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: 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 $a$:
$$
\text{erf}_a(x)= \text{erf}(a) + e^{-a^2} \sum_{n=1}^\infty (-1)^n \frac{H_n(a)}{n!}(x-a)^n
$$ 
A: Context for using the Taylor Series of $$e^{-t^2}$$ to find the Taylor expansion of the ERF function is found at Robert Ghrist/UPenn's Calculus wiki. 
