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Parte de uma série de artigos sobre:
a constante matemática π
3.1415926535 89793 23846 26433...
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Esta é uma lista de fórmulas significantes envolvendo a constantes matemática
π
{\displaystyle \scriptstyle {\pi }}
π
=
C
/
d
{\displaystyle \pi =C/d\!}
sendo C perímetro da circunferência e d
A
=
π
r
2
{\displaystyle A=\pi r^{2}\!}
sendo A a área da circunferência e r
V
=
4
3
π
r
3
{\displaystyle V={4 \over 3}\pi r^{3}\!}
sendo V esfera e r
S
A
=
4
π
r
2
{\displaystyle SA=4\pi r^{2}\!}
sendo SA r
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
μ
ν
−
1
2
g
μ
ν
R
+
Λ
g
μ
ν
=
8
π
G
c
4
T
μ
ν
{\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }\!}
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}} \!}
Período de um pêndulo com pequena amplitude:
T
≈
2
π
L
g
{\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}\!}
F
=
π
2
E
I
L
2
{\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}}
∫
−
∞
∞
sech
(
x
)
d
x
=
π
{\displaystyle \int \limits _{-\infty }^{\infty }{\text{sech}}(x)dx=\pi \!}
∫
−
∞
∞
∫
t
∞
e
−
1
/
2
t
2
−
x
2
+
x
t
d
x
d
t
=
∫
−
∞
∞
∫
t
∞
e
−
t
2
−
1
/
2
x
2
+
x
t
d
x
d
t
=
π
{\displaystyle \int \limits _{-\infty }^{\infty }\int \limits _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt}dxdt=\int \limits _{-\infty }^{\infty }\int \limits _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt}dxdt=\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 \!}
arctan sobre o domínio dos inteiros, dando o período da tan ).
∫
−
∞
∞
e
−
x
2
d
x
=
π
{\displaystyle \int \limits _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}\!}
Integral Gaussiana ).
∮
d
z
z
=
2
π
i
{\displaystyle \oint {\frac {dz}{z}}=2\pi i\!}
fórmula integral de Cauchy )
∫
−
∞
∞
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 \!}
prova de que 22/7 é maior que π ).
∑
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}}\!}
duplo fatorial )
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 }}\!}
algoritmo de Chudnovsky )
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 }}\!}
Séries de Ramanujan–Sato )
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]
As seguintes são eficientes para calcular dígitos binários arbitrários de pi:
∑
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 \!}
fórmula BBP )
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 \!}
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}}\!}
problema de Basileia e função zeta de Riemann )
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
)
=
∑
k
=
1
∞
1
k
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)=\sum _{k=1}^{\infty }{\frac {1}{k^{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)!}}\!}
B 2n um números de Bernoulli .
∑
n
=
1
∞
3
n
−
1
4
n
ζ
(
n
+
1
)
=
π
{\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi \!}
[ 2]
∑
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}}\!}
Fórmula de Leibniz para π )
∑
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
=
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 \!}
Após os primeiros dois termos, os sinais são determinados como segue: Se o denominador é primo da forma 4m - 1, o sinal é positivo; se o denominador é um primo da forma 4m + 1, o sinal é negativo; para números compostos, o sinal é igual ao produto dos seus sinais em seus fatores.[ 3] Também:
∑
n
=
1
∞
F
2
n
n
2
(
2
n
n
)
=
4
π
2
25
5
{\displaystyle \sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}}
onde
F
n
{\displaystyle F_{n}}
número de Fibonacci . Ver também fórmula de Machin .
(Unidades dos ângulos em radianos)
π
4
=
4
arctan
1
5
−
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}\!}
John Machin )
π
4
=
arctan
1
{\displaystyle {\frac {\pi }{4}}=\arctan 1}
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
=
6
arctan
1
8
+
2
arctan
1
57
+
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}\!}
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}}\!}
π
2
=
∑
n
=
0
∞
arctan
1
F
2
n
+
1
=
arctan
1
1
+
arctan
1
2
+
arctan
1
5
+
arctan
1
13
+
⋯
{\displaystyle {\frac {\pi }{2}}=\sum _{n=0}^{\infty }\arctan {\frac {1}{F_{2n+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots \!}
sendo
F
n
{\displaystyle F_{n}}
número de Fibonacci .
Algumas séries infinitas envolvendo pi são:[ 4]
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}}}\!}
onde
(
x
)
n
{\displaystyle (x)_{n}\!}
é o símbolo de Pochhammer .
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}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{16}}\cdot {\frac {19}{20}}\cdot {\frac {23}{24}}\cdot {\frac {29}{28}}\cdot {\frac {31}{32}}\cdots \!}
Euler )onde os numeradores são primos ímpares; cada denominador é um múltiplo de quatro próximo ao numerador.
∏
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 )
Fórmula da Viète :
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 }}}}}}}}}}\,}
Para mais informações sobre esta terceira identidade ver fração contínua de Euler .
(Ver também fração contínua e fração contínua generalizada .)
n
!
∼
2
π
n
(
n
e
)
n
{\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\!}
fórmula de Stirling )
e
i
π
+
1
=
0
{\displaystyle e^{i\pi }+1=0\!}
identidade de Euler )
∑
k
=
1
n
φ
(
k
)
∼
3
n
2
π
2
{\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}\!}
função totiente de Euler )
∑
k
=
1
n
φ
(
k
)
k
∼
6
n
π
2
{\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}\!}
função totiente de Euler )
Γ
(
1
2
)
=
π
{\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}\!}
função gama )
π
=
Γ
(
1
/
4
)
4
/
3
agm
(
1
,
2
)
2
/
3
2
{\displaystyle \pi ={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}\!}
média aritmética-geométrica )
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}}\!}
π
=
lim
n
→
∞
4
n
2
∑
k
=
1
n
n
2
−
k
2
{\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}}
soma de Riemann para avaliar a área da circunferência unitária)
π
=
lim
n
→
∞
2
4
n
n
(
2
n
n
)
2
{\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}}
fórmula de Stirling )
