含圓周率的公式列表

下面是一個涉及數學常數π的公式列表。

## 古典幾何

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

其中，$C$是一個圓的周長，$r$是半徑，$d$是直徑。

$$A = \pi r^2\!$$

其中$A$是一個圓的面積，$r$是半徑。

$$V = {4 \over 3}\pi r^3\!$$

其中，$V$是一個球體的體積，$r$是半徑。

$$A = 4\pi r^2\!$$

其中$A$是一個球體的表面積，$r$是半徑。

## 分析

### 積分

$$\int\limits_{-\infty}^{\infty} \text{sech}(x)dx = \pi \!$$

$$\int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x}} = \pi$$

$$\int\limits_{-1}^1 \sqrt{1-x^2}\,dx = \frac{\pi}{2}\!$$

$$\int\limits_{-1}^1\frac{dx}{\sqrt{1-x^2}} = \pi\!$$

$$\int\limits_{-\infty}^\infty\frac{dx}{1+x^2} = \pi\!$$

$\int\limits_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\!$ (參見 正態分布)

$\oint\frac{dz}{z}=2\pi i\!$ (參見 柯西積分公式)

$$\int\limits_{-\infty}^{\infty} \frac{\sin(x)}{x}\,dx=\pi \!$$

$\int\limits_0^1 {x^4(1-x)^4 \over 1+x^2}\,dx = {22 \over 7} - \pi\!$ (參見 證明22/7大於π)

### 高效的無窮級數

$\frac{\pi}{2}\!=\sum_{k=0}^\infty\frac{k!}{(2k+1)

}=\sum_{k=0}^\infty\frac{2^k k!^2}{(2k+1)!}$ (參見 雙階乘)

$\frac{1}{\pi}\!=12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + \frac{3}{2}}}$ (參見 楚德諾夫斯基演算法)

$\frac{1}{\pi}\!=\frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$ (參見拉馬努金)

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

以下是任意位的二進位的π計算：:

$\pi\!=\sum_{k = 0}^{\infty} \frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6}\right)$ (參見 貝利－波爾溫－普勞夫公式)

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

### 其他無窮級數

$\zeta(2) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}\!$ (參見巴塞爾問題和黎曼ζ函數)

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

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

$\frac{\pi}{4}\!=\sum_{n=0}^{\infty} {\left[ \frac{(-1)^{n}}{2n+1} \right] }^1 = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \arctan{1} = \int_{0}^{1} \frac{1}{1+x^2}dx$ (參見Π的萊布尼茨公式)

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

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

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

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

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

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

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

### 梅欽公式

參見梅欽公式.

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

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

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

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

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

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

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

### 無窮級數

一些涉及圓周率的無窮級數:

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

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

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

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

$\pi=\frac{27}{4Z}\!$｜$Z=\sum_{n=0}^{\infty } \left (\frac{2}{27} \right )^n \frac{(15n+2)\left ( \frac{1}{2} \right )_n \left ( \frac{1}{3} \right )_n \left ( \frac{2}{3} \right )_n} {(n!)^3}\!$

$\pi=\frac{15\sqrt{3}}{2Z}\!$｜$Z=\sum_{n=0}^{\infty } \left ( \frac{4}{125} \right )^n \frac{(33n+4)\left ( \frac{1}{2} \right )_n \left ( \frac{1}{3} \right )_n \left ( \frac{2}{3} \right )_n} {(n!)^3}\!$

$\pi=\frac{85\sqrt{85}}{18\sqrt{3}Z}\!$｜$Z=\sum_{n=0}^{\infty } \left ( \frac{4}{85} \right )^n \frac{(133n+8)\left ( \frac{1}{2} \right )_n \left ( \frac{1}{6} \right )_n \left ( \frac{5}{6} \right )_n} {(n!)^3}\!$

$\pi=\frac{5\sqrt{5}}{2\sqrt{3}Z} \!$｜$Z=\sum_{n=0}^{\infty } \left ( \frac{4}{125} \right )^n \frac{(11n+1)\left ( \frac{1}{2} \right )_n \left ( \frac{1}{6} \right )_n \left ( \frac{5}{6} \right )_n} {(n!)^3}\!$

$\pi=\frac{2\sqrt{3}}{Z} \!$｜$Z=\sum_{n=0}^{\infty } \frac{(8n+1)\left ( \frac{1}{2} \right )_n \left ( \frac{1}{4} \right )_n \left ( \frac{3}{4} \right )_n} {(n!)^3{9}^{n}}\!$

$\pi=\frac{\sqrt{3}}{9Z} \!$｜$Z=\sum_{n=0}^{\infty } \frac{(40n+3)\left ( \frac{1}{2} \right )_n \left ( \frac{1}{4} \right )_n \left ( \frac{3}{4} \right )_n} {(n!)^3{49}^{2n+1}}\!$

$\pi=\frac{2\sqrt{11}}{11Z} \!$｜$Z=\sum_{n=0}^{\infty } \frac{(280n+19)\left ( \frac{1}{2} \right )_n \left ( \frac{1}{4} \right )_n \left ( \frac{3}{4} \right )_n} {(n!)^3{99}^{2n+1}}\!$

$\pi=\frac{\sqrt{2}}{4Z} \!$｜$Z=\sum_{n=0}^{\infty } \frac{(10n+1) \left ( \frac{1}{2} \right )_n \left ( \frac{1}{4} \right )_n \left ( \frac{3}{4} \right )_n} {(n!)^3{9}^{2n+1}}\!$

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

$\pi=\frac{4\sqrt{3}}{3Z} \!$｜$Z=\sum_{n=0}^{\infty } \frac{(-1)^n(28n+3) \left ( \frac{1}{2} \right )_n \left ( \frac{1}{4} \right )_n \left ( \frac{3}{4} \right )_n} { (n!)^3{3^n}{4}^{n+1}}\!$

$\pi=\frac{4}{Z}\!$｜$Z=\sum_{n=0}^{\infty } \frac{(-1)^n(20n+3) \left ( \frac{1}{2} \right )_n \left ( \frac{1}{4} \right )_n \left ( \frac{3}{4} \right )_n} { (n!)^3{2}^{2n+1}}\!$

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

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

$(x)_n \!$是階乘冪中下降階乘冪的符號。

$\prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{4}{3} \cdot \frac{16}{15} \cdot \frac{36}{35} \cdot \frac{64}{63} \cdots = \frac{\pi}{2} \!$ (參見沃利斯乘積)

弗朗索瓦·韋達的公式:

$$\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots = \frac2\pi\!$$

### 連分數

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

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

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

(參見連分數。)

### 雜項

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

$e^{i \pi}+1=0$ (歐拉恆等式)

$$\sum_{k=1}^{n} \varphi (k) \approx \frac{3n^2}{\pi^2}\!$$

$$\sum_{k=1}^{n} \frac {\varphi (k)} {k} \approx \frac{6n}{\pi^2}\!$$

$\Gamma\left({1 \over 2}\right)=\sqrt{\pi}\!$ (伽瑪函數)

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

$$\lim_{n\rightarrow \infty}\frac{1}{n^2} \sum_{k=1}^n (n\;\bmod\;k) = 1-\frac{\pi^2}{12}\!$$

$$\lim_{n\rightarrow \infty} 10^{n+2}\cdot \sin\left(\frac{1^\circ}{\underbrace{55\cdots55^\circ}_{\mathrm{n\; digits}}}\right) = \pi\!$$

$$\lim_{n\rightarrow \infty} n\cdot \sin\left(\frac{180^\circ}{n}\right) = \pi$$

$$\lim_{n\rightarrow \infty} \frac n{\sqrt 2} \cdot \sqrt{1-\cos \left(\frac{360^\circ}n\right)} = \pi$$

## 物理

$$\Lambda = {{8\pi G} \over {3c^2}} \rho\!$$

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

$$R_{ik} - {g_{ik} R \over 2} + \Lambda g_{ik} = {8 \pi G \over c^4} T_{ik} \!$$

$$F = \frac{\left|q_1q_2\right|}{4 \pi \varepsilon_0 r^2}\!$$

$$\mu_0 = 4 \pi \cdot 10^{-7}\,(\mathrm{N/A^2})\!$$

- 單擺周期

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

## 拓展閱讀

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