taylor series for inverse of error function

Question

I was asked the following question:

Determine the Taylor Series degree 3 around $0$ of the inverse function of $erf(x)$.

I took the first derivative of the function $erf'(x) = \frac{2}{\sqrt{\pi}} e^{-x^2}$. When $erf(x)=0 \implies x=0$, thus I would have that $erf'(0)=\frac{2}{\sqrt{\pi}}$. The derivative of the inverse function is given by the $\frac{1}{erf'(0)}=\frac{\sqrt{\pi}}{2}$ and the first term of the Taylor Series is $\frac{\sqrt{\pi}}{2} x$. How would I proceed to get the 2nd and 3rd one. For some reason, I got a diferent 3rd derivative than what appears in the solution, and I can't understand why, specifically, why a $\pi^{\frac{3}{2}}$ appears in the 3rd derivative.

Answer 1

What you can do is to start with $$y=\text{erf}(x)=\frac{2 x}{\sqrt{\pi }}-\frac{2 x^3}{3 \sqrt{\pi }}+\frac{x^5}{5 \sqrt{\pi }}-\frac{x^7}{21 \sqrt{\pi }}+O\left(x^9\right)$$ and use series reversion to get $$x=\frac{\sqrt{\pi } }{2}y+\frac{\pi ^{3/2}}{24} y^3+\frac{7\pi ^{5/2}}{960} y^5+\frac{127 \pi ^{7/2} }{80640}y^7+O\left(y^9\right)$$

Answer 2

$$erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$ $$=c\int_0^xe^{-t^2}dt$$ We have, around $ t=0$, $$e^{-t^2}=1-t^2+o(t^2)\implies$$ by integration, $$erf(x)=c(x-\frac{x^3}{3})+o(x^3)$$

If $ g(x)=f^{-1}(x)=$

$$a_0+a_1x+a_2x^2+a_3x^3+o(x^3)$$ then write that $$g(f(x))=x$$ to get the coefficient $ a_i$. you will find that $$g(f(x))=a_0+a_1c(x-\frac{x^3}{3})+a_2c^2(x-\frac{x^3}{3})^2+a_3c^3(x-\frac{x^3}{3})^3+o(x^3)=$$ $$a_0+a_1cx+a_2c^2x^2-(\frac{a_1c}{3}-a_3c^3)x^3+o(x^3)$$ thus

$$a_0=0\;,\; a_1=\frac{\sqrt{\pi}}{2},a_2=0, a_3=\frac{a_1}{3c^2}$$

So the expansion you look for is $$erf^{-1}(x)=\frac{\sqrt{\pi}}{2}x+\frac{\pi\sqrt{\pi}x^3}{24}+o(x^3)$$