Maclaurin Series Error Function
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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 x e − t 2 d t = 2 π
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∫ 0 x e − t 2 d t . {\displaystyle {\begin − 6\operatorname error function table − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − 1 t\\&={\frac − 0{\sqrt {\pi }}}\int _ inverse error function 9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted erfc, is defined as erfc ( x ) = 1 − erf ( x ) = 2 π
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∫ 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 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]).
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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 }}}e^ − 1}D(x),\end − 0}} where D(x) is the Dawson function (which can be used instead of erfi to avoid arithmetic overflow[3]). Despite the name "imaginary error function", erfi ( x ) {\displaystyle \operatorname 8 (x)} is real when x is real. When the error function is evaluated for arbitrary complex arguments z, the resulting complex error function is usu
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Error Function Properties
us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people nyse erf studying math at any level and professionals in related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are https://en.wikipedia.org/wiki/Error_function 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 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 http://math.stackexchange.com/questions/125328/taylor-expansion-of-error-function 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. 53k5118254 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 $H_n$: $$ H_0(x)=1,\, H_1(x)=2x,\, H_2(x)=4x^2-2,\, H_3(x)=8x^3-12x,\, H_4(x)=16x^4-48x^2+12,\, H_5(x)=32x^5-160x^3+120x,\, H_6(x)=64x^6-480x^4+720x^2-120,\dots\, $$ You recognize that $H_{2n-1}(0)=0$, which gives the power series for $e^{-x^2}$ at $a=0$: $$ e^{-x^2} = 1 - \frac{2}{2!}x^2+\frac{12}{4!}x^4-\frac{120}{6!}x^6+\cdots $$ (see here). After multiplying by $2/\sqrt{\pi}$, this integrates to $$ \operatorname{erf}(z) =\frac{2}{\sqrt{\pi}} \left(z-\frac{z^3}{3}+\frac{z^5}{10}-\frac{z^7}{42}+\frac{z^9}{216}-\ \cdots\right) . $$ EDIT: Since $\displaystyle \frac{d^n}{dx^n}e^{-x^2}= (-1)^n e^{-x^2} H_n(x), $ one can do a Taylor Series for every $a$: $$ \text{erf}_a(x)= e^{-a^2} \sum_{n=0}^\infty (-1)^n \frac{H_n(a)}{n!}(x-a)^n $$ share|cite|improve this answer edite
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