2変数関数のテイラー（Taylor）の定理の導出

(1)式

${\displaystyle {\displaystyle {\int }_a^{a+h}\frac{\partial }{\partial x}{f}_y\left(x,b\right)dx}}$

${\displaystyle ={\displaystyle {\int }_a^{a+h}1\cdot f_{yx}\left(x,b\right)dx}}$

${\displaystyle ={\displaystyle {\int }_a^{a+h}-{\left(a+h-x\right)}^{\prime }\cdot f_{yx}\left(x,b\right)dx}}$

部分積分法を用いると

${\displaystyle ={\left[-\left(a+h-x\right)f_{yx}\left(x,b\right)\right]}_a^{a+h}}$${\displaystyle -\underset{a}{\overset{a+h}{{\displaystyle \int }}}-\left(a+h-x\right)\frac{\partial }{\partial x}{f}_{yx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left(a+h-x\right)f_{yxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}{\left\{-\frac{1}{2}{\left(a+h-x\right)}^2\right\}}^{\prime }{f}_{yxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yx}\left(a,b\right)}$${\displaystyle +{\left[-\frac{1}{2}{\left(a+h-x\right)}^2{f}_{yxx}\left(x,b\right)\right]}_a^{a+h}}$ ${\displaystyle -\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{-\frac{1}{2}{\left(a+h-x\right)}^2\right\}\frac{\partial }{\partial x}{f}_{yxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yx}\left(a,b\right)+\frac{1}{2}{h}^2{f}_{yxx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{\frac{1}{2}{\left(a+h-x\right)}^2\right\}{f}_{yxxx}\left(x,b\right)dx}$

積分を繰り返すと

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

(2)式

${\displaystyle {\displaystyle {\int }_a^{a+h}\frac{\partial }{\partial x}{f}_{yy}\left(x,b\right)dx}}$

${\displaystyle ={\displaystyle {\int }_a^{a+h}1\cdot f_{yyx}\left(x,b\right)dx}}$

${\displaystyle ={\displaystyle {\int }_a^{a+h}-{\left(a+h-x\right)}^{\prime }\cdot f_{yyx}\left(x,b\right)dx}}$

部分積分法を用いると

${\displaystyle ={\left[-\left(a+h-x\right)f_{yyx}\left(x,b\right)\right]}_a^{a+h}}$${\displaystyle -\underset{a}{\overset{a+h}{{\displaystyle \int }}}-\left(a+h-x\right)\frac{\partial }{\partial x}{f}_{yyx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yyx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}{\left\{-\frac{1}{2}{\left(a+h-x\right)}^2\right\}}^{\prime }{f}_{yyxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yyx}\left(a,b\right)}$${\displaystyle +{\left[-\frac{1}{2}{\left(a+h-x\right)}^2{f}_{yyxx}\left(x,b\right)\right]}_a^{a+h}}$${\displaystyle -\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{-\frac{1}{2}{\left(a+h-x\right)}^2\right\}\frac{\partial }{\partial x}{f}_{yyxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yyx}\left(a,b\right)+\frac{1}{2}{h}^2{f}_{yyxx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{\frac{1}{2}{\left(a+h-x\right)}^2\right\}{f}_{yyxxx}\left(x,b\right)dx}$

(1)式と同様に，積分を繰り返すと

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

(3)式

${\displaystyle {\displaystyle {\int }_a^{a+h}\frac{\partial }{\partial x}{f}_{yyy}\left(x,b\right)dx}}$

${\displaystyle ={\displaystyle {\int }_a^{a+h}1\cdot f_{yyyx}\left(x,b\right)dx}}$

${\displaystyle ={\displaystyle {\int }_a^{a+h}-{\left(a+h-x\right)}^{\prime }{f}_{yyyx}\left(x,b\right)dx}}$

${\displaystyle ={\left[-\left(a+h-x\right)f_{yyyx}\left(x,b\right)\right]}_a^{a+h}}$${\displaystyle -\underset{a}{\overset{a+h}{{\displaystyle \int }}}-\left(a+h-x\right)\frac{\partial }{\partial x}{f}_{yyyx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yyyx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}{\left\{-\frac{1}{2}{\left(a+h-x\right)}^2\right\}}^{\prime }{f}_{yyyxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yyyx}\left(a,b\right)}$${\displaystyle +{\left[-\frac{1}{2}{\left(a+h-x\right)}^2{f}_{yyyxx}\left(x,b\right)\right]}_a^{a+h}}$${\displaystyle -\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{-\frac{1}{2}{\left(a+h-x\right)}^2\right\}\frac{\partial }{\partial x}{f}_{yyyxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_{yyyx}\left(a,b\right)+\frac{1}{2}{h}^2{f}_{yyyxx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{\frac{1}{2}{\left(a+h-x\right)}^2\right\}{f}_{yyyxxx}\left(x,b\right)dx}$

積分を繰り返すと

${\displaystyle =h{f}_{yyyx}\left(a,b\right)+\frac{1}{2}{h}^2{f}_{yyyxx}\left(a,b\right)}$${\displaystyle +\cdots }$${\displaystyle +\frac{1}{\left(n-3\right)!}{h}^{n-3}{\left(\frac{\partial }{\partial x}\right)}^{n-3}{\left(\frac{\partial }{\partial y}\right)}^{3}f\left(a,b\right)}$

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最終更新日： 2023年1月21日