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：

• Kyoto-University
• Improper-Integral
• Double-Integral
• Taylor-Series
• Constrained-Optimization

Author

Isidore, Casablanca, 祭音Myyura

Description

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。

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)

Perform a Taylor series expansion of the function $f(x) = \log_e(x+ \triangle x)$, then insert $x=1, \triangle x = 0.02$. Calculating until the 5th term could lead to a result $0.01980256$, and then round it to $0.019803$

(2)

(solution by Isidore)

Perform the same expansion as (1) will directly prove

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

Let $f(x) = 1-\frac{x^2}{2}+\frac{x^3}{3} - \log_e(1+x)$, then its derivative is

$$f'(x) = \frac{x^3-x-1}{x+1}$$

We use Newton-Raphson's method to calculate the root of $x^3-x-1 = 0$.

Starting at $x_0 = 1.5$, the value of $x_1$ can be calculated as following

$$x_{1} = x_0 - \frac{x_0^3-x_0-1}{3x_0^2-1} \approx 1.3478$$

Therefore, insert $x=1.3478$,

$$\min \{f\} \approx f(1.3478) = 0.05 > 0$$

Q.E.D

---

(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

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

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

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

And consider the intersection $(x_0,y_0)$ of $y = x+1$ and $y = \frac 34 ex - \frac e8$, we have

$$x_0 = \frac{8+e}{6e} < 1$$

thus for $x > x_0, \frac 34 ex - \frac e8 > x+1$, hence for $x > 1$,

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

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

設問3

Perform the Lagrange multipliers method, we get

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

Calculate its derivative with the roots, the answer is

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