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Define a sequence by $b_1=\sqrt{2}, b_2=\sqrt{2+\sqrt{2}}$

and in general $b_{n+1}=\sqrt{2+b_n}$

I'm having a hard time solving what $b_n$ is using recurrence relations.

Blex
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ChrisV
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2 Answers2

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Set $b_n=2a_n$. Then we have: $$ a_{n+1}=\sqrt{\frac{1+a_n}{2}}$$ and assuming $a_n=\cos\theta$ it follows that $a_{n+1}=\cos\frac{\theta}{2}$. Since $a_1=\cos\frac{\pi}{4}$,

$$ b_n = 2 \cos\frac{\pi}{2^{n+1}}$$

is easily proved by induction.

Jack D'Aurizio
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    This actually answers the question, in contrast to the other attempts so far posted, which answer a different question. – Mark Bennet May 29 '15 at 20:03
  • how did you go about setting this up im still a bit lost on how you went about this to end up getting b_n – ChrisV May 29 '15 at 20:06
  • @ChrisV: the point is that, with a minor tweak (i.e. switching to $a_n$s), the recurrence relation for the $b_n$s is just the half-angle formula for the cosine function. See also http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula – Jack D'Aurizio May 29 '15 at 20:27
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What you are looking for is the limit of the sequence $b_n$. It is $2$. When learning about a sequence of real numbers, you often want to find a thing or two about it: if it ever converges, and if so what is the limit? your sequence is increasing, and bounded above by $2$, and thus there exists a limit called L. It must be true that $L = \sqrt{2+L} \to L = 2$.

DeepSea
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  • what would be the general formula for b_n? – ChrisV May 29 '15 at 20:01
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    If the question is to find an explicit formula for $b_n$, this is not an answer. – Jack D'Aurizio May 29 '15 at 20:04
  • why is it that we can just use our general formula for the next term and solve for the limit as you did? – ChrisV May 29 '15 at 20:30
  • It happens to be that your sequence can be solved using a "closed form" formula like above, but most of the times, sequences that holds more "secret" infor-mation can at most be discussed via its limit. – DeepSea May 29 '15 at 20:42