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=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<x_0<(2n+1)\pi for some n=0,\pm1,\pm2,\dots

In this case I will show that x_0<x_1<\dots and that \lim_{i\to\infty}x_i=(2n+1)\pi.

First, notice that if 2n\pi<x_i<(2n+1)\pi, then we can write x_i=2n\pi+t for some t with 0<t<\pi 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


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

  • Case 3. (2n+1)\pi<x_0<(2n+2)\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<x_i<(2n+2)\pi, then \sin(x_i)<0 so x_{i+1}<x_i, 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.


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