How can I prove that x_{n+1} = x_n \sin(x_n) converges

Mlejd5

Mlejd5

Answered question

2022-01-27

How can I prove that xn+1=xnsin(xn) converges for any x0?

Answer & Explanation

Anabelle Miller

Anabelle Miller

Beginner2022-01-28Added 12 answers

Let yn=|xn|
Then yn+1=ynsinynynn
So yn is decreasing and bounded below by 0. Therefore it converges to some limit l. By passing to the limit in (*) we get that either l=0 or sinl=1 giving the family of solutions discussed in the comments, namely l=2kπ+π2
Now note that if yn converges to 0 then xn also converges to 0. If yn converges to a nonzero value l then a little more work will also prove that xn also converges to either l or -l because for n large enough yn will be between l and l+π2 hence xn will have constant sign
Either way, xn will also be convergent

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