统计

重要值

(加权)平均数

$$\overline{X}=\frac1N \sum^n_{i=1}f_ix_i$$

方差(标准差)

$$s^2=\frac1N\sum^n_{i=1}(x_i-\overline x)^2\\ s=\sqrt{\frac1N\sum^n_{i=1}(x_i-\overline x)^2}$$

趣题

组合数学

魔法少女小圆第十话

$$(1+n)^p-n^p-1\;\;\text{mod}\;\;p\equiv 0\quad p为素数$$ $$C_p^0+ \underbrace {C_p^1n^1+C_p^2n^2+\cdots+C_p^{p-1}n^{p-1}} _{有p这个因子} +C_p^pn^p-n^p-1\;\;\text{mod}\;\;p\equiv 0$$

解析几何

椭圆

第一定义

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1(c^2 = a^2 - {b^2} )$$ $$e =\frac ca= \sqrt {1 - \frac{b^2}{a^2} } \in (0,1)$$

第二定义

$$准线\\\left\{ \begin{gathered} {l_1} = \frac{a^2}{c} > a \hfill \\ {l_2} = - \frac{a^2}{c} < - a \hfill \\ \end{gathered} \right.$$ $$\left\{ \begin{gathered} \left| {P{F_1} } \right| = a - e \cdot {x_P} \hfill \\ \left| {P{F_2} } \right| = a + e \cdot {x_P} \hfill \\ \end{gathered} \right.$$ $$通径H=\frac{2b^2}{a^2}$$

$$S_{\Delta P{F_1}{F_2} } = {b^2} \cdot \tan \frac{\ang F_1PF_2}{2}$$ $$\Delta P{F_1}{F_2}的外切圆B切x轴于顶点A$$

第三定义

$$k_{PF_1} \cdot k_{PF_2} = - \frac{b^2}{a^2}$$

$$中点弦\\ k_{OM} \cdot k_{PB}=k_{PA} \cdot k_{PB} = - \frac{b^2}{a^2}$$

$$e\cos \theta = \left| \frac{1 - \lambda }{1 + \lambda } \right|,(\lambda = \frac{AF_1}{BF_1})$$

极坐标

$$\rho = \frac{ep}{1 - e\cos \theta },(焦准距p = \frac{b^2}{c})$$ $$L = \left| \frac{H}{ {1 - {e^2}\cos ^2}\theta } \right|,(H=\frac{2b^2}{a}=2ep)$$

参数方程

$$\left\{ \begin{gathered} x = a\cos t \hfill \\ y = b\sin t \hfill \\ \end{gathered} \right.$$

双曲线

第一定义

$$\frac{ { {x^2} } } { { {a^2 } } }- \frac{ { {y^2} } }{ { {b^2} } } = 1(c = \sqrt { {a^2} + {b^2 } } )$$ $$\left| \left| {P{F_2} } \right| - \left| {P{F_1} } \right| \right| = 2a \Leftrightarrow \left| {P{F_2} } \right| - \left| {P{F_1} } \right| = \pm 2a$$

第二定义

$$e = \frac{c}{a} = \sqrt {1 + \frac{b^2}{a^2} } > 1$$ $$\left\{ \begin{gathered} \left| {P{F_1} } \right| = e{x_P} - a \hfill \\ \left| {P{F_2} } \right| = e{x_P} + a \hfill \\ \end{gathered} \right.$$

性质

$$渐近线\\ k = \pm \frac{b}{a},O{E_1} \bot {E_1}{F_1}$$ $$O{E_1} = a,{E_1}{F_1} = b$$

$${b^2} = (c - a)(c + a) = {c^2} - {a^2}$$

$$\Delta P{F_1}{F_2}的内切圆M切x轴于顶点A$$ $$S_\Delta P{F_1}{F_2} = \frac{ {b^2} }{ \tan \frac{ \angle {F_1}P{F_2} }{2} }$$

$$|P\overline P | = |Q\overline Q |,{k_{OM} } \cdot {k_{PQ } } = \frac{ { {b^2} } }{ { {a^2} } }\\\left\{ {\begin{array}{*{20}{l} } {x = \frac{a}{ {\cos t} } = a\sec t,} \\ {y = \pm b\tan t} \end{array} } \right.$$

抛物线

$$\left\{ {\begin{array}{*{20}{l} } { {y^2} = 2px,(x为对称轴)} \\ { {x^2} = 2py,(y为对称轴)} \end{array} } \right.$$ $$|PF|=|Pl|$$

焦点弦

$$\left\{ \begin{gathered} \left| {FP} \right| = \frac{p}{ {1 - \cos \theta } } \hfill \\ \left| {FQ} \right| = \frac{p}{ {1 + \cos \theta } } \hfill \\ \end{gathered} \right. \Rightarrow \left| {PQ} \right| = \frac{ {2p} }{ { { {\sin }^2}\theta } }$$ $${S_{\Delta PFQ} } = \frac{ { {p^2} } }{ {2\sin \theta } }$$ $$\left\{ \begin{gathered} {x_p} \cdot {x_Q} = \frac{ { {p^2} } }{4} \hfill \\ {y_P} \cdot {y_Q} = - {p^2} \hfill \\ \end{gathered} \right.$$ $$\frac{1}{ {\left| {FP} \right|} } + \frac{1}{ {\left| {FQ} \right|} } = \frac{2}{p}$$ $${k_{PQ} } \cdot {y_M} = p$$

$$\begin{gathered} MN = NP = NQ \hfill \\ AM = AN \hfill \\ \left\{ \begin{gathered} \left| {PF} \right| = \left| {AP} \right| = {x_P} + \frac{p}{2} \hfill \\ \left| {QF} \right| = \left| {BQ} \right| = {x_Q} + \frac{p}{2} \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered}$$ $$MP,MQ为切线$$

参数方程

$$\left\{ \begin{gathered} x = 2p{t^2} \hfill \\ y = 2pt \hfill \\ \end{gathered} \right.$$

切线

$$\frac {x^2} {a^2}\pm \frac {y^2} {b^2}=1\\ \frac{ {2{x_0} } }{ { {a^2} } } + \frac{ {2{y_0} } }{ { {b^2} } } \cdot \frac{ { {\text{d} }y} }{ { {\text{d} }x} } = 0 \Rightarrow \frac{ { {\text{d} }y} }{ { {\text{d} }x} } = - \frac{ { {b^2}{x_0} } }{ { {a^2}{y_0} } }$$

切线方程为

$$\frac{ {x_0}x }{ {a^2} } + \frac{ {y_0}y }{ {b^2} } = 1$$ $$y=kx\pm\sqrt{a^2k^2+b^2}$$

抛物线

$$\begin{gathered} {y^2} = 2px \hfill \\ 2{y_0} \cdot \frac{ { {\text{d} }y} }{ { {\text{d} }x} } = 2p \Rightarrow \frac{ { {\text{d} }y} }{ { {\text{d} }x} } = \frac{p}{ { {y_0} } } \hfill \\ \end{gathered}$$

双曲线

$$\frac{ {x_0}x }{ {a^2} } - \frac{ {y_0}y }{ {b^2} } = 1$$ $$k=\frac{ { {b^2}{x_0} } }{ { {a^2}{y_0} } }\Rightarrow x_0=\frac{ak}{\sqrt{k^2-1}}$$ $$y=kx\pm\sqrt{a^2k^2-b^2}$$ $$\frac{ {x_0}x }{ {a^2} } \pm \frac{ {y_0}y }{ {b^2} } = 1\\ y=kx\pm\sqrt{a^2k^2\pm b^2}$$

椭圆切线

$$y=kx\pm\sqrt{a^2k^2+ b^2}$$

双曲线切线

$$y=kx\pm\sqrt{a^2k^2- b^2}\textcolor{red} {(|k| > \frac{b}{a} )}$$

不包括斜率不存在时

$P({x_0},{y_0})$在切线上，解出$k$

$${y_0} - k{x_0} = \pm \sqrt {a^2k^2 - b^2}\\ ({x_0}^2 - {a^2}){k^2} - 2{x_0}{y_0}k + ({y_0}^2 - {b^2}) = 0$$

若$k^2$前系数为$0$，则为一次方程，有一条斜率存在的切线，还有一条不存在的切线

若$k^2$前的系数不为$0$，则该方程是一个关于$k$的一元二次方程，可能有$0$个，$1$个，$2$个解

$$\begin{array}{l} \Delta &= 4x_0y_0 - 4(x_0^2 - a^2)(y_0^2 + b^2)\\ &=4{a^2}{b^2}\left[\textcolor{red}{ 1 - \left(\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \right)}\right] \end{array}$$

若$\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2}=0$，有一根为$k=\frac{b}{a}/ - \frac{b}{a}$，故舍去

​

集合

$$\left\{ \begin{gathered} {\text{N} } \to 0,1,2 \cdots \hfill \\ { {\text{N} }^*}{\text{/} }{ {\text{N} }_ + } \to 1,2 \cdots \hfill \\ \Z \to - 2, - 1,0,1,2 \cdots \hfill \\ \Q \to - 1.3,\frac{5}{3},6 \cdots \hfill \\ \R \to \sqrt 3 ,\pi ,3 \cdots \hfill \\ \end{gathered} \right.$$ $$\emptyset \subseteq A,A \subseteq A$$ $$\left\{ \begin{gathered} A \cap B = \left\{ {x|x \in A,且x \in B} \right\}\\ A \cup B = \left\{ {x|x \in A,或x \in B} \right\} \end{gathered} \right.$$ $${\complement _U}A = \left\{ {x|x \in U,且x \notin A} \right\}$$ $$\left\{ \begin{gathered} {\complement _U}(A \cap B) = ({\complement _U}A) \cup ({\complement _U}B) \hfill \\ {\complement _U}(A \cup B) = ({\complement _U}A) \cap ({\complement _U}B) \hfill \\ \end{gathered} \right.$$ $$p\Rightarrow q\Longleftrightarrow p是q的充分条件\\ p\Leftarrow q\Longleftrightarrow p是q的必要条件\\ \left\{ \begin{gathered} p \Rightarrow q \hfill \\ p \Leftarrow q \hfill \\ \end{gathered} \right.\Longleftrightarrow p 是q的充要条件$$ $$\tan x:\left\{ {x \in \mathbb{R},x \ne \frac{\pi }{2} + k\pi ,k \in \mathbb{Z}} \right\}\\ \frac{1}{x}: x \ne 0 \quad\sqrt[{2k}]{x} :\left\{ {x \in \mathbb{R},x \geqslant 0} \right\}\\ {\log _a}x:a > 0,a \ne 1,x > 0\;x^0:x\ne0\\ \cot x:\left\{ {x \in \mathbb{R},x \ne k\pi ,k \in \mathbb{Z}} \right\} \\$$ $$\left\{ {\begin{array}{*{20}{l}} {\frac{1}{6} = 0.1666 \ldots } \\ {\frac{5}{6} = 0.8333 \ldots } \end{array}} \right.\\\left\{\begin{array}{l} \frac{2}{6}=0.3333 \ldots \\ \frac{4}{6}=0.6666 \ldots \end{array}\right.$$ $$\left\{\begin{array}{c} \frac{1}{9}=0.1111 \ldots \\ \frac{2}{9}=0.2222 \ldots \\ \frac{3}{9}=0 . \dot{3} \\ \ldots \end{array}\right.$$ $$\left\{\begin{aligned} \frac{1}{7} &=0.142857,142857,14 \ldots \\ \frac{2}{7} &=0.2857,142857,1428 \ldots \\ \frac{4}{7} &=0.57,142857,142857 \ldots \end{aligned}\right.$$ $$\left\{\begin{array}{c} \frac{1}{11}=0.090909 \ldots \\ \frac{2}{11}=\overline{0 . \dot{1} \dot{8}} \\ \frac{3}{11}=\overline{0 . \dot{2} \dot{7}} \\ \ldots \end{array}\right.$$

热学

$$m_{\text{partical}}\overset {N_A} \longleftrightarrow m_{\text{mol}}\overset{n}{\longleftrightarrow}m$$ $$\begin{gathered} \hfill \\ \left\{ \begin{gathered} {p_1}{V_1} = {p_2}{V_2} \hfill \\ pV = C \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered}$$

$$\begin{array}{|c|c|} \hline &p&V\\ \hline 原& p_1V_{1}+p_2V_{2}&\backslash \\ \hline 后&p&V_1+V_2\\ \hline \end{array}$$ $$p_1V_1+p_2V_2=p(V _1+V_2)$$

$$\frac{4}{3}\pi {r^3} = \frac{1}{6}\pi {d^3} = \frac{m}{n \cdot {N_A} \cdot \rho } \Rightarrow d = \sqrt[3]{\frac{6m}{n{N_A}\pi \rho }}$$ $$\left\{ \begin{gathered} solid \to V = \frac{ m_{partical} } {\rho } = \frac{m} { n\rho {N_A} } \hfill \\ gas \to V = \frac{ m_{partical} } {\rho }(\frac{m} { n\rho {N_A} } = distance) \hfill \\ \end{gathered} \right.$$

$$S_Ap_A+p_0S_B=mg\sin\theta+S_Bp_B+p_0S_A$$

打气问题

$$\sum\limits_{i = 1}^n {\frac{p_iV_i}{T_i}} = \frac{pV}{T}=nR$$ $$p_0V_0+p_0\triangle V =p_1V_0\\ p_1V_0+p_0\triangle V=p_2V_0\\ \cdots\\ p_{n-1}V_0+p_0\triangle V=p_nV_0 \\\Rightarrow p_n=\left(1+\frac{n\triangle V}{V_0}\right)p_0$$

抽气问题

$$p_0V_0=p_1(V_0+\triangle V)\\ p_1V_0=p_2(V_0+\triangle V)\\ \cdots\\ p_{n-1}V_0=p_n(V_0+\triangle V)\\ \Rightarrow p_n=\left(\frac{\triangle V}{\triangle V+V_0}\right)^np_0$$ $$mg\sin \theta - \mu (Eq + mg\cos \theta ) = ma$$

运动学

基本公式

$$\begin{array}{|c|c|} \hline v_{0},v,a, t & v=v_{0}+a t \\ \hline v_{0}, a, t, x & x=v_{0} t+\frac{1}{2} a t^{2} \\ \hline v_{0}, v, a, x & v^{2}-v_{0}^{2}=2 a x \\ \hline v_{0}, v, t, x & \color{red}{x=\frac{v+v_{0}}{2} t} \\ \hline \end{array}$$

推论

$$\bar{v}=\frac{v_{0}+v}{2}=v_{\frac{t}{2}}$$ $$v_{\frac{x}{2}}=\sqrt{\frac{v_{0}^{2}+v^{2}}{2}}$$ $$\triangle x=x_n-x_{n-1}=at^2$$ $$x_{1}: x_{2}: x_{3}: \cdots: x_{n}=1^{2}: 2^{2}:3^{2}: \cdots: n^{2}$$ $$x_{\mathrm{I}}: x_{\mathrm{II}}: x_{\mathrm{III}} : \cdots: x_{N}=1: 3: 5: \cdots:(2 n-1)$$ $$t_{1}: t_{2}: \cdots: t_{n}=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2}): \cdots:(\sqrt{n}-\sqrt{n-1})$$ $$a=\frac{(x_6+x_5+x_4)-(x_3+x_2+x_1)}{9T^2}=\frac{\left(\frac{d}{\triangle t_2}\right)^2-\left(\frac{d}{\triangle t_1}\right)^2}{2s}$$

平抛运动

$$\begin{cases} x=v_0t\\ y=\frac12gt^2 \end{cases}\Rightarrow y=\frac g{2v_0^2}x^2$$ $$\tan\varphi=2\tan\theta=\frac{gt}{v_0}$$

斜抛运动

$$\begin{cases} x=v_0\cos \theta t\\ y=v_0\sin \theta t-\frac12gt^2 \end{cases} \Rightarrow y=-\frac g{2v_0\cos\theta}x^2+\tan\theta x$$

圆周运动

$$\begin{aligned}a&=\omega v=\omega^2r=\frac{v^2}r\\&=\left(\frac{2\pi}T\right)^2r=(2\pi f)^2r\\&=(2\pi n)^2r\end{aligned}$$ $$\begin{cases} v=s/t\rightarrow \text{m/s}\\ \omega=\theta /t\rightarrow \text{rad/s}\\ T=2\pi/\omega\rightarrow \text{s}\\ f=1/T \rightarrow\text{r/s}\\ n=1/T \rightarrow\text{Hz} \end{cases}$$

$$\sum_{i=1}^n m_iv_i=\sum_{i=1}^n m_iv'_i\\ \sum F\triangle t=\sum_{i=1}^n m_i\triangle v_i\Rightarrow F=\frac{\text{d}(mv)}{\text{d}t}$$ $$\sum F\triangle x=\frac12mv^2-\frac12mv_0^2$$ $$v_1=\frac{m_1-m_2}{m_1+m_2}v_0\\v_2=\frac{2m_1}{m_1+m_2}v_0$$ $$x_{重心}=\frac{\sum G_ix_i}{\sum G_i}\\x_{质心}=\frac{\sum m_ix_i}{\sum m_i}$$ $$\begin{align} \text{E}_p&=\int_0^{\Delta x} kx\text{d}x \\&=\left.\frac12kx^2\right|_0^{\Delta x} \\&=\frac12k{\Delta x}^2 \end{align}$$ $$F = G\frac{m_1m_2}{r^2}\\ \begin{cases} r \to 距离 \\ R \to 半径 \\ h = r - R \hfill \\ \end{cases}$$

电磁学

$$F=qvB\sin \theta$$

$$mv_0=m(v_1+v_2)$$ $$E=Blv\sin\theta$$ $$F-\frac{B^2L^2(v_1-v_2)}{R}=ma_1\\ \frac{B^2L^2(v_1-v_2)}{R}=ma_2$$ $$E=\frac{\triangle \varphi }{\triangle t}=\frac{\triangle B }{\triangle t}S+B'lv=\frac12B\omega r^2=Blv$$

有外力电容

$$\\\left \{ \begin{array}{l}I=\frac{\triangle q}{\triangle t}\\ \triangle q=C\triangle U\\\triangle U=Bl\triangle v\\ a=\frac{\triangle v}{\triangle t} \end{array}\right.\Rightarrow I=BlCa\\ ma=F-BIl=F-B^2L^2Ca\Rightarrow a=\frac{F}{m-B^2L^2C}\triangle$$ $$无外力电容\\E=\frac QC=BLv\Rightarrow v=\frac{Q}{BLC} \\BIL\triangle t=mv-mv_0=BLQ$$ $$对于u=U_m\sin(\omega t)\\ U=\frac{U_m}{\sqrt{2}}\\Q=\frac{U^2}{R}T=\sum_{i=1}^n\frac{U_i}{R}t_i$$

经典分析

$$F\triangle x-mg\sin\theta\triangle x-k\triangle x-\mu mg\cos\theta\triangle x-Q=\frac12mv_1^2-\frac12mv_0^2$$ $$F\triangle t-mg\sin\theta\triangle t-\mu mg\cos\theta\triangle t-BLq(or\quad\frac{B^2L^2\triangle x}{R+r})=mv_1-mv_0$$ $$\frac{B^2L^2v}{R+r}=mg\sin\theta+\mu mg\cos\theta$$

近代物理

$$\textrm{Photoelectri Effect}\\\begin{cases}E_{k_0}=h\nu -W_0=U_ce \\E_{k_1}=E_{k_0}+U_{正}e\\E_{光强}=nh\nu \\W_0=h\nu _c(截止频率) \end{cases}$$ $$\begin{cases} E_{pn}=-k\frac{e^2}{r_n} \\E_{kn}=k\frac{e^2}{2r_n} \end{cases}\Longrightarrow E_n=E_{pn}+E_{kn}=-k\frac{e^2}{2r_n}$$ $$\tilde{\nu}=\frac1\lambda=R(\frac{1}{2^2}-\frac{1}{n^2})\quad n=3,4,5,\cdots$$ $$E_n=\frac{1}{n^2}E_1\quad (E_1=-13.6\text{eV})\;\;r_n=n^2r_1$$

$$r=\frac{mv}{Bq}\\ m_\alpha v_\alpha=m_{Th}v_{Th}$$ $$\begin{cases} E=h\nu \\p=\frac{h}{\lambda}\end{cases} \triangle p\cdot\triangle x\ge\frac{h}{4\pi}$$ $$\rm\alpha 衰变\\ 2 _{1}^{1}H +2 _{0}^{1}n \rightarrow _{2}^{4}He \\ _{92}^{238}U \rightarrow_{90}^{234}Th +_{2}^{4}He\\\rm \beta衰变\\_{0}^{1}n \rightarrow _{1}^{1}H +_{-1}^{0}e\\ _{90}^{234}Th \rightarrow _{91}^{234}Pa + _{-1}^{0}e \\+\beta衰变\\_{1}^{1}H \rightarrow _{0}^{1}n +_{1}^{0}e\\ _{15}^{30}P \rightarrow _{14}^{30}Si + _{1}^{0}e$$ $$人工转变\\ \begin{cases}\rm _{2}^{4}He+ _{7}^{14}N \rightarrow _{8}^{17}O + _{1}^{1}H(Rutherford)\\\rm _{2}^{4}He+ _{4}^{8}Be \rightarrow _{6}^{11}C + _{0}^{1}n(Chadwick)\\\rm _{2}^{4}He+ _{13}^{27}Al \rightarrow _{15}^{30}P + _{0}^{1}n \\\rm _{15}^{30}P \rightarrow _{14}^{30}Si + _{1}^{0}e(I. Joliot-Curie) \end{cases}$$

前川定理(The Maekawa-Justin Theorem)

简介

MAEKAWA Jun (前川淳)，日本的软件工程师，数学家，折纸艺术家。

平顶点折叠的例子

让$M(\text{Mountain})$和$V(\text{Valley})$分别代表平顶点折叠中山折和谷折的数量，则前川定理可以表示为

$M = V +2 \quad \text{or}\quad V = M + 2$

即$|M-V|=2$

$\text{Proof}\;1:$

由于我们只关心顶点$x$和周围的折痕，所以可以以$x$为圆心做一个圆$(a)$，按折痕折叠后形成$(b)$

从下往上看向顶点x，可以发现圆环形成了一个闭合回路$(c)$

想象有一个蚂蚁从$p$点出发在这个闭合回路上爬行，遇到山折便逆时针旋转$180^{\circ}$，遇到谷折便顺时针旋转$180^{\circ}$，最后回到原点，方向和开始一样，由于沿着闭合回路走了一周，相当于旋转了$360^{\circ}$度，即

$M · 180^{\circ} +V ·(−180^{\circ}) = 360^{\circ}$

$M − V = 2$

因为纸有两面，如果从另一面看，原来的山折变成了谷折，原来的谷折变成了山折，所以有

$V − M = 2$

这样便证明了前川定理

$\text{Proof}\;2:$

这个证明是由$\text{Jan Siwanowicz}$在他还是个高中生的时候提出的

将此前的闭合回路看作一个多边形，把山折看成内角等于$0$，谷折看成内角等于$360^{\circ}$

由多边形内角和定理

$\sum\limits_{i=1}^n\theta_i=(n-2)×180^{\circ}$

推得在这个多边形中，内角和为

$M · 0^{\circ} +V · 360^{\circ}$

所以$V · 360^{\circ} = (M +V-2)180^{\circ}$

$\therefore M = V +2 \quad \text{or}\quad V = M + 2$

推广：

$M + V = 2(V+1) \quad \text{or} \quad2(V − 1)$

得到偶数定理：单顶点折叠中折痕总数必为偶数，角的总数也必为偶数