General term of sequence $a_0=2$ and $a_{n+1}=2a_n^2-1$

Question

Is it possible to find general term of this sequence? $a_0=2$ and $a_{n+1}=2a_n^2-1$

Answer 1

set $\alpha = \cosh^{-1}(2)$ then : $a_n=\cosh(2^n \alpha)$.

It is easy to prove it by induction using $\cosh(2x)=2\cosh^2(x)-1$.

Why $\cosh$ and not $\cos$ ? because $a_0>1$, if $a_0=0.5$ for example I'll use $\cos$.

Answer 2

With $b_{n} =2 a_n$, we get $b_{n+1} = b_n^2-2$. Notice that $(x+\frac{1}{x})^2 -2 = x^2 + \frac{1}{x^2}$. Therefore, if $b_0 = t+\frac{1}{t}$, then

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

In our case $b_0 = 2\cdot 2= (2+\sqrt{3}) + (2-\sqrt{3})$, and so we get

$$a_n = \frac{1}{2}\left(\, (2+\sqrt{3})^{2^n} + (2-\sqrt{3})^{2^n}\right)$$

Note: like with trig or hyp trig subs, perhaps easier to fathom.