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 $c_n$?

  • 1
    What does the arcsin have to do with the arctan?2012-04-02
  • 3
    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.2012-04-02
  • 1
    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.2012-04-02
  • 0
    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 question2012-04-02
  • 0
    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.2012-04-02
  • 0
    When you write \sum_{n=0}^\infty in a "displayed" setting, it looks like this: $\displaystyle\sum_{n=0}^\infty$. But when you write \Sigma_{n=0}^\infty, it looks like this: $\displaystyle\Sigma_{n=0}^\infty$. Both the positioning of the subscripts and the size of the $\Sigma$ and probably in some contexts the spacing, fail to follow the conventions used for sums. (I edited the question accordingly.)2012-04-02
  • 0
    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.2012-04-02
  • 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).

  • 0
    Nice, great answer.2012-04-03
  • 1
    Don't we have to establish uniform convergence first in order for differentiation to make sense?2013-09-08