IS Math-2018-02

题目来源：Problem 2 日期：2024-07-21 题目主题：Math-常微分方程-序列函数收敛

具体题目

Let $f_1$ be a positive constant function on $[0,1]$ with $f_1(x)=c$, and let $p$ and $q$ be positive real numbers with $1/p+1/q=1$. Moreover, let $\{f_n\}$ be the sequence of functions on $[0,1]$ defined by

$$f_{n+1}(x)=p{\int }_0^x(f_n(t){)}^{1/q}\mathrm{d}t(n=1,2,\dots )$$

Answer the following questions.

1. Let $\{a_n\}$ and $\{c_n\}$ be the sequences of real numbers defined by $a_1=0$, $c_1=c$ and

$$\begin{array}{rl}{a}_{n+1}& =q^{-1}{a}_n+1(n=1,2,\dots ),\\ c_{n+1}& =\frac{p(c_n{)}^{1/q}}{a_{n+1}}(n=1,2,\dots ).\end{array}$$

Show that $f_n(x)=c_n{x}^{a_n}$.

2. Let $g_n$ be the function on $[0,1]$ defined by $g_n(x)=x^{a_n}-x^p$ for $n\ge 2$. Noting that $a_n\ge 1$ holds true for $n\ge 2$, show that $g_n$ attains its maximum at a point $x=x_n$, and find the value of $x_n$.

3. Show that ${\lim }_{n\to \infty }{g}_n(x)=0$ for any $x\in [0,1]$.

4. Let $d_n$ be defined by $d_n=(c_n{)}^{g_n}$. Show that $d_{n+1}/d_n$ converges to a finite positive value as $n\to \infty$. You may use the fact that ${\lim }_{t\to \infty }(1-1/t{)}^t=1/e$.

5. Find the value of ${\lim }_{n\to \infty }{c}_n$.

6. Show that ${\lim }_{n\to \infty }{f}_n(x)=x^p$ for any $x\in [0,1]$.

正确解答

1. Showing $f_n(x)=c_n{x}^{a_n}$

We start with the base case. For $n=1$,

$$f_1(x)=c=c_1{x}^0$$

So, $c_1=c$ and $a_1=0$ as given. Assume that for some $k$, $f_k(x)=c_k{x}^{a_k}$. Then,

$$f_{k+1}(x)=p{\int }_0^x(f_k(t){)}^{1/q}\mathrm{d}t=p{\int }_0^x(c_k{t}^{a_k}{)}^{1/q}\mathrm{d}t=p{c}_k^{1/q}{\int }_0^x{t}^{a_k/q}\mathrm{d}t$$

Evaluating the integral,

$${\int }_0^x{t}^{a_k/q}\mathrm{d}t=\frac{x^{a_k/q+1}}{a_k/q+1}$$

Thus,

$$f_{n+1}(x)=p{c}_n^{1/q}\frac{x^{a_n/q+1}}{a_n/q+1}=\frac{p{c}_n^{1/q}}{a_n/q+1}{x}^{a_n/q+1}$$

Now, we compare this with $f_{n+1}(x)=c_{n+1}{x}^{a_{n+1}}$. From the recurrence relations, we have:

$$a_{n+1}=\frac{a_n}{q}+1$$ $$c_{n+1}=\frac{p{c}_n^{1/q}}{a_{n+1}}$$

Thus, we have shown that if $f_n(x)=c_n{x}^{a_n}$, then $f_{n+1}(x)=c_{n+1}{x}^{a_{n+1}}$. By induction, $f_n(x)=c_n{x}^{a_n}$ for all $n$.

2. Finding the maximum of $g_n(x)$

To find the general term $a_n$ using the method of undetermined coefficients, we start with the recurrence relation:

$$a_{n+1}=\frac{a_n}{q}+1$$

Assume that $b_n=a_n+k$ for some constant $k$, and $b_{n+1}=q^{-1}{b}_n$.

$$a_{n+1}+k=q^{-1}(a_n+k)$$

Since $a_{n+1}=q^{-1}{a}_n+1$,

$$q^{-1}{a}_n+1+k=q^{-1}(a_n+k)$$

Solving for $k$,

$$k(1-q^{-1})=-1$$

Thus,

$$k=\frac{1}{q^{-1}-1}=-p$$

Since $a_1=0$, $b_1=-p$, and $b_{n+1}=q^{-1}{b}_n$,

$$b_n=-p{q}^{1-n}$$

Thus,

$$a_n=-p{q}^{1-n}+p$$

Now, we find the maximum of $g_n(x)=x^{a_n}-x^p$. The derivative of $g_n(x)$ is:

$$g_n^{\prime }(x)=a_n{x}^{a_n-1}-p{x}^{p-1}$$

Setting $g_n^{\prime }(x)=0$,

$$a_n{x}^{a_n-p}=p$$ $$x={\left(\frac{p}{a_n}\right)}^{\frac{1}{a_n-p}}=(1-q^{1-n}{)}^{\frac{1}{p{q}^{1-n}}}$$

Since $a_n\ge 1$ for $n\ge 2$,

$$g_n^{\prime \prime }(x)=a_n(a_n-1)x^{a_n-2}-p(p-1)x^{p-2}=x^{a_n-2}(a_n(a_n-1)-p(p-1)x^{p-a_n})=x^{a_n-2}{a}_n(a_n-p)\le 0.$$

Thus, $x=x_n$ is the maximum of $g_n(x)$.

In conclusion, $x_n=(1-q^{1-n}{)}^{\frac{1}{p{q}^{1-n}}}$ is the point where $g_n(x)$ attains its maximum.

3. Showing ${\lim }_{n\to \infty }{g}_n(x)=0$

First, we note that $0\le g_n(x)\le g_n(x_n)$ for all $x\in [0,1]$.

$$g_n(x_n)=x_n^{a_n}-x_n^p=x_n^p(x_n^{a_n-p}-1)=(1-q^{1-n}{)}^{\frac{1}{q^{1-n}}}((1-q^{1-n})-1)$$

Since ${\lim }_{n\to \infty }(1-q^{1-n}{)}^{\frac{1}{q^{1-n}}}=e^{-1}$, ${\lim }_{n\to \infty }((1-q^{1-n})-1)=0$,

$$\underset{n\to \infty }{\lim }{g}_n(x_n)=0$$

Thus, ${\lim }_{n\to \infty }{g}_n(x)=0$ for any $x\in [0,1]$.

4. Showing $\frac{d_{n+1}}{d_n}$ converges

Given $d_n=(c_n{)}^{g_n}$,

$$\frac{d_{n+1}}{d_n}={\left(\frac{c_{n+1}}{c_n}\right)}^{g_{n+1}}={\left(\frac{p}{a_{n+1}}\right)}^{q^{n+1}}=\frac{p^{q^{n+1}}}{p^{q^{n+1}}(1-q^{-n}{)}^{q^{n+1}}}={\left(1-\frac{1}{q^n}\right)}^{q^{n+1}}$$ $$\underset{n\to \infty }{\lim }\frac{d_{n+1}}{d_n}={\left(1-\frac{1}{q^n}\right)}^{q^{n+1}}=e^q$$

Hence, $\frac{d_{n+1}}{d_n}$ converges to a finite positive value as $n\to \infty$.

5. Finding ${\lim }_{n\to \infty }{c}_n$

Since $a_n\to p$ as $n\to \infty$,

$$\underset{n\to \infty }{\lim }{c}_n=\underset{n\to \infty }{\lim }\frac{p(c_n{)}^{1/q}}{a_{n+1}}=\frac{p}{p}=1$$

6. Showing ${\lim }_{n\to \infty }{f}_n(x)=x^p$

Given $f_n(x)=c_n{x}^{a_n}$ and $c_n\to 1$, $a_n\to p$,

$$\underset{n\to \infty }{\lim }{f}_n(x)=\underset{n\to \infty }{\lim }{c}_n{x}^{a_n}=x^p$$

Hence,

$$\underset{n\to \infty }{\lim }{f}_n(x)=x^p$$

知识点

常微分方程积分递归极限函数序列

难点解题思路

对于证明 $f_n(x)=c_n{x}^{a_n}$ 和找到函数 $g_n(x)$ 的最大值，关键是要正确理解递推关系，并且准确计算积分和导数。对于极限问题，利用递推关系的收敛性质是解决的关键。

解题技巧和信息

1. 对于递推关系，特别是涉及积分和导数的递推，明确每一步的变化

规律是至关重要的。

1. 在处理极限问题时，考虑函数和参数的收敛性，利用已知极限公式或性质可以简化计算。

2. 求解最大值问题时，通过导数求极值点是常用的方法，需要注意是否需要检查边界点。

重点词汇

sequence 序列

recurrence relation 递推关系

integral 积分

limit 极限

convergence 收敛

critical point 临界点

参考资料

1. Walter Rudin, Principles of Mathematical Analysis, Chapter 8, Sequences and Series of Functions.
2. Serge Lang, Undergraduate Analysis, Chapter 5, Integration and Differentiation.
3. Tom Apostol, Mathematical Analysis, Chapter 7, Sequences and Series of Functions.