Iterated Function System for Cosine

Question

So, if we let $f_1(x) = 2x^2-1$ and say $f_{n+1}(x) = f(f_{n}(x))$ then some graphing and experimenting seems to indicate that $\lim_{n\to\infty} f_n(2^{-n}x) = \cos(x)$. They converge pretty quickly for small values of x, according to wolfram $|f_3(2^{-3})-\cos(1)| < .0023$. Is there anyway to prove that this system will converge to Cosine?

Answer 1

Note that $f_n(\cos t)=\cos(2^n t)$. Indeed this is true for $n=1$, and if it is true for $n$ then $f_{n+1}(\cos t)= f_1(\cos(2^nt))=\cos(2^{n+1}t)$. Thus, $$f_n(2^{-n}x)=\cos(2^n\arccos(2^{-n}x))$$ So, for large $n$ we have $$\eqalign{f_n(2^{-n}x)&=\cos\left(2^n\left(\frac{\pi}{2}-\arcsin(2^{-n}x)\right)\right)=\cos(2^n\arcsin(2^{-n}x))\cr &=\cos\left(2^n\left(2^{-n}x+\mathcal{O}(2^{-3n})\right)\right)\cr &=\cos\left(x+\mathcal{O}(2^{-2n})\right)=\cos(x)+\mathcal{O}(2^{-2n}) }$$ This proves that $\lim_{n\to\infty}f_n(2^{-n}x)=\cos x$, and explains why this conergence is fast.