How find $a_{n}$ if $a_{n+1}=\sqrt{2a_n+1}$

Question

let $a_{1}=\dfrac{\sqrt{2}}{4}$ and such $$a_{n+1}=\sqrt{2a_{n}+1}$$ find $a_{n}$

my idea:let $a_{n}=\dfrac{1}{2}\cos{x_{n}}$ $$\Longrightarrow \dfrac{1}{2}\cos{x_{n+1}}=\sqrt{\cos{x_{n}}+1}=\sqrt{2}\cos{\dfrac{x_{n}}{2}}$$

following I can't work.

But I have see this $$a_{1}=\sqrt{2},a_{n+1}=\sqrt{2+a_{n}}$$ This can let $a_{n}=2\cos{x_{n}}$ can work

bceause $$\Longrightarrow 2\cos{x_{n+1}}=\sqrt{2+2\cos{x_{n}}}=2\cos{\dfrac{x_{n}}{2}}$$ so $$\cos{x_{n+1}}=\cos{x_{n}/2},x_{1}=\dfrac{\pi}{4}$$

so $$x_{n}=2\cos{\dfrac{\pi}{2^{n-1}}}$$

This question ought to be rephrased in order to receive a satisfactory answer. You've given a completely satisfactory definition of $a_n$ as the general term of a well-defined sequence. Are you looking specifically for a non-recursive expression? — , Jul 06 '13 at 02:33
this problem can use this methods? http://math.stackexchange.com/questions/336615/how-can-find-this-sequence-a-n1-a-nna-n-1?rq=1 — math110, Jul 06 '13 at 03:03
Of course this isn't a "stamp of death," as Wolfram Alpha can be wrong, but Wolfram Alpha says no solutions exist. — apnorton, Jul 06 '13 at 03:06
No,@anorton,But Wolfram says the second no solutions exist:http://www.wolframalpha.com/input/?i=solve+a_1%3Dsqrt%282%29%2C+a_%28n%2B1%29%3Dsqrt%28%28a_%28n%29%29%2B2%29 — math110, Jul 06 '13 at 03:12

score 3 · Answer 1 · answered Jul 06 '13 at 05:01

Note that if $a_n \gt 0$ then $a_{n+1}\gt 1$. Also if $a_n \lt 3$ then $a_{n+1} \lt 3$. So that bounds the sequence.

The fixed point of the sequence satisfies $a=\sqrt {2a+1}$ - square and solve to obtain $a=1+\sqrt 2$.

It is easy to prove that $a_n$ is increasing to this fixed point.

If you are hoping to guess a general expression for $a_n$ then it has to meet these constraints. However even calculating $a_2$ by hand things look to be getting messy - note: when stuck do work out the first few values.

How find $a_{n}$ if $a_{n+1}=\sqrt{2a_n+1}$

1 Answers1