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Cheatsheet – Python & R codes for common Machine Learning Algorithms：https://www.analyticsvidhya.com/blog/2015/09/full-cheatsheet-machine-learning-algorithms/

机器学习算法代码汇总：http://www.csuldw.com/2015/11/21/2015-11-21-machine-learning-algorithms/

Essentials of Machine Learning Algorithms (with Python and R Codes)：https://www.analyticsvidhya.com/blog/2015/08/common-machine-learning-algorithms/

10 种机器学习算法的要点（附 Python 和 R 代码）：http://blog.jobbole.com/92021/

机器学习算法比较：http://www.csuldw.com/2016/02/26/2016-02-26-choosing-a-machine-learning-classifier/

读书心得-Do we Need Hundreds of Classifiers to Solve Real World Classification Problems：http://blog.csdn.net/vivian_ll/article/details/61199941

1、用LaTeX高效排版化学方程式http://www.latexstudio.net/archives/9998

2、张敬信老师的LaTeX入门讲座http://www.latexstudio.net/archives/5325

3、用LaTeX制作私房菜谱http://www.latexstudio.net/archives/9988

4、我自己排版的《笑傲江湖》

5、时尚的图http://www.latexstudio.net/archives/6556

6、颜色不一样的烟火http://www.latexstudio.net/archives/5103

7、特殊符号http://www.latexstudio.net/archives/4719

8、图片跨页显示http://www.latexstudio.net/archives/4683

1.设$\displaystyle f_n\in O(D),f_n\rightrightarrows f$, $\displaystyle S_n=\cup_{k\geq n}f_k(D)$,令$\displaystyle S=\cap_{n=1}^\infty S_n$.问$S$的内部$\mathring{S}$和$f(D)$有什么关系?

2.Schwarz引理: $f:\triangle\to \triangle$解析, $f(0)=0$,则$|f'(0)|\leq 1$,并且$|f'(0)|= 1\Leftrightarrow f\in \mathrm{Aut} (\triangle)$.

一般地,固定$z_0\in\triangle$,定义$\mathscr{F}_{z_0}=\{f:\triangle\to \triangle\,\text{解析}|f(z_0)=0\}$,令$\lambda=\sup\{|f'(z_0)|f\in \mathscr{F}_{z_0}\}$.由Cauchy不等式, $\lambda<+\infty$.当然易证$\lambda>0$.取$f_n\in \mathscr{F}_{z_0}$,使得$\lim_{n\to\infty}|f'_n(z_0)|=\lambda$.由Montel定理,不妨设$f_n\rightrightarrows f$,则$f\in \mathscr{F}_{z_0}$且$|f'(z_0)|=\lambda$.

(1)求出$\lambda$;

(2)若$f\in \mathscr{F}_{z_0}$,则$f\in \mathrm{Aut} (\triangle) \Leftrightarrow |f'(z_0)|=\lambda$.

3.设$f_n,g_n\in O(\triangle),n\geq 1$,设$|f_n(z)|\leq 2,|g_n(z)|\leq 2,|f_n(z)g_n(z)|\leq\frac1n,\forall z,\forall n$.设$f_n(0)=0,g_n(0)=1,\forall n$.设$\{z_n\}$是$\triangle$中子列,若满足$f_n(z_n)=1,g_n(z_n)=0,\forall n$,则$\lim\limits_{n\to\infty}|z_n|=1$.

4.对$r\in (0,1)$,记$A_r=\{z\in \mathbb{C}|r<|z|<1\}$

(1)对$\forall r\in (0,1)$,证明$A_r,\triangle^\ast,\mathbb{C}^\ast$不共形等价;

(2)设$r_1,r_2\in (0,1)$,若$A_{r_1}\cong A_{r_2}$,则$r_1=r_2$;

(3)计算$A_r$的自同构群$\mathrm{Aut}(A_r)$.

5.证明$\mathbb{C}^\ast,\triangle^\ast,A_r$同胚.

Let $q>0$ such that $[q[qn]]+1=[q^2n],n=1,2,\cdots$,solve $q$.

球面$S^n$的Levi-Civita联络是$\mathbb{R}^{n+1}$的联络到球面的切空间的投影，也就是$$\nabla_{X}Y(\vec{x})=D_{X}Y(\vec{x})-\langle D_{X}Y(\vec{x}) , \vec{x}\rangle\vec{x},$$其中$D$是欧氏空间的联络，也就是普通的方向导数，设$\gamma$是球面的弧长参数的测地线，于是测地线方程是\begin{align*}0&=D_{\dot{\gamma}}\dot{\gamma}(\gamma)-\langle D_{\dot{\gamma}}\dot{\gamma}(\gamma) , \gamma\rangle\gamma=\ddot{\gamma}-\langle \ddot{\gamma} , \gamma\rangle\gamma\\&=\ddot{\gamma}-(\frac{d}{dt}\langle \dot{\gamma} , \gamma\rangle-\langle\dot{\gamma} , \dot{\gamma}\rangle)\gamma=\ddot{\gamma}-(\frac{d}{dt}0-1)\gamma.\end{align*}

也就是$\ddot{\gamma}=-\gamma$, 容易解得$\gamma(t)=\cos t\cdot\gamma(0)+\sin t\cdot\dot{\gamma}(0)$.

一、(30分)设$\displaystyle {S^n}\left( r \right) = \left\{ {x = \left( {{x^1}, \cdots ,{x^{n + 1}}} \right) \in {\mathbb{R}^{n + 1}}\left| {\sum\limits_{A = 1}^{n + 1} {{{\left( {{x^A}} \right)}^2}}  = {r^2},r > 0} \right.} \right\}$是$\mathbb{R}^{n + 1}$的半径为$r$的球面, $g_0=\sum_{A=1}^{n+1}\left(d x^A\right)^2$是$\mathbb{R}^{n + 1}$的标准黎曼度量.令$i:S^n (r)\to \mathbb{R}^{n + 1}$是包含映射.

发表在博士数学论坛。

仿照北大未名BBS上的文章，我在这个帖子也不定期更新数理学家的趣闻逸事。内容一般来源于书籍和网络，由于数量之多，恕我不一一注明出处！

Math8.钟开莱给人的印象那是牛逼哄哄的。甫列华氏门墙，一天华老爷子啰里八嗦讲了一大堆东西，钟开莱不服气回家折腾了十页的纲领性文字，第二天上课的时候扔给华罗庚，意思是你拿去看吧。华老爷子很不爽，说你不就是个毛头孩子吗，跟我玩这套，华罗庚也折腾了一晚上，硬是把这10页压缩成3页，回敬给钟开莱。当时是西南联大吧，学习氛围可见一斑。

1.（高等代数）证明:实对称矩阵${A_n} = \left[ {\begin{array}{*{20}{c}}{\frac{1}{1}}&{\frac{1}{2}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\\vdots & \vdots & \vdots &{}& \vdots \\{\frac{1}{n}}&{\frac{1}{n}}&{\frac{1}{n}}& \cdots &{\frac{1}{n}}\end{array}} \right]$的特征值都大于$0$,且小于等于$3+2\sqrt{2}$.

2.（高等概率论）Let the stochastic processes $\{X_k,1\leq k\leq n\}$ and $\{X'_k,1\leq k\leq n\}$ be independent of one another and have the same joint distributions. If $m_k$ is a median of $X_k,1\leq k\leq n$. Prove: for $\lambda>0$,

求一个可逆矩阵$Q$使得$AQ=QB$.

证.(SUCCEME)记$H=A-B,G=P/2$,则我们有$HP=PH,BG-GB=H$,只需取一个可逆矩阵$Q$使得$(B+H)Q=QB$.

事实上,我们可取\[Q = {e^{ - G}} = 1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots \]

由$P$幂零可知上面的和为有限和.由$e^G\cdot e^{-G}=I$可知$Q$可逆.又由

\begin{align*}B{G^n} - {G^n}B&= \left( {B{G^n} - GB{G^{n - 1}}} \right) + \left( {GB{G^{n - 1}} - {G^2}B{G^{n - 1}}} \right) + \cdots + \left( {{G^{n - 1}}BG - {G^n}B} \right)\\&= H{G^{n - 1}} + GH{G^{n - 2}} + \cdots + {G^{n - 1}}H = nH{G^{n - 1}},\end{align*}

可知

\[B{G^n} - {G^n}B=nH{G^{n - 1}}.\]

因此

\begin{align*}\left( {B + H} \right)Q - QB &= \left( {B + H} \right)\left( {1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots } \right) - \left( {1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots } \right)B\\&= \sum\limits_{n \ge 0} {\left[ {\left( {B + H} \right)\frac{{{{\left( { - 1} \right)}^n}{G^n}}}{{n!}} - \frac{{{{\left( { - 1} \right)}^n}{G^n}}}{{n!}}B} \right]} = \sum\limits_{n \ge 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\left( {B{G^n} - {G^n}B + H{G^n}} \right)} \\&= \sum\limits_{n \ge 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\left( {nH{G^{n - 1}} + H{G^n}} \right)} = \sum\limits_{n \ge 0} {{{\left( { - 1} \right)}^n}\left( {\frac{{H{G^{n - 1}}}}{{\left( {n - 1} \right)!}} + \frac{{H{G^n}}}{{n!}}} \right)} = 0.\end{align*}

这里约定$n=-1$时, $HP^n/n!=0$.所以此时$\left( {B + H} \right)Q = QB$成立.

假设$\displaystyle x_0=1,x_n=x_{n-1}+\cos x_{n-1}(n=1,2,\cdots )$,证明:当$x\rightarrow \infty $时, $\displaystyle x_n-\frac{\pi }{2}=o\left(\frac{1}{n^n}\right)$.

证.先证$1\leq x_n<\frac\pi /2$,得到$x_n-x_{n-1}>0$,由单调有界定理可知$x_n$极限存在且$\lim_{n\to\infty}=\frac\pi2$.下面用归纳法证明$\lim_{n\rightarrow\infty}n^n\left(x_n-\frac{\pi}{2}\right)=0$.假设

空间中四点$O,A,B,C$使得\[\angle AOB=\frac{\pi}{2},\angle BOC=\frac{\pi}{3},\angle COA=\frac{\pi}{4}\]
设$AOB$决定的平面为$\pi_1$,$BOC$决定的平面为$\pi_2$,求$\pi_1,\pi_2$二面角.求出二面角的余弦值即可.

解.

误差函数(entire function)

1.int e^(-x^2-2y^2),  y=0..x, x=0..\infty用WolframAlpha计算二重积分.

\[\int_0^\infty  {\int_0^x {{e^{ - {x^2} - 2{y^2}}}} } dydx = \frac{{\arctan \left( {\sqrt 2 } \right)}}{{2\sqrt 2 }}.\]

求\[I\left( b \right) = \int_0^\infty  {\int_0^x {{e^{ - {a^2}{x^2} - {b^2}{y^2}}}} } dydx,\qquad a,b\geq 0.\]

解.事实上,我们有

\begin{align*}I\left( b \right) &= \int_0^\infty  {\int_0^x {{e^{ - {a^2}{x^2} - {b^2}{y^2}}}} } dydx = \int_0^\infty  {{e^{ - {a^2}{x^2}}}dx\int_0^x {{e^{ - {b^2}{y^2}}}} } dy\\&= \frac{1}{b}\int_0^\infty  {{e^{ - {a^2}{x^2}}}dx\int_0^{bx} {{e^{ - {y^2}}}} } dy = \frac{{\sqrt \pi  }}{{2b}}\int_0^\infty  {{e^{ - {a^2}{x^2}}} \mathrm{erf}\left( {bx} \right)dx}. \end{align*}

法医秦明，完美案件

求解ODE$$\frac{{{d^2}r}}{{d{t^2}}} + 2h\frac{{dr}}{{dt}} + w_0^2r =  - \frac{e}{m}E.$$

参考：http://wenku.baidu.com/link?url=1z4AQfIr20D9-JVBbLLwyrYOwG2nVBubGuEVfCW8Urqz1rR2ABfF0lewEsWfWBwCp7W-0Z9fh4LRFh6wbEHhFEQwTiTtB0Nd51SBE1lTiFC