Quadratic Recurrences

Question

is there any way to solve 'simple' recurrences like $a_0=2$, $a_{n+1}=(2a_n-1)^2$ or even $a_0=1$, $a_{n+1}=(2a_n+1)^2$ ? Any help much welcomed ! Regards, Knud

Answer 1

Let us denote $b_n = 2a_n -1$, then $$b_0 = 3$$ $$b_{n+1} = 2b_n^2 -1$$

We seek $x$ satisfying $b_0 = \text{cosh}(x) := \frac{e^x+e^{-x}}{2}$ then $$x = \text{arcosh}(3) = \ln(3+2\sqrt{2})=2\ln(\sqrt{2}+1)$$ Now, it is easy to prove that $b_n = \text{cosh}(2^nx)$, indeed $$b_{n+1} = 2\left(\frac{e^{2^n x} +e^{-2^n x}}{2} \right)^2 -1 =\frac{e^{2^{n+1} x} +e^{-2^{n+1} x}}{2} = \text{cosh}(2^{n+1}x)$$

Conclusion:

$$a_n =\frac{\text{cosh}(2^{n+1}\ln(\sqrt{2}+1)) +1}{2} $$