# 積分を用いたテイラーの定理の導出

$f\left(b\right)=f\left(a\right)+{\int }_{a}^{b}{f}^{\prime }\left(x\right)dx$ ･･････(1)

と表すことができる．

$\underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)dx=\underset{a}{\overset{b}{\int }}1·{f}^{\prime }\left(x\right)dx$

$={\int }_{b}^{a}{\left\{-\left(b-x\right)\right\}}^{\text{'}}\cdot {f}^{\text{'}}\left(x\right)dx$

（∵1を${\left\{-\left(b-x\right)\right\}}^{\prime }$と考えている注）

$={\left[-\left(b-x\right){f}^{\prime }\left(x\right)\right]}_{a}^{b}-\underset{a}{\overset{b}{\int }}\left\{-\left(b-x\right)\right\}\cdot {f}^{″}\left(x\right)dx$

（∵部分積分

$=\frac{{f}^{\prime }\left(a\right)}{1!}\left(b-a\right)+\underset{a}{\overset{b}{\int }}\frac{{f}^{″}\left(x\right)}{1!}·\left(b-a\right)dx$ ･･････(2)

(2)を(1)に代入すると

$f\left(b\right)=f\left(a\right)+\frac{{f}^{\prime }\left(a\right)}{1!}\left(b-a\right)+{\int }_{a}^{b}\frac{{f}^{″}\left(x\right)}{1!}·\left(b-x\right)dx$     ･･････(3)

となる．(3)の右辺の積分を部分積分すると

${\int }_{a}^{b}\frac{{f}^{″}\left(x\right)}{1!}·\left(b-x\right)dx$

$={\int }_{a}^{b}\frac{{f}^{″}\left(x\right)}{1!}·{\left\{-\frac{1}{2}{\left(b-x\right)}^{2}\right\}}^{\prime }dx$

$={\left[\frac{{f}^{″}\left(x\right)}{1!}·\left\{-\frac{1}{2}{\left(b-x\right)}^{2}\right\}\right]}_{a}^{b}-{\int }_{a}^{b}\frac{{f}^{\left(3\right)}\left(x\right)}{1!}·\left\{-\frac{1}{2}{\left(b-x\right)}^{2}\right\}dx$

$=\frac{{f}^{″}\left(a\right)}{2!}{\left(b-a\right)}^{2}+{\int }_{a}^{b}\frac{{f}^{\left(3\right)}\left(x\right)}{2!}·{\left(b-x\right)}^{2}dx$     ･･････(4)

(4)を(3)に代入すると

$f\left(b\right)=f\left(a\right)+\frac{{f}^{\prime }\left(a\right)}{1!}\left(b-a\right)+\frac{{f}^{″}\left(a\right)}{2!}{\left(b-a\right)}^{2}+{\int }_{a}^{b}\frac{{f}^{\left(3\right)}\left(x\right)}{2!}·{\left(b-x\right)}^{2}dx$

となる．同様な操作を繰り返すと

$f\left(b\right)=f\left(a\right)+\frac{{f}^{\prime }\left(a\right)}{1!}\left(b-a\right)+\frac{{f}^{″}\left(a\right)}{2!}{\left(b-a\right)}^{2}+\cdots \cdots +\frac{{f}^{\left(n-1\right)}\left(a\right)}{\left(n-1\right)!}{\left(b-a\right)}^{n-1}+\underset{a}{\overset{b}{\int }}\frac{{f}^{\left(n\right)}\left(x\right)}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx$

が得られる．

${\int }_{a}^{b}\frac{m}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx\leqq {\int }_{a}^{b}\frac{{f}^{\left(n\right)}\left(x\right)}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx\leqq {\int }_{a}^{b}\frac{M}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx$

と表すことができる．

${\int }_{a}^{b}\frac{m}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx=\frac{m}{n!}{\left(b-a\right)}^{n}$

${\int }_{a}^{b}\frac{M}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx=\frac{M}{n!}{\left(b-a\right)}^{n}$

より

$\frac{m}{n!}{\left(b-a\right)}^{n}\leqq {\int }_{a}^{b}\frac{{f}^{\left(n\right)}\left(x\right)}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx\leqq \frac{M}{n!}{\left(b-a\right)}^{n}$

よって

${\int }_{a}^{b}\frac{{f}^{\left(n\right)}\left(x\right)}{\left(n-1\right)!}·{\left(b-x\right)}^{n-1}dx=\frac{{f}^{\left(n\right)}\left(c\right)}{n!}{\left(b-a\right)}^{n}$

となる， ${f}^{\left(n\right)}\left(c\right)\text{\hspace{0.17em}}\left(m\leqq {f}^{\left(n\right)}\left(c\right)\leqq M\text{\hspace{0.17em}},a が存在する．

$f\left(b\right)=f\left(a\right)+\frac{{f}^{\prime }\left(a\right)}{1!}\left(b-a\right)+\frac{{f}^{″}\left(a\right)}{2!}{\left(b-a\right)}^{2}+\cdots \cdots +\frac{{f}^{\left(n-1\right)}\left(a\right)}{\left(n-1\right)!}{\left(b-a\right)}^{n-1}+\frac{{f}^{\left(n\right)}\left(c\right)}{n!}{\left(b-a\right)}^{n}$

となり， テイラーの定理 が導かれる．

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