Eufisky - The lost book

## 幸子解答的李炯生习题

$\left\{ {\begin{array}{*{20}{c}}{v\left( x \right) = {v_1}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)}\\{w\left( x \right) = {w_1}\left( x \right) - {{\left( {x - 1} \right)}^n}\alpha \left( x \right)}\end{array}} \right.,$

$$\left\{ {\begin{array}{*{20}{c}}{v\left( x \right) = {v_0}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)}\\{w\left( x \right) = {w_0}\left( x \right) - {{\left( {x - 1} \right)}^n}\alpha \left( x \right)}\end{array}} \right.$$

\begin{align*}{v^{\left( i \right)}}\left( { - 1} \right) &= \sum\limits_{k = 0}^i {{{\left[ {1 - w\left( x \right){{\left( {x + 1} \right)}^n}} \right]}^{\left( k \right)}}{{\left[ {{{\left( {x - 1} \right)}^{ - n}}} \right]}^{\left( {i - k} \right)}}} {|_{x = - 1}} = {\left( { - 1} \right)^n}{2^{ - n - i}}n \cdots \left( {n + i - 1} \right)\\{w^{\left( i \right)}}\left( 1 \right) &= \sum\limits_{k = 0}^i {{{\left[ {1 - v\left( x \right){{\left( {x - 1} \right)}^n}} \right]}^{\left( k \right)}}{{\left[ {{{\left( {x + 1} \right)}^{ - n}}} \right]}^{\left( {i - k} \right)}}} {|_{x = 1}} = {\left( { - 1} \right)^i}{2^{ - n - i}}n \cdots \left( {n + i - 1} \right).\end{align*}

$f\left( x \right) = 2\left[ {{{\left( {x - 1} \right)}^n}\alpha \left( x \right) - {w_0}\left( x \right)} \right]{\left( {x + 1} \right)^n} + 1= 2\left[ {{v_0}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)} \right]{\left( {x - 1} \right)^n} - 1$均满足题设.

$T = UA{U^H} = \frac{1}{2}\left( {\begin{array}{*{20}{c}}{B + G + \left( {C - D} \right)i}&{B - G - \left( {C + D} \right)i}\\{B - G + \left( {C + D} \right)i}&{B + G - \left( {C - D} \right)i}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}P&Q\\{\bar Q}&{\bar P}\end{array}} \right).$

$\left( {\begin{array}{*{20}{c}}P&Q\\{\bar Q}&{\bar P}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{I_n}}&O\\O&{ - {I_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{P^H}}&{{Q^T}}\\{{Q^H}}&{{P^T}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{I_n}}&O\\O&{ - {I_n}}\end{array}} \right).$

## 曲线积分的计算

$\left\{ \begin{array}{l}{x^2} + {y^2} + {z^2} = 1\\x + y + z = 1\end{array} \right. \Rightarrow {\left( {x - \frac{1}{2} + \frac{y}{2}} \right)^2} + \frac{3}{4}{\left( {y - \frac{1}{3}} \right)^2} = \frac{1}{3}.$
$\left\{ \begin{array}{l}x - \frac{1}{2} + \frac{y}{2} = \frac{1}{{\sqrt 3 }}\cos \theta \\\frac{{\sqrt 3 }}{2}\left( {y - \frac{1}{3}} \right) = \frac{1}{{\sqrt 3 }}\sin \theta \end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = \frac{1}{3} - \frac{1}{3}\sin \theta + \frac{1}{{\sqrt 3 }}\cos \theta \\y = \frac{1}{3} + \frac{2}{3}\sin \theta \\z = 1 - x - y = \frac{1}{3} - \frac{1}{3}\sin \theta - \frac{1}{{\sqrt 3 }}\cos \theta \end{array} \right.$

\begin{align*}&\oint_C {\left( {{{\left( {x + 1} \right)}^2} + {{\left( {y - 2} \right)}^2}} \right)ds} \\= &\frac{{\sqrt 6 }}{3}\int_0^{2\pi } {\left( {\frac{5}{9}{{\sin }^2}\theta  + \frac{1}{3}{{\cos }^2}\theta  - \frac{{28}}{9}\sin \theta  + \frac{8}{{3\sqrt 3 }}\cos \theta  - \frac{2}{{3\sqrt 3 }}\sin \theta \cos \theta  + \frac{{41}}{9}} \right)d\theta } \\= &\frac{{\sqrt 6 }}{3} \times 5 = \frac{{5\sqrt 6 }}{3}.\end{align*}

## 二重积分难题荟萃

$\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {r,\theta } \right)}} = \left| {\begin{array}{*{20}{c}}{\cos \theta {e^{r\cos \theta }}}&{ - r\sin \theta {e^{r\cos \theta }}}\\{\sin \theta {e^{r\sin \theta }}}&{r\cos \theta {e^{r\sin \theta }}}\end{array}} \right| = r{e^{r\sin \theta }}{e^{r\cos \theta }}.$

$\left\{ \begin{array}{l}{e^{r\cos \theta }} + {e^{r\sin \theta }} = 1,\\{e^{2r\cos \theta }} + {e^{2r\sin \theta }} = 1.\end{array} \right.$

$I = \iint_\Delta {\frac{{drd\theta }}{r}} = \int_\pi ^{\frac{3}{2}\pi } {d\theta } \int_{\frac{1}{2}r\left( \theta \right)}^{r\left( \theta \right)} {\frac{{dr}}{r}} = \frac{\pi }{2}\ln 2.$

## 2015年浙大数分题

1. 求 $\lim\limits_{n\rightarrow+\infty} \dfrac{(n^2+1)(n^2+2)..(n^2+n)}{(n^2-1)(n^2-2)..(n^2-n)}$.

2. 求$\displaystyle\lim\limits_{x \rightarrow 0+} \dfrac{1}{x^5}\int_0^x e^{-t^2}dt+\dfrac{1}{3}\dfrac{1}{x^2}-\dfrac{1}{x^4}$.

3. $\displaystyle I(r)=\oint \dfrac{y}{x^2+y^2}\mathrm{d}x-\frac{x}{x^2+y^2}\mathrm{d}y$, 其中曲线方程为$x^2+y^2+xy=r^2$, 取正方向, 求$\lim\limits_{r\rightarrow\infty}I(r)$.

4. 求$\displaystyle\int_{e^{-2n\pi}}^0 \sin\ln\dfrac{1}{x}\mathrm{d}x$.

5. 考察黎曼函数的连续性, 可微性, 黎曼可积性.

6. 在$\mathbb{R}^n$中, $f$为定义在某个区域上的一个函数, 有一阶连续偏导, 且偏导数有界.

(1) 若$D$为凸区域证明$f$一致连续.
(2)考察$D$不是凸区域的情况.

7. $\{f_n\}$为一个连续函数列, 且对于任意给定的$x$, $\{f_n(x)\}$有界, 证明存在一个小区间在此小区间内$f_n$一致有界.

8. (1) 证明$\Gamma(s)$在 $(0,\infty$内无穷次可微.
(2) 证明$\Gamma(s)$ , $\ln(\Gamma)$都是严格凸函数.

9. $f$ 二阶可微, 且$f$, $f'$, $f''$ 都大于等于$0$, 且存在一个正数$c$, $f''(x)\leq cf(x)$. 证明:
(1) $\displaystyle\lim\limits_{x\rightarrow-\infty}f(x)=0$;
(2) 证明存在正数$a$, 有$f'\leq af$,并求出$a$.

10. 证明 Fejer定理.

11. 设$f$在$[A, B]$上黎曼可积, $0<f<1$, 对于任意的$\varepsilon$, 构造一个函数$g$, 满足
(1) $g$是一个阶梯函数, 且取值只能为$0$或$1$.
(2) $\displaystyle\left| \int_a^b f-g \mathrm{d}x\right|<\varepsilon$, $a$, $b$ 属于$[A,B]$不等号关于$a$, $b$是一致的.

## 浙大14年数分题：Dirichlet引理的证明

Riemann-Lebesgue引理大家都很熟悉,那Dirichlet引理呢?

$f(x)$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*}

## 级数中的一些反例

\begin{align*}\sum_{n=1}^{\infty}\frac{a_{n}}{n}&=\sum_{k=0}^{\infty}\left[\frac{1}{(3k+3)\log(3k+3)}-\frac{1}{\sqrt[3]{2}}\left(\frac{1}{(3k+1)\log(3k+1)}+\frac{1}{(3k+2)\log(3k+2)}\right)\right]\\&\sim\sum_{k=0}^{\infty}\frac{1}{k\log k}\end{align*}