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The recurrence $x_{n+1} = \cos(x_n) $ seems to converge to a value around .739 no matter what number is chosen for $x_0$, even complex numbers. What is the exact number that this recurrence converges to?

Thomas
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  • If the limit exists, then $L=\cos(L)$. You can't solve this analytically.https://www.wolframalpha.com/input/?i=x%3Dcos(x) –  Jan 08 '17 at 19:10
  • Is there a proof that $L=cos(L)$ can't be solved analytically? – Thomas Jan 08 '17 at 19:27
  • http://math.stackexchange.com/questions/46934/what-is-the-solution-of-cosx-x –  Jan 08 '17 at 19:28

2 Answers2

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If the limit exists, then $L=\cos(L)$. The solution to this equation is the Dottie number.

0

Hints:

  • $f: x\mapsto \cos(\cos x)$ is increasing on $[0,1]$.

  • For all $n\ge 1$, $x_n \in [0,1]$

  • Let $a_n = x_{2n}$ and $b_n = x_{2n+1}$. Use $f$ to prove that $(a_n)$ and $(b_n)$ both converge to the same value. Conclude that $(x_n)$ converges (the value to which it converges is the unique fixed point of $\cos$).