第3章 3.16

(1)

$e^{x^2-x}$

$f = e^{x^2-x}$ とおくと

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

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

$f''' = 2(2x-1)e^{x^2-x} + 4(2x-1)e^{x^2-x} + (2x-1)^3 e^{x^2-x} \qquad f'''_{(0)} = -2-4-1 = -7$

これから

$f \cong 1 - x + \frac{3}{2}x^2 - \frac{7}{6}x^3$

(2)

$\tan x$

$f = \tan x$ とおくと

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

$f'' = \frac{2\cos x \cdot \sin x}{\cos^4 x} = \frac{2\sin x}{\cos^3 x} \qquad f''_{(0)} = 0$

$f''' = \frac{2\cos^4 x - 2\sin x \cdot 3\cos^2 x \cdot (-\sin x)}{\cos^6 x} \qquad f'''_{(0)} = 2$

これから

$f \cong x + \frac{1}{3}x^3$

(3)

$e^{\sin x}$

$f = e^{\sin x}$ とおくと

$f' = \cos x \cdot e^{\sin x} \qquad f'(0) = 1$

$f'' = -\sin x \cdot e^{\sin x} + \cos^2 x \cdot e^{\sin x} \qquad f''(0) = 1$

$f''' = -\cos x \cdot e^{\sin x} - \sin x \cdot \cos x \cdot e^{\sin x} - 2\cos x \cdot \sin x \cdot e^{\sin x} + \cos^3 x \cdot e^{\sin x} \qquad f'''(0) = 0$

これから

$f \cong 1 + x + \frac{1}{2}x^2$

(4)

$\log(\cos x)$

$f = \log(\cos x)$ とおくと

$f' = \frac{-\sin x}{\cos x} \qquad f'(0) = 0$

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

$f''' = \frac{2\sin x}{\cos^3 x} \qquad f'''(0) = 0$

これから

$f \cong -\frac{1}{2}x^2$

(5)

$\log(1+e^x)$

$f = \log(1+e^x)$ とおくと

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

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

$f''' = \frac{e^x(1+e^x)^2 - 2e^{2x}(1+e^x)}{(1+e^x)^4} \qquad f'''_{(0)} = \frac{4-4}{16} = 0$

これから

$f \cong \log 2 + \frac{1}{2}x + \frac{1}{8}x^2$