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Is there any general method to solve a non linear recurrence relation. For example:

$$ \begin{eqnarray}f_{n+1}&=&2f_{n}^2-1 \\ f_1&=&4 \end{eqnarray}$$

Is there any way to solve it?

dshin
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Amit Kumar Basistha
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  • For the linear case, closed solutions are available. For the non linear case, closed form solutions may be found only in particular situations. – the_candyman Apr 02 '22 at 19:05
  • No general method exists. See here for a discussion – HallaSurvivor Apr 02 '22 at 20:14
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    the two nice monic quadratic without linear terms are $a_{n+1} = a_n^2 $ and $b_{n+1} = b_n^2 - 2 .$ Maybe there is a useful way to express your $f_n$ in terms of $b_n$ – Will Jagy Apr 02 '22 at 21:05
  • This problem has solution $f_n=\cosh(2^{n-1}\operatorname{arcosh}4)$ (see here if those functions are unfamiliar). – J.G. Apr 02 '22 at 21:21

1 Answers1

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TO SAVE WRITING, I am going to change the initial index, instead of $f_1 = 4$ I am switching to $\color{red}{f_0=4}$

that worked. Let $ g_n = 2 f_n, $ so that $f_n = \frac{g_n}{2}.$ This leads to $\color{red}{g_0=8}$

We get to the solvable $$ g_{n+1} = g_n^2 - 2 $$

I like to write this as $$g_0 = G + \frac{1}{G}, $$ then $$g_1 = G^2 + \frac{1}{G^2}, $$ $$g_2 = G^4 + \frac{1}{G^4}, $$ $$g_3 = G^8 + \frac{1}{G^8}, $$ generally $$g_n = G^{2^n} + \frac{1}{G^{2^n}}, $$

From $\color{red}{g_0=8}$ we need $ G + \frac{1}{G}=8, $ we may take $G = 4 + \sqrt{15}$ and $\frac{1}{G} = 4 - \sqrt{15}$

Will Jagy
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