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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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    $\begingroup$ 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. $\endgroup$ Apr 2, 2012 at 4:41
  • $\begingroup$ 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 $\endgroup$
    – William
    Apr 2, 2012 at 5:09
  • $\begingroup$ 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. $\endgroup$ Apr 2, 2012 at 7:14
  • $\begingroup$ Seems to me the pattern is given to you by the first method (whatever it was) of finding the coefficients. Then when you apply the second method, you just need to write a proof by induction that the coefficients emerging from it are the same ones that you got by the first method. $\endgroup$ Apr 2, 2012 at 17:19
  • $\begingroup$ @Michael Hardy: thanks, I wasn't sure how to format that properly. Also, since they already know the answer a priori you're probably right that an inductive argument is the most straightforward. I just wish I could think of a clever counting argument, so that you don't need to already know the answer. $\endgroup$
    – William
    Apr 2, 2012 at 18:16

1 Answer 1

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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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  • $\begingroup$ Nice, great answer. $\endgroup$
    – William
    Apr 3, 2012 at 2:54
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    $\begingroup$ Don't we have to establish uniform convergence first in order for differentiation to make sense? $\endgroup$
    – user70520
    Sep 8, 2013 at 1:01

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