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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?

EDIT :

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).

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  • $\begingroup$ 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

3 Answers 3

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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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There exists a general conclusion:

For $k\in\mathbb{N}$ and $|x|<1$, the function $\bigl(\frac{\arcsin x}{x}\bigr)^{k}$, whose value at $x=0$ is defined to be $1$, has Maclaurin's series expansion \begin{equation}\label{arcsin-series-expansion-unify} \biggl(\frac{\arcsin x}{x}\biggr)^{k} =1+\sum_{m=1}^{\infty} (-1)^m\frac{Q(k,2m)}{\binom{k+2m}{k}}\frac{(2x)^{2m}}{(2m)!}, \end{equation} where \begin{equation}\label{Q(m-k)-sum-dfn} Q(k,m)=\sum_{\ell=0}^{m} \binom{k+\ell-1}{k-1} s(k+m-1,k+\ell-1)\biggl(\frac{k+m-2}{2}\biggr)^{\ell} \end{equation} for $k\in\mathbb{N}$ and $m\ge2$.

This general result and its special cases such as $k=1,2,3,4,5$ have been established, reviewed, and surveyed in the papers [2, 3, 5] below, recovered in the paper [1] below, and generalized in the paper [4] below.

References

  1. Feng Qi, Taylor's series expansions for real powers of two functions containing squares of inverse cosine function, closed-form formula for specific partial Bell polynomials, and series representations for real powers of Pi, Demonstratio Mathematica 55 (2022), no. 1, 710--736; available online at https://doi.org/10.1515/dema-2022-0157.
  2. Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Applicable Analysis and Discrete Mathematics 16 (2022), no. 2, 427--466; available online at https://doi.org/10.2298/AADM210401017G.
  3. Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Series expansions of powers of arcsine, closed forms for special values of Bell polynomials, and series representations of generalized logsine functions, AIMS Mathematics 6 (2021), no. 7, 7494--7517; available online at https://doi.org/10.3934/math.2021438.
  4. F. Qi, Explicit formulas for partial Bell polynomials, Maclaurin's series expansions of real powers of inverse (hyperbolic) cosine and sine, and series representations of powers of Pi, Research Square (2021), available online at https://doi.org/10.21203/rs.3.rs-959177/v3.
  5. https://math.stackexchange.com/a/4657809.
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Starting from the well-known power series of $\arcsin$, \begin{equation} \newcommand{\where}[1]{\qquad(#1)} \newcommand{\abs}[1]{\lvert#1\rvert} \newcommand{\bigparens}[1]{\bigl(#1\bigr)} \newcommand{\N}{\mathbb{N}} \arcsin(x) = \sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n+1}}{4^n(2n+1)} =: \sum_{n=0}^\infty a_n x^{2n+1} \where{\abs{x}\le 1}, \end{equation} implies as a Cauchy product of two power series \begin{equation} \bigparens{\arcsin(x)}^2 = \sum_{n=0}^\infty b_n x^{2n+2} \where{\abs{x}\le 1}, \end{equation} with \begin{equation} b_n := \sum_{k=0}^n a_ka_{n-k} = 2^{2n+1}\frac{(n!)^2}{(2n+2)!} \where{n\in\N_0}, \end{equation} as can be shown by induction (or looked up in a table of series, e.g., eq. 1.645 2 in Gradshteyn).

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