Prove $a_n$ is within a specific range for all $n\in\mathbb{N}$

Question

Let $a_n = \cos (a_{n-1})$ and $a_1 = {\pi\over 4}$.

How to prove the range of $a_n$ is within the closed interval: $\left[ {{1 \over {\sqrt 2 }},{\pi \over 4}} \right]$?

I thought about analyzing the inequality $\cos(x) < x$, but I don't want to make it very complicated since it's part of a bigger proof.

Thanks.

Answer 1

You only need to show that cos(x) is a contraction on $[\frac{1}{\sqrt{2}}, \frac{\pi}{4}]$, i.e.

$\exists 0\le k < 1, |\cos(y) - \cos(x)| \le k |y - x|, \forall x,y \in [\frac{1}{\sqrt{2}}, \frac{\pi}{4}]$.

This can be proved by the mean value theorem, because the derivative is bounded and is less than $1$ in the interval concerned.

Answer 2

You can use the induction, $$a_n \in \left[ {{1 \over {\sqrt 2 }},{\pi \over 4}} \right]\implies a_{n+1} = \cos(a_n)$$ If $a_n \le \frac \pi4$, using the inequality $\displaystyle \cos(x) \le 1 - \frac{x^2}{2}$ you get $a_{n+1} = \cos(a_n) \le 1-\frac{a_n^2}{2} \le \pi/4$.

Also using the inequality $\cos x \ge x$ in interval $0\to \pi/2$ you get $a_{n+1}\ge \cos(a_n) \ge a_{n} \ge \frac{1}{\sqrt 2}$ which completes the induction.