May 2

Prove that for any finite sequence $\{a_{k}\}_{k=1}^{n}$ of real numbers there is an index $m\in\{0,\cdots,n\}$ such that
\[ \left|\sum_{1\leq k\leq m}a_{k}-\sum_{m<k\leq n}a_{k}\right|\leq \max_{1\leq k\leq n}|a_{k}| \]

Proof:


\[ S_{m}=\sum_{1\leq k\leq m}a_{k}-\sum_{m<k\leq n}a_{k}\]

\[ S_{0}=-S_{n} \]
而数列$\{S_{m}\}$将会改变符号,于是可以选择一个$p$,使得$S_{p}$和$S_{p-1}$不同号,必然存在一个$p$使得
$|S_{p}|$或者$|S_{p-1}|$不超过$\displaystyle\max_{1\leq k\leq n}|a_{k}|$,要是不会这样,就是说对所有能让$S_{p}$和$S_{p-1}$ 不同号的那些$p$,都有
\[ |S_{p}|>\max_{1\leq k\leq n}|a_{k}|,|S_{p-1}|>\max_{1\leq k\leq n}|a_{k}| \]
那么
\[ 2\max_{1\leq k\leq n}|a_{k}|<|S_{p}|+|S_{p-1}|=|S_{p}-S_{p-1}|=2|a_{p}|\leq 2\max_{1\leq k\leq n}|a_{k}|\]
便得到了矛盾,所以结论成立。

May 2

问题设$A,B\in M_{n}(C)$,求证:
\[ ||AB-BA||_{F}\leq \sqrt{2}||A||_{F}||B||_{F}  \]
证明:设
\[ A=\text{diag}(a_{1},a_{2},\cdots,a_{n})\qquad B=(b_{ij}) \]
\[ AB-BA=\text{diag}(a_{1},a_{2},\cdots,a_{n})B-B\text{diag}(a_{1},a_{2},\cdots,a_{n})=((a_{i}-a_{j})b_{ij})\]

\begin{align*}
 ||AB-BA||_{F}^{2}&=\sum_{i,j}|a_{i}-a_{j}||b_{ij}|^{2}\\
 &\leq \sum_{i,j}2\left(|a_{1}|^2+|a_{2}|^2+\cdots+|a_{n}|^{2}\right)|b_{ij}|^{2}\\
 &=2 ||A||_{F}^{2}||B||_{F}^{2}
\end{align*}
一般地,用A的奇异值SVD分解为
\[ A=U\text{diag}(a_{1},a_{2},\cdots,a_{n})V \]
其中$U,V$为酉矩阵,则变成上面的情形。

May 2

求极限
\[ \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx \]
解:
首先注意到极限
\[ \lim_{x\to 0}x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)=\frac{2}{3} \]
则有
\[ \frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx=\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}(\sin{nx})^{4}\cdot\frac{1}{x}\cdot x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)dx+\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}\frac{(\sin{nx})^{4}}{x^{3}}dx=A_{1}+A_{2} \]
由于$F(x)=x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)$可以通过补充定义让它连续,则在$\left[0,\frac{\pi}{2}\right]$上$F(x)$有最大值$M$,而$(\sin{nx})^{4}\leq |\sin{nx}|\leq nx $,则
\[ A_{1}\leq \frac{M}{n}\to 0\qquad (n\to+\infty) \]

\begin{align*}
 \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}\frac{\sin^{4}nx}{x^{3}}dx&=\lim_{n\to+\infty}\int_{0}^{\frac{n\pi}{2}}\frac{\sin^{4}{x}}{x^{3}}dx\\
&=\int_{0}^{+\infty}\frac{\sin^{4}{x}}{x^{3}}dx\\
&=\int_{0}^{+\infty}\frac{2\sin^{3}{x}\cos{x}}{x^{2}}dx\\
&=\int_{0}^{+\infty}\frac{6\sin^{2}x\cos^2{x}-2\sin^{4}{x}}{x}dx\\
&=\int_{0}^{+\infty}\frac{\cos{2x}-\cos{4x}}{x}dx\\
&=\ln{2}
\end{align*}
因此
\[ \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx=\ln{2} \]

May 2

证明
\[\lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(\sum_{k=0}^{n}{\frac{n^{k}}{k!}}-\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}  \right)}=\frac{4}{3}\]

(西西)
我们有
\[ e^{n}=\sum_{k=0}^{n}{\frac{n^{k}}{k!}}+\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}=\sum_{k=0}^{n}{\frac{n^{k}}{k!}}+\frac{1}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt\]
所以
\[ \sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}=\frac{1}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt\]
\[ \sum_{k=0}^{n}{\frac{n^{k}}{k!}}=e^{n}-\frac{1}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt\]
因此,只要计算
\[ \lim_{n\to\infty}\frac{n!}{n^{n}}\left(e^{n}-\frac{2}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt \right)\]
下面来估计
\[ \int_{0}^{n}e^{t}(n-t)^{n}dt \]
我们有
\begin{align*}
 \int_{0}^{n}e^{t}(n-t)^{n}dt&=n^{n+1}\int_{0}^{1}e^{nz}(1-z)^{n}dz\\
&=n^{n+1}\int_{0}^{1}e^{n(z+\ln(1-z))}dz\\
&=n^{n+1}\int_{0}^{1}e^{-\frac{1}{2}nz^{2}-\frac{1}{3}nz^{3}+o(nz^{3})}dz\\
&=n^{n+1}\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(1-\frac{1}{3}nz^{3}+o(nz^{3}) \right)dz
\end{align*}
\begin{align*}
 \frac{n!}{n^{n}}\left(e^{n}-\frac{2}{n!}\int_{0}^{n}e^{t}(n-t)^{n}dt \right)&= \frac{n!e^{n}}{n^{n}}-2n\left[\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(1-\frac{1}{3}nz^{3}+o(nz^{3}) \right)dz \right]\\
&=\left(\sqrt{2\pi{n}}e^{\frac{\theta_{n}}{12n}}-2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}dz\right)+2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(\frac{1}{3}nz^{3}+o(nz^{3})  \right)dz
\end{align*}
其中$\theta_{n}\in(0,1) $
显然有
\[ \lim_{n\to\infty}\left(\sqrt{2\pi{n}}e^{\frac{\theta_{n}}{12n}}-2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}dz\right)=0 \]
\[ \lim_{n\to\infty}2n\int_{0}^{1}e^{-\frac{1}{2}nz^{2}}\left(\frac{1}{3}nz^{3}+o(nz^{3})\right)dz=\lim_{n\to\infty}\frac{4}{3}\left(\int_{0}^{\frac{n}{2}}e^{-z}zdz+o\left( \frac{1}{n}\right)\right)=\frac{4}{3}\]
所以
\[\lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(\sum_{k=0}^{n}{\frac{n^{k}}{k!}}-\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}  \right)}=\frac{4}{3}\]
 

Mar 27

第五届全国大学生数学竞赛决赛数学类三、四年级试题答案