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}
圆 的周长 ，
r
{\displaystyle r}
半径 ，
d
{\displaystyle d}
直径 。
A
=
π
r
2
{\displaystyle A=\pi r^{2}\!}
其中
A
{\displaystyle A}
面积 ，
r
{\displaystyle r}
V
=
4
3
π
r
3
{\displaystyle V={4 \over 3}\pi r^{3}\!}
其中，
V
{\displaystyle V}
球体 的体积 ，
r
{\displaystyle r}
A
=
4
π
r
2
{\displaystyle A=4\pi r^{2}\!}
其中
A
{\displaystyle A}
表面积 ，
r
{\displaystyle r}
∫
−
∞
∞
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 }\!}
正态分布 )
∮
d
z
z
=
2
π
i
{\displaystyle \oint {\frac {dz}{z}=2\pi i\!}
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 \!}
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)!}
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}
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}
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)}
[ 1]
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)}
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}\!}
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}
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 \!}
[ 2]
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}\!}
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:[ 3]
π
=
1
Z
{\displaystyle \pi ={\frac {1}{Z}\!}
Z
=
∑
n
=
0
∞
[
(
2
n
)
!
]
3
(
42
n
+
5
)
(
n
!
)
6
16
3
n
+
1
{\displaystyle Z=\sum _{n=0}^{\infty }{\frac {[(2n)!]^{3}(42n+5)}{(n!)^{6}{16}^{3n+1}\!}
π
=
4
Z
{\displaystyle \pi ={\frac {4}{Z}\!}
Z
=
∑
n
=
0
∞
(
−
1
)
n
(
4
n
)
!
(
21460
n
+
1123
)
(
n
!
)
4
441
2
n
+
1
2
10
n
+
1
{\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}{441}^{2n+1}{2}^{10n+1}
π
=
4
Z
{\displaystyle \pi ={\frac {4}{Z}\!}
Z
=
∑
n
=
0
∞
(
6
n
+
1
)
(
1
2
)
n
3
4
n
(
n
!
)
3
{\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(6n+1)\left({\frac {1}{2}\right)_{n}^{3}{4^{n}(n!)^{3}\!}
π
=
32
Z
{\displaystyle \pi ={\frac {32}{Z}\!}
Z
=
∑
n
=
0
∞
(
5
−
1
2
)
8
n
(
42
n
5
+
30
n
+
5
5
−
1
)
(
1
2
)
n
3
64
n
(
n
!
)
3
{\displaystyle 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)_{n}^{3}{64^{n}(n!)^{3}\!}
π
=
27
4
Z
{\displaystyle \pi ={\frac {27}{4Z}\!}
Z
=
∑
n
=
0
∞
(
2
27
)
n
(
15
n
+
2
)
(
1
2
)
n
(
1
3
)
n
(
2
3
)
n
(
n
!
)
3
{\displaystyle 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}\!}
π
=
15
3
2
Z
{\displaystyle \pi ={\frac {15{\sqrt {3}{2Z}\!}
Z
=
∑
n
=
0
∞
(
4
125
)
n
(
33
n
+
4
)
(
1
2
)
n
(
1
3
)
n
(
2
3
)
n
(
n
!
)
3
{\displaystyle 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}\!}
π
=
85
85
18
3
Z
{\displaystyle \pi ={\frac {85{\sqrt {85}{18{\sqrt {3}Z}\!}
Z
=
∑
n
=
0
∞
(
4
85
)
n
(
133
n
+
8
)
(
1
2
)
n
(
1
6
)
n
(
5
6
)
n
(
n
!
)
3
{\displaystyle 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}\!}
π
=
5
5
2
3
Z
{\displaystyle \pi ={\frac {5{\sqrt {5}{2{\sqrt {3}Z}\!}
Z
=
∑
n
=
0
∞
(
4
125
)
n
(
11
n
+
1
)
(
1
2
)
n
(
1
6
)
n
(
5
6
)
n
(
n
!
)
3
{\displaystyle 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}\!}
π
=
2
3
Z
{\displaystyle \pi ={\frac {2{\sqrt {3}{Z}\!}
Z
=
∑
n
=
0
∞
(
8
n
+
1
)
(
1
2
)
n
(
1
4
)
n
(
3
4
)
n
(
n
!
)
3
9
n
{\displaystyle 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}\!}
π
=
3
9
Z
{\displaystyle \pi ={\frac {\sqrt {3}{9Z}\!}
Z
=
∑
n
=
0
∞
(
40
n
+
3
)
(
1
2
)
n
(
1
4
)
n
(
3
4
)
n
(
n
!
)
3
49
2
n
+
1
{\displaystyle 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}\!}
π
=
2
11
11
Z
{\displaystyle \pi ={\frac {2{\sqrt {11}{11Z}\!}
Z
=
∑
n
=
0
∞
(
280
n
+
19
)
(
1
2
)
n
(
1
4
)
n
(
3
4
)
n
(
n
!
)
3
99
2
n
+
1
{\displaystyle 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}\!}
π
=
2
4
Z
{\displaystyle \pi ={\frac {\sqrt {2}{4Z}\!}
Z
=
∑
n
=
0
∞
(
10
n
+
1
)
(
1
2
)
n
(
1
4
)
n
(
3
4
)
n
(
n
!
)
3
9
2
n
+
1
{\displaystyle 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}\!}
π
=
4
5
5
Z
{\displaystyle \pi ={\frac {4{\sqrt {5}{5Z}\!}
Z
=
∑
n
=
0
∞
(
644
n
+
41
)
(
1
2
)
n
(
1
4
)
n
(
3
4
)
n
(
n
!
)
3
5
n
72
2
n
+
1
{\displaystyle 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!)^{3}5^{n}{72}^{2n+1}\!}
π
=
4
3
3
Z
{\displaystyle \pi ={\frac {4{\sqrt {3}{3Z}\!}
Z
=
∑
n
=
0
∞
(
−
1
)
n
(
28
n
+
3
)
(
1
2
)
n
(
1
4
)
n
(
3
4
)
n
(
n
!
)
3
3
n
4
n
+
1
{\displaystyle 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}\!}
π
=
4
Z
{\displaystyle \pi ={\frac {4}{Z}\!}
Z
=
∑
n
=
0
∞
(
−
1
)
n
(
20
n
+
3
)
(
1
2
)
n
(
1
4
)
n
(
3
4
)
n
(
n
!
)
3
2
2
n
+
1
{\displaystyle 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}\!}
π
=
72
Z
{\displaystyle \pi ={\frac {72}{Z}\!}
Z
=
∑
n
=
0
∞
(
−
1
)
n
(
4
n
)
!
(
260
n
+
23
)
(
n
!
)
4
4
4
n
18
2
n
{\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(260n+23)}{(n!)^{4}4^{4n}18^{2n}\!}
π
=
3528
Z
{\displaystyle \pi ={\frac {3528}{Z}\!}
Z
=
∑
n
=
0
∞
(
−
1
)
n
(
4
n
)
!
(
21460
n
+
1123
)
(
n
!
)
4
4
4
n
882
2
n
{\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}4^{4n}882^{2n}\!}
(
x
)
n
{\displaystyle (x)_{n}\!}
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}\!}
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 }
Λ
=
8
π
G
3
c
2
ρ
{\displaystyle \Lambda ={8\pi G} \over {3c^{2}\rho \!}
Δ
x
Δ
p
≥
h
4
π
{\displaystyle \Delta x\,\Delta p\geq {\frac {h}{4\pi }\!}
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}\!}
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}\!}
μ
0
=
4
π
⋅
10
−
7
(
N
/
A
2
)
{\displaystyle \mu _{0}=4\pi \cdot 10^{-7}\,(\mathrm {N/A^{2} )\!}
T
=
2
π
L
g
{\displaystyle T=2\pi {\sqrt {\frac {L}{g}\!}
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![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/399b4d2be3f36939933cbe1a4c5a86c2bd8be332)
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![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b671be5afec6ba1326ec4e081683afcd47836574)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd6ca07d5af6efbde89bce24360afd81ed0156ad)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/09ac41eaebdc8d66b6ce0457c518b902eb936d36)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a76463a9b8ab58d66003d0bb505099977ba3ec5e)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/39666a115f5f2adda13526f937853874a10d7b57)
![{\displaystyle Z=\sum _{n=0}^{\infty }{\frac {[(2n)!]^{3}(42n+5)}{(n!)^{6}{16}^{3n+1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3dd2d4f85dcbd50f54ad20084eb939e79743dd5)
