明治大学 先端数理科学研究科 現象数理学専攻 2020年8月実施 微积分

标签：

• Meiji-University
• Taylor-Series
• Limit
• Integration

Author

思齐塾, 祭音Myyura

Description

(1) 次の関数を3次の項までマクローリン展開せよ。

(a) $x \cos x$

(b) $\log(1+3x)$

(2) 極限 $\lim_{x \to 0} \left(\frac{1}{\log(1+3x)} - \frac{1}{3x \cos x}\right)$ を求めよ。

(3) 次の積分を計算せよ。

$\iiint_V z dx dy dz, V = \{(x, y, z) | 0 \leq x \leq 1, 0 \leq y \leq 1-x, 0 \leq z \leq 1-x-y\}.$

题目描述

(1) 将下列函数作 Maclaurin 展开，写到三次项为止。

(a) $x\cos x$

(b) $\log(1+3x)$

(2) 求极限

$$\lim_{x\to0}\left( \frac{1}{\log(1+3x)}-\frac{1}{3x\cos x} \right).$$

(3) 计算三重积分

$$\iiint_V z\,dx\,dy\,dz,$$

其中

$$V=\left\{(x,y,z)\,\middle|\, 0\leq x\leq1,\; 0\leq y\leq1-x,\; 0\leq z\leq1-x-y \right\}.$$

Kai

(1)(a) $f(x) = x\cos x$ $f'(x) = \cos x - x\sin x$ $f''(x) = -\sin x - \sin x - x\cos x = -2\sin x - x\cos x$ $f'''(x) = -2\cos x - \cos x + x\sin x = -3\cos x + x\sin x$

$f(0) = 0, f'(0) = 1, f''(0) = 0, f'''(0) = -3$

$x\cos x = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ...$ $= x + \frac{-3}{6}x^3 + ...$ $= x - \frac{1}{2}x^3 + O(x^5)$

(b) $\log(1+3x) = 3x - \frac{(3x)^2}{2} + \frac{(3x)^3}{3} + O(x^4)$ $= 3x - \frac{9}{2}x^2 + 9x^3 + O(x^4)$

(2) $\lim_{x \to 0} \left(\frac{1}{\log(1+3x)} - \frac{1}{3x \cos x}\right) = \lim_{x \to 0} \frac{3x \cos x - \log(1+3x)}{3x\cos x \log(1+3x)}$ $\log(1+3x) = 3x - \frac{(3x)^2}{2} + \frac{(3x)^3}{3} + O(x^4) = 3x - \frac{9x^2}{2} + 9x^3 + O(x^4)$ $\cos x = 1 - \frac{x^2}{2} + O(x^4)$ $3x\cos x = 3x(1-\frac{x^2}{2} + O(x^4)) = 3x - \frac{3}{2}x^3 + O(x^5)$ $3x\cos x - \log(1+3x) = 3x - \frac{3}{2}x^3 - (3x - \frac{9x^2}{2} + 9x^3) + O(x^4) = \frac{9}{2}x^2 - \frac{21}{2}x^3 + O(x^4)$ $3x\cos x \log(1+3x) = (3x - \frac{3}{2}x^3)(3x - \frac{9}{2}x^2 + 9x^3) + O(x^5) = 9x^2 - \frac{27}{2}x^3 + 27x^4 - \frac{9}{2}x^4 + O(x^5) = 9x^2 - \frac{27}{2}x^3 + \frac{45}{2}x^4 + O(x^5)$ $\lim_{x \to 0} \frac{\frac{9}{2}x^2 - \frac{21}{2}x^3}{9x^2 - \frac{27}{2}x^3} = \lim_{x \to 0} \frac{\frac{9}{2} - \frac{21}{2}x}{9 - \frac{27}{2}x} = \frac{\frac{9}{2}}{9} = \frac{1}{2}$

(3) $\iiint_V z dx dy dz = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} z dz dy dx = \int_0^1 \int_0^{1-x} \frac{1}{2}(1-x-y)^2 dy dx = \int_0^1 [-\frac{1}{6}(1-x-y)^3]_0^{1-x} dx = \int_0^1 \frac{1}{6}(1-x)^3 dx = [-\frac{1}{24}(1-x)^4]_0^1 = \frac{1}{24}$