原先我们自己给李炯生习题作出的解答略显繁琐，这里贴出幸子的一些解答。

设$f(x)$是$2n+1$次多项式, $n$为正整数, $f(x)+1$被${(x-1)}^n$整除,而$f(x)-1$被${(x+1)}^n$整除,求$f(x)$.

解:由题意,存在$2u\left( x \right),2v\left( x \right) \in \mathbb{F}\left[ x \right]$,使得\[f\left( x \right) = 2u\left( x \right){\left( {x + 1} \right)^n} + 1 = 2v\left( x \right){\left( {x - 1} \right)^n} - 1.\]即得\[v\left( x \right){\left( {x - 1} \right)^n} + w\left( x \right){\left( {x + 1} \right)^n} = 1,\tag{$\ast$}\]其中$\deg v\left( x \right) = \deg w\left( x \right) = n + 1,w\left( x \right) = - u\left( x \right)$.

若$(\ast)$存在特解$v_1(x)$和$w_1(x)$,则${\left( {x + 1} \right)^n}|v\left( x \right) - {v_1}\left( x \right)$且${\left( {x - 1} \right)^n}|w\left( x \right) - {w_1}\left( x \right)$.

存在多项式$\alpha(x)$使得

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

而对于任意的$\alpha(x)\in \mathbb{F}[x]$,由

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

定义的$v(x),w(x)$均满足$(\ast)$.由上式定义的$v(x)$和$w(x)$为$(\ast)$通解,于是$v\left( x \right) = \left[ {1 - w\left( x \right){{\left( {x + 1} \right)}^n}} \right]{\left( {x - 1} \right)^{ - n}},w\left( x \right) = \left[ {1 - v\left( x \right){{\left( {x - 1} \right)}^n}} \right]{\left( {x + 1} \right)^{ - n}},v\left( { - 1} \right) = {\left( { - 2} \right)^{ - n}},w\left( 1 \right) = {2^{ - n}}$,

当$1\leq i\leq n-1$时,利用Leibniz公式

\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*}

由Taylor公式,存在唯一的次数小于$n$的多项式\[{v_0}\left( x \right) = \frac{1}{{{{\left( { - 2} \right)}^n}}}\sum\limits_{i = 0}^{n - 1} {\frac{1}{{{2^i}}}C_{n + i - 1}^i{{\left( {x + 1} \right)}^i}} ,{w_0}\left( x \right) = \frac{1}{{{2^n}}}\sum\limits_{i = 0}^{n - 1} {\frac{1}{{{{\left( { - 2} \right)}^i}}}C_{n + i - 1}^i{{\left( {x - 1} \right)}^i}} \]使得$(\ast)$成立.

因此对任意的一次多项式$\alpha \left( x \right)$,

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

设$A\in \mathbb{R}^{2n\times 2n}$,且$A\left( {\begin{array}{*{20}{c}}0&{{I_{\left( n \right)}}}\\{ - {I_{\left( n \right)}}}&0\end{array}} \right){A^T} = \left( {\begin{array}{*{20}{c}}0&{{I_{\left( n \right)}}}\\{ - {I_{\left( n \right)}}}&0\end{array}} \right)$.证明: $\det A=1$.

证:令$U = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}}{{I_n}}&{ - i{I_n}}\\{{I_n}}&{i{I_n}}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}B&C\\D&G\end{array}} \right)$,

由$A\left( {\begin{array}{*{20}{c}}O&{{I_n}}\\{ - {I_n}}&O\end{array}} \right){A^T} = \left( {\begin{array}{*{20}{c}}O&{{I_n}}\\{{I_n}}&O\end{array}} \right)$知

即$P{P^H} - Q{Q^H} = {I_n}$及$P{Q^T} = Q{P^T}$,注意到$P{P^H} = {I_n} + Q{Q^H} > 0$,即$\det P\neq0$,且$\det A=\det T$.而$\det \left( {\begin{array}{*{20}{c}}P&Q\\{\bar Q}&{\bar P}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{P^H}}&O\\{ - {Q^H}}&{{I_n}}\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}{{I_n}}&Q\\O&{\bar P}\end{array}} \right)$.因此$\det A=1$.

张元博问了这么一道题：计算曲线积分\[\oint_C {\left( {{{\left( {x + 1} \right)}^2} + {{\left( {y - 2} \right)}^2}} \right)ds} ,\]其中$C$表示曲面$x^2+y^2+z^2=1$与$x+y+z=1$的交线.

解:我是利用常规的三角换元解决的.联立方程有

第一个是09年西北大学的考研题,也在史济怀老师的数分书上找得到.这里的解答就是史老爷子自己的.

计算积分\[\iint_D {\frac{1}{{xy\left( {\ln^2 x + {{\ln }^2}y} \right)}}} dxdy,D = \left\{ {\left( {x,y} \right)\left| {x + y \ge 1,{x^2} + {y^2} \le 1} \right.} \right\}.\]

解.作变换$x=e^{r\cos\theta},y=e^{r\sin\theta}$,则

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

原积分变为\[I = \iint_\Delta {\frac{{drd\theta }}{r}} .\]

这里的$\Delta$是变换以后的积分区域.注意$x+y=1$和$x^2+y^2=1$分别被变为

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

现在来分析由上述两条曲线所围成的$(r,\theta)$平面上的区域是什么形状.从第一个式子可以看出$\theta$的变化范围必须使$\cos \theta$和$\sin \theta$都取负值,故$\theta$只能在$\left[\pi,\frac32\pi \right]$中取值.

假设由第一个式子确定的函数为$r=r(\theta)$,则由第二个式子确定的函数便为$r=\frac12 r(\theta)$.因此

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

苹果公司总部

本文主要给出史济怀老师复变函数的某些习题解答，主要来自陶哲轩小弟。

然后对后面的函数用最大模原理.

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

$f(x)$在$[0,1]$单增,证明:

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

这是Dirichlet引理,菲赫金哥尔茨的《微积分教程》第三卷P358有详细的证明.另外,汪林的《数学分析问题研究与评注》P147上有他的推广及其证明.

对任意给出的$\varepsilon>0$, $\exists 0<\delta<1$,使得对于$0<t\leq \delta$,

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

其中$M_1$是任意给定的常数.

考察积分

\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*}

对于$I_1$,运用积分第二中值定理,我们有

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

其中第二个因子对于一切值$y$一致有界.事实上,由反常积分$\displaystyle \int_0^\infty {\frac{{\sin z}}{z}dz}$的收敛性,可见当$z\to \infty$时, $z(z\geq 0)$的连续函数$\displaystyle \int_0^z {\frac{{\sin z}}{z}dz} $有有限的极限,并且对于一切值$z$有界

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

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

对于第一个因子,取$M_1=\frac{1 }{{4L}}$,则有$f\left( \delta \right) - f\left( {{0_ + }} \right) < \frac{\varepsilon }{{4L}}$.

因此\[\left| {{I_1}} \right| \le \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| < \frac{\varepsilon }{{4L}} \cdot 2L = \frac{\varepsilon }{2}.\]

至于$I_2$,由于$\displaystyle \int_\delta ^1 {\frac{{f\left( x \right) - f\left( {{0_ + }} \right)}}{x}dx} $存在,由Riemann-Lebesgue引理可知$\mathop {\lim }\limits_{y \to \infty } {I_2} = 0$,即对$\varepsilon >0,\exists M_2>0$,使得$y>M_2$时,有$\left| {{I_2}} \right| < \frac{\varepsilon }{2}$.

因此\[\left| {\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} } \right| \le \left| {{I_1}} \right| + \left| {{I_2}} \right| < \varepsilon .\]

即\[\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} = 0.\]

从而

\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*}

计算\[\mathop {\lim }\limits_{n \to  + \infty } \frac{{1 + n + \frac{{{n^2}}}{{2!}} +  \cdots  + \frac{{{n^n}}}{{n!}}}}{{{e^n}}} = \frac{1}{2}.\]

解.

法二.(小米)利用概率, 就是$n$个独立的参数为1的指数分布(泊松分布)和小于等于$n$的概率根据中心极限定理概率收敛到$1/2$.

源自：http://www.math.org.cn/forum.php?mod=viewthread&tid=31895&highlight=%E6%AD%A6%E6%B1%89%E5%A4%A7%E5%AD%A6

判断题：若级数$\displaystyle \sum_{n=1}^{+\infty}a_n^3$ 收敛, 则$\displaystyle \sum_{n=1}^{+\infty}\frac{a_n}{n}$亦收敛.

反例：(老骥伏枥)令$a_{n}=\frac{1}{\log n}\sqrt[3]{\cos\frac{2n\pi}{3}}$,所以$a_{n}^3=\frac{1}{\log^3n}\cos\frac{2n\pi}{3}$.用狄利克雷判别法级数$\sum_{n=1}^{\infty}a_{n}^3$收敛.(因为$\frac{1}{\log^3n}$递减趋于零, $\cos\frac{2n\pi}{3}$部分和有界).而

\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*}

发散。（柯西积分判别法）

如果是正项级数,根据Holder不等式能够说明$\sum_{n=1}^{\infty}\frac{a_{n}}{n}$收敛.