The following is a list of significant formulae involving the
mathematical constant
π . Many of these formulae can be found in the article
Pi , or the article
Approximations of π .
π
=
C
d
=
C
2
r
{\displaystyle \pi ={\frac {C}{d}}={\frac {C}{2r}}}
where C is the
circumference of a
circle , d is the
diameter , and r is the
radius . More generally,
π
=
L
w
{\displaystyle \pi ={\frac {L}{w}}}
where L and w are, respectively, the
perimeter and the width of any
curve of constant width .
A
=
π
r
2
{\displaystyle A=\pi r^{2}}
where A is the
area of a circle . More generally,
A
=
π
a
b
{\displaystyle A=\pi ab}
where A is the area enclosed by an
ellipse with semi-major axis a and semi-minor axis b .
C
=
2
π
agm
(
a
,
b
)
(
a
1
2
−
∑
n
=
2
∞
2
n
−
1
(
a
n
2
−
b
n
2
)
)
{\displaystyle C={\frac {2\pi }{\operatorname {agm} (a,b)}}\left(a_{1}^{2}-\sum _{n=2}^{\infty }2^{n-1}(a_{n}^{2}-b_{n}^{2})\right)}
where C is the circumference of an ellipse with semi-major axis a and semi-minor axis b and
a
n
,
b
n
{\displaystyle a_{n},b_{n}}
are the arithmetic and geometric iterations of
agm
(
a
,
b
)
{\displaystyle \operatorname {agm} (a,b)}
, the
arithmetic-geometric mean of a and b with the initial values
a
0
=
a
{\displaystyle a_{0}=a}
and
b
0
=
b
{\displaystyle b_{0}=b}
.
A
=
4
π
r
2
{\displaystyle A=4\pi r^{2}}
where A is the area between the
witch of Agnesi and its asymptotic line; r is the radius of the defining circle.
A
=
Γ
(
1
/
4
)
2
2
π
r
2
=
π
r
2
agm
(
1
,
1
/
2
)
{\displaystyle A={\frac {\Gamma (1/4)^{2}}{2{\sqrt {\pi }}}}r^{2}={\frac {\pi r^{2}}{\operatorname {agm} (1,1/{\sqrt {2}})}}}
where A is the area of a
squircle with minor radius r ,
Γ
{\displaystyle \Gamma }
is the
gamma function .
A
=
(
k
+
1
)
(
k
+
2
)
π
r
2
{\displaystyle A=(k+1)(k+2)\pi r^{2}}
where A is the area of an
epicycloid with the smaller circle of radius r and the larger circle of radius kr (
k
∈
N
{\displaystyle k\in \mathbb {N} }
), assuming the initial point lies on the larger circle.
A
=
(
−
1
)
k
+
3
8
π
a
2
{\displaystyle A={\frac {(-1)^{k}+3}{8}}\pi a^{2}}
where A is the area of a
rose with angular frequency k (
k
∈
N
{\displaystyle k\in \mathbb {N} }
) and amplitude a .
L
=
Γ
(
1
/
4
)
2
π
c
=
2
π
c
agm
(
1
,
1
/
2
)
{\displaystyle L={\frac {\Gamma (1/4)^{2}}{\sqrt {\pi }}}c={\frac {2\pi c}{\operatorname {agm} (1,1/{\sqrt {2}})}}}
where L is the perimeter of the
lemniscate of Bernoulli with focal distance c .
V
=
4
3
π
r
3
{\displaystyle V={4 \over 3}\pi r^{3}}
where V is the volume of a
sphere and r is the radius.
S
A
=
4
π
r
2
{\displaystyle SA=4\pi r^{2}}
where SA is the surface area of a sphere and r is the radius.
H
=
1
2
π
2
r
4
{\displaystyle H={1 \over 2}\pi ^{2}r^{4}}
where H is the hypervolume of a
3-sphere and r is the radius.
S
V
=
2
π
2
r
3
{\displaystyle SV=2\pi ^{2}r^{3}}
where SV is the surface volume of a 3-sphere and r is the radius.
Sum S of internal angles of a
regular convex polygon with n sides:
S
=
(
n
−
2
)
π
{\displaystyle S=(n-2)\pi }
Area A of a regular convex polygon with n sides and side length s :
A
=
n
s
2
4
cot
π
n
{\displaystyle A={\frac {ns^{2}}{4}}\cot {\frac {\pi }{n}}}
Inradius r of a regular convex polygon with n sides and side length s :
r
=
s
2
cot
π
n
{\displaystyle r={\frac {s}{2}}\cot {\frac {\pi }{n}}}
Circumradius R of a regular convex polygon with n sides and side length s :
R
=
s
2
csc
π
n
{\displaystyle R={\frac {s}{2}}\csc {\frac {\pi }{n}}}
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 {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}}
μ
0
≈
4
π
⋅
10
−
7
N
/
A
2
{\displaystyle \mu _{0}\approx 4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}}
Approximate period of a simple
pendulum with small amplitude:
T
≈
2
π
L
g
{\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}}
Exact period of a simple pendulum with amplitude
θ
0
{\displaystyle \theta _{0}}
(
agm
{\displaystyle \operatorname {agm} }
is the
arithmetic–geometric mean ):
T
=
2
π
agm
(
1
,
cos
(
θ
0
/
2
)
)
L
g
{\displaystyle T={\frac {2\pi }{\operatorname {agm} (1,\cos(\theta _{0}/2))}}{\sqrt {\frac {L}{g}}}}
R
3
T
2
=
G
M
4
π
2
{\displaystyle {\frac {R^{3}}{T^{2}}}={\frac {GM}{4\pi ^{2}}}}
F
=
π
2
E
I
L
2
{\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}}
A puzzle involving "colliding billiard balls":
⌊
b
N
π
⌋
{\displaystyle \lfloor {b^{N}\pi }\rfloor }
is the number of collisions made (in ideal conditions, perfectly elastic with no friction) by an object of mass m initially at rest between a fixed wall and another object of mass b 2N m , when struck by the other object.
[1] (This gives the digits of π in base b up to N digits past the radix point.)
2
∫
−
1
1
1
−
x
2
d
x
=
π
{\displaystyle 2\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx=\pi }
(integrating two halves
y
(
x
)
=
1
−
x
2
{\displaystyle y(x)={\sqrt {1-x^{2}}}}
to obtain the area of the unit circle)
∫
−
∞
∞
sech
x
d
x
=
π
{\displaystyle \int _{-\infty }^{\infty }\operatorname {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 _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt}\,dx\,dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt}\,dx\,dt=\pi }
∫
−
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 }
[2]
[note 2] (see also
Cauchy distribution )
∫
−
∞
∞
sin
x
x
d
x
=
π
{\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi }
(see
Dirichlet integral )
∫
−
∞
∞
e
−
x
2
d
x
=
π
{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}
(see
Gaussian integral ).
∮
d
z
z
=
2
π
i
{\displaystyle \oint {\frac {dz}{z}}=2\pi i}
(when the path of integration winds once counterclockwise around 0. See also
Cauchy's integral formula ).
∫
0
∞
ln
(
1
+
1
x
2
)
d
x
=
π
{\displaystyle \int _{0}^{\infty }\ln \left(1+{\frac {1}{x^{2}}}\right)\,dx=\pi }
[3]
∫
−
∞
∞
sin
x
x
d
x
=
π
{\displaystyle \int _{-\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 _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi }
(see also
Proof that 22/7 exceeds π ).
∫
0
∞
x
α
−
1
x
+
1
d
x
=
π
sin
π
α
,
0
<
α
<
1
{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{x+1}}\,dx={\frac {\pi }{\sin \pi \alpha }},\quad 0<\alpha <1}
∫
0
∞
d
x
x
(
x
+
a
)
(
x
+
b
)
=
π
agm
(
a
,
b
)
{\displaystyle \int _{0}^{\infty }{\frac {dx}{\sqrt {x(x+a)(x+b)}}}={\frac {\pi }{\operatorname {agm} ({\sqrt {a}},{\sqrt {b}})}}}
(where
agm
{\displaystyle \operatorname {agm} }
is the
arithmetic–geometric mean ;
[4] see also
elliptic integral )
Note that with symmetric integrands
f
(
−
x
)
=
f
(
x
)
{\displaystyle f(-x)=f(x)}
, formulas of the form
∫
−
a
a
f
(
x
)
d
x
{\textstyle \int _{-a}^{a}f(x)\,dx}
can also be translated to formulas
2
∫
0
a
f
(
x
)
d
x
{\textstyle 2\int _{0}^{a}f(x)\,dx}
.
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}}}
(see also
Double factorial )
∑
k
=
0
∞
k
!
(
2
k
)
!
(
25
k
−
3
)
(
3
k
)
!
2
k
=
π
2
{\displaystyle \sum _{k=0}^{\infty }{\frac {k!\,(2k)!\,(25k-3)}{(3k)!\,2^{k}}}={\frac {\pi }{2}}}
∑
k
=
0
∞
(
−
1
)
k
(
6
k
)
!
(
13591409
+
545140134
k
)
(
3
k
)
!
(
k
!
)
3
640320
3
k
=
4270934400
10005
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}}={\frac {4270934400}{{\sqrt {10005}}\pi }}}
(see
Chudnovsky algorithm )
∑
k
=
0
∞
(
4
k
)
!
(
1103
+
26390
k
)
(
k
!
)
4
396
4
k
=
9801
2
2
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {9801}{2{\sqrt {2}}\pi }}}
(see
Srinivasa Ramanujan ,
Ramanujan–Sato series )
The following are efficient for calculating arbitrary binary digits of π :
∑
k
=
0
∞
(
−
1
)
k
4
k
(
2
4
k
+
1
+
2
4
k
+
2
+
1
4
k
+
3
)
=
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{4^{k}}}\left({\frac {2}{4k+1}}+{\frac {2}{4k+2}}+{\frac {1}{4k+3}}\right)=\pi }
[5]
∑
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 }
(see
Bailey–Borwein–Plouffe formula )
∑
k
=
0
∞
1
16
k
(
8
8
k
+
2
+
4
8
k
+
3
+
4
8
k
+
4
−
1
8
k
+
7
)
=
2
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {8}{8k+2}}+{\frac {4}{8k+3}}+{\frac {4}{8k+4}}-{\frac {1}{8k+7}}\right)=2\pi }
∑
k
=
0
∞
(
−
1
)
k
2
10
k
(
−
2
5
4
k
+
1
−
1
4
k
+
3
+
2
8
10
k
+
1
−
2
6
10
k
+
3
−
2
2
10
k
+
5
−
2
2
10
k
+
7
+
1
10
k
+
9
)
=
2
6
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {{(-1)}^{k}}{2^{10k}}}\left(-{\frac {2^{5}}{4k+1}}-{\frac {1}{4k+3}}+{\frac {2^{8}}{10k+1}}-{\frac {2^{6}}{10k+3}}-{\frac {2^{2}}{10k+5}}-{\frac {2^{2}}{10k+7}}+{\frac {1}{10k+9}}\right)=2^{6}\pi }
Plouffe's series for calculating arbitrary decimal digits of π :
[6]
∑
k
=
1
∞
k
2
k
k
!
2
(
2
k
)
!
=
π
+
3
{\displaystyle \sum _{k=1}^{\infty }k{\frac {2^{k}k!^{2}}{(2k)!}}=\pi +3}
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}}}
(see also
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
)
=
∑
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)!}}}
, where B 2n is a
Bernoulli number .
∑
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 }
[7]
∑
n
=
2
∞
2
(
3
/
2
)
n
−
3
n
(
ζ
(
n
)
−
1
)
=
ln
π
{\displaystyle \sum _{n=2}^{\infty }{\frac {2(3/2)^{n}-3}{n}}(\zeta (n)-1)=\ln \pi }
∑
n
=
1
∞
ζ
(
2
n
)
x
2
n
n
=
ln
π
x
sin
π
x
,
0
<
|
x
|
<
1
{\displaystyle \sum _{n=1}^{\infty }\zeta (2n){\frac {x^{2n}}{n}}=\ln {\frac {\pi x}{\sin \pi x}},\quad 0<|x|<1}
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
=
1
−
1
3
+
1
5
−
1
7
+
1
9
−
⋯
=
arctan
1
=
π
4
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}}
(see
Leibniz formula for pi )
∑
n
=
0
∞
(
−
1
)
(
n
2
−
n
)
/
2
2
n
+
1
=
1
+
1
3
−
1
5
−
1
7
+
1
9
+
1
11
−
⋯
=
π
2
2
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{(n^{2}-n)/2}}{2n+1}}=1+{\frac {1}{3}}-{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}-\cdots ={\frac {\pi }{2{\sqrt {2}}}}}
(
Newton , Second Letter to Oldenburg , 1676)
[8]
∑
n
=
0
∞
(
−
1
)
n
3
n
(
2
n
+
1
)
=
1
−
1
3
1
⋅
3
+
1
3
2
⋅
5
−
1
3
3
⋅
7
+
1
3
4
⋅
9
−
⋯
=
3
arctan
1
3
=
π
2
3
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}(2n+1)}}=1-{\frac {1}{3^{1}\cdot 3}}+{\frac {1}{3^{2}\cdot 5}}-{\frac {1}{3^{3}\cdot 7}}+{\frac {1}{3^{4}\cdot 9}}-\cdots ={\sqrt {3}}\arctan {\frac {1}{\sqrt {3}}}={\frac {\pi }{2{\sqrt {3}}}}}
(
Madhava series )
∑
n
=
1
∞
(
−
1
)
n
+
1
n
2
=
1
1
2
−
1
2
2
+
1
3
2
−
1
4
2
+
⋯
=
π
2
12
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}}
∑
n
=
1
∞
1
(
2
n
)
2
=
1
2
2
+
1
4
2
+
1
6
2
+
1
8
2
+
⋯
=
π
2
24
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}}
∑
n
=
0
∞
(
1
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}{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
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}{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
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}{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}}}
In general,
∑
n
=
0
∞
(
−
1
)
n
(
2
n
+
1
)
2
k
+
1
=
(
−
1
)
k
E
2
k
2
(
2
k
)
!
(
π
2
)
2
k
+
1
,
k
∈
N
0
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{2k+1}}}=(-1)^{k}{\frac {E_{2k}}{2(2k)!}}\left({\frac {\pi }{2}}\right)^{2k+1},\quad k\in \mathbb {N} _{0}}
where
E
2
k
{\displaystyle E_{2k}}
is the
2
k
{\displaystyle 2k}
th
Euler number .
[9]
∑
n
=
0
∞
(
1
2
n
)
(
−
1
)
n
2
n
+
1
=
1
−
1
6
−
1
40
−
⋯
=
π
4
{\displaystyle \sum _{n=0}^{\infty }{\binom {\frac {1}{2}}{n}}{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{6}}-{\frac {1}{40}}-\cdots ={\frac {\pi }{4}}}
∑
n
=
0
∞
1
(
4
n
+
1
)
(
4
n
+
3
)
=
1
1
⋅
3
+
1
5
⋅
7
+
1
9
⋅
11
+
⋯
=
π
8
{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}}
∑
n
=
1
∞
(
−
1
)
(
n
2
+
n
)
/
2
+
1
|
G
(
(
−
1
)
n
+
1
+
6
n
−
3
)
/
4
|
=
|
G
1
|
+
|
G
2
|
−
|
G
4
|
−
|
G
5
|
+
|
G
7
|
+
|
G
8
|
−
|
G
10
|
−
|
G
11
|
+
⋯
=
3
π
{\displaystyle \sum _{n=1}^{\infty }(-1)^{(n^{2}+n)/2+1}\left|G_{\left((-1)^{n+1}+6n-3\right)/4}\right|=|G_{1}|+|G_{2}|-|G_{4}|-|G_{5}|+|G_{7}|+|G_{8}|-|G_{10}|-|G_{11}|+\cdots ={\frac {\sqrt {3}}{\pi }}}
(see
Gregory coefficients )
∑
n
=
0
∞
(
1
/
2
)
n
2
2
n
n
!
2
∑
n
=
0
∞
n
(
1
/
2
)
n
2
2
n
n
!
2
=
1
π
{\displaystyle \sum _{n=0}^{\infty }{\frac {(1/2)_{n}^{2}}{2^{n}n!^{2}}}\sum _{n=0}^{\infty }{\frac {n(1/2)_{n}^{2}}{2^{n}n!^{2}}}={\frac {1}{\pi }}}
(where
(
x
)
n
{\displaystyle (x)_{n}}
is the
rising factorial )
[10]
∑
n
=
1
∞
(
−
1
)
n
+
1
n
(
n
+
1
)
(
2
n
+
1
)
=
π
−
3
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(2n+1)}}=\pi -3}
(
Nilakantha series)
∑
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}}}}}
(where
F
n
{\displaystyle F_{n}}
is the n -th
Fibonacci number )
∑
n
=
1
∞
σ
(
n
)
e
−
2
π
n
=
1
24
−
1
8
π
{\displaystyle \sum _{n=1}^{\infty }\sigma (n)e^{-2\pi n}={\frac {1}{24}}-{\frac {1}{8\pi }}}
(where
σ
{\displaystyle \sigma }
is the
sum-of-divisors function )
π
=
∑
n
=
1
∞
(
−
1
)
ϵ
(
n
)
n
=
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 =\sum _{n=1}^{\infty }{\frac {(-1)^{\epsilon (n)}}{n}}=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 }
(where
ϵ
(
n
)
{\displaystyle \epsilon (n)}
is the number of prime factors of the form
p
≡
1
(
m
o
d
4
)
{\displaystyle p\equiv 1\,(\mathrm {mod} \,4)}
of
n
{\displaystyle n}
)
[11]
[12]
π
2
=
∑
n
=
1
∞
(
−
1
)
ε
(
n
)
n
=
1
+
1
2
−
1
3
+
1
4
+
1
5
−
1
6
−
1
7
+
1
8
+
1
9
+
⋯
{\displaystyle {\frac {\pi }{2}}=\sum _{n=1}^{\infty }{\frac {(-1)^{\varepsilon (n)}}{n}}=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}}+\cdots }
(where
ε
(
n
)
{\displaystyle \varepsilon (n)}
is the number of prime factors of the form
p
≡
3
(
m
o
d
4
)
{\displaystyle p\equiv 3\,(\mathrm {mod} \,4)}
of
n
{\displaystyle n}
)
[13]
π
=
∑
n
=
−
∞
∞
(
−
1
)
n
n
+
1
/
2
{\displaystyle \pi =\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{n+1/2}}}
π
2
=
∑
n
=
−
∞
∞
1
(
n
+
1
/
2
)
2
{\displaystyle \pi ^{2}=\sum _{n=-\infty }^{\infty }{\frac {1}{(n+1/2)^{2}}}}
[14]
The last two formulas are special cases of
π
sin
π
x
=
∑
n
=
−
∞
∞
(
−
1
)
n
n
+
x
(
π
sin
π
x
)
2
=
∑
n
=
−
∞
∞
1
(
n
+
x
)
2
{\displaystyle {\begin{aligned}{\frac {\pi }{\sin \pi x}}&=\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{n+x}}\\\left({\frac {\pi }{\sin \pi x}}\right)^{2}&=\sum _{n=-\infty }^{\infty }{\frac {1}{(n+x)^{2}}}\end{aligned}}}
which generate infinitely many analogous formulas for
π
{\displaystyle \pi }
when
x
∈
Q
∖
Z
.
{\displaystyle x\in \mathbb {Q} \setminus \mathbb {Z} .}
Some formulas relating π and harmonic numbers are given
here . Further infinite series involving π are:
[15]
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}}}}
where
(
x
)
n
{\displaystyle (x)_{n}}
is the Pochhammer symbol for the rising factorial. See also
Ramanujan–Sato series .
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
=
4
arctan
1
5
−
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}
(the original
Machin's formula)
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
4
=
(
∏
p
≡
1
(
mod
4
)
p
p
−
1
)
⋅
(
∏
p
≡
3
(
mod
4
)
p
p
+
1
)
=
3
4
⋅
5
4
⋅
7
8
⋅
11
12
⋅
13
12
⋯
,
{\displaystyle {\frac {\pi }{4}}=\left(\prod _{p\equiv 1{\pmod {4}}}{\frac {p}{p-1}}\right)\cdot \left(\prod _{p\equiv 3{\pmod {4}}}{\frac {p}{p+1}}\right)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,}
(Euler)
where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
3
π
6
=
(
∏
p
≡
1
(
mod
6
)
p
∈
P
p
p
−
1
)
⋅
(
∏
p
≡
5
(
mod
6
)
p
∈
P
p
p
+
1
)
=
5
6
⋅
7
6
⋅
11
12
⋅
13
12
⋅
17
18
⋯
,
{\displaystyle {\frac {{\sqrt {3}}\pi }{6}}=\left(\displaystyle \prod _{p\equiv 1{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p-1}}\right)\cdot \left(\displaystyle \prod _{p\equiv 5{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p+1}}\right)={\frac {5}{6}}\cdot {\frac {7}{6}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{18}}\cdots ,}
π
2
=
∏
n
=
1
∞
(
2
n
)
(
2
n
)
(
2
n
−
1
)
(
2
n
+
1
)
=
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
6
5
⋅
6
7
⋅
8
7
⋅
8
9
⋯
{\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+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 }
(see also
Wallis product )
π
2
=
∏
n
=
1
∞
(
1
+
1
n
)
(
−
1
)
n
+
1
=
(
1
+
1
1
)
+
1
(
1
+
1
2
)
−
1
(
1
+
1
3
)
+
1
⋯
{\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{n}}\right)^{(-1)^{n+1}}=\left(1+{\frac {1}{1}}\right)^{+1}\left(1+{\frac {1}{2}}\right)^{-1}\left(1+{\frac {1}{3}}\right)^{+1}\cdots }
(another form of Wallis product)
Viète's formula :
2
π
=
2
2
⋅
2
+
2
2
⋅
2
+
2
+
2
2
⋅
⋯
{\displaystyle {\frac {2}{\pi }}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \cdots }
A double infinite product formula involving the
Thue–Morse sequence :
π
2
=
∏
m
≥
1
∏
n
≥
1
(
(
4
m
2
+
n
−
2
)
(
4
m
2
+
2
n
−
1
)
2
4
(
2
m
2
+
n
−
1
)
(
4
m
2
+
n
−
1
)
(
2
m
2
+
n
)
)
ϵ
n
,
{\displaystyle {\frac {\pi }{2}}=\prod _{m\geq 1}\prod _{n\geq 1}\left({\frac {(4m^{2}+n-2)(4m^{2}+2n-1)^{2}}{4(2m^{2}+n-1)(4m^{2}+n-1)(2m^{2}+n)}}\right)^{\epsilon _{n}},}
where
ϵ
n
=
(
−
1
)
t
n
{\displaystyle \epsilon _{n}=(-1)^{t_{n}}}
and
t
n
{\displaystyle t_{n}}
is the Thue–Morse sequence (
Tóth 2020 ).
π
2
k
+
1
=
arctan
2
−
a
k
−
1
a
k
,
k
≥
2
{\displaystyle {\frac {\pi }{2^{k+1}}}=\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},\qquad \qquad k\geq 2}
π
4
=
∑
k
≥
2
arctan
2
−
a
k
−
1
a
k
,
{\displaystyle {\frac {\pi }{4}}=\sum _{k\geq 2}\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},}
where
a
k
=
2
+
a
k
−
1
{\displaystyle a_{k}={\sqrt {2+a_{k-1}}}}
such that
a
1
=
2
{\displaystyle a_{1}={\sqrt {2}}}
.
π
2
=
∑
k
=
0
∞
arctan
1
F
2
k
+
1
=
arctan
1
1
+
arctan
1
2
+
arctan
1
5
+
arctan
1
13
+
⋯
{\displaystyle {\frac {\pi }{2}}=\sum _{k=0}^{\infty }\arctan {\frac {1}{F_{2k+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots }
where
F
k
{\displaystyle F_{k}}
is the k -th Fibonacci number.
π
=
arctan
a
+
arctan
b
+
arctan
c
{\displaystyle \pi =\arctan a+\arctan b+\arctan c}
whenever
a
+
b
+
c
=
a
b
c
{\displaystyle a+b+c=abc}
and
a
{\displaystyle a}
,
b
{\displaystyle b}
,
c
{\displaystyle c}
are positive real numbers (see
List of trigonometric identities ). A special case is
π
=
arctan
1
+
arctan
2
+
arctan
3.
{\displaystyle \pi =\arctan 1+\arctan 2+\arctan 3.}
e
i
π
+
1
=
0
{\displaystyle e^{i\pi }+1=0}
(
Euler's identity )
The following equivalences are true for any
complex
z
{\displaystyle z}
:
e
z
∈
R
↔
ℑ
z
∈
π
Z
{\displaystyle e^{z}\in \mathbb {R} \leftrightarrow \Im z\in \pi \mathbb {Z} }
e
z
=
1
↔
z
∈
2
π
i
Z
{\displaystyle e^{z}=1\leftrightarrow z\in 2\pi i\mathbb {Z} }
[16]
Also
1
e
z
−
1
=
lim
N
→
∞
∑
n
=
−
N
N
1
z
−
2
π
i
n
−
1
2
,
z
∈
C
.
{\displaystyle {\frac {1}{e^{z}-1}}=\lim _{N\to \infty }\sum _{n=-N}^{N}{\frac {1}{z-2\pi in}}-{\frac {1}{2}},\quad z\in \mathbb {C} .}
Suppose a
lattice
Ω
{\displaystyle \Omega }
is generated by two periods
ω
1
,
ω
2
{\displaystyle \omega _{1},\omega _{2}}
. We define the quasi-periods of this lattice by
η
1
=
ζ
(
z
+
ω
1
;
Ω
)
−
ζ
(
z
;
Ω
)
{\displaystyle \eta _{1}=\zeta (z+\omega _{1};\Omega )-\zeta (z;\Omega )}
and
η
2
=
ζ
(
z
+
ω
2
;
Ω
)
−
ζ
(
z
;
Ω
)
{\displaystyle \eta _{2}=\zeta (z+\omega _{2};\Omega )-\zeta (z;\Omega )}
where
ζ
{\displaystyle \zeta }
is the
Weierstrass zeta function (
η
1
{\displaystyle \eta _{1}}
and
η
2
{\displaystyle \eta _{2}}
are in fact independent of
z
{\displaystyle z}
). Then the periods and quasi-periods are related by the Legendre identity :
η
1
ω
2
−
η
2
ω
1
=
2
π
i
.
{\displaystyle \eta _{1}\omega _{2}-\eta _{2}\omega _{1}=2\pi i.}
4
π
=
1
+
1
2
2
+
3
2
2
+
5
2
2
+
7
2
2
+
⋱
{\displaystyle {\frac {4}{\pi }}=1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}
[17]
ϖ
2
π
=
2
+
1
2
4
+
3
2
4
+
5
2
4
+
7
2
4
+
⋱
{\displaystyle {\frac {\varpi ^{2}}{\pi }}={2+{\cfrac {1^{2}}{4+{\cfrac {3^{2}}{4+{\cfrac {5^{2}}{4+{\cfrac {7^{2}}{4+\ddots \,}}}}}}}}}\quad }
(
Ramanujan ,
ϖ
{\displaystyle \varpi }
is the
lemniscate constant )
[18]
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 \,}}}}}}}}}}
[17]
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 }}}}}}}}}}}
2
π
=
6
+
2
2
12
+
6
2
12
+
10
2
12
+
14
2
12
+
18
2
12
+
⋱
{\displaystyle 2\pi ={6+{\cfrac {2^{2}}{12+{\cfrac {6^{2}}{12+{\cfrac {10^{2}}{12+{\cfrac {14^{2}}{12+{\cfrac {18^{2}}{12+\ddots }}}}}}}}}}}}
For more on the fourth identity, see
Euler's continued fraction formula .
(See also
Continued fraction and
Generalized continued fraction .)
a
0
=
1
,
a
n
+
1
=
(
1
+
1
2
n
+
1
)
a
n
,
π
=
lim
n
→
∞
a
n
2
n
{\displaystyle a_{0}=1,\,a_{n+1}=\left(1+{\frac {1}{2n+1}}\right)a_{n},\,\pi =\lim _{n\to \infty }{\frac {a_{n}^{2}}{n}}}
a
1
=
0
,
a
n
+
1
=
2
+
a
n
,
π
=
lim
n
→
∞
2
n
2
−
a
n
{\displaystyle a_{1}=0,\,a_{n+1}={\sqrt {2+a_{n}}},\,\pi =\lim _{n\to \infty }2^{n}{\sqrt {2-a_{n}}}}
(closely related to Viète's formula)
ω
(
i
n
,
i
n
−
1
,
…
,
i
1
)
=
2
+
i
n
2
+
i
n
−
1
2
+
⋯
+
i
1
2
=
ω
(
b
n
,
b
n
−
1
,
…
,
b
1
)
,
i
k
∈
{
−
1
,
1
}
,
b
k
=
{
0
if
i
k
=
1
1
if
i
k
=
−
1
,
π
=
lim
n
→
∞
2
n
+
1
2
h
+
1
ω
(
10
…
0
⏟
n
−
m
g
m
,
h
+
1
)
{\displaystyle \omega (i_{n},i_{n-1},\dots ,i_{1})=2+i_{n}{\sqrt {2+i_{n-1}{\sqrt {2+\cdots +i_{1}{\sqrt {2}}}}}}=\omega (b_{n},b_{n-1},\dots ,b_{1}),\,i_{k}\in \{-1,1\},\,b_{k}={\begin{cases}0&{\text{if }}i_{k}=1\\1&{\text{if }}i_{k}=-1\end{cases}},\,\pi ={\displaystyle \lim _{n\rightarrow \infty }{\frac {2^{n+1}}{2h+1}}{\sqrt {\omega \left(\underbrace {10\ldots 0} _{n-m}g_{m,h+1}\right)}}}}
(where
g
m
,
h
+
1
{\displaystyle g_{m,h+1}}
is the h+1-th entry of
m-bit Gray code ,
h
∈
{
0
,
1
,
…
,
2
m
−
1
}
{\displaystyle h\in \left\{0,1,\ldots ,2^{m}-1\right\}}
)
[19]
∀
k
∈
N
,
a
1
=
2
−
k
,
a
n
+
1
=
a
n
+
2
−
k
(
1
−
tan
(
2
k
−
1
a
n
)
)
,
π
=
2
k
+
1
lim
n
→
∞
a
n
{\displaystyle \forall k\in \mathbb {N} ,\,a_{1}=2^{-k},\,a_{n+1}=a_{n}+2^{-k}(1-\tan(2^{k-1}a_{n})),\,\pi =2^{k+1}\lim _{n\to \infty }a_{n}}
(quadratic convergence)
[20]
a
1
=
1
,
a
n
+
1
=
a
n
+
sin
a
n
,
π
=
lim
n
→
∞
a
n
{\displaystyle a_{1}=1,\,a_{n+1}=a_{n}+\sin a_{n},\,\pi =\lim _{n\to \infty }a_{n}}
(cubic convergence)
[21]
a
0
=
2
3
,
b
0
=
3
,
a
n
+
1
=
hm
(
a
n
,
b
n
)
,
b
n
+
1
=
gm
(
a
n
+
1
,
b
n
)
,
π
=
lim
n
→
∞
a
n
=
lim
n
→
∞
b
n
{\displaystyle a_{0}=2{\sqrt {3}},\,b_{0}=3,\,a_{n+1}=\operatorname {hm} (a_{n},b_{n}),\,b_{n+1}=\operatorname {gm} (a_{n+1},b_{n}),\,\pi =\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}
(
Archimedes' algorithm, see also
harmonic mean and
geometric mean )
[22]
For more iterative algorithms, see the
Gauss–Legendre algorithm and
Borwein's algorithm .
(
2
n
n
)
∼
4
n
π
n
{\displaystyle {\binom {2n}{n}}\sim {\frac {4^{n}}{\sqrt {\pi n}}}}
(asymptotic growth rate of the
central binomial coefficients )
C
n
∼
4
n
π
n
3
{\displaystyle C_{n}\sim {\frac {4^{n}}{\sqrt {\pi n^{3}}}}}
(asymptotic growth rate of the
Catalan numbers )
n
!
∼
2
π
n
(
n
e
)
n
{\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}
(
Stirling's approximation )
log
n
!
≃
(
n
+
1
2
)
log
n
−
n
+
log
2
π
2
{\displaystyle \log n!\simeq \left(n+{\frac {1}{2}}\right)\log n-n+{\frac {\log 2\pi }{2}}}
∑
k
=
1
n
φ
(
k
)
∼
3
n
2
π
2
{\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}}
(where
φ
{\displaystyle \varphi }
is
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}}}}
The symbol
∼
{\displaystyle \sim }
means that the ratio of the left-hand side and the right-hand side tends to one as
n
→
∞
{\displaystyle n\to \infty }
.
The symbol
≃
{\displaystyle \simeq }
means that the difference between the left-hand side and the right-hand side tends to zero as
n
→
∞
{\displaystyle n\to \infty }
.
Hypergeometric inversions
With
2
F
1
{\displaystyle {}_{2}F_{1}}
being the
hypergeometric function :
∑
n
=
0
∞
r
2
(
n
)
q
n
=
2
F
1
(
1
2
,
1
2
,
1
,
z
)
{\displaystyle \sum _{n=0}^{\infty }r_{2}(n)q^{n}={}_{2}F_{1}\left({\frac {1}{2}},{\frac {1}{2}},1,z\right)}
where
q
=
exp
(
−
π
2
F
1
(
1
/
2
,
1
/
2
,
1
,
1
−
z
)
2
F
1
(
1
/
2
,
1
/
2
,
1
,
z
)
)
,
z
∈
C
∖
{
0
,
1
}
{\displaystyle q=\exp \left(-\pi {\frac {{}_{2}F_{1}(1/2,1/2,1,1-z)}{{}_{2}F_{1}(1/2,1/2,1,z)}}\right),\quad z\in \mathbb {C} \setminus \{0,1\}}
and
r
2
{\displaystyle r_{2}}
is the
sum of two squares function .
Similarly,
1
+
240
∑
n
=
1
∞
σ
3
(
n
)
q
n
=
2
F
1
(
1
6
,
5
6
,
1
,
z
)
4
{\displaystyle 1+240\sum _{n=1}^{\infty }\sigma _{3}(n)q^{n}={}_{2}F_{1}\left({\frac {1}{6}},{\frac {5}{6}},1,z\right)^{4}}
where
q
=
exp
(
−
2
π
2
F
1
(
1
/
6
,
5
/
6
,
1
,
1
−
z
)
2
F
1
(
1
/
6
,
5
/
6
,
1
,
z
)
)
,
z
∈
C
∖
{
0
,
1
}
{\displaystyle q=\exp \left(-2\pi {\frac {{}_{2}F_{1}(1/6,5/6,1,1-z)}{{}_{2}F_{1}(1/6,5/6,1,z)}}\right),\quad z\in \mathbb {C} \setminus \{0,1\}}
and
σ
3
{\displaystyle \sigma _{3}}
is a
divisor function .
More formulas of this nature can be given, as explained by
Ramanujan's theory of
elliptic functions to alternative bases.
Perhaps the most notable hypergeometric inversions are the following two examples, involving the
Ramanujan tau function
τ
{\displaystyle \tau }
and the Fourier coefficients
j
{\displaystyle \mathrm {j} }
of the
J-invariant (
OEIS :
A000521 ):
∑
n
=
−
1
∞
j
n
q
n
=
256
(
1
−
z
+
z
2
)
3
z
2
(
1
−
z
)
2
,
{\displaystyle \sum _{n=-1}^{\infty }\mathrm {j} _{n}q^{n}=256{\dfrac {(1-z+z^{2})^{3}}{z^{2}(1-z)^{2}}},}
∑
n
=
1
∞
τ
(
n
)
q
n
=
z
2
(
1
−
z
)
2
256
2
F
1
(
1
2
,
1
2
,
1
,
z
)
12
{\displaystyle \sum _{n=1}^{\infty }\tau (n)q^{n}={\dfrac {z^{2}(1-z)^{2}}{256}}{}_{2}F_{1}\left({\frac {1}{2}},{\frac {1}{2}},1,z\right)^{12}}
where in both cases
q
=
exp
(
−
2
π
2
F
1
(
1
/
2
,
1
/
2
,
1
,
1
−
z
)
2
F
1
(
1
/
2
,
1
/
2
,
1
,
z
)
)
,
z
∈
C
∖
{
0
,
1
}
.
{\displaystyle q=\exp \left(-2\pi {\frac {{}_{2}F_{1}(1/2,1/2,1,1-z)}{{}_{2}F_{1}(1/2,1/2,1,z)}}\right),\quad z\in \mathbb {C} \setminus \{0,1\}.}
Furthermore, by expanding the last expression as a power series in
1
2
1
−
(
1
−
z
)
1
/
4
1
+
(
1
−
z
)
1
/
4
{\displaystyle {\dfrac {1}{2}}{\dfrac {1-(1-z)^{1/4}}{1+(1-z)^{1/4}}}}
and setting
z
=
1
/
2
{\displaystyle z=1/2}
, we obtain a rapidly convergent series for
e
−
2
π
{\displaystyle e^{-2\pi }}
:
[note 3]
e
−
2
π
=
w
2
+
4
w
6
+
34
w
10
+
360
w
14
+
4239
w
18
+
⋯
,
w
=
1
2
2
1
/
4
−
1
2
1
/
4
+
1
.
{\displaystyle e^{-2\pi }=w^{2}+4w^{6}+34w^{10}+360w^{14}+4239w^{18}+\cdots ,\quad w={\dfrac {1}{2}}{\dfrac {2^{1/4}-1}{2^{1/4}+1}}.}
Γ
(
s
)
Γ
(
1
−
s
)
=
π
sin
π
s
{\displaystyle \Gamma (s)\Gamma (1-s)={\frac {\pi }{\sin \pi s}}}
(Euler's reflection formula, see
Gamma function )
π
−
s
/
2
Γ
(
s
2
)
ζ
(
s
)
=
π
−
(
1
−
s
)
/
2
Γ
(
1
−
s
2
)
ζ
(
1
−
s
)
{\displaystyle \pi ^{-s/2}\Gamma \left({\frac {s}{2}}\right)\zeta (s)=\pi ^{-(1-s)/2}\Gamma \left({\frac {1-s}{2}}\right)\zeta (1-s)}
(the functional equation of the Riemann zeta function)
e
−
ζ
′
(
0
)
=
2
π
{\displaystyle e^{-\zeta '(0)}={\sqrt {2\pi }}}
e
ζ
′
(
0
,
1
/
2
)
−
ζ
′
(
0
,
1
)
=
π
{\displaystyle e^{\zeta '(0,1/2)-\zeta '(0,1)}={\sqrt {\pi }}}
(where
ζ
(
s
,
a
)
{\displaystyle \zeta (s,a)}
is the
Hurwitz zeta function and the derivative is taken with respect to the first variable)
π
=
B
(
1
/
2
,
1
/
2
)
=
Γ
(
1
/
2
)
2
{\displaystyle \pi =\mathrm {B} (1/2,1/2)=\Gamma (1/2)^{2}}
(see also
Beta function )
π
=
Γ
(
3
/
4
)
4
agm
(
1
,
1
/
2
)
2
=
Γ
(
1
/
4
)
4
/
3
agm
(
1
,
2
)
2
/
3
2
{\displaystyle \pi ={\frac {\Gamma (3/4)^{4}}{\operatorname {agm} (1,1/{\sqrt {2}})^{2}}}={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}}
(where agm is the
arithmetic–geometric mean )
π
=
agm
(
θ
2
2
(
1
/
e
)
,
θ
3
2
(
1
/
e
)
)
{\displaystyle \pi =\operatorname {agm} \left(\theta _{2}^{2}(1/e),\theta _{3}^{2}(1/e)\right)}
(where
θ
2
{\displaystyle \theta _{2}}
and
θ
3
{\displaystyle \theta _{3}}
are the
Jacobi theta functions
[23] )
agm
(
1
,
2
)
=
π
ϖ
{\displaystyle \operatorname {agm} (1,{\sqrt {2}})={\frac {\pi }{\varpi }}}
(due to
Gauss ,
[24]
ϖ
{\displaystyle \varpi }
is the
lemniscate constant )
N
(
2
ϖ
)
=
e
2
π
,
N
(
ϖ
)
=
e
π
/
2
{\displaystyle \operatorname {N} (2\varpi )=e^{2\pi },\quad \operatorname {N} (\varpi )=e^{\pi /2}}
(where
N
{\displaystyle \operatorname {N} }
is the
Gauss N-function )
i
π
=
Log
(
−
1
)
=
lim
n
→
∞
n
(
(
−
1
)
1
/
n
−
1
)
{\displaystyle i\pi =\operatorname {Log} (-1)=\lim _{n\to \infty }n\left((-1)^{1/n}-1\right)}
(where
Log
{\displaystyle \operatorname {Log} }
is the principal value of the
complex logarithm )
[note 4]
1
−
π
2
12
=
lim
n
→
∞
1
n
2
∑
k
=
1
n
(
n
mod
k
)
{\displaystyle 1-{\frac {\pi ^{2}}{12}}=\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n{\bmod {k}})}
(where
n
mod
k
{\textstyle n{\bmod {k}}}
is the
remainder upon division of n by k )
π
=
lim
r
→
∞
1
r
2
∑
x
=
−
r
r
∑
y
=
−
r
r
{
1
if
x
2
+
y
2
≤
r
0
if
x
2
+
y
2
>
r
{\displaystyle \pi =\lim _{r\to \infty }{\frac {1}{r^{2}}}\sum _{x=-r}^{r}\;\sum _{y=-r}^{r}{\begin{cases}1&{\text{if }}{\sqrt {x^{2}+y^{2}}}\leq r\\0&{\text{if }}{\sqrt {x^{2}+y^{2}}}>r\end{cases}}}
(summing a circle's area)
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}}}}
(
Riemann sum to evaluate the area of the unit circle)
π
=
lim
n
→
∞
2
4
n
n
!
4
n
(
2
n
)
!
2
=
lim
n
→
∞
2
4
n
n
(
2
n
n
)
2
=
lim
n
→
∞
1
n
(
(
2
n
)
!
!
(
2
n
−
1
)
!
!
)
2
{\displaystyle \pi =\lim _{n\to \infty }{\frac {2^{4n}n!^{4}}{n(2n)!^{2}}}=\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}=\lim _{n\rightarrow \infty }{\frac {1}{n}}\left({\frac {(2n)!!}{(2n-1)!!}}\right)^{2}}
(by combining Stirling's approximation with Wallis product)
π
=
lim
n
→
∞
1
n
ln
16
λ
(
n
i
)
{\displaystyle \pi =\lim _{n\to \infty }{\frac {1}{n}}\ln {\frac {16}{\lambda (ni)}}}
(where
λ
{\displaystyle \lambda }
is the
modular lambda function )
[25]
[note 5]
π
=
lim
n
→
∞
24
n
ln
(
2
1
/
4
G
n
)
=
lim
n
→
∞
24
n
ln
(
2
1
/
4
g
n
)
{\displaystyle \pi =\lim _{n\to \infty }{\frac {24}{\sqrt {n}}}\ln \left(2^{1/4}G_{n}\right)=\lim _{n\to \infty }{\frac {24}{\sqrt {n}}}\ln \left(2^{1/4}g_{n}\right)}
(where
G
n
{\displaystyle G_{n}}
and
g
n
{\displaystyle g_{n}}
are
Ramanujan's class invariants )
[26]
[note 6]
^ The relation
μ
0
=
4
π
⋅
10
−
7
N
/
A
2
{\displaystyle \mu _{0}=4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}}
was valid until the
2019 redefinition of the SI base units .
^ (integral form of
arctan over its entire domain, giving the period of
tan )
^ The coefficients can be obtained by
reversing the
Puiseux series of
z
↦
z
∑
n
=
0
∞
z
2
n
2
+
2
n
∑
n
=
−
∞
∞
z
2
n
2
{\displaystyle z\mapsto {\sqrt {z}}{\dfrac {\sum _{n=0}^{\infty }z^{2n^{2}+2n}}{\sum _{n=-\infty }^{\infty }z^{2n^{2}}}}}
at
z
=
0
{\displaystyle z=0}
.
^ The
n
{\displaystyle n}
th root with the smallest positive
principal argument is chosen.
^ When
n
∈
Q
+
{\displaystyle n\in \mathbb {Q} ^{+}}
, this gives
algebraic approximations to
Gelfond's constant
e
π
{\displaystyle e^{\pi }}
.
^ When
n
∈
Q
+
{\displaystyle {\sqrt {n}}\in \mathbb {Q} ^{+}}
, this gives algebraic approximations to Gelfond's constant
e
π
{\displaystyle e^{\pi }}
.
^ Galperin, G. (2003).
"Playing pool with π (the number π from a billiard point of view)" (PDF) . Regular and Chaotic Dynamics . 8 (4): 375–394.
doi :
10.1070/RD2003v008n04ABEH000252 .
^
Rudin, Walter (1987). Real and Complex Analysis (Third ed.). McGraw-Hill Book Company.
ISBN
0-07-100276-6 . p. 4
^
A000796 – OEIS
^ Carson, B. C. (2010),
"Elliptic Integrals" , in
Olver, Frank W. J. ; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.),
NIST Handbook of Mathematical Functions , Cambridge University Press,
ISBN
978-0-521-19225-5 ,
MR
2723248 .
^ Arndt, Jörg; Haenel, Christoph (2001). π Unleashed . Springer-Verlag Berlin Heidelberg.
ISBN
978-3-540-66572-4 . page 126
^ Gourdon, Xavier.
"Computation of the n-th decimal digit of π with low memory" (PDF) . Numbers, constants and computation . p. 1.
^
Weisstein, Eric W. "Pi Formulas", MathWorld
^ Chrystal, G. (1900). Algebra, an Elementary Text-book: Part II . p. 335.
^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi . American Mathematical Society.
ISBN
0-8218-3246-8 . p. 112
^ Cooper, Shaun (2017). Ramanujan's Theta Functions (First ed.). Springer.
ISBN
978-3-319-56171-4 . (page 647)
^ Euler, Leonhard (1748). Introductio in analysin infinitorum (in Latin). Vol. 1. p. 245
^
Carl B. Boyer , A History of Mathematics , Chapter 21., pp. 488–489
^ Euler, Leonhard (1748). Introductio in analysin infinitorum (in Latin). Vol. 1. p. 244
^ Wästlund, Johan.
"Summing inverse squares by euclidean geometry" (PDF) . The paper gives the formula with a minus sign instead, but these results are equivalent.
^ Simon Plouffe / David Bailey.
"The world of Pi" . Pi314.net. Retrieved 2011-01-29 .
"Collection of series for π " . Numbers.computation.free.fr. Retrieved 2011-01-29 .
^
Rudin, Walter (1987). Real and Complex Analysis (Third ed.). McGraw-Hill Book Company.
ISBN
0-07-100276-6 . p. 3
^
a
b Loya, Paul (2017). Amazing and Aesthetic Aspects of Analysis . Springer. p. 589.
ISBN
978-1-4939-6793-3 .
^
Perron, Oskar (1957). Die Lehre von den Kettenbrüchen: Band II (in German) (Third ed.). B. G. Teubner. p. 36, eq. 24
^ Vellucci, Pierluigi; Bersani, Alberto Maria (2019-12-01).
"$$\pi $$-Formulas and Gray code" . Ricerche di Matematica . 68 (2): 551–569.
arXiv :
1606.09597 .
doi :
10.1007/s11587-018-0426-4 .
ISSN
1827-3491 .
S2CID
119578297 .
^ Abrarov, Sanjar M.; Siddiqui, Rehan; Jagpal, Rajinder K.; Quine, Brendan M. (2021-09-04).
"Algorithmic Determination of a Large Integer in the Two-Term Machin-like Formula for π" . Mathematics . 9 (17): 2162.
arXiv :
2107.01027 .
doi :
10.3390/math9172162 .
^ Arndt, Jörg; Haenel, Christoph (2001). π Unleashed . Springer-Verlag Berlin Heidelberg.
ISBN
978-3-540-66572-4 . page 49
^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi . American Mathematical Society.
ISBN
0-8218-3246-8 . p. 2
^ Borwein, Jonathan M.; Borwein, Peter B. (1987). Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity (First ed.). Wiley-Interscience.
ISBN
0-471-83138-7 . page 225
^ Gilmore, Tomack.
"The Arithmetic-Geometric Mean of Gauss" (PDF) . Universität Wien . p. 13.
^ Borwein, J.; Borwein, P. (2000).
"Ramanujan and Pi" . Pi: A Source Book . Springer Link. pp. 588–595.
doi :
10.1007/978-1-4757-3240-5_62 .
ISBN
978-1-4757-3242-9 .
^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi . American Mathematical Society.
ISBN
0-8218-3246-8 . p. 248
Borwein, Peter (2000).
"The amazing number π " (PDF) . Nieuw Archief voor Wiskunde . 5th series. 1 (3): 254–258.
Zbl
1173.01300 .
Kazuya Kato, Nobushige Kurokawa, Saito Takeshi: Number Theory 1: Fermat's Dream. American Mathematical Society, Providence 1993,
ISBN
0-8218-0863-X .