含圆周率的公式列表

下面是一个涉及数学常数π的公式列表。 C = 2 π r = π d {\displaystyle C=2\pi r=\pi d\!} 其中， C {\displaystyle C} 是一个圆的周长， r {\displaystyle r} 是半径， d {\displaystyle d} 是直径。 A = π r 2 {\displaystyle A=\pi r^{2}\!} 其中 A {\displaystyle A} 是一个圆的面积， r {\displaystyle r} 是半径。 V = 4 3 π r 3 {\displaystyle V={4 \over 3}\pi r^{3}\!} 其中， V {\displaystyle V} 是一个球体的体积， r {\displaystyle r} 是半径。 A = 4 π r 2 {\displaystyle A=4\pi r^{2}\!} 其中 A {\displaystyle A} 是一个球体的表面积， r {\displaystyle r} 是半径。 ∫ − ∞ ∞ sech ( x ) d x = π {\displaystyle \int \limits _{-\infty }^{\infty }{\text{sech}}(x)dx=\pi \!} ∫ 0 ∞ d x ( x + 1 ) x = π {\displaystyle \int _{0}^{\infty }{\frac {dx}{(x+1){\sqrt {x}}}}=\pi } ∫ − 1 1 1 − x 2 d x = π 2 {\displaystyle \int \limits _{-1}^{1}{\sqrt {1-x^{2}}}\,dx={\frac {\pi }{2}}\!} ∫ − 1 1 d x 1 − x 2 = π {\displaystyle \int \limits _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi \!} ∫ − ∞ ∞ d x 1 + x 2 = π {\displaystyle \int \limits _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi \!} ∫ − ∞ ∞ e − x 2 d x = π {\displaystyle \int \limits _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}\!} (参见 正态分布) ∮ d z z = 2 π i {\displaystyle \oint {\frac {dz}{z}}=2\pi i\!} (参见 柯西积分公式) ∫ − ∞ ∞ sin ⁡ ( x ) x d x = π {\displaystyle \int \limits _{-\infty }^{\infty }{\frac {\sin(x)}{x}}\,dx=\pi \!} ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x = 22 7 − π {\displaystyle \int \limits _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi \!} (参见 證明22/7大於π) π 2 = ∑ k = 0 ∞ k ! ( 2 k + 1 ) ! ! = ∑ k = 0 ∞ 2 k k ! 2 ( 2 k + 1 ) ! {\displaystyle {\frac {\pi }{2}}\!=\sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}} (参见 双阶乘) 1 π = 12 ∑ k = 0 ∞ ( − 1 ) k ( 6 k ) ! ( 13591409 + 545140134 k ) ( 3 k ) ! ( k ! ) 3 640320 3 k + 3 2 {\displaystyle {\frac {1}{\pi }}\!=12\sum _{k=0}^{\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) 1 π = 2 2 9801 ∑ k = 0 ∞ ( 4 k ) ! ( 1103 + 26390 k ) ( k ! ) 4 396 4 k {\displaystyle {\frac {1}{\pi }}\!={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}} (参见拉马努金) π = 3 6 5 ∑ k = 0 ∞ [ ( 4 k ) ! ] 2 ( 6 k ) ! 9 k + 1 ( 12 k ) ! ( 2 k ) ! ( 127169 12 k + 1 − 1070 12 k + 5 − 131 12 k + 7 + 2 12 k + 11 ) {\displaystyle \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)} 以下是任意位的二进制的π计算：: π = ∑ k = 0 ∞ 1 16 k ( 4 8 k + 1 − 2 8 k + 4 − 1 8 k + 5 − 1 8 k + 6 ) {\displaystyle \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)} (参见 贝利－波尔温－普劳夫公式) π = 1 2 6 ∑ n = 0 ∞ ( − 1 ) n 2 10 n ( − 2 5 4 n + 1 − 1 4 n + 3 + 2 8 10 n + 1 − 2 6 10 n + 3 − 2 2 10 n + 5 − 2 2 10 n + 7 + 1 10 n + 9 ) {\displaystyle \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)} ζ ( 2 ) = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 {\displaystyle \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) ζ ( 4 ) = 1 1 4 + 1 2 4 + 1 3 4 + 1 4 4 + ⋯ = π 4 90 {\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}\!} ζ ( 2 n ) = 1 1 2 n + 1 2 2 n + 1 3 2 n + 1 4 2 n + ⋯ = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \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)!}}\!} π 4 = ∑ n = 0 ∞ [ ( − 1 ) n 2 n + 1 ] 1 = 1 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = arctan ⁡ 1 = ∫ 0 1 1 1 + x 2 d x {\displaystyle {\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}   (参见Π的莱布尼茨公式) π 2 8 = ∑ n = 0 ∞ [ ( − 1 ) n 2 n + 1 ] 2 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ {\displaystyle {\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 } π 3 32 = ∑ n = 0 ∞ [ ( − 1 ) n 2 n + 1 ] 3 = 1 1 3 − 1 3 3 + 1 5 3 − 1 7 3 + ⋯ {\displaystyle {\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 } π 4 96 = ∑ n = 0 ∞ [ ( − 1 ) n 2 n + 1 ] 4 = 1 1 4 + 1 3 4 + 1 5 4 + 1 7 4 + ⋯ {\displaystyle {\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 } 5 π 5 1536 = ∑ n = 0 ∞ [ ( − 1 ) n 2 n + 1 ] 5 = 1 1 5 − 1 3 5 + 1 5 5 − 1 7 5 + ⋯ {\displaystyle {\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 } π 6 960 = ∑ n = 0 ∞ [ ( − 1 ) n 2 n + 1 ] 6 = 1 1 6 + 1 3 6 + 1 5 6 + 1 7 6 + ⋯ {\displaystyle {\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 } π 4 = 3 4 × 5 4 × 7 8 × 11 12 × 13 12 × 17 16 × 19 20 × 23 24 × 29 28 × 31 32 × ⋯ {\displaystyle {\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) π = 1 + 1 2 + 1 3 + 1 4 − 1 5 + 1 6 + 1 7 + 1 8 + 1 9 − 1 10 + 1 11 + 1 12 − 1 13 + ⋯ {\displaystyle \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) 参见梅钦公式. π 4 = 4 arctan ⁡ 1 5 − arctan ⁡ 1 239 {\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}\!} (原始的梅钦公式.) π 4 = arctan ⁡ 1 2 + arctan ⁡ 1 3 {\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}\!} π 4 = 2 arctan ⁡ 1 2 − arctan ⁡ 1 7 {\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}\!} π 4 = 2 arctan ⁡ 1 3 + arctan ⁡ 1 7 {\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}\!} π 4 = 5 arctan ⁡ 1 7 + 2 arctan ⁡ 3 79 {\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}\!} π 4 = 12 arctan ⁡ 1 49 + 32 arctan ⁡ 1 57 − 5 arctan ⁡ 1 239 + 12 arctan ⁡ 1 110443 {\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}\!} π 4 = 44 arctan ⁡ 1 57 + 7 arctan ⁡ 1 239 − 12 arctan ⁡ 1 682 + 24 arctan ⁡ 1 12943 {\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}\!} 一些涉及圆周率的无穷级数: ( x ) n {\displaystyle (x)_{n}\!} 是阶乘幂中下降阶乘幂的符号。 ∏ n = 1 ∞ 4 n 2 4 n 2 − 1 = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ = 4 3 ⋅ 16 15 ⋅ 36 35 ⋅ 64 63 ⋯ = π 2 {\displaystyle \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}}\!} (参见沃利斯乘积) 弗朗索瓦·韦达的公式: 2 2 ⋅ 2 + 2 2 ⋅ 2 + 2 + 2 2 ⋅ ⋯ = 2 π {\displaystyle {\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 }}\!} π = 3 + 1 2 6 + 3 2 6 + 5 2 6 + 7 2 6 + ⋱ {\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}} π = 4 1 + 1 2 3 + 2 2 5 + 3 2 7 + 4 2 9 + ⋱ {\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}} π = 4 1 + 1 2 2 + 3 2 2 + 5 2 2 + 7 2 2 + ⋱ {\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}\,} (参见连分数。) n ! ≈ 2 π n ( n e ) n {\displaystyle n!\approx {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\!} (斯特灵公式) e i π + 1 = 0 {\displaystyle e^{i\pi }+1=0} (歐拉恆等式) ∑ k = 1 n φ ( k ) ≈ 3 n 2 π 2 {\displaystyle \sum _{k=1}^{n}\varphi (k)\approx {\frac {3n^{2}}{\pi ^{2}}}\!} ∑ k = 1 n φ ( k ) k ≈ 6 n π 2 {\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\approx {\frac {6n}{\pi ^{2}}}\!} Γ ( 1 2 ) = π {\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}\!} (伽玛函数) π = Γ ( 1 4 ) 4 3 a g m ( 1 , 2 ) 2 3 2 {\displaystyle \pi ={\frac {\Gamma \left({\frac {1}{4}}\right)^{\frac {4}{3}}\mathrm {agm} (1,{\sqrt {2}})^{\frac {2}{3}}}{2}}\!} lim n → ∞ 1 n 2 ∑ k = 1 n ( n mod k ) = 1 − π 2 12 {\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n\;{\bmod {\;}}k)=1-{\frac {\pi ^{2}}{12}}\!} lim n → ∞ 10 n + 2 ⋅ sin ⁡ ( 1 ∘ 55 ⋯ 55 ∘ ⏟ n d i g i t s ) = π {\displaystyle \lim _{n\rightarrow \infty }10^{n+2}\cdot \sin \left({\frac {1^{\circ }}{\underbrace {55\cdots 55^{\circ }} _{\mathrm {n\;digits} }}}\right)=\pi \!} lim n → ∞ n ⋅ sin ⁡ ( 180 ∘ n ) = π {\displaystyle \lim _{n\rightarrow \infty }n\cdot \sin \left({\frac {180^{\circ }}{n}}\right)=\pi } lim n → ∞ n 2 ⋅ 1 − cos ⁡ ( 360 ∘ n ) = π {\displaystyle \lim _{n\rightarrow \infty }{\frac {n}{\sqrt {2}}}\cdot {\sqrt {1-\cos \left({\frac {360^{\circ }}{n}}\right)}}=\pi } 宇宙常数: Λ = 8 π G 3 c 2 ρ {\displaystyle \Lambda ={{8\pi G} \over {3c^{2}}}\rho \!} 不确定性原理: Δ x Δ p ≥ h 4 π {\displaystyle \Delta x\,\Delta p\geq {\frac {h}{4\pi }}\!} 爱因斯坦场方程: R i k − g i k R 2 + Λ g i k = 8 π G c 4 T i k {\displaystyle R_{ik}-{g_{ik}R \over 2}+\Lambda g_{ik}={8\pi G \over c^{4}}T_{ik}\!} 库仑定律: F = | q 1 q 2 | 4 π ε 0 r 2 {\displaystyle F={\frac {\left|q_{1}q_{2}\right|}{4\pi \varepsilon _{0}r^{2}}}\!} 真空磁导率: μ 0 = 4 π ⋅ 10 − 7 ( N / A 2 ) {\displaystyle \mu _{0}=4\pi \cdot 10^{-7}\,(\mathrm {N/A^{2}} )\!} 单摆周期 T = 2 π L g {\displaystyle T=2\pi {\sqrt {\frac {L}{g}}}\!} Cetin Hakimoglu-Brown Derivation of Rapidly Converging Infinite Series （页面存档备份，存于互联网档案馆） Carl B. Boyer, A History of Mathematics, Chapter 21. Simon Plouffe / David Bailey. The world of Pi. Pi314.net. [2011-01-29]. （原始内容存档于2013-06-23）.  Collection of series for π. Numbers.computation.free.fr. [2011-01-29]. （原始内容存档于2013-06-23）.  Peter Borwein, The Amazing Number Pi（页面存档备份，存于互联网档案馆） Kazuya Kato, Nobushige Kurokawa, Saito Takeshi: Number Theory 1: Fermat's Dream. American Mathematical Society, Providence 1993, ISBN 0-8218-0863-X. 圓周率