查看“︁含圆周率的公式列表”︁的源代码

{{模板:Pi box}}
下面是一个涉及[[数学常数]][[圆周率|π]]的公式列表。
==古典几何==
:<math>C = 2 \pi r = \pi d\!</math>

其中，<math>C</math>是一个[[圆]]的[[周长]]，<math>r</math>是[[半径]]，<math>d</math>是[[直径]]。
:<math>A = \pi r^2\!</math>

其中<math>A</math>是一个圆的[[面积]]，<math>r</math>是半径。

:<math>V = {4 \over 3}\pi r^3\!</math>

其中，<math>V</math>是一个[[球体]]的[[体积]]，<math>r</math>是半径。

:<math>A = 4\pi r^2\!</math>

其中<math>A</math>是一个球体的[[表面积]]，<math>r</math>是半径。

==分析==
===积分===
:<math>\int\limits_{-\infty}^{\infty} \text{sech}(x)dx = \pi \!</math>
<br />
:<math>\int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x}} = \pi</math>
<br />
:<math>\int\limits_{-1}^1 \sqrt{1-x^2}\,dx = \frac{\pi}{2}\!</math>
<br />
:<math>\int\limits_{-1}^1\frac{dx}{\sqrt{1-x^2}} = \pi\!</math>
<br />
:<math>\int\limits_{-\infty}^\infty\frac{dx}{1+x^2} = \pi\!</math> 
<br />
:<math>\int\limits_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\!</math> (参见 [[正态分布]])
<br />
:<math>\oint\frac{dz}{z}=2\pi i\!</math> (参见 [[柯西积分公式]])
<br />
:<math>\int\limits_{-\infty}^{\infty} \frac{\sin(x)}{x}\,dx=\pi \!</math>
<br />
:<math>\int\limits_0^1 {x^4(1-x)^4 \over 1+x^2}\,dx = {22 \over 7} - \pi\!</math> (参见 [[證明22/7大於π]])
===高效的无穷级数===
{{See also|Category:圆周率算法}}
:<math>\frac{\pi}{2}\!=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}=\sum_{k=0}^\infty\frac{2^k k!^2}{(2k+1)!}</math> (参见 [[双阶乘]])
<br />
:<math>\frac{1}{\pi}\!=12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + \frac{3}{2}}}</math> (参见 [[楚德诺夫斯基算法]])
<br />
:<math>\frac{1}{\pi}\!=\frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}</math> (参见[[拉马努金#数学成就|拉马努金]])
<br />
:<math>\pi\!=\frac{\sqrt{3}}{6^5} \sum_{k = 0}^{\infty} \frac{[(4k)!]^2(6k)!}{9^{k+1}(12k)!(2k)!} \left( \frac{127169}{12k + 1} - \frac{1070}{12k + 5} - \frac{131}{12k + 7} + \frac{2}{12k + 11}\right)</math><ref>Cetin Hakimoglu-Brown [http://iamned.com/math/infiniteseries.pdf Derivation of Rapidly Converging Infinite Series] {{Wayback|url=http://iamned.com/math/infiniteseries.pdf |date=20170331204905 }}</ref>
<br />
以下是任意位的二进制的π计算：:
:<math>\pi\!=\sum_{k = 0}^{\infty} \frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6}\right)</math> (参见 [[贝利－波尔温－普劳夫公式]])
<br />
:<math>\pi=\frac{1}{2^6} \sum_{n=0}^{\infty} \frac{{(-1)}^n}{2^{10n}} \left( - \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}\right) </math>
<br />

===其他无穷级数===
:<math>\zeta(2) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}\!</math> &nbsp; (参见[[巴塞尔问题]]和[[黎曼ζ函數]])
<br />
:<math>\zeta(4)= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots = \frac{\pi^4}{90}\!</math>
<br />
:<math>\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)!}\!</math>
<br />
:<math> \frac{\pi}{4}\!=\sum_{n=0}^{\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} = \int_{0}^{1} \frac{1}{1+x^2}dx </math> &nbsp; (参见[[Π的莱布尼茨公式]])
<br />
:<math>\frac{\pi^2}{8}\!=\sum_{n=0}^{\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  </math>
<br />
:<math> \frac{\pi^3}{32}\!=\sum_{n=0}^{\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 </math>
<br />
:<math>\frac{\pi^4}{96}\!=\sum_{n=0}^{\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  </math>
<br />
:<math> \frac{5\pi^5}{1536}\!=\sum_{n=0}^{\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 </math>
<br />
:<math> \frac{\pi^6}{960}\!=\sum_{n=0}^{\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 </math>
<br />
:<math> \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 \! </math> ([[欧拉]])
<br />
:<math> \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 \!</math> (欧拉, 1748)<ref>[[Carl B. Boyer]], ''A History of Mathematics'', Chapter 21.</ref>
<br />

===梅钦公式===
参见[[梅钦公式]].
: <math>\frac{\pi}{4} = 4 \arctan\frac{1}{5} - \arctan\frac{1}{239} \!</math> (原始的梅钦公式.)
<br />
:<math>\frac{\pi}{4} = \arctan\frac{1}{2} + \arctan\frac{1}{3}\!</math>
<br />
:<math>\frac{\pi}{4} = 2 \arctan\frac{1}{2} - \arctan\frac{1}{7}\!</math>
<br />
:<math>\frac{\pi}{4} = 2 \arctan\frac{1}{3} + \arctan\frac{1}{7}\!</math>
<br />
:<math>\frac{\pi}{4} = 5 \arctan\frac{1}{7} + 2 \arctan\frac{3}{79}\!</math>
<br />
:<math>\frac{\pi}{4} = 12 \arctan\frac{1}{49} + 32 \arctan\frac{1}{57} - 5 \arctan\frac{1}{239} + 12 \arctan\frac{1}{110443}\!</math>
<br />
:<math>\frac{\pi}{4} = 44 \arctan\frac{1}{57} + 7 \arctan\frac{1}{239} - 12 \arctan\frac{1}{682} + 24 \arctan\frac{1}{12943}\!</math>

===无穷级数===
一些涉及圆周率的无穷级数:<ref>{{cite web |title=The world of Pi |url=http://www.pi314.net/eng/ramanujan.php |author=Simon Plouffe / David Bailey |publisher=Pi314.net |date= |accessdate=2011-01-29 |archive-date=2013-06-23 |archive-url=https://www.webcitation.org/6HZyTOr8m?url=http://www.pi314.net/eng/ramanujan.php |dead-url=no }}<br/>{{cite web |url=http://numbers.computation.free.fr/Constants/Pi/piSeries.html |title=Collection of series for {{pi}} |publisher=Numbers.computation.free.fr |date= |accessdate=2011-01-29 |archive-date=2013-06-23 |archive-url=https://www.webcitation.org/6HZyUsF3O?url=http://numbers.computation.free.fr/Constants/Pi/piSeries.html |dead-url=no }}</ref>
{| class="wikitable"
|-

| <math>\pi=\frac{1}{Z}\!</math>|| <math>Z=\sum_{n=0}^{\infty } \frac{[(2n)!]^3(42n+5)} {(n!)^6{16}^{3n+1}}\!</math>

|-

| <math>\pi=\frac{4}{Z}\!</math>|| <math>Z=\sum_{n=0}^{\infty } \frac{(-1)^n(4n)!(21460n+1123)} {(n!)^4{441}^{2n+1}{2}^{10n+1}}</math>

|-

| <math>\pi=\frac{4}{Z}\!</math>|| <math>Z=\sum_{n=0}^{\infty } \frac{(6n+1)\left ( \frac{1}{2} \right )^3_n} {{4^n}(n!)^3}\!</math>

|-

| <math>\pi=\frac{32}{Z}\!</math>|| <math>Z=\sum_{n=0}^{\infty } \left (\frac{\sqrt{5}-1}{2}  \right )^{8n} \frac{(42n\sqrt{5} +30n + 5\sqrt{5}-1) \left ( \frac{1}{2} \right )^3_n} {{64^n}(n!)^3}\!</math>

|-

| <math>\pi=\frac{27}{4Z}\!</math>|| <math>Z=\sum_{n=0}^{\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}\!</math>

|-

| <math>\pi=\frac{15\sqrt{3}}{2Z}\!</math>|| <math>Z=\sum_{n=0}^{\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}\!</math>

|-

| <math>\pi=\frac{85\sqrt{85}}{18\sqrt{3}Z}\!</math>|| <math>Z=\sum_{n=0}^{\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}\!</math>

|-

| <math>\pi=\frac{5\sqrt{5}}{2\sqrt{3}Z} \!</math>|| <math>Z=\sum_{n=0}^{\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}\!</math>

|-

| <math>\pi=\frac{2\sqrt{3}}{Z} \!</math>|| <math>Z=\sum_{n=0}^{\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}}\!</math>

|-

| <math>\pi=\frac{\sqrt{3}}{9Z} \!</math>|| <math>Z=\sum_{n=0}^{\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}}\!</math>

|-

| <math>\pi=\frac{2\sqrt{11}}{11Z} \!</math>|| <math>Z=\sum_{n=0}^{\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}}\!</math>

|-

| <math>\pi=\frac{\sqrt{2}}{4Z} \!</math>|| <math>Z=\sum_{n=0}^{\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}}\!</math>

|-

| <math>\pi=\frac{4\sqrt{5}}{5Z}  \!</math>|| <math>Z=\sum_{n=0}^{\infty } \frac{(644n+41) \left ( \frac{1}{2} \right )_n \left ( \frac{1}{4} \right )_n \left ( \frac{3}{4} \right )_n} {(n!)^35^n{72}^{2n+1}}\!</math>

|-

| <math>\pi=\frac{4\sqrt{3}}{3Z} \!</math>|| <math>Z=\sum_{n=0}^{\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}}\!</math>

|-

| <math> \pi=\frac{4}{Z}\!</math>|| <math>Z=\sum_{n=0}^{\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}}\!</math>

|-

| <math>\pi=\frac{72}{Z} \!</math>|| <math>Z=\sum_{n=0}^{\infty } \frac{(-1)^n(4n)!(260n+23)}{(n!)^44^{4n}18^{2n}}\!</math>

|-

| <math>\pi=\frac{3528}{Z} \!</math>|| <math>Z=\sum_{n=0}^{\infty } \frac{(-1)^n(4n)!(21460n+1123)}{(n!)^44^{4n}882^{2n}}\!</math>

|}
: <math>(x)_n \!</math>是[[阶乘幂]]中下降阶乘幂的符号。
:<math> \prod_{n=1}^{\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} \!</math> (参见[[沃利斯乘积]])

[[弗朗索瓦·韦达]]的公式:

:<math>\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots = \frac2\pi\!</math>
===连分数===
:<math>
\pi= {3 + \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \cfrac{7^2}{6 + \ddots\,}}}}}
</math>
:<math>
\pi = \cfrac{4}{1 + \cfrac{1^2}{3 + \cfrac{2^2}{5 + \cfrac{3^2}{7 + \cfrac{4^2}{9 + \ddots}}}}}
</math>
:<math>
\pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \cfrac{7^2}{2 + \ddots}}}}}\,
</math>

(参见[[连分数]]。)

===杂项===

:<math>n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\!</math> ([[斯特灵公式]])
<br />
:<math>e^{i \pi}+1=0 </math> ([[歐拉恆等式]])
<br />
:<math>\sum_{k=1}^{n} \varphi (k) \approx  \frac{3n^2}{\pi^2}\!</math>
<br />
:<math>\sum_{k=1}^{n} \frac {\varphi (k)} {k} \approx  \frac{6n}{\pi^2}\!</math>
<br />
:<math>\Gamma\left({1 \over 2}\right)=\sqrt{\pi}\!</math> ([[伽玛函数]])
<br />
:<math>\pi = \frac{\Gamma\left({\frac{1}{4}}\right)^{\frac{4}{3}} \mathrm{agm}(1, \sqrt{2})^{\frac{2}{3}}}{2}\!</math> 
<br />
:<math>\lim_{n\rightarrow \infty}\frac{1}{n^2} \sum_{k=1}^n (n\;\bmod\;k) = 1-\frac{\pi^2}{12}\!</math>
<br />
:<math>\lim_{n\rightarrow \infty} 10^{n+2}\cdot \sin\left(\frac{1^\circ}{\underbrace{55\cdots55^\circ}_{\mathrm{n\; digits}}}\right) = \pi\!</math> 
<br />
:<math>\lim_{n\rightarrow \infty} n\cdot \sin\left(\frac{180^\circ}{n}\right) = \pi</math> 
:
:
:<math>\lim_{n\rightarrow \infty} \frac n{\sqrt 2} \cdot \sqrt{1-\cos \left(\frac{360^\circ}n\right)} = \pi</math>

==物理==

*[[宇宙常数]]:
::<math>\Lambda = {{8\pi G} \over {3c^2}} \rho\!</math>
*[[不确定性原理]]:
::<math> \Delta x\, \Delta p \ge \frac{h}{4\pi} \!</math>
*[[爱因斯坦场方程]]:
::<math> R_{ik} - {g_{ik} R \over 2} + \Lambda g_{ik} = {8 \pi G \over c^4} T_{ik} \!</math>
*[[库仑定律]]:
::<math> F = \frac{\left|q_1q_2\right|}{4 \pi \varepsilon_0 r^2}\!</math>
*[[真空磁导率]]:
::<math> \mu_0 = 4 \pi \cdot 10^{-7}\,(\mathrm{N/A^2})\!</math>
*[[单摆]][[周期]]
::<math>T = 2\pi \sqrt\frac{L}{g}\!</math>

==参考来源==
{{reflist|2}}

==拓展阅读==
* Peter Borwein, ''[http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P159.pdf The Amazing Number Pi]{{Wayback|url=http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P159.pdf |date=20150923201439 }}''
* 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]]

[[Category:圆周率]]
[[Category:数学列表]]