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Error functions

Table of contents: Definitions - Illustrations - Integral representations - Connection formulas - Functional equations - Hypergeometric representations - Derivatives

Definitions

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Symbol: Erf erf(z)\operatorname{erf}(z) Error function
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Symbol: Erfc erfc(z)\operatorname{erfc}(z) Complementary error function
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Symbol: Erfi erfi(z)\operatorname{erfi}(z) Imaginary error function

Illustrations

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Image: X-ray of erf(z)\operatorname{erf}(z) on z[4,4]+[4,4]iz \in \left[-4, 4\right] + \left[-4, 4\right] i

Integral representations

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erf(z)=2π0zet2dt\operatorname{erf}(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} {e}^{-{t}^{2}} \, dt
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erfc(z)=2πzet2dt\operatorname{erfc}(z) = \frac{2}{\sqrt{\pi}} \int_{z}^{\infty} {e}^{-{t}^{2}} \, dt
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erfi(z)=2π0zet2dt\operatorname{erfi}(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} {e}^{{t}^{2}} \, dt

Connection formulas

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erf(z)+erfc(z)=1\operatorname{erf}(z) + \operatorname{erfc}(z) = 1
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erfc(z)=1erf(z)\operatorname{erfc}(z) = 1 - \operatorname{erf}(z)
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erfi(z)=ierf ⁣(iz)\operatorname{erfi}(z) = -i \operatorname{erf}\!\left(i z\right)

Functional equations

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erf ⁣(z)=erf(z)\operatorname{erf}\!\left(-z\right) = -\operatorname{erf}(z)
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erfi ⁣(z)=erfi(z)\operatorname{erfi}\!\left(-z\right) = -\operatorname{erfi}(z)
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erfc ⁣(z)=2erfc(z)\operatorname{erfc}\!\left(-z\right) = 2 - \operatorname{erfc}(z)

Hypergeometric representations

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erf(z)=2zπ1F1 ⁣(12,32,z2)\operatorname{erf}(z) = \frac{2 z}{\sqrt{\pi}} \,{}_1F_1\!\left(\frac{1}{2}, \frac{3}{2}, -{z}^{2}\right)
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erf(z)=2zez2π1F1 ⁣(1,32,z2)\operatorname{erf}(z) = \frac{2 z {e}^{-{z}^{2}}}{\sqrt{\pi}} \,{}_1F_1\!\left(1, \frac{3}{2}, {z}^{2}\right)
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erf(z)=zz2ez2zπU ⁣(12,12,z2)\operatorname{erf}(z) = \frac{z}{\sqrt{{z}^{2}}} - \frac{{e}^{-{z}^{2}}}{z \sqrt{\pi}} U^{*}\!\left(\frac{1}{2}, \frac{1}{2}, {z}^{2}\right)
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erfc(z)=ez2zπU ⁣(12,12,z2)\operatorname{erfc}(z) = \frac{{e}^{-{z}^{2}}}{z \sqrt{\pi}} U^{*}\!\left(\frac{1}{2}, \frac{1}{2}, {z}^{2}\right)

Derivatives

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erf(z)=2πez2\operatorname{erf}'(z) = \frac{2}{\sqrt{\pi}} {e}^{-{z}^{2}}
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erf(n)(z)=2π(1)n+1Hn1 ⁣(z)ez2{\operatorname{erf}}^{(n)}(z) = \frac{2}{\sqrt{\pi}} {\left(-1\right)}^{n + 1} H_{n - 1}\!\left(z\right) {e}^{-{z}^{2}}

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2019-10-05 13:11:19.856591 UTC