From Wikipedia, the free encyclopedia
The following is a list of significant formulae involving the mathematical constant 𝜏 . The list contains only formulae whose significance is established either in the article on the formula itself, the article on tau, or the one on numerical approximations of 𝜏 .
C
=
τ
r
=
τ
2
d
=
π
d
{\displaystyle C=\tau r={\frac {\tau }{2}}d=\pi d\!}
where C is the circumference of a circle , r is the radius and d is the diameter.
A
=
τ
2
r
2
{\displaystyle A={\frac {\tau }{2}}r^{2}\!}
where A is the area of a circle and r is the radius.
V
=
2
τ
3
r
3
{\displaystyle V={\frac {2\tau }{3}}r^{3}\!}
where V is the volume of a sphere and r is the radius.
S
A
=
2
τ
r
2
{\displaystyle SA=2\tau r^{2}\!}
where SA is the surface area of a sphere and r is the radius.
∫
−
∞
∞
sech
(
x
)
d
x
=
τ
2
{\displaystyle \int \limits _{-\infty }^{\infty }{\text{sech}}(x)dx={\frac {\tau }{2}}\!}
∫
−
∞
∞
∫
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
=
τ
2
{\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={\frac {\tau }{2}}\!}
∫
−
1
1
1
−
x
2
d
x
=
τ
4
{\displaystyle \int \limits _{-1}^{1}{\sqrt {1-x^{2}}}\,dx={\frac {\tau }{4}}\!}
∫
−
1
1
d
x
1
−
x
2
=
τ
2
{\displaystyle \int \limits _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}={\frac {\tau }{2}}\!}
∫
−
∞
∞
d
x
1
+
x
2
=
τ
2
{\displaystyle \int \limits _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}={\frac {\tau }{2}}\!}
(integral form of arctan over its entire domain, giving the period of tan ).
∫
−
∞
∞
e
−
x
2
d
x
=
τ
2
{\displaystyle \int \limits _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\frac {\tau }{2}}}\!}
(see gaussian integral ).
∮
d
z
z
=
τ
i
{\displaystyle \oint {\frac {dz}{z}}=\tau i\!}
(when the path of integration winds once counterclockwise around 0. See also Cauchy's integral formula )
∫
−
∞
∞
sin
x
x
d
x
=
τ
2
{\displaystyle \int \limits _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx={\frac {\tau }{2}}\!}
∫
0
1
x
4
(
1
−
x
)
4
1
+
x
2
d
x
=
22
7
−
τ
2
{\displaystyle \int \limits _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-{\frac {\tau }{2}}\!}
(see also Proof that 22/7 exceeds π ).
Efficient infinite series [ edit ]
∑
k
=
0
∞
k
!
(
2
k
+
1
)
!
!
=
∑
k
=
0
∞
2
k
k
!
2
(
2
k
+
1
)
!
=
τ
4
{\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\tau }{4}}\!}
(see also double factorial )
12
∑
k
=
0
∞
(
−
1
)
k
(
6
k
)
!
(
13591409
+
545140134
k
)
(
3
k
)
!
(
k
!
)
3
640320
3
k
+
3
/
2
=
2
τ
{\displaystyle 12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {2}{\tau }}\!}
(see Chudnovsky algorithm )
2
2
9801
∑
k
=
0
∞
(
4
k
)
!
(
1103
+
26390
k
)
(
k
!
)
4
396
4
k
=
2
τ
{\displaystyle {\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {2}{\tau }}\!}
(see Srinivasa Ramanujan )
1
6
5
3
∑
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
)
=
τ
{\displaystyle {\frac {1}{6^{5}{\sqrt {3}}}}\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)=\tau \!}
[ 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
)
=
τ
2
{\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)={\frac {\tau }{2}}\!}
(see Bailey-Borwein-Plouffe formula )
1
2
5
∑
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
)
=
τ
{\displaystyle {\frac {1}{2^{5}}}\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)=\tau \!}
Other infinite series [ edit ]
[ 2]
ζ
(
2
)
=
1
1
2
+
1
2
2
+
1
3
2
+
1
4
2
+
⋯
=
τ
2
24
{\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\tau ^{2}}{24}}\!}
(see also Basel problem and Riemann zeta function )
ζ
(
4
)
=
1
1
4
+
1
2
4
+
1
3
4
+
1
4
4
+
⋯
=
τ
4
1440
{\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\tau ^{4}}{1440}}\!}
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
τ
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}\tau ^{2n}}{2(2n)!}}\!}
, where B 2n is a Bernoulli number .
∑
n
=
1
∞
3
n
−
1
2
2
n
−
1
ζ
(
n
+
1
)
=
τ
{\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{2^{2n-1}}}\,\zeta (n+1)=\tau \!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
1
=
1
1
−
1
3
+
1
5
−
1
7
+
1
9
−
⋯
=
arctan
1
=
τ
8
{\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 {\tau }{8}}\!}
(see Leibniz formula for pi )
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
2
=
1
1
2
+
1
3
2
+
1
5
2
+
1
7
2
+
⋯
=
τ
2
32
{\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 {\tau ^{2}}{32}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
3
=
1
1
3
−
1
3
3
+
1
5
3
−
1
7
3
+
⋯
=
τ
3
256
{\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 {\tau ^{3}}{256}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
4
=
1
1
4
+
1
3
4
+
1
5
4
+
1
7
4
+
⋯
=
τ
4
1536
{\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 {\tau ^{4}}{1536}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
5
=
1
1
5
−
1
3
5
+
1
5
5
−
1
7
5
+
⋯
=
5
τ
5
49152
{\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\tau ^{5}}{49152}}\!}
∑
n
=
0
∞
(
(
−
1
)
n
2
n
+
1
)
6
=
1
1
6
+
1
3
6
+
1
5
6
+
1
7
6
+
⋯
=
τ
6
61440
{\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 {\tau ^{6}}{61440}}\!}
τ
8
=
3
4
×
5
4
×
7
8
×
11
12
×
13
12
×
17
16
×
19
20
×
23
24
×
29
28
×
31
32
×
⋯
{\displaystyle {\frac {\tau }{8}}={\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.
τ
2
=
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 {\frac {\tau }{2}}={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 \!}
(Euler, 1748)
After the first two terms, 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 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.[ 3]
Weisstein, Eric W. "Pi Formulas", MathWorld
See also Machin-like formula .
τ
8
=
4
arctan
1
5
−
arctan
1
239
{\displaystyle {\frac {\tau }{8}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}\!}
(the original Machin's formula)
τ
8
=
arctan
1
{\displaystyle {\frac {\tau }{8}}=\arctan 1}
τ
8
=
arctan
1
2
+
arctan
1
3
{\displaystyle {\frac {\tau }{8}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}\!}
τ
8
=
2
arctan
1
2
−
arctan
1
7
{\displaystyle {\frac {\tau }{8}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}\!}
τ
8
=
2
arctan
1
3
+
arctan
1
7
{\displaystyle {\frac {\tau }{8}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}\!}
τ
8
=
5
arctan
1
7
+
2
arctan
3
79
{\displaystyle {\frac {\tau }{8}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}\!}
τ
8
=
12
arctan
1
49
+
32
arctan
1
57
−
5
arctan
1
239
+
12
arctan
1
110443
{\displaystyle {\frac {\tau }{8}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}\!}
τ
8
=
44
arctan
1
57
+
7
arctan
1
239
−
12
arctan
1
682
+
24
arctan
1
12943
{\displaystyle {\frac {\tau }{8}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}\!}
Some infinite series involving tau are:[ 4]
τ
=
2
Z
{\displaystyle \tau ={\frac {2}{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}}}\!}
τ
=
8
Z
{\displaystyle \tau ={\frac {8}{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}}}}
τ
=
8
Z
{\displaystyle \tau ={\frac {8}{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}}}\!}
τ
=
64
Z
{\displaystyle \tau ={\frac {64}{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}}}\!}
τ
=
27
2
Z
{\displaystyle \tau ={\frac {27}{2Z}}\!}
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}}}\!}
τ
=
15
3
Z
{\displaystyle \tau ={\frac {15{\sqrt {3}}}{Z}}\!}
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}}}\!}
τ
=
85
85
9
3
Z
{\displaystyle \tau ={\frac {85{\sqrt {85}}}{9{\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}}}\!}
τ
=
5
5
3
Z
{\displaystyle \tau ={\frac {5{\sqrt {5}}}{{\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}}}\!}
τ
=
4
3
Z
{\displaystyle \tau ={\frac {4{\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}}}\!}
τ
=
2
3
9
Z
{\displaystyle \tau ={\frac {2{\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}}}\!}
τ
=
4
11
11
Z
{\displaystyle \tau ={\frac {4{\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}}}\!}
τ
=
2
2
Z
{\displaystyle \tau ={\frac {\sqrt {2}}{2Z}}\!}
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}}}\!}
τ
=
8
5
5
Z
{\displaystyle \tau ={\frac {8{\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}}}\!}
τ
=
8
3
3
Z
{\displaystyle \tau ={\frac {8{\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}}}\!}
τ
=
8
Z
{\displaystyle \tau ={\frac {8}{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}}}\!}
τ
=
144
Z
{\displaystyle \tau ={\frac {144}{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}}}\!}
τ
=
7056
Z
{\displaystyle \tau ={\frac {7056}{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}}}\!}
where
(
x
)
n
{\displaystyle (x)_{n}\!}
is the Pochhammer symbol for the falling factorial.
∏
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
⋯
=
τ
4
{\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 {\tau }{4}}\!}
(see also Wallis product )
Vieta 's formula:
2
2
⋅
2
+
2
2
⋅
2
+
2
+
2
2
⋅
⋯
=
4
τ
{\displaystyle {\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \cdots ={\frac {4}{\tau }}\!}
Three continued fractions [ edit ]
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 .)
n
!
∼
τ
n
(
n
e
)
n
{\displaystyle n!\sim {\sqrt {\tau n}}\left({\frac {n}{e}}\right)^{n}\!}
(Stirling's approximation )
e
i
τ
=
1
{\displaystyle e^{i\tau }=1\!}
(Euler's identity )
∑
k
=
1
n
φ
(
k
)
∼
12
n
2
τ
2
{\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {12n^{2}}{\tau ^{2}}}\!}
(see Euler's totient function )
∑
k
=
1
n
φ
(
k
)
k
∼
24
n
τ
2
{\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {24n}{\tau ^{2}}}\!}
(see Euler's totient function )
Γ
(
1
2
)
=
τ
2
{\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\tau \over 2}}\!}
(see also gamma function )
τ
=
Γ
(
1
/
4
)
4
/
3
a
g
m
(
1
,
2
)
2
/
3
{\displaystyle \tau =\Gamma \left({1/4}\right)^{4/3}\mathrm {agm} (1,{\sqrt {2}})^{2/3}\!}
(where agm is the arithmetic-geometric mean )
lim
n
→
∞
1
n
2
∑
k
=
1
n
(
n
mod
k
)
=
1
−
τ
2
48
{\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n\;{\bmod {\;}}k)=1-{\frac {\tau ^{2}}{48}}\!}
(where mod is the modulo function which gives the rest of a division this formula is getting better for higher n)
τ
2
=
lim
n
→
∞
4
n
2
∑
k
=
1
n
n
2
−
k
2
{\displaystyle {\frac {\tau }{2}}=\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}}
(Riemann sum to evaluate the area of the unit circle)
τ
2
=
lim
n
→
∞
2
4
n
n
(
2
n
n
)
2
{\displaystyle {\frac {\tau }{2}}=\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}}
(by Stirling's approximation )
Λ
=
4
τ
G
3
c
2
ρ
{\displaystyle \Lambda ={{4\tau G} \over {3c^{2}}}\rho \!}
Δ
x
Δ
p
≥
h
2
τ
{\displaystyle \Delta x\,\Delta p\geq {\frac {h}{2\tau }}\!}
R
μ
ν
−
1
2
g
μ
ν
R
+
Λ
g
μ
ν
=
4
τ
G
c
4
T
μ
ν
{\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={4\tau G \over c^{4}}T_{\mu \nu }\!}
F
=
|
q
1
q
2
|
2
τ
ε
0
r
2
{\displaystyle F={\frac {\left|q_{1}q_{2}\right|}{2\tau \varepsilon _{0}r^{2}}}\!}
μ
0
=
2
τ
⋅
10
−
7
N
/
A
2
{\displaystyle \mu _{0}=2\tau \cdot 10^{-7}\,\mathrm {N/A^{2}} \!}
Period of a simple pendulum with small amplitude
T
≈
τ
L
g
{\displaystyle T\approx \tau {\sqrt {\frac {L}{g}}}\!}
Category:Pi
Pi
Category:Pi algorithms