Finite-budget forward-difference Newton for a square vector residual

Use the same open domain, differentiability assumptions, initial point, residual tolerance and update budget as Finite-budget Newton for a square vector residual. Supply one increment vector $h=(h_0,\dots ,h_{n-1})$ with every $h_i\mathrm{\ne }0$ for $0\le i<n$. Output: the last valid iterate and one of root, invalid-stencil, singular-jacobian, invalid-trial, or iteration-budget-exhausted; every returned status is paired with the current last valid iterate.

For $k=0,1,\dots ,N$, in this order:

1. Evaluate $F(x_k)$. If $\mathrm{\parallel }F(x_k){\mathrm{\parallel }}_2\le \epsilon$, return $(x_k,\mathtt{\text{root}})$.
2. If $k=N$, return $(x_k,\mathtt{\text{iteration-budget-exhausted}})$.
3. Call Forward-difference Jacobian on an open domain at $(F,x_k,h)$. If it returns invalid-stencil, return $(x_k,\mathtt{\text{invalid-stencil}})$.
4. Otherwise call Newton correction for a square vector residual with $A=J_{h}F(x_k)$ and $r=F(x_k)$. If that correction is unavailable, return $(x_k,\mathtt{\text{singular-jacobian}})$. Otherwise let $s_k$ be the returned correction.
5. If the trial $x_k+s_k$ is outside $U$, return $(x_k,\mathtt{\text{invalid-trial}})$; otherwise set $x_{k+1}=x_k+s_k$ and continue.

The root status still means only $\mathrm{\parallel }F(x_k){\mathrm{\parallel }}_2\le \epsilon$. This approximate-Jacobian procedure has no quadratic-rate claim here.