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

Question

Luke Alvarez · Accepted Answer

Based on this power series expansion of #sin(x)#:
#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(sin(x))#. What we can do is treat the inner #sin(x)# as the argument for the outer #sin(x)#, similar to how #x# is the argument for #ln(x)#. Doing that, we can just replace #x# with #sin(x)#: #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(sin(x))#.  Unfortunately, it's not the Taylor series expansion (about #x=0#).  Taylor series expansions are power series (infinite sums of the form #\sum_{n=0}^{\infty}a_{n}x^{n}#.  To get the Taylor series expansion from the above approach, you'd have to replace each of the #sin^{2n+1}(x)# 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 #\sum_{n=0}^{\infty}a_{n}x^{n}#. Yikes! Probably a simpler approach overall is to just find the first few nonzero terms by differentiation: #f'(x)=cos(sin(x))\cdot cos(x)#, #f''(x)=-sin(x)cos(sin(x))-sin(sin(x))\cdot cos^{2}(x)#, 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(sin(x))=x-x^3/3+x^5/10-(8x^7)/315+(13x^9)/2520-(47x^11)/49896+\cdots#

Scarlett Allison · Answer

undefined The Taylor series expansion for sin(sin(x)) is:

sin(sin(x)) ≈ x - (1/3!)x^3 + (2/5!)x^5 - (5/7!)x^7 + ...