Error Function Expansions
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that occurs in probability, statistics, and partial differential equations describing diffusion. It is defined as:[1][2] erf ( x ) = 1 π ∫ − x complementary error function x e − t 2 d t = 2 π ∫ 0 error function calculator x e − t 2 d t . {\displaystyle {\begin − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi
Inverse Error Function
}}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − 1 t\\&={\frac − 0{\sqrt {\pi }}}\int _ 9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted
Error Function Table
erfc, is defined as erfc ( x ) = 1 − erf ( x ) = 2 π ∫ x ∞ e − t 2 d t = e − x 2 erfcx ( x ) , {\displaystyle {\begin 2\operatorname 1 (x)&=1-\operatorname 0 (x)\\&={\frac Φ 9{\sqrt {\pi }}}\int _ Φ 8^{\infty }e^{-t^ Φ 7}\,\mathrm error function matlab Φ 6 t\\&=e^{-x^ Φ 5}\operatorname Φ 4 (x),\end Φ 3}} which also defines erfcx, the scaled complementary error function[3] (which can be used instead of erfc to avoid arithmetic underflow[3][4]). Another form of erfc ( x ) {\displaystyle \operatorname 2 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc ( x | x ≥ 0 ) = 2 π ∫ 0 π / 2 exp ( − x 2 sin 2 θ ) d θ . {\displaystyle \operatorname 0 (x|x\geq 0)={\frac Φ 9{\pi }}\int _ Φ 8^{\pi /2}\exp \left(-{\frac Φ 7}{\sin ^ Φ 6\theta }}\right)d\theta \,.} The imaginary error function, denoted erfi, is defined as erfi ( x ) = − i erf ( i x ) = 2 π ∫ 0 x e t 2 d t = 2 π e x 2 D ( x ) , {\displaystyle {\begin Φ 0\operatorname − 9 (x)&=-i\operatorname − 8 (ix)\\&={\frac − 7{\sqrt {\pi }}}\int _ − 6^ − 5e^ − 4}\,\mathrm − 3 t\\&={\frac − 2{\sqrt {\pi }}
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Error Function Excel
and policies of this site About Us Learn more about Stack error function python Overflow the company Business Learn more about hiring developers or posting ads with us Mathematics Questions Tags Users erf(inf) Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it https://en.wikipedia.org/wiki/Error_function only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Taylor Expansion of Error Function up vote 4 down vote favorite 3 I am working on a question that involves finding the Taylor expansion of the error function. The question http://math.stackexchange.com/questions/125328/taylor-expansion-of-error-function is stated as follows The error function is defined by $\mathrm{erf}(x):=\frac {2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}}dt$. Find its Taylor expansion. I know that the Taylor series of the function $f$ at $a$ is given by $$f(x)=\sum_{n=0}^{\infty}\frac {f^{(n)}(a)}{n!}(x-a)^{n}.$$ However, the question doesn't give a point $a$ with which to center the Taylor series. How should I interpret this? May I use a Maclaurin series, with $a=0$? This appears to be what was done on the Wikipedia page here: http://en.wikipedia.org/wiki/Error_function Any explanations and advice would be appreciated. calculus special-functions taylor-expansion share|cite|improve this question edited Apr 28 '12 at 13:06 J. M. 52.8k5118254 asked Mar 28 '12 at 5:08 fitzgeraldo 14127 6 $a=0$ seems OK for me. I would expand $e^{-t^2}$ in a power series and integrate term by term. –marty cohen Mar 28 '12 at 5:38 add a comment| 1 Answer 1 active oldest votes up vote 2 down vote Elaborating a little on Marty's comment gives the following: $f^{(n)}(a)$ can be written in terms of Hermite polynomials $H_n$: $$ H_0(x)=1,\, H_1(x)=2x,\, H_2(x)=4x^2-2,\, H_3(x)=8x^3-12x,\, H_4(x)=16x^4-48x^2+12,\, H_5(x)=32x^5-160x^3+120x,\, H_6(x)=64x^6-480x^4+720x^2-120,\dots\
Permalink: http://dlmf.nist.gov/7.12 See also: info for 7 Contents §7.12(i) Complementary Error Function §7.12(ii) Fresnel Integrals http://dlmf.nist.gov/7.12 §7.12(iii) Goodwin–Staton Integral §7.12(i) Complementary Error Function Keywords: Stokes phenomenon, complementary error function, error functions Referenced by: §2.11(iv), §7.12(ii), Figure 7.3.6, Figure 7.3.6, 7.3.6 Permalink: http://dlmf.nist.gov/7.12.i See also: info for 7.12 As z→∞ 7.12.1 erfcz ∼e-z2π∑m=0∞(-1)m(12)mz2m+1, erfc(-z) ∼2-e-z2π∑m=0∞(-1)m(12)mz2m+1, Symbols: (a)n: Pochhammer’s symbol (or error function shifted factorial), ∼: Poincaré asymptotic expansion, erfcz: complementary error function, e: base of exponential function and z: complex variable A&S Ref: 7.1.23 (in different form) Referenced by: §3.5(ix), Other Changes Permalink: http://dlmf.nist.gov/7.12.E1 Encodings: TeX, TeX, pMML, pMML, png, png Notational Change (effective error function expansions with 1.0.9): Previously the RHS of these equations were written as e-z2πz∑m=0∞(-1)m1⋅3⋅5⋯(2m-1)(2z2)m and 2-e-z2πz∑m=0∞(-1)m1⋅3⋅5⋯(2m-1)(2z2)m. We have rewritten these sums more concisely using Pochhammer’s symbol. Reported 2014-03-13 by Giorgos Karagounis See also: info for 7.12(i) both expansions being valid when |phz|≤34π-δ (<34π). When |phz|≤14π the remainder terms are bounded in magnitude by the first neglected terms, and have the same sign as these terms when phz=0. When 14π≤|phz|<12π the remainder terms are bounded in magnitude by csc(2|phz|) times the first neglected terms. For these and other error bounds see Olver (1997b, pp. 109–112), with α=12 and z replaced by z2; compare (7.11.2). For re-expansions of the remainder terms leading to larger sectors of validity, exponential improvement, and a smooth interpretation of the Stokes p
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