1
$\begingroup$

How do I prove that iteration $x_{n+1}=\cos x_n$ converges for any $x_0\in \Bbb R$ ?

$\endgroup$

marked as duplicate by Watson, Rohan, Stefan4024, LutzL, Asaf Karagila Dec 25 '16 at 11:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ use the fixed point theorem of Banach $\endgroup$ – Dr. Sonnhard Graubner Dec 25 '16 at 11:08
  • $\begingroup$ There is a lot of duplicates, about 5 or so. $\endgroup$ – mvw Dec 25 '16 at 11:21
  • 3
    $\begingroup$ Come on folks, down votes without giving a comment what is wrong on christmas? $\endgroup$ – mvw Dec 25 '16 at 11:22
  • $\begingroup$ Yep, here is another duplicate math.stackexchange.com/questions/1701935/… $\endgroup$ – rtybase Dec 25 '16 at 12:28
5
$\begingroup$

Hint

WLOG, we can suppose $0\leq x_0\leq \pi$. then $\forall n\geq 2 \;\; |x_n|\leq 1$ and $\forall c\in[-1,1]\;| \sin(c)|\leq \sin(1)<1$. thus, by MVT

$$|x_{n+1}-x_n|<\sin(1)|x_n-x_{n-1}|.$$

From here, you prove that $(x_n)$ is Cauchy and converges to the fixed point.

$\endgroup$
  • $\begingroup$ $x_0\in\Bbb R$ gives $x_1\in[-1,1]$ and $x_2\in[\cos 1,1]$, which gives a little, but insignificantly so, more information than what you wrote. $\endgroup$ – LutzL Dec 25 '16 at 12:34
-2
$\begingroup$

Since $(\mathbf R, |\cdot|)$ is a Banach space and $\cos \colon \mathbf R \to \mathbf R$ fulfills $$|\cos x - \cos y | \leq |x-y|\qquad \text{for all } x,y \in \mathbf R$$ the Banach fixed-point theorem shows that the iterations converges for any $x_0 \in \mathbf R$. [The sequence $x_{n+1}= \cos (x_n)$ converges to the unique fixed point of $\cos$ that is $x_n \to x$ where $x$ is the solution of $x=\cos x$. Computing some $x_n$ gives $x\approx 0.7391$.]

$\endgroup$
  • 2
    $\begingroup$ Don't we need $|f(x)-f(y)|\le\alpha|x-y|$ for some $\alpha\lt1$? $\endgroup$ – robjohn Dec 25 '16 at 11:29
  • $\begingroup$ This is true only when we have the situation that robjohn is saying. However on compact metric space we can take alpha to be exactly 1 and instead of $≤$ we must have strict inequality $<$. But for non compact metric space this is not true. $\endgroup$ – Subhadip Majumder Apr 21 '18 at 15:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.