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Let $(a_n)_{n=1}^{\infty}$ be a sequence defined by

$a_1=x$ and $ a_{n+1}=\cos(a_n)$ then check the the convergence of sequence .

Here what Approached is like

$-1\leq \cos x \leq +1$ so we get $a_2 = cos(\mathbb{I_1})$ in general $a_{n+1} = cos(\mathbb{I_n})$ and cos being an even function $\mathbb{I_2}$ onwards it will contain elements $0 \leq x\leq c $ and $c<1$ hence it converges to $cos (0)=1$

It that a fair argument

uniquesolution
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  • Your notation is unclear. Why are replacing $a_n$ by $\mathbb{I_n}$?. Also, the final conclusion is not justified, so as it stands, the argument is not correct. – uniquesolution Aug 04 '17 at 09:14
  • No, the sequence $(a_n)$ indeed converges, for every $x$, but to $\ell\approx0.739$ the unique solution of $\cos\ell=\ell$. This was already explained in details on the site. – Did Aug 04 '17 at 09:15
  • I mean by that $\mathbb{I_n }$ is an interval –  Aug 04 '17 at 09:16
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    The equation $\cos x=x$ is known to have a unique solution ... see http://mathworld.wolfram.com/DottieNumber.html – Donald Splutterwit Aug 04 '17 at 09:17
  • (Already $\mathbb I_2$ does not contain $0$.) – Did Aug 04 '17 at 09:17
  • Can you make it more clear to me please –  Aug 04 '17 at 09:18
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    For $n\ge 2$, we will have $\cos 1\le a_n\le 1$. In that interval, the derivative of $\cos x$ is $\le \sin 1<1$ in absolute value, i.e., we have a contraction and convergence to the unique fixpoint. – Hagen von Eitzen Aug 04 '17 at 09:47
  • We can be perverse and argue that the OP did not say $x$ is real. – GEdgar Aug 04 '17 at 12:24
  • Numerical computation seems to suggest that the set $D\subseteq\mathbb{C}$ of $z$ for which the recurrence relation $a_{n+1} = \cos(a_n)$ with $a_1=z$ diverges is a fractal. We can be much more perverse and go further... – Sangchul Lee Aug 04 '17 at 13:10
  • Related question https://math.stackexchange.com/questions/1701935/let-a-n-cosa-n-1-l-a-1-a-2-a-n-is-there-an-a-0-such-that-l/ – rtybase Aug 04 '17 at 21:03

1 Answers1

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For the purposes of this answer, let superscript denote compositions. You want to evaluate: $$y=\lim_{n \to \infty} \cos^n{(x)}$$ $$y=\lim_{n \to \infty} \cos{\left(\cos^{n-1}{(x)}\right)}$$ $n-1$ goes to infinity the same as $n$ $$y=\lim_{n \to \infty} \cos^{n-1}{(x)}$$ $$y=\cos{y}$$ The unique solution to this equation is called the Dottie number $=D \approx 0.739085$. Now to prove convergence for all initial real $a_0$. After one composition, $-1 \leq a_n \leq 1$. Now that the sequence is bounded, we want to prove that it is either monotonically increasing or decreasing. If it is, it must converge to something within the interval. This cannot be done for the sequence as it is because it alternates about $D$. The operation must be changed from $\left (a_{n+1}=\cos{a_n}\right)$ to $\left(a_{n+1}=\cos{\left(\cos{a_n}\right)}\right)$ We want to prove that: $$a_n \leq a_{n+1} \leq D \quad \text{if} \quad a_n \leq D$$ $$a_n \geq a_{n+1} \geq D \quad \text{if} \quad a_n \geq D$$ $$\text{Define } f(x)=\cos{\left(\cos{x}\right)}-x \quad\quad f(D)=0$$ Since $D$ is the only zero of $f(x)$, $f(x)$ is continuous, and the end behaviors $\infty, -\infty$: $$f(x) \geq 0 \text{ when } x \leq D$$ $$f(x) \leq 0 \text{ when } x \geq D$$ The relationship between adjacent terms in the sequence is verified. Next, we must show that the sequence stays on the same side of $D$. Observe that after two compositions, $a_n$ is bounded: $0 \leq \cos{1} \leq a_n \leq 1 \quad$ since $\quad 1 \leq \frac{\pi}{2}$ $$\text{Define } f(x)=\cos{\left(\cos{x}\right)}-D \quad\quad f(D)=0$$ $$f'(x)=\sin{x}\sin{\left(\cos{x}\right)}$$ $$f'(D)=\sin^2{D}$$ Since the derivative is positive at $D$, $f(x)$ is continuous, and $D$ is the only zero of $f(x)$ in the interval $[0, 1]$: $$f(x) \geq 0 \text{ when } D \leq x \leq 1$$ $$f(x) \leq 0 \text{ when } 0 \leq x \leq D$$ Therefore, the two original inequalities are proven, and the sequence converges to $D$.

cvogt8
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