## 175 -A problem from Homework sets 9, 10

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. $x_0=npi$ for some $n=0,pm1,pm2,dots$

In this case $x_1=x_2=dots=npi$, so the sequence trivially converges.

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

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

First, notice that if $2npi, then we can write $x_i=2npi+t$ 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 $sintheta 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, $f'(theta)=cos(theta)-1le0$ for all $theta$, so $f$ is always decreasing, and the result follows.

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

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

We have shown (by induction) that the sequence ${x_i}_{ige0}$ is strictly increasing and bounded above (by $(2n+1)pi$). Thus, it converges. If $L$ is its limit, then

$L=lim_{itoinfty}x_{i+1}=lim_{itoinfty}x_i+sin(x_i)=L+sin(L),$

so $sin(L)=0$ and since $2npi, 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 $lim_{itoinfty}x_i=(2n+1)pi.$

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 $tin(0,pi)$, so $sin(x_i)=-sin(t)>-t$ and therefore $x_{i+1}>(2n+1)pi$. It follows that the sequence ${x_i}_{ige0}$ 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.