Let $x_0=0$. Define $x_{n+1}=\cos x_n$ for every $n\ge 0$

Question

Let $x_0=0$. Define $x_{n+1}=\cos x_n$ for every $n\ge0$. Then

A) $\{x_n\}$ is increasing and convergent

B) $\{x_n\}$ is decreasing and convergent

C) $\{x_n\}$ is convergent and $x_{2n}\lt\lim_{m\to\infty} x_m\lt x_{2n+1} $ for every $n\in\Bbb N$

D) $\{x_n\}$ is not convergent

Attempt: $x_0=0,x_1=1$, and $0\le x_n\le1$ for $n\gt 1$

So this sequence is non constant and bounded, Hence convergent But, I am unable to pick the right option.

A sequence can be non-constant and bounded and still be divergent. — Paul, Nov 23 '15 at 15:19
This sequence has been investigated (for example) here: http://math.stackexchange.com/questions/56770/what-does-recursive-cosine-sequence-converge-to. In particular, this answer http://math.stackexchange.com/a/56786/42969 describes the behaviour of the sequence precisely (if I understand it correctly). — Martin R, Nov 23 '15 at 15:20
@Paul you are right $(-1)^n$ is non constant and bounded but not convergent — Chiranjeev_Kumar, Nov 23 '15 at 15:21
@MartinR i have seen this discussion but i am not getting the part $x_{2n}\lt\lim_{m\to\infty} x_m\lt x_{2n+1}$ — Chiranjeev_Kumar, Nov 23 '15 at 15:23
If you already have seen a solution to the problem, why don't you mention that in your question? You could ask the author for clarification. — Martin R, Nov 23 '15 at 15:24
Sorry I was editing my confusion, but accidently i pressed the enter button and question is posted — Chiranjeev_Kumar, Nov 23 '15 at 15:26

score 1 · Answer 1 · answered Nov 23 '15 at 15:30

Finding the first $3$ values allows you to discount options A) and B). It now suffices to show that the sequence converges.

One helpful observation to that effect is that $\left| \frac d{dx} \cos(x) \right| \leq \alpha < 1$ on $[0,1]$ for some $\alpha$.

1 Answers1