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Remy J. Cano in his private email described the sequence of real numbers, recursively defined as $$a(n) = a(n-1)+\frac{2 \cdot \cos(\frac{a(n-1)}{2})}{2 \cdot \sin(\frac{a(n-1)}{2})-1},a(0)=0$$

This sequence converges to $-\pi$

that is for $n \rightarrow \infty $

$ a(n) \rightarrow -\pi$

Why this recursively defined sequence of real numbers converges to $-\pi$ ?

Alex
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    Taking the limit of both sides of the equation tells a lot. –  Jul 04 '14 at 17:25
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    Who is Remy J. Cano? Why is that relevant? – Ethan Splaver Jul 04 '14 at 18:47
  • I mentioned Remy J Cano as the source of this converging recurrence. Are there any other prior public references to it being known ? – Alex Jul 04 '14 at 20:43
  • Basically I posted this question with the purpose to make the record of this recursion; though it does not converge efficiently fast - it may still be worthwhile to make the record of its existence. – Alex Jul 05 '14 at 18:17
  • Here is an additional question (which is a natural extension of the original question) - what would be the method to try to convert this given recursive formula into explicit close form ? – Alex Jul 06 '14 at 15:03
  • Another additional related question towards bounty is - would it be possible to put some sensible upper limit - in the form of the integer numerator/denominator ratio - onto values of intermediate convergents, which arise in the discussed recursion. – Alex Jul 09 '14 at 16:49
  • @Alex - Why did you choose not to award the bounty? – Nathaniel Bubis Jul 15 '14 at 15:23
  • @nbubis But it went to you automatically (from how I understand the rules) - correct ? – Alex Jul 15 '14 at 15:46
  • @nbubis - I was hoping though that you will return to address additional questions (see my comments made on July 6 and July 9 ) towards which I really issued the bounty :-) – Alex Jul 15 '14 at 15:56
  • @Alex, for next time, the content of the bounty when offered is the one that counts, not whatever comments you added later on. – Nathaniel Bubis Jul 15 '14 at 17:28
  • @nbubis - thanks for the info - it worked out that way anyway. – Alex Jul 15 '14 at 19:53

2 Answers2

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Look at: $$T(x) = x+\frac{2 \cdot \cos\left(\frac{x}{2}\right)}{2 \cdot \sin\left(\frac{x}{2}\right)-1}$$ We know that $a_1=-2$. When $x \in X\equiv(-2\pi,0)$, $T(x)\in X$ as well. Moreover, since $T(x)$ has a continuous derivative in $X$, and: $$|T'(x)|<1\ \mid\ x\in X,$$ $T(x)$ is a contraction mapping. Thus, by the Banach Fixed Point Theorem, there exists a single stationary point. Since $x=-\pi$ is such a point, the series converges to $x=-\pi$.

Nathaniel Bubis
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The main reason is that $\cot(x)$ function have zeros at odd multiples of $\pi$. It is relatively similar but not the same than the original function I mailed. Almost any other thing (real number) used instead of your choosen $a_{0}$ will behave in the same way, since there are infinitely many multiples of $\pi$, therefore you always can set by convention any stop point (consisting in: To reach some desired precision), getting an approximation to an odd integer multiple of $\pi$. Computer Software with smart capabilities for dealing with trigonometric functions like PARI, have already pre-computed $\pi$ to any allowed precision, so it is not difficult to realize the described convergence by inspection dividing $a_{n}$ between $\pi$. (From: R. J. Cano, Physics student and CS amateur)

Remy Cano
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