# 2変数関数のテイラー（Taylor）の定理の証明

$f\left(a+h,b+k\right)=f\left(a,b\right)$$+\frac{1}{1!}\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)f\left(a,b\right)$$+\frac{1}{2!}\left(h\frac{\partial }{\partial x}+{k\frac{\partial }{\partial y}\right)}^{2}f\left(a,b\right)$$+\cdots$$+\frac{1}{n!}\left(h\frac{\partial }{\partial x}{+k\frac{\partial }{\partial y}\right)}^{n}f\left(a,b\right)$$+{R}_{n+1}$

${R}_{n+1}$$=\frac{1}{\left(n+1\right)!}\left(h\frac{\partial }{\partial x}{+k\frac{\partial }{\partial y}\right)}^{n+1}f\left(a+\theta h,b+\theta k\right)$ 　　（ただし, $0<\theta <1$

## ■導出

$z=f\left(x,y\right)$  ，$x=a+ht$$y=b+kt$　　( $a,b,h,k$ は定数)

のとき，$z\left(t\right)=f\left(a+ht,b+kt\right)$ とおく．

$z\left(t\right)$  を合成関数の微分法を用いて微分する．

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$  ･･････(1)

$\frac{dx}{dt}=\frac{d}{dt}\left(a+ht\right)=h$  ，$\frac{dy}{dt}=\frac{d}{dt}\left(b+kt\right)=k$ ･･････(2)

(2)を(1)に代入すると，

$\frac{dz}{dt}=\frac{\partial z}{\partial x}h+\frac{\partial z}{\partial y}k=\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)z$

$\frac{{d}^{2}z}{d{t}^{2}}$$=\frac{d}{dt}\left(\frac{dz}{dt}\right)$

$=\frac{d}{dt}\left\{\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)z\right\}$

$=\frac{d}{dt}\left(h\frac{\partial z}{\partial x}+k\frac{\partial z}{\partial y}\right)$

$=\frac{d}{dt}\left(h\frac{\partial z}{\partial x}\right)+\frac{d}{dt}\left(k\frac{\partial z}{\partial y}\right)$

$=h\frac{d}{dt}\left(\frac{\partial z}{\partial x}\right)+k\frac{d}{dt}\left(\frac{\partial z}{\partial y}\right)$

$\frac{{d}^{2}z}{d{t}^{2}}=h\left\{\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial x}\right)\frac{dx}{dt}$$+\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial x}\right)\frac{dy}{dt}\right\}$$+k\left\{\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)\frac{dx}{dt}$$+\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial y}\right)\frac{dy}{dt}\right\}$

(2)を代入して整理すると

$\frac{{d}^{2}z}{d{t}^{2}}=\left\{h\left(\frac{{\partial }^{2}z}{\partial {x}^{2}}\right)h$$+\left(\frac{{\partial }^{2}z}{\partial y\partial x}\right)k\right\}$$+k\left\{\left(\frac{{\partial }^{2}z}{\partial x\partial y}\right)h$$+\left(\frac{{\partial }^{2}z}{\partial {y}^{2}}\right)k\right\}$

$\frac{{d}^{2}z}{d{t}^{2}}={h}^{2}\left(\frac{{\partial }^{2}z}{\partial {x}^{2}}\right)+2hk\left(\frac{{\partial }^{2}z}{\partial y\partial x}\right)$$+{k}^{2}\left(\frac{{\partial }^{2}z}{\partial {y}^{2}}\right)$

よって

$\frac{{d}^{2}z}{d{t}^{2}}={\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{2}z$

$\frac{{d}^{n}z}{d{t}^{n}}={\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{n}z$

と推測できる．

ここで1変数関数$z\left(t\right)$マクローリンの定理を適用すると，

$z\left(t\right)=z\left(0\right)+\frac{1}{1!}{z}^{\prime }\left(0\right)t+\frac{1}{2!}{z}^{″}\left(0\right){t}^{2}$$+\cdots$ $+\frac{1}{n!}{z}^{\left(n\right)}\left(0\right){t}^{n}$ $+{R}_{n+1}$ ･･････(3)

${R}_{n+1}=\frac{1}{\left(n+1\right)!}{z}^{\left(n+1\right)}\left(\theta \right){t}^{n+1}$  $\left(0<\theta <1\right)$

$t=1$  とおくと，

$z\left(1\right)=z\left(0\right)+\frac{1}{1!}{z}^{\prime }\left(0\right)+\frac{1}{2!}{z}^{″}\left(0\right)+\cdots$$+\frac{1}{n!}{z}^{\left(n\right)}\left(0\right)+{R}_{n+1}$

${R}_{n+1}=\frac{1}{\left(n+1\right)!}{z}^{\left(n+1\right)}\left(\theta \right)$  $\left(0<\theta <1\right)$

$z\left(t\right)=f\left(a+ht,b+kt\right)$  なので

$z\left(0\right)=f\left(a,b\right)$

${z}^{\prime }\left(0\right)=\frac{dz\left(0\right)}{dt}=\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)z\left(0\right)$$=\left(h\frac{\partial }{\partial x}$$+k\frac{\partial }{\partial y}\right)f\left(a,b\right)$

${z}^{″}\left(0\right)=\frac{{d}^{2}z\left(0\right)}{d{t}^{2}}=\left(h\frac{\partial }{\partial x}{+k\frac{\partial }{\partial y}\right)}^{2}z\left(0\right)$$=\left(h\frac{\partial }{\partial x}$${+k\frac{\partial }{\partial y}\right)}^{2}f\left(a,b\right)$

${z}^{\left(n\right)}\left(0\right)=\frac{{d}^{n}z\left(0\right)}{d{t}^{n}}={\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{n}z\left(0\right)$$=\left(h\frac{\partial }{\partial x}$$+{k\frac{\partial }{\partial y}\right)}^{n}f\left(a,b\right)$

また

${z}^{\left(n+1\right)}\left(t\right)=\frac{{d}^{n+1}z\left(t\right)}{d{t}^{n+1}}={\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{n+1}z\left(t\right)$$=\left(h\frac{\partial }{\partial x}$${+k\frac{\partial }{\partial y}\right)}^{n+1}f\left(a+ht,b+kt\right)$

なので

${z}^{\left(n+1\right)}\left(\theta \right)=\frac{{d}^{n+1}z\left(\theta \right)}{d{t}^{n+1}}={\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)}^{n+1}z\left(\theta \right)$$=\left(h\frac{\partial }{\partial x}$${+k\frac{\partial }{\partial y}\right)}^{n+1}f\left(a+h\theta ,b+k\theta \right)$

となる．以上を(3)に代入すると

$f\left(a+h,b+k\right)=f\left(a,b\right)$$+\frac{1}{1!}\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)f\left(a,b\right)$$+\frac{1}{2!}\left(h\frac{\partial }{\partial x}+{k\frac{\partial }{\partial y}\right)}^{2}f\left(a,b\right)$$+\cdots$$+\frac{1}{n!}\left(h\frac{\partial }{\partial x}{+k\frac{\partial }{\partial y}\right)}^{n}f\left(a,b\right)$$+{R}_{n+1}$

${R}_{n+1}$$=\frac{1}{\left(n+1\right)!}\left(h\frac{\partial }{\partial x}{+k\frac{\partial }{\partial y}\right)}^{n+1}f\left(a+\theta h,b+\theta k\right)$ 　　（ただし, $0<\theta <1$

となる．

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