2変数関数のテイラー（Taylor）の定理の証明

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

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

■導出

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

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

${\displaystyle z\left(t\right)}$  を合成関数の微分法を用いて微分する．

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

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

(2)を(1)に代入すると，

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

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

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

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

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

合成関数の微分法より

(2)を代入して整理すると

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

偏微分は順序交換が可能であるので

${\displaystyle \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)}$${\displaystyle +k^2\left(\frac{{\partial }^{2}z}{\partial y^2}\right)}$

よって

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

以上より${\displaystyle t}$ での微分を${\displaystyle n}$ 回繰り返すと，

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

と推測できる．

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

${\displaystyle 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}$${\displaystyle +\cdots }$ ${\displaystyle +\frac{1}{n!}{z}^{\left(n\right)}\left(0\right)t^n}$ ${\displaystyle +R_{n+1}}$ ･･････(3)

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

${\displaystyle t=1}$  とおくと，

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

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

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

また

なので

となる．以上を(3)に代入すると

となる．

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最終更新日： 2023年10月16日