北京大学数学科学学院2015年直博生摸底考试试题解答

这份试题本来已经写好答案了，但因为电脑的事，里面文件都没了。下面重新给出解答：

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2.(30分) 判断级数$\sum\nolimits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  + {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}}}}$的敛散性,其中$[x]$表示$x$的取整.

证：\begin{align*}\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  + {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}}}}  &= \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}\left( {\sqrt n  - {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}} \right)}}{{n - 1}}} \\&= \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}\sqrt n }}{{n - 1}}}  - \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^{n + \left[ {\sqrt n } \right]}}}}{{n - 1}}}.\end{align*}

由Leibniz判别法知,级数\[\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}\sqrt n }}{{n - 1}}}  = \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  - \frac{1}{{\sqrt n }}}}} \]收敛.

当$k \le \sqrt n  < k + 1$,即${k^2} \le n < {\left( {k + 1} \right)^2}$时, ${\left[ {\sqrt n } \right]}=k$,则

\begin{align*}&\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^{n + \left[ {\sqrt n } \right]}}}}{{n - 1}}}  =  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\sum\limits_{n = {k^2}}^{{k^2} + 2k} {\frac{{{{\left( { - 1} \right)}^{n + k}}}}{{n - 1}}} }  =  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {{{\left( { - 1} \right)}^k}\sum\limits_{n = {k^2}}^{{k^2} + 2k} {\frac{{{{\left( { - 1} \right)}^n}}}{{n - 1}}} } \\&=  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {{{\left( { - 1} \right)}^k}\sum\limits_{n = {k^2}}^{{k^2} + 2k} {{{\left( { - 1} \right)}^{{k^2}}}\left[ {\left( {\frac{1}{{{k^2} - 1}} + \frac{1}{{{k^2} + 1}} +  \cdots  + \frac{1}{{{k^2} + 2k - 1}}} \right)} \right.} } \\&\left. { - \left( {\frac{1}{{{k^2}}} + \frac{1}{{{k^2} + 2}} +  \cdots  + \frac{1}{{{k^2} + 2k - 2}}} \right)} \right] =  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\left[ {\left( {\frac{1}{{{k^2} - 1}} + \frac{1}{{{k^2} + 1}} +  \cdots  + \frac{1}{{{k^2} + 2k - 1}}} \right)} \right.} \\&\left. { - \left( {\frac{1}{{{k^2}}} + \frac{1}{{{k^2} + 2}} +  \cdots  + \frac{1}{{{k^2} + 2k - 2}}} \right)} \right] \le  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\left[ {\left( {\frac{{k + 1}}{{{k^2} - 1}} - \frac{k}{{{k^2} + 2k - 2}}} \right)} \right.} \\&\le  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\left[ {\left( {\frac{1}{{k - 1}} - \frac{1}{{k + 2}}} \right)} \right.}  =  - \frac{1}{2} + 1 + \frac{1}{2} + \frac{1}{3} = \frac{4}{3}\end{align*}

且

\begin{align*}{a_n} &= \left( {\frac{1}{{{k^2} - 1}} + \frac{1}{{{k^2} + 1}} +  \cdots  + \frac{1}{{{k^2} + 2k - 1}}} \right) - \left( {\frac{1}{{{k^2}}} + \frac{1}{{{k^2} + 2}} +  \cdots  + \frac{1}{{{k^2} + 2k - 2}}} \right)\\&= \left( {\frac{1}{{{k^2} - 1}} - \frac{1}{{{k^2}}}} \right) + \left( {\frac{1}{{{k^2} + 1}} - \frac{1}{{{k^2} + 2}}} \right) +  \cdots  + \left( {\frac{1}{{{k^2} + 2k - 3}} - \frac{1}{{{k^2} + 2k - 2}}} \right) + \frac{1}{{{k^2} + 2k - 1}}\\&\ge \frac{1}{{{k^2} + 2k - 1}} > 0.\end{align*}

故\[\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^{n + \left[ {\sqrt n } \right]}}}}{{n - 1}}} \]收敛,从而数列$\sum\nolimits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  + {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}}}}$亦收敛.

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3.(30分) 证明\[\int_0^1 {\int_0^1 {\frac{1}{{{{\left( {xy} \right)}^{xy}}}}dxdy} }  = \int_0^1 {\frac{1}{{{x^x}}}dx}  = \sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^n}}}} .\]

证：令$u = xy,v = x$,则$x=v,y=\frac uv$.由$0\leq x,y\leq 1$可知$0\leq u\leq v,0\leq v\leq 1$,

\begin{align*}\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \left| {\begin{array}{*{20}{c}}0&1\\{\frac{1}{v}}&{ - \frac{u}{{{v^2}}}}\end{array}} \right| = - \frac{1}{v}\,,\end{align*}

那么有

\begin{align*}&\int_0^1 {\int_0^1 {\frac{1}{{{{\left( {xy} \right)}^{xy}}}}dxdy} } = \int_0^1 {dv} \int_0^v {\frac{1}{{{u^u}v}}du} \\&= \int_0^1 {du} \int_u^1 {\frac{1}{{{u^u}v}}dv} = \int_0^1 {\frac{{ - \ln u}}{{{u^u}}}du} \\&= \int_0^1 {\frac{{ - \ln u - 1}}{{{u^u}}}du} + \int_0^1 {\frac{1}{{{u^u}}}du} \\&= \left[ {\frac{1}{{{u^u}}}} \right]_0^1 + \int_0^1 {\frac{1}{{{u^u}}}du} = \int_0^1 {\frac{1}{{{x^x}}}dx}.\end{align*}

而

\begin{align*}&\int_0^1 {\frac{1}{{{x^x}}}dx} = \int_0^1 {{e^{ - x\ln x}}dx} = \int_0^1 {{e^{ - x\ln x}}dx} \\&= \int_0^1 {\sum\limits_{n = 0}^{+\infty} {\frac{{{{\left( { - x\ln x} \right)}^n}}}{{n!}}} dx} = \sum\limits_{n = 0}^{+\infty} {\frac{1}{{n!}}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} }.\end{align*}

令$t=-(n+1)\ln x$,有

\begin{align*}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} &= \frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}\int_0^{ + +\infty } {{t^n}{e^{ - t}}dt} \\&= \frac{{\Gamma \left( {n + 1} \right)}}{{{{\left( {n + 1} \right)}^{n + 1}}}} = \frac{{n!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}.\end{align*}

因此有

\begin{align*}\int_0^1 {\frac{1}{{{x^x}}}dx} = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{n!}}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} } = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}} =\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^n}}}} .\end{align*}

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4.(25分) 设$A$是一个$n$阶方阵,且$n\geq3$.$A^\ast$是$A$的伴随矩阵(即$A$的代数余子式所组成的矩阵).试证明,若${(A^\ast)}^\ast\neq O$ (零矩阵),则$A$可逆,且此时${(A^\ast)}^\ast$是$A$的一个纯量倍.

证：由于$A$可逆时,$A^\ast$必可逆,从而${(A^\ast)}^\ast$亦可逆;当$A$不可逆时,$A^\ast$的秩不大于$1$,从而${(A^\ast)}^\ast$必为零矩阵.

由此可知,当${(A^\ast)}^\ast\neq O$ 时,$A$可逆.再由\[A{A^ * } = \left| A \right|{I_n} \Rightarrow \left| {{A^ * }} \right| = {\left| A \right|^{n - 1}},{\left( {{A^ * }} \right)^{ - 1}} = \frac{1}{{\left| A \right|}}A\]及\[{A^ * }{\left( {{A^ * }} \right)^*} = \left| {{A^ * }} \right|{I_n}\]可知

\[{\left( {{A^ * }} \right)^*} = \left| {{A^ * }} \right|{\left( {{A^ * }} \right)^{ - 1}} = {\left| A \right|^{n - 1}} \cdot \frac{1}{{\left| A \right|}}A = {\left| A \right|^{n - 2}}A.\]由于${\left| A \right|^{n - 2}}$是个定值,我们得知${(A^\ast)}^\ast$是$A$的一个纯量倍.

解：\[f\left( \lambda \right) = \left| {\lambda {I_n} - A} \right| = \left( {\lambda + 1} \right){\left( {\lambda - 1} \right)^2}.\]

由Cayley---Hamilton定理可知,\[f\left( A \right) = \left( {A + 1} \right){\left( {A - 1} \right)^2} = {A^3} - {A^2} - A + {I_n} = 0.\]

我们有

\[A = {A^3} + \left( {{A^2} - {A^4}} \right).\]取\[f\left( x \right) = {x^3},\varphi \left( x \right) = {x^2} - {x^4}\]即可.

证：（1）记$A(1,0,0)$,设直线上有一点$B(x,y,z)$,则$\overrightarrow {AB} = \left( {x - 1,y,z} \right)=t(0,1,1)$,则$B$为$(1,t,t)$.对于给定$z=t$,其绕$z$轴旋转形成的图形为\[{x^2} + {y^2} = {t^2} + 1 = {z^2} + 1,\]故该二次曲面方程为\[{x^2} + {y^2} - {z^2} - 1 = 0.\]

（2）设固定平面$\Sigma$的方程为$Ax+By+Cz-a_t=0$

• 若$A^2+B^2=0$即$A=B=0$时,方程退化成

$z = \frac{{{a_t}}}{C}$,\[\left\{ \begin{array}{l}{x^2} + {y^2} - {z^2} = 1\\z = \frac{{{a_t}}}{C}\end{array} \right. \Rightarrow \frac{{{x^2}}}{{\frac{{a_t^2}}{{{C^2}}} + 1}} + \frac{{{y^2}}}{{\frac{{a_t^2}}{{{C^2}}} + 1}} = 1,\]

可知此时$\Sigma_t\cap S$为圆,当然可以看成是椭圆.

• 若$A^2+B^2\neq0$时,作坐标系旋转

\[\left\{ \begin{array}{l}{x_1} = \frac{B}{{\sqrt {{A^2} + {B^2}} }}x - \frac{A}{{\sqrt {{A^2} + {B^2}} }}y\\{y_1} = - \frac{{AC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}x - \frac{{BC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}y + \frac{{{A^2} + {B^2}}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}z\\{z_1} = \frac{A}{{\sqrt {{A^2} + {B^2} + {C^2}} }}x + \frac{B}{{\sqrt {{A^2} + {B^2} + {C^2}} }}y + \frac{C}{{\sqrt {{A^2} + {B^2} + {C^2}} }}z\end{array} \right.\]

即

\[\left\{ \begin{array}{l}x = \frac{B}{{\sqrt {{A^2} + {B^2}} }}{x_1} - \frac{{AC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}{y_1} + \frac{A}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{z_1}\\y = - \frac{A}{{\sqrt {{A^2} + {B^2}} }}{x_1} - \frac{{BC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}{y_1} + \frac{B}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{z_1}\\z = \frac{{\sqrt {{A^2} + {B^2}} }}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{y_1} + \frac{C}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{z_1}\end{array} \right.,\]

该平面方程化为${z_1} = \frac{{{a_t}}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$, $S$化为

\[x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}z_1^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0,\]

则截面方程为

\[x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{{\left( {{A^2} + {B^2} + {C^2}} \right)}^2}}}a_t^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0.\]显然$\Sigma_t\cap S$为椭圆.

综上可知, $\Sigma_t\cap S$总是椭圆.

（3）由(2)可知,在$x_1y_1z_1$坐标系中,椭圆的中心为

\[\left( {0,0,\frac{{{a_t}}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right),\]即中心在$z_1$轴上,\[\left\{ \begin{array}{l}{x_1} = \frac{B}{{\sqrt {{A^2} + {B^2}} }}x - \frac{A}{{\sqrt {{A^2} + {B^2}} }}y = 0\\{y_1} = - \frac{{AC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}x - \frac{{BC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}y + \frac{{{A^2} + {B^2}}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}z = 0\end{array} \right.,\]从而在$xyz$坐标系中,椭圆中心落在过原点的定直线

\[L: \left\{ \begin{array}{l}Bx - Ay = 0\\- ACx - BCy + \left( {{A^2} + {B^2}} \right)z = 0\end{array} \right.\]

上.

（4）设\[S: x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}z_1^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0,\]上一点$P\left( {{x_{1,0}},{y_{1,0}},{z_{1,0}}} \right)$,则曲面$S$在$P$点处的切面为

\[\left( {{x_1} - {x_{1,0}}} \right) \cdot \left( {2{x_{1,0}}} \right) + \left( {{y_1} - {y_{1,0}}} \right) \cdot \left( {2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,0}}} \right) + \left( {{z_1} - {z_{1,0}}} \right) \cdot \left( { - 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}}} \right) = 0,\]记坐标系$xyz$中,直线$L$上任一点$p$在$x_1,y_1,z_1$中的坐标为$p_1(0,0,m)$,则$p_1$满足切面方程,即\[\left( {0 - {x_{1,0}}} \right) \cdot \left( {2{x_{1,0}}} \right) + \left( {0 - {y_{1,0}}} \right) \cdot \left( {2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,0}}} \right) + \left( {m - {z_{1,0}}} \right) \cdot \left( { - 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}}} \right) = 0,\]\[ - 2m\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}} - \frac{{8C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 2 = 0 \Rightarrow {z_{1,0}} = \frac{{\left( {{A^2} + {B^2} + {C^2}} \right) + 4C\sqrt {{A^2} + {B^2}} }}{{m\left( {{A^2} + {B^2} - {C^2}} \right)}},\]故此时的$P$的$z_1$坐标为定值, 在$x_1,y_1,z_1$中$P$形成的轨迹为椭圆,且对应在$x,y,z$中$\Sigma_t\cap S$的一个椭圆.

（5）设$x_1y_1z_1$坐标系中, $\Gamma_p: \left\{ \begin{array}{l}{z_1} = \frac{{\left( {{A^2} + {B^2} + {C^2}} \right) + 4C\sqrt {{A^2} + {B^2}} }}{{m\left( {{A^2} + {B^2} - {C^2}} \right)}}\\x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}z_1^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0\end{array} \right.$所在平面落在$S$外部的一点$\hat p_1$的坐标为$\left( {{x_{1,1}},{y_{1,1}},{z_{1,1}}} \right)$, $\hat p_1$满足(4)中$P$点处的切面方程,即\[\left( {{x_{1,1}} - {x_{1,0}}} \right) \cdot \left( {2{x_{1,0}}} \right) + \left( {{y_{1,1}} - {y_{1,0}}} \right) \cdot \left( {2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,0}}} \right) + \left( {{z_{1,1}} - {z_{1,0}}} \right) \cdot \left( { - 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}}} \right) = 0,\]

则有

\[2{x_{1,1}}{x_{1,0}} + 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,1}}{y_{1,0}} - \frac{{8C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 2 + \frac{{2\left( {{A^2} + {B^2} + {C^2}} \right) + 8C\sqrt {{A^2} + {B^2}} }}{{m\left( {{A^2} + {B^2} - {C^2}} \right)}}{z_{1,0}} = 0.\]显然其经过$p_1$.进一步地,经过与之前类似的坐标轴旋转,我们知道$P$的轨迹落在一条双曲线上.