Recursive sequence defined using $\cos x$

Question

For any $x\in \mathbb{R}$ , the sequence $\{a_n\}$ , where $a_1=x$ and $a_{n+1}=\cos (a_n)$ for all $n$ is convergent . True/False

I think this is true. Well , my plan is to show $\cos x$ is a contractive mapping on the real line (complete metric space) and then the above sequence would converge to the unique fixed point of $\cos x$ by Banach Fixed Point Theorem.

But I am facing a little problem :

$|\cos x-\cos y|=2|\sin \big( \frac {x+y}2\big)||\sin \big( \frac {x-y}2\big)|$

using $|\sin x| \le |x|$ and $|\sin x|\le 1$

$\Rightarrow |\cos x-\cos y| \le |x-y|$ which doesn't fit the definition of contractive mapping.

Any way to tackle this.? Thanks for your time and attention.

Answer 1

Hint

Claim 1: $g(x) = \cos x - x$ has a unique solution $x^*$ and $x^* \in [\cos 1, 1]$

Proof: $g$ is strictly decreasing on $\mathbb R$, $g(0) = 1 > 0$ and $g\left(\frac{\pi}{2}\right) =-\frac{\pi}{2} < 0$.

Claim 2: $u_2 \in [\cos 1, 1]$

Proof: You have $u_1 \in [-1,1]$, $\cos x$ is even and therefore $\cos [[-1,1]] = \cos [[0,1]] = [\cos 1, 1]$ as $\cos x$ is strictly decreasing on $[0,\pi]$.

Claim 3: $\cos x$ is a contracting map from $[\cos 1, 1]$ to $[\cos 1,1]$

Proof: follows from the facts that $\cos x$ is strictly decreasing on $[\cos 1, 1]$ and $\cos (\cos 1) < 1$.

Recall that for a strictly decreasing map $f[[a,b]] = [f(b),f(a)]$.

Answer 2

You may use the Edelstein's version of the Banach fixed point theorem: if $X$ is a complete, compact metric space and a continuous $f:X\to X$ is such that $|f(x)-f(y)|< |x-y|$ holds for any $x\neq y$, then $f^{(n)}(x)$ converges to a fixed point of $f$.

In your case you may take $X=[-1,1]$ since $a_2\in[-1,1]$ regardless of the value of $a_1$.

---

Proof of the theorem: let $\Phi(x)=d(x,f(x))$. $\Phi$ is a continuous function: let $z\in X$ and $\{x_n\}_{n\geq 1}$ a sequence of elements of $X$ convergent to $z$. By the triangle inequality $$ \Phi(x_n)\leq d(x_n,z)+d(z,f(z))+d(f(z),f(x_n)) $$ i.e. $$ \Phi(x_n)-\Phi(z)\leq d(x_n,z)+d(f(x_n),f(z)) $$ where the RHS converges to zero as $n\to +\infty$, implying $$ \lim_{n\to +\infty}\Phi(x_n) = \Phi(z). $$ Crucial part: Since $X$ is compact, $\Phi$ has a minimum at some $y\in X$. Assuming $y\neq f(y)$ we have $$ \Phi(f(y))=d(f(y),f(f(y)))<d(y,f(y))=\Phi(y) $$ contradicting the minimality. This gives that $y$ is a fixed point of $f$.
Such fixed point is unique: assuming that $x\neq y$ is a different fixed point, $$ d(x,y)=d(f(x),f(y))<d(x,y) $$ leads to a contradiction.

Answer 3

You needn't worry too much. As the figure shows, past the first one, the function iterates (order blue, magenta, green, black) remain in $[\cos1,1]$ where the absolute slope of the cosine is below $1$.