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General analytic functions

Table of contents: Taylor series - Quadrature - Euler-Maclaurin formula

Taylor series

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f ⁣(z+x)=k=0f(k)(z)k!xkf\!\left(z + x\right) = \sum_{k=0}^{\infty} \frac{{f}^{(k)}(z)}{k !} {x}^{k}
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f(k)(z)k!CRk   where C=suptC,tz=Rf ⁣(t)\left|\frac{{f}^{(k)}(z)}{k !}\right| \le \frac{C}{{R}^{k}}\; \text{ where } C = \mathop{\operatorname{sup}}\limits_{t \in \mathbb{C},\,\left|t - z\right| = R} \left|f\!\left(t\right)\right|
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f ⁣(z+x)k=0N1f(k)(z)k!xkCDN1D   where C=suptC,tz=Rf ⁣(t),D=xR\left|f\!\left(z + x\right) - \sum_{k=0}^{N - 1} \frac{{f}^{(k)}(z)}{k !} {x}^{k}\right| \le \frac{C {D}^{N}}{1 - D}\; \text{ where } C = \mathop{\operatorname{sup}}\limits_{t \in \mathbb{C},\,\left|t - z\right| = R} \left|f\!\left(t\right)\right|,\,D = \frac{\left|x\right|}{R}

Quadrature

Related topics: Gaussian quadrature

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abf ⁣(t)dtba2k=1nwn,kf ⁣(ba2xn,k+a+b2)ba264M15(1ρ2)ρ2n   where M=suptEρf ⁣(ba2t+a+b2)\left|\int_{a}^{b} f\!\left(t\right) \, dt - \frac{b - a}{2} \sum_{k=1}^{n} w_{n,k} f\!\left(\frac{b - a}{2} x_{n,k} + \frac{a + b}{2}\right)\right| \le \frac{\left|b - a\right|}{2} \frac{64 M}{15 \left(1 - {\rho}^{-2}\right) {\rho}^{2 n}}\; \text{ where } M = \mathop{\operatorname{sup}}\limits_{t \in \mathcal{E}_{\rho}} \left|f\!\left(\frac{b - a}{2} t + \frac{a + b}{2}\right)\right|

Euler-Maclaurin formula

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k=NUf ⁣(k)=NUf ⁣(t)dt+f ⁣(N)+f ⁣(U)2+k=1MB2k(2k)!(f(2k1)(U)f(2k1)(N))+NUB2M ⁣(tt)(2M)!f(2M)(t)dt\sum_{k=N}^{U} f\!\left(k\right) = \int_{N}^{U} f\!\left(t\right) \, dt + \frac{f\!\left(N\right) + f\!\left(U\right)}{2} + \sum_{k=1}^{M} \frac{B_{2 k}}{\left(2 k\right)!} \left({f}^{(2 k - 1)}(U) - {f}^{(2 k - 1)}(N)\right) + \int_{N}^{U} \frac{B_{2 M}\!\left(t - \left\lfloor t \right\rfloor\right)}{\left(2 M\right)!} {f}^{(2 M)}(t) \, dt
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k=NUf ⁣(k)(NUf ⁣(t)dt+f ⁣(N)+f ⁣(U)2+k=1MB2k(2k)!(f(2k1)(U)f(2k1)(N)))4(2π)2MNUf(2M)(t)dt\left|\sum_{k=N}^{U} f\!\left(k\right) - \left(\int_{N}^{U} f\!\left(t\right) \, dt + \frac{f\!\left(N\right) + f\!\left(U\right)}{2} + \sum_{k=1}^{M} \frac{B_{2 k}}{\left(2 k\right)!} \left({f}^{(2 k - 1)}(U) - {f}^{(2 k - 1)}(N)\right)\right)\right| \le \frac{4}{{\left(2 \pi\right)}^{2 M}} \int_{N}^{U} \left|{f}^{(2 M)}(t)\right| \, dt

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2019-06-18 07:49:59.356594 UTC