The Contraction Cobweb

$|f(x) - f(y)| = \left|\frac{x-y}{2}\right| = \frac12|x - y|$, so $f$ is a contraction with $q = \frac12$.

Solving $x = \frac{x}{2} + 1$ gives $x^* = 2$.

Since $|x_{n+1} - 2| = \frac12|x_n - 2|$, induction gives $|x_n - 2| = \left(\frac12\right)^n|x_0 - 2|$. For $x_0 = 0$, $n = 8$, the error is $\frac{2}{256} = 0.0078125$.

$\cos$ is continuous on $[0,1]$ and differentiable on $(0,1)$, so the Mean Value Theorem applies: some $\xi$ between $x$ and $y$ gives $|\cos x - \cos y| = |-\sin\xi|\cdot|x - y|$.

Since $\xi \in [0,1] \subset [0, \frac{\pi}{2})$ and $\sin$ is increasing on $[0, \frac{\pi}{2})$, $|\sin\xi| \le \sin(1)$, so $|\cos x - \cos y| \le \sin(1)\,|x - y|$, and $\sin(1) \approx 0.8415 < 1$, so $\cos$ is a contraction.

Also $\cos$ is decreasing on $[0,1]$ (since $(\cos)' = -\sin < 0$), so $\cos([0,1]) = [\cos 1, \cos 0] = [\cos 1, 1] \subset [0,1]$ (with $\cos 1 \approx 0.5403 > 0$) — a contraction mapping $[0,1]$ into itself.

$[0,1]$ is closed in $\mathbb{R}$, hence complete — the completeness hypothesis of Banach's theorem holds. Banach's theorem then gives a unique fixed point in $[0,1]$ — the Dottie number $x^* \approx 0.739085$ (the value is from iteration).

Suppose $x^*$ and $y^*$ are both fixed points.

Then $|x^* - y^*| = |f(x^*) - f(y^*)| \le q|x^* - y^*|$.

Since $q < 1$, this forces $|x^* - y^*| = 0$, i.e. $x^* = y^*$.