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How to find the $m^\text{th}$ term for the following expression:

$$ \left.\frac{\partial^m}{\partial s^m}e^{a s^2}\right|_{s=0}$$

Is there any analytical approach?

I computed first few terms and used mathematica "FindSequenceFunction", which yielded the $m^\text{th}$ term as (edited expression):

$$ \frac{2^{m-1} \left(1+(-1)^{m}\right) a^{\frac{m}{2}} \Gamma \left(\frac{m+1}{2}\right)}{\sqrt{\pi }}$$

The first few terms:

$1,0,2 a,0,12 a^2,0,120 a^3,0,1680 a^4,0,30240 a^5,0,665280 a^6,0,17297280 a^7,0,518918400 a^8,0,17643225600 a^9,0,670442572800 a^{10}$

Any help will be appreciated.

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Since$$e^{as^2}=1+as^2+\frac{a^2}{2!}s^4+\frac{a^3}{3!}s^6+\cdots$$you have$$\left.\frac{\mathrm d^m}{\mathrm d^ms}e^{as^2}\right|_{s=0}=\begin{cases}m!\dfrac{a^{\frac m2}}{\left(\frac m2\right)!}&\text{ if $m$ is even}\\0&\text{ if $m$ is odd.}\end{cases}$$

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  • $\begingroup$ Why doesn't this coincide with Mathematica's result ? In particular, the exponent of $a$. $\endgroup$ – Yves Daoust May 10 at 6:42
  • $\begingroup$ @YvesDaoust I made a mistake. $\endgroup$ – Mark Robinson May 10 at 6:44
  • $\begingroup$ Jose's expression requires a factor of $(-1)^{m/2}$ , for a correction. $\endgroup$ – Dr Zafar Ahmed DSc May 10 at 7:31
  • $\begingroup$ @DrZafarAhmedDSc Jose's expression seems to be correct. Please see the edits I have made in the question section. $\endgroup$ – Mark Robinson May 10 at 7:38
  • $\begingroup$ Oh! sorry I bothered about $e^{-ax^2}$. $\endgroup$ – Dr Zafar Ahmed DSc May 10 at 7:44

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