Iterative scheme involving cosine.

Question

Define a sequence of real numbers $\{x_k : k \ge 0\}$ by $x_0 = 0$ and $x_k = \cos(x_{k-1})$. Does the limit $\lim_{k \to \infty} x_k = z$ exist?

Answer 1

If $z$ exists, then$$\cos z = \cos\left( \lim_{k \to \infty} x_k\right) = \lim_{k \to \infty} \cos x_k = \lim_{k \to \infty} x_{k+1} = z.$$This is enough to uniquely identify $z$, since $x - \cos x$ is strictly increasing on $\mathbb{R}$.

The combination of "iterative scheme" and "fixed point" strongly suggests the contraction mapping principle, an idea hampered by the fact that the cosine function is not a contraction on $\mathbb{R}$. However, we can restrict our attention to the interval $I = [0, 1]$. We have that $\cos$ is decreasing and positive on $I$, and $\cos 0 = 1$, so there is a function $f: I \to I$, $f(x) = \cos x$. Moreover, $f$ is a contraction. To see this, let $0 \le x < y \le 1$; then by the mean value theorem, there is some $w \in (x, y)$ where$$|f(y) - f(x)| = |f'(w)| \cdot |y - x| = |\sin w| \cdot |y - x| \le K|y - x|,$$where $K = \sin 1 < 1$. It follows that $f$ has a unique fixed point, and that $z_k \to z$ for any sequence $z_k \in I$ where $z_{k+1} = f(z_k)$ for all $k$. We know that $x_k$ as defined is such a sequence, so $x_k \to z$.

Answer 2

If the limit $\lim_{n\to\infty} x_n$ exists then this limit has to be a solution of the equation $\cos x=x$. A look at the graph of $\cos$ shows that there is indeed a unique solution $\xi\in\>]0,1[\>$ to this equation. This leads us to conjecture that in fact $\lim_{n\to\infty} x_n=\xi$. Experimenting with a pocket calculator we see that the $x_n$ are oscillating, so that we cannot use a monotonicity argument. Therefore let's look at the distances $|x_n-\xi|$. One has $$x_{n+1}-\xi=\cos x_n-\xi=\cos x_n-\cos\xi=-2\sin{x_n+\xi\over2}\>\sin{x_n-\xi\over2}\ ,$$ and this implies
$$|x_{n+1}-\xi|\leq\left|\sin{x_n+\xi\over2}\right|\>|x_n-\xi|\ .$$ Since $0\leq x_n\leq1<{\pi\over2}$ for all $n\geq0$ one has $$0\leq\left|\sin{x_n+\xi\over2}\right|<\sin 1=:q<1$$ for all $n$. The inequality $$|x_{n+1}-\xi|<q\>|x_n-\xi|$$ then shows that the $x_n$ converge "linearly" to $\xi$ with $n\to\infty$.