# List of formulae involving π

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.

## Euclidean geometry

${\displaystyle \pi ={\frac {C}{d}}={\frac {C}{2r}}}$

where C is the circumference of a circle, d is the diameter, and r is the radius. More generally,

${\displaystyle \pi ={\frac {L}{w}}}$

where L and w are, respectively, the perimeter and the width of any curve of constant width.

${\displaystyle A=\pi r^{2}}$

where A is the area of a circle. More generally,

${\displaystyle A=\pi ab}$

where A is the area enclosed by an ellipse with semi-major axis a and semi-minor axis b.

${\displaystyle A=4\pi r^{2}}$

where A is the area between the witch of Agnesi and its asymptotic line; r is the radius of the defining circle.

${\displaystyle A={\frac {\Gamma (1/4)^{2}}{2{\sqrt {\pi }}}}r^{2}={\frac {\pi r^{2}}{\operatorname {agm} (1,1/{\sqrt {2}})}}}$

where A is the area of a squircle with minor radius r, ${\displaystyle \Gamma }$ is the gamma function and ${\displaystyle \operatorname {agm} }$ is the arithmetic–geometric mean.

${\displaystyle A=(k+1)(k+2)\pi r^{2}}$

where A is the area of an epicycloid with the smaller circle of radius r and the larger circle of radius kr (${\displaystyle k\in \mathbb {N} }$), assuming the initial point lies on the larger circle.

${\displaystyle A={\frac {(-1)^{k}+3}{8}}\pi a^{2}}$

where A is the area of a rose with angular frequency k (${\displaystyle k\in \mathbb {N} }$) and amplitude a.

${\displaystyle L={\frac {\Gamma (1/4)^{2}}{\sqrt {\pi }}}c={\frac {2\pi c}{\operatorname {agm} (1,1/{\sqrt {2}})}}}$

where L is the perimeter of the lemniscate of Bernoulli with focal distance c.

${\displaystyle V={4 \over 3}\pi r^{3}}$

where V is the volume of a sphere and r is the radius.

${\displaystyle SA=4\pi r^{2}}$

where SA is the surface area of a sphere and r is the radius.

${\displaystyle H={1 \over 2}\pi ^{2}r^{4}}$

where H is the hypervolume of a 3-sphere and r is the radius.

${\displaystyle SV=2\pi ^{2}r^{3}}$

where SV is the surface volume of a 3-sphere and r is the radius.

### Regular convex polygons

Sum S of internal angles of a regular convex polygon with n sides:

${\displaystyle S=(n-2)\pi }$

Area A of a regular convex polygon with n sides and side length s:

${\displaystyle A={\frac {ns^{2}}{4}}\cot {\frac {\pi }{n}}}$

Inradius r of a regular convex polygon with n sides and side length s:

${\displaystyle r={\frac {s}{2}}\cot {\frac {\pi }{n}}}$

Circumradius R of a regular convex polygon with n sides and side length s:

${\displaystyle R={\frac {s}{2}}\csc {\frac {\pi }{n}}}$

## Physics

${\displaystyle \Lambda ={{8\pi G} \over {3c^{2}}}\rho }$
${\displaystyle \Delta x\,\Delta p\geq {\frac {h}{4\pi }}}$
${\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }}$
${\displaystyle F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}}$
${\displaystyle \mu _{0}\approx 4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}}$
• Approximate period of a simple pendulum with small amplitude:
${\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}}$
• Exact period of a simple pendulum with amplitude ${\displaystyle \theta _{0}}$ (${\displaystyle \operatorname {agm} }$ is the arithmetic–geometric mean):
${\displaystyle T={\frac {2\pi }{\operatorname {agm} (1,\cos(\theta _{0}/2))}}{\sqrt {\frac {L}{g}}}}$
${\displaystyle {\frac {R^{3}}{T^{2}}}={\frac {GM}{4\pi ^{2}}}}$
${\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}}$

A puzzle involving "colliding billiard balls":

${\displaystyle \lfloor {b^{N}\pi }\rfloor }$

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 b2Nm, when struck by the other object.[1] (This gives the digits of π in base b up to N digits past the radix point.)

## Formulae yielding π

### Integrals

${\displaystyle 2\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx=\pi }$ (integrating two halves ${\displaystyle y(x)={\sqrt {1-x^{2}}}}$ to obtain the area of the unit circle)
${\displaystyle \int _{-\infty }^{\infty }\operatorname {sech} x\,dx=\pi }$
${\displaystyle \int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt}\,dx\,dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt}\,dx\,dt=\pi }$
${\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi }$
${\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi }$[2][note 2] (see also Cauchy distribution)
${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}$ (see Gaussian integral).
${\displaystyle \oint {\frac {dz}{z}}=2\pi i}$ (when the path of integration winds once counterclockwise around 0. See also Cauchy's integral formula).
${\displaystyle \int _{0}^{\infty }\ln \left(1+{\frac {1}{x^{2}}}\right)\,dx=\pi }$[3]
${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi }$
${\displaystyle \int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi }$ (see also Proof that 22/7 exceeds π).
${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{x+1}}\,dx={\frac {\pi }{\sin \pi \alpha }},\quad 0<\alpha <1}$
${\displaystyle \int _{0}^{\infty }{\frac {dx}{\sqrt {x(x+a)(x+b)}}}={\frac {\pi }{\operatorname {agm} ({\sqrt {a}},{\sqrt {b}})}}}$ (where ${\displaystyle \operatorname {agm} }$ is the arithmetic–geometric mean;[4] see also elliptic integral)

Note that with symmetric integrands ${\displaystyle f(-x)=f(x)}$, formulas of the form ${\textstyle \int _{-a}^{a}f(x)\,dx}$ can also be translated to formulas ${\textstyle 2\int _{0}^{a}f(x)\,dx}$.

### Efficient infinite series

${\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}}$ (see also Double factorial)
${\displaystyle \sum _{k=0}^{\infty }{\frac {k!\,(2k)!\,(25k-3)}{(3k)!\,2^{k}}}={\frac {\pi }{2}}}$
${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}}={\frac {4270934400}{{\sqrt {10005}}\pi }}}$ (see Chudnovsky algorithm)
${\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {9801}{2{\sqrt {2}}\pi }}}$ (see Srinivasa Ramanujan, Ramanujan–Sato series)

The following are efficient for calculating arbitrary binary digits of π:

${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{4^{k}}}\left({\frac {2}{4k+1}}+{\frac {2}{4k+2}}+{\frac {1}{4k+3}}\right)=\pi }$[5]
${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)=\pi }$ (see Bailey–Borwein–Plouffe formula)
${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {8}{8k+2}}+{\frac {4}{8k+3}}+{\frac {4}{8k+4}}-{\frac {1}{8k+7}}\right)=2\pi }$
${\displaystyle \sum _{k=0}^{\infty }{\frac {{(-1)}^{k}}{2^{10k}}}\left(-{\frac {2^{5}}{4k+1}}-{\frac {1}{4k+3}}+{\frac {2^{8}}{10k+1}}-{\frac {2^{6}}{10k+3}}-{\frac {2^{2}}{10k+5}}-{\frac {2^{2}}{10k+7}}+{\frac {1}{10k+9}}\right)=2^{6}\pi }$

Plouffe's series for calculating arbitrary decimal digits of π:[6]

${\displaystyle \sum _{k=1}^{\infty }k{\frac {2^{k}k!^{2}}{(2k)!}}=\pi +3}$

### Other infinite series

${\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}}$ (see also Basel problem and Riemann zeta function)
${\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}}$
${\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}}$ , where B2n is a Bernoulli number.
${\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi }$[7]
${\displaystyle \sum _{n=2}^{\infty }{\frac {2(3/2)^{n}-3}{n}}(\zeta (n)-1)=\ln \pi }$
${\displaystyle \sum _{n=1}^{\infty }\zeta (2n){\frac {x^{2n}}{n}}=\ln {\frac {\pi x}{\sin \pi x}},\quad 0<|x|<1}$
${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}}$ (see Leibniz formula for pi)
${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{(n^{2}-n)/2}}{2n+1}}=1+{\frac {1}{3}}-{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}-\cdots ={\frac {\pi }{2{\sqrt {2}}}}}$ (Newton, Second Letter to Oldenburg, 1676)[8]
${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}(2n+1)}}=1-{\frac {1}{3^{1}\cdot 3}}+{\frac {1}{3^{2}\cdot 5}}-{\frac {1}{3^{3}\cdot 7}}+{\frac {1}{3^{4}\cdot 9}}-\cdots ={\sqrt {3}}\arctan {\frac {1}{\sqrt {3}}}={\frac {\pi }{2{\sqrt {3}}}}}$ (Madhava series)
${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}}$
${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}}$
${\displaystyle \sum _{n=0}^{\infty }\left({\frac {1}{2n+1}}\right)^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots ={\frac {\pi ^{2}}{8}}}$
${\displaystyle \sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots ={\frac {\pi ^{3}}{32}}}$
${\displaystyle \sum _{n=0}^{\infty }\left({\frac {1}{2n+1}}\right)^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots ={\frac {\pi ^{4}}{96}}}$
${\displaystyle \sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots ={\frac {5\pi ^{5}}{1536}}}$
${\displaystyle \sum _{n=0}^{\infty }\left({\frac {1}{2n+1}}\right)^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}}$

In general,

${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{2k+1}}}=(-1)^{k}{\frac {E_{2k}}{2(2k)!}}\left({\frac {\pi }{2}}\right)^{2k+1},\quad k\in \mathbb {N} _{0}}$

where ${\displaystyle E_{2k}}$ is the ${\displaystyle 2k}$th Euler number.[9]

${\displaystyle \sum _{n=0}^{\infty }{\binom {\frac {1}{2}}{n}}{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{6}}-{\frac {1}{40}}-\cdots ={\frac {\pi }{4}}}$
${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}}$
${\displaystyle \sum _{n=1}^{\infty }(-1)^{(n^{2}+n)/2+1}\left|G_{\left((-1)^{n+1}+6n-3\right)/4}\right|=|G_{1}|+|G_{2}|-|G_{4}|-|G_{5}|+|G_{7}|+|G_{8}|-|G_{10}|-|G_{11}|+\cdots ={\frac {\sqrt {3}}{\pi }}}$ (see Gregory coefficients)
${\displaystyle \sum _{n=0}^{\infty }{\frac {(1/2)_{n}^{2}}{2^{n}n!^{2}}}\sum _{n=0}^{\infty }{\frac {n(1/2)_{n}^{2}}{2^{n}n!^{2}}}={\frac {1}{\pi }}}$ (where ${\displaystyle (x)_{n}}$ is the rising factorial)[10]
${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(2n+1)}}=\pi -3}$ (Nilakantha series)
${\displaystyle \sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}}$ (where ${\displaystyle F_{n}}$ is the n-th Fibonacci number)
${\displaystyle \pi =\sum _{n=1}^{\infty }{\frac {(-1)^{\epsilon (n)}}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots }$   (where ${\displaystyle \epsilon (n)}$ is the number of prime factors of the form ${\displaystyle p\equiv 1\,(\mathrm {mod} \,4)}$ of ${\displaystyle n}$)[11][12]
${\displaystyle {\frac {\pi }{2}}=\sum _{n=1}^{\infty }{\frac {(-1)^{\varepsilon (n)}}{n}}=1+{\frac {1}{2}}-{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}-{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}+\cdots }$   (where ${\displaystyle \varepsilon (n)}$ is the number of prime factors of the form ${\displaystyle p\equiv 3\,(\mathrm {mod} \,4)}$ of ${\displaystyle n}$)[13]
${\displaystyle \pi =\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{n+1/2}}}$
${\displaystyle \pi ^{2}=\sum _{n=-\infty }^{\infty }{\frac {1}{(n+1/2)^{2}}}}$[14]

The last two formulas are special cases of

{\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 ${\displaystyle \pi }$ when ${\displaystyle x\in \mathbb {Q} \setminus \mathbb {Z} .}$

Some formulas relating π and harmonic numbers are given here. Further infinite series involving π are:[15]

 ${\displaystyle \pi ={\frac {1}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {((2n)!)^{3}(42n+5)}{(n!)^{6}{16}^{3n+1}}}}$ ${\displaystyle \pi ={\frac {4}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}{441}^{2n+1}{2}^{10n+1}}}}$ ${\displaystyle \pi ={\frac {4}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(6n+1)\left({\frac {1}{2}}\right)_{n}^{3}}{{4^{n}}(n!)^{3}}}}$ ${\displaystyle \pi ={\frac {32}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {{\sqrt {5}}-1}{2}}\right)^{8n}{\frac {(42n{\sqrt {5}}+30n+5{\sqrt {5}}-1)\left({\frac {1}{2}}\right)_{n}^{3}}{{64^{n}}(n!)^{3}}}}$ ${\displaystyle \pi ={\frac {27}{4Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {2}{27}}\right)^{n}{\frac {(15n+2)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}}$ ${\displaystyle \pi ={\frac {15{\sqrt {3}}}{2Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(33n+4)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}}$ ${\displaystyle \pi ={\frac {85{\sqrt {85}}}{18{\sqrt {3}}Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {4}{85}}\right)^{n}{\frac {(133n+8)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}}$ ${\displaystyle \pi ={\frac {5{\sqrt {5}}}{2{\sqrt {3}}Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(11n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}}$ ${\displaystyle \pi ={\frac {2{\sqrt {3}}}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(8n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{n}}}}$ ${\displaystyle \pi ={\frac {\sqrt {3}}{9Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(40n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{49}^{2n+1}}}}$ ${\displaystyle \pi ={\frac {2{\sqrt {11}}}{11Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(280n+19)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{99}^{2n+1}}}}$ ${\displaystyle \pi ={\frac {\sqrt {2}}{4Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(10n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{2n+1}}}}$ ${\displaystyle \pi ={\frac {4{\sqrt {5}}}{5Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(644n+41)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}5^{n}{72}^{2n+1}}}}$ ${\displaystyle \pi ={\frac {4{\sqrt {3}}}{3Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(28n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{3^{n}}{4}^{n+1}}}}$ ${\displaystyle \pi ={\frac {4}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(20n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{2}^{2n+1}}}}$ ${\displaystyle \pi ={\frac {72}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(260n+23)}{(n!)^{4}4^{4n}18^{2n}}}}$ ${\displaystyle \pi ={\frac {3528}{Z}}}$ ${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}4^{4n}882^{2n}}}}$

where ${\displaystyle (x)_{n}}$ is the Pochhammer symbol for the rising factorial. See also Ramanujan–Sato series.

### Machin-like formulae

${\displaystyle {\frac {\pi }{4}}=\arctan 1}$
${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}}$
${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}}$
${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}}$
${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}$ (the original Machin's formula)
${\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}}$
${\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}}$
${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}$
${\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}$

### Infinite products

${\displaystyle {\frac {\pi }{4}}=\left(\prod _{p\equiv 1{\pmod {4}}}{\frac {p}{p-1}}\right)\cdot \left(\prod _{p\equiv 3{\pmod {4}}}{\frac {p}{p+1}}\right)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {9}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,}$ (Euler)
where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
${\displaystyle {\frac {{\sqrt {3}}\pi }{6}}=\left(\displaystyle \prod _{p\equiv 1{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p-1}}\right)\cdot \left(\displaystyle \prod _{p\equiv 5{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p+1}}\right)={\frac {5}{6}}\cdot {\frac {7}{6}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{18}}\cdots ,}$
${\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+1)}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots }$ (see also Wallis product)
${\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{n}}\right)^{(-1)^{n+1}}=\left(1+{\frac {1}{1}}\right)^{+1}\left(1+{\frac {1}{2}}\right)^{-1}\left(1+{\frac {1}{3}}\right)^{+1}\cdots }$ (another form of Wallis product)
${\displaystyle {\frac {2}{\pi }}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \cdots }$

A double infinite product formula involving the Thue–Morse sequence:

${\displaystyle {\frac {\pi }{2}}=\prod _{m\geq 1}\prod _{n\geq 1}\left({\frac {(4m^{2}+n-2)(4m^{2}+2n-1)^{2}}{4(2m^{2}+n-1)(4m^{2}+n-1)(2m^{2}+n)}}\right)^{\epsilon _{n}},}$
where ${\displaystyle \epsilon _{n}=(-1)^{t_{n}}}$ and ${\displaystyle t_{n}}$ is the Thue–Morse sequence (Tóth 2020).

### Arctangent formulas

${\displaystyle {\frac {\pi }{2^{k+1}}}=\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},\qquad \qquad k\geq 2}$
${\displaystyle {\frac {\pi }{4}}=\sum _{k\geq 2}\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},}$

where ${\displaystyle a_{k}={\sqrt {2+a_{k-1}}}}$ such that ${\displaystyle a_{1}={\sqrt {2}}}$.

${\displaystyle {\frac {\pi }{2}}=\sum _{k=0}^{\infty }\arctan {\frac {1}{F_{2k+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots }$

where ${\displaystyle F_{k}}$ is the k-th Fibonacci number.

${\displaystyle \pi =\arctan a+\arctan b+\arctan c}$

whenever ${\displaystyle a+b+c=abc}$ and ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$ are positive real numbers (see List of trigonometric identities). A special case is

${\displaystyle \pi =\arctan 1+\arctan 2+\arctan 3.}$

### Complex exponential formulas

${\displaystyle e^{i\pi }+1=0}$ (Euler's identity)

The following equivalences are true for any complex ${\displaystyle z}$:

${\displaystyle e^{z}\in \mathbb {R} \leftrightarrow \Im z\in \pi \mathbb {Z} }$
${\displaystyle e^{z}=1\leftrightarrow z\in 2\pi i\mathbb {Z} }$[16]

Also

${\displaystyle {\frac {1}{e^{z}-1}}=\lim _{N\to \infty }\sum _{n=-N}^{N}{\frac {1}{z-2\pi in}}-{\frac {1}{2}},\quad z\in \mathbb {C} .}$

### Continued fractions

${\displaystyle {\frac {4}{\pi }}=1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}$[17]
${\displaystyle {\frac {\varpi ^{2}}{\pi }}={2+{\cfrac {1^{2}}{4+{\cfrac {3^{2}}{4+{\cfrac {5^{2}}{4+{\cfrac {7^{2}}{4+\ddots \,}}}}}}}}}\quad }$ (Ramanujan, ${\displaystyle \varpi }$ is the lemniscate constant)[18]
${\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}}$[17]
${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}}$
${\displaystyle 2\pi ={6+{\cfrac {2^{2}}{12+{\cfrac {6^{2}}{12+{\cfrac {10^{2}}{12+{\cfrac {14^{2}}{12+{\cfrac {18^{2}}{12+\ddots }}}}}}}}}}}}$

For more on the fourth identity, see Euler's continued fraction formula.

(See also Continued fraction and Generalized continued fraction.)

### Iterative algorithms

${\displaystyle a_{0}=1,\,a_{n+1}=\left(1+{\frac {1}{2n+1}}\right)a_{n},\,\pi =\lim _{n\to \infty }{\frac {a_{n}^{2}}{n}}}$
${\displaystyle a_{1}=0,\,a_{n+1}={\sqrt {2+a_{n}}},\,\pi =\lim _{n\to \infty }2^{n}{\sqrt {2-a_{n}}}}$ (closely related to Viète's formula)
${\displaystyle \omega (i_{n},i_{n-1},\dots ,i_{1})=2+i_{n}{\sqrt {2+i_{n-1}{\sqrt {2+\cdots +i_{1}{\sqrt {2}}}}}}=\omega (b_{n},b_{n-1},\dots ,b_{1}),\,i_{k}\in \{-1,1\},\,b_{k}={\begin{cases}0&{\text{if }}i_{k}=1\\1&{\text{if }}i_{k}=-1\end{cases}},\,\pi ={\displaystyle \lim _{n\rightarrow \infty }{\frac {2^{n+1}}{2h+1}}{\sqrt {\omega \left(\underbrace {10\ldots 0} _{n-m}g_{m,h+1}\right)}}}}$ (where ${\displaystyle g_{m,h+1}}$ is the h+1-th entry of m-bit Gray code, ${\displaystyle h\in \left\{0,1,\ldots ,2^{m}-1\right\}}$)[19]
${\displaystyle \forall k\in \mathbb {N} ,\,a_{1}=2^{-k},\,a_{n+1}=a_{n}+2^{-k}(1-\tan(2^{k-1}a_{n})),\,\pi =2^{k+1}\lim _{n\to \infty }a_{n}}$ (quadratic convergence)[20]
${\displaystyle a_{1}=1,\,a_{n+1}=a_{n}+\sin a_{n},\,\pi =\lim _{n\to \infty }a_{n}}$ (cubic convergence)[21]
${\displaystyle a_{0}=2{\sqrt {3}},\,b_{0}=3,\,a_{n+1}=\operatorname {hm} (a_{n},b_{n}),\,b_{n+1}=\operatorname {gm} (a_{n+1},b_{n}),\,\pi =\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}$ (Archimedes' algorithm, see also harmonic mean and geometric mean)[22]

For more iterative algorithms, see the Gauss–Legendre algorithm and Borwein's algorithm.

### Asymptotics

${\displaystyle {\binom {2n}{n}}\sim {\frac {4^{n}}{\sqrt {\pi n}}}}$ (asymptotic growth rate of the central binomial coefficients)
${\displaystyle C_{n}\sim {\frac {4^{n}}{\sqrt {\pi n^{3}}}}}$ (asymptotic growth rate of the Catalan numbers)
${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}$ (Stirling's approximation)
${\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}}$ (where ${\displaystyle \varphi }$ is Euler's totient function)
${\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}}$

### Miscellaneous

${\displaystyle \Gamma (s)\Gamma (1-s)={\frac {\pi }{\sin \pi s}}}$ (Euler's reflection formula, see Gamma function)
${\displaystyle \pi ^{-s/2}\Gamma \left({\frac {s}{2}}\right)\zeta (s)=\pi ^{-(1-s)/2}\Gamma \left({\frac {1-s}{2}}\right)\zeta (1-s)}$ (the functional equation of the Riemann zeta function)
${\displaystyle e^{-\zeta '(0)}={\sqrt {2\pi }}}$
${\displaystyle e^{\zeta '(0,1/2)-\zeta '(0,1)}={\sqrt {\pi }}}$ (where ${\displaystyle \zeta (s,a)}$ is the Hurwitz zeta function and the derivative is taken with respect to the first variable)
${\displaystyle \pi =\mathrm {B} (1/2,1/2)=\Gamma (1/2)^{2}}$ (see also Beta function)
${\displaystyle \pi ={\frac {\Gamma (3/4)^{4}}{\operatorname {agm} (1,1/{\sqrt {2}})^{2}}}={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}}$ (where agm is the arithmetic–geometric mean)
${\displaystyle \pi =\operatorname {agm} \left(\theta _{2}^{2}(1/e),\theta _{3}^{2}(1/e)\right)}$ (where ${\displaystyle \theta _{2}}$ and ${\displaystyle \theta _{3}}$ are the Jacobi theta functions[23])
${\displaystyle \pi =-{\frac {\operatorname {K} (k)}{\operatorname {K} \left({\sqrt {1-k^{2}}}\right)}}\ln q,\quad k={\frac {\theta _{2}^{2}(q)}{\theta _{3}^{2}(q)}}}$ (where ${\displaystyle q\in (0,1)}$ and ${\displaystyle \operatorname {K} (k)}$ is the complete elliptic integral of the first kind with modulus ${\displaystyle k}$; reflecting the nome-modulus inversion problem)[24]
${\displaystyle \pi =-{\frac {\operatorname {agm} \left(1,{\sqrt {1-k'^{2}}}\right)}{\operatorname {agm} (1,k')}}\ln q,\quad k'={\frac {\theta _{4}^{2}(q)}{\theta _{3}^{2}(q)}}}$ (where ${\displaystyle q\in (0,1)}$)[24]
${\displaystyle \operatorname {agm} (1,{\sqrt {2}})={\frac {\pi }{\varpi }}}$ (due to Gauss,[25] ${\displaystyle \varpi }$ is the lemniscate constant)
${\displaystyle i\pi =\operatorname {Log} (-1)=\lim _{n\to \infty }n\left((-1)^{1/n}-1\right)}$ (where ${\displaystyle \operatorname {Log} }$ is the principal value of the complex logarithm)[note 3]
${\displaystyle 1-{\frac {\pi ^{2}}{12}}=\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n{\bmod {k}})}$ (where ${\textstyle n{\bmod {k}}}$ is the remainder upon division of n by k)
${\displaystyle \pi =\lim _{r\to \infty }{\frac {1}{r^{2}}}\sum _{x=-r}^{r}\;\sum _{y=-r}^{r}{\begin{cases}1&{\text{if }}{\sqrt {x^{2}+y^{2}}}\leq r\\0&{\text{if }}{\sqrt {x^{2}+y^{2}}}>r\end{cases}}}$ (summing a circle's area)
${\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}}$ (Riemann sum to evaluate the area of the unit circle)
${\displaystyle \pi =\lim _{n\to \infty }{\frac {2^{4n}n!^{4}}{n(2n)!^{2}}}=\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}=\lim _{n\rightarrow \infty }{\frac {1}{n}}\left({\frac {(2n)!!}{(2n-1)!!}}\right)^{2}}$ (by combining Stirling's approximation with Wallis product)
${\displaystyle \pi =\lim _{n\to \infty }{\frac {1}{n}}\ln {\frac {16}{\lambda (ni)}}}$ (where ${\displaystyle \lambda }$ is the modular lambda function)[26][note 4]
${\displaystyle \pi =\lim _{n\to \infty }{\frac {24}{\sqrt {n}}}\ln \left(2^{1/4}G_{n}\right)=\lim _{n\to \infty }{\frac {24}{\sqrt {n}}}\ln \left(2^{1/4}g_{n}\right)}$ (where ${\displaystyle G_{n}}$ and ${\displaystyle g_{n}}$ are Ramanujan's class invariants)[27][note 5]

## References

### Notes

1. ^ The relation ${\displaystyle \mu _{0}=4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}}$ was valid until the 2019 redefinition of the SI base units.
2. ^ (integral form of arctan over its entire domain, giving the period of tan)
3. ^ The ${\displaystyle n}$th root with the smallest positive principal argument is chosen.
4. ^ When ${\displaystyle n\in \mathbb {Q} ^{+}}$, this gives algebraic approximations to Gelfond's constant ${\displaystyle e^{\pi }}$.
5. ^ When ${\displaystyle {\sqrt {n}}\in \mathbb {Q} ^{+}}$, this gives algebraic approximations to Gelfond's constant ${\displaystyle e^{\pi }}$.

### Other

1. ^ 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.
2. ^ Rudin, Walter (1987). Real and Complex Analysis (Third ed.). McGraw-Hill Book Company. ISBN 0-07-100276-6. p. 4
3. ^ A000796 – OEIS
4. ^ 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.
5. ^ Arndt, Jörg; Haenel, Christoph (2001). π Unleashed. Springer-Verlag Berlin Heidelberg. ISBN 978-3-540-66572-4. page 126
6. ^ Gourdon, Xavier. "Computation of the n-th decimal digit of π with low memory" (PDF). Numbers, constants and computation. p. 1.
7. ^ Weisstein, Eric W. "Pi Formulas", MathWorld
8. ^ Chrystal, G. (1900). Algebra, an Elementary Text-book: Part II. p. 335.
9. ^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi. American Mathematical Society. ISBN 0-8218-3246-8. p. 112
10. ^ Cooper, Shaun (2017). Ramanujan's Theta Functions (First ed.). Springer. ISBN 978-3-319-56171-4. (page 647)
11. ^ Euler, Leonhard (1748). Introductio in analysin infinitorum (in Latin). Vol. 1. p. 245
12. ^ Carl B. Boyer, A History of Mathematics, Chapter 21., pp. 488–489
13. ^ Euler, Leonhard (1748). Introductio in analysin infinitorum (in Latin). Vol. 1. p. 244
14. ^ Wästlund, Johan. "Summing inverse squares by euclidean geometry" (PDF). The paper gives the formula with a minus sign instead, but these results are equivalent.
15. ^ 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.
16. ^ Rudin, Walter (1987). Real and Complex Analysis (Third ed.). McGraw-Hill Book Company. ISBN 0-07-100276-6. p. 3
17. ^ a b Loya, Paul (2017). Amazing and Aesthetic Aspects of Analysis. Springer. p. 589. ISBN 978-1-4939-6793-3.
18. ^ Perron, Oskar (1957). Die Lehre von den Kettenbrüchen: Band II (in German) (Third ed.). B. G. Teubner. p. 36, eq. 24
19. ^ 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.
20. ^ 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.
21. ^ Arndt, Jörg; Haenel, Christoph (2001). π Unleashed. Springer-Verlag Berlin Heidelberg. ISBN 978-3-540-66572-4. page 49
22. ^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi. American Mathematical Society. ISBN 0-8218-3246-8. p. 2
23. ^ 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
24. ^ a b 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 41
25. ^ Gilmore, Tomack. "The Arithmetic-Geometric Mean of Gauss" (PDF). Universität Wien. p. 13.
26. ^ 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.
27. ^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi. American Mathematical Society. ISBN 0-8218-3246-8. p. 248