= Solution
The native paper prints $n\geq1$ while specifying $x_0$ alone. This does not determine $x_1$. We use the intended convention $x_{n+1}=\sin x_n$ for $n\geq0$; even with an independently chosen real $x_1$, the same proof gives the requested convergence.
For the <sine fixed-point iteration>, after one step $x_1\in[-1,1]$. On this <interval>, the <sine> has the same sign as its argument, and
$$
0<\sin t<t\qquad(0<t\leq1).
$$
For example, $t-\sin t=\int_0^t(1-\cos s)\,ds>0$, since the <integrand> is positive for $0<s\leq1$. By oddness of the <sine>, $a_n=|x_n|$ consequently satisfies $a_{n+1}=\sin a_n\leq a_n$ for $n\geq1$. This <monotone sequence> is bounded below by zero, so it has a limit $a\in[0,1]$. The <continuity> of the <sine> gives $a=\sin a$, and the strict inequality above forces $a=0$.
Hence $x_n\to0$, and $0$ is a <fixed point> of the <sine>. It is the only real <fixed point>: every such point has absolute value at most one, where the preceding strict inequality applies. Thus
$$
\boxed{\lim_{n\to\infty}x_n=0\quad\text{for every real initial value}.}
$$
The <Banach fixed-point theorem> does not directly apply to the <sine> on $[-1,1]$: the ratios $|\sin t-\sin0|/|t-0|$ tend to one as $t\to0$, so no uniform <contraction mapping> constant smaller than one exists there.
Back to article page
