175 -A problem from Homework sets 9, 10

November 18, 2008

I want to show here how to solve the following problem from this week’s homework set:

Starting with a given $x_0$, define the subsequent terms of a sequence by setting $x_{n+1}=x_n+\sin(x_n).$ Determine whether the sequence $\{x_n\}$ converges, and if it does, find its limit.

This is a nice simple example of a (discrete) dynamical system in one variable. It turns out that the sequence always converges, but the limit depends on the value of $x_0.$

• Case 1. $x_0=n\pi$ for some $n=0,\pm1,\pm2,\dots$

In this case $x_1=x_2=\dots=n\pi$, so the sequence trivially converges.

• Case 2. $2n\pi for some $n=0,\pm1,\pm2,\dots$

In this case I will show that $x_0 and that $\lim_{i\to\infty}x_i=(2n+1)\pi.$

First, notice that if $2n\pi, then we can write $x_i=2n\pi+t$ for some $t$ with $0 and $\sin(x_i)=\sin(t)>0$, so $x_{i+1}=x_i+\sin(x_i)>x_i.$

Second, recall that $\sin\theta<\theta$ for all $\theta>0$. You are probably familiar with this inequality from Calculus I; if not, one can prove it easily as follows: Let $f(\theta)=\sin(\theta)-\theta$, so $f(0)=0$. Also, $f'(\theta)=\cos(\theta)-1\le0$ for all $\theta$, so $f$ is always decreasing, and the result follows.

Also, recall that $\sin(t)=\sin(\pi-t)$, so

$x_{i+1}=x_i+\sin(x_i)=2n\pi+t+\sin(t)$ $=2n\pi+t+\sin(\pi-t)<2n\pi+t+(\pi-t)=(2n+1)\pi.$

We have shown (by induction) that the sequence $\{x_i\}_{i\ge0}$ is strictly increasing and bounded above (by $(2n+1)\pi$). Thus, it converges. If $L$ is its limit, then

$L=\lim_{i\to\infty}x_{i+1}=\lim_{i\to\infty}x_i+\sin(x_i)=L+\sin(L),$

so $\sin(L)=0$ and since $2n\pi, it follows that $L=(2n+1)\pi.$

• Case 3. $(2n+1)\pi for some $n=0,\pm1,\pm2,\dots$

In this case, $x_0>x_1>\dots$ and $\lim_{i\to\infty}x_i=(2n+1)\pi.$

The argument is very similar to the one for Case 2: If $(2n+1)\pi, then $\sin(x_i)<0$ so $x_{i+1}, and $x_i=(2n+1)\pi+t$ for some $t\in(0,\pi)$, so $\sin(x_i)=-\sin(t)>-t$ and therefore $x_{i+1}>(2n+1)\pi$. It follows that the sequence $\{x_i\}_{i\ge0}$ is decreasing and bounded below (by $(2n+1)\pi$), so it converges. As before, the limit must in fact be $(2n+1)\pi$, and we are done.