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is it possible to find general solution of the recurrence relation such as

$a_{n}=a_{n-1}+A\centerdot \cos(a_{n-1})$

where $a_{0}=0$ and $A \ll 1$

EDIT: At least for

$a_{n}=a_{n-1}+A - \frac{A}{2!}\centerdot a^2_{n-1} + \frac{A}{4!}\centerdot a^4_{n-1}$

where cosine is expanded with Taylor Series with 3 terms

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    I'd be surprised.2012-04-10
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    Hi, @joriki so it is not possible?2012-04-10
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    If I knew how to prove that it's not possible, I would have said so. I would just be rather surprised because the relation has $a_{n-1}$ both inside and outside a transcendental function, which is usually a sign of intractability. But I've been surprised before :-)2012-04-10
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    But what if I expand cosine with Taylor series and use only first 3-5 terms? Is is possible to solve recurrence relation in this form? Ex: $a_{n}=a_{n-1}+A-A\centerdot {a_{n-1}}^2/2!+A\centerdot {a_{n-1}}^4/4!$2012-04-10
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    I'd still be surprised.2012-04-10
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    Then this is not the recurrence relation which you consider in your post. // On the other hand, to show that, for every $0\lt A\leqslant1$, $a_n\to\pi/2$ when $n\to\infty$ is standard.2012-04-10
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    Nonlinear recurrences such as yours are so poorly understood that it would be quite a shock (at least to me) if your recurrence has a closed form. For related fun, look up the Dottie number.2012-04-10
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    To understand @joriki's and J.M.'s cautious words, consider the (nonlinear) logistic recurrence relation $x_{n+1}=4x_n(1-x_n)$ with $0\leqslant x_0\leqslant1$. But this is *the exception that proves the rule*, as the saying goes.2012-04-10
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    @Didier actually $a_n\to\pi/2$ for me. J.M. thanks, I'm going to look up the Dottie number.2012-04-10
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    Do you mean that $a_n\to\pi/2$ (which is what I said), or that you know how to prove that $a_n\to\pi/2$, or that your question is to know how to prove that $a_n\to\pi/2$?2012-04-10
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    [Here's a link](http://en.wikipedia.org/wiki/Logistic_map#Solution_in_some_cases) for @Didier's "counterexample".2012-04-10
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    I mean that for my application of this recurrence relation $a_n\to\pi/2$ :)2012-04-10
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    Thanks for link @joriki, I've read that article..2012-04-10
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    I do not understand your answer. Once again: are you interested in the proof that $a_n\to\pi/2$, or are you not?2012-04-10
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    @Didier no, I'm not interested in proof of that. I'm interested in general solution or good approximation if possible.2012-04-10

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