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Let $\mathrm{erf}(x) := \frac{2}{\sqrt{\pi}} \int_{-\infty}^x \exp(-t^2) \, dt$ be the error function $\mathrm{erf}: \mathbb{R} \to (-1,1)$. It is monotonously increasing and therefore has an inverse $\mathrm{erf}^{-1}: (-1,1) \to \mathbb{R}$.

Now I have two questions:

1) On https://en.wikipedia.org/wiki/Error_function#Inverse_functions it is said that $\mathrm{erf}^{-1}$ can be extended to the open unit disc. Unfortunately, there is no reference given for this. Anyone knows a reference for the complete derivation of the McLaurin series?

2) There are so-called Hardy spaces $H^p$ of functions analytic on the open disc which satisfy $$\sup_{0<r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right|^p \; \mathrm{d}\theta\right)^\frac{1}{p}<\infty.$$ see https://en.wikipedia.org/wiki/Hardy_space#Hardy_spaces_for_the_unit_disk.

I wonder wether $\mathrm{erf}^{-1}$ belongs to $H^p$ with $p<\infty$?

Thanks!

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Are you asking for a reference for the recursion formula for the coefficients, or for the convergence on the unit disk? Note that the recursion imples that the odd coefficients are all positive (the even coefficients are $0$ by symmetry). By a result of Pringsheim, if such a series $g(z)$ has radius of convergence $r$ we must have $\lim_{z \to r-} g(z) = \infty$. Of course $\text{erf}^{-1}$ goes to $\infty$ as $z \to 1-$, so the radius of convergence must be $1$.

It appears (just from a plot of the first few hundred terms, I don't have a proof) that $c_j = O(1/j)$, which would imply $\text{erf}^{-1} \in H^2$.

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  • $\begingroup$ I am looking for a reference of the recursion formula. But your argument is interesting too. can you give me a reference for this Pringsheim result? Thanks! $\endgroup$ – User133713 Nov 26 '15 at 18:17

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