2変数関数のテイラー（Taylor）の定理の証明

$3$\displaystyle{ f $ a+h,b+k $ =f $ a,b $ }$ $3$\displaystyle{ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1! \hspace{2}} $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ f $ a,b $ }$ $3$\displaystyle{ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2! \hspace{2}} $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} +{ k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}f $ a,b $ }$ $3$\displaystyle{ +\cdots }$ $3$\displaystyle{ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} n! \hspace{2}} $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} { +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{n}f $a,b$ }$ $3$\displaystyle{ +\hspace{1}{R}_{ n+1 } }$

$3$\displaystyle{ \hspace{1}{R}_{ n+1 } }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} $ n+1 $ ! \hspace{2}} $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} { +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{ n+1 } f $ a+{\theta}h,b+{\theta}k $ }$ 　　（ただし, $3$\displaystyle{ 0< {\theta}< 1 }$ ）

■導出

$3$\displaystyle{z=f $ x,y $ }$ ， $3$\displaystyle{x=a+ht}$ ， $3$\displaystyle{y=b+kt}$ 　　( $3$\displaystyle{a,b,h,k}$ は定数)

のとき， $3$\displaystyle{z $t$ =f $ a+ht,b+kt $ }$ とおく．

$3$\displaystyle{z $t$ }$ を合成関数の微分法を用いて微分する．

$3$\displaystyle{\frac{\hspace{2} dz \hspace{2}}{\hspace{2} dt \hspace{2}}=\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}}+\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}}}$ ･･････(1)

$3$\displaystyle{\frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}}=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ a+ht $ =h}$ ， $3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}}=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ b+kt $ =k}$ ･･････(2)

(2)を(1)に代入すると，

$3$\displaystyle{\frac{\hspace{2} dz \hspace{2}}{\hspace{2} dt \hspace{2}}=\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}h+\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}k= $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ z}$

$3$\displaystyle{\frac{\hspace{2} {d}^{2}z \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ \frac{\hspace{2} dz \hspace{2}}{\hspace{2} dt \hspace{2}} $ }$

$3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} \{ $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ z \} }$

$3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ h\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }$

$3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ h\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ +\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ k\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }$

$3$\displaystyle{=h\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ +k\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }$

合成関数の微分法より

$3$\displaystyle{ \frac{\hspace{2} {d}^{2}z \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}=h \{ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}} }$ $3$\displaystyle{ +\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}} \} }$ $3$\displaystyle{ +k \{ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}} }$ $3$\displaystyle{ +\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}} \} }$

(2)を代入して整理すると

$3$\displaystyle{ \frac{\hspace{2} {d}^{2}z \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}= \{ h $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}} $ h }$ $3$\displaystyle{ + $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}} $ k \} }$ $3$\displaystyle{ +k \{ $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}} $ h }$ $3$\displaystyle{ + $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}} $ k \} }$

偏微分は順序交換が可能であるので

$3$\displaystyle{ \frac{\hspace{2} {d}^{2}z \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}={h}^{2} $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}} $ +2hk $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}} $ }$ $3$\displaystyle{ +{k}^{2} $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}} $ }$

よって

$3$\displaystyle{\frac{\hspace{2} {d}^{2}z \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}={ $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}z}$

以上より $3$\displaystyle{t}$ での微分を $3$\displaystyle{n}$ 回繰り返すと，

$3$\displaystyle{\frac{\hspace{2} {d}^{n}z \hspace{2}}{\hspace{2} d{t}^{n} \hspace{2}}={ $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{n}z}$

と推測できる．

ここで1変数関数 $3$\displaystyle{ z $t$ }$ にマクローリンの定理を適用すると，

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

$3$\displaystyle{\hspace{1}{R}_{ n+1 }=\frac{\hspace{2}1\hspace{2}}{\hspace{2} $ n+1 $ ! \hspace{2}}{z}^{ $ n+1 $ } ${\theta}$ {t}^{ n+1 }}$ $3$\displaystyle{ $ 0< {\theta}< 1 $ }$

$3$\displaystyle{t=1}$ とおくと，

$3$\displaystyle{ z $1$ =z $0$ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1! \hspace{2}}{z}^{\prime } $0$ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2! \hspace{2}}{z}^{\prime\prime } $0$ +\cdots }$ $3$\displaystyle{ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} n! \hspace{2}}{z}^{ $n$ } $0$ +\hspace{1}{R}_{ n+1 } }$

$3$\displaystyle{\hspace{1}{R}_{ n+1 }=\frac{\hspace{2}1\hspace{2}}{\hspace{2} $ n+1 $ ! \hspace{2}}{z}^{ $ n+1 $ } ${\theta}$ }$ $3$\displaystyle{ $ 0< {\theta}< 1 $ }$

$3$\displaystyle{z $t$ =f $ a+ht,b+kt $ }$ なので

$3$\displaystyle{z $0$ =f $ a,b $ }$

$3$\displaystyle{ {z}^{\prime } $0$ =\frac{\hspace{2} dz $0$ \hspace{2}}{\hspace{2} dt \hspace{2}}= $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ z $0$ }$ $3$\displaystyle{ = $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} }$ $3$\displaystyle{ +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ f$a,b$ }$

$3$\displaystyle{ {z}^{\prime\prime } $0$ =\frac{\hspace{2} {d}^{2}z $0$ \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}= $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} { +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}z $0$ }$ $3$\displaystyle{ = $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} }$ $3$\displaystyle{ { +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}f$a,b$ }$

$3$\displaystyle{ {z}^{ $n$ } $0$ =\frac{\hspace{2} {d}^{n}z $0$ \hspace{2}}{\hspace{2} d{t}^{n} \hspace{2}}={ $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{n}z $0$ }$ $3$\displaystyle{ = $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} }$ $3$\displaystyle{ +{ k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{n}f $ a,b $ }$

また

$3$\displaystyle{ {z}^{ $ n+1 $ } $t$ =\frac{\hspace{2} {d}^{ n+1 }z $t$ \hspace{2}}{\hspace{2} d{t}^{ n+1 } \hspace{2}}={ $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{ n+1 }z $t$ }$ $3$\displaystyle{ = $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} }$ $3$\displaystyle{ { +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{ n+1 }f $ a+ht,b+kt $ }$

なので

$3$\displaystyle{ {z}^{ $ n+1 $ } ${\theta}$ =\frac{\hspace{2} {d}^{ n+1 }z ${\theta}$ \hspace{2}}{\hspace{2} d{t}^{ n+1 } \hspace{2}}={ $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{ n+1 }z ${\theta}$ }$ $3$\displaystyle{ = $ h\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} }$ $3$\displaystyle{ { +k\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{ n+1 }f $ a+h{\theta},b+k{\theta} $ }$

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