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Find the Taylor Series expansion of $e^{a\arcsin x}$ and hence deduce, the Taylor series expansion of $\arcsin x$.

I could find the Taylor Series expansion of $e^{a\arcsin x}$ as $$1+ax+\frac{a^2x^2}{2}+\frac{a^3+a}{6}x^3+\cdots $$

However, I have no idea how to deduce the series for $\arcsin x$ from the above expansion. I tried writing, $e^{a\arcsin x}$ as $$1+a\arcsin x+\frac{(a\arcsin x)^2}{2!}+\cdots$$ and comparing the coefficients with the one above, but it was not of any help because , I only got the inference that, $x=\arcsin x$ which was quite obvious. Any help regarding solving this issue will be highly appreciated.

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    $\begingroup$ coefficient of $a$. $\endgroup$ Commented Sep 24, 2023 at 5:45
  • $\begingroup$ Take the logarithm and continue with Taylor series. $\endgroup$ Commented Sep 24, 2023 at 9:19
  • $\begingroup$ @ClaudeLeibovici I tried it, but it doesn't seem to help me in any way. Am I missing something? $\endgroup$ Commented Sep 24, 2023 at 12:25
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    $\begingroup$ @ThomasFinley You have the Taylor series for $\exp(a\arcsin x)$, which you can regard as a series in $a$ with coefficient being a series in $x$. So look the coefficient of $a$, which gives you $\arcsin x=x+\frac16 x^3+\dots$ $\endgroup$ Commented Sep 24, 2023 at 12:54
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    $\begingroup$ What is $\frac{\partial}{\partial a}e^{a\arcsin(x)}$ at $a=0$? Apply that to the power series you've found. $\endgroup$
    – robjohn
    Commented Sep 24, 2023 at 14:15

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