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I was working the other day in the Math Help Centre, trying to help some first years with a calculus problem. The problem involved investigating the Taylor series of $\arcsin(x)$. Once the students had derived

$$\arcsin(x)=\sum_{n=0}^\infty \frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}$$

they were asked to rederive it in a different way:

Determine the sequence $\{c_n\}_{n\in\mathbb{N}}$ such that $$x=\sum_{n=0}^\infty c_n \left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^n.$$

Here's what I got: I recognized that the object in parentheses is the Taylor series of $\sin(x)$, so the idea is to let $\arcsin(x)=\sum_{n=0}^\infty c_nx^n$ and note that $\arcsin(\sin(x))=x$. After that it's just an issue of computing the $c_n$'s.

There's an obvious brute-force way to do it, where for each $n$ you say "indices larger than $n$ don't matter, so now it's a finite problem." Expand the relevant terms to get relations involving the $c_i$'s that you've already worked out. The problem is that this will only work for finitely many values, and it was hard to determine a pattern.

Is there and obvious pattern I'm missing? It occurred to me that the "relevant terms" that contribute to $c_n$ depend on the divisors on $n$, is this intuition correct?

Most importantly, what is the best way to solve the problem of computing $c_n$?

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What does the arcsin have to do with the arctan? –  marty cohen Apr 2 '12 at 4:15
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That’s not a rederivation of the Taylor series of $\arctan x$; that’s a derivation of the Taylor series of $\arcsin x$, which is $$\sum_{n\ge 0}\frac1{4^n(2n+1)}\binom{2n}nx^{2n+1}\;.$$ And I agree that the pattern $$\frac1{4^n(2n+1)}\binom{2n}n$$ isn’t terribly obvious. –  Brian M. Scott Apr 2 '12 at 4:17
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A way to do $\arcsin(x)$ that does show you the pattern is to find the series for $(1 - x^2)^{-1/2}$ and integrate. –  Robert Israel Apr 2 '12 at 4:41
    
OOPS! Bad memory I guess, $\arcsin(x)$ is the correct function, I edited the post. @Robert Isreal that is how the students derive the series in the first part of the question –  you Apr 2 '12 at 5:09
    
In that case, I think you're right: there's no obvious way to get the general formula from $\arcsin(\sin(x))=x$, you just get a finite number of terms. –  Robert Israel Apr 2 '12 at 7:14
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1 Answer

up vote 6 down vote accepted

Here's a concise argument, but it uses $\sin^2 x+\cos^2x=1$, which isn't exactly obvious from the power series. Differentiating

$$x=\sum_{n=0}^\infty c_n\sin^nx$$

twice yields

$$ \begin{eqnarray} 0 &=& \sum_{n=0}^\infty c_n\left(n(n-1)\sin^{n-2}x\cos^2x-n\sin^nx\right) \\ &=& \sum_{n=0}^\infty c_n\left(n(n-1)\sin^{n-2}x-n^2\sin^nx\right)\;, \end{eqnarray} $$

which gives the recurrence

$$ \begin{eqnarray} c_{n+2} &=& \frac{n^2}{(n+1)(n+2)}c_n \\ &=& \frac{n}{n+2}\frac{n}{n+1}c_n\;. \end{eqnarray}$$

The factors of $n$ and $n+2$ cancel except for the final $n+2$, and with $c_0=0$ and $c_1=1$ this leads to

$$c_{2n+1}=\frac{(2n)!}{(2^nn!)^2(2n+1)}\;.$$

I got the idea for this here (item 4.3, third proof).

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Nice, great answer. –  you Apr 3 '12 at 2:54
    
Don't we have to establish uniform convergence first in order for differentiation to make sense? –  user70520 Sep 8 '13 at 1:01
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