13
$\begingroup$

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 $c_n$?

  • 0
    @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.2012-04-02

1 Answers 1

8

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).

  • 1
    Don't we have to establish uniform convergence first in order for differentiation to make sense?2013-09-08