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Find asymptotic of recurrence sequence $x_{n+1} = (x_n)^2 + 1$.

How this can be done and if there some general method? For example, for a linear sequence one could find a generating function.

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    Quadratic recursions such as $x\to x^2+1$ notoriously have no general explicit solution. Obviously, $x_n\to\infty$ super exponentially fast in the sense that $$\frac{x_{n+1}}{x_n}\to\infty$$ The most one could do to precise this behaviour could be to determine the limit of $$\frac1{2^n}\log x_n$$ – 2017-02-15
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    @Did, could you please tell more about the last limit? – 2017-02-15
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    You mean, the limit of $(\log x_n)/2^n$? Sure: even to show its existence is not so obvious, but computing it is most probably hopeless. You might have missed the word "notoriously" in my first comment, which refers that the fact that the study of the iteration of quadratic maps $z\to z^2+c$ is in itself a whole topic in mathematics. I am curious, what is the context in which your question arose? – 2017-02-15
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    @Did, I came across it in some undergraduate problems set. The precise formulation is "Find as accurate as possible asymptotic of a sequence $x_{n+1} = x_n{}^2 + 5, x_0 = 0$". – 2017-02-15
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    maybe a few links are helpful here: https://en.wikipedia.org/wiki/Mandelbrot_set and http://oeis.org/search?q=0%2C+1%2C+2%2C+5%2C+26&language=english&go=Search – 2017-02-15
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    Elementary computations yield that the sequence $$y_n=2^{-n}\log x_n$$ solves the recursion $$y_n=y_{n-1}+2^{-n}\log(1+x_{n-1}^{-2})$$ hence $$y_n=\log x_0+\sum_{k=0}^{n-1}2^{-k-1}\log(1+x_k^{-2})$$ In particular, $x_k\geqslant x_0$ for every $k$ hence $$\log x_0$n$ and $(y_n)$ is increasing hence $y_n\to\log\ell(x_0)$ with $$\ell(x_0)=x_0\prod_{n=0}^\infty(1+x_n^{-2})^{1/2^{n+1}}$$ and $$x_0<\ell(x_0)$$x_n\approx\ell(x_0)^{2^n}$$ – 2017-02-15
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    The comment above applies to the recursion $x\to x^2+1$ and to $x_0>0$. If $x_0=0$, use $x_1>0$ as starting point. The same approach yields the asymptotics of the recursion $x\to x^2+c$ for every $c>0$. – 2017-02-15

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