京都大学 情報学研究科 知能情報学専攻 2024年2月実施 基礎科目 F1-2

标签：

• Kyoto-University
• Continuity-and-Differentiability
• Multivariable-Differentiation
• Double-Integral

Author

祭音Myyura (co-authored with GPT 5.6 SOL)

Description

Q.1

Consider the following functions $f_1(x)$ and $f_2(x)$.

$$f_1(x) = \begin{cases} x\sin\left(\frac{1}{x}\right) & (x \neq 0) \\ 0 & (x = 0) \end{cases}$$

$$f_2(x) = \begin{cases} x^2\sin\left(\frac{1}{x}\right) & (x \neq 0) \\ 0 & (x = 0) \end{cases}$$

(1) Evaluate whether $f_1(x)$ and $f_2(x)$ are differentiable, respectively.

(2) Evaluate whether their derivatives ${f_1}'(x)$ and ${f_2}'(x)$ are continuous at $x = 0$, respectively.

Q.2

Evaluate the directional differentiability and continuity of the following function.

$$f(x, y) = \begin{cases} \frac{xy^2}{x^2 + y^4} & (\text{Either } x \neq 0 \text{ or } y \neq 0) \\ 0 & (\text{Both } x = 0 \text{ and } y = 0) \end{cases}$$

Q.3

Compute the volume in the $xyz$-space for each of the following conditions.

(1) Under the surface $z = e^x \cos(y)$ and over the rectangle defined by $D = [0, 1] \times \left[0, \frac{\pi}{2}\right]$ on the $xy$-plane.

(2) Under the surface of the paraboloid $z = 1 - x^2 - y^2$ and over the $xy$-plane.

题目描述

1. 定义

   $$f_1(x)=\begin{cases}x\sin(1/x),&x\ne0\\0,&x=0,\end{cases} \qquad f_2(x)=\begin{cases}x^2\sin(1/x),&x\ne0\\0,&x=0.\end{cases}$$
   1. 分别判断 $f_1,f_2$ 是否可微；
   2. 分别判断导函数 $f_1',f_2'$ 在 $x=0$ 是否连续。

2. 对

   $$f(x,y)=\begin{cases} \dfrac{xy^2}{x^2+y^4},&(x,y)\ne(0,0),\\ 0,&(x,y)=(0,0), \end{cases}$$

   判断方向可微性与连续性。

3. 求下列 $xyz$ 空间体积：

   1. $xy$ 平面矩形 $D=[0,1]\times[0,\pi/2]$ 上、曲面 $z=e^x\cos y$ 下方的体积；
   2. 抛物面 $z=1-x^2-y^2$ 与 $xy$ 平面之间的体积。

Kai

Q.1

(1)

$$\lim_{h\to 0}{f_1(h)-f_1(0)\over h}=\lim_{h\to 0}{h\sin(\frac1h)\over h}=\lim_{h\to 0}{\sin(\frac1h)}$$

does not exist. Hence $f_1$ is differentiable for $x\ne0$ but not at $x=0$.

$$\lim_{h\to 0}{f_2(h)-f_2(0)\over h}=\lim_{h\to 0}{h^2\sin(\frac1h)\over h}=\lim_{h\to 0}h\sin(\frac1h)=0$$

by the squeeze theorem. Hence $f_2$ is differentiable at $0$ with $f_2'(0)=0$; it is smooth for $x\ne0$, so it is differentiable on all of $\mathbb R$.

(2)

At $x=0$, $f_1'(x)$ does not exist, so it is not continuous.

From (1), $f'_2(0)=0$ and $\lim_{x\to 0}f_2'(x)=\lim_{x\to 0}(2x\sin{\frac1x}-\cos{\frac1x})$ does not exist, so $f'_2$ is also not continuous at $x=0$.

Q.2

The function is smooth away from $(0,0)$, so only the origin requires examination.

For a direction $(a,b)\ne(0,0)$,

$$\lim_{h\to0}\frac{f(ha,hb)-f(0,0)}h =\begin{cases} \dfrac{b^2}{a},&a\ne0,\\[4pt] 0,&a=0. \end{cases}$$

Thus every directional derivative at $(0,0)$ exists. In particular, the derivatives along the $x$- and $y$-axes are both $0$.

Along two paths approaching the origin,

$$f(y^2,y)=\frac12,\qquad f(0,y)=0.$$

Thus the limit does not exist. Hence $f$ is not continuous at $(0,0)$ (and therefore is not Fréchet differentiable there).

Q.3

(1)

$$\begin{aligned} V&=\int_0^1\int_0^{\pi/2}e^x\cos (y) dy dx \\ &=\int_0^1e^x dx\int_0^{\pi/2}\cos y\,dy \\ &=e-1. \end{aligned}$$

(2)

The partial surface to be integrated is

$$S=\{(x,y)|x^2+y^2\le 1\}.$$

Let $x=r\cos \theta,y=r\sin\theta$, $r\in[0,1]$ and $\theta\in[0,2\pi)$.

$$\begin{aligned} V&=\iint_{(x,y)\in S}z\,dx\,dy\\ &=\int_0^{2\pi}\int_0^1(1-r^2)r\,dr\,d\theta\\ &=2\pi \int_0^1(r-r^3)\,dr\\ &=2\pi \left(\frac12-\frac14\right)=\boxed{\frac{\pi}{2}}. \end{aligned}$$