Showing monotone convergence of recursive relation with $ x_{n+1} = \sin(x_{n}) $

Question

how I can show the following monotonic relation : $x_{n}-x_{n+1}>=0$ with $x_{n+1}=sin(x_{n})$:

My Idea: with $x_{n}-x_{n+1}=x_{n}-sin(x_{n})=....$here I am stuck. it were nice if someone can help me at.

greetings

Answer 1

This requires an additional ssumption. For example if $x_1=-\frac {\pi} 2$ then $x_2=-1 >x_1$.

If $x_n \geq 0$ for all $x$ then this result is true and it follows from the inequality $\sin x \leq x$ for all $x \geq 0$.

Proof of $\sin x \leq x$ for $x \geq 0$: let $f(x)=x-\sin x$. Then $f(0)=0$ and $f'(x)=1-\cos x \geq 0$. Hence $f$ is montonically increasing so $f(x) \geq f(0)=0$ for $x \geq 0$.