Eufisky - The lost book

## 2018年考研试题

$\frac{1}{5}<\int_0^1\frac{xe^xdx}{\sqrt{x^2-x+25}}<\frac{2}{\sqrt{99}}.$

$x^2-x+25=\left(x-\frac{1}{2}\right)^2+\frac{99}{4}>\frac{99}{4}, a.e. x\in [0,1],$

$\int_0^1\frac{xe^xdx}{\sqrt{x^2-x+25}}<\frac{2}{\sqrt{99}}\int_0^1xe^xdx=\frac{2}{\sqrt{99}}.$

$\int_0^1\frac{xe^xdx}{\sqrt{x^2-x+25}}=\frac{(x-1)e^x}{\sqrt{x^2-x+25}}\Big|_0^1+\int_0^1\frac{(x-1)(x-\frac{1}{2})e^x}{\sqrt{(x^2-x+25)^3}}dx=\frac{1}{5}+\int_0^1\frac{(x-1)(x-\frac{1}{2})e^x}{\sqrt{(x^2-x+25)^3}}dx.$
$f(x)=\frac{x-\frac{1}{2}}{\sqrt{(x^2-x+25)^3}},g(x)=(x-1)e^x,$
$f'(x)=\frac{-2x^2+2x+\frac{97}{4}}{\sqrt{(x^2-x+25)^5}}>0,\forall x\in [0,1],\int_0^1 f(x)dx=0, g'(x)=xe^x,$

$\int_0^1\frac{(x-1)(x-\frac{1}{2})e^x}{\sqrt{(x^2-x+25)^3}}dx=\int_0^1f(x)g(x)dx>\int_0^1 f(x)dx\int_0^1g(x)dx=0.$
$\int_0^1\frac{xe^xdx}{\sqrt{x^2-x+25}}>\frac{1}{5}.$

$\frac{1}{5}<\int_0^1\frac{xe^xdx}{\sqrt{x^2-x+25}}<\frac{2}{\sqrt{99}}.$

2018年武汉大学653数学分析

2.计算极限$$\lim_{n\rightarrow\infty}\frac{\int_0^\mathrm\pi\sin^nx\cos^6x\operatorname dx}{\int_0^\mathrm\pi\sin^nx\operatorname dx}$$
3.已知$x_{n+1}=\ln\left(1+x_n\right)$,且$x_1>0$,计算$$\lim_{n\rightarrow\infty}nx_n$$

$$\oint\limits_S\frac{\partial^2 u}{\partial x_i\partial x_j}{\rm d}S，i，j=1,2,3$$,其中$S:x_1^2+x_2^2+x_3^2=R^2$

2.在适当的条件下，证明牛顿切线法收敛

1.存在阶梯函数$g_\varepsilon(x)$使得$$\int_a^b\left|f(x)-g_\varepsilon(x)\right|\operatorname dx<\frac\varepsilon2$$
2.计算$$\lim_{n\rightarrow\infty}\int_a^b\varphi(nx)\operatorname dx$$
3.证明$$\lim_{n\rightarrow\infty}\int_a^bf(x)\varphi(nx)\operatorname dx=\frac1T\int_0^T\varphi(x)\operatorname dx\int_a^bf(x)\operatorname dx$$
4.计算$$\lim_{n\rightarrow\infty}\frac1{\ln n}\int_0^T\frac{\varphi(nx)}xdx,其中函数\frac{\varphi(nx)}x收敛$$

$\int_0^1\frac{\sin x^n}{x^n}dx=\frac{1}{n}\int_0^1\frac{\sin y}{y^{2-\frac{1}{n}}}dy$
$\int_0^1 y^{\frac{1}{n}-2}\sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)!}y^{2k+1}dy=\int_0^1 y^{\frac{1}{n}-1}dy+\int_0^1\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}y^{2k-1+\frac{1}{n}}dy$

$\int_0^1 y^{\frac{1}{n}-1}dy=n,\left|\frac{(-1)^k}{(2k+1)!}y^{2k-1+\frac{1}{n}}\right|\le \frac{1}{(2k+1)!}$

$=n+\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\int_0^1 y^{2k-1+\frac{1}{n}}dy=n+\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+\frac{1}{n}}$

$\int_0^1\frac{\sin x^n}{x^n}dx=1+\frac{1}{n}\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+\frac{1}{n}}$

$\lim_{n\to+\infty}n\ln \left(1+\frac{1}{n}\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+\frac{1}{n}}\right)=\sum_{k=1}^{+\infty}\left(\frac{(-1)^k}{2k(2k+1)!}\right)$
$\left|\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+\frac{1}{n}}\right|\le \sum_{k=1}^{+\infty}\frac{1}{(2k+1)!}<+\infty$
$\lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+\frac{1}{n}}=0$
$\lim_{n\to+\infty}n\ln \left(1+\frac{1}{n}\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+\frac{1}{n}}\right)=\lim_{n\to+\infty}\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+\frac{1}{n}}$

$f(x)=\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k+x}$

$=\lim_{n\to +\infty}f(\frac{1}{n})=f(0)=\sum_{k=1}^{+\infty}\frac{(-1)^k}{(2k+1)!}\frac{1}{2k}$

$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^n\ln(1+\frac{k^2-k}{n^2})=\ln 2-2+\frac{\pi}{2}$

$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^n\ln(1+\frac{k^2}{n^2})=\int_0^1\ln(1+x^2)dx=x\ln(1+x^2)\left|_0^1\right.-\int_0^1\frac{2x^2}{1+x^2}dx=\ln 2-2+\frac{\pi}{2}$

$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^n\ln(1+\frac{k^2-k}{n^2})=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^n\ln(1+\frac{k^2}{n^2})$

$x,y\ge 0,\left|\ln(1+x)-\ln(1+y)\right|=\left|\frac{x-y}{1+\xi}\right|\le|x-y|$
$\left|\frac{1}{n}\sum_{k=1}^n\ln(1+\frac{k^2-k}{n^2})-\frac{1}{n}\sum_{k=1}^n\ln(1+\frac{k^2}{n^2})\right|\le \frac{1}{n}\sum_{k=1}^n\left|\ln(1+\frac{k^2-k}{n^2})-\ln(1+\frac{k^2}{n^2})\right|$
$\le \frac{1}{n}\sum_{k=1}^n\frac{k}{n^2}=\frac{n+1}{2n^2}\to 0$

$f(x)$在$[1,+\infty)$上二次可导,$\forall x\in [1,+\infty),f(x)>0,f''(x)\le 0,f(+\infty)=+\infty$

$\lim_{s\to 0^+}\sum_{n=1}^{+\infty}\frac{(-1)^n}{f^s(n)}$

$\forall x\ge x_0,f'(x)\le f'(x_0)\le 0,f(x)\le f(x_0)$与$f(+\infty)=+\infty$矛盾.因此$f$在$[1,+\infty)$严增.

$S_{2n}(s)=\sum_{k=1}^n \left(\frac{1}{f^s(2k)}-\frac{1}{f^s(2k-1)}\right)$

$\lim_{n\to +\infty}S_{2n}(s)$存在且等于
$\sum_{n=1}^{+\infty}\frac{(-1)^n}{f^s(n)}$

$\frac{1}{f^s(2k)}-\frac{1}{f^s(2k-1)}=\frac{-sf'(\xi)}{f^{s+1}(\xi)},\xi\in (2k-1,2k)$

$\frac{-sf'(2k-1)}{f^{s+1}(2k-1)}\le\frac{1}{f^s(2k)}-\frac{1}{f^s(2k-1)}\le \frac{-sf'(2k)}{f^{s+1}(2k)}$
$\sum_{k=1}^n\frac{-sf'(2k-1)}{f^{s+1}(2k-1)}\le S_{2n}(s)\le \sum_{k=1}^n\frac{-sf'(2k)}{f^{s+1}(2k)}$

$\frac{f'(2k-1)}{f^{s+1}(2k-1)}\le \frac{1}{2}\int_{2k-3}^{2k-1} \frac{f'(t)}{f^{s+1}(t)}dt$
$\sum_{k=2}^{+\infty}\frac{f'(2k-1)}{f^{s+1}(2k-1)}\le\frac{1}{2}\int_{1}^{+\infty} \frac{f'(t)}{f^{s+1}(t)}dt=\frac{1}{2}\frac{-1}{sf^s(t)}\left|_{t=1}^{t=+\infty}\right.=\frac{1}{2sf^s(1)}$

$S_{2n}(s)$有下界故极限存在.再次利用面积原理
$k\ge 1$时
$\frac{f'(2k)}{f^{s+1}(2k)}\ge \frac{1}{2}\int_{2k}^{2k+2} \frac{f'(t)}{f^{s+1}(t)}dt$
$\sum_{k=1}^{+\infty}\frac{f'(2k)}{f^{s+1}(2k)}\ge\frac{1}{2}\int_{2}^{+\infty} \frac{f'(t)}{f^{s+1}(t)}dt=\frac{1}{2}\frac{-1}{sf^s(t)}\left|_{t=2}^{t=+\infty}\right.=\frac{1}{2sf^s(2)}$

$-s\left(\frac{f'(1)}{f^{s+1}(1)}+\frac{1}{2sf^s(1)}\right)\le \lim_{n\to +\infty}S_{2n}(s)\le -s\frac{1}{2sf^s(2)}$

$-s\left(\frac{f'(1)}{f^{s+1}(1)}+\frac{1}{2sf^s(1)}\right)\le \sum_{n=1}^{+\infty}\frac{(-1)^n}{f^s(n)}\le -s\frac{1}{2sf^s(2)}$

$\lim_{s\to 0^+}\sum_{n=1}^{+\infty}\frac{(-1)^n}{f^s(n)}=-\frac{1}{2}$

01. (15pt) 计算极限
$\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^{x}\text{．}$
02. (15pt) 计算极限
$\lim_{x\to 0} \left(\frac{4+\mathrm{e}^{\frac1x}}{2+\mathrm{e}^{\frac4x}}+\frac{\sin x}{|x|} \right)\text{．}$
03. (15pt) 判断 (并证明) 函数 $f(x,y)=\sqrt{|{xy}|}$ 在点 $(0,0)$ 处的可微性．

04. (15pt) 求三个实常数 $a,b,c$，使得下式成立
$\lim_{x\to 0}\frac1{\tan x -ax}\int_b^x\frac{s^2}{\sqrt{1-s^2}}\,\mathrm{d}s =c\text{．}$
05. (15pt) 计算不定积分
$\int\frac{\mathrm{d}x}{\sin^6 x+\cos^6 x}\text{．}$
06. (15pt) 设函数 $f(x)$ 在 $[-1,1]$ 上二次连续可微，$f(0)=0$，证明：
$\left|\int_{-1}^1 f(x)\,\mathrm{d}x\right|\leq\frac{M}{3},\quad \text{其中 }M=\max_{x\in[-1,1]}\left|f''(x)\right|\text{．}$
07. (15pt) 求曲线 $y=\dfrac12x^2$ 上的点，使得曲线在该点处的法线被曲线所截得的线段长度最短．

08. (15pt) 设 $x>0$，证明
$\sqrt{x+1}-\sqrt{x}=\frac1{2\sqrt{x+\theta}}\text{，}$其中 $\theta=\theta(x)>0$，并且 $\lim\limits_{x\to 0}\theta(x)=\dfrac 14$．

09. (15pt) 设
$u_n(x)=\frac{(-1)^n}{(n^2-n+1)^x}\quad (n\geq 0)\text{，}$求函数 $f(x)=\sum\limits_{n=0}^{\infty}u_n(x)$ 的绝对收敛、条件收敛以及发散的区域．

10. (15pt) 证明
$\frac15<\int_0^1\frac{x\mathrm{e}^x}{\sqrt{x^2-x+25}}\,\mathrm{d}x<\frac{2\sqrt{11}}{33}\text{．}$

$\frac{r(x)}{p(x)q(x)}=\frac{u(x)}{p(x)}+\frac{v(x)}{q(x)}\text{．}$

(1) 计算 $D_4$；
(2) 证明 $D_n$ 满足递推关系式 $D_n=-4D_{n-1}-4D_{n-2}$；
(3) 求 $n$ 阶方阵 $A_n=\left(\left|\frac1i-\frac1j\right|^{\llap{\phantom{b}}}\right)_{1\leq i,j \leq n}$ 的行列式 $\mathrm{det}(A_n)$．

(1) $\mathrm{rank}(A)+\mathrm{rank}(B) \leq n$；
(2) 对于方阵 $A$ 和正整数 $k\,(\mathrm{rank}(A) \leq k \leq n)$，必存在方阵 $B$，使得
$\mathrm{rank}(A)+\mathrm{rank}(B)=k\text{．}$

$q(x_1,x_2,x_3)=5x_1^2+5x_2^2+5x_3^2-2x_1x_2-2x_2x_3-2x_1x_3\text{．}$

$V_1=\left\{\,x\in\mathbb{C}^n\,\middle|\,A_1 x=0\,\right\},\quad V_1=\left\{\,x\in\mathbb{C}^n\,\middle|\,A_2 x=0\,\right\}\text{，}$证明：矩阵 $A$ 可逆的充分必要条件是向量空间 $\mathbb{C}^n$ 能够表示成子空间 $V_1$ 与 $V_2$ 的直和：$\mathbb{C}^n=V_1 \oplus V_2$．

$f(A)=A+A'\quad \forall A\in V\text{，}$其中 $A'$ 为 $A$ 的转置．求 $f$ 的特征值、特征子空间、极小多项式．

9.  设 $B_R=\{(x,y): x^2+y^2< R^2\},u\in C^2( B_R)\cap C(\overline {B_R})$ .

1) 若$\Delta u\geqslant 0$, 证明
$\max_{(x,y)\in\overline {B_R}} u(x,y)= \max_{(x,y) \in \partial B_R} u(x,y).$

$\Delta v_\varepsilon (x,y)=\Delta u(x,y)+4\varepsilon\geqslant 4\varepsilon.$

$\max_{(x,y)\in\overline {B_R}} v_\varepsilon (x,y)= \max_{(x,y) \in \partial B_R} v_\varepsilon(x,y).$

$\max_{(x,y)\in\overline {B_R}} u(x,y)= \max_{(x,y) \in \partial B_R} u(x,y).$

2).  若$\Delta u(x,y)=0$, 则
$\frac{d}{dr}\left(\frac{1}{2\pi r}\int_{\partial B_r}u(x,y)ds\right)=0, 0\leqslant r\leqslant R.$

$\frac{1}{2\pi r}\int_{\partial B_r}u(x,y)ds=\frac{1}{2\pi}\int_0^{2\pi}u(r\cos\theta,r\sin\theta)d\theta= \int_{\partial B_1}u(rx,ry)ds.$从而根据Gauss公式, 得到

\begin{align*}\frac{d}{dr}\left(\frac{1}{2\pi r}\int_{\partial
B_r}u(x,y)ds\right)&=\frac{1}{2\pi}\int_{\partial B_1}(u_x(rx,ry)x+u_y(rx,ry)y)ds\\
&=\frac{1}{2\pi}\int_{\partial B_1} \frac{\partial
u(rx,ry)}{\partial
\nu}ds\\
&=\frac{1}{2\pi}\iint\limits_{\overline B_1}\Delta
u(rx,ry)dxdy\\
&=0.\end{align*}
3).  证明 若$\Delta u(x,y)=0$, 则
$u(0,0)=\frac{1}{2\pi r}\int_{\partial B_r}u(x,y)ds.$

$\frac{1}{2\pi r}\int_{\partial B_r}u(x,y)ds=\lim_{r\to 0^+}\frac{1}{2\pi r}\int_{\partial B_r}u(x,y)ds=u(0,0).$

2017-2018学年北京大学高等代数实验班期末试题2018.1.9
2018.1.9 上午8：30--10：30\\

$$T(f)(t)=f(-t), \ \ U(f)(t)=f(t+1)-f(t), \ \ \forall \ f\in V, t\in F_p.$$

$$T_W(\alpha)=T(\alpha), \alpha\in W,$$
$$T_{V/W}(\alpha+W)=T(\alpha)+W, \alpha\in V.$$

(a) 对任意的$a\in V-\{0\}$, 存在$\beta\in W$使得$\alpha A\beta^t\neq 0$.

(b) 对任意的$\gamma\in F^n$, 存在$\beta\in W$使得对任意的$\alpha\in V$有$\alpha V\beta^t=\alpha\gamma^t$.

$$V\cap M=\{0\}, \ \ \ V+M=F[x].$$

## 2017考研题解

$$\lim_{n\rightarrow\infty}n^n\left(x_n-\frac{\pi}{2}\right)=0.$$

\begin{align*}\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_{n+1}-\frac{\pi}{2}\right)&=\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_n+\cos x_n-\frac{\pi}{2}\right)\\&=\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_n+\sin\left(\frac{\pi}{2}-x_n\right)-\frac{\pi}{2}\right)\\&=\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_n+\left(\frac{\pi}{2}-x_n\right)-\frac{1}{6}\left(\frac{\pi}{2}-x_n\right)^3-\frac{\pi}{2}\right)\\&=-\frac{1}{6}\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(\frac{\pi}{2}-x_n\right)^3\\&=-\frac{1}{6}\lim_{n\rightarrow\infty}\frac{\left(n+1\right)^{n+1}}{n^{3n}}\left[n^n\left(x_n-\frac{\pi}{2}\right)\right]^3=0.\end{align*}

TangSong：令$y_n=\frac\pi2-x_n$,得到$y_n=y_{n-1}-\sin y_{n-1}$.可以证明$$\lim_{n\rightarrow\infty}\frac{y_{n+1}}{y_{n}^{3}}=\frac{1}{6}.$$因此当$n>N$时,我们有$$\frac{y_{n+1}}{y_{n}^{3}}<\frac{1}{2}.$$因此$$0<y_n<\frac{1}{2}y_{n-1}^{3}<\left(\frac{1}{2}\right)^{1+3}y_{n-2}^{3^2}<\cdots <\left(\frac{1}{2}\right)^{1+3+\cdots +3^{n-N-2}}y_{N+1}^{3^{n-N-1}},$$

\begin{align*}\left( {\begin{array}{*{20}{c}}{{x_{3n - 3}}}\\{{x_{3n - 2}}}\\{{x_{3n - 1}}}\end{array}} \right) &= {\left( {\begin{array}{*{20}{c}}3&{ - 2}&1\\4&{ - 1}&0\\4&{ - 3}&2\end{array}} \right)^{n - 1}}\left( {\begin{array}{*{20}{c}}{{x_0}}\\{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{1/2}&3\\2&0&4\\2&0&5\end{array}} \right){\left( {\begin{array}{*{20}{c}}1&1&0\\0&1&0\\0&0&2\end{array}} \right)^{n - 1}}\left( {\begin{array}{*{20}{c}}0&{5/2}&{ - 2}\\2&1&{ - 2}\\0&{ - 1}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}5\\7\\8\end{array}} \right)\\&= \left( {\begin{array}{*{20}{c}}1&{1/2}&3\\2&0&4\\2&0&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&{n - 1}&0\\0&1&0\\0&0&{{2^{n - 1}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&{5/2}&{ - 2}\\2&1&{ - 2}\\0&{ - 1}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}5\\7\\8\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{n + 1 + 3 \times {2^{n - 1}}}\\{2n + 1 + {2^{n + 1}}}\\{2n + 1 + 5 \times {2^{n - 1}}}\end{array}} \right).\end{align*}

## 武汉大学2016年基础数学复试笔试试题

1.设$f(x)$在$(a,b)$上可微,且$f(x)$在$a$点右连续,试证:

(1) 若导函数$f'(x)$的右函数极限存在且为$A$,证明导函数$f'(x)$在$a$点的右侧存在且${{f'}_ + }\left( a \right) = \mathop {\lim }\limits_{x \to {a^ + }} f'\left( x \right) = A.$

(2) $f'(x)$在$(a,b)$上不存在第一类间断点.

2.若级数$\sum_{n=1}^\infty a_n$收敛,证明级数$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{\sqrt {{r_{n - 1}}} + \sqrt {{r_n}} }}}$收敛,其中${r_n} = \sum_{k = n + 1}^\infty {{a_k}}$.

3.讨论$\lambda$取何值时, $y''+\lambda y=0$有非零的初值解,其中$y(0)=y(1)=0$.

4.$A$为正定矩阵, $A-B$为半正定矩阵,试证明:

(1) 方程$|\lambda B-A|=0$关于根$\lambda\geq1$;

(2) $|B|\leq |A|$.

5.讨论积分$\int_0^1 x^{p-1}\ln^2 xdx$在下列情况下的一致收敛性.

(1) $p\geq p_0>0$;

(2) $p>0$..

6. 设非负函数$f(x,y)$在区域$D$上可积,证明积分$\iint\limits_D f(x,y)dx=0$充分必要条件为$f(x,y)$在$D$上的连续点上等于$0$.

## 武汉大学2015年基础数学复试笔试试题

1.导函数极限定理, $f'(0)$存在, 而$\lim_{x\to0} f'(x)$不存在的例子.

$f\left( x \right) = \begin{cases}{x^2}\sin \frac{1}{x}, &x \ne 0\\0, &x = 0\end{cases}.$

2.$\{a_n\}$是正项数列且单增.证明: $\sum_{n = 1}^\infty {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}} - 1} \right)}$收敛$\Leftrightarrow$ $\{a_n\}$有界.

3.设$A$是$n$阶可逆复方阵，证明存在分解

$A=UT,$

4.讨论微分方程过点y=0的解的存在性和唯一性，其中$\alpha>0$.

$\frac{dy}{dx}=|y|^{\alpha}.$

5.证明含参变量积分

$\int_{0}^{+\infty}\frac{\sin{xy}}{y(1+x)}dy$

6.利用数学归纳法证明$n$维空间中的$n+1$面体${B_{n + 1}}:\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^{1/\alpha }}} \le 1,\alpha > 0$的体积为$V = \frac{{{2^n}{\alpha ^{n - 1}}{{\left[ {\Gamma \left( \alpha \right)} \right]}^{n - 1}}\Gamma \left( {\alpha + 1} \right)}}{{\Gamma \left( {\alpha n + 1} \right)}},$

1.山东大学---徐老师：

2.

## 某家公司的笔试题

1.设实数列$\{a_n\}$满足$a_{n+p}\leq a_{n}+a_{p}$对于任意的正整数$p,n$，证明：$$\lim\limits_{n\to +\infty}\dfrac{a_n}{n}=\inf\limits_n \dfrac{a_n}{n}.$$
2.设实函数$f(x)$在$(0,1)$内一阶可导且满足$f(1)=1,f(0)=0$，设
\begin{equation*}\int_0^1|f'(x)-f(x)|\geq {u}.\end{equation*}求$u$的最大值。
3.给定一个圆求在这个圆里面随机选择四个点围成一个凸集的概率。

## 2013武大数分压轴题

（13年武大数分）求$\displaystyle I = \iint\limits_\Sigma {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{3}{2}}}{{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)}^{ - \frac{1}{2}}}dS}$,其中$\sum$为椭球面: $\displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1(a,b,c>0)$.

$\frac{{\partial \left( {y,z} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = bc{\sin ^2}\varphi \cos \theta ,\frac{{\partial \left( {z,x} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ac{\sin ^2}\varphi \sin \theta ,\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ab\sin \varphi \cos \varphi ,$

\begin{align*}EG - {F^2} &= {\left( {\frac{{\partial \left( {y,z} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2} + {\left( {\frac{{\partial \left( {z,x} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2} + {\left( {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2}\\& = {\left( {abc} \right)^2}{\sin ^2}\varphi \left( {\frac{{{{\sin }^2}\varphi {{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\varphi {{\sin }^2}\theta }}{{{b^2}}} + \frac{{{{\cos }^2}\varphi }}{{{c^2}}}} \right).\end{align*}

\begin{align*}&{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)^{ - \frac{1}{2}}}\\= &{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)^{ - \frac{3}{2}}}\\&{\left( {\frac{{{{\sin }^2}\varphi {{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\varphi {{\sin }^2}\theta }}{{{b^2}}} + \frac{{{{\cos }^2}\varphi }}{{{c^2}}}} \right)^{ - \frac{1}{2}}}.\end{align*}

$I = abc\iint\limits_{\left[ {0,\pi } \right] \times \left[ {0,2\pi } \right]} {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}\sin \varphi d\varphi d\theta }$

$\int {{{\left( {M + N{x^2}} \right)}^{ - 3/2}}dx} = \frac{1}{M} \cdot \frac{x}{{\sqrt {M + N{x^2}} }} + C.$

\begin{align*}I &= abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}\sin \varphi d\varphi } \\&=  - abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}d\left( {\cos \varphi } \right)} \\&=  - abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left[ {\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right) + \left( {{c^2} - {a^2}{{\cos }^2}\theta  - {b^2}{{\sin }^2}\theta } \right){{\cos }^2}\varphi } \right]}^{ - \frac{3}{2}}}d\left( {\cos \varphi } \right)} \\&= abc\int_0^{2\pi } {d\theta } \int_{ - 1}^1 {{{\left[ {\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right) + \left( {{c^2} - {a^2}{{\cos }^2}\theta  - {b^2}{{\sin }^2}\theta } \right){x^2}} \right]}^{ - \frac{3}{2}}}dx} \\&= abc\int_0^{2\pi } {\frac{2}{{\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right)c}}d\theta }  = 4ab\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta } .\end{align*}
\begin{align*}&\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  = \int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  + \int_{\frac{\pi }{2}}^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta } \\= &\int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  + \int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta }}d\theta } \\= &\int_0^{ + \infty } {\frac{1}{{{a^2} + {b^2}{x^2}}}d}  + \int_0^{ + \infty } {\frac{1}{{{a^2}{x^2} + {b^2}}}dx}  = \frac{1}{{ab}}\left. {\arctan \left( {\frac{b}{a}x} \right)} \right|_0^{ + \infty } + \frac{1}{{ab}}\left. {\arctan \left( {\frac{a}{b}x} \right)} \right|_0^{ + \infty }\\= &\frac{\pi }{{ab}}.\end{align*}

$I = 4ab\int_0^\pi {\frac{1}{{{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta }}d\theta } = 4\pi .$

$$n=\frac{\left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\right)}{\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}},$$

$$\iint\limits_\Sigma \frac{xdydz+yd zd x+zdxdy}{\sqrt{(x^2+y^2+z^2)^3}}.$$

$$\iint\limits_\Sigma \frac{xd yd z+yd zd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}} =\iint\limits_{S_\varepsilon} \frac{xdyd z+ydzd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}}=4\pi.$$ 即
$$\iint\limits_\Sigma\frac{d S}{\sqrt{(x^2+y^2+z^2)^3}\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}}=4\pi.$$

## 2014年浙江大学数学分析考研试题解答

6.设空间体积为$V$的任意$\Omega,X_0\in \Omega ,0<\alpha<3$.证明

$\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le C{V^{\alpha /3}}, \text{其中C只与\alpha有关}.$

\begin{align*}\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \\\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} .\end{align*}

\begin{align*}\int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\xi - {X_0}} \right|^{\alpha - 3}}{V_{{D_2}}},\xi \in {D_2}\\\int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\eta - {X_0}} \right|^{\alpha - 3}}{V_{{\Omega_2}}},\eta \in {\Omega _2}.\end{align*}

7.$f(x)$在$[0,1]$单增,证明:

$\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right).$

$0 \le g\left( t \right) - g\left( {{0_ + }} \right) < M_1\varepsilon ,$

\begin{align*}\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} &= \left( {\int_0^\delta {} + \int_\delta ^1 {} } \right)\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx\\&= {I_1} + {I_2}.\end{align*}

${I_1} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_\eta ^\delta {\frac{{\sin xy}}{x}dx} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} ,$

$\left| {\int_0^z {\frac{{\sin z}}{z}dz} } \right| \le L\left( L \text{为常数}\right),$从而

$\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| = \left| {\int_0^{y\delta } {} + \int_0^{y\eta } {} } \right| \le 2L.$

\begin{align*}&\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right)\\=& \mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} + f\left( {{0_ + }} \right)\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\frac{{\sin xy}}{x}dx} \\= &0 + f\left( {{0_ + }} \right)\int_0^{ + \infty } {\frac{{\sin z}}{z}dz} = \frac{\pi }{2}f\left( {{0_ + }} \right).\end{align*}