Apr 4

If $a,b,c\ge 0$ such that $ab+bc+ca=3$, then

$$4(a^3+b^3+c^3)+7abc+125\ge 48(a+b+c).$$

Proof(Vo Quoc Ba Can)

The inequality is equivalent to \[4(a^3+b^3+c^3)+7abc +125 \ge 16(ab+bc+ca)(a+b+c),\]

or

\[125 \ge A,\]

where $A=16(a+b+c)(ab+bc+ca)-4(a^3+b^3+c^3) -7abc.$

From the equivalence, it is obvious that the inequality holds for $A \le 0,$ so we only have to consider it when $A >0.$ In that case, it is equivalent to \[125^2 \ge A^2,\] and there are two possible cases can occur (assume that $a \ge b \ge c$):
1、Case $6(a+c) \ge 7b.$

Using the AM-GM inequality, we have

\[A^2 =(2a+2c+b)^2\cdot \frac{A}{2a+2c+b}\cdot \frac{A}{2a+2c+b} \le \left[\frac{(2a+2c+b)^2+2\cdot \frac{A}{2a+2c+b}}{3}\right]^3.\]

Thus, it suffices to prove that

\[25\cdot 3 \ge (2a+2c+b)^2+\frac{2A}{2a+2c+b},\]

or

\[25(ab+bc+ca) \ge (2a+2c+b)^2+\frac{2A}{2a+2c+b}.\]

An easy computation shows that, for the last inequality, we have

\[LHS-RHS =\frac{(a-b)(b-c)(6a+6c-7b)}{2a+2c+b},\]

which is obviously nonnegative.
2、Case $6(a+c) \le 7b.$

Applying the AM-GM inequality in the same way as the first case, but this time we change the positions of $a$ and $b$ ($a\to b$ and $b\to a$), it leads us to the inequality

\[25(ab+bc+ca) \ge (2b+2c+a)^2+\frac{2A}{2b+2c+a},\]

which is equivalent to

\[\frac{(a-b)(a-c)(7a-6b-6c)}{2b+2c+a} \ge 0.\]

This is true because

$$(a-b)(a-c) \ge 0$$

and

\[7a-6b-6c=13a-6(a+b+c) \ge 13b-6(a+b+c)=7b-(6a+6c) \ge 0.\]
The proof is completed. $\blacksquare$

 

Mar 22

这里感谢风碎便士和西哥的提示。

1.计算
\[ \int_{0}^{\infty}{\frac{(1-e^{-6x})e^{-x}}{x(1+e^{-2x}+e^{-4x}+e^{-6x}+e^{-8x})}dx} \]

\begin{align*}
 &\int_{0}^{\infty}{\frac{(1-e^{-6x})e^{-x}}{x(1+e^{-2x}+e^{-4x}+e^{-6x}+e^{-8x})}dx}\\
 &= \int_{0}^{\infty}{\frac{(1-e^{-6x})e^{-x}(1-e^{-2x})}{x(1-e^{-10x})}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{ \frac{(1-e^{-6x})e^{-(10n+1)x}(1-e^{-2x})}{x}}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{\frac{(e^{-(10n+1)x}-e^{-(10n+7)x})-(e^{-(10n+3)x}-e^{-(10n+9)x})}{x}}dx}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(10n+7)(10n+3)}{(10n+1)(10n+9)}}} \qquad \text{(这里套Froullani公式)}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(n+\frac{7}{10})(n+\frac{3}{10})}{(n+\frac{1}{10})(n+\frac{9}{10})}}}\\
 &=\lim_{n\rightarrow\infty}{\ln{\frac{\Gamma{(\frac{1}{10})}\Gamma{(\frac{9}{10})}\Gamma{(\frac{3}{10}+n)}\Gamma{(\frac{7}{10}+n)}}{\Gamma{(\frac{3}{10})}\Gamma{(\frac{7}{10})}\Gamma{(n+\frac{1}{10})}\Gamma{(n+\frac{9}{10})}}}}\qquad \text{(这里套Euler-Guess公式,也就是伽马函数的无穷乘积公式)}\\
 &=\ln{\frac{\sin{\frac{3\pi}{10}}}{\sin{\frac{\pi}{10}}}}\qquad \text{(这里套余元公式)}\\
 &=\ln{\frac{\sqrt{5}+1}{\sqrt{5}+1}}
\end{align*}


 

2.计算
\[ \int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)}dx} \]
仿照上题的手段,我们有
\begin{align*}
 & \int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)}dx}\\
 &=\int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})(1+e^{-2x})(1-e^{-6x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)(1+e^{-2x})(1-e^{-6x})} dx}\\
 &=\int_{0}^{\infty}{\frac{e^{-x}(1-e^{-4x})(1-e^{-6x})}{x(1-e^{-24x})}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{\frac{(1-e^{-4x})e^{-(24n+1)x}(1-e^{-6x})}{x}}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{ \frac{(e^{-(24n+1)x}-e^{-(24n+7)x})-(e^{-(24n+5)x}-e^{-(24n+11)x})}{x}}        dx}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(24n+7)(24n+5)}{(24n+1)(24n+11)}}} \qquad \text{(这里套Froullani公式)}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(n+\frac{7}{24})(n+\frac{5}{24})}{(n+\frac{1}{24})(n+\frac{11}{24})}}}\\
 &=\lim_{n\rightarrow\infty}{\ln{\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}\Gamma{(\frac{7}{24}+n)}\Gamma{(\frac{5}{24}+n)}}{\Gamma{(\frac{7}{24})}\Gamma{(\frac{5}{24})}\Gamma{(n+\frac{1}{24})}\Gamma{(n+\frac{11}{24})}}}}\qquad \text{(这里套Euler-Guess公式,也就是伽马函数的无穷乘积公式)}\\
 &=\ln{\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}}{\Gamma{(\frac{7}{24})}\Gamma{(\frac{5}{24})}}}
\end{align*}
另一方面
\begin{align*}
\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{7}{24})}}
&=\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{23}{24})}\Gamma{(\frac{11}{24})}\Gamma{(\frac{13}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{23}{24})}}\\
&=\frac{\pi^{2}}{\sin{(\frac{\pi}{24})}\cos{(\frac{\pi}{24})}}\cdot\frac{1}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{23}{24})}}\\
&=\frac{2\pi^2}{\sin{(\frac{\pi}{12})}}\cdot\frac{\Gamma{(\frac{15}{24})}\Gamma{(\frac{21}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{21}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{15}{24})}\Gamma{(\frac{23}{24})}}
\end{align*}
再利用公式
\[ \Gamma{(z)}\Gamma{\left(z+\frac{1}{3}\right)}\Gamma{\left(z+\frac{2}{3}\right)}=\frac{2\pi\sqrt{3}}{3^{2\pi}}\Gamma{(3z)} \]
我们有
\[ \Gamma{\left(\frac{5}{24}\right)}\Gamma{\left(\frac{13}{24}\right)}\Gamma{\left(\frac{21}{24}\right)}=\frac{2\pi\sqrt{3}}{3^{\frac{15}{24}}}\Gamma{\left(\frac{15}{24}\right)} \]
\[\Gamma{\left(\frac{7}{24}\right)}\Gamma{\left(\frac{15}{24}\right)}\Gamma{\left(\frac{23}{24}\right)}=\frac{2\pi\sqrt{3}}{3^{\frac{21}{24}}}\Gamma{\left(\frac{21}{24}\right)} \]

\[ \frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{7}{24})}}=\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{23}{24})}\Gamma{(\frac{11}{24})}\Gamma{(\frac{13}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{23}{24})}}=\frac{3\sqrt{2}+\sqrt{6}}{2}
\]
所以
\[  \int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)}dx}=\ln{\left(\frac{3\sqrt{2}+\sqrt{6}}{2} \right)} \]

Mar 21

Let $a,b,c>0$,with $a+b+c=3$,Prove that
\[ \sqrt{\frac{a^{3}}{a^{2}+3b^{2}}}+\sqrt{\frac{b^{3}}{b^{2}+3c^{2}}}+\sqrt{\frac{c^{3}}{c^{2}+3a^{2}}}\geq \frac{3}{2}\]

Proof:
\begin{align*}
\sum_{cyc}{\sqrt{\frac{a^{3}}{a^{2}+3b^{2}}}}&=6\sum_{cyc}{\frac{a^{2}}{\sqrt{4a(a+b+c)\cdot3(a^2+3b^2)}}}\\
&\geq 12\sum_{cyc}{\frac{a^2}{4a(a+b+c)+3(a^2+3b^2)}}\\
&=12\sum_{cyc}{\frac{a^{2}}{7a^{2}+9b^2+4ab+4ca}} 
\end{align*}
Now,Using the Awesome CYH technology,we have
\begin{align*}
&\left(\sum_{cyc}{\frac{a^{2}}{7a^{2}+9b^2+4ab+4ca}}\right)\left[\sum_{cyc}{(2a+c)^2(7a^2+9b^2+4ab+4ca)}\right]\\
&\geq \left[\sum_{cyc}{a(c+2a)}\right]^{2}\\
&=\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2 
\end{align*}
Thus,it's suffice to check
\[ 8\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2\geq \sum_{cyc}{(2a+c)^2(7a^2+9b^2+4ab+4ca)}\]
Or
\[ \sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2}+3\sum_{cyc}{a^3b}-3\sum_{cyc}{ab^3}-2abc\sum_{cyc}{a}\geq 0 \]
Which is not a problem for us now,so we are done!
$\square$

Mar 20

前几天到成都考试,经过几天漫长的火车,终于回到了学校。今天特地编辑了下考试的题目,大家可以在下面的链接中下载。

说起这次考试,我个人觉得题目很难,有些还是从来没有见到过的,可能是因为自己水平太差的关系,看来以后还得多多努力才行。至于我的得奖情况,这里就不好意思说了。blush(听领队老师说31-49分是二等奖,50-71是一等奖,剩下的是三等奖,最后在参赛的94位同学中,有19个一等奖,28个二等奖)

第四届全国大学生数学竞赛决赛(数学类)试题

关于试题的解答本帖会陆续更新。大家有什么好的解法也可以写在下面的评论中,Thank you in advance。

顺便插入图片几张

 

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解答区

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 一、设$A$为正整数,直线$L$与双曲线$x^2-y^2=2\ (x>0)$所围成的面积为$A$,证明:
(i)上述$L$被双曲线$x^2-y^2=2\ (x>0)$所截线段的中点的轨迹为双曲线。
(ii) $L$总是(i)中轨迹曲线的切线。


解:(西西提供)
作坐标旋转,令$x=\frac{X+Y}{\sqrt{2}},y=\frac{X-Y}{\sqrt{2}}$,代入双曲线方程$x^2-y^2=2$,得
\[ \left( \frac{X+Y}{\sqrt{2}}\right)^{2}-\left( \frac{X-Y}{\sqrt{2}}\right)^{2}=2XY=2 \]

\[ Y=\frac{1}{X}\]
(i)设在新坐标系中,$L$与双曲线 $ Y=\frac{1}{X}$的两个交点坐标为$(s,\frac{1}{s}),(t,\frac{1}{t})$,其中$0<s<t$.这时$L$的方程为
\[ Y=\frac{\frac{1}{t}-\frac{1}{s}}{t-s}(X-s)+\frac{1}{s}=\frac{s+t-X}{st} \]
$L$与双曲线$Y=\frac{1}{X}$所围成的有限部分的面积为
\[ A=\int_{s}^{t}{\left(\frac{s+t-X}{st}-\frac{1}{X}  \right)dX}=\frac{t^2-s^2}{2st}-\ln{\left( \frac{t}{s} \right)}=\frac{1}{2}\left(\frac{t}{s}-\frac{s}{t}\right)-\ln{\left( \frac{t}{s} \right)} \]
因为$A$是一个常数,所以由上述方程确定的$\frac{t}{s}$也是一个常数,设$\frac{t}{s}=k,t=ks $.
$L$被双曲线所截线段的中点的横坐标为\[
X=\frac{s+t}{2}=\frac{s+ks}{2}=\frac{1+k}{2}\cdot s\]
纵坐标为
\[ Y=\frac{1}{2}\left(\frac{1}{s}+\frac{1}{t}\right)=\frac{1}{2}\left(\frac{1}{s}+\frac{1}{ks}\right) \]
所以,中点的轨迹方程为
\[ Y=\frac{1+k}{2}\cdot s=\frac{(1+k)^2}{4k}\cdot\frac{2ks}{1+k}=\frac{(1+k)^2}{4kX} \]
可见中点的轨迹为双曲线。
(ii)对$Y=\frac{(1+k)^2}{4kX} $求导,得
\[ Y'(X)=-\frac{(1+k)^2}{4kX^{2}} \]
这是中点轨迹曲线上横坐标为$X$点处曲线切线的斜率。
当$X=\frac{1+k}{2}s$时,切线斜率为
\[Y'\left(\frac{1+k}{2}s\right)=-\frac{(1+k)^2}{4k}\cdot\frac{4}{(1+k)^2s^2}=-\frac{1}{ks^2}\]
而通过点$\left(\frac{1+k}{2}s,\frac{1+k}{2ks}  \right)$的直线$L$的方程为
\[ Y=\frac{s+t-X}{st}=\frac{s+ks-X}{ks^{2}} \]
它的斜率就是$-\frac{1}{ks^2}$,可见直线$L$是双曲线$Y=\frac{(1+k)^2}{4kX} $的切线
$\square$

二、设函数$f(x)$满足条件:
(1)$-\infty<a\leq f(x)\leq b<+\infty$,  $a\leq x\leq b$;
(2)对任意$x,y\in[a,b]$有$|f(x)-f(y)|<L|x-y|$,其中$L$是大于$0$小于$1$的常数。
设$x_{1}\in[a,b]$,令$x_{n+1}=\frac{1}{2}\left[x_{n}+f(x_{n})\right],\quad n=1,2,\dots$
证明$\displaystyle \lim_{n\rightarrow\infty}{x_{n}}=x$存在,且$f(x)=x$。

 

证明:首先,我们证明$f(x)$是$[a,b]$上的连续函数
对$\forall \varepsilon>0$,$\exists \delta=\frac{\varepsilon}{L}$,对$\forall x,y\in[a,b]$满足$|x-y|<\delta$有
\[ |f(x)-f(y)|<L|x-y|<\varepsilon \]
这表明$f(x)$是$[a,b]$上的一致连续连续函数,显然$f(x)$是$[a,b]$上的连续函数。
其次,我们证明$x_{n}\in[a,b]$
用数学归纳法,当$n=1$时,由条件可知命题成立。
假设$n=m$时命题成立,也即$x_{m}\in[a,b]$
那么当$n=m+1$时,
\[ a\leq x_{m+1}=\frac{1}{2}(x_{m}+f(x_{m}))\leq b \]
故对$x_{n}$有$x_{n}\in[a,b],\quad n=1,2,\dots$
接着,我们注意到
\[ |x_{n+1}-x_{n}|\leq \frac{1}{2}\left(|x_{n}-x_{n-1}|+|f(x_n)-f(x_{n-1})|\right)<\frac{1+L}{2}|x_{n}-x_{n-1}| \]
为了方便起见,这里我们设$M=\frac{1+L}{2}$,显然有$0<M<1$
由上面的式子可知
\[  |x_{n+1}-x_{n}|<M|x_{n}-x_{n-1}|<\cdots<M^{n-1}|x_{2}-x_{1}| \]

\begin{align*}
|x_{n+p}-x_{n}|&\leq |x_{n+p}-x_{n+p-1}|+\cdots+|x_{n+1}-x_{n}|\\
&<(M^{n+p-2}+M^{n+p-3}+\cdots+M^{n-1})|x_{2}-x_{1}| \\
&=\frac{M^{n-1}(1-M^{p})}{1-M}|x_{2}-x_{1}|\\
&<\frac{M^{n-1}}{1-M}|x_{2}-x_{1}| 
 \end{align*}
这样,对$\forall \varepsilon>0$,存在$N=\left[\log_{M}{\frac{\varepsilon(1-M)}{|x_{2}-x_{1}|}}\right]+2 $,对$\forall n>N$及$\forall p\in N^{+}$有
\[ |x_{n+p}-x_{n}|<\varepsilon \]
由序列的Cauchy收敛准则知$x_{n}$收敛,设$\displaystyle \lim_{n\rightarrow\infty}{x_{n}}=x$

\[ x_{n+1}=\frac{1}{2}\left[x_{n}+f(x_{n})\right] \]
两边令$n\rightarrow \infty$,(这样做是合理的,由于我们事先证明了$f(x)$是连续函数)
故得
\[f(x)=x\]

$\square$

三、设实$n$阶方阵$A$的每个元素的绝对值为$2$,证明:
当$n\geq 3$时,$|A|\leq \frac{1}{3}\cdot 2^{n+1}n! $

证明:由于
\begin{align*}
 |A|=\begin{vmatrix}
  x_{11} & x_{12} & \cdots & x_{1n}\\
  x_{21} & x_{22} & \cdots & x_{2n}\\
   \vdots & \vdots &        &\vdots\\
   x_{n1} & x_{n2} & \cdots & x_{nn}\\
\end{vmatrix} 
\end{align*}
其中$x_{ij}=\pm 2,\quad i,j=1,2\cdots,n$,这样,对每一列提出$2$来,就是
\begin{align*} |A|=2^{n}\cdot\begin{vmatrix}
  y_{11} & y_{12} & \cdots & y_{1n}\\
  y_{21} & y_{22} & \cdots & y_{2n}\\
  \vdots & \vdots &        &\vdots\\
  y_{n1} & y_{n2} & \cdots & y_{nn}\\
\end{vmatrix}
\end{align*}
其中$y_{ij}=\pm 1,\quad i,j=1,2\cdots,n$,于是问题转化为证明
\begin{align*}
 |B|=\begin{vmatrix}
  y_{11} & y_{12} & \cdots & y_{1n}\\
  y_{21} & y_{22} & \cdots & y_{2n}\\
   \vdots & \vdots &        &\vdots\\
   y_{n1} & y_{n2} & \cdots & y_{nn}\\
   \end{vmatrix}\leq \frac{2}{3}n!
  \end{align*}
对$n\geq 3$成立。
我们先验证$n=3$的情况,当$n=3$时
\begin{align*}
|B_{3}|=
\begin{vmatrix}
y_{11} & y_{12} & y_{13}\\
y_{21} & y_{22} & y_{23}\\
y_{31} & y_{32} & y_{33}\\
\end{vmatrix}
\end{align*}
按第一行展开就是
\begin{align*}
|B_{3}|=y_{11}
\begin{vmatrix}
y_{22} & y_{23} \\
y_{32} & y_{33} \\
\end{vmatrix}
-y_{12}\begin{vmatrix}
y_{21} & y_{23} \\
y_{31} & y_{33} \\
\end{vmatrix}
+y_{13}\begin{vmatrix}
y_{21} & y_{22} \\
y_{31} & y_{32} \\
\end{vmatrix}
\end{align*}
 注意到由于在这三个列向量
\begin{align*}
\left[\begin{array}{c}
y_{21} \\
y_{31} \\
\end{array}\right],\
\left[\begin{array}{c}
y_{22} \\
y_{32} \\
\end{array}\right],\
\left[\begin{array}{c}
y_{23} \\
y_{33} \\
\end{array}\right] 
\end{align*}
  只  只可能有以下4种情况
  \begin{align*}
\left[\begin{array}{c}
1\\
1\\
\end{array}\right],\
\left[\begin{array}{c}
-1\\
-1\\
\end{array}\right],\
\left[\begin{array}{c}
1\\
-1\\
\end{array}\right],\
\left[\begin{array}{c}
-1\\
1\\
\end{array}\right]
\end{align*}
    由抽屉原理知必有2个列向量要么完全相同,要么互为负向量。这样
 \begin{align*}
\begin{vmatrix}
y_{22} & y_{23} \\
y_{32} & y_{33} \\
\end{vmatrix},\
\begin{vmatrix}
y_{21} & y_{23} \\
y_{31} & y_{33} \\
\end{vmatrix},\
\begin{vmatrix}
y_{21} & y_{22} \\
y_{31} & y_{32} \\
\end{vmatrix}
\end{align*}
  在这3个行列式中必有一个为0,不妨设
\begin{align*}   
\begin{vmatrix}
y_{21} & y_{23} \\
y_{31} & y_{33} \\
\end{vmatrix}=0
\end{align*}
  因此
\begin{align*}        
|B_{3}|=y_{11}\begin{vmatrix}
y_{22} & y_{23} \\
y_{32} & y_{33} \\
\end{vmatrix}
+y_{13}\begin{vmatrix}
y_{21} & y_{22} \\
y_{31} & y_{32} \\
\end{vmatrix}\leq 4=\frac{2}{3}\cdot 3!
\end{align*}
  故$n=3$的情况成立,我们假设$n=m$的情况成立。$(m>3)$即有
  \[ |B_{m}|\leq \frac{2}{3}\cdot m! \]
  则当$n=m+1$时,对$|B_{m+1}|$按第一行展开
  \[       |B_{m+1}|=\sum_{i=1}^{m+1}{b_{1i}A_{1i}} \]
  其中$A_{1i},i=1,2,\cdots m+1$ 是$ b_{1i},i=1,2,\cdots m+1$的代数余子式,显然是$m$阶行列式,故有
  \[    |B_{m+1}|=\sum_{i=1}^{m+1}{b_{1i}A_{1i}}\leq (m+1)\frac{2}{3}m!=\frac{2}{3}(m+1)! \]
  因此,对$n\geq 3$有
\begin{align*} |B|=\begin{vmatrix}
y_{11} & y_{12} & \cdots & y_{1n}\\
y_{21} & y_{22} & \cdots & y_{2n}\\
\vdots & \vdots &        &\vdots\\
y_{n1} & y_{n2} & \cdots & y_{nn}\\
\end{vmatrix}\leq \frac{2}{3}n!
\end{align*}
也即\[ |A|\leq \frac{1}{3}\cdot 2^{n+1}n! \]
等号成立当$A$为正交矩阵。
$\square$

 

四、设$f(x)$为区间$(a,b)$上的可导函数。对$x_{0}\in(a,b)$,若存在$x_{0}$的邻域$U$使得任意的$x\in U\setminus\{x_{0}\}$有$f(x)>f(x_{0})+f'(x_{0})(x-x_{0})$,则称$x_{0}$为$f(x)$的凹点。类似地,若存在$x_{0}$的邻域$U$使得任意的$x\in U\setminus\{x_{0}\}$有$f(x)<f(x_{0})+f'(x_{0})(x-x_{0})$,则称$x_{0}$为$f(x)$的凸点。
求证:若$f(x)$是区间$(a,b)$上的可导函数且不是一次函数,则$f(x)$一定存在凹点或凸点。

证明:(原解答有误,Sorry :( 还在思考中。。。)

 

 

 

 

 

 

 

 

 

 

五、(本题20分)设$\displaystyle A=\left(\begin{array}{ccc}
a_{11}& a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{array} \right)$
为实对称矩阵,$A^{*}$为$A$的伴随矩阵, 记$\displaystyle f(x_1,x_2,x_3,x_4)=\begin{vmatrix}
x_{1}^{2} & x_{2} & x_{3} & x_{4} \\
-x_{2} & a_{11} & a_{12} & a_{13} \\
-x_{3} & a_{21} & a_{22} & a_{23} \\
-x_{4} & a_{31} & a_{32} & a_{33}
\end{vmatrix}$.若$A$的行列式为$-12$,$A$的所有特征值的和为$1$,且$(1,0,-2)^{T}$为$(A^{*}-4I)x=0$的一个解。
试给出一正交变换$\displaystyle\left(
\begin{array}{c}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{array}
\right)=Q\left(\begin{array}{c}
y_1 \\
y_2 \\
y_3 \\
y_4 \\
\end{array}\right)$使得$f(x_1,x_2,x_3,x_4)$化为标准型。

解 (西西提供)

设$A$的三个特征值为$\lambda_{1},\lambda_{2},\lambda_{3}$,则有$$\lambda_{1}\lambda_{2}\lambda_{3}=|A|=-12,\lambda_{1}+\lambda_{2}+\lambda_{3}=1$$
设$$v_{1}=\begin{bmatrix}
1\\
0\\
-2
\end{bmatrix}$$

$$(A^*-4I)v_{1}=0$$,
可得
$$(A+3I)v_{1}=0$$
即$v_{1}$是$A$关于特征值-3的特征向量.
不妨设$\lambda_{1}=-3,$,则显然有$\lambda_{2}=\lambda_{3}=2$,由于$A$的不同特征值的特征向量相互正交,故$\lambda_{2}=\lambda_{3}=2$的特征向量满足
\[x_{1}-2x_{3}=0\]
由此可得特征向量
\[v_{2}=(0,1,0)^T,v_{3}=(2,0,1)^T\]将
$v_{1},v_{2},v_{3}$单位化得到正交阵
$$P=\begin{bmatrix}
\dfrac{1}{\sqrt{5}}&0&\dfrac{2}{\sqrt{5}}\\
0&1&0\\
-\dfrac{2}{\sqrt{5}}&0&\dfrac{1}{\sqrt{5}}
\end{bmatrix}$$
使得
$$P'AP=\begin{bmatrix}
-3&\cdots&\cdots\\
\cdots&2&\cdots\\
\cdots&\cdots&2
\end{bmatrix}$$


\[A=P\begin{bmatrix}
-3&\cdots&\cdots\\
\cdots&2&\cdots\\
\cdots&\cdots&2
\end{bmatrix}P'\]

所以
$$ A=\begin{bmatrix}
1&0&2\\
0&2&0\\
2&0&-2
\end{bmatrix}$$
将A的元素带入得
\[f(x_{1},x_{2},x_{3},x_{4})=-12x^2_{1}-4x^2_{2}-6x^2_{3}+2x^2_{4}-8x_{2}x_{4}\]
仅考虑$x_{2},x_{4}$对应的对称阵
$$\begin{bmatrix}
-4&-4\\
-4&2
\end{bmatrix}$$,
得到特征值为
\[x_{1}=4,x_{2}=-6\]
对应的特征向量为
\[u_{1}=(1,-2)^T,u_{2}=(2,1)^T\],
单位化后可得正交矩阵
$$\begin{bmatrix}
1&0&0&0\\
0&\dfrac{1}{\sqrt{5}}&0&\dfrac{2}{\sqrt{5}}\\
0&0&1&0\\
0&-\dfrac{2}{\sqrt{5}}&0&\dfrac{1}{\sqrt{5}}
\end{bmatrix}$$
使得作变换$x=Qy$后可得
\[f(y_{1},y_{2},y_{3},y_{4})=-12y^2_{1}+4y^2_{2}-6y^2_{3}-6y^2_{4}\]
证毕.$\square$
后面标准化简单做法

由降阶公式得
\[f=|A|(x^2_{1}+(x_{2},x_{3},x_{4})A^{-1}(x_{2},x_{3},x_{4})^T)\]
因此正交阵Q就是正交阵P左上方加1变成4阶方阵,得到的对角阵就是
diag$\{-12(1,-\dfrac{1}{3},\dfrac{1}{2},\dfrac{1}{2})\}$

 

六、令$\mathbf{R}$为实数域,$n$为给定的正整数,$A$表示所有$n$次首一实系数多项式组成的集合。
证明:
\[ \inf_{b\in\mathbf{R},c>0,P(x)\in A}{\dfrac{\int_{b}^{b+c}{|P(x)|dx}}{c^{n+1}}}>0 \]
证明(刘飞提供)
注意到对于首一多项式$P$如果含有虚数根则必然成对出现,比如$g\pm ih$,实根为$l$,于是
\[ P(x)=\prod{(x-l)}\prod{(x^2-2gx+g^2+h^2)} \]
于是,当$x$是实数时有
\[ |P(x)|=\left|\prod{(x-l)}\prod{(x^2-2gx+g^2+h^2)}\right|\geq \prod{(x-l)}\prod{(x-g)^2}\equiv Q(x) \]
而$Q(x)$的根都是实数,重新标号为$s_{1}\leq s_{2}\leq \cdots \leq s_{n}$,现在,利用一个平移,不妨设积分区间为$[0,c]$,而$Q(x)$的零点一般来说不会全部落入此区间,譬如只有$r$个$(0\leq r\leq n)$在此区间,那么这$r$个点将会把这个区间分割成若干个小区间,其中至少有一个长度不小于$\frac{c}{n+1}$,记此区间为$[u,v]((n+1)(v-u)\geq c)$,则$Q(x)$在区间$[u,v]$上保持定号。这时$Q(x)$的零点不会落入$(u,v)$中,设$Q(x)$的根有$p$个小于或等于$u$,记为$t_{k}$,$q$个大于或等于$v$,记为$f_{k}$,
于是当$x\in[u,v]$时,有
\[ |P(x)|\geq |Q(x)|=\prod{|x-l|}\prod{|x-g|^2}\geq (x-u)^{p}(v-x)^{q} \]
\[ \frac{\int_{0}^{c}{|P(x)|dx}}{c^{n+1}}\geq \frac{\int_{u}^{v}{|P(x)|dx}}{[(n+1)(v-u)]^{n+1}}\geq \frac{\int_{u}^{v}{|(x-u)^p(v-x)^q|dx}}{[(n+1)(v-u)]^{n+1}}=\frac{B(p+1,q+1)}{(n+1)^{n+1}} \]
\[ \frac{\int_{0}^{c}{|P(x)|dx}}{c^{n+1}}\geq \min_{p+q=n}{\frac{B(p+1,q+1)}{(n+1)^{n+1}}}>0 \]

$\square$

另外一个解答

 1. The map $\tau_b\colon A\to A$, $\tau_b(P)(x):=P(x+b)$ is bijective, hence it's enough to consider the case $b=0$.
 2. Making the substitution $ct=x$, we are reduced to show that
$$\inf_{a_0,\dots,a_{n-1},c}\int_0^1\left|t^n+\sum_{j=0}^{n-1}a_jt^jc^{j-n}\right|dt>0.$$
 3. Putting $b_j:=a_jc^{j-n}$, we actually have to show that
$$\inf_{b_0,\dots,b_{n-1}}\int_0^1\left|t^n+\sum_{j=0}^{n-1}b_jt^j\right|dt>0.$$
 4. To conclude, we use functional analysis as notes Julien in the comments, that is, we view the latest infimum as the distance of $t^n$ to the finite-dimensional hence closed subspace of polynomials of degree smaller than $n-1$ for the norm $N(f):=\int_0^1|f(x)|dx$.

 

Mar 12

209 设$f:[0,1]\rightarrow R$是连续函数,且满足
\[ \int_{0}^{1}{f(x)dx}=\int_{0}^{1}{xf(x)dx} \]
求证:存在$\xi\in(0,1)$满足
$$f(\xi)=f'(\xi)\int_{0}^{\xi}{f(x)dx} $$
证明:
我们设\[ F(x)=\int_{0}^{x}{f(t)dt} \]
可得
\[ \int_{0}^{1}{F(x)dx}=0 \]
也就是存在
\[ F(c)=0 \]
构造辅助函数
\[ G(x)=e^{-f(x)}\int_{0}^{x}{f(t)dt} \]
于是有
\[ G(0)=G(c)=0 \]
由Rolle定理知,存在$\xi$,使得
\[ f(\xi)=f'(\xi)\int_{0}^{\xi}{f(t)dt} \]
$\square$

210-212 设$f:[0,1]\rightarrow R$是连续函数,且满足
\[ \int_{0}^{1}{f(x)dx}=\int_{0}^{1}{xf(x)dx} \]
求证:存在$\xi\in(0,1)$满足
(1) \[ \xi f(\xi)=\int_{0}^{\xi}{xf(x)dx} \]
(2) \[ \xi f(\xi)=2\int_{\xi}^{0}{xf(x)dx} \]
(3) \[ \xi^2f(\xi)=\int_{0}^{\xi}{xf(x)dx} \]
证明:很容易看见这3个问题构造的辅助函数为
\[ G_{1}(x)=e^{-x}\int_{0}^{x}{tf(t)dt} \]
\[ G_{2}(x)=e^{2x}\int_{0}^{x}{tf(t)dt} \]
\[ G_{3}(x)=\frac{\int_{0}^{x}{tf(t)dt}}{x} \]
并且都有
$$ G_{i}(0)=0 \qquad i=1,2,3 $$
这3个辅助函数都和$\displaystyle \int_{0}^{x}{tf(t)dt} $有关,也就是说我们剩下的工作只要找到一点$c\in(0,1)$使得
\[ \int_{0}^{c}{tf(t)dt}=0 \]
然后再用Rolle定理,则上面3个问题自明。
为此,设
\[ F(x)=\int_{0}^{x}{f(t)dt} \]
问题变成
要找到一点$c\in(0,1)$使得
\[ cF(c)=\int_{0}^{c}{F(x)dx} \]
我们有条件
\[ F(0)=0, \int_{0}^{1}{F(x)dx}=0 \]
要怎么证明上面的问题呢?
自然想到构造辅助函数
\[ H(x)=\frac{\int_{0}^{x}{F(t)dt}}{x} \]
并做连续性开拓
\[  H(x)=\left\{
     \begin{array}{ll}
      \frac{\int_{0}^{x}{F(t)dt}}{x} & \hbox{$x\in(0,1]$;} \\
       F(0) & \hbox{$x=0$.}
     \end{array}
   \right. \]
容易发现
\[ H(0)=H(1)=0 \]
所以对$H(x)$用Rolle定理有$c\in(0,1)$使得
\[ cF(c)=\int_{0}^{c}{F(x)dx} \]
也就是
\[ \int_{0}^{c}{tf(t)dt}=0 \]
因此,有
\[ G_{i}(0)=G_{i}(c)=0 \]
再用Rolle定理,就全部搞定了。
$\square$