How to solve the following recurrence relation, assuming that $x_0 > 1$:
$$x_{n+1} = \frac{x^2_n + 1}{x_n}$$ Am I allowed to divide the fraction, that is $x_{n+1} = x_n + \frac{1}{x_n}$?
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I doubt there is a closed form for an arbitrary $x_0$. But you can investigate convergence/divergence. Is this what you are asking? And yes, you can separate the fraction like that. Provided the first one makes sense in the first place ($x_n\neq 0$). – Julien Jul 02 '13 at 06:01
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@julien: There is a closed form :-). – copper.hat Jul 02 '13 at 06:20
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@copper.hat Ah, ah, right. I meant for $x_n$... – Julien Jul 02 '13 at 06:24
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@julien: That would be interesting... – copper.hat Jul 02 '13 at 06:28
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Hello, welcome to Math.SE. Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – Lord_Farin Jul 02 '13 at 06:35
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Special case of https://math.stackexchange.com/questions/1130082 – doraemonpaul Aug 31 '17 at 16:04
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If $x_n \neq 0$ you can write $x_{n+1} = x_n + \frac{1}{x_n}$. Since $x_n$ and $\frac{1}{x_n}$ have the same sign, it is clear that if $x_n>0$ then $x_{n+1}>x_n$. Hence $x_n$ is increasing.
There are two possibilities, if $x_n$ has an upper bound, then, since it is increasing, we have $x_n \uparrow x$, and since $x_{n+1} - x_n = \frac{1}{x_n}$, we see the right hand side converges to $\frac{1}{x}$ and the left hand side converges to $0$, a contradiction. Hence $x_n$ has no upper bound, that is, $x_n \uparrow +\infty$.
A more or less identical argument shows that if $x_0 <0$, then $x_n \downarrow -\infty$.
Hence we have $\lim_n x_n = \begin{cases} -\infty, & x_0 <0 \\ +\infty & x_0 >0 \end{cases}$.
copper.hat
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