# Computing the inverse error function

I need a formula/series to compute the inverse error function, which is the inverse of $$\operatorname{erf}(x) = \dfrac{2}{\sqrt{\pi}} \int\limits_{0}^{x} \mathrm{e}^{-t^2} \,\mathrm{d}t.$$

Apparently the Maclaurin series for the inverse error function is $$\operatorname{erf}^{-1}(x) = \sqrt{\pi} \left( \frac{1}{2}x + \frac{1}{24}\pi x^3 + \frac{7}{960}\pi^2 x^5 + \frac{127}{80\,640}\pi^3 x^7 + \ldots \right),$$ but I have no idea how many terms I need for it to converge (something like 4 decimal places would be fine for me, but I'd like to know the general answer). Alternatively, is there a better way of computing this function?

I'm no expert on the error function, but I can offer you this. To match the series you give, your $\mathrm{erf}(x)$ should be defined with $\frac{2}{\sqrt{\pi}}$ out front.
Also, the error in any series, will be on the order of $t^{n+1}$, where $n$ is the exponent of the last term you included in your series. Since the even-powered terms in this series always have $0$ coefficient, you may as well assume that you included a final zero term, and so your convergence error will be on the order of $t$ to the next odd integer. Since the inverse error function blows up to $\pm\infty$ at $\pm 1$, and you want $\lvert t \rvert^{n+2}<10^{-4}$, the series will converge accurately enough for you in the range: $\lvert t \rvert < 10^{-4/(n+2)}.$ Unfortunately, this doesn't converge very fast and you will need more and more terms the larger and larger $\lvert t \rvert$ gets.