积分
编辑
∫
−
∞
∞
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 证明22/7大于π)
高效的无穷级数
编辑
参见：Category:圆周率算法
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)}
[1]
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)}
其他无穷级数
编辑
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函数)
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)[2]
梅钦公式
编辑
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}}\!}
无穷级数
编辑
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}\!}
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 }}\!}
连分数
编辑
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。)
杂项
编辑
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}
(欧拉恒等式)
∑
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 }