九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2024年1月実施 解析学・微積分

标签：

• 九州大学
• 约束最优化
• 一阶常微分方程
• 幂级数
• 泰勒级数与收敛半径

Author

祭音Myyura (co-authored with ChatGPT 5.5)

Description

(1) Find the maximum and the minimum of the function $f(x,y)=x^2+2xy+2y^2$ over the curve $x^2 + xy + y^2 = 1$, and find the points which achieve them.

(2) Find the solution to the following differential equation defined for $x \ge 0$, where $y(0)=0$,

$$\frac{dy}{dx} + y = \sin x$$

(3) Find the radius of convergence of the series

$$\sum_{n=1}^{\infty} \frac{(z+1)^{n-1}}{(n+1)^3 2^n}$$

where $z$ is a complex number.

题目描述

1. 在约束曲线
   $$x^2+xy+y^2=1$$
   上，求函数
   $$f(x,y)=x^2+2xy+2y^2$$
   的最大值和最小值，并求出所有达到这些值的点。
2. 在 $x\ge0$ 上求初值问题
   $$\frac{dy}{dx}+y=\sin x,\qquad y(0)=0$$
   的解。
3. 设 $z$ 为复数，求幂级数
   $$\sum_{n=1}^{\infty}\frac{(z+1)^{n-1}}{(n+1)^3\,2^n}$$
   的收敛半径。

Kai

(1)

We want to optimize

$$f(x,y)=x^2+2xy+2y^2$$

subject to

$$x^2+xy+y^2=1.$$

Use Lagrange multipliers:

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

where

$$g(x,y)=x^2+xy+y^2.$$

Then

$$\frac{\partial f}{\partial x}=2x+2y, \qquad \frac{\partial f}{\partial y}=2x+4y,$$

and

$$\frac{\partial g}{\partial x}=2x+y, \qquad \frac{\partial g}{\partial y}=x+2y.$$

So

$$2x+2y=\lambda(2x+y),$$

$$2x+4y=\lambda(x+2y).$$

Solving this generalized eigenvalue problem gives

$$\lambda=2 \qquad \text{or} \qquad \lambda=\frac{2}{3}.$$

Because the constraint satisfies $g(x,y)=1$, the value of $f$ equals $\lambda$.

Case 1: $\lambda=2$

This gives

$$x=0.$$

Using the constraint,

$$y^2=1,$$

so

$$y=\pm 1.$$

Thus the maximum value is

$$\boxed{2}$$

at the points

$$\boxed{(0,1),\ (0,-1)}.$$

Case 2: $\lambda=\frac{2}{3}$

This gives

$$x+2y=0,$$

so

$$x=-2y.$$

Substitute into the constraint:

$$x^2+xy+y^2=1.$$

Then

$$4y^2-2y^2+y^2=1,$$

so

$$3y^2=1.$$

Hence

$$y=\pm \frac{1}{\sqrt{3}},$$

and

$$x=\mp \frac{2}{\sqrt{3}}.$$

Thus the minimum value is

$$\boxed{\frac{2}{3}}$$

at the points

$$\boxed{\left(-\frac{2}{\sqrt{3}},\frac{1}{\sqrt{3}}\right), \left(\frac{2}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)}.$$

Therefore,

$$\boxed{\max f=2}$$

and

$$\boxed{\min f=\frac{2}{3}}.$$

(2)

We solve

$$\frac{dy}{dx}+y=\sin x,$$

with initial condition

$$y(0)=0.$$

This is a first-order linear differential equation. The integrating factor is

$$e^x.$$

Multiplying both sides by $e^x$, we get

$$e^x\frac{dy}{dx}+e^x y=e^x\sin x.$$

The left side becomes

$$\frac{d}{dx}(e^x y).$$

So

$$\frac{d}{dx}(e^x y)=e^x\sin x.$$

Integrate both sides:

$$e^x y=\int e^x\sin x\,dx.$$

Using

$$\int e^x\sin x\,dx = \frac{e^x(\sin x-\cos x)}{2},$$

we have

$$e^x y=\frac{e^x(\sin x-\cos x)}{2}+C.$$

Divide by $e^x$:

$$y=\frac{\sin x-\cos x}{2}+Ce^{-x}.$$

Use the initial condition $y(0)=0$:

$$0=\frac{\sin 0-\cos 0}{2}+C$$

$$0=\frac{0-1}{2}+C$$

$$C=\frac{1}{2}.$$

Therefore,

$$\boxed{ y(x)=\frac{\sin x-\cos x+e^{-x}}{2} }$$

for $x\ge 0$.

(3) Radius of convergence

Consider the power series

$$\sum_{n=1}^{\infty} \frac{(z+1)^{n-1}}{(n+1)^3 2^n}.$$

Let

$$w=z+1.$$

Then the series becomes

$$\sum_{n=1}^{\infty} \frac{w^{n-1}}{(n+1)^3 2^n}.$$

We examine the general term

$$a_n=\frac{w^{n-1}}{(n+1)^3 2^n}.$$

Using the root test,

$$\lim_{n\to\infty} |a_n|^{1/n} = \lim_{n\to\infty} \left( \frac{|w|^{n-1}}{(n+1)^3 2^n} \right)^{1/n}.$$

This equals

$$\frac{|w|}{2} = \frac{|z+1|}{2}.$$

The root test proves absolute convergence when

$$\frac{|z+1|}{2}<1.$$

Thus

$$|z+1|<2.$$

Therefore, the radius of convergence is

$$\boxed{2}.$$

On $|z+1|=2$, the absolute value of the $n$-th term is $1/[2(n+1)^3]$, so the series also converges absolutely on the boundary.