京都大学 情報学研究科 知能情報学専攻 2020年2月実施 情報学基礎 F1-2

标签：

• 京都大学
• 泰勒级数
• 极限
• 微分

Author

思齐塾, 祭音Myyura

Description

Answer the following questions. In the following, n is a positive integer and $\log x$ denotes $\log_e x$ .

(1) Let $f(x) = e^x (x^2 + x)$ . Derive $\frac{d^n f(x)}{dx^n}$ .

(2) (i) Find the Maclaurin series of $\log(1 - x)$ .

(ii) Derive the following limit:

$$\lim_{x \to 0} \frac{x + \log(1 - x)}{x^2}$$

(3) The Maclaurin series of $e^x$ is given by $e^x = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + \cdots$ . Derive the following limit:

$$\lim_{x \to 0} \frac{x \sin x}{e^{x^2} - \cos x}$$

题目描述

回答下列问题。以下 $n$ 为正整数，$\log x$ 表示自然对数。

1. 设

$$f(x)=e^x(x^2+x).$$

求

$$\frac{d^nf(x)}{dx^n}.$$

2. 完成：

   1. 求 $\log(1-x)$ 的 Maclaurin 级数。
   2. 利用该展开求极限

$$\lim_{x\to0} \frac{x+\log(1-x)}{x^2}.$$

3. 已知 $e^x$ 的 Maclaurin 级数为

$$e^x = 1+x+\frac{x^2}{2!} +\cdots+\frac{x^n}{n!}+\cdots.$$

利用级数展开求

$$\lim_{x\to0} \frac{x\sin x}{e^{x^2}-\cos x}.$$

Kai

(1) Let $f(x) = e^x(x^2+x)$ . We have: $f'(x) = e^x(x^2+x) + e^x(2x+1) = e^x(x^2 + 3x + 1)$ $f''(x) = e^x(x^2+3x+1) + e^x(2x+3) = e^x(x^2 + 5x + 4)$ $f'''(x) = e^x(x^2+5x+4) + e^x(2x+5) = e^x(x^2+7x+9)$ In general, $\frac{d^n f(x)}{dx^n} = e^x(x^2 + (2n+1)x + n^2)$. Differentiating this expression once gives $e^x(x^2+(2n+3)x+(n+1)^2)$, so induction proves the formula.

(2) (i) For $|x|<1$, the Maclaurin series of $\log(1-x)$ is as follows. The resulting series also converges to $\log2$ at $x=-1$; it diverges at $x=1$. Recall that the Maclaurin series of $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}$ . So, $\log(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots = - \sum_{n=1}^{\infty} \frac{x^n}{n}$ .

(ii) $\lim_{x \to 0} \frac{x + \log(1 - x)}{x^2} = \lim_{x \to 0} \frac{x - x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots}{x^2} = \lim_{x \to 0} \frac{-\frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots}{x^2} = \lim_{x \to 0} (-\frac{1}{2} - \frac{x}{3} - \frac{x^2}{4} - \cdots) = -\frac{1}{2}$ .

(3) $\lim_{x \to 0} \frac{x \sin x}{e^{x^2} - \cos x}$ . $e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots$ $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$ $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ $x \sin x = x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \frac{x^8}{7!} + \cdots$ $e^{x^2} - \cos x = (1 + x^2 + \frac{x^4}{2!} + \cdots) - (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots) = x^2 + \frac{x^2}{2} + \frac{x^4}{2} - \frac{x^4}{24} + \cdots = \frac{3x^2}{2} + \frac{11x^4}{24} + \cdots$ $\lim_{x \to 0} \frac{x \sin x}{e^{x^2} - \cos x} = \lim_{x \to 0} \frac{x^2 - \frac{x^4}{6} + \cdots}{\frac{3x^2}{2} + \frac{11x^4}{24} + \cdots} = \lim_{x \to 0} \frac{1 - \frac{x^2}{6} + \cdots}{\frac{3}{2} + \frac{11x^2}{24} + \cdots} = \frac{1}{\frac{3}{2}} = \frac{2}{3}$ .