部分和分の証明

$\text{合成関数の前進差分の公式}$
$\frac{{∆}_h(f\left(x\right)g\left(x\right))}{{∆}_{h}x}=f\left(x\right)\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}+\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}g\left(x\right)+h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}$
$\text{の両辺を和分して}$
$f\left(x\right)g\left(x\right)={∆}_h^{-1}(f\left(x\right)\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}){∆}_{h}x+{∆}_h^{-1}(\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}g\left(x\right)){∆}_{h}x+{∆}_h^{-1}(h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}){∆}_{h}x$
$\text{変形して}$
${∆}_h^{-1}(f\left(x\right)\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}){∆}_{h}x=f\left(x\right)g\left(x\right)-{∆}_h^{-1}(\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}g\left(x\right)){∆}_{h}x-{∆}_h^{-1}(h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}){∆}_{h}x$

$\text{合成関数の後退差分の公式}$
$\frac{{\nabla }_h(f\left(x\right)g\left(x\right))}{{\nabla }_{h}x}=\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}g\left(x\right)+f\left(x\right)\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}-h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}$
$\text{の両辺を和分して}$
$f\left(x\right)g\left(x\right)={\nabla }_h^{-1}(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}g\left(x\right)){\nabla }_{h}x+{\nabla }_h^{-1}(f\left(x\right)\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}){\nabla }_{h}x-{\nabla }_h^{-1}(h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}){\nabla }_{h}x$
$\text{変形して}$
${\nabla }_h^{-1}(f\left(x\right)\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}){\nabla }_{h}x=f\left(x\right)g\left(x\right)-{\nabla }_h^{-1}(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}g\left(x\right)){\nabla }_{h}x+{\nabla }_h^{-1}(h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}){\nabla }_{h}x$

$\text{合成関数の中心差分の公式}$
$\frac{{\delta }_h(f\left(x\right)g\left(x\right))}{{\delta }_{h}x}=M_{x}f\left(x\right)\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}+\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}{M}_{x}g\left(x\right)$
$\text{の両辺を和分して}$
$f\left(x\right)g\left(x\right)={\delta }_h^{-1}(M_{x}f\left(x\right)\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}){\delta }_{h}x+{\delta }_h^{-1}(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}{M}_{x}g\left(x\right)){\delta }_{h}x$
$\text{変形して}$
${\delta }_h^{-1}(M_{x}f\left(x\right)\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}){\delta }_{h}x=f\left(x\right)g\left(x\right)-{\delta }_h^{-1}(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}{M}_{x}g\left(x\right)){\delta }_{h}x$

${∆}_h^{-1}(f\left(x\right)\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}){∆}_{h}x=f\left(x\right)g\left(x\right)-{∆}_h^{-1}(\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}g\left(x\right)){∆}_{h}x-{∆}_h^{-1}(h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}){∆}_{h}x$
${\nabla }_h^{-1}(f\left(x\right)\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}){\nabla }_{h}x=f\left(x\right)g\left(x\right)-{\nabla }_h^{-1}(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}g\left(x\right)){\nabla }_{h}x+{\nabla }_h^{-1}(h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}){\nabla }_{h}x$
${\delta }_h^{-1}(M_{x}f\left(x\right)\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}){\delta }_{h}x=f\left(x\right)g\left(x\right)-{\delta }_h^{-1}(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}{M}_{x}g\left(x\right)){\delta }_{h}x$
$\text{ただし}$
$M_{x}f\left(x\right)=\frac{f(x+\frac{h}{2})+f(x-\frac{h}{2})}{2}$
$M_{x}g\left(x\right)=\frac{g(x+\frac{h}{2})+g(x-\frac{h}{2})}{2}$