逆数の差分の証明

$\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}=\frac{f(x+h)-f\left(x\right)}{h}$
$\text{変形して}$
$f(x+h)=h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}-f\left(x\right)\text{…①}$

$\text{①より}$
$\frac{{∆}_h\frac{1}{f\left(x\right)}}{{∆}_{h}x}=\frac{\frac{1}{f(x+h)}-\frac{1}{f\left(x\right)}}{h}$
$=\frac{f\left(x\right)-f(x+h)}{hf(x+h)f\left(x\right)}=-\frac{\left(\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}\right)}{f(x+h)f\left(x\right)}$
$=-\frac{\left(\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}\right)}{(h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}-f\left(x\right))f\left(x\right)}$
$=\frac{\left(\frac{∆f\left(x\right)}{∆x}\right)}{(f\left(x\right)-h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x})f\left(x\right)}$
$=\frac{\left(\frac{∆f\left(x\right)}{∆x}\right)}{{\left(f\left(x\right)\right)}^2-h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}f\left(x\right)}$

$\text{よって}$
$\frac{{∆}_h\frac{1}{f\left(x\right)}}{{∆}_{h}x}=-\frac{\left(\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}\right)}{{\left(f\left(x\right)\right)}^2-h\frac{{∆}_{h}f\left(x\right)}{{∆}_{h}x}f\left(x\right)}$

$\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}=\frac{f\left(x\right)-f(x-h)}{h}$
$\text{変形して}$
$f(x-h)=f\left(x\right)-h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\text{…①}$

$\text{①より}$
$\frac{{\nabla }_h\frac{1}{f\left(x\right)}}{{\nabla }_{h}x}=\frac{\frac{1}{f\left(x\right)}-\frac{1}{f(x-h)}}{h}$
$=\frac{f(x-h)-f\left(x\right)}{hf\left(x\right)f(x-h)}$
$=-\frac{\left(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\right)}{f\left(x\right)f(x-h)}$
$=-\frac{\left(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\right)}{f\left(x\right)(f\left(x\right)-h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x})}$
$=\frac{\left(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\right)}{f\left(x\right)(h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}-f\left(x\right))}$
$=\frac{\left(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\right)}{h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}f\left(x\right)-{\left(f\left(x\right)\right)}^2}$
$\text{よって}$
$\frac{{\nabla }_h\frac{1}{f\left(x\right)}}{{\nabla }_{h}x}=\frac{\left(\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}\right)}{h\frac{{\nabla }_{h}f\left(x\right)}{{\nabla }_{h}x}f\left(x\right)-{\left(f\left(x\right)\right)}^2}$

$M_{x}f\left(x\right)=\frac{f(x+\frac{h}{2})+f(x-\frac{h}{2})}{2}$
$=\frac{2f(x+\frac{h}{2})-f(x+\frac{h}{2})+f(x-\frac{h}{2})}{2}$
$=f(x+\frac{h}{2})-\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}$
$\text{変形して}$
$f(x+\frac{h}{2})=M_{x}f\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}\text{…①}$

$M_{x}f\left(x\right)=\frac{f(x+\frac{h}{2})+f(x-\frac{h}{2})}{2}$
$=\frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})+2f(x-\frac{h}{2})}{2}$
$=f(x-\frac{h}{2})+\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}$
$\text{変形して}$
$f(x-\frac{h}{2})=M_{x}f\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}\text{…②}$

$\text{①と②より}$
$\frac{{\delta }_h\frac{1}{f\left(x\right)}}{{\delta }_{h}x}=\frac{\frac{1}{f(x+\frac{h}{2})}-\frac{1}{f(x-\frac{h}{2})}}{h}$
$=\frac{f(x-\frac{h}{2})-f(x+\frac{j}{2})}{hf(x+\frac{h}{2})f(x-\frac{h}{2})}$
$=-\frac{\left(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}\right)}{f(x+\frac{h}{2})f(x-\frac{h}{2})}$
$=-\frac{\left(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}\right)}{(M_{x}f\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x})(M_{x}f\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x})}$
$=-\frac{\left(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}\right)}{{\left(M_{x}f\left(x\right)\right)}^2-{(\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x})}^2}$
$=\frac{\left(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}\right)}{{(\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x})}^2-{\left(M_{x}f\left(x\right)\right)}^2}$
$\text{よって}$
$\frac{{\delta }_h\frac{1}{f\left(x\right)}}{{\delta }_{h}x}=\frac{\left(\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x}\right)}{{(\frac{h}{2}\frac{{\delta }_{h}f\left(x\right)}{{\delta }_{h}x})}^2-{\left(M_{x}f\left(x\right)\right)}^2}$

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