第3章 3.5

(1)

$(1+x)^5$

$f = (1+x)^5$ とおくと

$f' = 5(1+x)^4 \qquad f'' = 20(1+x)^3$

１次 $f = f_{(a)} + f'_{(a)}(x-a)$ $a=0$

$= 1+5x$ $x=-0.02$から

$= 1-5\times0.02 = 0.9$

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

$= 1+5x+10x^2$ $x=-0.02$から

$= 1-5\times0.02+10\times(0.02)^2 = 0.904$

(2)

$\sqrt[3]{x}$

$f = \sqrt[3]{x} = x^{\frac{1}{3}}$ とおくと

$f' = \frac{1}{3}x^{-\frac{2}{3}} \qquad f'' = -\frac{2}{9}x^{-\frac{5}{3}}$

１次 $f = f_{(a)} + f'_{(a)}(x-a)$ $a=8$

$= 2+\dfrac{1}{12}(x-8)$ $x=8.01$から

$= 2+\dfrac{1}{12}(8.01-8) = 2.000833$

２次 $f = f_{(a)} + f'_{(a)}(x-a) + \dfrac{f''_{(a)}}{2}(x-a)^2$ $a=8$

$= 2+\dfrac{1}{12}(x-8)-\dfrac{1}{288}(x-8)^2$ $x=8.01$から

$= 2+\dfrac{1}{12}(8.01-8)-\dfrac{1}{288}(8.01-8)^2 = 2.000832$

(3)

$\log(1+x)$

$f = \log(1+x)$ とおくと

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

１次 $f = f_{(a)} + f'_{(a)}(x-a)$ $a=0$

$= 0 + x \qquad x=0.04\text{から}$

$= 0.04$

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

$= 0 + x - \frac{x^2}{2} \qquad x=0.04\text{から}$

$= 0.04 - \frac{(0.04)^2}{2} = 0.0392$

(4)

$\cos x$

$f = \cos x$ とおくと

$f' = -\sin x \qquad f'' = -\cos x$

１次 $f = f_{(a)} + f'_{(a)}(x-a)$ $a = \dfrac{\pi}{3}$

$= \frac{1}{2} - \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right) \qquad x=56°\text{から} \qquad x = \frac{\pi}{3} - \frac{4\pi}{180}$

$= \frac{1}{2} - \frac{\sqrt{3}}{2}\left(\frac{\pi}{3} - \frac{4\pi}{180} - \frac{\pi}{3}\right) = 0.560459$

２次 $f = f_{(a)} + f'_{(a)}(x-a) + \dfrac{f''_{(a)}}{2}(x-a)^2$ $a = \dfrac{\pi}{3}$

$= \frac{1}{2} - \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right) - \frac{1}{4}\left(x - \frac{\pi}{3}\right)^2 \qquad x=56°\text{から} \qquad x = \frac{\pi}{3} - \frac{4\pi}{180}$

$= \frac{1}{2} - \frac{\sqrt{3}}{2}\left(\frac{\pi}{3} - \frac{4\pi}{180} - \frac{\pi}{3}\right) - \frac{1}{4}\left(\frac{\pi}{3} - \frac{4\pi}{180} - \frac{\pi}{3}\right)^2 = 0.559241$