List of formulae involving π

Part of a series of articles on the
mathematical constant π
3.1415926535897932384626433...
Uses
Area of a circle Circumference Use in other formulae
Properties
Irrationality Transcendence
Value
Less than 22/7 Approximations Milü Madhava's correction term Memorization
People
Archimedes Liu Hui Zu Chongzhi Aryabhata Madhava Jamshīd al-Kāshī Ludolph van Ceulen François Viète Seki Takakazu Takebe Kenko William Jones John Machin William Shanks Srinivasa Ramanujan John Wrench Chudnovsky brothers Yasumasa Kanada
History
Chronology A History of Pi
In culture
Indiana pi bill Pi Day
Related topics
Squaring the circle Basel problem Six nines in π Other topics related to π
v t e

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.

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

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.

${\displaystyle C={\frac {2\pi }{\operatorname {agm} (a,b)}}\left(a_{1}^{2}-\sum _{n=2}^{\infty }2^{n-1}(a_{n}^{2}-b_{n}^{2})\right)}$

where C is the circumference of an ellipse with semi-major axis a and semi-minor axis b and ${\displaystyle a_{n},b_{n}}$ are the arithmetic and geometric iterations of ${\displaystyle \operatorname {agm} (a,b)}$, the arithmetic-geometric mean of a and b with the initial values ${\displaystyle a_{0}=a}$ and ${\displaystyle b_{0}=b}$.

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

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.

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

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.

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:

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

• The cosmological constant:

  ${\displaystyle \Lambda ={{8\pi G} \over {3c^{2}}}\rho }$

• Heisenberg's uncertainty principle:

  ${\displaystyle \Delta x\,\Delta p\geq {\frac {h}{4\pi }}}$

• Einstein's field equation of general relativity:

  ${\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }}$

• Coulomb's law for the electric force in vacuum:

  ${\displaystyle F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}}$

• Magnetic permeability of free space:^{[note 1]}

  ${\displaystyle \mu _{0}\approx 4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}}$

• 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):

  ${\displaystyle T={\frac {2\pi }{\operatorname {agm} (1,\cos(\theta _{0}/2))}}{\sqrt {\frac {L}{g}}}}$

• Period of a spring-mass system with spring constant ${\displaystyle k}$ and mass ${\displaystyle m}$:

  ${\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}}$

• Kepler's third law of planetary motion:

  ${\displaystyle {\frac {R^{3}}{T^{2}}}={\frac {GM}{4\pi ^{2}}}}$

• The buckling formula:

  ${\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}}$

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 b^2Nm, when struck by the other object.^[1] (This gives the digits of π in base b up to N digits past the radix point.)

${\displaystyle 2\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx=\pi }$ (integrating two halves ${\displaystyle y(x)={\sqrt {1-x^{2}}}}$ to obtain the area of the unit circle)

${\displaystyle \int _{0}^{2}{\sqrt {4-x^{2}}}\,dx=\pi }$ (integrating a quarter of a circle with a radius of two ${\displaystyle x^{2}+y^{2}=4}$ to obtain ${\displaystyle {4\pi }/4}$)

${\displaystyle \int _{-\infty }^{\infty }\operatorname {sech} x\,dx=\pi }$

${\displaystyle \int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt}\,dx\,dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt}\,dx\,dt=\pi }$

${\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi }$

${\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi }$^{[2][note 2]} (see also Cauchy distribution)

${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi }$ (see Dirichlet integral)

${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}$ (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]

${\displaystyle \int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi }$ (see also Proof that 22/7 exceeds π).

${\displaystyle \int _{0}^{1}{x^{2}(1+x)^{4} \over 1+x^{2}}\,dx=\pi -{17 \over 15}}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{x+1}}\,dx={\frac {\pi }{\sin \pi \alpha }},\quad 0<\alpha <1}$

${\displaystyle \int _{0}^{\infty }{\frac {dx}{\sqrt {x(x+a)(x+b)}}}={\frac {\pi }{\operatorname {agm} ({\sqrt {a}},{\sqrt {b}})}}}$ (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 ${\textstyle 2\int _{0}^{a}f(x)\,dx}$.

${\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}}$ (see also Double factorial)

${\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{2^{k}(2k+1)!!}}={\frac {2\pi }{3{\sqrt {3}}}}}$

${\displaystyle \sum _{k=0}^{\infty }{\frac {k!\,(2k)!\,(25k-3)}{(3k)!\,2^{k}}}={\frac {\pi }{2}}}$

${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}}={\frac {4270934400}{{\sqrt {10005}}\pi }}}$ (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]

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

${\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}}$ (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 B_2n is a Bernoulli number.

${\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi }$^[7]

${\displaystyle \sum _{n=1}^{\infty }{\frac {7^{n}-1}{8^{n}}}\,\zeta (n+1)=(1+{\sqrt {2}})\pi }$

${\displaystyle \sum _{n=2}^{\infty }{\frac {2(3/2)^{n}-3}{n}}(\zeta (n)-1)=\ln \pi }$

${\displaystyle \sum _{n=1}^{\infty }\zeta (2n){\frac {x^{2n}}{n}}=\ln {\frac {\pi x}{\sin \pi x}},\quad 0<|x|<1}$

${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}}$ (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 ${\displaystyle F_{2}n}$ is the ${\displaystyle 2n}$th Fibonacci number)

${\displaystyle \sum _{n=1}^{\infty }{\frac {L_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {\pi ^{2}}{5}}}$ (where ${\displaystyle L_{n}}$ is the ${\displaystyle n}$th Lucas number)

${\displaystyle \sum _{n=1}^{\infty }\sigma (n)e^{-2\pi n}={\frac {1}{24}}-{\frac {1}{8\pi }}}$ (where ${\displaystyle \sigma }$ is the sum-of-divisors function)

${\displaystyle \pi =\sum _{n=1}^{\infty }{\frac {(-1)^{\varepsilon (n)}}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots }$   (where ${\displaystyle \varepsilon (n)}$ 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 ${\displaystyle x\in \mathbb {Q} \setminus \mathbb {Z} .}$

${\displaystyle \pi ={\sqrt {\sum _{n=1}^{\infty }{\frac {6}{n^{2}}}}}}$ (derived from Euler's solution to the Basel problem)

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.

${\displaystyle {\frac {\pi }{4}}=\arctan 1}$

${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}}$

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}}$

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}}$

${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}$ (the original Machin's formula)

${\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}}$

${\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}}$

${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}$

${\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}$

${\displaystyle {\frac {\pi }{4}}=\left(\prod _{p\equiv 1{\pmod {4}}}{\frac {p}{p-1}}\right)\cdot \left(\prod _{p\equiv 3{\pmod {4}}}{\frac {p}{p+1}}\right)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,}$ (Euler)

where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.

${\displaystyle {\frac {{\sqrt {3}}\pi }{6}}=\left(\displaystyle \prod _{p\equiv 1{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p-1}}\right)\cdot \left(\displaystyle \prod _{p\equiv 5{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p+1}}\right)={\frac {5}{6}}\cdot {\frac {7}{6}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{18}}\cdots }$

${\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+1)}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots }$ (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)

Viète's formula:

${\displaystyle {\frac {2}{\pi }}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \cdots }$

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).

Infinite product representation :

${\displaystyle {\frac {\pi }{4}}=\prod _{n=2}^{\infty }{\frac {(n-1)^{n(2n-1)}(n+1)^{n(2n+1)}}{n^{4n^{2}}}}}$^[16]

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

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

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 ${\displaystyle k}$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

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

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

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

${\displaystyle e^{z}\in \mathbb {R} \leftrightarrow \Im z\in \pi \mathbb {Z} }$

${\displaystyle e^{z}=1\leftrightarrow z\in 2\pi i\mathbb {Z} }$^[17]

Also

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

Suppose a lattice ${\displaystyle \Omega }$ is generated by two periods ${\displaystyle \omega _{1},\omega _{2}}$. We define the quasi-periods of this lattice by ${\displaystyle \eta _{1}=\zeta (z+\omega _{1};\Omega )-\zeta (z;\Omega )}$ and ${\displaystyle \eta _{2}=\zeta (z+\omega _{2};\Omega )-\zeta (z;\Omega )}$ where ${\displaystyle \zeta }$ is the Weierstrass zeta function (${\displaystyle \eta _{1}}$ and ${\displaystyle \eta _{2}}$ are in fact independent of ${\displaystyle z}$). Then the periods and quasi-periods are related by the Legendre identity:

${\displaystyle \eta _{1}\omega _{2}-\eta _{2}\omega _{1}=2\pi i.}$

${\displaystyle {\frac {4}{\pi }}=1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}$^[18]

${\displaystyle {\frac {\varpi ^{2}}{\pi }}={2+{\cfrac {1^{2}}{4+{\cfrac {3^{2}}{4+{\cfrac {5^{2}}{4+{\cfrac {7^{2}}{4+\ddots \,}}}}}}}}}\quad }$ (Ramanujan, ${\displaystyle \varpi }$ is the lemniscate constant)^[19]

${\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}}$^[18]

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}}$

${\displaystyle 2\pi ={6+{\cfrac {2^{2}}{12+{\cfrac {6^{2}}{12+{\cfrac {10^{2}}{12+{\cfrac {14^{2}}{12+{\cfrac {18^{2}}{12+\ddots }}}}}}}}}}}}$

${\displaystyle \pi =4-{\cfrac {2}{1+{\cfrac {1}{1-{\cfrac {1}{1+{\cfrac {2}{1-{\cfrac {2}{1+{\cfrac {3}{1-{\cfrac {3}{\ddots }}}}}}}}}}}}}}}$

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

${\displaystyle a_{0}=1,\,a_{n+1}=\left(1+{\frac {1}{2n+1}}\right)a_{n},\,\pi =\lim _{n\to \infty }{\frac {a_{n}^{2}}{n}}}$

${\displaystyle a_{1}=0,\,a_{n+1}={\sqrt {2+a_{n}}},\,\pi =\lim _{n\to \infty }2^{n}{\sqrt {2-a_{n}}}}$ (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, ${\displaystyle h\in \left\{0,1,\ldots ,2^{m}-1\right\}}$)^[20]

${\displaystyle \forall k\in \mathbb {N} ,\,a_{1}=2^{-k},\,a_{n+1}=a_{n}+2^{-k}(1-\tan(2^{k-1}a_{n})),\,\pi =2^{k+1}\lim _{n\to \infty }a_{n}}$ (quadratic convergence)^[21]

${\displaystyle a_{1}=1,\,a_{n+1}=a_{n}+\sin a_{n},\,\pi =\lim _{n\to \infty }a_{n}}$ (cubic convergence)^[22]

${\displaystyle a_{0}=2{\sqrt {3}},\,b_{0}=3,\,a_{n+1}=\operatorname {hm} (a_{n},b_{n}),\,b_{n+1}=\operatorname {gm} (a_{n+1},b_{n}),\,\pi =\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}$ (Archimedes' algorithm, see also harmonic mean and geometric mean)^[23]

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

${\displaystyle {\binom {2n}{n}}\sim {\frac {4^{n}}{\sqrt {\pi n}}}}$ (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)

${\displaystyle \log n!\simeq \left(n+{\frac {1}{2}}\right)\log n-n+{\frac {\log 2\pi }{2}}}$

${\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}}$ (where ${\displaystyle \varphi }$ is Euler's totient function)

${\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}}$

The symbol ${\displaystyle \sim }$ means that the ratio of the left-hand side and the right-hand side tends to one as ${\displaystyle n\to \infty }$.

The symbol ${\displaystyle \simeq }$ means that the difference between the left-hand side and the right-hand side tends to zero as ${\displaystyle n\to \infty }$.

With ${\displaystyle {}_{2}F_{1}}$ being the hypergeometric function:

${\displaystyle \sum _{n=0}^{\infty }r_{2}(n)q^{n}={}_{2}F_{1}\left({\frac {1}{2}},{\frac {1}{2}},1,z\right)}$

where

${\displaystyle q=\exp \left(-\pi {\frac {{}_{2}F_{1}(1/2,1/2,1,1-z)}{{}_{2}F_{1}(1/2,1/2,1,z)}}\right),\quad z\in \mathbb {C} \setminus \{0,1\}}$

and ${\displaystyle r_{2}}$ is the sum of two squares function.

Similarly,

${\displaystyle 1+240\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}_3(n)q^n={}_2{F}_1{(\frac{1}{6},\frac{5}{6}}^{}}$