2変数関数のテイラー（Taylor）の定理の導出

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

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

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

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

${\displaystyle =h{f}_x\left(a,b\right)+{\displaystyle {\int }_a^{a+h}\left(a+h-x\right)f_{xx}\left(x,b\right)dx}}$

${\displaystyle =h{f}_x\left(a,b\right)+{\left[-\frac{1}{2}{\left(a+h-x\right)}^2{f}_{xx}\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}_{xx}\left(x,b\right)dx}$

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

${\displaystyle =h{f}_x\left(a,b\right)+\frac{1}{2}{h}^2{f}_{xx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}{\left\{-\frac{1}{6}{\left(a+h-x\right)}^3\right\}}^{\prime }{f}_{xxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_x\left(a,b\right)+\frac{1}{2}{h}^2{f}_{xx}\left(a,b\right)}$${\displaystyle +{\left[-\frac{1}{6}{\left(a+h-x\right)}^3{f}_{xxx}\left(x,b\right)\right]}_a^{a+h}}$${\displaystyle -\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{-\frac{1}{6}{\left(a+h-x\right)}^3\right\}\frac{\partial }{\partial x}{f}_{xxx}\left(x,b\right)dx}$

${\displaystyle =h{f}_x\left(a,b\right)+\frac{1}{2}{h}^2{f}_{xx}\left(a,b\right)+\frac{1}{6}{h}^3{f}_{xxx}\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}\left\{\frac{1}{6}{\left(a+h-x\right)}^3\right\}{f}_{xxxx}\left(x,b\right)dx}$

このことから，積分を続けていくと

${\displaystyle =h{f}_x\left(a,b\right)+\frac{1}{2}{h}^2{f}_{xx}\left(a,b\right)+\frac{1}{6}{h}^3{f}_{xxx}\left(a,b\right)}$

${\displaystyle +\cdots $ {\displaystyle +\frac{1}{\left(n-1\right)!}{h}^{n-1}{\left(\frac{\partial }{\partial x}\right)}^{n-1}f\left(a,b\right)}$}$${\displaystyle +\frac{1}{n!}{h}^n{\left(\frac{\partial }{\partial x}\right)}^{n}f\left(a,b\right)}$${\displaystyle +\underset{a}{\overset{a+h}{{\displaystyle \int }}}\frac{1}{\left(n+1\right)!}{\left\{-{\left(a+h-x\right)}^{n+1}\right\}}^{\prime }{\left(\frac{\partial }{\partial x}\right)}^{n+1}f\left(x,b\right)dx}$

となる．

${\displaystyle {\displaystyle {\int }_b^{b+k}{f}_y\left(a+h,b\right)dy}}$

${\displaystyle ={\displaystyle {\int }_b^{b+k}1\cdot f_y\left(a+h,y\right)dy}}$

${\displaystyle ={\int }_b^{b+k}{\left\{-\left(b+k-y\right)\right\}}^{\prime }{f}_y\left(a+h,y\right)dy}$

${\displaystyle ={\left[-\left(b+k-y\right)f_y\left(a+h,y\right)\right]}_b^{b+k}}$${\displaystyle -\underset{b}{\overset{b+k}{{\displaystyle \int }}}\left\{-\left(b+k-y\right)\right\}\frac{\partial }{\partial y}{f}_y\left(a+h,y\right)dy}$

${\displaystyle =k{f}_y\left(a+h,b\right)+{\left[-\frac{1}{2}{\left(b+k-y\right)}^2{f}_{yy}\left(a+h,y\right)\right]}_b^{b+k}}$${\displaystyle -\underset{b}{\overset{b+k}{{\displaystyle \int }}}\left\{-\frac{1}{2}{\left(b+k-y\right)}^2\right\}\frac{\partial }{\partial y}{f}_{yy}\left(a+h,y\right)dy}$

${\displaystyle =k{f}_y\left(a+h,b\right)+\frac{1}{2}{k}^2{f}_{yy}\left(a+h,b\right)}$${\displaystyle +\underset{b}{\overset{b+k}{{\displaystyle \int }}}\left\{\frac{1}{2}{\left(b+k-y\right)}^2\right\}{f}_{yyy}\left(a+h,y\right)dy}$

${\displaystyle =k{f}_y\left(a+h,b\right)+\frac{1}{2}{k}^2{f}_{yy}\left(a+h,b\right)}$${\displaystyle +\underset{b}{\overset{b+k}{{\displaystyle \int }}}{\left\{-\frac{1}{6}{\left(b+k-y\right)}^3\right\}}^{\prime }{f}_{yyy}\left(a+h,y\right)dy}$

${\displaystyle =k{f}_y\left(a+h,b\right)+\frac{1}{2}{k}^2{f}_{yy}\left(a+h,b\right)}$${\displaystyle +{\left[-\frac{1}{6}{\left(b+k-y\right)}^3{f}_{yyy}\left(a+h,y\right)\right]}_b^{b+k}}$${\displaystyle -\underset{b}{\overset{b+k}{{\displaystyle \int }}}\left\{-\frac{1}{6}{\left(b+k-y\right)}^3\right\}\frac{\partial }{\partial y}{f}_{yyy}\left(a+h,y\right)dy}$

${\displaystyle =k{f}_y\left(a+h,b\right)+\frac{1}{2}{k}^2{f}_{yy}\left(a+h,b\right)}$${\displaystyle +\frac{1}{6}{k}^3{f}_{yyy}\left(a+h,b\right)}$${\displaystyle +\underset{b}{\overset{b+k}{{\displaystyle \int }}}\left\{\frac{1}{6}{\left(b+k-y\right)}^3\right\}{f}_{yyyy}\left(a+h,y\right)dy}$

このことから，積分を続けていくと

${\displaystyle =k{f}_y\left(a+h,b\right)}$${\displaystyle +\frac{1}{2}{k}^2{f}_{yy}\left(a+h,b\right)}$${\displaystyle +\frac{1}{6}{k}^3{f}_{yyy}\left(a+h,b\right)}$

${\displaystyle +\cdots }$${\displaystyle +\frac{1}{\left(n-1\right)!}{k}^{n-1}{\left(\frac{\partial }{\partial y}\right)}^{n-1}f\left(a+h,b\right)}$${\displaystyle +\frac{1}{n!}{k}^n{\left(\frac{\partial }{\partial y}\right)}^{n}f\left(a+h,b\right)}$${\displaystyle +\underset{b}{\overset{b+k}{{\displaystyle \int }}}\frac{1}{\left(n+1\right)!}{\left\{-{\left(b+k-y\right)}^{n+1}\right\}}^{\prime }{\left(\frac{\partial }{\partial y}\right)}^{n+1}f\left(a+h,y\right)dy}$

となる．

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最終更新日： 2019年9月17日