1. ## A constant function!

True or False:

$\displaystyle x \ + \ \sin x \ + \ \sin(x \ + \ \sin x) \ + \ \sin(x \ + \ \sin x \ + \ \sin(x \ + \ \sin x)) \ + \ \cdots=\pi$ for all $\displaystyle 0 < x < 2\pi.$

2. Let $\displaystyle f(x)=x+\sin x$ for $\displaystyle 0<x<2\pi$. Then $\displaystyle f'(x)=1+\cos x\geq0$, so $\displaystyle f(x)$ is one-one, mapping $\displaystyle I=(0,2\pi)$ to itself.

Define $\displaystyle f_0(x)=f(x)$, $\displaystyle f_n(x)=f(f_{n-1}(x))$. The given function is $\displaystyle G(x)=\lim_{n\to\infty}f_n(x)$, where the limit exists, and $\displaystyle G:I\to I$.

It follows that $\displaystyle f(G(x))=G(x)$, and so $\displaystyle G(x)+\sin G(x)=G(x)$.

Therefore $\displaystyle \sin G(x)=0$, which implies that $\displaystyle G(x)=\pi$ on $\displaystyle I$.

As for the limit, we know that $\displaystyle 0<\frac{\sin x}x\leq 1$ for $\displaystyle -\pi<x<\pi$, and so $\displaystyle 0\leq 1-\frac{\sin(\pi-x)}{\pi-x}=1-\frac{\sin x}{\pi-x}<1$ for $\displaystyle x\in I$.

Given $\displaystyle \epsilon>0$ we may choose $\displaystyle k<1$ depending on $\displaystyle \epsilon$ such that $\displaystyle 1-\frac{\sin x}{\pi-x}<k$ for all $\displaystyle x\in[\epsilon,2\pi-\epsilon]$.

It follows that $\displaystyle |\pi-x-\sin x|<k|\pi-x|$ for all $\displaystyle x\in[\epsilon,2\pi-\epsilon]$.

This means that $\displaystyle |\pi-f(x)|<k|\pi-x|$ in this interval. Also if $\displaystyle f$ maps $\displaystyle [\epsilon,2\pi-\epsilon]$ to itself, since $\displaystyle f(x)$ is closer to $\displaystyle \pi$ than $\displaystyle x$ is. Thus

$\displaystyle |\pi-f_n(x)|<k|\pi-f_{n-1}(x)|<k^2|\pi-f_{n-2}(x)|<\dots<k^n|\pi-f_0(x)|$. It follows that $\displaystyle G(x)=\lim_{n\to\infty}f_n(x)=\pi$ for all $\displaystyle x$ in the interval $\displaystyle [\epsilon,2\pi-\epsilon]$.

Since $\displaystyle \epsilon$ is arbitrary (get this!) we have $\displaystyle G(x)=\pi$ for all $\displaystyle x\in I$.

If this works I will definitely LOL.