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For the Euclidean norm on ${\mathbb{R}}^d$, the regularized radial function $f(z)=\sqrt{1+\Vert z{\Vert }^2}$ has gradient $z/f$ and Hessian $I/f-z{z}^T/f^3$. The Cauchy-Schwarz inequality makes its Hessian a positive-definite matrix, so $f$ is strictly convex with unique minimum at zero. Its inverse matrix is $f(I+z{z}^T)$: multiplication verifies this using $(z{z}^T{)}^2=\Vert z{\Vert }^{2}z{z}^T$ and $f^2=1+\Vert z{\Vert }^2$. Substitution in Newton's method in optimization therefore gives $z_{k+1}=-\Vert z_k{\Vert }^2{z}_k$. Consequently $\Vert z_{k+1}\Vert =\Vert z_k{\Vert }^3$: nonzero starts in the open unit ball have exact cubic convergence, starts on the unit sphere alternate between opposite points, and starts outside it diverge. This example separates strict convexity from global convergence of an undamped iteration.

Define the relative reciprocal error ${\epsilon }_k=1-p{x}_k$. The iteration has the exact identitysoConsequently $|1-p{x}_0|<1$ guarantees $x_k\to 1/p$. For real $p,x_0$, this condition is $0<p{x}_0<2$. A starting guess can be found without division by choosing $x_0$ with the sign of $p$ and repeatedly halving its magnitude until $0<p{x}_0<2$.

For the absolute signed error $e_k=x_k-1/p$, the same identity givesThus the method has cubic convergence, of order of convergence three, with asymptotic error constant $p^2>0$ for a nonexact convergent start. Convergence requires a suitable guess: $p{x}_0=0$ and $p{x}_0=2$ correspond to ${\epsilon }_0=1$ and ${\epsilon }_0=-1$, respectively, so both are fixed failures. For $|1-p{x}_0|>1$ the error magnitude grows without bound. The exact initial value $p{x}_0=1$ is already the answer.

For iterates approaching ${\mathbf{x}}_{\ast }$, write $e_k=\Vert {\mathbf{x}}_k-{\mathbf{x}}_{\ast }\Vert$. Exact order of convergence $p$ means $e_{k+1}/e_k^p\to C$ for a finite constant $C>0$, with nonzero errors. Thus quadratic convergence has $p=2$, and cubic convergence has $p=3$. The weaker eventual bound $e_{k+1}\le C{e}_k^p$ means at least order $p$. Finite termination is a separate case.^[ref]

The assertion about being always faster is false. For example, the fixed-point iterations with errors $e_{k+1}=e_k^2$ and $d_{k+1}=50d_k^3$ have exact quadratic convergence and cubic convergence, respectively. Starting both at $0.1$, their next errors are $0.01$ and $0.05$: the cubic convergence method is worse on that step, although both converge. Error constants, starting points, and work per step all matter. The higher order of convergence describes asymptotic behavior within a convergence region, not a uniform comparison of all computations.

The update in Newton's method in optimization is ${\mathbf{z}}_{k+1}={\mathbf{z}}_k-H_f({\mathbf{z}}_k{)}^{-1}\nabla f({\mathbf{z}}_k)$. Substituting the previous formulas givesThus the distance from the unique minimum obeys $r_{k+1}=r_k^3$, where $r_k=\Vert {\mathbf{z}}_k\Vert$. In particular,For $0<r_0<1$, the error ratio $r_{k+1}/r_k^3$ is exactly $1$, proving exact cubic convergence. If $r_0=0$, the initial point already attains the minimum. All updates are defined because the Hessian is invertible everywhere. This is Newton minimization of the square-root norm; its special update also shows why the given initial condition matters: $r_0=1$ produces an alternating two-cycle, and $r_0>1$ diverges.