Numerical differentiation and derivative error control — Examples

The distinct candidate Jacobian rows $J=(1,2)$ and $K=(3,0)$ are realized at $(0,0)$ by the linear maps $F(z_0,z_1)=z_0+2z_1$ and $G(z_0,z_1)=3z_0$. Both supports lie in $S_0$, but [F1] gives $Js=1+2=3$ and $Ks=3+0=3$.

Since the same observation is compatible with different entries in columns $0$ and $1$, this colliding seed does not identify the row. The collision violates the proper-coloring hypothesis of [F2], so the unconditional statement is false.

On a real neighborhood of $1$, $f(t)=\mathrm{\mid }t\mathrm{\mid }=t$, so the ordinary derivative of the real restriction at $1$ is $1$.

By [F1], $f(1+ih)=\sqrt{1+h^2}$ is real. Thus the quotient in the Statement is $0$ for every $h>0$, which is not the real derivative $1$.

The complex difference quotient along real increments $s$ is $(\mathrm{\mid }1+s\mathrm{\mid }-1)\mathrm{/}s=1$ for $\mathrm{\mid }s\mathrm{\mid }<1$ and $s\mathrm{\ne }0$. Along imaginary increments it is $(\sqrt{1+s^2}-1)\mathrm{/}(is)=-is\mathrm{/}(\sqrt{1+s^2}+1)\to 0$. The limits differ, so $f$ is not complex differentiable at $1$ and the formally defined estimator in [F2] and guarantee in [F3] do not apply; step 1.2 instead refutes the unqualified quotient claim.

Differentiating the four components in coordinate order $(a,b,c,d)$, indexed $0,1,2,3$, gives $J=\left(\begin{array}{cccc}-k_1& 0& 0& 0\\ k_1& 0& 0& 0\\ 0& 0& -k_2& 0\\ 0& 0& k_2& 0\end{array}\right)$. The row supports are $\{0\},\{0\},\{2\},\{2\}$, so no row contains two columns and the coloring that assigns all four columns one color is proper.

Its seed is $s=(1,1,1,1{)}^T$ and $Js=(-k_1,k_1,-k_2,k_2{)}^T$. Each row has at most one supported column of this color, so [F3] recovers the nonzero entries from these four values; the remaining entries are zero by the declared pattern.

Substitution gives $D_h^{+}f(1)=((1+h{)}^2-1)\mathrm{/}h=2+h$, while $f^{\mathrm{\prime }}(1)=2$; the truncation error is exactly $h=M_{2}h\mathrm{/}2$ because $f^{\mathrm{\prime }\mathrm{\prime }}=2$.

With the stated errors, ${\widehat{D}}_h^{+}f(1)=((1+h{)}^2+\delta -(1-\delta ))\mathrm{/}h=2+h+2\delta \mathrm{/}h$, so the error bound is attained exactly.

Set $\delta =1\mathrm{/}2$. Then [F2] gives $h=1$; the truncation and evaluation contributions are each $1$, so the total error is $2$, equal to the minimum $2\sqrt{2\delta }$.