# What is the Taylor Series Expansion for sin(sin(x))?

May 18, 2015

Based on this power series expansion of $\sin \left(x\right)$:
sin(x) = x-x^3/(3!)+x^5/(5!)-x^7/(7!)+...
= (-1)^0[x^(2*0+1)/((2*0+1)!)]+(-1)^1[x^(2*1+1)/((2*1+1)!)]+(-1)^2[x^(2*2+1)/((2*2+1)!)]+(-1)^3[x^(2*3+1)/((2*3+1)!)]+...
= sum_(n=0)^(\infty)((-1)^nx^(2n+1))/((2n+1)!)
we can derive it for $\sin \left(\sin \left(x\right)\right)$.

What we can do is treat the inner $\sin \left(x\right)$ as the argument for the outer $\sin \left(x\right)$, similar to how $x$ is the argument for $\ln \left(x\right)$. Doing that, we can just replace $x$ with $\sin \left(x\right)$:

sin(sin(x)) = (sinx)/(1!)-(sinx)^3/(3!)+(sinx)^5/(5!)-(sinx)^7/(7!)+...

= sum_(n=0)^(\infty)((-1)^n(sinx)^(2n+1))/((2n+1)!)

or

sum_(n=0)^(\infty)((-1)^nsin^(2n+1)(x))/((2n+1)!)

This is a series expansion for $\sin \left(\sin \left(x\right)\right)$. Unfortunately, it's not the Taylor series expansion (about $x = 0$). Taylor series expansions are power series (infinite sums of the form $\setminus {\sum}_{n = 0}^{\setminus \infty} {a}_{n} {x}^{n}$. To get the Taylor series expansion from the above approach, you'd have to replace each of the ${\sin}^{2 n + 1} \left(x\right)$ terms with, unfortunately, (\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!})^{2n+1}, which results in

sin(sin(x))=sum_(n=0)^(\infty)((-1)^n)/((2n+1)!)((\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!})^{2n+1}). You'd then have to simplify this into the form $\setminus {\sum}_{n = 0}^{\setminus \infty} {a}_{n} {x}^{n}$. Yikes!

Probably a simpler approach overall is to just find the first few nonzero terms by differentiation:

$f ' \left(x\right) = \cos \left(\sin \left(x\right)\right) \setminus \cdot \cos \left(x\right)$, $f ' ' \left(x\right) = - \sin \left(x\right) \cos \left(\sin \left(x\right)\right) - \sin \left(\sin \left(x\right)\right) \setminus \cdot {\cos}^{2} \left(x\right)$, etc...

Then the Taylor series (about $x = 0$) is:

f(0)+f'(0)x+\frac{1}{2!}f''(0)x^{2}+\cdots=x+\cdots

If you felt like finding more nonzero terms, you could use Wolfram Alpha to help you take more derivatives, or enter: Series[sin[sin[x]],{x,0,10}] into it and get:

$\sin \left(\sin \left(x\right)\right) = x - {x}^{3} / 3 + {x}^{5} / 10 - \frac{8 {x}^{7}}{315} + \frac{13 {x}^{9}}{2520} - \frac{47 {x}^{11}}{49896} + \setminus \cdots$