Maclaurin expansion of Sin(Sin(x))

Question

I want to calculate the limit, $$ \lim_{x\to 0} \frac{sin(sin x) - sin x}{x^3}$$ and doing so using Maclaurin expansion.

Now $sin x$ expands to $x -\frac{x^3}{3!}x^3 + O(x^5)$

Which would give $sin(sinx)= (x -\frac{1}{3!}x^3 + O(x^5)) -\frac{1}{3!}(x -\frac{x^3}{3!}x^3 + O(x^5))^3 + O(x^5)$

This expression should simplify to (I have the answer) $x - \frac{1}{3}x^3 + O(x^5)$ But i cannot see how this step comes about. Particulary, how do I handle the term $(x -\frac{x^3}{3!}x^3 + O(x^5))^3$?

Thanks a lot! Alexander

Why not expand like this $\sin(\sin(x))=\sin(x)-(\sin(x))^3/6+\ldots$ ? — Raskolnikov, May 07 '13 at 09:56

score 1 · Accepted Answer · answered May 07 '13 at 09:46

1

You are on the right track.

Since $\sin x=x-\frac{x^3}{6}+O(x^5)$, we have $$ \begin{align} \sin(\sin x)&=x-\frac{x^3}{6}+O(x^5)-\frac{1}{6}(x-\frac{x^3}{6}+O(x^5))^3+O((x-\frac{x^3} {6}+O(x^5))^5)\\ &\text{Expanding $(x-\frac{x^3}{6}+O(x^5))^3=x^3+O(x^5)$, you get}\\ &=x-\frac{x^3}{6}-\frac{1}{6}x^3+O(x^5). \end{align} $$ Now I believe you can continue.

answered May 07 '13 at 09:46

Ma Ming

7,482

Yes I solved it. Thank you! – Alexander May 07 '13 at 11:02

Maclaurin expansion of Sin(Sin(x))

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