# 2変数関数のテイラー（Taylor）の定理の導出

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

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

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

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

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

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

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

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

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

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

$=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)$$+\underset{a}{\overset{a+h}{\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

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

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

${\int }_{b}^{b+k}{f}_{y}\left(a+h,b\right)dy$

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

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

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

$=k{f}_{y}\left(a+h,b\right)+{\int }_{b}^{b+k}\left(b+k-y\right){f}_{yy}\left(a+h,y\right)dy$

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

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

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

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

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

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

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

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