含圆周率的公式列表

Template:Pi box 下面是一个涉及数学常数π的公式列表。

古典几何

$C=2\pi r=\pi d$

其中，$C$是一个圆的周长，$r$是半径，$d$是直径。

$A=\pi r^2$

其中$A$是一个圆的面积，$r$是半径。

$V=\frac{4}{3}\pi r^3$

其中，$V$是一个球体的体积，$r$是半径。

$A=4\pi r^2$

其中$A$是一个球体的表面积，$r$是半径。

分析

积分

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\text{sech}(x)dx=\pi$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{(x+1)\sqrt{x}}=\pi$

$\underset{-1}{\overset{1}{\int }}\sqrt{1-x^2}dx=\frac{\pi }{2}$

$\underset{-1}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^2}}=\pi$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{1+x^2}=\pi$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\sqrt{\pi }$ (参见 正态分布)

$\oint \frac{dz}{z}=2\pi i$ (参见 柯西积分公式)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin (x)}{x}dx=\pi$

$\underset{0}{\overset{1}{\int }}\frac{x^4(1-x{)}^4}{1+x^2}dx=\frac{22}{7}-\pi$ (参见 證明22/7大於π)

高效的无穷级数

$\frac{\pi }{2}=\sum _{k=0}^{\mathrm{\infty }}\frac{k!}{(2k+1)!!}=\sum _{k=0}^{\mathrm{\infty }}\frac{2^{k}k{!}^2}{(2k+1)!}$ (参见 双阶乘)

$\frac{1}{\pi }=12\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k(6k)!(13591409+545140134k)}{(3k)!(k!{)}^{3}640320^{3k+\frac{3}{2}}}$ (参见 楚德诺夫斯基算法)

$\frac{1}{\pi }=\frac{2\sqrt{2}}{9801}\sum _{k=0}^{\mathrm{\infty }}\frac{(4k)!(1103+26390k)}{(k!{)}^{4}396^{4k}}$ (参见拉马努金)

$\pi =\frac{\sqrt{3}}{6^5}\sum _{k=0}^{\mathrm{\infty }}\frac{[(4k)!{]}^2(6k)!}{9^{k+1}(12k)!(2k)!}(\frac{127169}{12k+1}-\frac{1070}{12k+5}-\frac{131}{12k+7}+\frac{2}{12k+11})$

以下是任意位的二进制的π计算：:

$\pi =\sum _{k=0}^{\mathrm{\infty }}\frac{1}{16^k}(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6})$ (参见 贝利－波尔温－普劳夫公式)

$\pi =\frac{1}{2^6}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^{10n}}(-\frac{2^5}{4n+1}-\frac{1}{4n+3}+\frac{2^8}{10n+1}-\frac{2^6}{10n+3}-\frac{2^2}{10n+5}-\frac{2^2}{10n+7}+\frac{1}{10n+9})$

其他无穷级数

$\zeta (2)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\frac{{\pi }^2}{6}$   (参见巴塞尔问题和黎曼ζ函數)

$\zeta (4)=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\cdots =\frac{{\pi }^4}{90}$

$\zeta (2n)=\frac{1}{1^{2n}}+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\frac{1}{4^{2n}}+\cdots =(-1{)}^{n+1}\frac{B_{2n}(2\pi {)}^{2n}}{2(2n)!}$

$\frac{\pi }{4}=\sum _{n=0}^{\mathrm{\infty }}{\left[\frac{(-1{)}^n}{2n+1}\right]}^1=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots =\arctan 1=\underset{0}{\overset{1}{\int }}\frac{1}{1+x^2}dx$   (参见Π的莱布尼茨公式)

$\frac{{\pi }^2}{8}=\sum _{n=0}^{\mathrm{\infty }}{\left[\frac{(-1{)}^n}{2n+1}\right]}^2=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots$

$\frac{{\pi }^3}{32}=\sum _{n=0}^{\mathrm{\infty }}{\left[\frac{(-1{)}^n}{2n+1}\right]}^3=\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\cdots$

$\frac{{\pi }^4}{96}=\sum _{n=0}^{\mathrm{\infty }}{\left[\frac{(-1{)}^n}{2n+1}\right]}^4=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\frac{1}{7^4}+\cdots$

$\frac{5{\pi }^5}{1536}=\sum _{n=0}^{\mathrm{\infty }}{\left[\frac{(-1{)}^n}{2n+1}\right]}^5=\frac{1}{1^5}-\frac{1}{3^5}+\frac{1}{5^5}-\frac{1}{7^5}+\cdots$

$\frac{{\pi }^6}{960}=\sum _{n=0}^{\mathrm{\infty }}{\left[\frac{(-1{)}^n}{2n+1}\right]}^6=\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6}+\frac{1}{7^6}+\cdots$

$\frac{\pi }{4}=\frac{3}{4}\times \frac{5}{4}\times \frac{7}{8}\times \frac{11}{12}\times \frac{13}{12}\times \frac{17}{16}\times \frac{19}{20}\times \frac{23}{24}\times \frac{29}{28}\times \frac{31}{32}\times \cdots$ (欧拉)

$\pi =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\cdots$ (欧拉, 1748)

梅钦公式

参见梅钦公式.

$\frac{\pi }{4}=4\arctan \frac{1}{5}-\arctan \frac{1}{239}$ (原始的梅钦公式.)

$\frac{\pi }{4}=\arctan \frac{1}{2}+\arctan \frac{1}{3}$

$\frac{\pi }{4}=2\arctan \frac{1}{2}-\arctan \frac{1}{7}$

$\frac{\pi }{4}=2\arctan \frac{1}{3}+\arctan \frac{1}{7}$

$\frac{\pi }{4}=5\arctan \frac{1}{7}+2\arctan \frac{3}{79}$

$\frac{\pi }{4}=12\arctan \frac{1}{49}+32\arctan \frac{1}{57}-5\arctan \frac{1}{239}+12\arctan \frac{1}{110443}$

$\frac{\pi }{4}=44\arctan \frac{1}{57}+7\arctan \frac{1}{239}-12\arctan \frac{1}{682}+24\arctan \frac{1}{12943}$

无穷级数

一些涉及圆周率的无穷级数:

$\pi =\frac{1}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{[(2n)!{]}^3(42n+5)}{(n!{)}^6{16}^{3n+1}}$
$\pi =\frac{4}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(4n)!(21460n+1123)}{(n!{)}^4{441}^{2n+1}{2}^{10n+1}}$
$\pi =\frac{4}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(6n+1){\left(\frac{1}{2}\right)}_n^3}{4^n(n!{)}^3}$
$\pi =\frac{32}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{\sqrt{5}-1}{2}\right)}^{8n}\frac{(42n\sqrt{5}+30n+5\sqrt{5}-1){\left(\frac{1}{2}\right)}_n^3}{64^n(n!{)}^3}$
$\pi =\frac{27}{4Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{2}{27}\right)}^n\frac{(15n+2){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{(n!{)}^3}$
$\pi =\frac{15\sqrt{3}}{2Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{4}{125}\right)}^n\frac{(33n+4){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{(n!{)}^3}$
$\pi =\frac{85\sqrt{85}}{18\sqrt{3}Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{4}{85}\right)}^n\frac{(133n+8){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{(n!{)}^3}$
$\pi =\frac{5\sqrt{5}}{2\sqrt{3}Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{4}{125}\right)}^n\frac{(11n+1){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{(n!{)}^3}$
$\pi =\frac{2\sqrt{3}}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(8n+1){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n{\left(\frac{3}{4}\right)}_n}{(n!{)}^3{9}^n}$
$\pi =\frac{\sqrt{3}}{9Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(40n+3){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n{\left(\frac{3}{4}\right)}_n}{(n!{)}^3{49}^{2n+1}}$
$\pi =\frac{2\sqrt{11}}{11Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(280n+19){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n{\left(\frac{3}{4}\right)}_n}{(n!{)}^3{99}^{2n+1}}$
$\pi =\frac{\sqrt{2}}{4Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(10n+1){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n{\left(\frac{3}{4}\right)}_n}{(n!{)}^3{9}^{2n+1}}$
$\pi =\frac{4\sqrt{5}}{5Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(644n+41){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n{\left(\frac{3}{4}\right)}_n}{(n!{)}^3{5}^n{72}^{2n+1}}$
$\pi =\frac{4\sqrt{3}}{3Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(28n+3){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n{\left(\frac{3}{4}\right)}_n}{(n!{)}^3{3}^n{4}^{n+1}}$
$\pi =\frac{4}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(20n+3){\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n{\left(\frac{3}{4}\right)}_n}{(n!{)}^3{2}^{2n+1}}$
$\pi =\frac{72}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(4n)!(260n+23)}{(n!{)}^4{4}^{4n}18^{2n}}$
$\pi =\frac{3528}{Z}$	$Z=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(4n)!(21460n+1123)}{(n!{)}^4{4}^{4n}882^{2n}}$

$(x{)}_n$是阶乘幂中下降阶乘幂的符号。

$\prod _{n=1}^{\mathrm{\infty }}\frac{4n^2}{4n^2-1}=\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \frac{8}{7}\cdot \frac{8}{9}\cdots =\frac{4}{3}\cdot \frac{16}{15}\cdot \frac{36}{35}\cdot \frac{64}{63}\cdots =\frac{\pi }{2}$ (参见沃利斯乘积)

弗朗索瓦·韦达的公式:

$\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot \cdots =\frac{2}{\pi }$

连分数

$\pi =3+\frac{1^2}{6+\frac{3^2}{6+\frac{5^2}{6+\frac{7^2}{6+\ddots }}}}$

$\pi =\frac{4}{1+\frac{1^2}{3+\frac{2^2}{5+\frac{3^2}{7+\frac{4^2}{9+\ddots }}}}}$

$\pi =\frac{4}{1+\frac{1^2}{2+\frac{3^2}{2+\frac{5^2}{2+\frac{7^2}{2+\ddots }}}}}$

(参见连分数。)

杂项

$n!\approx \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$ (斯特灵公式)

$e^{i\pi }+1=0$ (歐拉恆等式)

$\sum _{k=1}^n\phi (k)\approx \frac{3n^2}{{\pi }^2}$

$\sum _{k=1}^n\frac{\phi (k)}{k}\approx \frac{6n}{{\pi }^2}$

$\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$ (伽玛函数)

$\pi =\frac{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^{\frac{4}{3}}\mathrm{a}\mathrm{g}\mathrm{m}(1,\sqrt{2}{)}^{\frac{2}{3}}}{2}$

$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n^2}\sum _{k=1}^n(nmodk)=1-\frac{{\pi }^2}{12}$

$\underset{n\to \mathrm{\infty }}{\lim }10^{n+2}\cdot \sin \left(\frac{1^{\circ }}{\underset{\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{g}\mathrm{i}\mathrm{t}\mathrm{s}}{\underbrace{55\cdots 55^{\circ }}}}\right)=\pi$

$\underset{n\to \mathrm{\infty }}{\lim }n\cdot \sin \left(\frac{180^{\circ }}{n}\right)=\pi$

$\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{\sqrt{2}}\cdot \sqrt{1-\cos \left(\frac{360^{\circ }}{n}\right)}=\pi$

物理

• 宇宙常数:

$\mathrm{\Lambda }=\frac{8\pi G}{3c^2}\rho$

• 不确定性原理:

$\mathrm{\Delta }x\mathrm{\Delta }p\ge \frac{h}{4\pi }$

• 爱因斯坦场方程:

$R_{ik}-\frac{g_{ik}R}{2}+\mathrm{\Lambda }{g}_{ik}=\frac{8\pi G}{c^4}{T}_{ik}$

• 库仑定律:

$F=\frac{\left|q_1{q}_2\right|}{4\pi {\epsilon }_0{r}^2}$

• 真空磁导率:

${\mu }_0=4\pi \cdot 10^{-7}(\mathrm{N}/{\mathrm{A}}^{\mathrm{2}})$

• 单摆周期

$T=2\pi \sqrt{\frac{L}{g}}$

拓展阅读

• Peter Borwein, The Amazing Number PiTemplate:Wayback
• Kazuya Kato, Nobushige Kurokawa, Saito Takeshi: Number Theory 1: Fermat's Dream. American Mathematical Society, Providence 1993, ISBN 0-8218-0863-X.

参见

• 圓周率