京都大学 情報学研究科 知能情報学専攻 2023年8月実施 情報学基礎 F1-2

标签：

• 京都大学
• 反常积分
• 二重积分
• 泰勒级数
• 约束最优化

Author

Isidore, Casablanca, 祭音Myyura

Description

大学公表の原題

設問1

以下の積分を求めよ。計算過程も明示すること。

(1) $\int_0^{\infty} \frac{1}{(x^2 + 1)^2} \text{d}x$

(2) $D = \left \{ (x, y) \mid x^2 + \frac{y^2}{4} \leq 1 \right \}$ としたときに、

$$\iint_D x^2 y^2 \text{d}x \text{d}y$$

設問2

以下の問いに答えよ。計算過程も明示すること。

(1) $\log_e (1.02)$ の小数第 $7$ 位を四捨五入し、小数第 $6$ 位まで求めよ。

(2) $x > 0$ に対して、次の不等式が成り立つことを示せ。

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

設問3

$3x^2 + 2y^2 + z^2 = 1$ の条件の下で、 $xyz$ の最大値と最小値を求めよ。

题目描述

回答下列三题，并明确写出计算过程。

1. 计算：

   1. 计算

$$\int_0^\infty\frac{1}{(x^2+1)^2}\,dx;$$

2. 对椭圆区域

$$D=\left\{(x,y)\ \middle|\ x^2+\frac{y^2}{4}\leq1\right\},$$

计算

$$\iint_Dx^2y^2\,dx\,dy.$$

2. 完成下列对数近似与不等式问题：

   1. 计算 $\log_e(1.02)$，将小数点后第 7 位四舍五入，给出保留 6 位小数的结果。
   2. 证明对每个 $x>0$ 都有

$$x-\frac{x^2}{2} < \log_e(1+x) < 1-\frac{x^2}{2}+\frac{x^3}{3}.$$

3. 在约束

$$3x^2+2y^2+z^2=1$$

下，求 $xyz$ 的最大值和最小值。

Kai

設問1

(1)

Let $x = \tan \theta$ , we have $dx=\frac{d\theta}{\cos^{2} \theta}$ . Then

$$\begin{aligned} \int_{0}^{\infty}\frac{1}{(1+x^{2})^{2}}dx &= \int_{0}^{\pi/2}\cos^{4} \theta~\frac{d\theta}{\cos^{2} \theta} = \int_{0}^{\pi/2}\cos^{2}\theta~d\theta \\ &= \frac{1}{2} \int_{0}^{\pi/2} (1 + \cos 2\theta) d\theta = \frac{\pi}{4} \end{aligned}$$

(2)

Let $x = r \cos \theta, y = 2r \sin \theta$ , the Jacobian determinant

$$\begin{aligned} J &= \begin{vmatrix} \cos\theta & -r\sin\theta\\ 2\sin\theta & 2r\cos\theta \end{vmatrix} = 2r\cos^{2}\theta+2r\sin^{2}\theta = 2r \end{aligned}$$

Then we have

$$\begin{aligned} \iint_{D}x^{2}y^{2}dxdy &= \int_{0}^{1}\int_{0}^{2\pi}4r^{4}\sin^{2}\theta\cos^{2}\theta~(|2r|dr)d\theta = \int_{0}^{1}2r^{5}~dr\int_{0}^{2\pi}\sin^{2}2\theta~d\theta\\[0.7em] &= \int_{0}^{1}2r^{5}~dr\frac{1}{2}\int_{0}^{2\pi}(1-\cos 4\theta)~d\theta = \left[\frac{r^{6}}{3}\right]_{0}^{1}\cdot \frac{1}{2}\left[\theta-\frac{1}{4}\sin 4\theta\right]_{0}^{2\pi} = \frac{\pi}{3} \end{aligned}$$

設問2

(1)

Using $\log(1+t)=\sum_{k=1}^{\infty}(-1)^{k+1}t^k/k$ with $t=0.02$,

$$\sum_{k=1}^{5}(-1)^{k+1}\frac{0.02^k}{k} =0.019802627306\ldots .$$

The alternating-series remainder is less than $0.02^6/6<1.1\times10^{-11}$, so the required value is $0.019803$.

(2)

(solution by Isidore)

For the lower bound, let

$$g(x)=\log(1+x)-x+\frac{x^2}{2}.$$

Then $g(0)=0$ and $g'(x)=x^2/(1+x)>0$ for $x>0$, so $x-x^2/2<\log(1+x)$.

For the upper bound, first note that

$$h(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\log(1+x)$$

satisfies $h(0)=0$ and $h'(x)=x^3/(1+x)>0$. Thus, for $0<x\le1$,

$$\log(1+x)<x-\frac{x^2}{2}+\frac{x^3}{3} \le1-\frac{x^2}{2}+\frac{x^3}{3}.$$

For $x\ge1$, write $y=x-1\ge0$. Concavity gives $\log(1+x)\le\log2+y/2$, so

$$\begin{aligned} 1-\frac{x^2}{2}+\frac{x^3}{3}-\log(1+x) &\ge\frac56-\log2+\frac{y^2-y}{2}+\frac{y^3}{3}\\ &\ge\frac{17}{24}-\log2>\frac1{120}>0. \end{aligned}$$

Here $\log2<7/10$, since $e^{7/10}>\sum_{k=0}^4(7/10)^k/k!>2$.

---

(solution by Casablanca)

Easy to see that we only need to prove that:

$$1+x < e^{1- \frac{x^2}{2}+ \frac{x^3}{3}}$$

Let $f(x) = e^{1- \frac{x^2}{2}+ \frac{x^3}{3}}$ . Then we have $f(x) \geq f(1) = e^{\frac 56}$ .

• (i) for $x \in (0, 1], x+1 \leq 2 < e^{\frac 56} \leq f(x)$ $\quad$ (note: $e^x > 1+ x + \frac{x^2}{2}, e^{\frac 56} > \frac{157}{72}>2$ )
• (ii) for $x \in (1, + \infty)$ , $f'(x) = (x^2 - x)f(x) > 0$ , $f(x)$ increases and we consider the point $(\frac{3}{2},e)$ on $(x,f(x))$ , $f'(\frac 32) = \frac 34 e$ .

Let

$$F(x) = f(x) - (\frac 34 e(x - \frac 32) + e) = f(x) - \frac 34 ex + \frac e8,$$

where

$$x \in (1, +\infty)$$

and we know that: $F(\frac 32) = 0, F'(x) = (x^2-x)f(x) - \frac 34 e$ , it's obvious that $F'(x)$ increases for $x > 1$ , and $F'(\frac 32) = 0$ .

Hence $F(x) \geq F(\frac 32 ) = 0$ for $x > 1$ .

Thus

$$f(x) \geq \frac 34 ex - \frac e8, x \in (1,+\infty)$$

Consider the intersection $(x_0,y_0)$ of $y = x+1$ and $y = \frac 34 ex - \frac e8$ . We have

$$x_0 = \frac{8+e}{6e-8}\approx1.290.$$

Since $e>65/24>46/17$, we have $x_0<13/10$. Also $e^{5/6}>\sum_{k=0}^6(5/6)^k/k!>23/10$. Thus, for $1<x\leq x_0$, $f(x)\geq f(1)=e^{5/6}>x_0+1\geq x+1$. For $x>x_0$, the tangent bound gives

$$f(x) \geq \frac 34 ex - \frac e8 > x+1$$

and from (i) and (ii), finally ,we know that $f(x) > 1+x$

設問3

Perform the Lagrange multipliers method, we get

$$L(x,y,z;\lambda) = xyz - \lambda(3x^2+2y^2+z^2-1)$$

At a nonzero extremum, the stationary equations give

$$3x^2=2y^2=z^2=\frac13.$$

Thus $|x|=1/3$, $|y|=1/\sqrt6$, $|z|=1/\sqrt3$, and both signs of the product occur. Therefore,

$$\min \{xyz\} = -\frac{1}{18}\sqrt{2}, \quad \max \{xyz\} = \frac{1}{18}\sqrt{2}$$