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商の差分の証明です。

商の差分の証明

商の差分の証明です。

更新日:

前進差分

hf(x)g(x)hx=h(f(x)1g(x))hx
=f(x)h1g(x)hx+hf(x)hx1g(x)+hhf(x)hxh1g(x)hx
=f(x)(hg(x)hx)(g(x))2-hhg(x)hxg(x)+hf(x)hx1g(x)+hhf(x)hx(hg(x)hx)(g(x))2-hhg(x)hxg(x)
=f(x)hg(x)hx+hf(x)hx(g(x)-hhg(x)hx)+hhf(x)hxhg(x)hx(g(x))2-hhg(x)hxg(x)
=f(x)hg(x)hx+hf(x)hxg(x)-hhf(x)hxhg(x)hx+hhf(x)hxhg(x)hx(g(x))2-hhg(x)hxg(x)
=f(x)hg(x)hx+hf(x)hxg(x)(g(x))2-hhg(x)hxg(x)
よって
hf(x)g(x)hx=f(x)hg(x)hx+hf(x)hxg(x)(g(x))2-hhg(x)hxg(x)

後退差分

hf(x)g(x)hx=h(f(x)1g(x))hx
=f(x)h1g(x)hx+hf(x)hx1g(x)-hhf(x)hxh1g(x)hx
=f(x)(hg(x)hx)hhg(x)hxg(x)-(g(x))2+hf(x)hx1g(x)-hhf(x)hx(hg(x)hx)hhg(x)hxg(x)-(g(x))2
=f(x)hg(x)hx+hf(x)hx(hhg(x)hx-g(x))-hhf(x)hxhg(x)hxhhg(x)hxg(x)-(g(x))2
=f(x)hg(x)hx-hf(x)hxg(x)+hhf(x)hxhg(x)hx-hhf(x)hxhg(x)hxhhg(x)hxg(x)-(g(x))2
=f(x)hg(x)hx-hf(x)hxg(x)hhg(x)hxg(x)-(g(x))2
よって
hf(x)g(x)hx=f(x)hg(x)hx-hf(x)hxg(x)hhg(x)hxg(x)-(g(x))2

中心差分

Mx1g(x)=1g(x+h2)+1g(x-h2)2
=g(x-h2)+g(x+h2)2g(x+h2)g(x-h2)
=Mxg(x)g(x+h2)g(x-h2)
=Mxg(x)(Mxg(x))2-(h2δhg(x)δhx)2
=-Mxg(x)(h2δhg(x)δhx)2-(Mxg(x))2…①

δhf(x)g(x)δhx=δh(f(x)1g(x))δhx
=δhf(x)δhxMx1g(x)+Mxf(x)δh1g(x)δhx
=Mxf(x)(δhg(x)δhx)(h2δhg(x)δhx)2-(Mxg(x))2-δhf(x)δhxMxg(x)(h2δhg(x)δhx)2-(Mxg(x))2
=Mxf(x)δhg(x)δhx-δhf(x)δhxMxg(x)(h2δhg(x)δhx)2-(Mxg(x))2
よって
δhf(x)g(x)δhx=Mxf(x)δhg(x)δhx-δhf(x)δhxMxg(x)(h2δhg(x)δhx)2-(Mxg(x))2

完成した公式

hf(x)g(x)hx=f(x)hg(x)hx+hf(x)hxg(x)(g(x))2-hhg(x)hxg(x)
hf(x)g(x)hx=f(x)hg(x)hx-hf(x)hxg(x)hhg(x)hxg(x)-(g(x))2
δhf(x)g(x)δhx=Mxf(x)δhg(x)δhx-δhf(x)δhxMxg(x)(h2δhg(x)δhx)2-(Mxg(x))2
ただし
Mxf(x)=f(x+h2)+f(x-h2)2
Mxg(x)=g(x+h2)+g(x-h2)2