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Let $\beta\in[-1,1]$ and let $a_n$ defined by $ \begin{cases} a_1=\beta, \\ a_{n+1}=\cos(a_n) \end{cases}$

Let $c\in \mathbb{R}$, s.a $\cos(c)=c$.
Prove that: $\lim\limits_{n\to\infty}a_n=c$

My work so far: I know that $c\in(0,1)$.

$|\cos(a_n)-c|=|\cos(a_n)-\cos(c)|=|-2\sin(\frac{a_n+c}{2})\sin(\frac{a_n-c}{2})|\le\frac{1}{2}|a_n+c||a_n-c|$

But I don't know how to continue from here.

Guy Fsone
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Ro168
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  • This is an immediate consequence of the Banach fixpoint theorem. The fixpoint is also unique, if you wondered. –  Jan 29 '18 at 14:31
  • @Math_QED why is that true? cos is not a contraction – Guy Fsone Jan 29 '18 at 14:53
  • @GuyFsone as written below… $\cos$ IS da contraction mapping on the given interval $[-1,1]$ – Gono Jan 29 '18 at 15:27
  • Why isn't cosine a contraction? –  Jan 29 '18 at 15:49
  • @Math_QED to be sure for you what is a contraction? for me it has a lipschitz constant $k<1$ This is totally impossible with cosine you can even prove it I have done this in one post here on MSE – Guy Fsone Jan 29 '18 at 15:53
  • @GuyFsone Are you reading other comments as well? As written TWICE(!) now… cosine on $[-1,1]$ IS a contraction with lipschitz constant $\sin(1) < 1$! – Gono Jan 29 '18 at 17:20
  • @Gono can you prove that ? that is completely wrong take $x=0$ and $x=\pi/2$ then you see it is not true – Guy Fsone Jan 29 '18 at 19:00
  • OFC I can… and Fred did at his answer below. $\frac{\pi}{2} \not\in [-1,1]$ hence your example won't work and $\cos$ is still a contraction on $[-1,1]$ – Gono Jan 29 '18 at 19:02

2 Answers2

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Define $f: [-1,1] \to [-1,1]$ by $f(x)= \cos x$. If $x,y \in [-1,1]$, then there is $t$ between $x$ and $y$ such that

$|f(x)-f(y)| = | \sin t| \cdot |x-y| \le q|x-y|$,

where $q= | \sin 1| <1$.

Now apply Banach's fix point theorem

Fred
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  • How do you get the midle inequality? Not that cos is not a contraction – Guy Fsone Jan 29 '18 at 14:53
  • @GuyFsone, is a contraction if the interval is small enough. – Martín-Blas Pérez Pinilla Jan 29 '18 at 15:15
  • @Martín-BlasPérezPinilla that is not that obvious but let metell you that it is enough to show that $\cos^2x$ is a contraction but not $\cos x$ Moreover I dont get the meaning of contraction in small intervall – Guy Fsone Jan 29 '18 at 15:19
  • @GuyFsone the equality is just mean value theorem for a $t\in (x,y)$ and the inequality holds because $\sin(x)$ is increasing on $[-1,1]$ with $|\sin(-1)| = |\sin(1)| $ and we are done… and as you can see, $\cos$ is a contraction on $[-1,1]$ – Gono Jan 29 '18 at 15:23
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Elaborating a bit more on an answer already given, note that $\cos$ has $-\sin$ as derivative, and note that in the interval $[-1, 1]$, the derivative of $\cos$ in absolute value never exceeds $1$, given that $|(\cos x)'| = |\sin x| < 1$ for $x \in [-1, 1]$.

On the other hand, the mean value theorem gives that $\forall x,y\ \cos(x) - \cos(y) = (\cos a)'(x - y), a\in]x,y[$ (provided $y>x$) which translates into

$$\forall x, y\ |\cos(x) - \cos(y)| = |\sin(a)||x - y|$$

Now the truth is that $|\sin(a)| < 1$ when $a \in [-1, 1]$. Let $L$ be the maximum value $|\sin(a)|$ can attain in the interval $[-1, 1]$. Then,

$$|\cos(x) - \cos(y)| \leq L|x - y| < |x - y|$$

Now set $x = c$ and $y = \beta$:

$$|\cos(c) - \cos(\beta)| = |c - a_1| \leq L|c - \beta|$$

Now do the same thing, but with $x = c$ and $y = a_1$:

$$|\cos(c) - \cos(a_1)| = |c - a_2| \leq L|c - a_1| \leq L(L|c - \beta|) = L^2|c - \beta|$$

See where this is going? Setting $y = a_n$ gives

$$|\cos(c) - \cos(a_n)| = |c - a_{n+1}| \leq L|c - a_n| \leq L(L^n|c - \beta|) = L^{n+1}|c - \beta|$$

Now just take the limit $n \to \infty$. Because $|L| < 1, \lim_{n\to\infty}$ will give $|c - a_n| \to 0$.

RGS
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