第3章 3.15

(1)

$\frac{2}{2-x}$

$f = \dfrac{2}{2-x}$ とおくと

$f' = \dfrac{2}{(2-x)^2}$ $f'_{(0)} = \dfrac{1}{2}$

$f'' = \dfrac{4}{(2-x)^3}$ $f''_{(0)} = \dfrac{1}{2}$

$f''' = \dfrac{12}{(2-x)^4}$ $f'''_{(0)} = \dfrac{3}{4}$

$f^{(4)} = \dfrac{48}{(2-x)^5}$ $f^{(4)}_{(0)} = \dfrac{3}{2}$

これから

$f \cong 1 + \frac{1}{2}\cdot\frac{x}{1!} + \frac{1}{2}\cdot\frac{x^2}{2!} + \frac{3}{4}\cdot\frac{x^3}{3!} + \frac{3}{2}\cdot\frac{x^4}{4!} = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16}$

(2)

$\frac{1-x}{1+x}$

$f = \dfrac{1-x}{1+x}$ とおくと

$f' = \dfrac{-1-x-1+x}{(1+x)^2} = \dfrac{-2}{(1+x)^2}$ $f'_{(0)} = -2$

$f'' = \dfrac{4}{(1+x)^3}$ $f''_{(0)} = 4$

$f''' = \dfrac{-12}{(1+x)^4}$ $f'''_{(0)} = -12$

$f^{(4)} = \dfrac{48}{(1+x)^5}$ $f^{(4)}_{(0)} = 48$

これから

$f \cong 1 - \frac{2}{1!}x + \frac{4}{2!}x^2 - \frac{12}{3!}x^3 + \frac{48}{4!}x^4 = 1 - 2x + 2x^2 - 2x^3 + 2x^4$

(3)

$(1-x)^{\frac{3}{2}}$

$f = (1-x)^{\frac{3}{2}}$ とおくと

$f' = -\frac{3}{2}(1-x)^{\frac{1}{2}} \qquad f'_{(0)} = -\frac{3}{2}$

$f'' = \frac{3}{4}(1-x)^{-\frac{1}{2}} \qquad f''_{(0)} = \frac{3}{4}$

$f''' = \frac{3}{8}(1-x)^{-\frac{3}{2}} \qquad f'''_{(0)} = \frac{3}{8}$

$f^{(4)} = \frac{9}{16}(1-x)^{-\frac{5}{2}} \qquad f^{(4)}{}_{(0)} = \frac{9}{16}$

これから

$f \cong 1 - \frac{3}{2}x + \frac{3}{4}\cdot\frac{x^2}{2!} + \frac{3}{8}\cdot\frac{x^3}{3!} + \frac{9}{16}\cdot\frac{x^4}{4!} = 1 - \frac{3}{2}x + \frac{3}{8}x^2 + \frac{1}{16}x^3 + \frac{3}{128}x^4$