テイラー展開の計算方法

$$l:y=f(a)+f^{\prime }(a)(x-a)$$

$$f(x)\simeq f(a)+f^{\prime }(a)(x-a)$$

$$\begin{array}{}\text{(1)}& f(x)\simeq f(a)+f^{\prime }(a)(x-a)+p{x}^2+qx+r\end{array}$$

$$\begin{array}{}\text{(2)}& f^{\prime }(x)\simeq f^{\prime }(a)+2px+q\end{array}$$

$$\begin{array}{}\text{(3)}& f^{\prime }{}^{\prime }(x)\simeq 2p\end{array}$$

$$\begin{array}{rl}f(a)& \simeq f(a)+p{a}^2+qa+r\\ \\ f^{\prime }(a)& \simeq f^{\prime }(a)+2pa+q\\ \\ f^{\prime }{}^{\prime }(a)& \simeq 2p\end{array}$$

$$\begin{array}{l}p{a}^2+qa+r=0\\ \\ 2pa+q=0\\ \\ 2p=f^{\prime }{}^{\prime }(a)\end{array}$$

$$\begin{array}{rl}p& ={\displaystyle \frac{1}{2}{f}^{\prime }{}^{\prime }(a)}\\ \\ q& =-a{f}^{\prime }{}^{\prime }(a)\\ \\ r& ={\displaystyle \frac{1}{2}{a}^2{f}^{\prime }{}^{\prime }(a)}\end{array}$$

$$p{x}^2+qx+r={\displaystyle \frac{1}{2}{f}^{\prime }{}^{\prime }(a)(x-a{)}^2}$$

$$f(x)\simeq f(a)+f^{\prime }(a)(x-a)+{\displaystyle \frac{1}{2}{f}^{\prime }{}^{\prime }(a)(x-a{)}^2}$$

$$f(x)\simeq f(a)+f^{\prime }(a)(x-a)+{\displaystyle \frac{1}{2}{f}^{\prime }{}^{\prime }(a)(x-a{)}^2+{\displaystyle \frac{1}{6}{f}^{\prime }{}^{\prime }{}^{\prime }(a)(x-a{)}^3}}$$

$$f(x)\simeq {\displaystyle \sum _{k=0}^n{\displaystyle \frac{1}{k!}{f}^{(k)}(a)(x-a{)}^k}}$$

$$f(x)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{n!}{f}^{(n)}(a)(x-a{)}^n}}$$

$$f(x)={\displaystyle \frac{x^2}{1-x}}$$

$$\begin{array}{ll}{f}^{(n)}(x)={\displaystyle \frac{x(2-x)}{(1-x{)}^2}}& (n=1)\\ \\ f^{(n)}(x)=n!(1-x{)}^{-n-1}& (n\ge 2)\end{array}$$

$$\begin{array}{rl}{f}_1(x)& ={\displaystyle \sum _{n=0}^1{\displaystyle \frac{1}{n!}{f}^{(n)}(2)(x-2{)}^n}}\\ \\ & =f(2)+f^{(1)}(2)(x-2)\\ \\ & =-4\end{array}$$

$$\begin{array}{rl}{f}_2(x)& ={\displaystyle \sum _{n=0}^2{\displaystyle \frac{1}{n!}{f}^{(n)}(2)(x-2{)}^n}}\\ \\ & =f_1(x)+{\displaystyle \frac{1}{2}{f}^{(2)}(2)(x-2{)}^2}\\ \\ & =-x^2+4x-8\end{array}$$

$$\begin{array}{rl}{f}_3(x)& ={\displaystyle \sum _{n=0}^3{\displaystyle \frac{1}{n!}{f}^{(n)}(2)(x-2{)}^n}}\\ \\ & =f_2(x)+{\displaystyle \frac{1}{6}{f}^{(3)}(2)(x-2{)}^3}\\ \\ & =x^3-7x^2+16x-16\end{array}$$

$$\begin{array}{rl}{f}_4(x)& ={\displaystyle \sum _{n=0}^4{\displaystyle \frac{1}{n!}{f}^{(n)}(2)(x-2{)}^n}}\\ \\ & =f_3(x)+{\displaystyle \frac{1}{24}{f}^{(4)}(2)(x-2{)}^4}\\ \\ & =-x^4+9x^3-31x^2+48x-32\end{array}$$

$$\begin{array}{rl}{f}_5(x)& ={\displaystyle \sum _{n=0}^5{\displaystyle \frac{1}{n!}{f}^{(n)}(2)(x-2{)}^n}}\\ \\ & =f_4(x)+{\displaystyle \frac{1}{120}{f}^{(5)}(2)(x-2{)}^5}\\ \\ & =x^5-11x^4+49x^3-111x^2+128x-64\end{array}$$

$$f(x)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{n!}{f}^{(n)}(0)x^n}}$$

$$f(x)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{n!}{f}^{(n)}(a)(x-a{)}^n}}$$

$$f(x)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{n!}{f}^{(n)}(0)x^n}}$$