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(f,"class","math math-inline"),e(I,"class","math math-display"),e(S,"class","math math-inline"),e(Q,"class","math math-inline"),e(ps,"class","math math-display"),e(U,"class","math math-inline"),e(O,"class","math math-inline"),e(F,"class","math math-inline"),e(Z,"class","math math-inline"),e(k,"class","math math-display"),e(gs,"class","math math-display")},m(y,w){l(y,i,w),n(i,b),n(i,d),d.innerHTML=C,n(i,x),n(i,g),g.innerHTML=H,n(i,P),n(i,f),f.innerHTML=M,n(i,A),l(y,_,w),l(y,I,w),I.innerHTML=G,l(y,L,w),l(y,z,w),n(z,D),n(z,S),S.innerHTML=j,n(z,K),n(z,Q),Q.innerHTML=X,n(z,ns),l(y,V,w),l(y,ps,w),ps.innerHTML=ts,l(y,q,w),l(y,T,w),n(T,ss),n(T,U),U.innerHTML=N,n(T,fs),n(T,O),O.innerHTML=R,n(T,xs),n(T,F),F.innerHTML=rs,n(T,as),l(y,B,w),l(y,cs,w),n(cs,bs),n(cs,Z),Z.innerHTML=hs,n(cs,Ps),l(y,$,w),l(y,k,w),k.innerHTML=ds,l(y,ms,w),l(y,ws,w),n(ws,Y),l(y,J,w),l(y,gs,w),gs.innerHTML=Ls},p:Os,d(y){y&&s(i),y&&s(_),y&&s(I),y&&s(L),y&&s(z),y&&s(V),y&&s(ps),y&&s(q),y&&s(T),y&&s(B),y&&s(cs),y&&s($),y&&s(k),y&&s(ms),y&&s(ws),y&&s(J),y&&s(gs)}}}function ue(W){let i,b,d,C,x='wi=∫−11(π(x))2=2(1−xi2)[Pn′(xi)]2w_i = \\int_{-1}^1 (\\pi(x))^2 = \\dfrac{2}{(1 - x_i^2)[P_n'(x_i)]^2}';return{c(){i=p("p"),b=r("The weights of Legendre-Gauss Quadrature can be calculated by the following formula."),d=u(),C=p("div"),this.h()},l(g){i=t(g,"P",{});var H=m(i);b=c(H,"The weights of Legendre-Gauss Quadrature can be calculated by the following formula."),H.forEach(s),d=v(g),C=t(g,"DIV",{class:!0});var P=m(C);P.forEach(s),this.h()},h(){e(C,"class","math math-display")},m(g,H){l(g,i,H),n(i,b),l(g,d,H),l(g,C,H),C.innerHTML=x},p:Os,d(g){g&&s(i),g&&s(d),g&&s(C)}}}function ve(W){let i,b,d,C='π′(x)\\pi'(x)',x,g,H='π′(xi)\\pi'(x_i)',P,f,M='i=1,2,⋯ ,ni = 1, 2, \\cdots, n',A,_,I,G=`π(x)=∏j=1n(x−xj)=(x−xi)∏j=1,j≠in(x−xj)π′(x)=∏j=1,j≠in(x−xj)+(x−xi)(∏j=1,j≠in(x−xj))′π′(xi)=∏j=1,j≠in(xi−xj)+(xi−xi)(∏j=1,j≠in(xi−xj))′=∏j=1,j≠in(xi−xj)\\pi(x) = \\prod_{j=1}^n (x - x_j) = (x - x_i) \\prod_{j=1, j \\neq i}^n (x - x_j)\\\\ -\\pi'(x) = \\prod_{j=1, j \\neq i}^n (x - x_j) + (x - x_i)(\\prod_{j=1, j \\neq i}^n (x - x_j))'\\\\ -\\pi'(x_i) = \\prod_{j=1, j \\neq i}^n (x_i - x_j) + (x_i - x_i)(\\prod_{j=1, j \\neq i}^n (x_i - x_j))' = \\prod_{j=1, j \\neq i}^n (x_i - x_j)`,L,z,D,S,j='li(x)l_i(x)',K,Q,X='π(x)\\pi(x)',ns,V,ps='xix_i',ts,q,T,ss,U,N,fs='li(x)=π(x)(x−xi)π′(xi)=Pn(x)(x−xi)Pn′(xi)l_i(x) = \\dfrac{\\pi(x)}{(x - x_i)\\pi'(x_i)} = \\dfrac{P_n(x)}{(x - x_i)P_n'(x_i)}',O,R,xs,F,rs='li(x)Pn′(x)l_i(x)P_n'(x)',as,B,cs='2n−22n - 2',bs,Z,hs,Ps='∫−11li(x)Pn′(x)dx=∑j=1nwjli(xj)Pn′(xj)\\int_{-1}^1 l_i(x)P_n'(x) dx = \\sum_{j=1}^n w_j l_i(x_j) P_n'(x_j)',$,k,ds,ms,ws='li(xj)=0l_i(x_j) = 0',Y,J,gs='j≠ij \\neq i',Ls,y,w='li(xi)=1l_i(x_i) = 1',ls,es,zs,Ss='∫−11li(x)Pn′(x)dx=wili(xi)Pn′(xi)=wiPn′(xi)\\int_{-1}^1 l_i(x)P_n'(x) dx = w_i l_i(x_i) P_n'(x_i) = w_i P_n'(x_i)',_s,Ns,ys,na,is,Vs='∫−11li(x)Pn′(x)dx=∫−11Pn(x)Pn′(x)(x−xi)Pn′(xi)dx\\int_{-1}^1 l_i(x)P_n'(x) dx = \\int_{-1}^1 \\dfrac{P_n(x)P_n'(x)}{(x - x_i)P_n'(x_i)} dx',pa,us,Es,Ms,os,$s='wi=1(Pn′(x))2∫−11Pn(x)Pn′(x)x−xjdxw_i = \\dfrac{1}{(P_n'(x))^2} \\int_{-1}^1 \\dfrac{P_n(x)P_n'(x)}{x - x_j} dx',Rs,Hs,Ma,Ks,Us,un=`∫−11(li(x))2=∫−11(Pn(x))2(x−xi)2(Pn′(xi))2dx=1(Pn′(xi))2∫−11(Pn(x))2(x−xi)2dx=1(Pn′(xi))2∫−11−(Pn(x))2d1x−xi=1(Pn′(xi))2((−(Pn(x))2x−xi)∣−11+∫−111x−xid(Pn(x))2)=1(Pn′(xi))2((−(Pn(1))21−xi−−(Pn(−1))2−1−xi)+2∫−11Pn(x)Pn′(x)x−xidx)=−1(Pn′(xi))2(11+xi+11−xi)+2(Pn′(xi))2∫−11Pn(x)Pn′(x)x−xidx=−2(1−xi2)(Pn′(xi))2+2wi\\int_{-1}^1 (l_i(x))^2\\\\ -= \\int_{-1}^1 \\dfrac{(P_n(x))^2}{(x - x_i)^2(P_n'(x_i))^2} dx\\\\ -= \\dfrac{1}{(P_n'(x_i))^2}\\int_{-1}^1 \\dfrac{(P_n(x))^2}{(x - x_i)^2} dx\\\\ -= \\dfrac{1}{(P_n'(x_i))^2}\\int_{-1}^1 -(P_n(x))^2 d \\dfrac{1}{x - x_i}\\\\ -= \\dfrac{1}{(P_n'(x_i))^2} ((\\dfrac{-(P_n(x))^2}{x - x_i})|_{-1}^1 + \\int_{-1}^1 \\dfrac{1}{x - x_i} d(P_n(x))^2)\\\\ -= \\dfrac{1}{(P_n'(x_i))^2} ((\\dfrac{-(P_n(1))^2}{1 - x_i} - \\dfrac{-(P_n(-1))^2}{-1 - x_i}) + 2\\int_{-1}^1 \\dfrac{P_n(x)P_n'(x)}{x - x_i} dx)\\\\ -= -\\dfrac{1}{(P_n'(x_i))^2} (\\dfrac{1}{1 + x_i} + \\dfrac{1}{1 - x_i}) + \\dfrac{2}{(P_n'(x_i))^2}\\int_{-1}^1 \\dfrac{P_n(x)P_n'(x)}{x - x_i} dx\\\\ -= -\\dfrac{2}{(1 - x_i^2)(P_n'(x_i))^2} + 2w_i\\\\`,Gs,vs,da,ta,vn='(li(x))2(l_i(x))^2',ma,la,dn='2n−22n - 2',Xs,fn,Bs,ea='∫−11(li(x))2dx=∑j=1nwj(li(xj))2=wi(li(xi))2=wi\\int_{-1}^1 (l_i(x))^2 dx = \\sum_{j=1}^n w_j (l_i(x_j))^2 = w_i (l_i(x_i))^2 = w_i',fa,Ts,ka,Ws,ks,ln=`wi=∫−11(li(x))2dx=−2(1−xi2)(Pn′(xi))2+2wiwi=2(1−xi2)(Pn′(xi))2w_i = \\int_{-1}^1 (l_i(x))^2 dx = -\\dfrac{2}{(1 - x_i^2)(P_n'(x_i))^2} + 2w_i\\\\ -w_i = \\dfrac{2}{(1 - x_i^2)(P_n'(x_i))^2}`;return{c(){i=p("p"),b=r("We first find "),d=p("span"),x=r(" and "),g=p("span"),P=r(" for "),f=p("span"),A=r("."),_=u(),I=p("div"),L=u(),z=p("p"),D=r("Consider the Lagrange Interpolation Polynomial "),S=p("span"),K=r(" of "),Q=p("span"),ns=r(" at "),V=p("span"),ts=r("."),q=u(),T=p("p"),ss=r("It can be expressed as"),U=u(),N=p("div"),O=u(),R=p("p"),xs=r("Since "),F=p("span"),as=r(" is a polynomial of degree "),B=p("span"),bs=r(", we have"),Z=u(),hs=p("div"),$=u(),k=p("p"),ds=r("Since "),ms=p("span"),Y=r(" for "),J=p("span"),Ls=r(", and "),y=p("span"),ls=r(", we have"),es=u(),zs=p("div"),_s=u(),Ns=p("p"),ys=r("On the other hand,"),na=u(),is=p("div"),pa=u(),us=p("p"),Es=r("Hence,"),Ms=u(),os=p("div"),Rs=u(),Hs=p("p"),Ma=r("Similarly, we move on to the next step."),Ks=u(),Us=p("div"),Gs=u(),vs=p("p"),da=r("On the other hand, since "),ta=p("span"),ma=r(" is a polynomial of degree "),la=p("span"),Xs=r(", we have"),fn=u(),Bs=p("div"),fa=u(),Ts=p("p"),ka=r("Combine the two equations, we get"),Ws=u(),ks=p("div"),this.h()},l(h){i=t(h,"P",{});var E=m(i);b=c(E,"We first find "),d=t(E,"SPAN",{class:!0});var ia=m(d);ia.forEach(s),x=c(E," and "),g=t(E,"SPAN",{class:!0});var sp=m(g);sp.forEach(s),P=c(E," for "),f=t(E,"SPAN",{class:!0});var Pa=m(f);Pa.forEach(s),A=c(E,"."),E.forEach(s),_=v(h),I=t(h,"DIV",{class:!0});var ut=m(I);ut.forEach(s),L=v(h),z=t(h,"P",{});var Zs=m(z);D=c(Zs,"Consider the Lagrange Interpolation Polynomial "),S=t(Zs,"SPAN",{class:!0});var xn=m(S);xn.forEach(s),K=c(Zs," of "),Q=t(Zs,"SPAN",{class:!0});var xa=m(Q);xa.forEach(s),ns=c(Zs," at "),V=t(Zs,"SPAN",{class:!0});var vt=m(V);vt.forEach(s),ts=c(Zs,"."),Zs.forEach(s),q=v(h),T=t(h,"P",{});var en=m(T);ss=c(en,"It can be expressed as"),en.forEach(s),U=v(h),N=t(h,"DIV",{class:!0});var ra=m(N);ra.forEach(s),O=v(h),R=t(h,"P",{});var ca=m(R);xs=c(ca,"Since "),F=t(ca,"SPAN",{class:!0});var La=m(F);La.forEach(s),as=c(ca," is a polynomial of degree "),B=t(ca,"SPAN",{class:!0});var dt=m(B);dt.forEach(s),bs=c(ca,", we have"),ca.forEach(s),Z=v(h),hs=t(h,"DIV",{class:!0});var ap=m(hs);ap.forEach(s),$=v(h),k=t(h,"P",{});var Fs=m(k);ds=c(Fs,"Since "),ms=t(Fs,"SPAN",{class:!0});var oa=m(ms);oa.forEach(s),Y=c(Fs," for "),J=t(Fs,"SPAN",{class:!0});var ha=m(J);ha.forEach(s),Ls=c(Fs,", and "),y=t(Fs,"SPAN",{class:!0});var rn=m(y);rn.forEach(s),ls=c(Fs,", we have"),Fs.forEach(s),es=v(h),zs=t(h,"DIV",{class:!0});var np=m(zs);np.forEach(s),_s=v(h),Ns=t(h,"P",{});var cn=m(Ns);ys=c(cn,"On the other hand,"),cn.forEach(s),na=v(h),is=t(h,"DIV",{class:!0});var ga=m(is);ga.forEach(s),pa=v(h),us=t(h,"P",{});var bn=m(us);Es=c(bn,"Hence,"),bn.forEach(s),Ms=v(h),os=t(h,"DIV",{class:!0});var Ea=m(os);Ea.forEach(s),Rs=v(h),Hs=t(h,"P",{});var dp=m(Hs);Ma=c(dp,"Similarly, we move on to the next step."),dp.forEach(s),Ks=v(h),Us=t(h,"DIV",{class:!0});var pp=m(Us);pp.forEach(s),Gs=v(h),vs=t(h,"P",{});var sa=m(vs);da=c(sa,"On the other hand, since "),ta=t(sa,"SPAN",{class:!0});var ba=m(ta);ba.forEach(s),ma=c(sa," is a polynomial of degree "),la=t(sa,"SPAN",{class:!0});var ft=m(la);ft.forEach(s),Xs=c(sa,", we have"),sa.forEach(s),fn=v(h),Bs=t(h,"DIV",{class:!0});var wn=m(Bs);wn.forEach(s),fa=v(h),Ts=t(h,"P",{});var aa=m(Ts);ka=c(aa,"Combine the two equations, we get"),aa.forEach(s),Ws=v(h),ks=t(h,"DIV",{class:!0});var tp=m(ks);tp.forEach(s),this.h()},h(){e(d,"class","math math-inline"),e(g,"class","math math-inline"),e(f,"class","math math-inline"),e(I,"class","math 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de(W){let i,b,d,C,x,g,H,P,f,M,A,_,I,G,L=`f(x)≈If(x)∫abf(x)dx≈∫abIf(x)dxf(x) \\approx I f(x)\\\\ -\\int_a^b f(x) dx \\approx \\int_a^b I f(x) dx`,z,D,S,j,K='nn',Q,X,ns,V,ps,ts,q=`f(x)=∑i=1nf(xi)∏j=1,j≠inx−xjxi−xj+f(n)(ξ)n!∏i=1n(x−xi)∫abf(x)dx=∑i=1nf(xi)∫ab∏j=1,j≠inx−xjxi−xjdx+∫abf(n)(ξ)n!∏i=1n(x−xi)dxf(x) = \\sum_{i=1}^n f(x_i) \\prod_{j=1, j \\neq i}^n\\dfrac{x-x_j}{x_i-x_j} + \\dfrac{f^{(n)}(\\xi)}{n!} \\prod_{i=1}^n (x-x_i)\\\\ -\\int_a^b f(x) dx = \\sum_{i=1}^n f(x_i) \\int_a^b \\prod_{j=1, j \\neq i}^n\\dfrac{x-x_j}{x_i-x_j} dx + \\int_a^b \\dfrac{f^{(n)}(\\xi)}{n!} \\prod_{i=1}^n (x-x_i) dx`,T,ss,U,N,fs='∫abf(n)(ξ)n!∏i=1n(x−xi)dx\\int_a^b \\dfrac{f^{(n)}(\\xi)}{n!} \\prod_{i=1}^n (x-x_i) dx',O,R,xs='E(f)E(f)',F,rs,as,B,cs,bs='∫ab∏j=1,j≠inx−xjxi−xjdx\\int_a^b \\prod_{j=1, j \\neq i}^n\\dfrac{x-x_j}{x_i-x_j} dx',Z,hs,Ps='f(x)f(x)',$,k,ds,ms,ws,Y,J,gs,Ls,y,w,ls,es,zs,Ss,_s='1,x,x2,⋯1, x, x^2, \\cdots',Ns,ys,na='1,x,x2,⋯ ,xn1, x, x^2, \\cdots, 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\\dfrac{f^{(5)}(\\xi)}{2880}(b-a)^7',wn,aa,tp,zn,Jm='55',xt,fp,$a,Ha,mp,bt,xp,Ta,wt,_n,Km='99',zt,bp,Mn,_t,wp,on,Um='∫abf(x)dx=∑i=0(n−1)/2b−a3n(f(x2i)+4f(x2i+1)+f(x2i+2))−f(3)(ξ)90(b−a)5\\int_a^b f(x) dx = \\sum_{i=0}^{(n - 1)/2} \\dfrac{b - a}{3n}(f(x_{2i}) + 4f(x_{2i+1}) + f(x_{2i+2})) - \\dfrac{f^{(3)}(\\xi)}{90}(b-a)^5',zp,kn,Mt,_p,Sa,Na,lp,kt,Mp,Ys,Pt,Pn,Xm='sin⁡xx\\dfrac{\\sin x}{x}',Lt,Ln,sl='00',Et,En,al='11',$t,kp,$n,Ht,Pp,Hn,Tt,Lp,hn,nl='∫abf(x)dx=(b−a)f(a+b2)−f′′(ξ)24(b−a)3\\int_a^b f(x) dx = (b - a)f(\\dfrac{a + b}{2}) - \\dfrac{f''(\\xi)}{24}(b-a)^3',Ep,Aa,ja,ep,St,$p,Tn,Nt,Hp,Sn,At,Tp,Ia,qa,ip,jt,Sp,ya,It,Nn,pl='xix_i',qt,An,tl='wiw_i',Ct,Np,Ca,Ap,Da,jp,Qa,Ip,Va,Ra,rp,Dt,qp,Js,Qt,jn,ml='xix_i',Vt,In,ll='Pn(x)P_n(x)',Rt,qn,el='2n−12n-1',Gt,Cp,Ga,Wt,Cn,il='[−1,1][-1, 1]',Ft,Dp,Wa,Qp,Dn,Ot,Vp,gn,rl='E(f)=f(2n)(ξ)(2n)!∫−11∏i=1n(x−xi)2dx=f(2n)(ξ)(2n)!∫ab(π(x))2dxE(f) = \\dfrac{f^{(2n)}(\\xi)}{(2n)!}\\int_{-1}^1 \\prod_{i=1}^n (x - x_i)^2 dx = \\dfrac{f^{(2n)}(\\xi)}{(2n)!}\\int_a^b(\\pi(x))^2dx',Rp,Fa,Bt,Qn,cl='π(x)=∏i=1n(x−xi)\\pi(x) = \\prod_{i=1}^n (x - x_i)',Zt,Gp,Oa,Wp,Ba,Za,cp,Yt,Fp,Ya,Op,Ja,Bp,Ka,Ua,op,Jt,Zp,ua,Kt,Vn,ol='∫abρ(x)f(x)dx\\int_a^b \\rho(x)f(x) dx',Ut,Rn,hl='ρ(x)\\rho(x)',Xt,Yp,Xa,sm,Gn,gl=`ρ(x)=11+x2\\rho(x) = \\dfrac{1}{\\sqrt {1 + x^2}}`,am,Jp,sn,nm,Wn,yl='xix_i',pm,Kp,an,nn,hp,tm,Up,As,mm,Fn,ul='aa',lm,On,vl='bb',em,Bn,dl='Pn−2′(x)P'_{n-2}(x)',im,Zn,fl='xix_i',rm,Xp,Yn,cm,st;return w=new yn({props:{variant:"definition",$$slots:{default:[te]},$$scope:{ctx:W}}}),Ms=new yn({props:{variant:"theorem",$$slots:{default:[me]},$$scope:{ctx:W}}}),$s=new Xn({props:{$$slots:{default:[le]},$$scope:{ctx:W}}}),Hs=new yn({props:{variant:"theorem",$$slots:{default:[ee]},$$scope:{ctx:W}}}),ea=new yn({props:{variant:"collary",$$slots:{default:[ie]},$$scope:{ctx:W}}}),Ts=new Xn({props:{$$slots:{default:[re]},$$scope:{ctx:W}}}),Ca=new yn({props:{variant:"theorem",$$slots:{default:[ce]},$$scope:{ctx:W}}}),Da=new 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Let "),j=p("span"),Q=r(" be "),X=p("strong"),ns=r("the number of points we use"),V=r(". (Not the degree of the polynomial)"),ps=u(),ts=p("div"),T=u(),ss=p("p"),U=r("The part "),N=p("span"),O=r(" is called the remainder term, denoted as "),R=p("span"),F=r("."),rs=u(),as=p("p"),B=r("The "),cs=p("span"),Z=r(" part is unrelated to "),hs=p("span"),$=r(", so we can precompute it and store it in a table."),k=u(),ds=p("p"),ms=r("As for the remainder term, sadly we have no way to find a closed form. So we just keep the ugly form."),ws=u(),Y=p("h3"),J=p("a"),gs=p("span"),Ls=r("Degree of Precision"),y=u(),js(w.$$.fragment),ls=u(),es=p("p"),zs=r("To find out a formula’s degree of precision, we can verify it on "),Ss=p("span"),Ns=r(". Since integration is a linear operator, if the formula is exact for "),ys=p("span"),is=r(", then it’s exact for their linear combination. If they are not exact, of course its degree of precision is less than "),Vs=p("span"),us=r(" since the counterexample exists."),Es=u(),js(Ms.$$.fragment),os=u(),js($s.$$.fragment),Rs=u(),js(Hs.$$.fragment),Ma=u(),Ks=p("p"),Us=r("This is obvious since the points are symmetric, so the odd degree terms are cancelled out."),un=u(),Gs=p("h3"),vs=p("a"),da=p("span"),ta=r("Trapezoidal Rule"),vn=u(),ma=p("p"),la=r("If we just take the two end points, we get the Trapezoidal Rule."),dn=u(),Xs=p("div"),Bs=u(),js(ea.$$.fragment),fa=u(),js(Ts.$$.fragment),ka=u(),Ws=p("h3"),ks=p("a"),ln=p("span"),h=r("Simpson’s Rule"),E=u(),ia=p("p"),sp=r("If we take the "),Pa=p("span"),Zs=r(" points, we get the Simpson’s Rule."),xn=u(),xa=p("div"),en=u(),ra=p("p"),ca=r("The degree of precision is "),La=p("span"),ap=r(". Proof is trivial."),Fs=u(),oa=p("h3"),ha=p("a"),rn=p("span"),np=r("Boole’s Rule"),cn=u(),ga=p("p"),bn=r("If we take the "),Ea=p("span"),pp=r(" points, we get the Boole’s Rule."),sa=u(),ba=p("div"),wn=u(),aa=p("p"),tp=r("The degree of precision is "),zn=p("span"),xt=r(". Proof is trivial."),fp=u(),$a=p("h3"),Ha=p("a"),mp=p("span"),bt=r("Composite Newton-Cotes Formula"),xp=u(),Ta=p("p"),wt=r("When the number of points exceeds "),_n=p("span"),zt=r(", some of the coefficients become negative. This is not good for numerical computation. So we can divide the interval into subintervals, and apply the Newton-Cotes formula on each subinterval."),bp=u(),Mn=p("p"),_t=r("Take composite Simpson’s Rule as an example."),wp=u(),on=p("div"),zp=u(),kn=p("p"),Mt=r("The error term is the sum of the error terms of each subinterval, whose form is exactly the same as the error term of the original Simpson’s Rule."),_p=u(),Sa=p("h3"),Na=p("a"),lp=p("span"),kt=r("Open Newton-Cotes Formula"),Mp=u(),Ys=p("p"),Pt=r("Some functions are not defined at the end points, such as "),Pn=p("span"),Lt=r(". If we want to integrate it from "),Ln=p("span"),Et=r(" to "),En=p("span"),$t=r(", we can use the Open Newton-Cotes Formula."),kp=u(),$n=p("p"),Ht=r("The idea of Open Newton-Cotes Formula is simply avoiding the end points."),Pp=u(),Hn=p("p"),Tt=r("Midpoint Rule is the simplest Open Newton-Cotes Formula."),Lp=u(),hn=p("div"),Ep=u(),Aa=p("h3"),ja=p("a"),ep=p("span"),St=r("Adaptive Quadrature"),$p=u(),Tn=p("p"),Nt=r("Some part of the function changes rapidly, while some part changes slowly. If we use the same number of points in the whole interval, we may get Time Limit Exceeded. So we can use Adaptive Quadrature to use more points in the rapidly changing part, and less points in the slowly changing part."),Hp=u(),Sn=p("p"),At=r("We first start recursion with the whole interval, then divide the interval into two subintervals, and recurse on each subinterval. The key is comparing the difference between the integral of the whole interval and the sum of the integrals of the two subintervals. If the difference is small enough, we can stop the recursion."),Tp=u(),Ia=p("h2"),qa=p("a"),ip=p("span"),jt=r("Gaussian Quadrature"),Sp=u(),ya=p("p"),It=r("The idea of Gaussian Quadrature is to choose the points "),Nn=p("span"),qt=r(" and weights "),An=p("span"),Ct=r(" such that the degree of precision is as high as possible."),Np=u(),js(Ca.$$.fragment),Ap=u(),js(Da.$$.fragment),jp=u(),js(Qa.$$.fragment),Ip=u(),Va=p("h3"),Ra=p("a"),rp=p("span"),Dt=r("Legendre-Gauss Quadrature"),qp=u(),Js=p("p"),Qt=r("The idea is picking "),jn=p("span"),Vt=r(" as the roots of Legendre polynomial "),In=p("span"),Rt=r(". In this way, the degree of precision is exactly "),qn=p("span"),Gt=r("."),Cp=u(),Ga=p("p"),Wt=r("To make life easier, we can normalize the interval to "),Cn=p("span"),Ft=r(". This do no harm to the generality of the formula."),Dp=u(),js(Wa.$$.fragment),Qp=u(),Dn=p("p"),Ot=r("The error term is"),Vp=u(),gn=p("div"),Rp=u(),Fa=p("p"),Bt=r("where "),Qn=p("span"),Zt=r("."),Gp=u(),js(Oa.$$.fragment),Wp=u(),Ba=p("h3"),Za=p("a"),cp=p("span"),Yt=r("Weights of Legendre-Gauss Quadrature"),Fp=u(),js(Ya.$$.fragment),Op=u(),js(Ja.$$.fragment),Bp=u(),Ka=p("h3"),Ua=p("a"),op=p("span"),Jt=r("Weighted Gaussian Quadrature"),Zp=u(),ua=p("p"),Kt=r("Consider we want to integrate "),Vn=p("span"),Ut=r(", where "),Rn=p("span"),Xt=r(" is a weight function. We can use the same idea to get the weighted Gaussian Quadrature."),Yp=u(),Xa=p("p"),sm=r("Here we only discuss the case where "),Gn=p("span"),am=r("."),Jp=u(),sn=p("p"),nm=r("In this case, we use Chebyshev Zeroes as the points "),Wn=p("span"),pm=r(". The proof is similar to the proof of Gauss-Legendre Quadrature, we omit it here."),Kp=u(),an=p("h3"),nn=p("a"),hp=p("span"),tm=r("Lobatto Quadrature"),Up=u(),As=p("p"),mm=r("Lobatto’s idea is use "),Fn=p("span"),lm=r(", "),On=p("span"),em=r(" and the roots of "),Bn=p("span"),im=r(" as the points "),Zn=p("span"),rm=r("."),Xp=u(),Yn=p("p"),cm=r("In this way, it uses the end points as information."),this.h()},l(a){i=t(a,"H1",{id:!0});var o=m(i);b=t(o,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var gp=m(b);d=t(gp,"SPAN",{class:!0}),m(d).forEach(s),gp.forEach(s),C=c(o,"Numeric Integration"),o.forEach(s),x=v(a),g=t(a,"H2",{id:!0});var Jn=m(g);H=t(Jn,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var yp=m(H);P=t(yp,"SPAN",{class:!0}),m(P).forEach(s),yp.forEach(s),f=c(Jn,"Newton-Cotes Formula"),Jn.forEach(s),M=v(a),A=t(a,"P",{});var up=m(A);_=c(up,"The general idea of Newton-Cotes formula is to first interpolate the function, then integrate the interpolation."),up.forEach(s),I=v(a),G=t(a,"DIV",{class:!0});var at=m(G);at.forEach(s),z=v(a),D=t(a,"P",{});var wa=m(D);S=c(wa,"We substitute Lagrange interpolation polynomial into the formula. Let "),j=t(wa,"SPAN",{class:!0});var nt=m(j);nt.forEach(s),Q=c(wa," be "),X=t(wa,"STRONG",{});var vp=m(X);ns=c(vp,"the number of points we use"),vp.forEach(s),V=c(wa,". (Not the degree of the polynomial)"),wa.forEach(s),ps=v(a),ts=t(a,"DIV",{class:!0});var pt=m(ts);pt.forEach(s),T=v(a),ss=t(a,"P",{});var za=m(ss);U=c(za,"The part "),N=t(za,"SPAN",{class:!0});var tt=m(N);tt.forEach(s),O=c(za," is called the remainder term, denoted as "),R=t(za,"SPAN",{class:!0});var mt=m(R);mt.forEach(s),F=c(za,"."),za.forEach(s),rs=v(a),as=t(a,"P",{});var _a=m(as);B=c(_a,"The "),cs=t(_a,"SPAN",{class:!0});var xl=m(cs);xl.forEach(s),Z=c(_a," part is unrelated to "),hs=t(_a,"SPAN",{class:!0});var bl=m(hs);bl.forEach(s),$=c(_a,", so we can precompute it and store it in a table."),_a.forEach(s),k=v(a),ds=t(a,"P",{});var Mm=m(ds);ms=c(Mm,"As for the remainder term, sadly we have no way to find a closed form. So we just keep the ugly form."),Mm.forEach(s),ws=v(a),Y=t(a,"H3",{id:!0});var om=m(Y);J=t(om,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var km=m(J);gs=t(km,"SPAN",{class:!0}),m(gs).forEach(s),km.forEach(s),Ls=c(om,"Degree of Precision"),om.forEach(s),y=v(a),Is(w.$$.fragment,a),ls=v(a),es=t(a,"P",{});var pn=m(es);zs=c(pn,"To find out a formula’s degree of precision, we can verify it on "),Ss=t(pn,"SPAN",{class:!0});var wl=m(Ss);wl.forEach(s),Ns=c(pn,". Since integration is a linear operator, if the formula is exact for "),ys=t(pn,"SPAN",{class:!0});var zl=m(ys);zl.forEach(s),is=c(pn,", then it’s exact for their linear combination. If they are not exact, of course its degree of precision is less than "),Vs=t(pn,"SPAN",{class:!0});var _l=m(Vs);_l.forEach(s),us=c(pn," since the counterexample exists."),pn.forEach(s),Es=v(a),Is(Ms.$$.fragment,a),os=v(a),Is($s.$$.fragment,a),Rs=v(a),Is(Hs.$$.fragment,a),Ma=v(a),Ks=t(a,"P",{});var Pm=m(Ks);Us=c(Pm,"This is obvious since the points are symmetric, so the odd degree terms are cancelled out."),Pm.forEach(s),un=v(a),Gs=t(a,"H3",{id:!0});var hm=m(Gs);vs=t(hm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Lm=m(vs);da=t(Lm,"SPAN",{class:!0}),m(da).forEach(s),Lm.forEach(s),ta=c(hm,"Trapezoidal Rule"),hm.forEach(s),vn=v(a),ma=t(a,"P",{});var Em=m(ma);la=c(Em,"If we just take the two end points, we get the Trapezoidal Rule."),Em.forEach(s),dn=v(a),Xs=t(a,"DIV",{class:!0});var Ml=m(Xs);Ml.forEach(s),Bs=v(a),Is(ea.$$.fragment,a),fa=v(a),Is(Ts.$$.fragment,a),ka=v(a),Ws=t(a,"H3",{id:!0});var gm=m(Ws);ks=t(gm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var $m=m(ks);ln=t($m,"SPAN",{class:!0}),m(ln).forEach(s),$m.forEach(s),h=c(gm,"Simpson’s Rule"),gm.forEach(s),E=v(a),ia=t(a,"P",{});var lt=m(ia);sp=c(lt,"If we take the "),Pa=t(lt,"SPAN",{class:!0});var kl=m(Pa);kl.forEach(s),Zs=c(lt," points, we get the Simpson’s Rule."),lt.forEach(s),xn=v(a),xa=t(a,"DIV",{class:!0});var Pl=m(xa);Pl.forEach(s),en=v(a),ra=t(a,"P",{});var et=m(ra);ca=c(et,"The degree of precision is "),La=t(et,"SPAN",{class:!0});var Ll=m(La);Ll.forEach(s),ap=c(et,". Proof is trivial."),et.forEach(s),Fs=v(a),oa=t(a,"H3",{id:!0});var ym=m(oa);ha=t(ym,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Hm=m(ha);rn=t(Hm,"SPAN",{class:!0}),m(rn).forEach(s),Hm.forEach(s),np=c(ym,"Boole’s Rule"),ym.forEach(s),cn=v(a),ga=t(a,"P",{});var it=m(ga);bn=c(it,"If we take the "),Ea=t(it,"SPAN",{class:!0});var El=m(Ea);El.forEach(s),pp=c(it," points, we get the Boole’s Rule."),it.forEach(s),sa=v(a),ba=t(a,"DIV",{class:!0});var $l=m(ba);$l.forEach(s),wn=v(a),aa=t(a,"P",{});var rt=m(aa);tp=c(rt,"The degree of precision is "),zn=t(rt,"SPAN",{class:!0});var Hl=m(zn);Hl.forEach(s),xt=c(rt,". Proof is trivial."),rt.forEach(s),fp=v(a),$a=t(a,"H3",{id:!0});var um=m($a);Ha=t(um,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Tm=m(Ha);mp=t(Tm,"SPAN",{class:!0}),m(mp).forEach(s),Tm.forEach(s),bt=c(um,"Composite Newton-Cotes Formula"),um.forEach(s),xp=v(a),Ta=t(a,"P",{});var ct=m(Ta);wt=c(ct,"When the number of points exceeds "),_n=t(ct,"SPAN",{class:!0});var Tl=m(_n);Tl.forEach(s),zt=c(ct,", some of the coefficients become negative. This is not good for numerical computation. So we can divide the interval into subintervals, and apply the Newton-Cotes formula on each subinterval."),ct.forEach(s),bp=v(a),Mn=t(a,"P",{});var Sm=m(Mn);_t=c(Sm,"Take composite Simpson’s Rule as an example."),Sm.forEach(s),wp=v(a),on=t(a,"DIV",{class:!0});var Sl=m(on);Sl.forEach(s),zp=v(a),kn=t(a,"P",{});var Nm=m(kn);Mt=c(Nm,"The error term is the sum of the error terms of each subinterval, whose form is exactly the same as the error term of the original Simpson’s Rule."),Nm.forEach(s),_p=v(a),Sa=t(a,"H3",{id:!0});var vm=m(Sa);Na=t(vm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Am=m(Na);lp=t(Am,"SPAN",{class:!0}),m(lp).forEach(s),Am.forEach(s),kt=c(vm,"Open Newton-Cotes Formula"),vm.forEach(s),Mp=v(a),Ys=t(a,"P",{});var tn=m(Ys);Pt=c(tn,"Some functions are not defined at the end points, such as "),Pn=t(tn,"SPAN",{class:!0});var Nl=m(Pn);Nl.forEach(s),Lt=c(tn,". If we want to integrate it from "),Ln=t(tn,"SPAN",{class:!0});var Al=m(Ln);Al.forEach(s),Et=c(tn," to "),En=t(tn,"SPAN",{class:!0});var jl=m(En);jl.forEach(s),$t=c(tn,", we can use the Open Newton-Cotes Formula."),tn.forEach(s),kp=v(a),$n=t(a,"P",{});var jm=m($n);Ht=c(jm,"The idea of Open Newton-Cotes Formula is simply avoiding the end points."),jm.forEach(s),Pp=v(a),Hn=t(a,"P",{});var Im=m(Hn);Tt=c(Im,"Midpoint Rule is the simplest Open Newton-Cotes Formula."),Im.forEach(s),Lp=v(a),hn=t(a,"DIV",{class:!0});var Il=m(hn);Il.forEach(s),Ep=v(a),Aa=t(a,"H3",{id:!0});var dm=m(Aa);ja=t(dm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var qm=m(ja);ep=t(qm,"SPAN",{class:!0}),m(ep).forEach(s),qm.forEach(s),St=c(dm,"Adaptive Quadrature"),dm.forEach(s),$p=v(a),Tn=t(a,"P",{});var Cm=m(Tn);Nt=c(Cm,"Some part of the function changes rapidly, while some part changes slowly. If we use the same number of points in the whole interval, we may get Time Limit Exceeded. So we can use Adaptive Quadrature to use more points in the rapidly changing part, and less points in the slowly changing part."),Cm.forEach(s),Hp=v(a),Sn=t(a,"P",{});var Dm=m(Sn);At=c(Dm,"We first start recursion with the whole interval, then divide the interval into two subintervals, and recurse on each subinterval. The key is comparing the difference between the integral of the whole interval and the sum of the integrals of the two subintervals. If the difference is small enough, we can stop the recursion."),Dm.forEach(s),Tp=v(a),Ia=t(a,"H2",{id:!0});var fm=m(Ia);qa=t(fm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Qm=m(qa);ip=t(Qm,"SPAN",{class:!0}),m(ip).forEach(s),Qm.forEach(s),jt=c(fm,"Gaussian Quadrature"),fm.forEach(s),Sp=v(a),ya=t(a,"P",{});var Kn=m(ya);It=c(Kn,"The idea of Gaussian Quadrature is to choose the points "),Nn=t(Kn,"SPAN",{class:!0});var ql=m(Nn);ql.forEach(s),qt=c(Kn," and weights "),An=t(Kn,"SPAN",{class:!0});var Cl=m(An);Cl.forEach(s),Ct=c(Kn," such that the degree of precision is as high as possible."),Kn.forEach(s),Np=v(a),Is(Ca.$$.fragment,a),Ap=v(a),Is(Da.$$.fragment,a),jp=v(a),Is(Qa.$$.fragment,a),Ip=v(a),Va=t(a,"H3",{id:!0});var xm=m(Va);Ra=t(xm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Vm=m(Ra);rp=t(Vm,"SPAN",{class:!0}),m(rp).forEach(s),Vm.forEach(s),Dt=c(xm,"Legendre-Gauss Quadrature"),xm.forEach(s),qp=v(a),Js=t(a,"P",{});var mn=m(Js);Qt=c(mn,"The idea is picking "),jn=t(mn,"SPAN",{class:!0});var Dl=m(jn);Dl.forEach(s),Vt=c(mn," as the roots of Legendre polynomial "),In=t(mn,"SPAN",{class:!0});var Ql=m(In);Ql.forEach(s),Rt=c(mn,". In this way, the degree of precision is exactly "),qn=t(mn,"SPAN",{class:!0});var Vl=m(qn);Vl.forEach(s),Gt=c(mn,"."),mn.forEach(s),Cp=v(a),Ga=t(a,"P",{});var ot=m(Ga);Wt=c(ot,"To make life easier, we can normalize the interval to "),Cn=t(ot,"SPAN",{class:!0});var Rl=m(Cn);Rl.forEach(s),Ft=c(ot,". This do no harm to the generality of the formula."),ot.forEach(s),Dp=v(a),Is(Wa.$$.fragment,a),Qp=v(a),Dn=t(a,"P",{});var Rm=m(Dn);Ot=c(Rm,"The error term is"),Rm.forEach(s),Vp=v(a),gn=t(a,"DIV",{class:!0});var Gl=m(gn);Gl.forEach(s),Rp=v(a),Fa=t(a,"P",{});var ht=m(Fa);Bt=c(ht,"where "),Qn=t(ht,"SPAN",{class:!0});var Wl=m(Qn);Wl.forEach(s),Zt=c(ht,"."),ht.forEach(s),Gp=v(a),Is(Oa.$$.fragment,a),Wp=v(a),Ba=t(a,"H3",{id:!0});var bm=m(Ba);Za=t(bm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Gm=m(Za);cp=t(Gm,"SPAN",{class:!0}),m(cp).forEach(s),Gm.forEach(s),Yt=c(bm,"Weights of Legendre-Gauss Quadrature"),bm.forEach(s),Fp=v(a),Is(Ya.$$.fragment,a),Op=v(a),Is(Ja.$$.fragment,a),Bp=v(a),Ka=t(a,"H3",{id:!0});var wm=m(Ka);Ua=t(wm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Wm=m(Ua);op=t(Wm,"SPAN",{class:!0}),m(op).forEach(s),Wm.forEach(s),Jt=c(wm,"Weighted Gaussian Quadrature"),wm.forEach(s),Zp=v(a),ua=t(a,"P",{});var Un=m(ua);Kt=c(Un,"Consider we want to integrate "),Vn=t(Un,"SPAN",{class:!0});var Fl=m(Vn);Fl.forEach(s),Ut=c(Un,", where "),Rn=t(Un,"SPAN",{class:!0});var Ol=m(Rn);Ol.forEach(s),Xt=c(Un," is a weight function. We can use the same idea to get the weighted Gaussian Quadrature."),Un.forEach(s),Yp=v(a),Xa=t(a,"P",{});var gt=m(Xa);sm=c(gt,"Here we only discuss the case where "),Gn=t(gt,"SPAN",{class:!0});var Bl=m(Gn);Bl.forEach(s),am=c(gt,"."),gt.forEach(s),Jp=v(a),sn=t(a,"P",{});var yt=m(sn);nm=c(yt,"In this case, we use Chebyshev Zeroes as the points "),Wn=t(yt,"SPAN",{class:!0});var Zl=m(Wn);Zl.forEach(s),pm=c(yt,". The proof is similar to the proof of Gauss-Legendre Quadrature, we omit it here."),yt.forEach(s),Kp=v(a),an=t(a,"H3",{id:!0});var zm=m(an);nn=t(zm,"A",{"aria-hidden":!0,tabindex:!0,href:!0});var Fm=m(nn);hp=t(Fm,"SPAN",{class:!0}),m(hp).forEach(s),Fm.forEach(s),tm=c(zm,"Lobatto Quadrature"),zm.forEach(s),Up=v(a),As=t(a,"P",{});var va=m(As);mm=c(va,"Lobatto’s idea is use "),Fn=t(va,"SPAN",{class:!0});var Yl=m(Fn);Yl.forEach(s),lm=c(va,", "),On=t(va,"SPAN",{class:!0});var Jl=m(On);Jl.forEach(s),em=c(va," and the roots of "),Bn=t(va,"SPAN",{class:!0});var Kl=m(Bn);Kl.forEach(s),im=c(va," as the points "),Zn=t(va,"SPAN",{class:!0});var Ul=m(Zn);Ul.forEach(s),rm=c(va,"."),va.forEach(s),Xp=v(a),Yn=t(a,"P",{});var Om=m(Yn);cm=c(Om,"In this way, it uses the end points as information."),Om.forEach(s),this.h()},h(){e(d,"class","icon icon-link"),e(b,"aria-hidden","true"),e(b,"tabindex","-1"),e(b,"href","#numeric-integration"),e(i,"id","numeric-integration"),e(P,"class","icon 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P=l(y);P.forEach(s),E=c(k," is a monic polynomial with degree "),T=m(k,"SPAN",{class:!0});var S=l(T);S.forEach(s),$=c(k,"."),k.forEach(s),this.h()},h(){r(u,"class","math math-inline"),r(y,"class","math math-inline"),r(T,"class","math math-inline")},m(_,k){e(_,o,k),a(o,b),a(o,u),u.innerHTML=z,a(o,f),a(o,y),y.innerHTML=x,a(o,E),a(o,T),T.innerHTML=M,a(o,$)},p:Ba,d(_){_&&s(o)}}}function up(R){let o,b,u,z,f=`121−1T1(x)=x122−1T2(x)=12(2x2−1)=x2−12\\dfrac{1}{2 ^ {1 - 1}} T_1(x) = x\\\\ -\\dfrac{1}{2 ^ {2 - 1}} T_2(x) = \\dfrac{1}{2} (2x^2 - 1) = x^2 - \\dfrac{1}{2}`,y,x,E,T,M,$,_,k='12n−2Tn−1(x)\\dfrac{1}{2 ^ {n - 2}} T_{n - 1}(x)',B,P,S='n−1n - 1',us,Y,O='12n−3Tn−2(x)\\dfrac{1}{2 ^ {n - 3}} T_{n - 2}(x)',C,ls,bs='n−2n - 2',V,js,Q,ts=`12n−1Tn(x)=12n−1⋅2xTn−1(x)−12n−1Tn−2(x)=x(12n−2Tn−1(x))−12n−1Tn−2(x)\\dfrac{1}{2 ^ {n - 1}} T_n(x) = \\dfrac{1}{2 ^ {n - 1}} \\cdot 2xT_{n - 1}(x) - \\dfrac{1}{2 ^ {n - 1}} T_{n - 2}(x)\\\\ -= x(\\dfrac{1}{2 ^ {n - 2}}T_{n - 1}(x)) - \\dfrac{1}{2 ^ {n - 1}} T_{n - 2}(x)`,A,H,W,Ms,_s='nn',q,es,ks='n−2n - 2',F,zs,Ts='nn',D;return{c(){o=p("p"),b=i("Basis Step:"),u=v(),z=p("div"),y=v(),x=p("p"),E=i("Inductive Step:"),T=v(),M=p("p"),$=i("Assume that "),_=p("span"),B=i(" is a monic polynomial with degree "),P=p("span"),us=i(", and "),Y=p("span"),C=i(" is a polynomial with degree "),ls=p("span"),V=i("."),js=v(),Q=p("div"),A=v(),H=p("p"),W=i("The first term provided the monic polynomial with degree "),Ms=p("span"),q=i(", and the second term provided the polynomial with degree "),es=p("span"),F=i(", so the sum of the two terms is a monic polynomial with degree "),zs=p("span"),D=i("."),this.h()},l(g){o=m(g,"P",{});var L=l(o);b=c(L,"Basis Step:"),L.forEach(s),u=w(g),z=m(g,"DIV",{class:!0});var Ls=l(z);Ls.forEach(s),y=w(g),x=m(g,"P",{});var Gs=l(x);E=c(Gs,"Inductive Step:"),Gs.forEach(s),T=w(g),M=m(g,"P",{});var N=l(M);$=c(N,"Assume that "),_=m(N,"SPAN",{class:!0});var Es=l(_);Es.forEach(s),B=c(N," is a monic polynomial with degree "),P=m(N,"SPAN",{class:!0});var aa=l(P);aa.forEach(s),us=c(N,", and "),Y=m(N,"SPAN",{class:!0});var Ps=l(Y);Ps.forEach(s),C=c(N," is a polynomial with degree "),ls=m(N,"SPAN",{class:!0});var J=l(ls);J.forEach(s),V=c(N,"."),N.forEach(s),js=w(g),Q=m(g,"DIV",{class:!0});var Ca=l(Q);Ca.forEach(s),A=w(g),H=m(g,"P",{});var j=l(H);W=c(j,"The first term provided the monic polynomial with degree "),Ms=m(j,"SPAN",{class:!0});var G=l(Ms);G.forEach(s),q=c(j,", and the second term provided the polynomial with degree "),es=m(j,"SPAN",{class:!0});var is=l(es);is.forEach(s),F=c(j,", so the sum of the two terms is a monic polynomial with degree "),zs=m(j,"SPAN",{class:!0});var vs=l(zs);vs.forEach(s),D=c(j,"."),j.forEach(s),this.h()},h(){r(z,"class","math math-display"),r(_,"class","math math-inline"),r(P,"class","math math-inline"),r(Y,"class","math math-inline"),r(ls,"class","math math-inline"),r(Q,"class","math math-display"),r(Ms,"class","math math-inline"),r(es,"class","math math-inline"),r(zs,"class","math math-inline")},m(g,L){e(g,o,L),a(o,b),e(g,u,L),e(g,z,L),z.innerHTML=f,e(g,y,L),e(g,x,L),a(x,E),e(g,T,L),e(g,M,L),a(M,$),a(M,_),_.innerHTML=k,a(M,B),a(M,P),P.innerHTML=S,a(M,us),a(M,Y),Y.innerHTML=O,a(M,C),a(M,ls),ls.innerHTML=bs,a(M,V),e(g,js,L),e(g,Q,L),Q.innerHTML=ts,e(g,A,L),e(g,H,L),a(H,W),a(H,Ms),Ms.innerHTML=_s,a(H,q),a(H,es),es.innerHTML=ks,a(H,F),a(H,zs),zs.innerHTML=Ts,a(H,D)},p:Ba,d(g){g&&s(o),g&&s(u),g&&s(z),g&&s(y),g&&s(x),g&&s(T),g&&s(M),g&&s(js),g&&s(Q),g&&s(A),g&&s(H)}}}function vp(R){let o,b,u,z='w(x)w(x)',f;return{c(){o=p("p"),b=i("No other zeros provide a narrower bound for "),u=p("span"),f=i(" than zeros of chebyshev polynomials."),this.h()},l(y){o=m(y,"P",{});var x=l(o);b=c(x,"No other zeros provide a narrower bound for "),u=m(x,"SPAN",{class:!0});var E=l(u);E.forEach(s),f=c(x," than zeros of chebyshev polynomials."),x.forEach(s),this.h()},h(){r(u,"class","math math-inline")},m(y,x){e(y,o,x),a(o,b),a(o,u),u.innerHTML=z,a(o,f)},p:Ba,d(y){y&&s(o)}}}function wp(R){let o,b,u,z='cos⁡\\cos',f,y,x,E='Tn(x)=cos⁡(narccos⁡x)∈[−1,1]T_n(x) = \\cos(n \\arccos x) \\in [-1, 1]',T,M,$,_,k='w(x)w(x)',B,P,S,us='w(x)=∏i=1n(x−xi)w(x) = \\prod_{i = 1}^n (x - x_i)',Y,O,C,ls='w(x)w(x)',bs,V,js='nn',Q,ts,A,H,W,Ms='w(x)w(x)',_s,q,es='Tn(x)T_n(x)',ks,F,zs='12n−1Tn(x)\\dfrac{1}{2 ^ {n - 1}} T_n(x)',Ts,D,g,L,Ls,Gs='n>1n > 1',N,Es,aa='12n−1Tn(x)\\dfrac{1}{2 ^ {n - 1}} T_n(x)',Ps,J,Ca='w(x)w(x)',j,G,is='nn',vs,$s,oa='nn',ss,Os,as,en='w(x)=12n−1Tn(x)w(x) = \\dfrac{1}{2 ^ {n - 1}} T_n(x)',Qs,Z,qa,Ss,ha='Tn(x)T_n(x)',cs,Is,K,ga='w(x)=12n−1Tn(x)∈[−12n−1,12n−1]w(x) = \\dfrac{1}{2 ^ {n - 1}} T_n(x) \\in [-\\dfrac{1}{2 ^ {n - 1}}, \\dfrac{1}{2 ^ {n - 1}}]',ps,I,rs,ws,Js='n+1n + 1',na,xs,n='w(x)w(x)',h,os,Ks='−12n−1-\\dfrac{1}{2 ^ {n - 1}}',Rs,ms,ta='12n−1\\dfrac{1}{2 ^ {n - 1}}',Vs,ya,fs,ja,Ws,cn='w(x)w(x)',As,da,Fs,rn='∣w(x)∣=∣12n−1Tn(x)∣=12n−1∣Tn(x)∣≤12n−1|w(x)| = |\\dfrac{1}{2 ^ {n - 1}} T_n(x)| = \\dfrac{1}{2 ^ {n - 1}} |T_n(x)| \\le \\dfrac{1}{2 ^ {n - 1}}',Us,Bs,Hs,ua,Xs,va,Ja,U,on,wa,Qn='p(x)p(x)',hn,xa,Jn='w(x)w(x)',gn,fa,Kn='w(x)w(x)',yn,ba,Un='p(x)p(x)',dn,Ma,Xn='nn',un,Ka,Ys,vn,_a,Yn='w(x)w(x)',wn,Ua,pa,xn,ka,Zn='∣p(x)∣<12n−1|p(x)| < \\dfrac{1}{2 ^ {n - 1}}',Xa,Cs,fn,za,st='w(x)=12n−1w(x) = \\dfrac{1}{2 ^ {n - 1}}',bn,Ta,at='w(x)−p(x)=12n−1−p(x)>0w(x) - p(x) = \\dfrac{1}{2 ^ {n - 1}} - p(x) > 0',Ya,qs,Mn,La,nt='w(x)=−12n−1w(x) = -\\dfrac{1}{2 ^ {n - 1}}',_n,Ea,tt='w(x)−p(x)=−11n−1−p(x)<0w(x) - p(x) = -\\dfrac{1}{1 ^ {n - 1}} - p(x) < 0',Za,ns,kn,Pa,pt='w(x)−p(x)w(x) - p(x)',zn,$a,mt='n+1n + 1',Tn,Sa,lt='nn',Ln,Aa,et='w(x)−p(x)w(x) - p(x)',En,sn,X,Pn,Ha,it='w(x)w(x)',$n,Na,ct='p(x)p(x)',Sn,Da,rt='nn',An,Ia,ot='w(x)−p(x)w(x) - p(x)',Hn,Ra,ht='n−1n - 1',Nn,an,hs,Dn,Va,gt='n−1n - 1',In,Wa,yt='nn',Rn,Fa,dt='p(x)p(x)',Vn;return{c(){o=p("p"),b=i("From the range of the function "),u=p("span"),f=i(", we have"),y=v(),x=p("div"),T=v(),M=p("p"),$=i("The "),_=p("span"),B=i(" in the expression of remainder of lagrange interpolation is:"),P=v(),S=p("div"),Y=v(),O=p("p"),C=p("span"),bs=i(" is a monic polynomial with degree "),V=p("span"),Q=i(". This is trivial."),ts=v(),A=p("p"),H=i("If we use zeros of chebyshev polynomials as interpolation points, then "),W=p("span"),_s=i(" shares the same zeros with "),q=p("span"),ks=i(", in another word, shares the same zeros with "),F=p("span"),Ts=i("."),D=v(),g=p("p"),L=i("When "),Ls=p("span"),N=i(", since both "),Es=p("span"),Ps=i(" and "),J=p("span"),j=i(" are monic polynomials with degree "),G=p("span"),vs=i(" and exactly same "),$s=p("span"),ss=i(" zeros, they must be the exactly same polynomial."),Os=v(),as=p("div"),Qs=v(),Z=p("p"),qa=i("From the range of "),Ss=p("span"),cs=i(", we have"),Is=v(),K=p("div"),ps=v(),I=p("p"),rs=i("And there exist exactly "),ws=p("span"),na=i(" extrema of "),xs=p("span"),h=i(", alternating between "),os=p("span"),Rs=i(" and "),ms=p("span"),Vs=i("."),ya=v(),fs=p("p"),ja=i("If we take absolute value of "),Ws=p("span"),As=i(" just as we do when we calculate the error bound of lagrange interpolation, we get:"),da=v(),Fs=p("div"),Us=v(),Bs=p("p"),Hs=i("Now we are going to prove that zeros of chebyshev polynomials are the best interpolation points."),ua=v(),Xs=p("p"),va=i("Proof by contradiction:"),Ja=v(),U=p("p"),on=i("Suppose that there exist a monic polynomial "),wa=p("span"),hn=i(" with same degree as "),xa=p("span"),gn=i(", which provide a narrower error bound than "),fa=p("span"),yn=i(", and "),ba=p("span"),dn=i(" has "),Ma=p("span"),un=i(" zeros."),Ka=v(),Ys=p("p"),vn=i("We consider the extrema of "),_a=p("span"),wn=i("."),Ua=v(),pa=p("p"),xn=i("Since "),ka=p("span"),Xa=v(),Cs=p("p"),fn=i("When "),za=p("span"),bn=i(", "),Ta=p("span"),Ya=v(),qs=p("p"),Mn=i("When "),La=p("span"),_n=i(", "),Ea=p("span"),Za=v(),ns=p("p"),kn=i("Since "),Pa=p("span"),zn=i(" alternates between positive and negative at "),$a=p("span"),Tn=i(" points, there must be at least "),Sa=p("span"),Ln=i(" zeros of "),Aa=p("span"),En=i("."),sn=v(),X=p("p"),Pn=i("Since "),Ha=p("span"),$n=i(" and "),Na=p("span"),Sn=i(" is monic polynomial with degree "),Da=p("span"),An=i(", "),Ia=p("span"),Hn=i(" is a polynomial with degree "),Ra=p("span"),Nn=i("."),an=v(),hs=p("p"),Dn=i("Polynomial with degree "),Va=p("span"),In=i(" can’t have "),Wa=p("span"),Rn=i(" zeros, so no such polynomial "),Fa=p("span"),Vn=i(" exists."),this.h()},l(t){o=m(t,"P",{});var d=l(o);b=c(d,"From the range of the function "),u=m(d,"SPAN",{class:!0});var ut=l(u);ut.forEach(s),f=c(d,", we have"),d.forEach(s),y=w(t),x=m(t,"DIV",{class:!0});var vt=l(x);vt.forEach(s),T=w(t),M=m(t,"P",{});var nn=l(M);$=c(nn,"The "),_=m(nn,"SPAN",{class:!0});var wt=l(_);wt.forEach(s),B=c(nn," in the expression of remainder of lagrange interpolation is:"),nn.forEach(s),P=w(t),S=m(t,"DIV",{class:!0});var xt=l(S);xt.forEach(s),Y=w(t),O=m(t,"P",{});var Ga=l(O);C=m(Ga,"SPAN",{class:!0});var ft=l(C);ft.forEach(s),bs=c(Ga," is a monic polynomial with degree "),V=m(Ga,"SPAN",{class:!0});var bt=l(V);bt.forEach(s),Q=c(Ga,". This is trivial."),Ga.forEach(s),ts=w(t),A=m(t,"P",{});var Zs=l(A);H=c(Zs,"If we use zeros of chebyshev polynomials as interpolation points, then "),W=m(Zs,"SPAN",{class:!0});var Mt=l(W);Mt.forEach(s),_s=c(Zs," shares the same zeros with "),q=m(Zs,"SPAN",{class:!0});var _t=l(q);_t.forEach(s),ks=c(Zs,", in another word, shares the same zeros with "),F=m(Zs,"SPAN",{class:!0});var kt=l(F);kt.forEach(s),Ts=c(Zs,"."),Zs.forEach(s),D=w(t),g=m(t,"P",{});var gs=l(g);L=c(gs,"When "),Ls=m(gs,"SPAN",{class:!0});var zt=l(Ls);zt.forEach(s),N=c(gs,", since both "),Es=m(gs,"SPAN",{class:!0});var Tt=l(Es);Tt.forEach(s),Ps=c(gs," and "),J=m(gs,"SPAN",{class:!0});var Lt=l(J);Lt.forEach(s),j=c(gs," are monic polynomials with degree "),G=m(gs,"SPAN",{class:!0});var Et=l(G);Et.forEach(s),vs=c(gs," and exactly same "),$s=m(gs,"SPAN",{class:!0});var Pt=l($s);Pt.forEach(s),ss=c(gs," zeros, they must be the exactly same polynomial."),gs.forEach(s),Os=w(t),as=m(t,"DIV",{class:!0});var $t=l(as);$t.forEach(s),Qs=w(t),Z=m(t,"P",{});var tn=l(Z);qa=c(tn,"From the range of "),Ss=m(tn,"SPAN",{class:!0});var St=l(Ss);St.forEach(s),cs=c(tn,", we have"),tn.forEach(s),Is=w(t),K=m(t,"DIV",{class:!0});var At=l(K);At.forEach(s),ps=w(t),I=m(t,"P",{});var Ns=l(I);rs=c(Ns,"And there exist exactly "),ws=m(Ns,"SPAN",{class:!0});var Ht=l(ws);Ht.forEach(s),na=c(Ns," extrema of "),xs=m(Ns,"SPAN",{class:!0});var Nt=l(xs);Nt.forEach(s),h=c(Ns,", alternating between "),os=m(Ns,"SPAN",{class:!0});var Dt=l(os);Dt.forEach(s),Rs=c(Ns," and "),ms=m(Ns,"SPAN",{class:!0});var It=l(ms);It.forEach(s),Vs=c(Ns,"."),Ns.forEach(s),ya=w(t),fs=m(t,"P",{});var pn=l(fs);ja=c(pn,"If we take absolute value of "),Ws=m(pn,"SPAN",{class:!0});var Rt=l(Ws);Rt.forEach(s),As=c(pn," just as we do when we calculate the error bound of lagrange interpolation, we get:"),pn.forEach(s),da=w(t),Fs=m(t,"DIV",{class:!0});var Vt=l(Fs);Vt.forEach(s),Us=w(t),Bs=m(t,"P",{});var Bn=l(Bs);Hs=c(Bn,"Now we are going to prove that zeros of chebyshev polynomials are the best interpolation points."),Bn.forEach(s),ua=w(t),Xs=m(t,"P",{});var Cn=l(Xs);va=c(Cn,"Proof by contradiction:"),Cn.forEach(s),Ja=w(t),U=m(t,"P",{});var ys=l(U);on=c(ys,"Suppose that there exist a monic polynomial "),wa=m(ys,"SPAN",{class:!0});var Wt=l(wa);Wt.forEach(s),hn=c(ys," with same degree as "),xa=m(ys,"SPAN",{class:!0});var Ft=l(xa);Ft.forEach(s),gn=c(ys,", which provide a narrower error bound than "),fa=m(ys,"SPAN",{class:!0});var Bt=l(fa);Bt.forEach(s),yn=c(ys,", and "),ba=m(ys,"SPAN",{class:!0});var Ct=l(ba);Ct.forEach(s),dn=c(ys," has "),Ma=m(ys,"SPAN",{class:!0});var qt=l(Ma);qt.forEach(s),un=c(ys," zeros."),ys.forEach(s),Ka=w(t),Ys=m(t,"P",{});var mn=l(Ys);vn=c(mn,"We consider the extrema of "),_a=m(mn,"SPAN",{class:!0});var jt=l(_a);jt.forEach(s),wn=c(mn,"."),mn.forEach(s),Ua=w(t),pa=m(t,"P",{});var Wn=l(pa);xn=c(Wn,"Since "),ka=m(Wn,"SPAN",{class:!0});var Gt=l(ka);Gt.forEach(s),Wn.forEach(s),Xa=w(t),Cs=m(t,"P",{});var 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{m_j}',Cs,Hs,Ds;return{c(){t=p("p"),u=r("Since "),y=p("span"),h=r(" are pairwise coprime, we have "),i=p("span"),C=r(" for all "),_=p("span"),H=r("."),E=f(),d=p("p"),U=r("Then for each "),k=p("span"),I=r(" from "),T=p("span"),ts=r(" to "),L=p("span"),N=r(", we have "),q=p("span"),ps=r(" is coprime to "),R=p("span"),F=r(", which means "),Q=p("span"),hs=r(" has a multiplicative inverse modulo "),D=p("span"),ms=r("."),es=f(),Y=p("p"),J=r("For each "),K=p("span"),cs=r(" from "),V=p("span"),W=r(" to "),w=p("span"),us=r(", we have"),as=f(),j=p("div"),o=f(),B=p("p"),os=r("Since "),is=p("span"),ys=r(" is divisible by "),ls=p("span"),gs=r(" for all "),rs=p("span"),qs=r(", we have"),bs=f(),ds=p("div"),Rs=f(),Ps=p("p"),Is=r("Hence,"),Vs=f(),_s=p("div"),Cs=f(),Hs=p("p"),Ds=r("So the equation holds."),this.h()},l(g){t=m(g,"P",{});var z=l(t);u=c(z,"Since "),y=m(z,"SPAN",{class:!0});var Xs=l(y);Xs.forEach(s),h=c(z," are pairwise coprime, we have "),i=m(z,"SPAN",{class:!0});var 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have"),fs.forEach(s),as=x(g),j=m(g,"DIV",{class:!0});var ha=l(j);ha.forEach(s),o=x(g),B=m(g,"P",{});var xs=l(B);os=c(xs,"Since "),is=m(xs,"SPAN",{class:!0});var oa=l(is);oa.forEach(s),ys=c(xs," is divisible by "),ls=m(xs,"SPAN",{class:!0});var ga=l(ls);ga.forEach(s),gs=c(xs," for all "),rs=m(xs,"SPAN",{class:!0});var va=l(rs);va.forEach(s),qs=c(xs,", we have"),xs.forEach(s),bs=x(g),ds=m(g,"DIV",{class:!0});var ua=l(ds);ua.forEach(s),Rs=x(g),Ps=m(g,"P",{});var Gs=l(Ps);Is=c(Gs,"Hence,"),Gs.forEach(s),Vs=x(g),_s=m(g,"DIV",{class:!0});var ya=l(_s);ya.forEach(s),Cs=x(g),Hs=m(g,"P",{});var Os=l(Hs);Ds=c(Os,"So the equation holds."),Os.forEach(s),this.h()},h(){v(y,"class","math math-inline"),v(i,"class","math math-inline"),v(_,"class","math math-inline"),v(k,"class","math math-inline"),v(T,"class","math math-inline"),v(L,"class","math math-inline"),v(q,"class","math math-inline"),v(R,"class","math math-inline"),v(Q,"class","math math-inline"),v(D,"class","math math-inline"),v(K,"class","math 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coprime, "),D=p("span"),ms=r(", which means "),es=p("span"),J=r("."),K=f(),G=p("p"),cs=r("Hence, "),V=p("span"),W=r("."),this.h()},l(w){t=m(w,"P",{});var $=l(t);u=c($,"Suppose there are two solutions "),y=m($,"SPAN",{class:!0});var us=l(y);us.forEach(s),h=c($," and "),i=m($,"SPAN",{class:!0});var as=l(i);as.forEach(s),C=c($,"."),$.forEach(s),_=x(w),b=m(w,"P",{});var j=l(b);H=c(j,"Then, for each "),E=m(j,"SPAN",{class:!0});var a=l(E);a.forEach(s),U=c(j," from "),k=m(j,"SPAN",{class:!0});var o=l(k);o.forEach(s),I=c(j," to "),T=m(j,"SPAN",{class:!0});var B=l(T);B.forEach(s),ts=c(j,", we have "),L=m(j,"SPAN",{class:!0});var os=l(L);os.forEach(s),N=c(j," or "),q=m(j,"SPAN",{class:!0});var is=l(q);is.forEach(s),ps=c(j,"."),j.forEach(s),R=x(w),A=m(w,"P",{});var X=l(A);F=c(X,"As "),Q=m(X,"SPAN",{class:!0});var ys=l(Q);ys.forEach(s),hs=c(X," are pairwise coprime, "),D=m(X,"SPAN",{class:!0});var ls=l(D);ls.forEach(s),ms=c(X,", which means "),es=m(X,"SPAN",{class:!0});var 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It’s easy to see that there are "),S=m(es,"SPAN",{class:!0});var Fs=e(S);Fs.forEach(s),Y=i(es," such numbers."),es.forEach(s),N=E(cs),J=m(cs,"P",{});var ls=e(J);as=i(ls,"Similarly, there are "),P=m(ls,"SPAN",{class:!0});var Us=e(P);Us.forEach(s),X=i(ls," numbers less than "),B=m(ls,"SPAN",{class:!0});var Ws=e(B);Ws.forEach(s),ms=i(ls," that are not relatively prime to "),H=m(ls,"SPAN",{class:!0});var hs=e(H);hs.forEach(s),rs=i(ls," and the union of these 2 sets is empty since there is no such "),Z=m(ls,"SPAN",{class:!0});var pa=e(Z);pa.forEach(s),is=i(ls," that satisfy "),ts=m(ls,"SPAN",{class:!0});var W=e(ts);W.forEach(s),us=i(ls,", "),G=m(ls,"SPAN",{class:!0});var ps=e(G);ps.forEach(s),fs=i(ls," and "),ys=m(ls,"SPAN",{class:!0});var Bs=e(ys);Bs.forEach(s),Ms=i(ls,"."),ls.forEach(s),bs=E(cs),js=m(cs,"P",{});var qs=e(js);os=i(qs,"Therefore, "),K=m(qs,"SPAN",{class:!0});var Vs=e(K);Vs.forEach(s),_s=i(qs,"."),qs.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(w,"class","math math-inline"),r(x,"class","math math-inline"),r(f,"class","math math-inline"),r(S,"class","math math-inline"),r(P,"class","math math-inline"),r(B,"class","math math-inline"),r(H,"class","math math-inline"),r(Z,"class","math math-inline"),r(ts,"class","math math-inline"),r(G,"class","math math-inline"),r(ys,"class","math math-inline"),r(K,"class","math math-inline")},m(cs,es){c(cs,n,es),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),a(n,y),y.innerHTML=_,a(n,L),a(n,w),w.innerHTML=h,a(n,T),a(n,x),x.innerHTML=A,a(n,q),a(n,f),f.innerHTML=k,a(n,F),a(n,S),S.innerHTML=V,a(n,Y),c(cs,N,es),c(cs,J,es),a(J,as),a(J,P),P.innerHTML=O,a(J,X),a(J,B),B.innerHTML=U,a(J,ms),a(J,H),H.innerHTML=C,a(J,rs),a(J,Z),Z.innerHTML=ns,a(J,is),a(J,ts),ts.innerHTML=ws,a(J,us),a(J,G),G.innerHTML=R,a(J,fs),a(J,ys),ys.innerHTML=Cs,a(J,Ms),c(cs,bs,es),c(cs,js,es),a(js,os),a(js,K),K.innerHTML=ss,a(js,_s)},p:Ss,d(cs){cs&&s(n),cs&&s(N),cs&&s(J),cs&&s(bs),cs&&s(js)}}}function Bn(D){let n,u,g='φ(pk)=pk−pk−1\\varphi(p^k) = p^k - p^{k - 1}',b,d,o='pp',I;return{c(){n=p("p"),u=p("span"),b=l(", if "),d=p("span"),I=l(" is prime"),this.h()},l(M){n=m(M,"P",{});var y=e(n);u=m(y,"SPAN",{class:!0});var _=e(u);_.forEach(s),b=i(y,", if "),d=m(y,"SPAN",{class:!0});var L=e(d);L.forEach(s),I=i(y," is prime"),y.forEach(s),this.h()},h(){r(u,"class","math math-inline"),r(d,"class","math math-inline")},m(M,y){c(M,n,y),a(n,u),u.innerHTML=g,a(n,b),a(n,d),d.innerHTML=o,a(n,I)},p:Ss,d(M){M&&s(n)}}}function Vn(D){let n,u,g,b='pp',d,o,I='gcd⁡(p,k)≠1\\operatorname{gcd}(p, k) \\neq 1',M,y,_='p∣kp | k',L,w,h,T,x,A='pkp^k',q,f,k='pkp^k',F,S,V='p,2p,…,(pk−1−1)pp, 2p, \\dots, (p^{k - 1} - 1)p',Y,N,J='⌊pkp⌋−1=pk−1−1\\lfloor \\dfrac{p ^{k}}{p} \\rfloor - 1 = p^{k - 1} - 1',as,P,O,X,B,U='φ(pk)=pk−1−(pk−1−1)=pk−pk−1\\varphi(p^k) = p^k - 1 - (p^{k - 1} - 1) = p^k - p^{k - 1}',ms;return{c(){n=p("p"),u=l("Since "),g=p("span"),d=l(" is prime, "),o=p("span"),M=l(" gives "),y=p("span"),L=l("."),w=z(),h=p("p"),T=l("Hence, numbers less than "),x=p("span"),q=l(" that are not relatively prime to "),f=p("span"),F=l(" are "),S=p("span"),Y=l(". It’s easy to see that there are "),N=p("span"),as=l(" such numbers."),P=z(),O=p("p"),X=l("Therefore, "),B=p("span"),ms=l("."),this.h()},l(H){n=m(H,"P",{});var C=e(n);u=i(C,"Since "),g=m(C,"SPAN",{class:!0});var rs=e(g);rs.forEach(s),d=i(C," is prime, "),o=m(C,"SPAN",{class:!0});var Z=e(o);Z.forEach(s),M=i(C," gives "),y=m(C,"SPAN",{class:!0});var ns=e(y);ns.forEach(s),L=i(C,"."),C.forEach(s),w=E(H),h=m(H,"P",{});var is=e(h);T=i(is,"Hence, numbers less than "),x=m(is,"SPAN",{class:!0});var ts=e(x);ts.forEach(s),q=i(is," that are not relatively prime to "),f=m(is,"SPAN",{class:!0});var ws=e(f);ws.forEach(s),F=i(is," are "),S=m(is,"SPAN",{class:!0});var us=e(S);us.forEach(s),Y=i(is,". It’s easy to see that there are "),N=m(is,"SPAN",{class:!0});var G=e(N);G.forEach(s),as=i(is," such numbers."),is.forEach(s),P=E(H),O=m(H,"P",{});var R=e(O);X=i(R,"Therefore, "),B=m(R,"SPAN",{class:!0});var fs=e(B);fs.forEach(s),ms=i(R,"."),R.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(x,"class","math math-inline"),r(f,"class","math math-inline"),r(S,"class","math math-inline"),r(N,"class","math math-inline"),r(B,"class","math math-inline")},m(H,C){c(H,n,C),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),a(n,y),y.innerHTML=_,a(n,L),c(H,w,C),c(H,h,C),a(h,T),a(h,x),x.innerHTML=A,a(h,q),a(h,f),f.innerHTML=k,a(h,F),a(h,S),S.innerHTML=V,a(h,Y),a(h,N),N.innerHTML=J,a(h,as),c(H,P,C),c(H,O,C),a(O,X),a(O,B),B.innerHTML=U,a(O,ms)},p:Ss,d(H){H&&s(n),H&&s(w),H&&s(h),H&&s(P),H&&s(O)}}}function Dn(D){let n,u,g='φ(n)\\varphi(n)',b;return{c(){n=p("p"),u=p("span"),b=l(" is multiplicative"),this.h()},l(d){n=m(d,"P",{});var o=e(n);u=m(o,"SPAN",{class:!0});var I=e(u);I.forEach(s),b=i(o," is multiplicative"),o.forEach(s),this.h()},h(){r(u,"class","math math-inline")},m(d,o){c(d,n,o),a(n,u),u.innerHTML=g,a(n,b)},p:Ss,d(d){d&&s(n)}}}function Fn(D){let n,u,g,b='aa',d,o,I='bb',M,y,_='φ(ab)=φ(a)φ(b)\\varphi(ab) = \\varphi(a)\\varphi(b)',L,w,h,T,x,A,q,f,k='m,nm, n',F,S,V='a,ba, b',Y,N,J='xx',as,P,O='0≤x<ab0 \\leq x < ab',X,B,U='x≡m(moda)x \\equiv m \\pmod{a}',ms,H,C='x≡n(modb)x \\equiv n \\pmod{b}',rs,Z,ns,is,ts,ws,us='gcd⁡(x,ab)=1  ⟺  gcd⁡(x,a)=1∧gcd⁡(x,b)=1\\operatorname{gcd}(x, ab) = 1 \\iff \\operatorname{gcd}(x, a) = 1 \\land \\operatorname{gcd}(x, b) = 1',G,R,fs,ys,Cs='xx',Ms,bs,js='abab',os,K,ss,_s,cs,es='xx',As,Hs,ta='a,ba, b',ks,Ns,na='(m,n)(m, n)',Fs,ls,Us='abab',Ws,hs,pa='0≤m<a0 \\leq m < a',W,ps,Bs='0≤n<b0 \\leq n < b',qs,Vs,Rs,Xs,Ys,la='φ(a)φ(b)\\varphi(a)\\varphi(b)',Zs,Gs,Qs='abab',Os,aa,Is='abab',ma;return{c(){n=p("p"),u=l("Let "),g=p("span"),d=l(" and "),o=p("span"),M=l(" be two arbitrary coprime positive integers. We want to show that "),y=p("span"),L=l("."),w=z(),h=p("p"),T=l("From "),x=p("a"),A=l("CRT"),q=l(", we know that for each pair of number "),f=p("span"),F=l(" that is coprime to "),S=p("span"),Y=l(", there is a unique number "),N=p("span"),as=l(" such that "),P=p("span"),X=l(" and "),B=p("span"),ms=l(" and "),H=p("span"),rs=l("."),Z=z(),ns=p("p"),is=l("Since"),ts=z(),ws=p("div"),G=z(),R=p("p"),fs=l("we know such "),ys=p("span"),Ms=l(" is also coprime to "),bs=p("span"),os=l("."),K=z(),ss=p("p"),_s=l("It’s easy to see echo "),cs=p("span"),As=l(" which is coprime to "),Hs=p("span"),ks=l(" can be mapped to a unique pair of numbers "),Ns=p("span"),Fs=l(" that is coprime to "),ls=p("span"),Ws=l(" s.t "),hs=p("span"),W=l(" and "),ps=p("span"),qs=l("."),Vs=z(),Rs=p("p"),Xs=l("Therefore, there are "),Ys=p("span"),Zs=l(" numbers less than "),Gs=p("span"),Os=l(" that are coprime to "),aa=p("span"),ma=l("."),this.h()},l($){n=m($,"P",{});var Q=e(n);u=i(Q,"Let "),g=m(Q,"SPAN",{class:!0});var j=e(g);j.forEach(s),d=i(Q," and "),o=m(Q,"SPAN",{class:!0});var gs=e(o);gs.forEach(s),M=i(Q," be two arbitrary coprime positive integers. We want to show that "),y=m(Q,"SPAN",{class:!0});var wa=e(y);wa.forEach(s),L=i(Q,"."),Q.forEach(s),w=E($),h=m($,"P",{});var ds=e(h);T=i(ds,"From "),x=m(ds,"A",{href:!0});var ea=e(x);A=i(ea,"CRT"),ea.forEach(s),q=i(ds,", we know that for each pair of number "),f=m(ds,"SPAN",{class:!0});var sa=e(f);sa.forEach(s),F=i(ds," that is coprime to "),S=m(ds,"SPAN",{class:!0});var ba=e(S);ba.forEach(s),Y=i(ds,", there is a unique number "),N=m(ds,"SPAN",{class:!0});var ca=e(N);ca.forEach(s),as=i(ds," such that "),P=m(ds,"SPAN",{class:!0});var oa=e(P);oa.forEach(s),X=i(ds," and "),B=m(ds,"SPAN",{class:!0});var Ds=e(B);Ds.forEach(s),ms=i(ds," and "),H=m(ds,"SPAN",{class:!0});var _a=e(H);_a.forEach(s),rs=i(ds,"."),ds.forEach(s),Z=E($),ns=m($,"P",{});var ha=e(ns);is=i(ha,"Since"),ha.forEach(s),ts=E($),ws=m($,"DIV",{class:!0});var ga=e(ws);ga.forEach(s),G=E($),R=m($,"P",{});var Js=e(R);fs=i(Js,"we know such "),ys=m(Js,"SPAN",{class:!0});var da=e(ys);da.forEach(s),Ms=i(Js," is also coprime to "),bs=m(Js,"SPAN",{class:!0});var Ma=e(bs);Ma.forEach(s),os=i(Js,"."),Js.forEach(s),K=E($),ss=m($,"P",{});var vs=e(ss);_s=i(vs,"It’s easy to see echo "),cs=m(vs,"SPAN",{class:!0});var ka=e(cs);ka.forEach(s),As=i(vs," which is coprime to "),Hs=m(vs,"SPAN",{class:!0});var Ks=e(Hs);Ks.forEach(s),ks=i(vs," can be mapped to a unique pair of numbers "),Ns=m(vs,"SPAN",{class:!0});var ua=e(Ns);ua.forEach(s),Fs=i(vs," that is coprime to "),ls=m(vs,"SPAN",{class:!0});var fa=e(ls);fa.forEach(s),Ws=i(vs," s.t "),hs=m(vs,"SPAN",{class:!0});var $a=e(hs);$a.forEach(s),W=i(vs," and "),ps=m(vs,"SPAN",{class:!0});var ia=e(ps);ia.forEach(s),qs=i(vs,"."),vs.forEach(s),Vs=E($),Rs=m($,"P",{});var xs=e(Rs);Xs=i(xs,"Therefore, there are "),Ys=m(xs,"SPAN",{class:!0});var ya=e(Ys);ya.forEach(s),Zs=i(xs," numbers less than "),Gs=m(xs,"SPAN",{class:!0});var xa=e(Gs);xa.forEach(s),Os=i(xs," that are coprime to "),aa=m(xs,"SPAN",{class:!0});var La=e(aa);La.forEach(s),ma=i(xs,"."),xs.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(x,"href","/note/chinese-remainder-theorem"),r(f,"class","math math-inline"),r(S,"class","math math-inline"),r(N,"class","math math-inline"),r(P,"class","math math-inline"),r(B,"class","math math-inline"),r(H,"class","math math-inline"),r(ws,"class","math math-display"),r(ys,"class","math math-inline"),r(bs,"class","math math-inline"),r(cs,"class","math math-inline"),r(Hs,"class","math math-inline"),r(Ns,"class","math math-inline"),r(ls,"class","math math-inline"),r(hs,"class","math math-inline"),r(ps,"class","math math-inline"),r(Ys,"class","math math-inline"),r(Gs,"class","math math-inline"),r(aa,"class","math math-inline")},m($,Q){c($,n,Q),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),a(n,y),y.innerHTML=_,a(n,L),c($,w,Q),c($,h,Q),a(h,T),a(h,x),a(x,A),a(h,q),a(h,f),f.innerHTML=k,a(h,F),a(h,S),S.innerHTML=V,a(h,Y),a(h,N),N.innerHTML=J,a(h,as),a(h,P),P.innerHTML=O,a(h,X),a(h,B),B.innerHTML=U,a(h,ms),a(h,H),H.innerHTML=C,a(h,rs),c($,Z,Q),c($,ns,Q),a(ns,is),c($,ts,Q),c($,ws,Q),ws.innerHTML=us,c($,G,Q),c($,R,Q),a(R,fs),a(R,ys),ys.innerHTML=Cs,a(R,Ms),a(R,bs),bs.innerHTML=js,a(R,os),c($,K,Q),c($,ss,Q),a(ss,_s),a(ss,cs),cs.innerHTML=es,a(ss,As),a(ss,Hs),Hs.innerHTML=ta,a(ss,ks),a(ss,Ns),Ns.innerHTML=na,a(ss,Fs),a(ss,ls),ls.innerHTML=Us,a(ss,Ws),a(ss,hs),hs.innerHTML=pa,a(ss,W),a(ss,ps),ps.innerHTML=Bs,a(ss,qs),c($,Vs,Q),c($,Rs,Q),a(Rs,Xs),a(Rs,Ys),Ys.innerHTML=la,a(Rs,Zs),a(Rs,Gs),Gs.innerHTML=Qs,a(Rs,Os),a(Rs,aa),aa.innerHTML=Is,a(Rs,ma)},p:Ss,d($){$&&s(n),$&&s(w),$&&s(h),$&&s(Z),$&&s(ns),$&&s(ts),$&&s(ws),$&&s(G),$&&s(R),$&&s(K),$&&s(ss),$&&s(Vs),$&&s(Rs)}}}function On(D){let n,u,g='φ(n)=n∏p∣n(1−1p)\\varphi(n) = n\\prod_{p|n}(1 - \\frac{1}{p})';return{c(){n=p("p"),u=p("span"),this.h()},l(b){n=m(b,"P",{});var d=e(n);u=m(d,"SPAN",{class:!0});var o=e(u);o.forEach(s),d.forEach(s),this.h()},h(){r(u,"class","math math-inline")},m(b,d){c(b,n,d),a(n,u),u.innerHTML=g},p:Ss,d(b){b&&s(n)}}}function Cn(D){let n,u,g,b='n=∏i=1kpiain = \\prod_{i = 1}^k p_i^{a_i}',d,o,I='pip_i',M,y,_='aia_i',L,w,h,T=`φ(n)=φ(∏i=1kpiai)=∏i=1kφ(piai)=∏i=1kpiai−piai−1=n∏i=1k(1−1pi)\\varphi(n) \\\\ -= \\varphi(\\prod_{i = 1}^k p_i^{a_i}) \\\\ -= \\prod_{i = 1}^k \\varphi(p_i^{a_i}) \\\\ -= \\prod_{i = 1}^k p_i^{a_i} - p_i^{a_i - 1} \\\\ -= n\\prod_{i = 1}^k(1 - \\frac{1}{p_i})`;return{c(){n=p("p"),u=l("Let a positive number "),g=p("span"),d=l(" where "),o=p("span"),M=l(" are distinct primes and "),y=p("span"),L=l(" are positive integers."),w=z(),h=p("div"),this.h()},l(x){n=m(x,"P",{});var A=e(n);u=i(A,"Let a positive number "),g=m(A,"SPAN",{class:!0});var q=e(g);q.forEach(s),d=i(A," where "),o=m(A,"SPAN",{class:!0});var f=e(o);f.forEach(s),M=i(A," are distinct primes and "),y=m(A,"SPAN",{class:!0});var k=e(y);k.forEach(s),L=i(A," are positive integers."),A.forEach(s),w=E(x),h=m(x,"DIV",{class:!0});var F=e(h);F.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(h,"class","math math-display")},m(x,A){c(x,n,A),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),a(n,y),y.innerHTML=_,a(n,L),c(x,w,A),c(x,h,A),h.innerHTML=T},p:Ss,d(x){x&&s(n),x&&s(w),x&&s(h)}}}function Rn(D){let n,u,g='∑d∣nφ(d)=n\\sum_{d|n} \\varphi(d) = n';return{c(){n=p("p"),u=p("span"),this.h()},l(b){n=m(b,"P",{});var d=e(n);u=m(d,"SPAN",{class:!0});var o=e(u);o.forEach(s),d.forEach(s),this.h()},h(){r(u,"class","math math-inline")},m(b,d){c(b,n,d),a(n,u),u.innerHTML=g},p:Ss,d(b){b&&s(n)}}}function Yn(D){let n,u,g,b='n−1n-1',d,o,I='1n,2n,…,n−1n\\dfrac{1}{n}, \\dfrac{2}{n}, \\dots, \\dfrac{n-1}{n}',M,y,_,L,w,h='nn',T,x,A='nn',q,f,k,F,S,V='∑d∣n,n≠1φ(d)=n−1\\sum_{d|n, n \\ne 1} \\varphi(d) = n - 1',Y,N,J='∑d∣nφ(d)=n\\sum_{d|n} \\varphi(d) = n',as;return{c(){n=p("p"),u=l("Consider "),g=p("span"),d=l(" fractions "),o=p("span"),M=l("."),y=z(),_=p("p"),L=l("After reducing each fraction to its simplest form, we get "),w=p("span"),T=l(" fractions, each of which has a denominator that is a divisor of "),x=p("span"),q=l(", and the numerator of each fraction is coprime to the denominator."),f=z(),k=p("p"),F=l("Therefore "),S=p("span"),Y=l(". And apply property 1, we get "),N=p("span"),as=l("."),this.h()},l(P){n=m(P,"P",{});var O=e(n);u=i(O,"Consider "),g=m(O,"SPAN",{class:!0});var X=e(g);X.forEach(s),d=i(O," fractions "),o=m(O,"SPAN",{class:!0});var B=e(o);B.forEach(s),M=i(O,"."),O.forEach(s),y=E(P),_=m(P,"P",{});var U=e(_);L=i(U,"After reducing each fraction to its simplest form, we get "),w=m(U,"SPAN",{class:!0});var ms=e(w);ms.forEach(s),T=i(U," fractions, each of which has a denominator that is a divisor of "),x=m(U,"SPAN",{class:!0});var H=e(x);H.forEach(s),q=i(U,", and the numerator of each fraction is coprime to the denominator."),U.forEach(s),f=E(P),k=m(P,"P",{});var C=e(k);F=i(C,"Therefore "),S=m(C,"SPAN",{class:!0});var rs=e(S);rs.forEach(s),Y=i(C,". And apply property 1, we get "),N=m(C,"SPAN",{class:!0});var Z=e(N);Z.forEach(s),as=i(C,"."),C.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(w,"class","math math-inline"),r(x,"class","math math-inline"),r(S,"class","math math-inline"),r(N,"class","math math-inline")},m(P,O){c(P,n,O),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),c(P,y,O),c(P,_,O),a(_,L),a(_,w),w.innerHTML=h,a(_,T),a(_,x),x.innerHTML=A,a(_,q),c(P,f,O),c(P,k,O),a(k,F),a(k,S),S.innerHTML=V,a(k,Y),a(k,N),N.innerHTML=J,a(k,as)},p:Ss,d(P){P&&s(n),P&&s(y),P&&s(_),P&&s(f),P&&s(k)}}}function Gn(D){let n,u,g,b='aa',d,o,I='nn',M,y,_,L='aφ(n)≡1(modn)a^{\\varphi(n)} \\equiv 1 \\pmod{n}';return{c(){n=p("p"),u=l("If "),g=p("span"),d=l(" and "),o=p("span"),M=l(" are coprime positive integers, then"),y=z(),_=p("div"),this.h()},l(w){n=m(w,"P",{});var h=e(n);u=i(h,"If "),g=m(h,"SPAN",{class:!0});var T=e(g);T.forEach(s),d=i(h," and "),o=m(h,"SPAN",{class:!0});var x=e(o);x.forEach(s),M=i(h," are coprime positive integers, then"),h.forEach(s),y=E(w),_=m(w,"DIV",{class:!0});var A=e(_);A.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(_,"class","math math-display")},m(w,h){c(w,n,h),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),c(w,y,h),c(w,_,h),_.innerHTML=L},p:Ss,d(w){w&&s(n),w&&s(y),w&&s(_)}}}function Qn(D){let n,u,g,b='B={b1,b2,…,bφ(n)}B = \\{b_1, b_2, \\dots, b_{\\varphi(n)}\\}',d,o,I='nn',M,y,_='nn',L,w,h,T,x,A='bib_i',q,f,k='bjb_j',F,S,V='abi≡abj(modn)a b_i \\equiv a b_j \\pmod{n}',Y,N,J='bi≠bjb_i \\neq b_j',as,P,O,X=`abi≡abj(modn)abi−abj≡0(modn)a(bi−bj)≡0(modn)a b_i \\equiv a b_j \\pmod{n} \\\\ -a b_i - a b_j \\equiv 0 \\pmod{n} \\\\ -a (b_i - b_j) \\equiv 0 \\pmod{n}`,B,U,ms,H,C='aa',rs,Z,ns='nn',is,ts,ws='n∣a(bi−bj)n | a (b_i - b_j)',us,G,R='n∣bi−bjn | b_i - b_j',fs,ys,Cs='bi≡bj(modn)b_i \\equiv b_j \\pmod{n}',Ms,bs,js='bi≠bjb_i \\neq b_j',os,K,ss,_s,cs,es='aa',As,Hs,ta='bib_i',ks,Ns,na='nn',Fs,ls,Us='abia b_i',Ws,hs,pa='nn',W,ps,Bs,qs,Vs,Rs='{ab1,ab2,…,abφ(n)}\\{a b_1, a b_2, \\dots, a b_{\\varphi(n)}\\}',Xs,Ys,la='BB',Zs,Gs,Qs,Os,aa,Is,ma=`∏i=1φ(n)abi≡∏i=1φ(n)bi(modn)aφ(n)∏i=1φ(n)bi≡∏i=1φ(n)bi(modn)aφ(n)≡1(modn)\\prod_{i = 1}^{\\varphi(n)} ab_i \\equiv \\prod_{i = 1}^{\\varphi(n)} b_i \\pmod{n} \\\\ -a^{\\varphi(n)} \\prod_{i = 1}^{\\varphi(n)} b_i \\equiv \\prod_{i = 1}^{\\varphi(n)} b_i \\pmod{n} \\\\ -a^{\\varphi(n)} \\equiv 1 \\pmod{n}`;return{c(){n=p("p"),u=l("Let "),g=p("span"),d=l(" be the set of numbers less than "),o=p("span"),M=l(" that are coprime to "),y=p("span"),L=l("."),w=z(),h=p("p"),T=l("Assume that there are two such numbers "),x=p("span"),q=l(" and "),f=p("span"),F=l(" such that "),S=p("span"),Y=l(" and "),N=p("span"),as=l("."),P=z(),O=p("div"),B=z(),U=p("p"),ms=l("Since "),H=p("span"),rs=l(" and "),Z=p("span"),is=l(" are coprime, "),ts=p("span"),us=l(" gives "),G=p("span"),fs=l(" or "),ys=p("span"),Ms=l(", which contradicts with the assumption that "),bs=p("span"),os=l("."),K=z(),ss=p("p"),_s=l("What’s more, since both "),cs=p("span"),As=l(" and "),Hs=p("span"),ks=l(" are coprime to "),Ns=p("span"),Fs=l(", "),ls=p("span"),Ws=l(" is also coprime to "),hs=p("span"),W=l("."),ps=z(),Bs=p("p"),qs=l("Therefore, "),Vs=p("span"),Xs=l(" is a permutation of "),Ys=p("span"),Zs=l("."),Gs=z(),Qs=p("p"),Os=l("Hence,"),aa=z(),Is=p("div"),this.h()},l($){n=m($,"P",{});var Q=e(n);u=i(Q,"Let "),g=m(Q,"SPAN",{class:!0});var j=e(g);j.forEach(s),d=i(Q," be the set of numbers less than "),o=m(Q,"SPAN",{class:!0});var gs=e(o);gs.forEach(s),M=i(Q," that are coprime to "),y=m(Q,"SPAN",{class:!0});var wa=e(y);wa.forEach(s),L=i(Q,"."),Q.forEach(s),w=E($),h=m($,"P",{});var ds=e(h);T=i(ds,"Assume that there are two such numbers "),x=m(ds,"SPAN",{class:!0});var ea=e(x);ea.forEach(s),q=i(ds," and "),f=m(ds,"SPAN",{class:!0});var sa=e(f);sa.forEach(s),F=i(ds," such that "),S=m(ds,"SPAN",{class:!0});var ba=e(S);ba.forEach(s),Y=i(ds," and "),N=m(ds,"SPAN",{class:!0});var ca=e(N);ca.forEach(s),as=i(ds,"."),ds.forEach(s),P=E($),O=m($,"DIV",{class:!0});var oa=e(O);oa.forEach(s),B=E($),U=m($,"P",{});var Ds=e(U);ms=i(Ds,"Since "),H=m(Ds,"SPAN",{class:!0});var _a=e(H);_a.forEach(s),rs=i(Ds," and "),Z=m(Ds,"SPAN",{class:!0});var ha=e(Z);ha.forEach(s),is=i(Ds," are coprime, "),ts=m(Ds,"SPAN",{class:!0});var ga=e(ts);ga.forEach(s),us=i(Ds," gives "),G=m(Ds,"SPAN",{class:!0});var Js=e(G);Js.forEach(s),fs=i(Ds," or "),ys=m(Ds,"SPAN",{class:!0});var da=e(ys);da.forEach(s),Ms=i(Ds,", which contradicts with the assumption that "),bs=m(Ds,"SPAN",{class:!0});var Ma=e(bs);Ma.forEach(s),os=i(Ds,"."),Ds.forEach(s),K=E($),ss=m($,"P",{});var vs=e(ss);_s=i(vs,"What’s more, since both "),cs=m(vs,"SPAN",{class:!0});var ka=e(cs);ka.forEach(s),As=i(vs," and "),Hs=m(vs,"SPAN",{class:!0});var Ks=e(Hs);Ks.forEach(s),ks=i(vs," are coprime to "),Ns=m(vs,"SPAN",{class:!0});var ua=e(Ns);ua.forEach(s),Fs=i(vs,", "),ls=m(vs,"SPAN",{class:!0});var fa=e(ls);fa.forEach(s),Ws=i(vs," is also coprime to "),hs=m(vs,"SPAN",{class:!0});var $a=e(hs);$a.forEach(s),W=i(vs,"."),vs.forEach(s),ps=E($),Bs=m($,"P",{});var ia=e(Bs);qs=i(ia,"Therefore, "),Vs=m(ia,"SPAN",{class:!0});var xs=e(Vs);xs.forEach(s),Xs=i(ia," is a permutation of "),Ys=m(ia,"SPAN",{class:!0});var ya=e(Ys);ya.forEach(s),Zs=i(ia,"."),ia.forEach(s),Gs=E($),Qs=m($,"P",{});var xa=e(Qs);Os=i(xa,"Hence,"),xa.forEach(s),aa=E($),Is=m($,"DIV",{class:!0});var La=e(Is);La.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(x,"class","math math-inline"),r(f,"class","math math-inline"),r(S,"class","math math-inline"),r(N,"class","math math-inline"),r(O,"class","math math-display"),r(H,"class","math math-inline"),r(Z,"class","math math-inline"),r(ts,"class","math math-inline"),r(G,"class","math math-inline"),r(ys,"class","math 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Jn(D){let n,u,g,b='pp',d,o,I='aa',M,y,_='pp',L,w,h,T='ap−1≡1(modp)a^{p - 1} \\equiv 1 \\pmod{p}';return{c(){n=p("p"),u=l("If "),g=p("span"),d=l(" is a prime and "),o=p("span"),M=l(" is a positive integer that is not divisible by "),y=p("span"),L=l(", then"),w=z(),h=p("div"),this.h()},l(x){n=m(x,"P",{});var A=e(n);u=i(A,"If "),g=m(A,"SPAN",{class:!0});var q=e(g);q.forEach(s),d=i(A," is a prime and "),o=m(A,"SPAN",{class:!0});var f=e(o);f.forEach(s),M=i(A," is a positive integer that is not divisible by "),y=m(A,"SPAN",{class:!0});var k=e(y);k.forEach(s),L=i(A,", then"),A.forEach(s),w=E(x),h=m(x,"DIV",{class:!0});var F=e(h);F.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(h,"class","math math-display")},m(x,A){c(x,n,A),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),a(n,y),y.innerHTML=_,a(n,L),c(x,w,A),c(x,h,A),h.innerHTML=T},p:Ss,d(x){x&&s(n),x&&s(w),x&&s(h)}}}function Kn(D){let n,u,g,b='pp',d,o,I='φ(p)=p−1\\varphi(p) = p - 1',M,y,_,L,w,h='aa',T,x,A='pp',q,f,k='pp',F,S,V='aa',Y,N,J='pp',as,P,O,X,B,U='ap−1≡1(modp)a^{p - 1} \\equiv 1 \\pmod{p}',ms;return{c(){n=p("p"),u=l("Since "),g=p("span"),d=l(" is a prime, "),o=p("span"),M=l("."),y=z(),_=p("p"),L=l("Since "),w=p("span"),T=l(" is less than "),x=p("span"),q=l(" and "),f=p("span"),F=l(" is a prime, "),S=p("span"),Y=l(" must be coprime to "),N=p("span"),as=l("."),P=z(),O=p("p"),X=l("Apply Euler’s Theorem and get "),B=p("span"),ms=l("."),this.h()},l(H){n=m(H,"P",{});var C=e(n);u=i(C,"Since "),g=m(C,"SPAN",{class:!0});var rs=e(g);rs.forEach(s),d=i(C," is a prime, "),o=m(C,"SPAN",{class:!0});var Z=e(o);Z.forEach(s),M=i(C,"."),C.forEach(s),y=E(H),_=m(H,"P",{});var ns=e(_);L=i(ns,"Since "),w=m(ns,"SPAN",{class:!0});var is=e(w);is.forEach(s),T=i(ns," is less than "),x=m(ns,"SPAN",{class:!0});var ts=e(x);ts.forEach(s),q=i(ns," and "),f=m(ns,"SPAN",{class:!0});var ws=e(f);ws.forEach(s),F=i(ns," is a prime, "),S=m(ns,"SPAN",{class:!0});var us=e(S);us.forEach(s),Y=i(ns," must be coprime to "),N=m(ns,"SPAN",{class:!0});var G=e(N);G.forEach(s),as=i(ns,"."),ns.forEach(s),P=E(H),O=m(H,"P",{});var R=e(O);X=i(R,"Apply Euler’s Theorem and get "),B=m(R,"SPAN",{class:!0});var fs=e(B);fs.forEach(s),ms=i(R,"."),R.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(w,"class","math math-inline"),r(x,"class","math math-inline"),r(f,"class","math math-inline"),r(S,"class","math math-inline"),r(N,"class","math math-inline"),r(B,"class","math math-inline")},m(H,C){c(H,n,C),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),c(H,y,C),c(H,_,C),a(_,L),a(_,w),w.innerHTML=h,a(_,T),a(_,x),x.innerHTML=A,a(_,q),a(_,f),f.innerHTML=k,a(_,F),a(_,S),S.innerHTML=V,a(_,Y),a(_,N),N.innerHTML=J,a(_,as),c(H,P,C),c(H,O,C),a(O,X),a(O,B),B.innerHTML=U,a(O,ms)},p:Ss,d(H){H&&s(n),H&&s(y),H&&s(_),H&&s(P),H&&s(O)}}}function Un(D){let n,u='(p−1)!≡−1(modp)  ⟺  p is prime(p - 1)! \\equiv -1 \\pmod{p} \\iff p \\text{ is prime}',g,b,d,o,I='pp',M,y,_='11',L;return{c(){n=p("div"),g=z(),b=p("p"),d=l("where "),o=p("span"),M=l(" is an integer greater than "),y=p("span"),L=l("."),this.h()},l(w){n=m(w,"DIV",{class:!0});var h=e(n);h.forEach(s),g=E(w),b=m(w,"P",{});var T=e(b);d=i(T,"where "),o=m(T,"SPAN",{class:!0});var x=e(o);x.forEach(s),M=i(T," is an integer greater than "),y=m(T,"SPAN",{class:!0});var A=e(y);A.forEach(s),L=i(T,"."),T.forEach(s),this.h()},h(){r(n,"class","math math-display"),r(o,"class","math math-inline"),r(y,"class","math math-inline")},m(w,h){c(w,n,h),n.innerHTML=u,c(w,g,h),c(w,b,h),a(b,d),a(b,o),o.innerHTML=I,a(b,M),a(b,y),y.innerHTML=_,a(b,L)},p:Ss,d(w){w&&s(n),w&&s(g),w&&s(b)}}}function Xn(D){let n,u,g='p is prime  ⟹  (p−1)!≡−1(modp)p \\text{ is prime} \\implies (p - 1)! \\equiv -1 \\pmod{p}',b,d,o,I,M='p=2p = 2',y,_,L='(p−1)!=1≡−1(modp)(p - 1)! = 1 \\equiv -1 \\pmod{p}',w,h,T='pp',x,A,q,f;return{c(){n=p("p"),u=p("span"),b=z(),d=p("p"),o=l("When "),I=p("span"),y=l(", "),_=p("span"),w=l(", which is obviously true, so we are only going to prove the case where "),h=p("span"),x=l(" is an odd prime."),A=z(),q=p("p"),f=l("We are going to prove a lemma first."),this.h()},l(k){n=m(k,"P",{});var F=e(n);u=m(F,"SPAN",{class:!0});var S=e(u);S.forEach(s),F.forEach(s),b=E(k),d=m(k,"P",{});var V=e(d);o=i(V,"When "),I=m(V,"SPAN",{class:!0});var Y=e(I);Y.forEach(s),y=i(V,", "),_=m(V,"SPAN",{class:!0});var N=e(_);N.forEach(s),w=i(V,", which is obviously true, so we are only going to prove the case where "),h=m(V,"SPAN",{class:!0});var J=e(h);J.forEach(s),x=i(V," is an odd prime."),V.forEach(s),A=E(k),q=m(k,"P",{});var as=e(q);f=i(as,"We are going to prove a lemma first."),as.forEach(s),this.h()},h(){r(u,"class","math math-inline"),r(I,"class","math math-inline"),r(_,"class","math math-inline"),r(h,"class","math math-inline")},m(k,F){c(k,n,F),a(n,u),u.innerHTML=g,c(k,b,F),c(k,d,F),a(d,o),a(d,I),I.innerHTML=M,a(d,y),a(d,_),_.innerHTML=L,a(d,w),a(d,h),h.innerHTML=T,a(d,x),c(k,A,F),c(k,q,F),a(q,f)},p:Ss,d(k){k&&s(n),k&&s(b),k&&s(d),k&&s(A),k&&s(q)}}}function Zn(D){let n,u,g='x2≡1(modp)x ^ 2 \\equiv 1 \\pmod{p}',b,d,o='pp',I,M,y='11',_,L,w='p−1p - 1';return{c(){n=p("p"),u=p("span"),b=l(" has only 2 solutions where "),d=p("span"),I=l(" is a odd prime, and they are "),M=p("span"),_=l(" and "),L=p("span"),this.h()},l(h){n=m(h,"P",{});var T=e(n);u=m(T,"SPAN",{class:!0});var x=e(u);x.forEach(s),b=i(T," has only 2 solutions where "),d=m(T,"SPAN",{class:!0});var A=e(d);A.forEach(s),I=i(T," is a odd prime, and they are "),M=m(T,"SPAN",{class:!0});var q=e(M);q.forEach(s),_=i(T," and "),L=m(T,"SPAN",{class:!0});var f=e(L);f.forEach(s),T.forEach(s),this.h()},h(){r(u,"class","math math-inline"),r(d,"class","math math-inline"),r(M,"class","math math-inline"),r(L,"class","math math-inline")},m(h,T){c(h,n,T),a(n,u),u.innerHTML=g,a(n,b),a(n,d),d.innerHTML=o,a(n,I),a(n,M),M.innerHTML=y,a(n,_),a(n,L),L.innerHTML=w},p:Ss,d(h){h&&s(n)}}}function st(D){let n,u=`x2≡1(modp)x2−1≡0(modp)(x−1)(x+1)≡0(modp)x ^ 2 \\equiv 1 \\pmod{p} \\\\ -x ^ 2 - 1 \\equiv 0 \\pmod{p} \\\\ -(x - 1)(x + 1) \\equiv 0 \\pmod{p}`,g,b,d,o,I='pp',M,y,_='p∣(x−1)(x+1)p | (x - 1)(x + 1)',L,w,h='p∣x−1p | x - 1',T,x,A='p∣x+1p | x + 1',q,f,k,F,S,V='11',Y,N,J='p−1p - 1',as;return{c(){n=p("div"),g=z(),b=p("p"),d=l("Since "),o=p("span"),M=l(" is a prime, "),y=p("span"),L=l(" gives "),w=p("span"),T=l(" or "),x=p("span"),q=l("."),f=z(),k=p("p"),F=l("Therefore, the only solutions are "),S=p("span"),Y=l(" and "),N=p("span"),as=l("."),this.h()},l(P){n=m(P,"DIV",{class:!0});var O=e(n);O.forEach(s),g=E(P),b=m(P,"P",{});var X=e(b);d=i(X,"Since "),o=m(X,"SPAN",{class:!0});var B=e(o);B.forEach(s),M=i(X," is a prime, "),y=m(X,"SPAN",{class:!0});var U=e(y);U.forEach(s),L=i(X," gives "),w=m(X,"SPAN",{class:!0});var ms=e(w);ms.forEach(s),T=i(X," or "),x=m(X,"SPAN",{class:!0});var H=e(x);H.forEach(s),q=i(X,"."),X.forEach(s),f=E(P),k=m(P,"P",{});var C=e(k);F=i(C,"Therefore, the only solutions are "),S=m(C,"SPAN",{class:!0});var rs=e(S);rs.forEach(s),Y=i(C," and "),N=m(C,"SPAN",{class:!0});var Z=e(N);Z.forEach(s),as=i(C,"."),C.forEach(s),this.h()},h(){r(n,"class","math math-display"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(w,"class","math math-inline"),r(x,"class","math math-inline"),r(S,"class","math math-inline"),r(N,"class","math math-inline")},m(P,O){c(P,n,O),n.innerHTML=u,c(P,g,O),c(P,b,O),a(b,d),a(b,o),o.innerHTML=I,a(b,M),a(b,y),y.innerHTML=_,a(b,L),a(b,w),w.innerHTML=h,a(b,T),a(b,x),x.innerHTML=A,a(b,q),c(P,f,O),c(P,k,O),a(k,F),a(k,S),S.innerHTML=V,a(k,Y),a(k,N),N.innerHTML=J,a(k,as)},p:Ss,d(P){P&&s(n),P&&s(g),P&&s(b),P&&s(f),P&&s(k)}}}function at(D){let n,u,g,b='{ab1,ab2,…,abφ(n)}\\{a b_1, a b_2, \\dots, a b_{\\varphi(n)}\\}',d,o,I='BB',M,y,_='aa',L,w,h='nn',T,x,A='nn',q,f,k='0<a<n0 < a < n',F,S,V,Y,N,J='BB',as,P,O='pp',X,B,U='11',ms,H,C='p−1p - 1',rs,Z,ns,is,ts,ws,us=`∏i=1p−1i≡1⋅(p−1)≡−1(modp)(p−1)!≡−1(modp)\\prod_{i = 1}^{p - 1} i \\equiv 1 \\cdot (p - 1) \\equiv -1 \\pmod{p} \\\\ -(p - 1)! \\equiv -1 \\pmod{p}`;return{c(){n=p("p"),u=l("Consider Euler’s Theorem’s proof. We know that "),g=p("span"),d=l(" is a permutation of "),o=p("span"),M=l(", which gives the inverse of "),y=p("span"),L=l(" modulo "),w=p("span"),T=l(" is unique if "),x=p("span"),q=l(" is prime and "),f=p("span"),F=l("."),S=z(),V=p("p"),Y=l("Thus, we can pair up each number in "),N=p("span"),as=l(" with its inverse modulo "),P=p("span"),X=l(" except "),B=p("span"),ms=l(" and "),H=p("span"),rs=l(", whose inverse is themselves."),Z=z(),ns=p("p"),is=l("Therefore, we have"),ts=z(),ws=p("div"),this.h()},l(G){n=m(G,"P",{});var R=e(n);u=i(R,"Consider Euler’s Theorem’s proof. We know that "),g=m(R,"SPAN",{class:!0});var fs=e(g);fs.forEach(s),d=i(R," is a permutation of "),o=m(R,"SPAN",{class:!0});var ys=e(o);ys.forEach(s),M=i(R,", which gives the inverse of "),y=m(R,"SPAN",{class:!0});var Cs=e(y);Cs.forEach(s),L=i(R," modulo "),w=m(R,"SPAN",{class:!0});var Ms=e(w);Ms.forEach(s),T=i(R," is unique if "),x=m(R,"SPAN",{class:!0});var bs=e(x);bs.forEach(s),q=i(R," is prime and "),f=m(R,"SPAN",{class:!0});var js=e(f);js.forEach(s),F=i(R,"."),R.forEach(s),S=E(G),V=m(G,"P",{});var os=e(V);Y=i(os,"Thus, we can pair up each number in "),N=m(os,"SPAN",{class:!0});var K=e(N);K.forEach(s),as=i(os," with its inverse modulo "),P=m(os,"SPAN",{class:!0});var ss=e(P);ss.forEach(s),X=i(os," except "),B=m(os,"SPAN",{class:!0});var _s=e(B);_s.forEach(s),ms=i(os," and "),H=m(os,"SPAN",{class:!0});var cs=e(H);cs.forEach(s),rs=i(os,", whose inverse is themselves."),os.forEach(s),Z=E(G),ns=m(G,"P",{});var es=e(ns);is=i(es,"Therefore, we have"),es.forEach(s),ts=E(G),ws=m(G,"DIV",{class:!0});var As=e(ws);As.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(w,"class","math math-inline"),r(x,"class","math math-inline"),r(f,"class","math math-inline"),r(N,"class","math math-inline"),r(P,"class","math math-inline"),r(B,"class","math math-inline"),r(H,"class","math math-inline"),r(ws,"class","math math-display")},m(G,R){c(G,n,R),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),a(n,y),y.innerHTML=_,a(n,L),a(n,w),w.innerHTML=h,a(n,T),a(n,x),x.innerHTML=A,a(n,q),a(n,f),f.innerHTML=k,a(n,F),c(G,S,R),c(G,V,R),a(V,Y),a(V,N),N.innerHTML=J,a(V,as),a(V,P),P.innerHTML=O,a(V,X),a(V,B),B.innerHTML=U,a(V,ms),a(V,H),H.innerHTML=C,a(V,rs),c(G,Z,R),c(G,ns,R),a(ns,is),c(G,ts,R),c(G,ws,R),ws.innerHTML=us},p:Ss,d(G){G&&s(n),G&&s(S),G&&s(V),G&&s(Z),G&&s(ns),G&&s(ts),G&&s(ws)}}}function nt(D){let n,u,g='(p−1)!≡−1(modp)  ⟹  p is prime(p - 1)! \\equiv -1 \\pmod{p} \\implies p \\text{ is prime}';return{c(){n=p("p"),u=p("span"),this.h()},l(b){n=m(b,"P",{});var d=e(n);u=m(d,"SPAN",{class:!0});var o=e(u);o.forEach(s),d.forEach(s),this.h()},h(){r(u,"class","math math-inline")},m(b,d){c(b,n,d),a(n,u),u.innerHTML=g},p:Ss,d(b){b&&s(n)}}}function tt(D){let n,u,g,b='pp',d,o,I='(p−1)!≡−1(modp)(p - 1)! \\equiv -1 \\pmod{p}',M,y,_,L,w,h='pp',T,x,A='44',q,f,k='p=4p = 4',F,S,V,Y,N,J='p>=4p >= 4',as,P,O='(p−1)!≡(p−1)(p−2)!(p - 1)! \\equiv (p - 1)(p - 2)!',X,B,U,ms,H='pp',C,rs,Z='dd',ns,is,ts,ws,us,G=`d≤p≤p−2d \\leq \\sqrt{p} \\leq p - 2`,R,fs,ys,Cs,Ms,bs='(p−2)!≡d⋅(p−2)!d(modp)(p - 2)! \\equiv d \\cdot \\dfrac{(p - 2)!}{d} \\pmod{p}',js,os,K,ss,_s,cs='dd',es,As,Hs='pp',ta,ks,Ns='d⋅(p−2)!d≢1(modp)d \\cdot \\dfrac{(p - 2)!}{d} \\not\\equiv 1 \\pmod{p}',na,Fs,ls,Us,Ws,hs='(p−1)!≡(p−1)(p−2)!≢−1(modp)(p - 1)! \\equiv (p - 1)(p - 2)! \\not\\equiv -1 \\pmod{p}',pa;return{c(){n=p("p"),u=l("We want to prove that non-prime "),g=p("span"),d=l(" can’t satisfy "),o=p("span"),M=l("."),y=z(),_=p("p"),L=l("Then minimum integer "),w=p("span"),T=l(" that is not a prime is "),x=p("span"),q=l(", so we start from "),f=p("span"),F=l("."),S=z(),V=p("p"),Y=l("When "),N=p("span"),as=l(", "),P=p("span"),X=z(),B=p("p"),U=l("Since "),ms=p("span"),C=l(" is composite, let its minimum divisor be "),rs=p("span"),ns=l("."),is=z(),ts=p("p"),ws=l("Then "),us=p("span"),R=l("."),fs=z(),ys=p("p"),Cs=l("We obtain "),Ms=p("span"),js=l("."),os=z(),K=p("p"),ss=l("Since "),_s=p("span"),es=l(" has no inverse modulo "),As=p("span"),ta=l(" (by applying CRT), we have "),ks=p("span"),na=l("."),Fs=z(),ls=p("p"),Us=l("Therefore, "),Ws=p("span"),pa=l("."),this.h()},l(W){n=m(W,"P",{});var ps=e(n);u=i(ps,"We want to prove that non-prime "),g=m(ps,"SPAN",{class:!0});var Bs=e(g);Bs.forEach(s),d=i(ps," can’t satisfy "),o=m(ps,"SPAN",{class:!0});var qs=e(o);qs.forEach(s),M=i(ps,"."),ps.forEach(s),y=E(W),_=m(W,"P",{});var Vs=e(_);L=i(Vs,"Then minimum integer "),w=m(Vs,"SPAN",{class:!0});var Rs=e(w);Rs.forEach(s),T=i(Vs," that is not a prime is "),x=m(Vs,"SPAN",{class:!0});var Xs=e(x);Xs.forEach(s),q=i(Vs,", so we start from "),f=m(Vs,"SPAN",{class:!0});var Ys=e(f);Ys.forEach(s),F=i(Vs,"."),Vs.forEach(s),S=E(W),V=m(W,"P",{});var la=e(V);Y=i(la,"When "),N=m(la,"SPAN",{class:!0});var Zs=e(N);Zs.forEach(s),as=i(la,", "),P=m(la,"SPAN",{class:!0});var Gs=e(P);Gs.forEach(s),la.forEach(s),X=E(W),B=m(W,"P",{});var Qs=e(B);U=i(Qs,"Since "),ms=m(Qs,"SPAN",{class:!0});var Os=e(ms);Os.forEach(s),C=i(Qs," is composite, let its minimum divisor be "),rs=m(Qs,"SPAN",{class:!0});var aa=e(rs);aa.forEach(s),ns=i(Qs,"."),Qs.forEach(s),is=E(W),ts=m(W,"P",{});var Is=e(ts);ws=i(Is,"Then "),us=m(Is,"SPAN",{class:!0});var ma=e(us);ma.forEach(s),R=i(Is,"."),Is.forEach(s),fs=E(W),ys=m(W,"P",{});var $=e(ys);Cs=i($,"We obtain "),Ms=m($,"SPAN",{class:!0});var Q=e(Ms);Q.forEach(s),js=i($,"."),$.forEach(s),os=E(W),K=m(W,"P",{});var j=e(K);ss=i(j,"Since "),_s=m(j,"SPAN",{class:!0});var gs=e(_s);gs.forEach(s),es=i(j," has no inverse modulo "),As=m(j,"SPAN",{class:!0});var wa=e(As);wa.forEach(s),ta=i(j," (by applying CRT), we have "),ks=m(j,"SPAN",{class:!0});var ds=e(ks);ds.forEach(s),na=i(j,"."),j.forEach(s),Fs=E(W),ls=m(W,"P",{});var ea=e(ls);Us=i(ea,"Therefore, "),Ws=m(ea,"SPAN",{class:!0});var sa=e(Ws);sa.forEach(s),pa=i(ea,"."),ea.forEach(s),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(w,"class","math math-inline"),r(x,"class","math math-inline"),r(f,"class","math math-inline"),r(N,"class","math math-inline"),r(P,"class","math math-inline"),r(ms,"class","math math-inline"),r(rs,"class","math math-inline"),r(us,"class","math math-inline"),r(Ms,"class","math math-inline"),r(_s,"class","math math-inline"),r(As,"class","math math-inline"),r(ks,"class","math math-inline"),r(Ws,"class","math math-inline")},m(W,ps){c(W,n,ps),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),c(W,y,ps),c(W,_,ps),a(_,L),a(_,w),w.innerHTML=h,a(_,T),a(_,x),x.innerHTML=A,a(_,q),a(_,f),f.innerHTML=k,a(_,F),c(W,S,ps),c(W,V,ps),a(V,Y),a(V,N),N.innerHTML=J,a(V,as),a(V,P),P.innerHTML=O,c(W,X,ps),c(W,B,ps),a(B,U),a(B,ms),ms.innerHTML=H,a(B,C),a(B,rs),rs.innerHTML=Z,a(B,ns),c(W,is,ps),c(W,ts,ps),a(ts,ws),a(ts,us),us.innerHTML=G,a(ts,R),c(W,fs,ps),c(W,ys,ps),a(ys,Cs),a(ys,Ms),Ms.innerHTML=bs,a(ys,js),c(W,os,ps),c(W,K,ps),a(K,ss),a(K,_s),_s.innerHTML=cs,a(K,es),a(K,As),As.innerHTML=Hs,a(K,ta),a(K,ks),ks.innerHTML=Ns,a(K,na),c(W,Fs,ps),c(W,ls,ps),a(ls,Us),a(ls,Ws),Ws.innerHTML=hs,a(ls,pa)},p:Ss,d(W){W&&s(n),W&&s(y),W&&s(_),W&&s(S),W&&s(V),W&&s(X),W&&s(B),W&&s(is),W&&s(ts),W&&s(fs),W&&s(ys),W&&s(os),W&&s(K),W&&s(Fs),W&&s(ls)}}}function pt(D){let n,u=`$'fn[INIT-PHI](n)$ [ - $p = 'varnothing$ - $'varphi = 'varnothing$ - $'varphi(1) = 1$ - for $i = 2$ to $n$ [ - $'varphi(i) = 0$ - ] - for $i = 2$ to $m$ [ - if $'varphi(i) = 0$ [ - $'varphi(i) = i - 1$ - $p(|p|) = i$ - ] - for $j$ in $p$ [ - if $i 'cdot j > n$ [ - break - ] - if $i 'bmod j = 0$ [ - $'varphi(i 'cdot j) = 'varphi(i) 'cdot j$ - break - ] - $'varphi(i 'cdot j) = 'varphi(i) 'cdot (j - 1)$ - ] - ] - return $'varphi, p$ -]`;return{c(){n=p("pre"),this.h()},l(g){n=m(g,"PRE",{class:!0});var b=e(n);b.forEach(s),this.h()},h(){r(n,"class","language-pesudo")},m(g,b){c(g,n,b),n.innerHTML=u},p:Ss,d(g){g&&s(n)}}}function mt(D){let n,u,g,b='φ(x)\\varphi(x)',d,o,I='x≤nx \\leq n',M,y,_='nn',L,w,h='O(n)O(n)',T,x,A,q;return A=new An({props:{$$slots:{default:[pt]},$$scope:{ctx:D}}}),{c(){n=p("p"),u=l("Euler’s Sieve is an algorithm to calculate "),g=p("span"),d=l(" for all "),o=p("span"),M=l(" and give the list of prime numbers less than "),y=p("span"),L=l(" in "),w=p("span"),T=l(" time. Notice that this algorithm can be use on any multiplicative function with slightly modification."),x=z(),$s(A.$$.fragment),this.h()},l(f){n=m(f,"P",{});var k=e(n);u=i(k,"Euler’s Sieve is an algorithm to calculate "),g=m(k,"SPAN",{class:!0});var F=e(g);F.forEach(s),d=i(k," for all "),o=m(k,"SPAN",{class:!0});var S=e(o);S.forEach(s),M=i(k," and give the list of prime numbers less than "),y=m(k,"SPAN",{class:!0});var V=e(y);V.forEach(s),L=i(k," in "),w=m(k,"SPAN",{class:!0});var Y=e(w);Y.forEach(s),T=i(k," time. Notice that this algorithm can be use on any multiplicative function with slightly modification."),k.forEach(s),x=E(f),Ls(A.$$.fragment,f),this.h()},h(){r(g,"class","math math-inline"),r(o,"class","math math-inline"),r(y,"class","math math-inline"),r(w,"class","math math-inline")},m(f,k){c(f,n,k),a(n,u),a(n,g),g.innerHTML=b,a(n,d),a(n,o),o.innerHTML=I,a(n,M),a(n,y),y.innerHTML=_,a(n,L),a(n,w),w.innerHTML=h,a(n,T),c(f,x,k),zs(A,f,k),q=!0},p(f,k){const F={};k&2&&(F.$$scope={dirty:k,ctx:f}),A.$set(F)},i(f){q||(Es(A.$$.fragment,f),q=!0)},o(f){Ps(A.$$.fragment,f),q=!1},d(f){f&&s(n),f&&s(x),Ts(A,f)}}}function et(D){let n,u,g,b='φ\\varphi',d,o,I='φ(x)=x−1\\varphi(x) = x - 1',M,y,_='xx',L,w,h,T,x,A='xx',q,f,k='dd',F,S,V,Y=`φ(x)={φ(xd)⋅dxmod  d=0φ(xd)⋅(d−1)xmod  d≠0\\varphi(x) = \\begin{cases} - \\varphi\\left(\\dfrac{x}{d}\\right) \\cdot d & x \\mod d = 0 \\\\ - \\varphi\\left(\\dfrac{x}{d}\\right) \\cdot (d - 1) & x \\mod d \\neq 0 -\\end{cases}`,N,J,as,P='φ(1)=1\\varphi(1)=1',O,X,B,U,ms,H='x>1x > 1',C,rs,Z,ns,is,ts='xx',ws,us,G='dd',R,fs,ys='d≤xdd \\leq \\dfrac{x}{d}',Cs,Ms,bs='xd\\dfrac{x}{d}',js,os,K,ss,_s,cs='dd',es,As,Hs='pp',ta,ks,Ns='xd\\dfrac{x}{d}',na,Fs,ls='φ(x)\\varphi(x)',Us,Ws,hs,pa,W,ps='xx',Bs,qs,Vs='φ(x)\\varphi(x)',Rs,Xs,Ys='xx',la,Zs,Gs='φ(x)=0\\varphi(x) = 0',Qs,Os,aa='φ(x)=x−1\\varphi(x) = x - 1',Is,ma,$,Q;return{c(){n=p("p"),u=l("By the property of "),g=p("span"),d=l(", we know that "),o=p("span"),M=l(" if "),y=p("span"),L=l(" is prime."),w=z(),h=p("p"),T=l("And for composite "),x=p("span"),q=l(", suppose its minimum divisor is "),f=p("span"),F=l(", then we have"),S=z(),V=p("div"),N=z(),J=p("p"),as=p("span"),O=l(" is satisfied by the algorithm at the beginning and does not change."),X=z(),B=p("p"),U=l("Then consider an "),ms=p("span"),C=l("."),rs=z(),Z=p("p"),ns=l("If "),is=p("span"),ws=l(" is composite, then there must be a minimum prime divisor "),us=p("span"),R=l(", such that "),fs=p("span"),Cs=l(". 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"),W=l("span"),J=d("."),us=z(),N=l("div"),ss=z(),I=l("p"),hs=d("Since "),U=l("span"),gs=d(" is multiplicative, "),K=l("span"),D=d("."),f=z(),M=l("div"),this.h()},l(x){n=e(x,"P",{});var T=i(n);h=u(T,"Suppose "),p=e(T,"SPAN",{class:!0});var _s=i(p);_s.forEach(a),T.forEach(a),b=k(x),r=e(x,"DIV",{class:!0});var Ts=i(r);Ts.forEach(a),E=k(x),y=e(x,"P",{});var H=i(y);_=u(H,"Since "),P=e(H,"SPAN",{class:!0});var G=i(P);G.forEach(a),V=u(H,", every divisor of "),j=e(H,"SPAN",{class:!0});var $s=i(j);$s.forEach(a),A=u(H," is a unique product of a divisor of "),Z=e(H,"SPAN",{class:!0});var Ms=i(Z);Ms.forEach(a),Ps=u(H," and a divisor of "),R=e(H,"SPAN",{class:!0});var X=i(R);X.forEach(a),cs=u(H,", because for any prime "),Y=e(H,"SPAN",{class:!0});var Hs=i(Y);Hs.forEach(a),C=u(H," that divides "),B=e(H,"SPAN",{class:!0});var s=i(B);s.forEach(a),rs=u(H,", it can only divide one of "),Q=e(H,"SPAN",{class:!0});var c=i(Q);c.forEach(a),xs=u(H," or "),W=e(H,"SPAN",{class:!0});var 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math-display")},m(x,T){m(x,n,T),t(n,h),t(n,p),p.innerHTML=g,m(x,b,T),m(x,r,T),r.innerHTML=w,m(x,E,T),m(x,y,T),t(y,_),t(y,P),P.innerHTML=$,t(y,V),t(y,j),j.innerHTML=fs,t(y,A),t(y,Z),Z.innerHTML=F,t(y,Ps),t(y,R),R.innerHTML=q,t(y,cs),t(y,Y),Y.innerHTML=ds,t(y,C),t(y,B),B.innerHTML=ks,t(y,rs),t(y,Q),Q.innerHTML=ns,t(y,xs),t(y,W),W.innerHTML=O,t(y,J),m(x,us,T),m(x,N,T),N.innerHTML=bs,m(x,ss,T),m(x,I,T),t(I,hs),t(I,U),U.innerHTML=os,t(I,gs),t(I,K),K.innerHTML=ys,t(I,D),m(x,f,T),m(x,M,T),M.innerHTML=Ns},p:vs,d(x){x&&a(n),x&&a(b),x&&a(r),x&&a(E),x&&a(y),x&&a(us),x&&a(N),x&&a(ss),x&&a(I),x&&a(f),x&&a(M)}}}function Da(L){let n,h,p,g='g(n)=∑d∣nf(d)g(n) = \\sum_{d | n} f(d)',b,r,w='f(n)f(n)',E;return{c(){n=l("p"),h=d("If "),p=l("span"),b=d(" is a multiplicative function, then so is "),r=l("span"),E=d("."),this.h()},l(y){n=e(y,"P",{});var _=i(n);h=u(_,"If "),p=e(_,"SPAN",{class:!0});var P=i(p);P.forEach(a),b=u(_," is a multiplicative function, then so is "),r=e(_,"SPAN",{class:!0});var $=i(r);$.forEach(a),E=u(_,"."),_.forEach(a),this.h()},h(){o(p,"class","math math-inline"),o(r,"class","math math-inline")},m(y,_){m(y,n,_),t(n,h),t(n,p),p.innerHTML=g,t(n,b),t(n,r),r.innerHTML=w,t(n,E)},p:vs,d(y){y&&a(n)}}}function qa(L){let n,h,p,g='n=1n = 1',b,r,w='g(1)=f(1)=f(1×1)=g(1)g(1)=g(1)2g(1) = f(1) = f(1 \\times 1) = g(1)g(1) = g(1)^2',E,y,_='g(1)=f(1)=0g(1) = f(1) = 0',P,$,V='g(1)=f(1)=1g(1) = f(1) = 1',j,fs,A,Z,F,Ps='nn',R,q,cs='g(n)=f(n)+f(1)g(n) = f(n) + f(1)',Y,ds,C,B,ks,rs='n=n1n2n = n_1n_2',Q,ns,xs='gcd⁡(n1,n2)=1\\gcd(n_1, n_2) = 1',W,O,J,us,N,bs='ff',ss,I,hs='d<nd < n',U,os,gs,K=`g(n)=g(n1n2)=∑d∣n1n2f(d)=∑d1∣n1∑d2∣n2f(d1d2)g(n) \\\\ -= g(n_1n_2) \\\\ -= \\sum_{d | n_1n_2} f(d) \\\\ -= \\sum_{d_1 | n_1} \\sum_{d_2 | n_2} f(d_1d_2) \\\\`,ys,D,f,M,Ns='f(n1n2)f(n_1n_2)',x,T,_s='ff',Ts,H,G,$s=`=∑d1∣n1∑d2∣n2f(d1)f(d2)−f(n1)(n2)+f(n1n2)=(∑d1∣n1f(d1))(∑d2∣n2f(d2))−f(n1)f(n2)+f(n1n2)=g(n1)g(n2)−f(n1)f(n2)+f(n1n2)= \\sum_{d_1 | n_1} \\sum_{d_2 | n_2} f(d_1)f(d_2) - f(n_1)(n_2) + f(n_1n_2) \\\\ -= \\left(\\sum_{d_1 | n_1} f(d_1)\\right) \\left(\\sum_{d_2 | n_2} f(d_2)\\right) - f(n_1)f(n_2) + f(n_1n_2) \\\\ -= g(n_1)g(n_2) - f(n_1)f(n_2) + f(n_1n_2) \\\\`,Ms,X,Hs,s,c='g(n1)g(n2)g(n_1)g(n_2)',Ls,ws,as,qs='g(n)−g(n1)g(n2)=f(n1n2)−f(n1)f(n2)g(n) - g(n_1)g(n_2) = f(n_1n_2) - f(n_1)f(n_2) \\\\',Es,zs,Is,Ss,Js='gg',js,Rs,Vs,Ks='0=f(n1n2)−f(n1)f(n2)0 = f(n_1n_2) - f(n_1)f(n_2) \\\\',Ws,As,Cs,Bs,Us='ff',aa;return{c(){n=l("p"),h=d("If "),p=l("span"),b=d(", then "),r=l("span"),E=d(", which gives either "),y=l("span"),P=d(" or "),$=l("span"),j=d("."),fs=z(),A=l("p"),Z=d("If "),F=l("span"),R=d(" is prime, then "),q=l("span"),Y=d("."),ds=z(),C=l("p"),B=d("If "),ks=l("span"),Q=d(" where "),ns=l("span"),W=d(", then we prove by induction."),O=z(),J=l("p"),us=d("Suppose "),N=l("span"),ss=d(" is multiplicative for all "),I=l("span"),U=d("."),os=z(),gs=l("div"),ys=z(),D=l("p"),f=d("Only "),M=l("span"),x=d(" is out of the induction hypothesis, so we apply the multiplicative 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oa=i(ks);oa.forEach(a),Q=u(Gs," where "),ns=e(Gs,"SPAN",{class:!0});var ga=i(ns);ga.forEach(a),W=u(Gs,", then we prove by induction."),Gs.forEach(a),O=k(v),J=e(v,"P",{});var Ys=i(J);us=u(Ys,"Suppose "),N=e(Ys,"SPAN",{class:!0});var ya=i(N);ya.forEach(a),ss=u(Ys," is multiplicative for all "),I=e(Ys,"SPAN",{class:!0});var va=i(I);va.forEach(a),U=u(Ys,"."),Ys.forEach(a),os=k(v),gs=e(v,"DIV",{class:!0});var da=i(gs);da.forEach(a),ys=k(v),D=e(v,"P",{});var Qs=i(D);f=u(Qs,"Only "),M=e(Qs,"SPAN",{class:!0});var ua=i(M);ua.forEach(a),x=u(Qs," is out of the induction hypothesis, so we apply the multiplicative property of "),T=e(Qs,"SPAN",{class:!0});var fa=i(T);fa.forEach(a),Ts=u(Qs," for all other terms."),Qs.forEach(a),H=k(v),G=e(v,"DIV",{class:!0});var ba=i(G);ba.forEach(a),Ms=k(v),X=e(v,"P",{});var Xs=i(X);Hs=u(Xs,"Move "),s=e(Xs,"SPAN",{class:!0});var wa=i(s);wa.forEach(a),Ls=u(Xs," to the left hand side."),Xs.forEach(a),ws=k(v),as=e(v,"DIV",{class:!0});var 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Ba(L){let n,h,p,g='μ(n)\\mu(n)',b,r,w,E='∑d∣nμ(d)=[n=1]\\sum_{d | n} \\mu(d) = [n = 1]';return{c(){n=l("p"),h=d("The Mobius Function, denoted as "),p=l("span"),b=d(", is defined by the following rules:"),r=z(),w=l("div"),this.h()},l(y){n=e(y,"P",{});var _=i(n);h=u(_,"The Mobius Function, denoted as "),p=e(_,"SPAN",{class:!0});var P=i(p);P.forEach(a),b=u(_,", is defined by the following rules:"),_.forEach(a),r=k(y),w=e(y,"DIV",{class:!0});var $=i(w);$.forEach(a),this.h()},h(){o(p,"class","math math-inline"),o(w,"class","math math-display")},m(y,_){m(y,n,_),t(n,h),t(n,p),p.innerHTML=g,t(n,b),m(y,r,_),m(y,w,_),w.innerHTML=E},p:vs,d(y){y&&a(n),y&&a(r),y&&a(w)}}}function Fa(L){let n,h='f(n)=∑d∣ng(d)  ⟺  g(n)=∑d∣nμ(d)f(nd)f(n) = \\sum_{d | n} g(d) \\iff g(n) = \\sum_{d | n} \\mu(d) f\\left(\\frac{n}{d}\\right)';return{c(){n=l("div"),this.h()},l(p){n=e(p,"DIV",{class:!0});var g=i(n);g.forEach(a),this.h()},h(){o(n,"class","math 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Ja(L){let n,h;const p=[L[0],ma];let g={$$slots:{default:[Qa]},$$scope:{ctx:L}};for(let b=0;b{h(0,n=na(na({},n),ta(p)))},n=ta(n),[n]}class an extends $a{constructor(n){super(),Ma(this,n,Ka,Ja,La,{})}}export{an as component}; diff --git a/docs/_app/version.json b/docs/_app/version.json deleted file mode 100644 index 97ac512..0000000 --- a/docs/_app/version.json +++ /dev/null @@ -1 +0,0 @@ -{"version":"1714752774009"} \ No newline at end of file diff --git a/docs/assets/common.css b/docs/assets/common.css deleted file mode 100644 index 8890c97..0000000 --- a/docs/assets/common.css +++ /dev/null @@ -1,76 +0,0 @@ -:root { - font-family: "Times New Roman", Times, serif; -} - -.algorithm-container, .graph-container { - /** - * The algorithm box is a container for the algorithm - * There is an inner element that contains the algorithm - **/ - background-color: #f9f9f9; - padding: 1em; - border-radius: 0.5em; - display: flex; - justify-content: center; -} - -@media screen and (min-width: 1200px) { - .katex-display>.katex { - /** - * Override the default text-align: center - * Since modern screens are wide - * If we place equations in the center - * It will be too far from the text - **/ - text-align: left !important; - padding-left: 10em !important; - } - - .algorithm-container, .graph-container { - text-align: left; - padding-left: 10em; - width: calc(100% - 10em); - display: block; - } -} - -.md-main { - line-height: 1.6em; -} - -.md-main a { - color: black; -} - -h1 { - font-size: 2em; - border-bottom: 1px solid #ccc; - padding: 0.5em 0; -} - -h2 { - font-size: 1.5em; - border-bottom: 1px solid #ccc; - padding: 0.5em 0; -} - -h3 { - font-size: 1.2em; - border-bottom: 1px solid #ccc; - padding: 0.5em 0; -} - -p { - margin: 0.5em 0; -} - -.code { - background-color: #eee; - padding: 0.2em 0.4em; - border-radius: 0.2em; -} - -/**Add more spaces between formulas*/ -.mspace.newline { - height: 1.5em; -} \ No newline at end of file diff --git a/docs/assets/creditchart/userscript.patch b/docs/assets/creditchart/userscript.patch deleted file mode 100644 index b5d8965..0000000 --- a/docs/assets/creditchart/userscript.patch +++ /dev/null @@ -1,83 +0,0 @@ -Index: MCBBS CreditAnalysis.user.js -=================================================================== ---- MCBBS CreditAnalysis.user.js -+++ MCBBS CreditAnalysis.user.js -@@ -9,9 +9,16 @@ - // @match https://*.mcbbs.net/home.php?mod=space&username=* - // @grant none - // ==/UserScript== - -+ - (function() { -+ // CaveNightingale - begin -+ const creditRegex = /
  • 积分<\/em>(-?[0-9]+)<\/li>
  • 人气<\/em>(-?[0-9]+) 点<\/li>\n
  • 金粒<\/em>(-?[0-9]+) 粒<\/li>\n
  • 金锭\[已弃用\]<\/em>(-?[0-9]+) 块<\/li>\n
  • 宝石<\/em>(-?[0-9]+) 颗<\/li>\n
  • 下界之星<\/em>(-?[0-9]+) 枚<\/li>\n
  • 贡献<\/em>(-?[0-9]+) 份<\/li>\n
  • 爱心<\/em>(-?[0-9]+) 心<\/li>\n
  • 钻石<\/em>(-?[0-9]+) 颗<\/li>/; -+ const creditList = ['积分', '人气', '金粒', '金锭', '宝石', '下界之心', '贡献', '爱心', '钻石']; -+ const postsRegex = / target="_blank">回帖数 (-?[0-9]+?)<\/a>/; -+ const threadsRegex = / target="_blank">主题数 (-?[0-9]+?)<\/a>/; -+ // CaveNightingale - end - - console.log(" %c MCBBS %c CreditAnalysis", "color: #fff; background: #f8981d; padding:5px;", "color:#fff; background: #000; padding:5px;"); - console.log(" %c Made by %c 快乐小方 ", "color: #fff; background: #815098; padding:5px;", "color:#fff; background: #000; padding:5px;"); - console.group('MCA Log'); -@@ -23,21 +30,39 @@ - console.log("[L] 用户UID：" + uid); - - console.log("[L] 开始获取用户数据"); - // 调用api -- jq.ajax({ -- type:'get', -- url:"https://www.mcbbs.net/api/mobile/index.php?module=profile&uid=" + uid, -- success:function(body,heads,status){ -+ // CaveNightingale - begin -+ // copyed form netlify/functions/get-mcbbs-credit.js -+ console.log("[L] 正在运行的CreditAnalysis是洞穴夜莺补丁版本"); -+ let data = document.body.innerHTML; -+ let result = {}; -+ let creditParsed = creditRegex.exec(data); -+ if (creditParsed) { -+ let credits = {}; -+ for (let i = 0; i < creditList.length; i++) -+ credits[creditList[i]] = parseInt(creditParsed[i + 1]); -+ result.credits = credits; -+ } -+ let postsParsed = postsRegex.exec(data); -+ if (postsParsed) -+ result.posts = parseInt(postsParsed[1]); -+ let threadsParsed = threadsRegex.exec(data); -+ if (threadsParsed) -+ result.threads = parseInt(threadsParsed[1]); -+ // CaveNightingale - end - console.log("[L] 成功获取用户数据"); -- var credits = body.Variables.space.credits; //积分总值 -- var extcredits1 = body.Variables.space.extcredits1; //人气 -- var extcredits2 = body.Variables.space.extcredits6; //贡献 -- var extcredits3 = body.Variables.space.extcredits7; //爱心 -- var extcredits4 = body.Variables.space.extcredits8; //钻石 -- var posts = body.Variables.space.posts; //发帖数 -- var threads = body.Variables.space.threads; //主题数 -- var digestposts = body.Variables.space.digestposts; //精华数 -+ // CaveNightingale - begin -+ var credits = result.credits['积分']; //积分总值 -+ var extcredits1 = result.credits['人气']; //人气 -+ var extcredits2 = result.credits['贡献']; //贡献 -+ var extcredits3 = result.credits['爱心']; //爱心 -+ var extcredits4 = result.credits['钻石']; //钻石 -+ var posts = result.posts + result.threads; //发帖数 -+ var threads = result.threads; //主题数 -+ var tempSum = Math.round(extcredits1*3 + extcredits2*10 + extcredits3*4 + extcredits4*2 + posts / 3 + threads * 2); -+ var digestposts = (credits - tempSum) / 45; //精华数 -+ // CaveNightingale - end - - //DeBug：console.log("[D] 人气分:" + extcredits1*3 + "\n贡献分:" + extcredits2*10 + "\n爱心分:" + extcredits3*4 + "\n钻石分:" + extcredits4*2 + "\n发帖分:" + Math.round(posts/3) + "\n主题分:" + threads*2 + "\n精华分:" + digestposts*45) - var json = [credits,extcredits1*3,extcredits2*10,extcredits3*4,extcredits4*2,posts/3,threads*2,digestposts*45]; - console.log("[D] 用户积分数：" + json[0] +"，公式计算得：" + Math.round(json[1] + json[2] + json[3] + json[4] + json[5] + json[6] + json[7])); -@@ -276,8 +301,5 @@ - console.log("[L] 绘制完成"); - - - `); -- console.groupEnd(); -- } -- }); - })(); -\ No newline at end of file diff --git a/docs/assets/freemutecode/block.gif b/docs/assets/freemutecode/block.gif deleted file mode 100644 index 3f39a73..0000000 Binary files a/docs/assets/freemutecode/block.gif and /dev/null differ diff --git a/docs/assets/icon/parent.svg b/docs/assets/icon/parent.svg deleted file mode 100644 index 6011492..0000000 --- a/docs/assets/icon/parent.svg +++ /dev/null @@ -1,56 +0,0 @@ - - - - - - - - - - - - - diff --git a/docs/assets/icon/subtree.svg b/docs/assets/icon/subtree.svg deleted file mode 100644 index 84d1f6e..0000000 --- a/docs/assets/icon/subtree.svg +++ /dev/null @@ -1,46 +0,0 @@ - - - - - - - - - - diff --git a/docs/favicon.png b/docs/favicon.png deleted file mode 100644 index 5cffcf9..0000000 Binary files a/docs/favicon.png and /dev/null differ diff --git a/docs/freegiftcode.html b/docs/freegiftcode.html deleted file mode 100644 index 2d9d0f2..0000000 --- a/docs/freegiftcode.html +++ /dev/null @@ -1,6 +0,0 @@ - - - - - - \ No newline at end of file diff --git a/docs/index.html b/docs/index.html deleted file mode 100644 index ca38f79..0000000 --- a/docs/index.html +++ /dev/null @@ -1,61 +0,0 @@ - - - - - - - - - - - - - - - - - 夜莺洞穴 - - - -
    - - - - -
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    - - diff --git a/docs/note.html b/docs/note.html deleted file mode 100644 index 65f6ac9..0000000 --- a/docs/note.html +++ /dev/null @@ -1,78 +0,0 @@ - - - - - - - - - - - - - - - - - - - 夜莺洞穴 - 目录 - - -
    - - - - -
    -
    目录
    -
    - -

    To-do list

    -
    • Network flow
      • -
    • Dirichlet convolution
      • -
    • Numerical integration (Romberg / Lobatto)
    -
    - Experiment
    - -
    - Graph Theory
    - -
    - Number Theory
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    - Numerical Analysis
    - -
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    - - - - -
    - - diff --git a/docs/note/chebyshev-polynomial.html b/docs/note/chebyshev-polynomial.html deleted file mode 100644 index 992d851..0000000 --- a/docs/note/chebyshev-polynomial.html +++ /dev/null @@ -1,164 +0,0 @@ - - - - - - - - - - - - - - - - - - - - - - 夜莺洞穴 - 笔记 - - - -
    - - - - -
    -
    笔记 -
    -
    - -

    Chebyshev Polynomial

    -

    Definition

    -
    Definition - - -

    Define Chebyshev polynomial of the first kind Tn(x)T_n(x) as

    -
    Tn(x)=cos⁡(narccos⁡x)T_n(x) = \cos(n \arccos x)
     -
    -
    Definition - - -

    Define a polynomial whose coeffient of highest degree is 1 as a monic polynomial.

     -
    -

    Recurrence Relation

    -

    Basis Step:

    -
    T0(x)=cos⁡(0arccos⁡x)=1T1(x)=cos⁡(1arccos⁡x)=xT_0(x) = \cos(0 \arccos x) = 1\\ -T_1(x) = \cos(1 \arccos x) = x
    -

    Recursive Step:

    -
    cos⁡(n−1)x=cos⁡xcos⁡nx+sin⁡xsin⁡nxcos⁡(n+1)x=cos⁡xcos⁡nx−sin⁡xsin⁡nx\cos (n - 1)x = \cos x \cos nx + \sin x \sin nx\\ -\cos (n + 1)x = \cos x \cos nx - \sin x \sin nx
    -

    Add the two equations above, we get:

    -
    cos⁡(n−1)x+cos⁡(n+1)x=2cos⁡xcos⁡nx\cos (n - 1)x + \cos (n + 1)x = 2 \cos x \cos nx
    -

    Replace xx with arccos⁡x\arccos x, we get:

    -
    cos⁡[(n−1)arccos⁡x]+cos⁡[(n+1)arccos⁡x]=2cos⁡(arccos⁡x)cos⁡(narccos⁡x)Tn−1(x)+Tn+1(x)=2xTn(x)Tn+1(x)=2xTn(x)−Tn−1(x)\cos \left[(n - 1)\arccos x\right] + \cos \left[(n + 1)\arccos x\right] = 2 \cos (\arccos x) \cos (n \arccos x) \\ -T_{n - 1}(x) + T_{n + 1}(x) = 2xT_n(x)\\ -T_{n + 1}(x) = 2xT_n(x) - T_{n - 1}(x)
    -

    Why its zeros help relieve Runge’s phenomenon

    -

    We only discuss the case when x∈[−1,1]x \in [-1, 1], so arccos⁡x\arccos x is real.

    -
    Lemma - - -

    For any positive integer nn, 12n−1Tn(x)\dfrac{1}{2^{n - 1}}T_n(x) is a monic polynomial with degree nn.

     -
    -
    Proof - -

    Basis Step:

    -
    121−1T1(x)=x122−1T2(x)=12(2x2−1)=x2−12\dfrac{1}{2 ^ {1 - 1}} T_1(x) = x\\ -\dfrac{1}{2 ^ {2 - 1}} T_2(x) = \dfrac{1}{2} (2x^2 - 1) = x^2 - \dfrac{1}{2}
    -

    Inductive Step:

    -

    Assume that 12n−2Tn−1(x)\dfrac{1}{2 ^ {n - 2}} T_{n - 1}(x) is a monic polynomial with degree n−1n - 1, and 12n−3Tn−2(x)\dfrac{1}{2 ^ {n - 3}} T_{n - 2}(x) is a polynomial with degree n−2n - 2.

    -
    12n−1Tn(x)=12n−1⋅2xTn−1(x)−12n−1Tn−2(x)=x(12n−2Tn−1(x))−12n−1Tn−2(x)\dfrac{1}{2 ^ {n - 1}} T_n(x) = \dfrac{1}{2 ^ {n - 1}} \cdot 2xT_{n - 1}(x) - \dfrac{1}{2 ^ {n - 1}} T_{n - 2}(x)\\ -= x(\dfrac{1}{2 ^ {n - 2}}T_{n - 1}(x)) - \dfrac{1}{2 ^ {n - 1}} T_{n - 2}(x)
    -

    The first term provided the monic polynomial with degree nn, and the second term provided the polynomial with degree n−2n - 2, so the sum of the two terms is a monic polynomial with degree nn.

     - ⯀ -
    -
    Theorem - - -

    No other zeros provide a narrower bound for w(x)w(x) than zeros of chebyshev polynomials.

     -
    -
    Proof - -

    From the range of the function cos⁡\cos, we have

    -
    Tn(x)=cos⁡(narccos⁡x)∈[−1,1]T_n(x) = \cos(n \arccos x) \in [-1, 1]
    -

    The w(x)w(x) in the expression of remainder of lagrange interpolation is:

    -
    w(x)=∏i=1n(x−xi)w(x) = \prod_{i = 1}^n (x - x_i)
    -

    w(x)w(x) is a monic polynomial with degree nn. This is trivial.

    -

    If we use zeros of chebyshev polynomials as interpolation points, then w(x)w(x) shares the same zeros with Tn(x)T_n(x), in another word, shares the same zeros with 12n−1Tn(x)\dfrac{1}{2 ^ {n - 1}} T_n(x).

    -

    When n>1n > 1, since both 12n−1Tn(x)\dfrac{1}{2 ^ {n - 1}} T_n(x) and w(x)w(x) are monic polynomials with degree nn and exactly same nn zeros, they must be the exactly same polynomial.

    -
    w(x)=12n−1Tn(x)w(x) = \dfrac{1}{2 ^ {n - 1}} T_n(x)
    -

    From the range of Tn(x)T_n(x), we have

    -
    w(x)=12n−1Tn(x)∈[−12n−1,12n−1]w(x) = \dfrac{1}{2 ^ {n - 1}} T_n(x) \in [-\dfrac{1}{2 ^ {n - 1}}, \dfrac{1}{2 ^ {n - 1}}]
    -

    And there exist exactly n+1n + 1 extrema of w(x)w(x), alternating between −12n−1-\dfrac{1}{2 ^ {n - 1}} and 12n−1\dfrac{1}{2 ^ {n - 1}}.

    -

    If we take absolute value of w(x)w(x) just as we do when we calculate the error bound of lagrange interpolation, we get:

    -
    ∣w(x)∣=∣12n−1Tn(x)∣=12n−1∣Tn(x)∣≤12n−1|w(x)| = |\dfrac{1}{2 ^ {n - 1}} T_n(x)| = \dfrac{1}{2 ^ {n - 1}} |T_n(x)| \le \dfrac{1}{2 ^ {n - 1}}
    -

    Now we are going to prove that zeros of chebyshev polynomials are the best interpolation points.

    -

    Proof by contradiction:

    -

    Suppose that there exist a monic polynomial p(x)p(x) with same degree as w(x)w(x), which provide a narrower error bound than w(x)w(x), and p(x)p(x) has nn zeros.

    -

    We consider the extrema of w(x)w(x).

    -

    Since ∣p(x)∣<12n−1|p(x)| < \dfrac{1}{2 ^ {n - 1}}

    -

    When w(x)=12n−1w(x) = \dfrac{1}{2 ^ {n - 1}}, w(x)−p(x)=12n−1−p(x)>0w(x) - p(x) = \dfrac{1}{2 ^ {n - 1}} - p(x) > 0

    -

    When w(x)=−12n−1w(x) = -\dfrac{1}{2 ^ {n - 1}}, w(x)−p(x)=−11n−1−p(x)<0w(x) - p(x) = -\dfrac{1}{1 ^ {n - 1}} - p(x) < 0

    -

    Since w(x)−p(x)w(x) - p(x) alternates between positive and negative at n+1n + 1 points, there must be at least nn zeros of w(x)−p(x)w(x) - p(x).

    -

    Since w(x)w(x) and p(x)p(x) is monic polynomial with degree nn, w(x)−p(x)w(x) - p(x) is a polynomial with degree n−1n - 1.

    -

    Polynomial with degree n−1n - 1 can’t have nn zeros, so no such polynomial p(x)p(x) exists.

     - ⯀ -
    -

    So this is why we use zeros of chebyshev polynomials as interpolation points.

    -
    - Experiment
    - -
    - Graph Theory
    - -
    - Number Theory
    - -
    - Numerical Analysis
    -
    - Chebyshev Polynomial
    - -
    - Numeric Differentiation
    - -
    - Numeric Integration
    - -
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    - - diff --git a/docs/note/chinese-remainder-theorem.html b/docs/note/chinese-remainder-theorem.html deleted file mode 100644 index 9e05630..0000000 --- a/docs/note/chinese-remainder-theorem.html +++ /dev/null @@ -1,151 +0,0 @@ - - - - - - - - - - - - - - - - - - - - - - 夜莺洞穴 - 笔记 - Chebyshev Polynomial - - -
    - - - - -
    返回 -
    笔记 - Chebyshev Polynomial
    -
    - -

    Chinese Remainder Theorem (CRT)

    -

    CRT is a theorem that gives a unique solution to simultaneous linear congruences with pairwise coprime moduli.

    -
    {x≡a1(modm1)x≡a2(modm2)⋮x≡an(modmn)\begin{cases} -x \equiv a_1 \pmod {m_1} \\ -x \equiv a_2 \pmod {m_2} \\ -\vdots \\ -x \equiv a_n \pmod {m_n} -\end{cases}
    -

    The theorem states that the solution is given by

    -
    x≡∑i=1naiMiyi(modM)x \equiv \sum_{i = 1}^{n} a_i M_i y_i \pmod M
    -

    where

    -
    M≡∏i=1nmiMi≡Mmjyi≡Mi−1(modmi)\begin{gather} -M \equiv \prod_{i = 1}^{n} m_i \\ -M_i \equiv {M \over m_j} \\ -y_i \equiv M_i^{-1} \pmod {m_i} -\end{gather}
    -

    To prove CRT, we are going to two things:

    -
    • x≡∑i=1naiMiyi(modM)x \equiv \sum_{i = 1}^{n} a_i M_i y_i \pmod M is a solution
      • -
    • There is no other solution under modulo MM
    -
    Lemma - - -

    aa has a multiplicative inverse modulo mm if and only if gcd⁡(a,m)=1\operatorname{gcd}(a, m) = 1

     -
    -
    Proof - -

    Let the inverse be rr

    -

    ra≡1(modm)ra \equiv 1 \pmod m implies ra+km=1ra + km = 1 for some integer kk.

    -

    According to Bezout’s Lemma, gcd⁡(a,m)=1\operatorname{gcd}(a, m) = 1.

     - ⯀ -
    -
    Proposition - - -

    x≡∑i=1naiMiyi(modM)x \equiv \sum_{i = 1}^{n} a_i M_i y_i \pmod M is a solution

     -
    -
    Proof - -

    Since m1,m2,...mnm_1, m_2, ... m_n are pairwise coprime, we have gcd⁡(mi,mj)=1\operatorname{gcd}(m_i, m_j) = 1 for all i≠ji \neq j.

    -

    Then for each jj from 11 to nn, we have Mj=∏i=1,i≠jnmiM_j = \prod_{i = 1, i \neq j}^{n} m_i is coprime to mjm_j, which means MjM_j has a multiplicative inverse modulo mjm_j.

    -

    For each jj from 11 to nn, we have

    -
    ∑i=1naiMiyi=∑i=1,i≠jnaiMiyi+ajMjyj\sum_{i = 1}^{n} a_i M_i y_i = \sum_{i = 1, i \neq j}^{n} a_i M_i y_i + a_j M_j y_j
    -

    Since M1=∏i=1,i≠jnmiM_1 = \prod_{i = 1, i \neq j}^{n} m_i is divisible by mim_i for all i≠ji \neq j, we have

    -
    ∑i=1,i≠jnaiMiyi≡0(modmj)\sum_{i = 1, i \neq j}^{n} a_i M_i y_i \equiv 0 \pmod {m_j}
    -

    Hence,

    -
    ∑i=1naiMiyi≡ajMjyj≡ajMjMj−1≡aj(modmj)\sum_{i = 1}^{n} a_i M_i y_i \equiv a_j M_j y_j \equiv a_j M_j M_j^{-1} \equiv a_j \pmod {m_j}
    -

    So the equation holds.

     - ⯀ -
    -
    Proposition - - -

    There is no other solution under modulo MM

     -
    -
    Proof - -

    Suppose there are two solutions x1x_1 and x2x_2.

    -

    Then, for each jj from 11 to nn, we have x1−x2≡0(modmj)x_1 - x_2 \equiv 0 \pmod {m_j} or mj∣x1−x2m_j | x_1 - x_2.

    -

    As m1,m2,...,mnm_1, m_2, ... , m_n are pairwise coprime, lcm⁡(m1,m2,...,mn)=M\operatorname{lcm}(m_1, m_2, ... ,m_n) = M, which means M∣x1−x2M | x_1 - x_2.

    -

    Hence, x1≡x2(modM)x_1 \equiv x_2 \pmod M.

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    Dirichlet convolution

    -
    Definition - - -

    The Dirichlet convolution of two arithmetic functions ff and gg is defined as

    -
    (f∗g)(n)=∑d∣nf(d)g(nd)(f*g)(n) = \sum_{d|n} f(d)g(\frac{n}{d})
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    Euler’s Totient Function

    -
    Definition - - -

    The Euler’s Totient Function, denoted as φ(n)\varphi(n), is defined as the number of natural number strictly less than to nn that are relatively prime to nn.

     -
    -

    Properties

    -
    Property - 1 - -

    φ(1)=1\varphi(1) = 1

     -
    -
    Proof - -

    Only 00 is less than 11 and relatively prime to 11.

     - ⯀ -
    -
    Property - 2 - -

    φ(n)=n−1\varphi(n) = n - 1, if nn is prime

     -
    -
    Proof - -

    It’s easy to see that all positive intergers less than nn are relatively prime to nn if nn is prime.

    -

    Therefore, φ(n)=n−1\varphi(n) = n - 1.

     - ⯀ -
    -
    Property - 3 - -

    φ(pq)=(p−1)(q−1)\varphi(pq) = (p - 1)(q - 1), if pp and qq are distinct primes

     -
    -
    Proof - -

    Since pp is prime, gcd⁡(p,k)≠1\operatorname{gcd}(p, k) \neq 1 gives p∣kp | k. Hence, numbers less than pqpq that are not relatively prime to pqpq are p,2p,…,(q−1)pp, 2p, \dots, (q - 1)p. It’s easy to see that there are q−1q - 1 such numbers.

    -

    Similarly, there are p−1p - 1 numbers less than pqpq that are not relatively prime to pqpq and the union of these 2 sets is empty since there is no such xx that satisfy p∣xp | x, q∣xq | x and x<pqx < pq.

    -

    Therefore, φ(pq)=pq−1−(p−1)−(q−1)=(p−1)(q−1)\varphi(pq) = pq - 1 - (p - 1) - (q - 1) = (p - 1)(q - 1).

    - ⯀ -
    -
    Property - 4 - -

    φ(pk)=pk−pk−1\varphi(p^k) = p^k - p^{k - 1}, if pp is prime

     -
    -
    Proof - -

    Since pp is prime, gcd⁡(p,k)≠1\operatorname{gcd}(p, k) \neq 1 gives p∣kp | k.

    -

    Hence, numbers less than pkp^k that are not relatively prime to pkp^k are p,2p,…,(pk−1−1)pp, 2p, \dots, (p^{k - 1} - 1)p. It’s easy to see that there are ⌊pkp⌋−1=pk−1−1\lfloor \dfrac{p ^{k}}{p} \rfloor - 1 = p^{k - 1} - 1 such numbers.

    -

    Therefore, φ(pk)=pk−1−(pk−1−1)=pk−pk−1\varphi(p^k) = p^k - 1 - (p^{k - 1} - 1) = p^k - p^{k - 1}.

    - ⯀ -
    -
    Property - 5 - -

    φ(n)\varphi(n) is multiplicative

     -
    -
    Proof - -

    Let aa and bb be two arbitrary coprime positive integers. We want to show that φ(ab)=φ(a)φ(b)\varphi(ab) = \varphi(a)\varphi(b).

    -

    From CRT, we know that for each pair of number m,nm, n that is coprime to a,ba, b, there is a unique number xx such that 0≤x<ab0 \leq x < ab and x≡m(moda)x \equiv m \pmod{a} and x≡n(modb)x \equiv n \pmod{b}.

    -

    Since

    -
    gcd⁡(x,ab)=1  ⟺  gcd⁡(x,a)=1∧gcd⁡(x,b)=1\operatorname{gcd}(x, ab) = 1 \iff \operatorname{gcd}(x, a) = 1 \land \operatorname{gcd}(x, b) = 1
    -

    we know such xx is also coprime to abab.

    -

    It’s easy to see echo xx which is coprime to a,ba, b can be mapped to a unique pair of numbers (m,n)(m, n) that is coprime to abab s.t 0≤m<a0 \leq m < a and 0≤n<b0 \leq n < b.

    -

    Therefore, there are φ(a)φ(b)\varphi(a)\varphi(b) numbers less than abab that are coprime to abab.

     - ⯀ -
    -
    Property - 6 - -

    φ(n)=n∏p∣n(1−1p)\varphi(n) = n\prod_{p|n}(1 - \frac{1}{p})

     -
    -
    Proof - -

    Let a positive number n=∏i=1kpiain = \prod_{i = 1}^k p_i^{a_i} where pip_i are distinct primes and aia_i are positive integers.

    -
    φ(n)=φ(∏i=1kpiai)=∏i=1kφ(piai)=∏i=1kpiai−piai−1=n∏i=1k(1−1pi)\varphi(n) \\ -= \varphi(\prod_{i = 1}^k p_i^{a_i}) \\ -= \prod_{i = 1}^k \varphi(p_i^{a_i}) \\ -= \prod_{i = 1}^k p_i^{a_i} - p_i^{a_i - 1} \\ -= n\prod_{i = 1}^k(1 - \frac{1}{p_i})
     - ⯀ -
    -
    Property - 7 - -

    ∑d∣nφ(d)=n\sum_{d|n} \varphi(d) = n

     -
    -
    Proof - -

    Consider n−1n-1 fractions 1n,2n,…,n−1n\dfrac{1}{n}, \dfrac{2}{n}, \dots, \dfrac{n-1}{n}.

    -

    After reducing each fraction to its simplest form, we get nn fractions, each of which has a denominator that is a divisor of nn, and the numerator of each fraction is coprime to the denominator.

    -

    Therefore ∑d∣n,n≠1φ(d)=n−1\sum_{d|n, n \ne 1} \varphi(d) = n - 1. And apply property 1, we get ∑d∣nφ(d)=n\sum_{d|n} \varphi(d) = n.

     - ⯀ -
    -

    Euler’s Theorem

    -
    Theorem - - (Euler's theorem) -

    If aa and nn are coprime positive integers, then

    -
    aφ(n)≡1(modn)a^{\varphi(n)} \equiv 1 \pmod{n}
     -
    -
    Proof - -

    Let B={b1,b2,…,bφ(n)}B = \{b_1, b_2, \dots, b_{\varphi(n)}\} be the set of numbers less than nn that are coprime to nn.

    -

    Assume that there are two such numbers bib_i and bjb_j such that abi≡abj(modn)a b_i \equiv a b_j \pmod{n} and bi≠bjb_i \neq b_j.

    -
    abi≡abj(modn)abi−abj≡0(modn)a(bi−bj)≡0(modn)a b_i \equiv a b_j \pmod{n} \\ -a b_i - a b_j \equiv 0 \pmod{n} \\ -a (b_i - b_j) \equiv 0 \pmod{n}
    -

    Since aa and nn are coprime, n∣a(bi−bj)n | a (b_i - b_j) gives n∣bi−bjn | b_i - b_j or bi≡bj(modn)b_i \equiv b_j \pmod{n}, which contradicts with the assumption that bi≠bjb_i \neq b_j.

    -

    What’s more, since both aa and bib_i are coprime to nn, abia b_i is also coprime to nn.

    -

    Therefore, {ab1,ab2,…,abφ(n)}\{a b_1, a b_2, \dots, a b_{\varphi(n)}\} is a permutation of BB.

    -

    Hence,

    -
    ∏i=1φ(n)abi≡∏i=1φ(n)bi(modn)aφ(n)∏i=1φ(n)bi≡∏i=1φ(n)bi(modn)aφ(n)≡1(modn)\prod_{i = 1}^{\varphi(n)} ab_i \equiv \prod_{i = 1}^{\varphi(n)} b_i \pmod{n} \\ -a^{\varphi(n)} \prod_{i = 1}^{\varphi(n)} b_i \equiv \prod_{i = 1}^{\varphi(n)} b_i \pmod{n} \\ -a^{\varphi(n)} \equiv 1 \pmod{n}
     - ⯀ -
    -

    Fermat’s Little Theorem

    -
    Theorem - - (Fermat's Little Theorem) -

    If pp is a prime and aa is a positive integer that is not divisible by pp, then

    -
    ap−1≡1(modp)a^{p - 1} \equiv 1 \pmod{p}
     -
    -
    Proof - -

    Since pp is a prime, φ(p)=p−1\varphi(p) = p - 1.

    -

    Since aa is less than pp and pp is a prime, aa must be coprime to pp.

    -

    Apply Euler’s Theorem and get ap−1≡1(modp)a^{p - 1} \equiv 1 \pmod{p}.

     - ⯀ -
    -

    Wilson’s Theorem

    -
    Theorem - - (Wilson's Theorem) -
    (p−1)!≡−1(modp)  ⟺  p is prime(p - 1)! \equiv -1 \pmod{p} \iff p \text{ is prime}
    -

    where pp is an integer greater than 11.

     -
    -

    We need to prove two goals:

    -
    1. p is prime  ⟹  (p−1)!≡−1(modp)p \text{ is prime} \implies (p - 1)! \equiv -1 \pmod{p}
      2. -
    2. (p−1)!≡−1(modp)  ⟹  p is prime(p - 1)! \equiv -1 \pmod{p} \implies p \text{ is prime}
    -
    Goal - 1 - -

    p is prime  ⟹  (p−1)!≡−1(modp)p \text{ is prime} \implies (p - 1)! \equiv -1 \pmod{p}

    -

    When p=2p = 2, (p−1)!=1≡−1(modp)(p - 1)! = 1 \equiv -1 \pmod{p}, which is obviously true, so we are only going to prove the case where pp is an odd prime.

    -

    We are going to prove a lemma first.

     -
    -
    Lemma - - -

    x2≡1(modp)x ^ 2 \equiv 1 \pmod{p} has only 2 solutions where pp is a odd prime, and they are 11 and p−1p - 1

     -
    -
    Proof - -
    x2≡1(modp)x2−1≡0(modp)(x−1)(x+1)≡0(modp)x ^ 2 \equiv 1 \pmod{p} \\ -x ^ 2 - 1 \equiv 0 \pmod{p} \\ -(x - 1)(x + 1) \equiv 0 \pmod{p}
    -

    Since pp is a prime, p∣(x−1)(x+1)p | (x - 1)(x + 1) gives p∣x−1p | x - 1 or p∣x+1p | x + 1.

    -

    Therefore, the only solutions are 11 and p−1p - 1.

     - ⯀ -
    -

    Then we go back to the proof of Wilson’s Theorem.

    -
    Proof - -

    Consider Euler’s Theorem’s proof. We know that {ab1,ab2,…,abφ(n)}\{a b_1, a b_2, \dots, a b_{\varphi(n)}\} is a permutation of BB, which gives the inverse of aa modulo nn is unique if nn is prime and 0<a<n0 < a < n.

    -

    Thus, we can pair up each number in BB with its inverse modulo pp except 11 and p−1p - 1, whose inverse is themselves.

    -

    Therefore, we have

    -
    ∏i=1p−1i≡1⋅(p−1)≡−1(modp)(p−1)!≡−1(modp)\prod_{i = 1}^{p - 1} i \equiv 1 \cdot (p - 1) \equiv -1 \pmod{p} \\ -(p - 1)! \equiv -1 \pmod{p}
     - ⯀ -
    -
    Goal - 2 - -

    (p−1)!≡−1(modp)  ⟹  p is prime(p - 1)! \equiv -1 \pmod{p} \implies p \text{ is prime}

     -
    -
    Proof - -

    We want to prove that non-prime pp can’t satisfy (p−1)!≡−1(modp)(p - 1)! \equiv -1 \pmod{p}.

    -

    Then minimum integer pp that is not a prime is 44, so we start from p=4p = 4.

    -

    When p>=4p >= 4, (p−1)!≡(p−1)(p−2)!(p - 1)! \equiv (p - 1)(p - 2)!

    -

    Since pp is composite, let its minimum divisor be dd.

    -

    Then d≤p≤p−2d \leq \sqrt{p} \leq p - 2.

    -

    We obtain (p−2)!≡d⋅(p−2)!d(modp)(p - 2)! \equiv d \cdot \dfrac{(p - 2)!}{d} \pmod{p}.

    -

    Since dd has no inverse modulo pp (by applying CRT), we have d⋅(p−2)!d≢1(modp)d \cdot \dfrac{(p - 2)!}{d} \not\equiv 1 \pmod{p}.

    -

    Therefore, (p−1)!≡(p−1)(p−2)!≢−1(modp)(p - 1)! \equiv (p - 1)(p - 2)! \not\equiv -1 \pmod{p}.

     - ⯀ -
    -

    Euler’s Sieve

    -
    Algorithm - - (Euler's Sieve) -

    Euler’s Sieve is an algorithm to calculate φ(x)\varphi(x) for all x≤nx \leq n and give the list of prime numbers less than nn in O(n)O(n) time. Notice that this algorithm can be use on any multiplicative function with slightly modification.

    - - - -
    - 
    $'fn[INIT-PHI](n)$ [
-	$p = 'varnothing$
-	$'varphi = 'varnothing$
-	$'varphi(1) = 1$
-	for $i = 2$ to $n$ [
-		$'varphi(i) = 0$
-	]	
-	for $i = 2$ to $m$ [
-		if $'varphi(i) = 0$ [
-			$'varphi(i) = i - 1$
-			$p(|p|) = i$
-		]
-		for $j$ in $p$ [
-			if $i 'cdot j > n$ [
-				break
-			]
-			if $i 'bmod j = 0$ [
-				$'varphi(i 'cdot j) = 'varphi(i) 'cdot j$
-				break
-			]
-			$'varphi(i 'cdot j) = 'varphi(i) 'cdot (j - 1)$
-		]
-	]
-	return $'varphi, p$
-]
     -
    -
    Proof - of Correctness -

    By the property of φ\varphi, we know that φ(x)=x−1\varphi(x) = x - 1 if xx is prime.

    -

    And for composite xx, suppose its minimum divisor is dd, then we have

    -
    φ(x)={φ(xd)⋅dxmod  d=0φ(xd)⋅(d−1)xmod  d≠0\varphi(x) = \begin{cases} - \varphi\left(\dfrac{x}{d}\right) \cdot d & x \mod d = 0 \\ - \varphi\left(\dfrac{x}{d}\right) \cdot (d - 1) & x \mod d \neq 0 -\end{cases}
    -

    φ(1)=1\varphi(1)=1 is satisfied by the algorithm at the beginning and does not change.

    -

    Then consider an x>1x > 1.

    -

    If xx is composite, then there must be a minimum prime divisor dd, such that d≤xdd \leq \dfrac{x}{d}. (If this is not the case, then we can make a smaller prime divisor from xd\dfrac{x}{d}, which leads to a contradiction.)

    -

    This means dd is in the prime list pp when we reach xd\dfrac{x}{d}. And by the condition of the loop, we know that φ(x)\varphi(x) has been calculated correctly.

    -

    If xx is prime, then φ(x)\varphi(x) cannot be reached by previous iterations, so when we reach xx, φ(x)=0\varphi(x) = 0 and we set φ(x)=x−1\varphi(x) = x - 1.

    -

    Therefore, the algorithm is correct.

     - ⯀ -
    -
    Proof - of Time Complexity -

    By the proof of correctness, we know that φ(x)\varphi(x) is assigned exactly once for each xx except for the initial assignment.

    -

    And there are assignment statement in all loops, so the other parts will have at most the same order of time complexity as the assignment statement.

    -

    Therefore, the time complexity is O(n)O(n).

     - ⯀ -
    -

    Miscellaneous

    -

    Disproof of 22n+12^{2^n} + 1 is prime for all nn

    -

    Take an anti-example n=5n = 5.

    -

    3225≡3029026160(mod225+1)3 ^ {2 ^ {2 ^ 5}} \equiv 3029026160 \pmod{2 ^ {2 ^ 5} + 1}

    -

    This disagrees with Fermat’s Little Theorem, so the proposition is false.

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    Pre-requisite Number Theory

    -
    Assertion - - -

    We can reorder the items in a summation if the region of summation is finate, or countable but the sum is absolutely convergent.

     -
    -

    This is a well-known calculus theorem so we are not going to prove it here.

    -
    Lemma - - -
    ∑d∣nf(d)=∑d∣nf(nd)\sum_{d | n} f(d) = \sum_{d | n} f\left(\frac{n}{d}\right)
     -
    -
    Proof - -

    Reverse the order of summation.

     - ⯀ -
    -
    Lemma - - -
    ∑d∣n∑e∣df(d,e)=∑e∣n∑k∣(n/e)f(ke,e)\sum_{d | n} \sum_{e | d} f(d, e) = \sum_{e | n} \sum_{k | (n / e)} f(ke, e)
     -
    -
    Proof - -

    We just proof the two regions are the same, since the order of summation makes no difference.

    -
    d∣n∧e∣d  ⟺  n=k1d∧d=k2e, where k1,k2 are integers  ⟺  n=k1k2e∧d=k2e  ⟺  e∣n∧k1k2=(n/e)∧d=k2e  ⟺  e∣n∧k2∣(n/e)∧d=k2ed | n \wedge e | d \\ -\iff n=k_1d \wedge d=k_2e, \text{ where }k_1, k_2 \text{ are integers} \\ -\iff n=k_1k_2e \wedge d = k_2e \\ -\iff e | n \wedge k_1k_2= (n / e) \wedge d = k_2e \\ -\iff e | n \wedge k_2 | (n / e) \wedge d = k_2e
     - ⯀ -
    -
    Lemma - - -

    If f(n)f(n) is a multiplicative function, then so is g(n)=∑d∣nf(d)g(n) = \sum_{d | n} f(d).

     -
    -
    Proof - -

    Suppose gcd⁡(n1,n2)=1\gcd(n_1, n_2) = 1

    -
    g(n1n2)=∑d∣n1n2f(d)g(n_1n_2) -= \sum_{d | n_1n_2} f(d)
    -

    Since gcd⁡(n1,n2)=1\gcd(n_1, n_2) = 1, every divisor of n1n2n_1n_2 is a unique product of a divisor of n1n_1 and a divisor of n2n_2, because for any prime pp that divides n1n2n_1n_2, it can only divide one of n1n_1 or n2n_2.

    -
    =∑d1∣n1∑d2∣n2f(d1d2)= \sum_{d_1 | n_1} \sum_{d_2 | n_2} f(d_1d_2)\\
    -

    Since f(n)f(n) is multiplicative, f(d1d2)=f(d1)f(d2)f(d_1d_2) = f(d_1)f(d_2).

    -
    =∑d1∣n1∑d2∣n2f(d1)f(d2)=(∑d1∣n1f(d1))(∑d2∣n2f(d2))=g(n1)g(n2)= \sum_{d_1 | n_1} \sum_{d_2 | n_2} f(d_1)f(d_2)\\ -= \left(\sum_{d_1 | n_1} f(d_1)\right) \left(\sum_{d_2 | n_2} f(d_2)\right)\\ -= g(n_1)g(n_2)
     - ⯀ -
    -
    Lemma - - -

    If g(n)=∑d∣nf(d)g(n) = \sum_{d | n} f(d) is a multiplicative function, then so is f(n)f(n).

     -
    -
    Proof - -

    If n=1n = 1, then g(1)=f(1)=f(1×1)=g(1)g(1)=g(1)2g(1) = f(1) = f(1 \times 1) = g(1)g(1) = g(1)^2, which gives either g(1)=f(1)=0g(1) = f(1) = 0 or g(1)=f(1)=1g(1) = f(1) = 1.

    -

    If nn is prime, then g(n)=f(n)+f(1)g(n) = f(n) + f(1).

    -

    If n=n1n2n = n_1n_2 where gcd⁡(n1,n2)=1\gcd(n_1, n_2) = 1, then we prove by induction.

    -

    Suppose ff is multiplicative for all d<nd < n.

    -
    g(n)=g(n1n2)=∑d∣n1n2f(d)=∑d1∣n1∑d2∣n2f(d1d2)g(n) \\ -= g(n_1n_2) \\ -= \sum_{d | n_1n_2} f(d) \\ -= \sum_{d_1 | n_1} \sum_{d_2 | n_2} f(d_1d_2) \\
    -

    Only f(n1n2)f(n_1n_2) is out of the induction hypothesis, so we apply the multiplicative property of ff for all other terms.

    -
    =∑d1∣n1∑d2∣n2f(d1)f(d2)−f(n1)(n2)+f(n1n2)=(∑d1∣n1f(d1))(∑d2∣n2f(d2))−f(n1)f(n2)+f(n1n2)=g(n1)g(n2)−f(n1)f(n2)+f(n1n2)= \sum_{d_1 | n_1} \sum_{d_2 | n_2} f(d_1)f(d_2) - f(n_1)(n_2) + f(n_1n_2) \\ -= \left(\sum_{d_1 | n_1} f(d_1)\right) \left(\sum_{d_2 | n_2} f(d_2)\right) - f(n_1)f(n_2) + f(n_1n_2) \\ -= g(n_1)g(n_2) - f(n_1)f(n_2) + f(n_1n_2) \\
    -

    Move g(n1)g(n2)g(n_1)g(n_2) to the left hand side.

    -
    g(n)−g(n1)g(n2)=f(n1n2)−f(n1)f(n2)g(n) - g(n_1)g(n_2) = f(n_1n_2) - f(n_1)f(n_2) \\
    -

    Apply the multiplicative property of gg.

    -
    0=f(n1n2)−f(n1)f(n2)0 = f(n_1n_2) - f(n_1)f(n_2) \\
    -

    Hence, ff is multiplicative.

     - ⯀ -
    -

    Definition

    -
    Definition - - -

    The Mobius Function, denoted as μ(n)\mu(n), is defined by the following rules:

    -
    ∑d∣nμ(d)=[n=1]\sum_{d | n} \mu(d) = [n = 1]
     -
    -

    Properties

    -
    Theorem - - (Inversion Principle) -
    f(n)=∑d∣ng(d)  ⟺  g(n)=∑d∣nμ(d)f(nd)f(n) = \sum_{d | n} g(d) \iff g(n) = \sum_{d | n} \mu(d) f\left(\frac{n}{d}\right)
     -
    -
    Proof - -

    Left to right:

    -
    ∑d∣mμ(d)f(md)=∑d∣mμ(md)f(d)=∑d∣mμ(md)∑k∣dg(k)=∑d∣m∑k∣dμ(md)g(k)=∑k∣m∑d∣(m/k)μ(mkd)g(k)=∑k∣mg(k)∑d∣(m/k)μ(mkd)=∑k∣mg(k)∑d∣(m/k)μ(mk/mkd)=∑k∣mg(k)∑d∣(m/k)μ(d)=∑k∣mg(k)[m/k=1]=g(m)\sum_{d | m} \mu(d) f\left(\frac{m}{d}\right) \\ -= \sum_{d | m} \mu\left(\frac{m}{d}\right) f(d) \\ -= \sum_{d | m} \mu\left(\frac{m}{d}\right) \sum_{k | d} g(k) \\ -= \sum_{d | m} \sum_{k | d} \mu\left(\frac{m}{d}\right) g(k) \\ -= \sum_{k | m} \sum_{d | (m / k)} \mu\left(\frac{m}{kd}\right) g(k) \\ -= \sum_{k | m} g(k) \sum_{d | (m / k)} \mu\left(\frac{m}{kd}\right) \\ -= \sum_{k | m} g(k) \sum_{d | (m / k)} \mu\left(\frac{m}{k} / \frac{m}{kd}\right) \\ -= \sum_{k | m} g(k) \sum_{d | (m / k)} \mu(d) \\ -= \sum_{k | m} g(k) [m / k = 1] \\ -= g(m)
    -

    Right to left:

    -
    ∑d∣mg(d)=∑d∣m∑k∣dμ(k)f(dk)=∑k∣m∑d∣(m/k)μ(k)f(d)=∑k∣mf(k)∑d∣(m/k)μ(k)=∑k∣mf(k)[m/k=1]=f(m)\sum_{d | m} g(d) \\ -= \sum_{d | m} \sum_{k | d} \mu(k) f\left(\frac{d}{k}\right) \\ -= \sum_{k | m} \sum_{d | (m / k)} \mu(k) f(d)\\ -= \sum_{k | m} f(k) \sum_{d | (m / k)} \mu(k) \\ -= \sum_{k | m} f(k) [m / k = 1] \\ -= f(m)
     - ⯀ -
    -
    Theorem - - (Multiplicative Property) -

    The Mobius Function is a multiplicative function.

     -
    -
    Proof - -

    [n=1][n = 1] is multiplicative. Immediately apply the lemma above.

     - ⯀ -
    -
    Theorem - - (General Term) -
    μ(n)={0if n has a squared prime factor1if n=1(−1)kif n is a product of k distinct primes\mu(n) = \begin{cases} - 0 & \text{if $n$ has a squared prime factor} \\ - 1 & \text{if $n = 1$} \\ - (-1)^k & \text{if $n$ is a product of $k$ distinct primes} -\end{cases}
     -
    -
    Proof - -

    If n=1n = 1

    -
    ∑d∣1μ(d)=μ(1)=[1=1]=1\sum_{d|1}\mu(d) = \mu(1) = [1 = 1] = 1
    -

    If n=pn = p, where pp is a prime number.

    -
    μ(p)=∑d∣pμ(d)−μ(1)=[p=1]−1=−1\mu(p) = \sum_{d|p}\mu(d) - \mu(1) = [p = 1] - 1 = -1\\
    -

    If n=pkn = p^k, where pp is a prime number and k>1k > 1.

    -
    μ(pk)=∑d∣pkμ(d)−∑d∣pk−1μ(d)=[pk=1]−[pk−1=1]=0\mu(p^k) = \sum_{d|p^k}\mu(d) - \sum_{d|p^{k-1}}\mu(d) = [p^k = 1] - [p^{k-1} = 1] = 0
    -

    If nn has more than one prime factors, we can apply multiplicative property. Therefore,

    -
    μ(n)=μ(p1a1p2a2⋯pkak)=μ(p1a1)μ(p2a2)⋯μ(pkak)={0μ(piai)=0 for some i(−1)kμ(piai)=−1 for all i={0max⁡ai>1(−1)kmax⁡ai=1\mu(n)\\ -= \mu(p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}) \\ -= \mu(p_1^{a_1})\mu(p_2^{a_2})\cdots\mu(p_k^{a_k})\\ -= \begin{cases} - 0 & \mu(p_i^{a_i}) = 0 \text{ for some }i \\ - (-1)^k & \mu(p_i^{a_i}) = -1 \text{ for all }i -\end{cases}\\ -= \begin{cases} - 0 & \max{a_i} > 1 \\ - (-1)^k & \max{a_i} = 1 -\end{cases}
    -

    Combining all cases, we get the general term.

     - ⯀ -
    -
    Theorem - - (Relation to Eular's Totient Function) -
    φ(n)=∑d∣nμ(d)nd\varphi(n) = \sum_{d | n} \mu(d)\frac{n}{d}
     -
    -
    Proof - -

    Immediate from applying the inversion principle on Propterty 7 of Euler’s Totient Function.

     - ⯀ -
    -
    - Experiment
    - -
    - Graph Theory
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    - Number Theory
    -
    - Chinese Remainder Theorem (CRT)
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    - Euler's Totient Function
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    - - diff --git a/docs/note/network-flow.html b/docs/note/network-flow.html deleted file mode 100644 index bc501d0..0000000 --- a/docs/note/network-flow.html +++ /dev/null @@ -1,540 +0,0 @@ - - - - - - - - - - - - - - - - - - - - - - - - 夜莺洞穴 - 笔记 - Mobius Function - - -
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    返回 -
    笔记 - Mobius Function
    -
    - -

    Network Flow

    -

    Network flow is an interesting but difficult topic in graph theory. Most algorithm have high time complexity.

    -

    Definition

    -
    Definition - - -

    A network is a simple directed graph G=(V,E)G = (V, E), where EE is anti-symmetric, with a capacity function c:E→R+c: E \to \mathbb{R}^+ and two distinguished vertices, the source ss and the sink tt.

     -
    -
    Example - - -
    - 
    node(50 + 150i, "s");
-node(400 + 150i, "t");
-node(150 + 150i, "v_1");
-node(300 + 150i, "v_2");
-node(200 + 250i, "v_3");
-node(200 + 50i, "v_4");
-edge(s, v_1, 10);
-edge(s, v_3, 5);
-edge(v_1, v_2, 5);
-edge(v_1, v_3, 15);
-edge(v_1, v_4, 10);
-edge(v_3, v_2, 5);
-edge(v_2, t, 10);
-edge(v_3, t, 10);
-edge(v_4, t, 10);
    -
     -
    -
    Definition - - -

    A flow in a graph (V,E)(V, E) is defined as a real-valued function f:E→R+f: E \to \mathbb{R}^+ that satisfies the following conditions:

    -
    • Capacity Contraint: 0≤f(e)≤c(e)0 \leq f(e) \leq c(e) for all e∈Ee \in E.
      • -
    • Flow conservation: ∑(u,v)∈Ef(u,v)=∑(v,w)∈Ef(v,w)\sum_{(u, v) \in E} f(u, v) = \sum_{(v, w) \in E} f(v, w) for all v∈(V−{s,t})v \in (V - \{s, t\}).
     -
    -
    Definition - - -

    The value of a flow ff is defined as the total flow out of the source ss minus the total flow into the source ss, noted as ∣f∣=∑(s,v)∈Ef(s,v)−∑(v,s)∈Ef(v,s)|f| = \sum_{(s, v) \in E} f(s, v) - \sum_{(v, s) \in E} f(v, s).

     -
    -

    Model Reduction

    -

    For non-simple digraphs, we can merge parallel edges into a single edge with the sum of the capacities and remove self-loops.

    -

    For non-anti-symmetric digraphs, we can replace (u,v),(v,u)(u, v), (v, u) by (u,v′),(v′,v),(v,u)(u, v'), (v', v), (v, u) where v′v' is a new vertex, setting the capacity of (u,v′),(v,v′)(u, v'), (v, v') to be the same as (u,v)(u, v).

    -

    For multiple sources and sinks, we can add a new source s′s', called supersource, and sink t′t', called supersink, connecting s′s' to all sources and all sinks to t′t'.

    -

    Agumenting Path and Max-Flow Min-Cut Theorem

    -
    Definition - - (Properly Oriented Edge) -

    Consider there is path P=(v0,v1,…,vn)P = (v_0, v_1, \dots, v_n) in the undirected copy of the netwrok GG, an edge e=(u,v)∈Ge = (u, v) \in G is called properly oriented with respect to PP if u=viu = v_i and v=vi+1v = v_{i + 1} for some ii and improperly oriented if u=vi+1u = v_{i + 1} and v=viv = v_i for some ii.

     -
    -
    Example - - -
    - 
    node(50 + 150i, "s");
-node(225 + 150i, "v_1");
-node(137 + 225i, "'text{properly oriented}", "text");
-node(400 + 150i, "t");
-node(317 + 225i, "'text{improperly oriented}", "text");
-node(225 + 75i, "P = (s, v_1, t)", "text");
-edge(s, v_1, 10);
-edge(t, v_1, 10);
    -
     -
    -
    Definition - - (Argumenting Path) -

    Consider a network GG with a flow ff. An argumenting path is a simple path P=(s,v0,v1,…,vn,t)P = (s, v_0, v_1, \dots, v_n, t) in the undirected copy of the network GG satisfying the following conditions:

    -
    • For all edges ee properly oriented with respect to PP, f(e)<c(e)f(e) < c(e).
      • -
    • For all edges ee improperly oriented with respect to PP, f(e)>0f(e) > 0.
     -
    -
    Theorem - - -

    Suppose there is an argumenting path PP in the network GG with respect to the flow ff.

    -
    d=min⁡e∈E{c(e)−f(e)if e is properly oriented with respect to Pf(e)if e is improperly oriented with respect to P+∞otherwised = \min_{e \in E} \begin{cases} -c(e) - f(e) & \text{if } e \text{ is properly oriented with respect to }P\\ -f(e) & \text{if } e \text{ is improperly oriented with respect to }P\\ -+\infty & \text{otherwise} -\end{cases}
    -

    Then we can construct a new flow f∗f^* with value ∣f∗∣=∣f∣+d|f^*| = |f| + d.

    -
    f∗(e)={f(e)+dif e is properly oriented with respect to Pf(e)−dif e is improperly oriented with respect to Pf(e)otherwisef^*(e) = \begin{cases} -f(e) + d & \text{if } e \text{ is properly oriented with respect to }P\\ -f(e) - d & \text{if } e \text{ is improperly oriented with respect to }P\\ -f(e) & \text{otherwise} -\end{cases}
     -
    -
    Proof - -

    First by the definition of argumenting path, we have d>0d > 0.

    -

    And by the formula of dd, for improperly oriented edges, f∗(e)=f(e)−d>0f^*(e) = f(e) - d > 0. For properly oriented edges, f∗(e)=f(e)+d<c(e)f^*(e) = f(e) + d < c(e). Other edges are unchanged. Therefore the adjusted flow f∗f^* is a valid flow.

    -

    The since the argument path must pass through the source ss and sink tt, the value of the new flow ∣f∗∣=∣f∣+d|f^*| = |f| + d.

     - ⯀ -
    -
    Example - - -
    - 
    node(225 + 20i, "a/b'text{ stands for flow } a 'text{ out of capacity } b", "text");
-node(10 + 100i, "f", "text");
-node(50 + 150i, "s");
-node(175 + 150i, "v_1");
-node(300 + 150i, "v_2");
-node(425 + 150i, "t");
-node(225 + 125i, "'dots", "text");
-node(225 + 175i, "'dots$", "text");
-edge(s, v_1, "5/10");
-edge(v_2, v_1, "3/5");
-edge(v_2, t, "0/5");
-node(225 + 250i, "d = 3", "text");
-node(10 + 300i, "f^*", "text");
-node(50 + 350i, "s$");
-node(175 + 350i, "v_1$");
-node(300 + 350i, "v_2$");
-node(425 + 350i, "t$");
-node(225 + 325i, "'dots$$", "text");
-node(225 + 375i, "'dots$$$", "text");
-edge(s$, v_1$, "8/10");
-edge(v_2$, v_1$, "0/5");
-edge(v_2$, t$, "3/5");
    -
     -
    -
    Definition - - (Cut) -

    A cut is a partition of the vertices VV into two sets PP and P‾=V−P\overline{P} = V - P such that s∈Ss \in S and t∈Tt \in T.

     -
    -
    Definition - - (Capacity of a Cut) -

    The capacity of a cut (P,P‾)(P, \overline{P}) is defined as the sum of the capacities of the edges from PP to P‾\overline{P}, denoted as C(P,P‾)=∑(u,v)∈E,u∈P,v∈P‾c(u,v)C(P, \overline{P}) = \sum_{(u, v) \in E, u \in P, v \in \overline{P}} c(u, v).

     -
    -
    Example - - -
    - 
    node(50 + 150i, "s");
-node(400 + 150i, "t", "light");
-node(150 + 150i, "v_1");
-node(300 + 150i, "v_2", "light");
-node(200 + 250i, "v_3", "light");
-node(200 + 50i, "v_4");
-edge(s, v_1, 10);
-edge(s, v_3, 5);
-edge(v_1, v_2, 5);
-edge(v_1, v_3, 15);
-edge(v_1, v_4, 10);
-edge(v_3, v_2, 5);
-edge(v_2, t, 10);
-edge(v_3, t, 10);
-edge(v_4, t, 10);
-node(200 + 290i, "C(P, 'overline{P}) = 35", "text");
    -
     -
    -
    Theorem - - -

    For any flow ff in the network GG, and any cut (P,P‾)(P, \overline{P}), we have

    -
    ∣f∣≤C(P,P‾)|f| \leq C(P, \overline{P})
     -
    -
    Proof - -

    Consider flows enter P−{s}P - \{s\} and leave P‾−{s}\overline{P} - \{s\}, we have

    -
    0=∑u∈(P−s)0=∑u∈(P−s)(∑v∈V,(u,v)∈Ef(u,v)−∑v∈V,(v,u)∈Ef(v,u))=∑u∈(P−s)(∑v∈P,(u,v)∈Ef(u,v)−∑v∈P,(v,u)∈Ef(v,u)+∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))=∑u∈(P−s)(∑v∈P,(u,v)∈Ef(u,v)−∑v∈P,(v,u)∈Ef(v,u))+∑u∈(P−s)(∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))=∑u∈P,v∈P,(u,v)∈Ef(u,v)−∑u∈P,v∈P,(v,u)∈Ef(v,u)+∑u∈(P−s)(∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))(Note: Simple graph.)=∑u∈(P−s)(∑v∈P‾,(u,v)∈Ef(u,v)−∑v∈P‾,(v,u)∈Ef(v,u))(Note: First 2 summations have same region.)=(∑u∈P,v∈P‾,(u,v)∈Ef(u,v)−∑u∈P,v∈P‾,(v,u)∈Ef(v,u))−(∑v∈P‾,(s,v)∈Ef(s,v)−∑v∈P‾,(v,s)∈Ef(v,s))≤(∑u∈P,v∈P‾,(u,v)∈Ef(u,v)−∑u∈P,v∈P‾,(v,u)∈Ef(v,u))−(∑v∈V,(s,v)∈Ef(s,v)−∑v∈V,(v,s)∈Ef(v,s))=∑u∈P,v∈P‾,(u,v)∈Ef(u,v)−∑u∈P,v∈P‾,(v,u)∈Ef(v,u)−∣f∣≤∑u∈P,v∈P‾,(u,v)∈Ec(u,v)−∑u∈P,v∈P‾,(v,u)∈Ec(v,u)−∣f∣=C(P,P‾)−∣f∣0\\ -= \sum_{u \in (P - {s})} 0\\ -= \sum_{u \in (P - {s})} \left(\sum_{v \in V, (u, v) \in E} f(u, v) - \sum_{v \in V, (v, u) \in E} f(v, u)\right)\\ -= \sum_{u \in (P - {s})} \left(\sum_{v \in P, (u, v) \in E} f(u, v) - \sum_{v \in P, (v, u) \in E} f(v, u) + \sum_{v \in \overline{P}, (u, v) \in E} f(u, v) - \sum_{v \in \overline{P}, (v, u) \in E} f(v, u)\right)\\ -= \sum_{u \in (P - {s})} \left(\sum_{v \in P, (u, v) \in E} f(u, v) - \sum_{v \in P, (v, u) \in E} f(v, u)\right) + \sum_{u \in (P - {s})}\left(\sum_{v \in \overline{P}, (u, v) \in E} f(u, v) - \sum_{v \in \overline{P}, (v, u) \in E} f(v, u)\right)\\ -= \sum_{u \in P, v \in P, (u, v) \in E} f(u, v) - \sum_{u \in P, v \in P, (v, u) \in E} f(v, u) + \sum_{u \in (P - {s})}\left(\sum_{v \in \overline{P}, (u, v) \in E} f(u, v) - \sum_{v \in \overline{P}, (v, u) \in E} f(v, u)\right) \text{(Note: Simple graph.)}\\ -= \sum_{u \in (P - {s})}\left(\sum_{v \in \overline{P}, (u, v) \in E} f(u, v) - \sum_{v \in \overline{P}, (v, u) \in E} f(v, u)\right) \text{(Note: First 2 summations have same region.)}\\ -= \left(\sum_{u \in P, v \in \overline{P}, (u, v) \in E} f(u, v) - \sum_{u \in P, v \in \overline{P}, (v, u) \in E} f(v, u)\right) - \left(\sum_{v \in \overline{P}, (s, v) \in E} f(s, v) - \sum_{v \in \overline{P}, (v, s) \in E} f(v, s)\right)\\ -\le \left(\sum_{u \in P, v \in \overline{P}, (u, v) \in E} f(u, v) - \sum_{u \in P, v \in \overline{P}, (v, u) \in E} f(v, u)\right) - \left(\sum_{v \in V, (s, v) \in E} f(s, v) - \sum_{v \in V, (v, s) \in E} f(v, s)\right) \\ -= \sum_{u \in P, v \in \overline{P}, (u, v) \in E} f(u, v) - \sum_{u \in P, v \in \overline{P}, (v, u) \in E} f(v, u) - |f|\\ -\le \sum_{u \in P, v \in \overline{P}, (u, v) \in E} c(u, v) - \sum_{u \in P, v \in \overline{P}, (v, u) \in E} c(v, u) - |f|\\ -= C(P, \overline{P}) - |f|
    -

    And move the ∣f∣|f| to the left side, we have ∣f∣≤C(P,P‾)|f| \leq C(P, \overline{P}).

     - ⯀ -
    -
    Theorem - - (Max-Flow Min-Cut Theorem) -

    If the equality ∣f∣=C(P,P‾)|f| = C(P, \overline{P}) holds for some flow ff and cut (P,P‾)(P, \overline{P}), then the flow ff is a maximum flow and the cut (P,P‾)(P, \overline{P}) is a minimum cut. And the equality hold if and only if f(i,j)=c(i,j)f(i, j) = c(i, j) for all (i,j)∈E(i, j) \in E with i∈Pi \in P and j∈P‾j \in \overline{P} and f(i,j)=0f(i, j) = 0 otherwise.

     -
    -
    Proof - -

    Obtained from the above reduction when two ≤\le is used in the proof.

     - ⯀ -
    -

    Ford-Fulkerson Method

    -
    Definition - - (Residual Network) -

    Resiual Network GfG_f of a network GG with respect to a flow ff is defined as the graph with the same vertices and edges as GG and the capacity function cf:E→R+c_f: E \to \mathbb{R}^+ defined as

    -
    cf(u,v)={c(u,v)−f(u,v)if (u,v)∈Ef(v,u)if (v,u)∈E0otherwisec_f(u, v) = \begin{cases} -c(u, v) - f(u, v) & \text{if } (u, v) \in E\\ -f(v, u) & \text{if } (v, u) \in E\\ -0 & \text{otherwise} -\end{cases}
     -
    -
    Algorithm - - (Ford-Fulkerson Method) -

    Coninuously find an argumenting path in the residual network GfG_f and adjust the flow ff until no argumenting path can be found.

     -
    -
    Theorem - - -

    After the Ford-Fulkerson Method, the flow ff is a maximum flow.

     -
    -
    Proof - -

    First, after the Ford-Fulkerson Method, the sink tt should not be reachable from the source ss in the residual network GfG_f. Otherwise, there is an argumenting path in the residual network GfG_f.

    -

    Therefore, the vertices reachable from the source ss in the residual network GfG_f form a cut (P,P‾)(P, \overline{P}) with s∈Ps \in P and t∈P‾t \in \overline{P}. There shouldn’t be a edge from PP to P‾\overline{P} in the residual network GfG_f, otherwise there exists a vertex in P‾\overline{P} reachable from the source ss, which leads to a contradiction.

    -

    Therefore, the flow ff is a maximum flow.

    -

    But this does not guarantee the termination of the algorithm. If the capacities are integers, the algorithm obviously terminates since the flow value increases and bounded. But if it is not, the algorithm may not terminate since we know there are some infinate increasing sequences in real numbers.

     - ⯀ -
    -
    Algorithm - - (Dinic's Implementation of Ford-Fulkerson Method) -

    This algorithm terminates. We will provide the proof later. The time complexity is O(V2E)O(V^2E).

    - - - -
    - 
    $'fn{bfs}(s, t, vertices)$ [
-	for $u$ in $vertices$ [
-		$u.level = -1$
-	]
-	$q = {s}$
-	$s.level = -1$
-	while $'neg 'fn{QUQUE-EMPTY}(q)$ [
-		$u = 'fn{QUQUE-POP}(q)$
-		for $e$ in $u.edges$ [
-			if $e.v.level = -1$ and $e.cap > 0$ [
-				$e.v.level = u.level + 1$
-				$'fn{QUQUE-PUSH}(q, e.v)$
-			]
-		]
-	]
-	return $t.level \neq -1$
-]
-
-$'fn{dfs}(u, f, t)$ [
-	if $u = t 'lor f 'le 0$ [
-		return $f$
-	]
-	$r = 0$
-	while $u.current < |u.edges|$ [
-		$e = u.edges(u.current)$
-		if $e.cap > 0 'land e.v.level = u.level + 1$ [
-			$delta = 'fn{dfs}(e.v, 'min(f - r, e.cap), t)$
-			$e.cap -= delta$
-			$e.rev.cap += delta$
-			$r += delta$
-			if $r = f$ [
-				return $r$
-			]
-		]
-		$u.current += 1$
-	]
-	return $r$
-]
-
-# G is the network, c is the capacity function, s is the source, t is the sink
-# Return the maximum flow and the flow function
-$'fn{dinic}(digraph, c, s, t)$ [
-	$vertices = 'varnothing$
-	$edges = 'varnothing$
-	for $u$ in $digraph.vertices$ [
-		$vertices(u) = $ new $'text{Vertex}'{level: -1, edges: 'varnothing, current: 0'}$
-	]
-	for $e$ in $digraph.edges$ [
-		$u = vertices(e.u)$
-		$v = vertices(e.v)$
-		$proper = $ new $'text{Edge}'{v: v, cap: c(e), rev: $null$'}$
-		$improper = $ new $'text{Edge}'{v: u, cap: 0, rev: proper'}$
-		$proper.rev = improper$
-		$u.edges(|u.edges|) = proper$
-		$v.edges(|v.edges|) = improper$
-		$edges(e) = proper$
-	]
-	$flow = 0$
-	$s = vertices(s)$
-	$t = vertices(t)$
-	while $'fn{bfs}(s, t, vertices)$ [
-		for $u$ in $vertices$ [
-			$u.current = 0$
-		]
-		$flow += 'fn{dfs}(s, +'infty, t)$
-	]
-	$f = 'varnothing$
-	for $e$ in $digraph.edges$ [
-		$f(e) = edges(e).rev.cap$
-	]
-	return $flow, f$
-]
     -
    -
    Proof - of Correctness -

    To be continued.

     - ⯀ -
    -

    Min-Cost Flow

    -

    To be continued.

    -

    Bipartite Matching

    -

    To be continued.

    -
    - Experiment
    - -
    - Graph Theory
    -
    - Network Flow
    - -
    -
    - Number Theory
    - -
    - Numerical Analysis
    - -
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    - - - - -
    - - diff --git a/docs/note/numerical-differentiation.html b/docs/note/numerical-differentiation.html deleted file mode 100644 index e288048..0000000 --- a/docs/note/numerical-differentiation.html +++ /dev/null @@ -1,265 +0,0 @@ - - - - - - - - - - - - - - - - - - - 夜莺洞穴 - 笔记 - Network Flow - - -
    - - - - -
    返回 -
    笔记 - Network Flow
    -
    - -

    Numeric Differentiation

    -

    Finite Difference

    -

    By defination of differentiation, we have

    -
    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
    -

    By Taylor’s Theorem, if ff is twice differentiable, we have

    -
    f(x+h)=f(x)+hf′(x)+h22f′′(ξ)f(x+h) = f(x) + hf'(x) + {h^2 \over 2} f''(\xi)
    -

    where ξ\xi is a real number between xx and x+hx + h.

    -

    Then we have

    -
    f′(x)=f(x+h)−f(x)h−h2f′′(ξ)≈f(x+h)−f(x)hf'(x) = \frac{f(x+h) - f(x)}{h} - {h \over 2} f''(\xi) \approx \frac{f(x+h) - f(x)}{h}
    -

    In this case −h2f′′(ξ)-{h \over 2} f''(\xi) become error.

    -

    Generally, if error is O(hk)O(h^k), we say the method is of order kk.

    -
    f(x+h)=f(x)+hf′(x)+h22f′′(x)+h36f′′′(ξ1)f(x−h)=f(x)−hf′(x)+h22f′′(x)−h36f′′′(ξ2)f(x + h) = f(x) + hf'(x) + {h^2 \over 2}f''(x) + {h^3 \over 6}f'''(\xi_1)\\ -f(x - h) = f(x) - hf'(x) + {h^2 \over 2}f''(x) - {h^3 \over 6}f'''(\xi_2)
    -

    where ξ1\xi_1 is a real number between xx and x+hx + h, and ξ2\xi_2 is a real number between x−hx - h and xx.

    -

    Then we have

    -
    f′(x)=f(x+h)−f(x−h)2h−h212f′′′(ξ1)−h212f′′′(ξ2)≈f(x+h)−f(x−h)2hf'(x) = \frac{f(x+h) - f(x - h)}{2h} - {h^2 \over 12} f'''(\xi_1) - {h^2 \over 12} f'''(\xi_2) \approx \frac{f(x+h) - f(x - h)}{2h}
    -

    So we obtain a method of order 2. And obviously this can be written as

    -
    f′(x)=f(x+h)−f(x−h)2h−h26f′′′(ξ)≈f(x+h)−f(x−h)2hf'(x) = \frac{f(x+h) - f(x - h)}{2h} - {h^2 \over 6} f'''(\xi) \approx \frac{f(x+h) - f(x - h)}{2h}
    -

    However, we wonder if we can get the derivative from f(x)f(x), f(x+h)f(x+h) and f(x+2h)f(x+2h).

    -

    Consider

    -
    f(x+2h)=f(x)+2hf′(x)+2h2f′′(x)+4h33f′′′(ξ1)f(x+h)=f(x)+hf′(x)+h22f′′(x)+h36f′′′(ξ2)\begin{gather} -f(x + 2h) = f(x) + 2hf'(x) + 2h^2 f''(x) + {4h^3 \over 3} f'''(\xi_1)\\ -f(x + h) = f(x) + hf'(x) + {h^2 \over 2} f''(x) + {h^3 \over 6} f'''(\xi_2) -\end{gather}
    -

    We want to remove f′′(x)f''(x) so we subtract 44 times the second equation from the first one and get

    -
    f(x+2h)−4f(x+h)=−3f(x)−2hf′(x)+2h33f′′′(ξ1)−2h33f′′′(ξ2)f′(x)=−3f(x)+4f(x+h)−f(x+2h)2h+h23f′′′(ξ1)−h23f′′′(ξ2)≈−3f(x)+4f(x+h)−f(x+2h)2hf(x + 2h) - 4 f(x + h) = -3 f(x) - 2h f'(x) + {2h^3 \over 3} f'''(\xi_1) - {2h^3 \over 3} f'''(\xi_2)\\ -f'(x) = \frac{- 3 f(x) + 4 f(x + h) - f(x + 2h) }{2h} + {h^2 \over 3} f'''(\xi_1) - {h^2 \over 3} f'''(\xi_2) \approx \frac{- 3 f(x) + 4 f(x + h) - f(x + 2h)}{2h}
    -

    The error is O(h2)O(h^2). The f(x−2h)f(x - 2h) case is symmetric, so we just skip it.

    -

    Notice that since hh is used as denominator, we can’t make it too small.

    -

    Higher Order Derivatives

    -

    Differentiation Matrix

    -

    While it’s possible to get higher order derivatives by Taylor’s Theorem, it’s often hard.

    -

    So we apply first order derivative many times.

    -

    Notice that in the previous section, the first order derivative is a linear combination of f(x)f(x) near xx.

    -

    The operation can be written as a matrix multiplication. Let xn=x+nhx_n = x + nh, then we have

    -
    (f′(x1)f′(x2)f′(x3)f′(x4)⋮f′(xn))≈12h(−34−10⋯01−210⋯001−21⋯0001−2⋯0⋮⋮⋮⋮⋱⋮0000⋯−3)(f(x1)f(x2)f(x3)f(x4)⋮f(xn))\begin{pmatrix} -f'(x_1)\\ -f'(x_2)\\ -f'(x_3)\\ -f'(x_4)\\ -\vdots\\ -f'(x_n) -\end{pmatrix} -\approx -\dfrac{1}{2h} -\begin{pmatrix} --3 & 4 & -1 & 0 & \cdots & 0\\ -1 & -2 & 1 & 0 & \cdots & 0\\ -0 & 1 & -2 & 1 & \cdots & 0\\ -0 & 0 & 1 & -2 & \cdots & 0\\ -\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ -0 & 0 & 0 & 0 & \cdots & -3 -\end{pmatrix} -\begin{pmatrix} -f(x_1)\\ -f(x_2)\\ -f(x_3)\\ -f(x_4)\\ -\vdots\\ -f(x_n) -\end{pmatrix}
    -

    So we can have a general formula

    -
    (f(k)(x1)f(k)(x2)f(k)(x3)f(k)(x4)⋮f(k)(xn))≈(12h)k(−34−10⋯01−210⋯001−21⋯0001−2⋯0⋮⋮⋮⋮⋱⋮0000⋯−3)k(f(x1)f(x2)f(x3)f(x4)⋮f(xn))\begin{pmatrix} -f^{(k)}(x_1)\\ -f^{(k)}(x_2)\\ -f^{(k)}(x_3)\\ -f^{(k)}(x_4)\\ -\vdots\\ -f^{(k)}(x_n) -\end{pmatrix} -\approx -(\dfrac{1}{2h})^k -\begin{pmatrix} --3 & 4 & -1 & 0 & \cdots & 0\\ -1 & -2 & 1 & 0 & \cdots & 0\\ -0 & 1 & -2 & 1 & \cdots & 0\\ -0 & 0 & 1 & -2 & \cdots & 0\\ -\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ -0 & 0 & 0 & 0 & \cdots & -3 -\end{pmatrix}^k -\begin{pmatrix} -f(x_1)\\ -f(x_2)\\ -f(x_3)\\ -f(x_4)\\ -\vdots\\ -f(x_n) -\end{pmatrix}
    -

    Interpolation and Differentiation

    -

    It’s a good idea to interpolate the function first, then differentiate the interpolation.

    -

    Typically, we can use Cubic Spline Interpolation for derivative less than 3, since it’s first and second order derivative converge to the original function.

    -

    For higher order derivative, we can use Chebyshev Polynomial Interpolation or Lagrange Polynomial Interpolation. They are polynomials and easy to differentiate.

    -
    - Experiment
    - -
    - Graph Theory
    - -
    - Number Theory
    - -
    - Numerical Analysis
    -
    - Chebyshev Polynomial
    - -
    - Numeric Differentiation
    - -
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    - -
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    -
    - - - - -
    - - diff --git a/docs/note/numerical-integration.html b/docs/note/numerical-integration.html deleted file mode 100644 index 7acf763..0000000 --- a/docs/note/numerical-integration.html +++ /dev/null @@ -1,268 +0,0 @@ - - - - - - - - - - - - - - - - - - - - - - 夜莺洞穴 - 笔记 - Numeric Differentiation - - -
    - - - - -
    返回 -
    笔记 - Numeric Differentiation
    -
    - -

    Numeric Integration

    -

    Newton-Cotes Formula

    -

    The general idea of Newton-Cotes formula is to first interpolate the function, then integrate the interpolation.

    -
    f(x)≈If(x)∫abf(x)dx≈∫abIf(x)dxf(x) \approx I f(x)\\ -\int_a^b f(x) dx \approx \int_a^b I f(x) dx
    -

    We substitute Lagrange interpolation polynomial into the formula. Let nn be the number of points we use. (Not the degree of the polynomial)

    -
    f(x)=∑i=1nf(xi)∏j=1,j≠inx−xjxi−xj+f(n)(ξ)n!∏i=1n(x−xi)∫abf(x)dx=∑i=1nf(xi)∫ab∏j=1,j≠inx−xjxi−xjdx+∫abf(n)(ξ)n!∏i=1n(x−xi)dxf(x) = \sum_{i=1}^n f(x_i) \prod_{j=1, j \neq i}^n\dfrac{x-x_j}{x_i-x_j} + \dfrac{f^{(n)}(\xi)}{n!} \prod_{i=1}^n (x-x_i)\\ -\int_a^b f(x) dx = \sum_{i=1}^n f(x_i) \int_a^b \prod_{j=1, j \neq i}^n\dfrac{x-x_j}{x_i-x_j} dx + \int_a^b \dfrac{f^{(n)}(\xi)}{n!} \prod_{i=1}^n (x-x_i) dx
    -

    The part ∫abf(n)(ξ)n!∏i=1n(x−xi)dx\int_a^b \dfrac{f^{(n)}(\xi)}{n!} \prod_{i=1}^n (x-x_i) dx is called the remainder term, denoted as E(f)E(f).

    -

    The ∫ab∏j=1,j≠inx−xjxi−xjdx\int_a^b \prod_{j=1, j \neq i}^n\dfrac{x-x_j}{x_i-x_j} dx part is unrelated to f(x)f(x), so we can precompute it and store it in a table.

    -

    As for the remainder term, sadly we have no way to find a closed form. So we just keep the ugly form.

    -

    Degree of Precision

    -
    Definition - - -

    The degree of precision of a numerical integration formula is the largest integer nn such that the formula is exact for all polynomials of degree nn or less.

     -
    -

    To find out a formula’s degree of precision, we can verify it on 1,x,x2,⋯1, x, x^2, \cdots. Since integration is a linear operator, if the formula is exact for 1,x,x2,⋯ ,xn1, x, x^2, \cdots, x^n, then it’s exact for their linear combination. If they are not exact, of course its degree of precision is less than nn since the counterexample exists.

    -
    Theorem - - -

    If we use nn points in Lagrange Interpolation, then the degree of precision is at least n−1n-1.

     -
    -
    Proof - -

    For any polynomial f(x)f(x) of degree n−1n-1, Lagrange Interpolation restores f(x)f(x) exactly. So the degree of precision is at least n−1n-1.

     - ⯀ -
    -
    Theorem - - -

    If we use nn equally spaced points, then the degree of precision is at least nn when nn is odd, and n−1n-1 when nn is even.

     -
    -

    This is obvious since the points are symmetric, so the odd degree terms are cancelled out.

    -

    Trapezoidal Rule

    -

    If we just take the two end points, we get the Trapezoidal Rule.

    -
    f(x)=f(a)x−ba−b+f(b)x−ab−a+f′′(ξ)2(x−a)(x−b)∫abf(x)dx=b−a2(f(a)+f(b))−f′′(ξ)12(b−a)3f(x) = f(a)\dfrac{x-b}{a-b} + f(b)\dfrac{x-a}{b-a} + \dfrac{f''(\xi)}{2}(x-a)(x-b)\\ -\int_a^b f(x) dx = \dfrac{b-a}{2}(f(a) + f(b)) - \dfrac{f''(\xi)}{12}(b-a)^3
    -
    Collary - - -

    The degree of precision is 11.

     -
    -
    Proof - -

    When f(x)=1f(x) = 1, ∫abf(x)dx=b−a\int_a^b f(x) dx = b-a. The formula is exact.

    -

    When f(x)=xf(x) = x, ∫abf(x)dx=b2−a22\int_a^b f(x) dx = \dfrac{b^2-a^2}{2}. The formula is exact.

    -

    When f(x)=x2f(x) = x^2, ∫abf(x)dx=b3−a33\int_a^b f(x) dx = \dfrac{b^3-a^3}{3}. The formula is not exact.

     - ⯀ -
    -

    Simpson’s Rule

    -

    If we take the 33 points, we get the Simpson’s Rule.

    -
    ∫abf(x)dx=b−a6(f(a)+4f(a+b2)+f(b))−f(3)(ξ)90(b−a)5\int_a^b f(x) dx = \dfrac{b-a}{6}(f(a) + 4f(\dfrac{a + b}{2}) + f(b)) - \dfrac{f^{(3)}(\xi)}{90}(b-a)^5
    -

    The degree of precision is 33. Proof is trivial.

    -

    Boole’s Rule

    -

    If we take the 55 points, we get the Boole’s Rule.

    -
    ∫abf(x)dx=b−a90(7f(a)+32f(3a+b4)+12f(a+b2)+32f(a+3b4)+7f(b))−f(5)(ξ)2880(b−a)7\int_a^b f(x) dx = \dfrac{b-a}{90}(7f(a) + 32f(\dfrac{3a + b}{4}) + 12f(\dfrac{a + b}{2}) + 32f(\dfrac{a + 3b}{4}) + 7f(b)) - \dfrac{f^{(5)}(\xi)}{2880}(b-a)^7
    -

    The degree of precision is 55. Proof is trivial.

    -

    Composite Newton-Cotes Formula

    -

    When the number of points exceeds 99, some of the coefficients become negative. This is not good for numerical computation. So we can divide the interval into subintervals, and apply the Newton-Cotes formula on each subinterval.

    -

    Take composite Simpson’s Rule as an example.

    -
    ∫abf(x)dx=∑i=0(n−1)/2b−a3n(f(x2i)+4f(x2i+1)+f(x2i+2))−f(3)(ξ)90(b−a)5\int_a^b f(x) dx = \sum_{i=0}^{(n - 1)/2} \dfrac{b - a}{3n}(f(x_{2i}) + 4f(x_{2i+1}) + f(x_{2i+2})) - \dfrac{f^{(3)}(\xi)}{90}(b-a)^5
    -

    The error term is the sum of the error terms of each subinterval, whose form is exactly the same as the error term of the original Simpson’s Rule.

    -

    Open Newton-Cotes Formula

    -

    Some functions are not defined at the end points, such as sin⁡xx\dfrac{\sin x}{x}. If we want to integrate it from 00 to 11, we can use the Open Newton-Cotes Formula.

    -

    The idea of Open Newton-Cotes Formula is simply avoiding the end points.

    -

    Midpoint Rule is the simplest Open Newton-Cotes Formula.

    -
    ∫abf(x)dx=(b−a)f(a+b2)−f′′(ξ)24(b−a)3\int_a^b f(x) dx = (b - a)f(\dfrac{a + b}{2}) - \dfrac{f''(\xi)}{24}(b-a)^3
    -

    Adaptive Quadrature

    -

    Some part of the function changes rapidly, while some part changes slowly. If we use the same number of points in the whole interval, we may get Time Limit Exceeded. So we can use Adaptive Quadrature to use more points in the rapidly changing part, and less points in the slowly changing part.

    -

    We first start recursion with the whole interval, then divide the interval into two subintervals, and recurse on each subinterval. The key is comparing the difference between the integral of the whole interval and the sum of the integrals of the two subintervals. If the difference is small enough, we can stop the recursion.

    -

    Gaussian Quadrature

    -

    The idea of Gaussian Quadrature is to choose the points xix_i and weights wiw_i such that the degree of precision is as high as possible.

    -
    Theorem - - -

    If we use nn points in Newton-Cotes formula, then the degree of precision is at most 2n−12n-1.

     -
    -
    Proof - -

    Proof by contradiction. Suppose that some one choose points xix_i and weights wiw_i such that the degree of precision is 2n2n.

    -

    Then let f(x)=∏i=1n(x−xi)2f(x) = \prod_{i = 1}^n (x - x_i)^2, which is a polynomial of degree 2n2n. Then the formula is exact for f(x)f(x).

    -

    Hence

    -
    ∫abf(x)dx=∑i=1nwif(xi)=∑i=1nwi⋅0=0\int_a^b f(x) dx = \sum_{i=1}^n w_i f(x_i) = \sum_{i = 1}^n w_i \cdot 0 = 0
    -

    However, f(x)=0f(x) = 0 holds only when x=xix = x_i for all ii, and f(x)>0f(x) > 0 for all other places. So the integral is positive. Contradiction.

     - ⯀ -
    -
    Definition - - -

    If nn points xix_i and weights wiw_i satisfy the degree of precision is 2n−12n-1, then the intergation formula is called Gaussian Quadrature.

     -
    -

    Legendre-Gauss Quadrature

    -

    The idea is picking xix_i as the roots of Legendre polynomial Pn(x)P_n(x). In this way, the degree of precision is exactly 2n−12n-1.

    -

    To make life easier, we can normalize the interval to [−1,1][-1, 1]. This do no harm to the generality of the formula.

    -
    Proof - -

    Let f(x)f(x) be a polynomial of degree 2n−12n-1. Then f(x)=Pn(x)g(x)+r(x)f(x) = P_n(x)g(x) + r(x), where g(x)g(x), r(x)r(x) are two polynomial of degree less than nn.

    -

    It’s trival that g(x)g(x) can be express as a linear combination of P0(x),P1(x),⋯ ,Pn−1(x)P_0(x), P_1(x), \cdots, P_{n-1}(x).

    -

    By the othogonality of Legendre polynomial and linearity of integration, we have

    -
    ∫−11f(x)dx=∫−11Pn(x)g(x)dx+∫−11r(x)dx=∫−11r(x)dx\int_{-1}^1 f(x) dx = \int_{-1}^1 P_n(x)g(x) dx + \int_{-1}^1 r(x) dx = \int_{-1}^1 r(x) dx
    -

    By the previous theorem, we know that the the integration formula is at least for r(x)r(x), so the degree of precision is at least 2n−12n-1.

    -

    Since the degree of precision is at most 2n−12n-1, the degree of precision is exactly 2n−12n-1.

     - ⯀ -
    -

    The error term is

    -
    E(f)=f(2n)(ξ)(2n)!∫−11∏i=1n(x−xi)2dx=f(2n)(ξ)(2n)!∫ab(π(x))2dxE(f) = \dfrac{f^{(2n)}(\xi)}{(2n)!}\int_{-1}^1 \prod_{i=1}^n (x - x_i)^2 dx = \dfrac{f^{(2n)}(\xi)}{(2n)!}\int_a^b(\pi(x))^2dx
    -

    where π(x)=∏i=1n(x−xi)\pi(x) = \prod_{i=1}^n (x - x_i).

    -
    Proof - -

    Consider Hermite Interpolation Polynomial H(x)H(x) of f(x)f(x) at xix_i.

    -
    H(xi)=f(xi)H′(xi)=f′(xi)H(x_i) = f(x_i)\\ -H'(x_i) = f'(x_i)
    -

    Then H(x)H(x) is a polynomial of degree at most 2n−12n-1.

    -
    f(x)=H(x)+f(2n)(ξ)(2n)!(π(x))2∫−11f(x)dx=∫−11H(x)dx+f(2n)(ξ)(2n)!∫−11(π(x))2dxf(x) = H(x) + \dfrac{f^{(2n)}(\xi)}{(2n)!}(\pi(x))^2 \\ -\int_{-1}^1 f(x) dx = \int_{-1}^1 H(x) dx + \dfrac{f^{(2n)}(\xi)}{(2n)!}\int_{-1}^1 (\pi(x))^2 dx
    -

    Since H(x)H(x) is a polynomial of degree at most 2n−12n-1, the integration formula is exact for H(x)H(x).

    -

    By the condition of Hermite Interpolation Polynomial, we know that H(xi)=f(xi)H(x_i) = f(x_i).

    -
    ∫−11H(x)dx=∑i=1nwiH(xi)=∑i=1nwif(xi)\int_{-1}^1 H(x) dx = \sum_{i=1}^n w_i H(x_i) = \sum_{i=1}^n w_i f(x_i)
    -

    Substitute it into the previous equation, we get

    -
    f(x)=H(x)+f(2n)(ξ)(2n)!(π(x))2∫−11f(x)dx=∑i=1nwif(xi)+f(2n)(ξ)(2n)!∫−11(π(x))2dxf(x) = H(x) + \dfrac{f^{(2n)}(\xi)}{(2n)!}(\pi(x))^2 \\ -\int_{-1}^1 f(x) dx = \sum_{i=1}^n w_i f(x_i) + \dfrac{f^{(2n)}(\xi)}{(2n)!}\int_{-1}^1 (\pi(x))^2 dx
     - ⯀ -
    -

    Weights of Legendre-Gauss Quadrature

    -
    Theorem - - -

    The weights of Legendre-Gauss Quadrature can be calculated by the following formula.

    -
    wi=∫−11(π(x))2=2(1−xi2)[Pn′(xi)]2w_i = \int_{-1}^1 (\pi(x))^2 = \dfrac{2}{(1 - x_i^2)[P_n'(x_i)]^2}
     -
    -
    Proof - -

    We first find π′(x)\pi'(x) and π′(xi)\pi'(x_i) for i=1,2,⋯ ,ni = 1, 2, \cdots, n.

    -
    π(x)=∏j=1n(x−xj)=(x−xi)∏j=1,j≠in(x−xj)π′(x)=∏j=1,j≠in(x−xj)+(x−xi)(∏j=1,j≠in(x−xj))′π′(xi)=∏j=1,j≠in(xi−xj)+(xi−xi)(∏j=1,j≠in(xi−xj))′=∏j=1,j≠in(xi−xj)\pi(x) = \prod_{j=1}^n (x - x_j) = (x - x_i) \prod_{j=1, j \neq i}^n (x - x_j)\\ -\pi'(x) = \prod_{j=1, j \neq i}^n (x - x_j) + (x - x_i)(\prod_{j=1, j \neq i}^n (x - x_j))'\\ -\pi'(x_i) = \prod_{j=1, j \neq i}^n (x_i - x_j) + (x_i - x_i)(\prod_{j=1, j \neq i}^n (x_i - x_j))' = \prod_{j=1, j \neq i}^n (x_i - x_j)
    -

    Consider the Lagrange Interpolation Polynomial li(x)l_i(x) of π(x)\pi(x) at xix_i.

    -

    It can be expressed as

    -
    li(x)=π(x)(x−xi)π′(xi)=Pn(x)(x−xi)Pn′(xi)l_i(x) = \dfrac{\pi(x)}{(x - x_i)\pi'(x_i)} = \dfrac{P_n(x)}{(x - x_i)P_n'(x_i)}
    -

    Since li(x)Pn′(x)l_i(x)P_n'(x) is a polynomial of degree 2n−22n - 2, we have

    -
    ∫−11li(x)Pn′(x)dx=∑j=1nwjli(xj)Pn′(xj)\int_{-1}^1 l_i(x)P_n'(x) dx = \sum_{j=1}^n w_j l_i(x_j) P_n'(x_j)
    -

    Since li(xj)=0l_i(x_j) = 0 for j≠ij \neq i, and li(xi)=1l_i(x_i) = 1, we have

    -
    ∫−11li(x)Pn′(x)dx=wili(xi)Pn′(xi)=wiPn′(xi)\int_{-1}^1 l_i(x)P_n'(x) dx = w_i l_i(x_i) P_n'(x_i) = w_i P_n'(x_i)
    -

    On the other hand,

    -
    ∫−11li(x)Pn′(x)dx=∫−11Pn(x)Pn′(x)(x−xi)Pn′(xi)dx\int_{-1}^1 l_i(x)P_n'(x) dx = \int_{-1}^1 \dfrac{P_n(x)P_n'(x)}{(x - x_i)P_n'(x_i)} dx
    -

    Hence,

    -
    wi=1(Pn′(x))2∫−11Pn(x)Pn′(x)x−xjdxw_i = \dfrac{1}{(P_n'(x))^2} \int_{-1}^1 \dfrac{P_n(x)P_n'(x)}{x - x_j} dx
    -

    Similarly, we move on to the next step.

    -
    ∫−11(li(x))2=∫−11(Pn(x))2(x−xi)2(Pn′(xi))2dx=1(Pn′(xi))2∫−11(Pn(x))2(x−xi)2dx=1(Pn′(xi))2∫−11−(Pn(x))2d1x−xi=1(Pn′(xi))2((−(Pn(x))2x−xi)∣−11+∫−111x−xid(Pn(x))2)=1(Pn′(xi))2((−(Pn(1))21−xi−−(Pn(−1))2−1−xi)+2∫−11Pn(x)Pn′(x)x−xidx)=−1(Pn′(xi))2(11+xi+11−xi)+2(Pn′(xi))2∫−11Pn(x)Pn′(x)x−xidx=−2(1−xi2)(Pn′(xi))2+2wi\int_{-1}^1 (l_i(x))^2\\ -= \int_{-1}^1 \dfrac{(P_n(x))^2}{(x - x_i)^2(P_n'(x_i))^2} dx\\ -= \dfrac{1}{(P_n'(x_i))^2}\int_{-1}^1 \dfrac{(P_n(x))^2}{(x - x_i)^2} dx\\ -= \dfrac{1}{(P_n'(x_i))^2}\int_{-1}^1 -(P_n(x))^2 d \dfrac{1}{x - x_i}\\ -= \dfrac{1}{(P_n'(x_i))^2} ((\dfrac{-(P_n(x))^2}{x - x_i})|_{-1}^1 + \int_{-1}^1 \dfrac{1}{x - x_i} d(P_n(x))^2)\\ -= \dfrac{1}{(P_n'(x_i))^2} ((\dfrac{-(P_n(1))^2}{1 - x_i} - \dfrac{-(P_n(-1))^2}{-1 - x_i}) + 2\int_{-1}^1 \dfrac{P_n(x)P_n'(x)}{x - x_i} dx)\\ -= -\dfrac{1}{(P_n'(x_i))^2} (\dfrac{1}{1 + x_i} + \dfrac{1}{1 - x_i}) + \dfrac{2}{(P_n'(x_i))^2}\int_{-1}^1 \dfrac{P_n(x)P_n'(x)}{x - x_i} dx\\ -= -\dfrac{2}{(1 - x_i^2)(P_n'(x_i))^2} + 2w_i\\
    -

    On the other hand, since (li(x))2(l_i(x))^2 is a polynomial of degree 2n−22n - 2, we have

    -
    ∫−11(li(x))2dx=∑j=1nwj(li(xj))2=wi(li(xi))2=wi\int_{-1}^1 (l_i(x))^2 dx = \sum_{j=1}^n w_j (l_i(x_j))^2 = w_i (l_i(x_i))^2 = w_i
    -

    Combine the two equations, we get

    -
    wi=∫−11(li(x))2dx=−2(1−xi2)(Pn′(xi))2+2wiwi=2(1−xi2)(Pn′(xi))2w_i = \int_{-1}^1 (l_i(x))^2 dx = -\dfrac{2}{(1 - x_i^2)(P_n'(x_i))^2} + 2w_i\\ -w_i = \dfrac{2}{(1 - x_i^2)(P_n'(x_i))^2}
     - ⯀ -
    -

    Weighted Gaussian Quadrature

    -

    Consider we want to integrate ∫abρ(x)f(x)dx\int_a^b \rho(x)f(x) dx, where ρ(x)\rho(x) is a weight function. We can use the same idea to get the weighted Gaussian Quadrature.

    -

    Here we only discuss the case where ρ(x)=11+x2\rho(x) = \dfrac{1}{\sqrt {1 + x^2}}.

    -

    In this case, we use Chebyshev Zeroes as the points xix_i. The proof is similar to the proof of Gauss-Legendre Quadrature, we omit it here.

    -

    Lobatto Quadrature

    -

    Lobatto’s idea is use aa, bb and the roots of Pn−2′(x)P'_{n-2}(x) as the points xix_i.

    -

    In this way, it uses the end points as information.

    -
    - Experiment
    - -
    - Graph Theory
    - -
    - Number Theory
    - -
    - Numerical Analysis
    -
    - Chebyshev Polynomial
    - -
    - Numeric Differentiation
    - -
    - Numeric Integration
    - -
    -
    -
    - - - - -
    - - diff --git a/src/routes/note/network-flow/+page.md b/src/routes/note/network-flow/+page.md index dce9ad6..0b7c7c1 100644 --- a/src/routes/note/network-flow/+page.md +++ b/src/routes/note/network-flow/+page.md @@ -341,15 +341,15 @@ $'fn{dfs}(u, f, t)$ [ return $r$ ] -# G is the network, c is the capacity function, s is the source, t is the sink +# d is the digraph of the network, c is the capacity function, s is the source, t is the sink # Return the maximum flow and the flow function -$'fn{dinic}(digraph, c, s, t)$ [ +$'fn{dinic}(d, c, s, t)$ [ $vertices = 'varnothing$ $edges = 'varnothing$ - for $u$ in $digraph.vertices$ [ + for $u$ in $d.vertices$ [ $vertices(u) = $ new $'text{Vertex}'{level: -1, edges: 'varnothing, current: 0'}$ ] - for $e$ in $digraph.edges$ [ + for $e$ in $d.edges$ [ $u = vertices(e.u)$ $v = vertices(e.v)$ $proper = $ new $'text{Edge}'{v: v, cap: c(e), rev: $null$'}$ @@ -369,7 +369,7 @@ $'fn{dinic}(digraph, c, s, t)$ [ $flow += 'fn{dfs}(s, +'infty, t)$ ] $f = 'varnothing$ - for $e$ in $digraph.edges$ [ + for $e$ in $d.edges$ [ $f(e) = edges(e).rev.cap$ ] return $flow, f$ diff --git a/svelte.config.js b/svelte.config.js index ebf40a2..7cb6a9d 100644 --- a/svelte.config.js +++ b/svelte.config.js @@ -3,6 +3,7 @@ import mdsvexConfig from "./mdsvex.config.js"; import adapter from "@sveltejs/adapter-static"; import { vitePreprocess } from "@sveltejs/kit/vite"; import fs from "fs"; +import path from "path"; /** @@ -48,6 +49,9 @@ function toSnakeCategory(category) { function scanNotes() { let root = 'src/routes/note'; let output = 'src/lib/generated/topic-list.json'; + if (!fs.existsSync(path.dirname(output))) { + fs.mkdirSync(path.dirname(output), { recursive: true }); + } let lastGeneratedTime = fs.existsSync(output) ? fs.statSync(output).mtimeMs : 0; let lastSourceTime = fs.statSync(root).mtimeMs;