京都大学 情報学研究科 知能情報学専攻 2025年8月実施 情報学基礎 F1-2

标签：

• Kyoto-University
• Constrained-Optimization
• Differentiation
• Multivariable-Differentiation
• Limit
• Infinite-Series

Author

祭音Myyura

Description

In the questions below, $\mathbb{R}$ is the set of all real numbers and $e$ denotes Napier's constant (the base of the natural logarithm function).

Q.1

Consider the real-valued function $f(x, y, z) = xyz$ defined on $\mathbb{R}^3$. Derive the maximum and minimum values of $f(x, y, z)$ subject to the constraint that $g(x, y, z) = x^2 + y^2 + z^2 - 12 = 0$. Derivations must be clearly shown.

Q.2

Answer the following questions about the real-valued functions $f(x) = x \sin(ax)$ and $g(x) = \sin(ax)$. Let $a$ be a non-zero real number and $n$ be an integer greater than $0$. Here, $f^{(n)}(x)$ and $g^{(n)}(x)$ denote the $n$-th derivatives of $f(x)$ and $g(x)$, respectively.

(1) Derive $f^{(1)}(x)$ and $f^{(2)}(x)$.

(2) Derive $g^{(n)}(x)$.

(3) Derive $f^{(n)}(x)$.

Q.3

Answer the following questions. Derivations must be clearly shown.

(1) Find the following limit, if it exists. If the limit does not exist, prove it.

$$\lim_{(x,y)\to(0,0)} \frac{x + y}{\sqrt{x^2 + y^2}}$$

(2) Determine the set of $x \in \mathbb{R}$ for which the following series of functions converges.

$$\sum_{n=1}^{\infty} \frac{e^{nx}}{\sqrt{2n^4 + 4} - \sqrt{n^4 + 1}}$$

题目描述

以下 $\mathbb R$ 表示实数集，$e$ 表示自然对数的底。

1. 考虑定义在 $\mathbb R^3$ 上的实值函数

   $$f(x,y,z)=xyz.$$

   在约束

   $$g(x,y,z)=x^2+y^2+z^2-12=0$$

   下求 $f(x,y,z)$ 的最大值与最小值，并清楚写出推导过程。

2. 对实值函数 $f(x)=x\sin(ax)$ 与 $g(x)=\sin(ax)$ 回答下列问题。其中 $a$ 为非零实数，$n$ 为正整数，$f^{(n)}(x)$ 与 $g^{(n)}(x)$ 分别表示 $n$ 阶导数。

   （1）求 $f^{(1)}(x)$ 与 $f^{(2)}(x)$。

   （2）求 $g^{(n)}(x)$。

   （3）求 $f^{(n)}(x)$。

3. 回答下列问题，并清楚写出推导过程。

   （1）求极限

   $$\lim_{(x,y)\to(0,0)}\frac{x+y}{\sqrt{x^2+y^2}}.$$

   若极限不存在，则证明其不存在。

   （2）确定使函数项级数

   $$\sum_{n=1}^{\infty} \frac{e^{nx}} {\sqrt{2n^4+4}-\sqrt{n^4+1}}$$

   收敛的所有 $x\in\mathbb R$。

考点

• 约束极值：使用 Lagrange 乘数法求球面上三元乘积的全局最大值与最小值。
• 高阶导数：结合三角函数导数周期与 Leibniz 公式推导 $x\sin(ax)$ 的一般阶导数。
• 二元函数极限：沿不同趋近路径检验多元极限是否存在。
• 无穷级数收敛性：分析一般项的渐近阶，并用根值或比值判别法确定参数范围。

Kai

Q.1

Let

$$f(x,y,z)=xyz, \qquad g(x,y,z)=x^2+y^2+z^2-12.$$

Since the constraint set is compact, $f$ attains both a maximum and a minimum.

Using Lagrange multipliers,

$$\nabla f=\lambda \nabla g,$$

so

$$yz=2\lambda x,\qquad xz=2\lambda y,\qquad xy=2\lambda z.$$

If $xyz\neq 0$, multiplying these equations by $x,y,z$, respectively, gives

$$x^2=y^2=z^2.$$

Together with

$$x^2+y^2+z^2=12,$$

we obtain

$$x^2=y^2=z^2=4.$$

Thus $x,y,z\in\{\pm2\}$, and hence

$$xyz=\pm8.$$

Points with one or more zero coordinates give $f=0$, so they are not global extrema.

Therefore,

$$\boxed{\max f=8}$$

at

$$(2,2,2),\quad (2,-2,-2),\quad (-2,2,-2),\quad (-2,-2,2),$$

and

$$\boxed{\min f=-8}$$

at the remaining four points with coordinates $\pm2$.

Q.2

Let

$$f(x)=x\sin(ax), \qquad g(x)=\sin(ax), \qquad a\neq0.$$

(1)

By direct differentiation,

$$\boxed{ f'(x)=\sin(ax)+ax\cos(ax) }$$

and

$$\boxed{ f''(x)=2a\cos(ax)-a^2x\sin(ax) }.$$

(2)

The derivatives of $\sin(ax)$ repeat with period four. Hence

$$\boxed{ g^{(n)}(x) = a^n\sin\left(ax+\frac{n\pi}{2}\right) }.$$

(3)

Using Leibniz's rule and the fact that derivatives of $x$ of order at least two vanish,

$$f^{(n)}(x) = xg^{(n)}(x)+n g^{(n-1)}(x).$$

Therefore,

$$\boxed{ f^{(n)}(x) = a^n x\sin\left(ax+\frac{n\pi}{2}\right) + na^{n-1}\sin\left(ax+\frac{(n-1)\pi}{2}\right) }.$$

Q.3

(1)

Consider the path $y=0$. Then

$$\frac{x+y}{\sqrt{x^2+y^2}} = \frac{x}{|x|}.$$

As $x\to0^+$,

$$\frac{x}{|x|}\to1,$$

whereas as $x\to0^-$,

$$\frac{x}{|x|}\to-1.$$

Since the two values are different, the limit does not exist:

$$\boxed{ \lim_{(x,y)\to(0,0)} \frac{x+y}{\sqrt{x^2+y^2}} \text{ does not exist} }.$$

(2)

Consider

$$\sum_{n=1}^{\infty} \frac{e^{nx}} {\sqrt{2n^4+4}-\sqrt{n^4+1}}.$$

Let

$$d_n=\sqrt{2n^4+4}-\sqrt{n^4+1}.$$

Then

$$\frac{d_n}{n^2} = \sqrt{2+\frac{4}{n^4}} - \sqrt{1+\frac{1}{n^4}} \longrightarrow \sqrt2-1.$$

Hence

$$d_n\sim(\sqrt2-1)n^2.$$

• If $x<0$, the root test gives

$$\lim_{n\to\infty} \sqrt[n]{ \frac{e^{nx}}{d_n} } =e^x<1,$$

so the series converges.

• If $x=0$, then

$$\frac{1}{d_n} \sim \frac{\sqrt2+1}{n^2},$$

so the series converges by comparison with $\sum 1/n^2$.

• If $x>0$, then

$$\frac{e^{nx}}{d_n} \sim (\sqrt2+1)\frac{e^{nx}}{n^2},$$

which does not tend to zero. Therefore, the series diverges.

Thus the set of convergence is

$$\boxed{(-\infty,0]}.$$