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積の差分の証明です。

積の差分の証明

積の差分の証明です。

更新日:

前進差分

h(f(x)g(x))hx=f(x+h)g(x+h)-f(x)g(x)(x+h)-x
=f(x+h)g(x+h)-f(x)g(x)h+f(x)g(x+h)-f(x)g(x+h)h
=f(x+h)g(x+h)-f(x)g(x+h)h+f(x)g(x+h)-f(x)g(x)h
=hf(x)hxg(x+h)+f(x)hg(x)hx
=hf(x)hxg(x+h)+f(x)hg(x)hx+(hf(x)hxg(x)-hf(x)hxg(x))
=f(x)hg(x)hx+hf(x)hxg(x)+hf(x)hx(g(x+h)-g(x))
=f(x)hg(x)hx+hf(x)hxg(x)+hhf(x)hxhg(x)hx
よって
h(f(x)g(x))hx=f(x)hg(x)hx+hf(x)hxg(x)+hhf(x)hxhg(x)hx

後退差分

h(f(x)g(x))hx=f(x)g(x)-f(x-h)g(x-h)h
=f(x)g(x)-f(x-h)g(x-h)h+f(x)g(x-h)-f(x)g(x-h)h
=f(x)g(x)-f(x)g(x-h)h+f(x)g(x-h)-f(x-h)g(x-h)h
=f(x)hg(x)hx+hf(x)hxg(x-h)
=f(x)hg(x)hx+hf(x)hxg(x-h)+(hf(x)hxg(x)-hf(x)hxg(x))
=f(x)hg(x)hx+hf(x)hxg(x)+hf(x)hx(g(x-h)-g(x))
=f(x)hg(x)hx+hf(x)hxg(x)-hhf(x)hxhg(x)hx

よって
h(f(x)g(x))hx=f(x)hg(x)hx+hf(x)hxg(x)-hhf(x)hxhg(x)hx

中心差分

Mxf(x)=f(x+h2)+f(x-h2)2
=2f(x+h2)-f(x+h2)+f(x-h2)2
=f(x+h2)-h2δhf(x)δhx
変形して
f(x+h2)=Mxf(x)+h2δhf(x)δhx…①

Mxf(x)=f(x+h2)+f(x-h2)2
=f(x+h2)-f(x-h2)+2f(x-h2)2
=f(x-h2)+h2δhf(x)δhx
変形して
f(x-h2)=Mxf(x)-h2δhf(x)δhx…②

Mxg(x)=g(x+h2)+g(x-h2)2
=2g(x+h2)-g(x+h2)+g(x-h2)2
=g(x+h2)-h2δhg(x)δhx
変形して
g(x+h2)=Mxg(x)+h2δhg(x)δhx…③

Mxg(x)=g(x+h2)+g(x-h2)2
=g(x+h2)-g(x-h2)+2g(x-h2)2
=g(x-h2)+h2δhg(x)δhx
変形して
g(x-h2)=Mxg(x)-h2δhg(x)δhx…④

①~④より
δh(f(x)g(x))δhx=f(x+h2)g(x+h2)-f(x-h2)g(x-h2)h
=(Mxf(x)+h2δhf(x)δhx)(Mxg(x)+h2δhg(x)δhx)h-(Mxf(x)-h2δhf(x)δhx)(Mxg(x)-h2δhg(x)δhx)h
=Mxf(x)Mxg(x)+h2δhf(x)δhxMxg(x)+Mxf(x)h2δhg(x)δhx+(h2)2δhf(x)δhxδhg(x)δhxh-Mxf(x)Mxg(x)-h2δhf(x)δhxMxg(x)-Mxf(x)h2δhg(x)δhx+(h2)2δhf(x)δhxδhg(x)δhxh
=h2δhf(x)δhxMxg(x)+Mxf(x)h2δhg(x)δhxh--h2δhf(x)δhxMxg(x)-Mxf(x)h2δhg(x)δhxh
=h2δhf(x)δhxMxg(x)+Mxf(x)h2δhg(x)δhxh+h2δhf(x)δhxMxg(x)+Mxf(x)h2δhg(x)δhxh
=hδhf(x)δhxMxg(x)+Mxf(x)hδhg(x)δhxh
=δhf(x)δhxMxg(x)+Mxf(x)δhg(x)δhx

よって
δh(f(x)g(x))δhx=Mxf(x)δhg(x)δhx+δhf(x)δhxMxg(x)

完成した公式

h(f(x)g(x))hx=f(x)hg(x)hx+hf(x)hxg(x)+hhf(x)hxhg(x)hx
h(f(x)g(x))hx=f(x)hg(x)hx+hf(x)hxg(x)-hhf(x)hxhg(x)hx
δh(f(x)g(x))δhx=Mxf(x)δhg(x)δhx+δhf(x)δhxMxg(x)
ただし
Mxf(x)=f(x+h2)+f(x-h2)2
Mxg(x)=g(x+h2)+g(x-h2)2