The map $x\mapsto (x+2\mathrm{/}x)\mathrm{/}2$ is a contraction of $[1,2]$ with fixed point $\sqrt{2}$, and the a priori bound gives the error after $n$ steps

Let $X:=[1,2]\subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ inherited from $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and define

$f(x):=\frac{x+2\mathrm{/}x}{2}=\frac{x}{2}+\frac{1}{x}\mathrm{.}$

Then:

1. $(X,d)$ is a nonempty complete metric space (Complete metric space: every Cauchy sequence converges in the space).
2. $f$ maps $X$ into $X$.
3. $f$ is a contraction with constant $q=1\mathrm{/}2$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction).
4. $f$ has exactly one fixed point in $X$, namely $\sqrt{2}$ (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$), and the iterates $x_{n+1}=f(x_n)$ converge to it from every starting point $x_0\in X$ (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point).
5. Starting from $x_0=1$ one has $x_1=3\mathrm{/}2$ and $d(x_1,x_0)=1\mathrm{/}2$, so the a priori bound of The a priori bound $d(x^{\ast },x_n)\le q^{n}d(x_1,x_0)\mathrm{/}(1-q)$ and the a posteriori bound $d(x^{\ast },x_{n+1})\le qd(x_{n+1},x_n)\mathrm{/}(1-q)$ reads $\mid \sqrt{2}-x_n\mid \le \frac{(1\mathrm{/}2{)}^n\cdot (1\mathrm{/}2)}{1-1\mathrm{/}2}=(\frac{1}{2}{)}^n,$ and the a posteriori bound reads $\mid \sqrt{2}-x_{n+1}\mid \le \mathrm{\mid }{x}_{n+1}-x_n\mathrm{\mid }$.

• No derivative is used anywhere. The contraction constant comes from the algebraic identity of step 1.4, $f(x)-f(y)=(x-y)(1\mathrm{/}2-1\mathrm{/}(xy))$, and from the bound $1\le xy\le 4$ that the interval supplies. This matters here: the mean value theorem, which is how this estimate is usually obtained, belongs to a later page of this library and is not available at this point.
• The lower endpoint is what makes the estimate close. The bound $\mathrm{\mid }1\mathrm{/}2-1\mathrm{/}(xy)\mathrm{\mid }\le 1\mathrm{/}2$ of step 2.2 needs $xy\ge 1$, and that is exactly what $x,y\ge 1$ supplies; on a set of small positive reals the factor $1\mathrm{/}(xy)$ is large and the same computation gives nothing. The self-mapping property uses both endpoints: step 1.3 gives the lower bound $\sqrt{2}$ and step 2.1 the upper bound $2$.
• What the bounds say numerically. From $x_0=1$ the a priori bound guarantees $\mathrm{\mid }\sqrt{2}-x_n\mathrm{\mid }\le 2^{-n}$, so ten steps give an error at most $2^{-10}$. This is a guarantee, not the truth: the iteration is Newton's method for $x^2-2$, and it converges far faster than the bound admits. The a posteriori bound of claim 5, which costs one subtraction, is what a computation would actually report.
• The restriction to $[1,2]$ is convenience, not necessity. On $[1,\mathrm{\infty })$ one still has $x,y\ge 1$, hence $0<1\mathrm{/}(xy)\le 1$ and $\mathrm{\mid }1\mathrm{/}2-1\mathrm{/}(xy)\mathrm{\mid }\le 1\mathrm{/}2$, so the same computation makes $f$ a contraction with the same constant there, and step 1.3 already shows $f$ maps $[1,\mathrm{\infty })$ into itself. The shorter interval is used only to keep the arithmetic of step 2.2 explicit.