I stumbled upon this particular expansion that was included in this post.

$$ \displaystyle \arcsin^{2}(x) = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2} \binom{2n}{n}} (2x)^{2n}$$

This caught my eye because I remember trying to derive a Taylor series for $\arcsin^{2}(x)$ a while ago without much success.

Can anyone prove this or point me to a material that would show a proof of this identity?


Feel free to use any mathematical apparatus at hand. I'm not interested in a proof fit for a certain level, nor am I looking for utmost elegance (though that would be lovely).

  • $\begingroup$ See Z. R. Melzak. Companion to Concrete Mathematics. Wiley–Interscience, New York, 1973 page 108 $\endgroup$
    – Norbert
    Sep 23, 2015 at 23:37
  • $\begingroup$ Sorry, my bad for not mentioning I'm looking for some material that can readily be accessed online. $\endgroup$
    – Victor
    Sep 23, 2015 at 23:46
  • $\begingroup$ You can find a derivation of the Taylor series of $\frac{\arcsin(x)}{\sqrt{1-x^2}}$ this answer. Notice that $\frac{d\arcsin^2(x)}{dx} = \frac{2\arcsin(x)}{\sqrt{1-x^2}}$ so a simple integration gives your series. $\endgroup$
    – Winther
    Sep 23, 2015 at 23:57
  • $\begingroup$ Thank you so much for the point, Now it seems plain obvious, to be honest... I'll work it out this way. However, I'm going to leave the question open in case someone thinks of a different method as well as it may be of use to others in the future. $\endgroup$
    – Victor
    Sep 24, 2015 at 0:02

1 Answer 1


You can find a derivation for the Taylor series of $\frac{\arcsin(x)}{\sqrt{1-x^2}}$ in this nice answer. Since $$\frac{d\arcsin^2(x)}{dx} = \frac{2\arcsin(x)}{\sqrt{1-x^2}}$$

the Taylor series for $\arcsin^2(x)$ follows by integration.


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