# Taylor Expansion of Error Function

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 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.

-
$a=0$ 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

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 $a$: $$\text{erf}_a(x)= e^{-a^2} \sum_{n=0}^\infty (-1)^n \frac{H_n(a)}{n!}(x-a)^n$$

-
"$f^{(n)}(a)$ can be written in terms of Hermite polynomials..." - of course, since the Rodrigues formula for the Hermite polynomials involves the repeated differentiation of the Gaussian function $\exp(-x^2)$... – J. M. Mar 28 '12 at 15:19
Also: the odd-order Hermite polynomials are odd (and thus $H_{2n+1}(0)=0$) precisely because the Gaussian weight function is even. – J. M. Mar 28 '12 at 15:21
@J.M. And so, one can do a Taylor Series for every $a$: $\text{erf}_a(x)=\sum_{n=0}^{\infty}(-1)^n e^{-a^2} \frac {H_n(a)}{n!}(x-a)^{n}$, right? – draks ... Mar 28 '12 at 17:38
Sure, the error function is analytic... – J. M. Mar 28 '12 at 18:38
I would not be able to solve for $erf(a)$ if I were to expand around $a$ - would I? – fitzgeraldo Mar 28 '12 at 22:05