2017

Problem 1

Suppose that three-dimensional vectors $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$ satisfy the equation

$$\left(\begin{array}{c}{x}_{n+1}\\ y_{n+1}\\ z_{n+1}\end{array}\right)=A\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)(n=0,1,2,\dots ),$$

where $x_0,y_0,z_0$ and $\alpha$ are real numbers, and

$$A=\left(\begin{array}{ccc}1-2\alpha & \alpha & \alpha \\ \alpha & 1-\alpha & 0\\ \alpha & 0& 1-\alpha \end{array}\right),0<\alpha <\frac{1}{3}.$$

Answer the following questions.

1. Express $x_n+y_n+z_n$ using $x_0,y_0$ and $z_0$.
2. Obtain the eigenvalues ${\lambda }_1,{\lambda }_2$ and ${\lambda }_3$, and their corresponding eigenvectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ and ${\mathbf{v}}_3$ of the matrix $A$.
3. Express the matrix $A$ using ${\lambda }_1,{\lambda }_2,{\lambda }_3,{\mathbf{v}}_1,{\mathbf{v}}_2$ and ${\mathbf{v}}_3$.
4. Express $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$ using $x_0,y_0,z_0$ and $\alpha$.
5. Obtain ${\lim }_{n\to \infty }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$.
6. Regard

$$f(x_0,y_0,z_0)=\frac{\left(\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right),\left(\begin{array}{c}{x}_{n+1}\\ y_{n+1}\\ z_{n+1}\end{array}\right)\right)}{\left(\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right),\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)\right)}$$

as a function of $x_0,y_0$ and $z_0$. Obtain the maximum and the minimum values of $f(x_0,y_0,z_0)$, where we assume that $x_0^2+y_0^2+z_0^2\ne 0$.

---

假设三维向量 $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$ 满足方程

$$\left(\begin{array}{c}{x}_{n+1}\\ y_{n+1}\\ z_{n+1}\end{array}\right)=A\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)(n=0,1,2,\dots ),$$

其中 $x_0,y_0,z_0$ 和 $\alpha$ 是实数，且

$$A=\left(\begin{array}{ccc}1-2\alpha & \alpha & \alpha \\ \alpha & 1-\alpha & 0\\ \alpha & 0& 1-\alpha \end{array}\right),0<\alpha <\frac{1}{3}.$$

回答以下问题。

1. 用 $x_0,y_0$ 和 $z_0$ 表示 $x_n+y_n+z_n$。
2. 求矩阵 $A$ 的特征值 ${\lambda }_1,{\lambda }_2$ 和 ${\lambda }_3$，以及它们对应的特征向量 ${\mathbf{v}}_1,{\mathbf{v}}_2$ 和 ${\mathbf{v}}_3$。
3. 用 ${\lambda }_1,{\lambda }_2,$\lambda_3, \mathbf{v}_1, \mathbf{v}_2$和$\mathbf{v}_3$表示矩阵$A$。
4. 用 $x_0,y_0,z_0$ 和 $\alpha$ 表示 $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$。
5. 求 ${\lim }_{n\to \infty }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$。
6. 将

$$f(x_0,y_0,z_0)=\frac{\left(\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right),\left(\begin{array}{c}{x}_{n+1}\\ y_{n+1}\\ z_{n+1}\end{array}\right)\right)}{\left(\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right),\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)\right)}$$

视为 $x_0,y_0$ 和 $z_0$ 的函数。在假设 $x_0^2+y_0^2+z_0^2\ne 0$ 的情况下，求 $f(x_0,y_0,z_0)$ 的最大值和最小值。

---

Problem 2

A real-valued function $u(x,t)$ is defined in $0\le x\le 1$ and $t\ge 0$. Here, $x$ and $t$ are independent. Suppose solving the following partial differential equation:

$$\frac{\partial u}{\partial t}=\frac{{\partial }^{2}u}{\partial x^2},$$

under the following conditions:

Boundary condition: $u(0,t)=u(1,t)=0$,

Initial condition: $u(x,0)=x-x^2$.

Since the constant function $u(x,t)=0$ is obviously a solution of the partial differential equation, consider the other solutions. Answer the following questions.

1. Calculate the following expression, where $n$ and $m$ are positive integers.

$${\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x$$

2. Suppose $u(x,t)=\xi (x)\tau (t)$, where $\xi (x)$ is a function only of $x$ and $\tau (t)$ is a function only of $t$. Express the ordinary differential equations for $\xi$ and $\tau$ using an arbitrary constant $C$. You may use that $f(x)$ and $g(t)$ are constant functions when $f(x)$ and $g(t)$ satisfy $f(x)=g(t)$ for arbitrary $x$ and $t$.

3. Solve the ordinary differential equations in question (2). Next, show that a solution of partial differential equation which satisfies the boundary condition is given by the following $u_n(x,t)$, and express $\alpha$ and $\beta$ using a positive integer $n$.

$$u_n(x,t)=e^{\alpha t}\sin (\beta x)$$

4. The solution of partial differential equation which satisfies the boundary and initial conditions is represented by the linear combination of $u_n(x,t)$ as shown below. Obtain $c_n$. You may use the result of question (1).

$$u(x,t)=\sum _{n=1}^{\infty }{c}_n{u}_n(x,t)$$

---

定义实值函数 $u(x,t)$ 在 $0\le x\le 1$ 和 $t\ge 0$。这里，$x$ 和 $t$ 是独立的。假设求解以下偏微分方程：

$$\frac{\partial u}{\partial t}=\frac{{\partial }^{2}u}{\partial x^2},$$

在以下条件下：

边界条件：$u(0,t)=u(1,t)=0$，

初始条件：$u(x,0)=x-x^2$。

由于常数函数 $u(x,t)=0$ 显然是偏微分方程的解，考虑其他解。回答以下问题。

1. 计算以下表达式，其中 $n$ 和 $m$ 是正整数。

$${\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x$$

2. 假设 $u(x,t)=\xi (x)\tau (t)$，其中 $\xi (x)$ 仅是 $x$ 的函数，$\tau (t)$ 仅是 $t$ 的函数。用任意常数 $C$ 表示 $\xi$ 和 $\tau$ 的常微分方程。可以假设当 $f(x)$ 和 $g(t)$ 满足 $f(x)=g(t)$ 对任意 $x$ 和 $t$ 时，$f(x)$ 和 $g(t)$ 是常数函数。

3. 求解问题 (2) 中的常微分方程。接下来，证明满足边界条件的偏微分方程的解由以下 $u_n(x,t)$ 给出，并用正整数 $n$ 表示 $\alpha$ 和 $\beta$。

$$u_n(x,t)=e^{\alpha t}\sin (\beta x)$$

4. 满足边界和初始条件的偏微分方程的解由 $u_n(x,t)$ 的线性组合表示如下。求 $c_n$。可以使用问题 (1) 的结果。

$$u(x,t)=\sum _{n=1}^{\infty }{c}_n{u}_n(x,t)$$

---

Problem 3

1. If the probability density function $f(t)$ of a continuous random variable $T$ is denoted by

$$f(t)=\{\begin{array}{ll}\lambda e^{-\lambda t}& (t\ge 0)\\ 0& (t<0)\end{array}$$

with a positive constant $\lambda$, then we say that $T$ follows an exponential distribution with parameter $\lambda$. Compute the average of this random variable. Also, derive the probability distribution function $F(t)=P(T\le t)$ of this exponential distribution, where $P(X)$ is the probability of the event $X$.

1. Show that the probability distribution given in question (1) is memoryless. Namely, show that

$$P(T>s+t\mid T>s)=P(T>t)$$

holds for any $s>0$ and $t>0$, where $P(X\mid Y)$ is the probability of the event $X$ conditioned on the event $Y$.

1. Let us call the time interval between the time when one starts solving a problem and the time when one finishes it “time required for solution”. Assume that the time required for solution of each of $n$ students follows the exponential distribution with the same parameter ${\lambda }_0$. Let all the $n$ students start solving the problem at the same time. Find the probability distribution function and the average of the time required for solution of the student who finishes solving the problem earliest. Here, the time required for solution of each student is mutually independent.

2. Assume that the times required for solution of a student A and a student B follow the exponential distributions with parameters ${\lambda }_A$ and ${\lambda }_B$, respectively. Let the two students start solving the problem at the same time. Find the probability that student A finishes solving the problem earlier than student B.

3. Let a smart student Hideo and other ten students start solving a problem at the same time. Assume that the time required for solution of each of all the students follows an exponential distribution, where the average time required for solution of each of all the students except Hideo is ten times longer than that of Hideo. Find the probabilities that Hideo finishes solving the problem first and fourth, respectively.

---

1. 如果连续随机变量 $T$ 的概率密度函数 $f(t)$ 表示为

$$f(t)=\{\begin{array}{ll}\lambda e^{-\lambda t}& (t\ge 0)\\ 0& (t<0)\end{array}$$

其中 $\lambda$ 是正常数，那么我们说 $T$ 服从参数为 $\lambda$ 的指数分布。计算该随机变量的平均值。同时，推导该指数分布的概率分布函数 $F(t)=P(T\le t)$，其中 $P(X)$ 是事件 $X$ 的概率。

1. 证明问题 (1) 给出的概率分布是无记忆的。即，证明

$$P(T>s+t\mid T>s)=P(T>t)$$

对任意 $s>0$ 和 $t>0$ 成立，其中 $P(X\mid Y)$ 是事件 $X$ 在事件 $Y$ 条件下的概率。

1. 我们称从开始解决问题到完成问题之间的时间间隔为“解决问题所需时间”。假设每个学生解决问题所需的时间均服从参数相同的指数分布 ${\lambda }_0$。令所有 $n$ 个学生同时开始解决问题。求最早完成问题的学生的解决问题所需时间的概率分布函数和平均值。这里，每个学生解决问题所需的时间是相互独立的。

2. 假设学生 A 和学生 B 解决问题所需的时间分别服从参数为 ${\lambda }_A$ 和 ${\lambda }_B$ 的指数分布。令两名学生同时开始解决问题。求学生 A 比学生 B 先完成问题的概率。

3. 令聪明的学生 Hideo 和其他十名学生同时开始解决问题。假设每个学生解决问题所需的时间均服从指数分布，其中除 Hideo 外的每个学生解决问题所需的平均时间是 Hideo 的十倍。求 Hideo 最先完成和第四个完成问题的概率。