From Wikipedia, the free encyclopedia
The following is a list of significant formulae involving the mathematical constant π . The list contains only formulae whose significance is established either in the article on the formula itself, the article on pi, or the one on numerical approximations of pi .
Classical geometry
C
=
2
π
r
=
π
d
{\displaystyle C=2\pi r=\pi d\!}
where C circle , r d
A
=
π
r
2
{\displaystyle A=\pi r^{2}\!}
where A is the area of a circle and r
V
=
4
3
π
r
3
{\displaystyle V={4 \over 3}\pi r^{3}\!}
where V sphere and r
A
=
4
π
r
2
{\displaystyle A=4\pi r^{2}\!}
where A r
Analysis Integrals
∫
−
∞
∞
sech
(
x
)
d
x
=
π
{\displaystyle \int _{-\infty }^{\infty }{\text{sech}}(x)dx=\pi \!}
∫
−
1
1
1
−
x
2
d
x
=
π
2
{\displaystyle \int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx={\frac {\pi }{2}}\!}
∫
−
1
1
d
x
1
−
x
2
=
π
{\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi \!}
∫
−
∞
∞
d
x
1
+
x
2
=
π
{\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi \!}
arctan over its entire domain, giving the period of tan ).
∫
−
∞
∞
e
−
x
2
d
x
=
π
{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}\!}
normal distribution ).
∮
d
z
z
=
2
π
i
{\displaystyle \oint {\frac {dz}{z}}=2\pi i\!}
Cauchy's integral formula )
∫
0
∞
sin
(
x
)
x
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}\!}
∫
0
1
x
4
(
1
−
x
)
4
1
+
x
2
d
x
=
22
7
−
π
{\displaystyle \int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi \!}
Proof that 22/7 exceeds π ).
Efficient infinite series
∑
k
=
0
∞
k
!
(
2
k
+
1
)
!
!
=
∑
k
=
0
∞
2
k
k
!
2
(
2
k
+
1
)
!
=
π
2
{\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}\!}
double factorial )
12
∑
k
=
0
∞
(
−
1
)
k
(
6
k
)
!
(
13591409
+
545140134
k
)
(
3
k
)
!
(
k
!
)
3
640320
3
k
+
3
/
2
=
1
π
{\displaystyle 12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {1}{\pi }}\!}
Chudnovsky algorithm )
2
2
9801
∑
k
=
0
∞
(
4
k
)
!
(
1103
+
26390
k
)
(
k
!
)
4
396
4
k
=
1
π
{\displaystyle {\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {1}{\pi }}\!}
Srinivasa Ramanujan )
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
)
=
π
{\displaystyle {\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)=\pi \!}
[1]
The following are good for calculating arbitrary binary digits of π :
∑
k
=
0
∞
1
16
k
(
4
8
k
+
1
−
2
8
k
+
4
−
1
8
k
+
5
−
1
8
k
+
6
)
=
π
{\displaystyle \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)=\pi \!}
Bailey-Borwein-Plouffe formula )
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
)
=
π
{\displaystyle {\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)=\pi \!}
Other infinite series
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}}\!}
Basel problem and Riemann zeta function )
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)!}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
1
=
1
1
−
1
3
+
1
5
−
1
7
+
1
9
−
⋯
=
arctan
1
=
π
4
{\displaystyle \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}={\frac {\pi }{4}}\!}
Leibniz formula for pi )
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
2
=
1
1
2
+
1
3
2
+
1
5
2
+
1
7
2
+
⋯
=
π
2
8
{\displaystyle \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 ^{2}}{8}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
3
=
1
1
3
−
1
3
3
+
1
5
3
−
1
7
3
+
⋯
=
π
3
32
{\displaystyle \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 ^{3}}{32}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
4
=
1
1
4
+
1
3
4
+
1
5
4
+
1
7
4
+
⋯
=
π
4
96
{\displaystyle \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 {\pi ^{4}}{96}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
5
=
1
1
5
−
1
3
5
+
1
5
5
−
1
7
5
+
⋯
=
5
π
5
1536
{\displaystyle \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 {5\pi ^{5}}{1536}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
6
=
1
1
6
+
1
3
6
+
1
5
6
+
1
7
6
+
⋯
=
π
6
960
{\displaystyle \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 ^{6}}{960}}\!}
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 \!}
Euler )where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
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 \!}
The signs are determined as follows: If the denominator is a prime of the form 4m - 1, the sign is positive; if the denominator is 2 or a prime of the form 4m + 1, the sign is negative; for composite numbers, the sign is equal the product of the signs of its factors.[2]
Machin-like formulae See also Machin-like formula .
π
4
=
4
arctan
1
5
−
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}\!}
Machin's formula)
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}}\!}
Infinite products
∏
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}}\!}
Wallis product )Vieta 's formula:
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 }}\!}
Three continued fractions
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 }}}}}}}}}}\,}
For more on this third identity, see Euler's continued fraction formula .
(See also continued fraction and generalized continued fraction .)
Miscellaneous
n
!
∼
2
π
n
(
n
e
)
n
{\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\!}
Stirling's approximation )
e
i
π
=
−
1
{\displaystyle e^{i\pi }=-1\!}
Euler's identity )
∑
k
=
1
n
φ
(
k
)
∼
3
n
2
π
2
{\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}\!}
Euler's totient function )
∑
k
=
1
n
φ
(
k
)
k
∼
6
n
π
2
{\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}\!}
Euler's totient function )
Γ
(
1
2
)
=
π
{\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}\!}
gamma function )
π
=
Γ
(
1
/
4
)
4
/
3
a
g
m
(
1
,
2
)
2
/
3
2
{\displaystyle \pi ={\frac {\Gamma \left({1/4}\right)^{4/3}\mathrm {agm} (1,{\sqrt {2}})^{2/3}}{2}}\!}
arithmetic-geometric mean )
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
5555
⏟
n
d
i
g
i
t
s
)
=
π
{\displaystyle \lim _{n\rightarrow \infty }10^{n+2}\cdot \sin({\frac {1}{\underbrace {5555} _{\mathrm {n\;digits} }}})=\pi \!}
degrees , not radians )
Physics
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}} \!}
Period of a simple pendulum with small amplitude
T
≈
2
π
L
g
{\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}\!}
References Further reading 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. See also 
