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$\newcommand{\erf}{\operatorname{erf}}$ The computation of $\frac{d^n}{dx^n}\int{(e^{-x²})}^{\erf(x)}dx$ with wolfram alpha we have for $n=1, n=2, ..n=4$ interesting expansion which seems present altern series such that exploit the formula for the constant $\pi$ , Now My attempt fails to get the series expansion of that integral at $x=0$ since I don't know any thing about it's closed form. Now my question here is :

Question :What is the series expansion of the $n$-th derivative of that function(integral) ? and what is the convergence rate (how many digits of $\pi$ can be computed at each increment in summation)?

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  • $\begingroup$ $n=1,2,3,4,...$ ? Why dont you write $$ \frac{{\rm d}^{n-1}}{{\rm d}x^{n-1}} \left(e^{-x^2}\right) ^{{\erf}(x)} \, ? $$ $\endgroup$ – Diger Aug 18 '18 at 11:05
  • $\begingroup$ Thanks , yes it is the same $\endgroup$ – zeraoulia rafik Aug 18 '18 at 11:06
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$\newcommand{\erf}{\operatorname{erf}}$I don't know what you expect from the series expansion, but maybe for the start you can just use the series for each functions and write $$ e^{-x^2 \erf(x)} = \sum_{m=0}^\infty x^m \substack{\sum_{k_1=0}^\infty \cdots \sum_{k_n=0}^\infty \\ \frac{m-2(k_1 + ... + k_n)}{3} \stackrel{!}{=} \, n \in {\mathbb N}} \frac{(-1)^{n+k_1+...+k_n}\left(\frac{2}{\sqrt{\pi}}\right)^n}{n! k_1!\cdots k_n! (2k_1+1)\cdots (2k_n+1)} \, . $$ The sums over $k_1 \cdots k_n$ are in fact finite sums, because $n\geq 0$ limited by $k_{\rm max}=\left\lfloor \frac{m}{2} \right \rfloor$.

Notation: The inner sum means, that the summation over $k_1 \cdots k_n$ is constrained by the requirement $$\frac{m-2(k_1 + ... + k_n)}{3} = \frac{m-2k}{3} = \, n \geq 0 \in {\mathbb N} \, .$$

For $m=0$ all the $k$ have to be zero in order for $n=0$ and this gives the trivial term $1$.

For $m=1$, $k$ values $>0$ lead to negative $n$ and $k=0$ leads to $n=\frac{1}{3} \notin {\mathbb N}$.

For $m=2$, $k=0$ or $k=1$. $k=0$ leads to a contradiction and $k=1$ leads to $n=0$ which is the empty sum (which starts at $k_1$ and not $k_0$).

$m=3$ is the first non-trivial case. $k$ must be either $0$ or $1$ for $n\geq 0$. $k=1$ again leads to $n=\frac{1}{3}$ so we only have $k=0$ in which case $n=1$ and the sum is actually just $$\sum_{\substack{k_1=0 \\ k_1 \stackrel{!}{=}0}}^1 \frac{(-1)^{1+k_1}\frac{2}{\sqrt{\pi}}}{1!k_1!(2k_1+1)} = -\frac{2}{\sqrt{\pi}} \, .$$

$m=4$ is again trivial, because $k=0,1,2$ leads to $n=\frac{4}{3},\frac{2}{3},0$.

The sums start to get more involved at $m=9$. Now $k=0,1,2,3,4$ leads to $n=3,\frac{7}{3},\frac{5}{3},1,\frac{1}{3}$ so there are 2 valid $n$ \begin{align} n=1: &\sum_{\substack{k_1=0 \\ k_1\stackrel{!}{=}3}}^{4} \frac{(-1)^{1+k_1}\frac{2}{\sqrt{\pi}}}{1!k_1!(2k_1+1)} = \frac{2}{\sqrt{\pi}} \frac{1}{3!(2\cdot3+1)} = \frac{1}{21 \sqrt{\pi}} \\ n=3: &\substack{\sum_{k_1=0}^4 \sum_{k_2=0}^4 \sum_{k_3=0}^4 \\ (k_1 + k_2 + k_3) \stackrel{!}{=} \, 0} \frac{(-1)^{3+k_1+k_2+k_3}\left(\frac{2}{\sqrt{\pi}}\right)^3}{3! k_1! k_2! k_3! (2k_1+1) (2k_2+1) (2k_3+1)} \\ &=-\frac{4}{3\pi^{3/2}} \, . \end{align}

There is a pattern in the $n$: For odd $m$ at $m=6i-3$ the number of allowed $n$ increases by $1$ of which there are $i$ in total. For instance at $m=9$ we had $n=1,3$. At $m=15$ we have $n=1,3,5$ and so on.

Similarly for even $m$ at $m=6i$ the number of allowed $n$ equals $i$ whose values take $n=2,4,6,8,\dots$.

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  • $\begingroup$ That is some pretty funky summation notation, and could probably use some explaining. $\endgroup$ – Theoretical Economist Aug 18 '18 at 12:08
  • $\begingroup$ @Diger , Thanks , could show more how you got that , I want some steps explanation $\endgroup$ – zeraoulia rafik Aug 18 '18 at 12:53
  • $\begingroup$ added some notation infos. $\endgroup$ – Diger Aug 18 '18 at 14:21
  • $\begingroup$ Thanks , it's stay to know about rate of convergence as stated above in my question $\endgroup$ – zeraoulia rafik Aug 18 '18 at 14:28
  • $\begingroup$ radius of convergence should be $\infty$. $\endgroup$ – Diger Aug 18 '18 at 15:49

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