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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

mergenchik
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  • I'd be surprised. – joriki Apr 10 '12 at 09:54
  • Hi, @joriki so it is not possible? – mergenchik Apr 10 '12 at 09:58
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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 :-) – joriki Apr 10 '12 at 10:02
  • 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: $a_{n}=a_{n-1}+A-A\centerdot {a_{n-1}}^2/2!+A\centerdot {a_{n-1}}^4/4!$ – mergenchik Apr 10 '12 at 10:10
  • I'd still be surprised. – joriki Apr 10 '12 at 10:25
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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. – Did Apr 10 '12 at 10:26
  • 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. – J. M. ain't a mathematician Apr 10 '12 at 10:27
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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. – Did Apr 10 '12 at 10:37
  • @Didier actually $a_n\to\pi/2$ for me. J.M. thanks, I'm going to look up the Dottie number. – mergenchik Apr 10 '12 at 10:38
  • 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 $a_n\to\pi/2$? – Did Apr 10 '12 at 10:40
  • Here's a link for @Didier's "counterexample". – joriki Apr 10 '12 at 10:40
  • I mean that for my application of this recurrence relation $a_n\to\pi/2$ :) – mergenchik Apr 10 '12 at 10:41
  • Thanks for link @joriki, I've read that article.. – mergenchik Apr 10 '12 at 10:43
  • I do not understand your answer. Once again: are you interested in the proof that $a_n\to\pi/2$, or are you not? – Did Apr 10 '12 at 11:11
  • @Didier no, I'm not interested in proof of that. I'm interested in general solution or good approximation if possible. – mergenchik Apr 10 '12 at 11:15

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Since the comments above led to the concluson that good approximation(s) was something the OP was after, here are some.

Let $f:a\mapsto a+A\cdot\cos(a)$ and assume that $0\lt A\leqslant1$. Then $f$ is increasing on $[0,\pi/2]$ and $f(a)\gt a$ for every $a$ in $[0,\pi/2)$ hence the sequence $(a_n)$ defined by $a_0=0$ and $a_{n+1}=f(a_n)$ for every $n\geqslant0$ is increasing with limit $\pi/2$.

Let $b_n=(\pi/2)-a_n$, then $b_0=\pi/2$, $(b_n)$ is decreasing to the limit $0$ and $b_{n+1}=b_n-A\cdot\sin(b_n)$.

Since $b_n=o(1)$, $\sin(b_n)=b_n+o(b_n)$ and $b_{n+1}=(1-A)\cdot b_n+o(b_n)$. Assume first that $\color{red}{0\lt A\lt1}$. Then a simple recursion yields $$ \frac{b_n}{(1-A)^n}=B-\frac{A}{1-A}B_n,\quad B=\sum_{k=0}^{+\infty}\frac{b_k-\sin(b_k)}{(1-A)^k},\quad B_n=\sum_{k=n}^{+\infty}\frac{b_k-\sin(b_k)}{(1-A)^k}. $$ Since $\sin(b)=b-\frac16b^3+o(b^3)$ when $b\to0$, one can show that $B$ converges and that $B_n$ is at most of order $(1-A)^{2n}$. Hence, with the finite $\color{red}{B\gt0}$ defined above, $$ \color{red}{a_n=(\pi/2)-B\cdot(1-A)^n+O((1-A)^{3n})}. $$ When $\color{blue}{A=1}$, $b_{n+1}=b_n-\sin(b_n)\sim\frac16b_n^3$ and one can show that $\log b_n=-c\cdot3^n+o(3^n)$ for some finite positive $c$, hence there exists some $\color{blue}{0\lt C\lt1}$ such that $$ \color{blue}{a_n=(\pi/2)-C^{3^n+o(3^n)}}. $$

Did
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  • couldn't catch after $b_{n+1} = b_{n} - A\cdot \sin(b_{n})$. Verified that $a_{n}+b_{n}=\pi/2$ for $0\lt A \lt 1$ with program :). Couldn't understand this line "Since $b_n=o(1), sin(b_n)=b_n+o(b_n)$" – mergenchik Apr 10 '12 at 14:19
  • If $a_{n+1}=a_n+A\cos(a_n)$ and $b_n=(\pi/2)-a_n$, $b_{n+1}=(\pi/2)-a_n-A\cos(a_n)$ and $\cos(a_n)=\sin(b_n)$ hence $b_{n+1}=b_n-A\sin(b_n)$. // How one can verify that $a_n+b_n=\pi/2$ (with a program!?) since this is the definition of $b_n$ somewhat baffles me. // When $x=o(1)$ (that is, $x\to0$), $\sin(x)=x+o(x)$. Since $b_n=o(1)$, $\sin(b_n)=b_n+o(b_n)$. – Did Apr 10 '12 at 14:43
  • :D sorry. My brain stopped since thinking for a long time for this problem. – mergenchik Apr 10 '12 at 14:47
  • will $B$ converge? As $k\to +\infty$ then $b_k - \sin(b_k) \to \pi /2$ but since $A\lt1$ then $(1-A)^k \to 0$. or I'm missing something? – mergenchik Apr 10 '12 at 15:06
  • Yes $B$ converges. Proof: Since $b_n\to0$ and $b_{n+1}=(1-A)b_n+o(b_n)$, for every $1-A\lt F\lt1$, $b_{n+1}\leqslant Fb_n$ for $n$ large enough. Hence $b_n\leqslant cF^n$ and $b_n-\sin(b_n)\leqslant\frac16b_n^3\leqslant c'F^{3n}$. Choosing $F\gt1-A$ such that $F^3\lt1-A$, one sees that $B$ converges. End of the proof. – Did Apr 10 '12 at 15:16
  • Thanks @Didier. Yes $B$ converges :). For large values of $n$ not bad results, but for small will use recursion. – mergenchik Apr 10 '12 at 16:30