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I am trying to find the general formula / taylor expansion for $\arcsin(x)$ at $x=0$. I am not allowed to integrate, otherwise I would have used the binomial theorem and integrated afterwards. I have proven a statement before, which says that if I have a taylor series $T_n$ for a function $f(x)$, then $T_n(x^2)$ is the series for $g(x) = f(x^2)$. So basically, I can just substitute $x^2$ in the taylor series.

I suppose that this could come in handy, since I could then just expand $\arcsin(\sqrt x)$, which would make the derivatives easier.

All in all, I am not sure, however, how to find the formula. I looked it up and when I start differentiating $\arcsin(\sqrt x)$ a lot, I can not really see any pattern. What can I do here?

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  • $\begingroup$ Do you mean you are not allowed to differentiate? $\endgroup$ – Thomas Andrews Mar 13 '17 at 17:41
  • $\begingroup$ @ThomasAndrews I made a typo, I can differentiate and use the formula for the taylor expansion, but I can not integrate $\endgroup$ – Jack4t3 Mar 13 '17 at 17:42
  • $\begingroup$ Do you know that $\sum_{k=0}^{\infty}\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$? $\endgroup$ – Thomas Andrews Mar 13 '17 at 17:48
  • $\begingroup$ Hmm I dont think I have seen that before... $\endgroup$ – Jack4t3 Mar 13 '17 at 17:50
  • $\begingroup$ Do you know inversion of series ? If you do, you can start from the Taylor series of $\sin(x)$. $\endgroup$ – Claude Leibovici Mar 13 '17 at 18:18
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Hint

If you look here, you will find the method for inversing a Taylor series.

So start using the Taylor series of $\sin(x)$ and apply the method. For sure, finding the general form of the coefficients would be a problem.

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