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<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_{itoinfty}x_i=(2n+1)pi.

First, notice that if 2npi<x_i<(2n+1)pi, then we can write x_i=2npi+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 sintheta<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)-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


so sin(L)=0 and since 2npi<Lle(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_{itoinfty}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 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.


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