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$a_n=\frac12 a_{n-1}^2-4\quad \forall\quad n\ge2,a_1=\frac{20}{3}.\text{Find } a_n$

My Working : claim: $a_n=2t^{2^n}+2t^{-2^n},\quad\text{where }t=\sqrt 3 $

Proof: let $a_k$ be $2t^{2^k}+2t^{-2^k}$ for some $k\in\mathbb{N}$

$\implies a_{k+1}=\frac12(2t^{2^k}+2t^{-2^k})^2-4=2t^{2^{k+1}}+2t^{-2^{k+1}}$

Is there any other way ? Also, how do I deal with $T_n=aT_{n-1}^2+b$ type recurrence relations in general?

Kamal Saleh
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Makar
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  • Only for some $a$, $b$ can we give an elementary solution. –  Jan 27 '23 at 13:59
  • If you divide both sides of the recursion by 2 and let $b_n=a_n/2$ for all $n$, the recurrence becomes $b_{n}=b_n^2-2$. That may be (slightly) easier to work with. – Semiclassical Jan 27 '23 at 14:13

1 Answers1

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There is no method for solving (or even predicting the number of solutions) of nonlinear recurrence relations $a_{n+1} = f(a_0,\ldots,a_n)$ in general. However, when $f = f(a_n) = \alpha a_n^2 + \beta a_n + \gamma$ is a quadratic map, closed forms are known to exist only when $\gamma = \frac{\beta^2-2\beta}{4\alpha}$ or $\gamma = \frac{\beta^2-2\beta-8}{4\alpha}$, with solution $a_n = \frac{2t^{2^n}-b}{2\alpha}$ and $a_n = \frac{2t^{2^n}+2t^{-2^n}-b}{2\alpha}$ respectively, where $t$ has to be determined with the help of an initial condition. Your problem corresponds to the second case for $(\alpha,\beta,\gamma) = (\frac{1}{2},0,-4)$.

Abezhiko
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