A constant function!

**NonCommAlg** · July 6th 2009, 12:00 PM

True or False:

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

**halbard** · July 6th 2009, 05:08 PM

Let $f(x)=x+\sin x$ for $0<x<2\pi$ . 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 $-\pi<x<\pi$ , 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 $1-\frac{\sin x}{\pi-x}<k$ for all $x\in[\epsilon,2\pi-\epsilon]$ .

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

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

$|\pi-f_n(x)|<k|\pi-f_{n-1}(x)|<k^2|\pi-f_{n-2}(x)|<\dots<k^n|\pi-f_0(x)|$ . 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.

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