对数加法
logax+logay=loga(x⋅y)log_ax+log_a y=log_a({x\cdotp y})logax+logay=loga(x⋅y)
等比数列求和
sn={n⋅a1 (q=1)a1⋅(1−qn)1−q(q≠1)s_n=\begin{cases} n\cdotp a_1\quad\qquad\ \ (q=1)\\ \frac{a_1\cdotp(1-q^n)}{1-q}\qquad(q\neq1) \end{cases}sn={n⋅a1 (q=1)1−qa1⋅(1−qn)(q̸=1)
三角函数
和差角
sin(a±b)=sin(a)⋅cos(b)±cos(a)⋅sin(b)sin(a\pm b)=sin(a)\cdotp cos(b)\pm cos(a)\cdotp sin(b)sin(a±b)=sin(a)⋅cos(b)±cos(a)⋅sin(b)
cos(a±b)=cos(a)⋅cos(b)∓sin(a)⋅sin(b)cos(a\pm b)=cos(a)\cdotp cos(b)\mp sin(a)\cdotp sin(b)cos(a±b)=cos(a)⋅cos(b)∓sin(a)⋅sin(b)
tan(a±b)=tan(a)±tan(b)1∓tan(a)⋅tan(b)tan(a\pm b)=\frac{tan(a)\pm tan(b)}{1\mp tan(a)\cdotp tan(b)}tan(a±b)=1∓tan(a)⋅tan(b)tan(a)±tan(b)
和差化积
sin(a)+sin(b)=2⋅sin(a+b2)cos(a−b2)sin(a)+sin(b)=2\cdotp sin(\frac{a+b}{2})cos(\frac{a-b}{2})sin(a)+sin(b)=2⋅sin(2a+b)cos(2a−b)
sin(a)−sin(b)=2⋅cos(a+b2)sin(a−b2)sin(a)-sin(b)=2\cdotp cos(\frac{a+b}{2})sin(\frac{a-b}{2})sin(a)−sin(b)=2⋅cos(2a+b)sin(2a−b)
cos(a)+cos(b)=2⋅cos(a+b2)cos(a−b2)cos(a)+cos(b)=2\cdotp cos(\frac{a+b}{2})cos(\frac{a-b}{2})cos(a)+cos(b)=2⋅cos(2a+b)cos(2a−b)
cos(a)−cos(b)=−2⋅sin(a+b2)sin(a−b2)cos(a)-cos(b)=-2\cdotp sin(\frac{a+b}{2})sin(\frac{a-b}{2})cos(a)−cos(b)=−2⋅sin(2a+b)sin(2a−b)
tan(a)±tan(b)=sin(a±b)cos(a)cos(b)tan(a)\pm tan(b)=\frac{sin(a\pm b)}{cos(a)cos(b)}tan(a)±tan(b)=cos(a)cos(b)sin(a±b)
积化和差
sin(a)cos(b)=12(sin(a+b)+sin(a−b))sin(a)cos(b)=\frac{1}{2}\big(sin(a+b)+sin(a-b)\big)sin(a)cos(b)=21(sin(a+b)+sin(a−b))
cos(a)sin(b)=12(sin(a+b)−sin(a−b))cos(a)sin(b)=\frac{1}{2}\big(sin(a+b)-sin(a-b)\big)cos(a)sin(b)=21(sin(a+b)−sin(a−b))
cos(a)cos(b)=12(cos(a+b)+cos(a−b))cos(a)cos(b)=\frac{1}{2}\big(cos(a+b)+cos(a-b)\big)cos(a)cos(b)=21(cos(a+b)+cos(a−b))
sin(a)sin(b)=−12(cos(a+b)−cos(a−b))sin(a)sin(b)=-\frac{1}{2}\big(cos(a+b)-cos(a-b)\big)sin(a)sin(b)=−21(cos(a+b)−cos(a−b))
常见数列求和
∑i=1ni2=n(n+1)(2n+1)6\sum\limits_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}i=1∑ni2=6n(n+1)(2n+1)
∑i=1n(2i−1)2=n(4n2−1)3\sum\limits_{i=1}^{n}(2i-1)^2=\frac{n(4n^2-1)}{3}i=1∑n(2i−1)2=3n(4n2−1)
∑i=1ni3=n2(n+1)24\sum\limits_{i=1}^{n}i^3=\frac{n^2(n+1)^2}{4}i=1∑ni3=4n2(n+1)2
∑i=1n(2i−1)3=n2(2n2−1)\sum\limits_{i=1}^{n}(2i-1)^3=n^2(2n^2-1)i=1∑n(2i−1)3=n2(2n2−1)
∑i=1nn(n+1)=n(n+1)(n+2)3\sum\limits_{i=1}^{n}n(n+1)=\frac{n(n+1)(n+2)}{3}i=1∑nn(n+1)=3n(n+1)(n+2)
本源勾股数
a=2⋅n+1(n≥1)a=2\cdotp n+1\quad(n\ge1)a=2⋅n+1(n≥1)
**那么: **{b=2⋅n2+2⋅nc=2⋅n2+2⋅n+1\begin{cases} b=2\cdotp n^2+2\cdotp n\\ c=2\cdotp n^2+2\cdotp n+1 \end{cases}{b=2⋅n2+2⋅nc=2⋅n2+2⋅n+1
a=2⋅n(n≥2)a=2\cdotp n\quad(n\ge2)a=2⋅n(n≥2)
那么: {b=n2−1c=n2+1\begin{cases} b=n^2-1\\ c=n^2+1\\ \end{cases}{b=n2−1c=n2+1
均值不等式
调和平均 Hn=n∑i=1n1xiH_n=\frac{n}{\sum\limits_{i=1}^n\frac{1}{x_i}}Hn=i=1∑nxi1n
几何平均 Gn=∏i=1nxinG_n=\sqrt[n] {\prod\limits_{i=1}^n x_i}Gn=ni=1∏nxi
算术平均 An=∑i=1nxinA_n=\frac{\sum\limits_{i=1}^n x_i}{n}An=ni=1∑nxi
平方平均 Qn=∑i=1nxi2nQ_n=\sqrt{\frac{\sum\limits_{i=1}^n x_i^2}{n}}Qn=ni=1∑nxi2
那么:Hn≤Gn≤An≤QnH_n\le G_n\le A_n \le Q_nHn≤Gn≤An≤Qn
错排
Dn=n!⋅∑i=0n(−1)ii!=⌊n!e+0.5⌋D_n=n!\cdot\sum\limits_{i=0}^n\frac{(-1)^i}{i!}=\Big\lfloor \frac{n!}{e}+0.5\Big\rfloorDn=n!⋅i=0∑ni!(−1)i=⌊en!+0.5⌋
Dn=(n−1)⋅(Dn−1+Dn−2)D_n=(n-1)\cdotp \big(D_{n-1}+D_{n-2}\big)Dn=(n−1)⋅(Dn−1+Dn−2)
D1=0D2=1D_{1}=0\quad D_2=1D1=0D2=1
斯特林公式
n!≈2πn(ne)nn!\approx \sqrt{2\pi n}(\frac{n}{e})^nn!≈2πn(en)n
矩阵树定理
a[x][y]={deg(x)x=y−1 x≠y⋂x与y相邻0 其他a[x][y]= \begin{cases} deg(x)\qquad x=y\\ -1\qquad\quad\ \ x\neq y \bigcap x与y相邻\\ 0\qquad\qquad\ 其他 \end{cases}a[x][y]=⎩⎪⎨⎪⎧deg(x)x=y−1 x̸=y⋂x与y相邻0 其他
基姆拉尔森计算公式
week=(day+2month+3(month+1)5+year+year4−year100+year400)%7+1week=(day+2month+\frac{3(month+1)}{5}+year+\frac{year}{4}-\frac{year}{100}+\frac{year}{400})\%7+1week=(day+2month+53(month+1)+year+4year−100year+400year)%7+1
1, 2月当作上一年的13, 14月
曲线区域分割
nnn条 直线 最多能把平面分成 12n(n−1)+1\frac{1}{2}n(n-1)+121n(n−1)+1 部分
nnn个 三角形 最多能把平面分成 3n2−3n+23n^2-3n+23n2−3n+2 部分
nnn个 四边形 最多能把平面分成 2⋅(3−2n)22\cdotp(3-2n)^22⋅(3−2n)2 部分
nnn个 圆 最多能把平面分成 n2−n+2n^2-n+2n2−n+2 部分
nnn个 椭圆 最多能把平面分成 2(n2−n+1)2(n^2-n+1)2(n2−n+1) 部分
nnn个 d−1d-1d−1维超平面 最多能把ddd维空间分成 ∑i=0dCni\sum\limits_{i=0}^dC_{n}^ii=0∑dCni
平面图欧拉公式
顶点数−-−边数+++面数=2=2=2
皮克公式
面积===内部点个数+12+\frac{1}{2}+21边界上点个数−1-1−1
莫比乌斯反演
{f(n)=∑d∣ng(d)g(n)=∑d∣nμ(d)f(nd)\begin{cases} f(n)=\sum\limits_{d|n}g(d)\\ g(n)=\sum\limits_{d|n}\mu(d)f(\frac{n}{d}) \end{cases}⎩⎪⎨⎪⎧f(n)=d∣n∑g(d)g(n)=d∣n∑μ(d)f(dn)
{f(n)=∑n∣dg(d)g(n)=∑n∣dμ(dn)f(d)\begin{cases} f(n)=\sum\limits_{n|d}g(d)\\g(n)=\sum\limits_{n|d}\mu(\frac{d}{n})f(d) \end{cases}⎩⎪⎨⎪⎧f(n)=n∣d∑g(d)g(n)=n∣d∑μ(nd)f(d)