# A constant function!

• July 6th 2009, 12:00 PM
NonCommAlg
A constant function!
True or False:

$x \ + \ \sin x \ + \ \sin(x \ + \ \sin x) \ + \ \sin(x \ + \ \sin x \ + \ \sin(x \ + \ \sin x)) \ + \ \cdots=\pi$ for all $0 < x < 2\pi.$
• July 6th 2009, 05:08 PM
halbard
Let $f(x)=x+\sin x$ for $0. Then $f'(x)=1+\cos x\geq0$, so $f(x)$ is one-one, mapping $I=(0,2\pi)$ to itself.

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

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

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

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

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

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

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

$|\pi-f_n(x)|. It follows that $G(x)=\lim_{n\to\infty}f_n(x)=\pi$ for all $x$ in the interval $[\epsilon,2\pi-\epsilon]$.

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

If this works I will definitely LOL.