(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let a_{0}=0 and a_{n+1}=cos(a_{n}) for n≥0.

a) prove that a_{2n}≤a_{2n+2}≤a_{2n+3}≤a_{2n+1}for all n≥0.

b) use mean value theorem to find a number r<1 such that |a_{n+2}-a_{n+1}| ≤ r|a_{n}-a_{n+1}| for all n≥0. Using this, prove that the sequence {a_{n}} is Cauchy.

2. Relevant equations

N/A

3. The attempt at a solution

a) I proved part (a) using induction. Furthermore, I showed that all terms of the sequence is bounded between 0 and 1.

b) Claim: |a_{n+2}-a_{n+1}| ≤ 0.85|a_{n+1}-a_{n}|

Assuming the claim is true, I have that

|a_{m}-a_{n}|=|a_{m}-a_{m-1}|+...+|a_{n+1}-a_{n}|≤0.85^{m-1}|a_{1}-a_{0}|+...+0.85^{n}|a_{1}-a_{0}|≤0.85^{n}/(1-0.85) ->0 as n->∞. Thus, by squeeze theorem, the sequence is Cauchy.

But I am not sure how to prove the claim. If I consider a special case a_{n+1}≥a_{n}(which is clearly not true in our case), i.e. assuming the sequence is nondecreasing, then by mean value theorem, a_{n+2}-a_{n+1}=-sin(c)(a_{n+1}-a_{n}) where c must be somewhere between 0 and 1, so |sin(c)|<0.85, and the claim is proved in this special case. But the trouble is that our sequence is actually not even monotone (as shown in part a), sohow can we prove the claim in our case?

Any help is much appreciated! :)

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# Homework Help: Mean value theorem & Cauchy sequence

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