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

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

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

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

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

$A=\pi r^{2}$

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

$A=\pi ab$

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

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

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

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

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

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

where A is the area of a squircle with minor radius r, $\Gamma$ is the gamma function.

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

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

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

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

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

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

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

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

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

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

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

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

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

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:

$S=(n-2)\pi$

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

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

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

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

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

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

Physics

• The cosmological constant:

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

• Heisenberg's uncertainty principle:

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

• Einstein's field equation of general relativity:

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

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

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

• Magnetic permeability of free space:

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

• Approximate period of a simple pendulum with small amplitude:

$T\approx 2\pi {\sqrt {\frac {L}{g}}}$

• Exact period of a simple pendulum with amplitude $\theta _{0}$ ($\operatorname {agm}$ is the arithmetic–geometric mean):

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

• Kepler's third law of planetary motion:

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

• The buckling formula:

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

A puzzle involving "colliding billiard balls":

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

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. (This gives the digits of π in base b up to N digits past the radix point.)

Formulae yielding π

Integrals

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

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

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

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

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

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

$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}$ (see Gaussian integral).

$\oint {\frac {dz}{z}}=2\pi i$ (when the path of integration winds once counterclockwise around 0. See also Cauchy's integral formula).

$\int _{0}^{\infty }\ln \left(1+{\frac {1}{x^{2}}}\right)\,dx=\pi$

$\int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi$

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

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

$\int _{0}^{\infty }{\frac {dx}{\sqrt {x(x+a)(x+b)}}}={\frac {\pi }{\operatorname {agm} ({\sqrt {a}},{\sqrt {b}})}}$ (where $\operatorname {agm}$ is the arithmetic–geometric mean; see also elliptic integral)

Note that with symmetric integrands $f(-x)=f(x)$, 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

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

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

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

$\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {9801}{2{\sqrt {2}}\pi }}$ (see Srinivasa Ramanujan, Ramanujan–Sato series)

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

$\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{4^{k}}}\left({\frac {2}{4k+1}}+{\frac {2}{4k+2}}+{\frac {1}{4k+3}}\right)=\pi$

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

$\sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {8}{8k+2}}+{\frac {4}{8k+3}}+{\frac {4}{8k+4}}-{\frac {1}{8k+7}}\right)=2\pi$

$\sum _{k=0}^{\infty }{\frac {{(-1)}^{k}}{2^{10k}}}\left(-{\frac {2^{5}}{4k+1}}-{\frac {1}{4k+3}}+{\frac {2^{8}}{10k+1}}-{\frac {2^{6}}{10k+3}}-{\frac {2^{2}}{10k+5}}-{\frac {2^{2}}{10k+7}}+{\frac {1}{10k+9}}\right)=2^{6}\pi$

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

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

Other infinite series

$\zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}$ (see also Basel problem and Riemann zeta function)

$\zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}$

$\zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}$ , where B_2n is a Bernoulli number.

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

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

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

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}$ (see Leibniz formula for pi)

$\sum _{n=0}^{\infty }{\frac {(-1)^{(n^{2}-n)/2}}{2n+1}}=1+{\frac {1}{3}}-{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}-\cdots ={\frac {\pi }{2{\sqrt {2}}}}$ (Newton, Second Letter to Oldenburg, 1676)

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}(2n+1)}}=1-{\frac {1}{3^{1}\cdot 3}}+{\frac {1}{3^{2}\cdot 5}}-{\frac {1}{3^{3}\cdot 7}}+{\frac {1}{3^{4}\cdot 9}}-\cdots ={\sqrt {3}}\arctan {\frac {1}{\sqrt {3}}}={\frac {\pi }{2{\sqrt {3}}}}$ (Madhava series)

$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}$

$\sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}$

$\sum _{n=0}^{\infty }\left({\frac {1}{2n+1}}\right)^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots ={\frac {\pi ^{2}}{8}}$

$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots ={\frac {\pi ^{3}}{32}}$

$\sum _{n=0}^{\infty }\left({\frac {1}{2n+1}}\right)^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots ={\frac {\pi ^{4}}{96}}$

$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots ={\frac {5\pi ^{5}}{1536}}$

$\sum _{n=0}^{\infty }\left({\frac {1}{2n+1}}\right)^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}$

In general,

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

where $E_{2k}$ is the $2k$th Euler number.

$\sum _{n=0}^{\infty }{\binom {\frac {1}{2}}{n}}{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{6}}-{\frac {1}{40}}-\cdots ={\frac {\pi }{4}}$

$\sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}$

$\sum _{n=1}^{\infty }(-1)^{(n^{2}+n)/2+1}\left|G_{\left((-1)^{n+1}+6n-3\right)/4}\right|=|G_{1}|+|G_{2}|-|G_{4}|-|G_{5}|+|G_{7}|+|G_{8}|-|G_{10}|-|G_{11}|+\cdots ={\frac {\sqrt {3}}{\pi }}$ (see Gregory coefficients)

$\sum _{n=0}^{\infty }{\frac {(1/2)_{n}^{2}}{2^{n}n!^{2}}}\sum _{n=0}^{\infty }{\frac {n(1/2)_{n}^{2}}{2^{n}n!^{2}}}={\frac {1}{\pi }}$ (where $(x)_{n}$ is the rising factorial)

$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(2n+1)}}=\pi -3$ (Nilakantha series)

$\sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}$ (where $F_{n}$ is the n-th Fibonacci number)

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

$\pi =\sum _{n=1}^{\infty }{\frac {(-1)^{\epsilon (n)}}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots$   (where $\epsilon (n)$ is the number of prime factors of the form $p\equiv 1\,(\mathrm {mod} \,4)$ of $n$)

${\frac {\pi }{2}}=\sum _{n=1}^{\infty }{\frac {(-1)^{\varepsilon (n)}}{n}}=1+{\frac {1}{2}}-{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}-{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}+\cdots$   (where $\varepsilon (n)$ is the number of prime factors of the form $p\equiv 3\,(\mathrm {mod} \,4)$ of $n$)

$\pi =\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{n+1/2}}$

$\pi ^{2}=\sum _{n=-\infty }^{\infty }{\frac {1}{(n+1/2)^{2}}}$

The last two formulas are special cases of

${\begin{aligned}{\frac {\pi }{\sin \pi x}}&=\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{n+x}}\\\left({\frac {\pi }{\sin \pi x}}\right)^{2}&=\sum _{n=-\infty }^{\infty }{\frac {1}{(n+x)^{2}}}\end{aligned}}$

which generate infinitely many analogous formulas for $\pi$ when $x\in \mathbb {Q} \setminus \mathbb {Z} .$

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

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

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

Machin-like formulae

See also: Machin-like formula

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

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

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

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

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

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

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

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

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

Infinite products

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

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

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

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

${\frac {\pi }{2}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{n}}\right)^{(-1)^{n+1}}=\left(1+{\frac {1}{1}}\right)^{+1}\left(1+{\frac {1}{2}}\right)^{-1}\left(1+{\frac {1}{3}}\right)^{+1}\cdots$ (another form of Wallis product)

Viète's formula:

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

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

${\frac {\pi }{2}}=\prod _{m\geq 1}\prod _{n\geq 1}\left({\frac {(4m^{2}+n-2)(4m^{2}+2n-1)^{2}}{4(2m^{2}+n-1)(4m^{2}+n-1)(2m^{2}+n)}}\right)^{\epsilon _{n}},$

where $\epsilon _{n}=(-1)^{t_{n}}$ and $t_{n}$ is the Thue–Morse sequence (Tóth 2020).

Arctangent formulas

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

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

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

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

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

$\pi =\arctan a+\arctan b+\arctan c$

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

$\pi =\arctan 1+\arctan 2+\arctan 3.$

Complex functions

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

The following equivalences are true for any complex $z$:

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

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

Also

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

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

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

Continued fractions

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

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

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

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

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

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

(See also Continued fraction and Generalized continued fraction.)

Iterative algorithms

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

$a_{1}=0,\,a_{n+1}={\sqrt {2+a_{n}}},\,\pi =\lim _{n\to \infty }2^{n}{\sqrt {2-a_{n}}}$ (closely related to Viète's formula)

$\omega (i_{n},i_{n-1},\dots ,i_{1})=2+i_{n}{\sqrt {2+i_{n-1}{\sqrt {2+\cdots +i_{1}{\sqrt {2}}}}}}=\omega (b_{n},b_{n-1},\dots ,b_{1}),\,i_{k}\in \{-1,1\},\,b_{k}={\begin{cases}0&{\text{if }}i_{k}=1\\1&{\text{if }}i_{k}=-1\end{cases}},\,\pi ={\displaystyle \lim _{n\rightarrow \infty }{\frac {2^{n+1}}{2h+1}}{\sqrt {\omega \left(\underbrace {10\ldots 0} _{n-m}g_{m,h+1}\right)}}}$ (where $g_{m,h+1}$ is the h+1-th entry of m-bit Gray code, $h\in \left\{0,1,\ldots ,2^{m}-1\right\}$)

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

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

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

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

Asymptotics

${\binom {2n}{n}}\sim {\frac {4^{n}}{\sqrt {\pi n}}}$ (asymptotic growth rate of the central binomial coefficients)

$C_{n}\sim {\frac {4^{n}}{\sqrt {\pi n^{3}}}}$ (asymptotic growth rate of the Catalan numbers)

$n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}$ (Stirling's approximation)

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

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

Hypergeometric inversions

With ${}_{2}F_{1}$ being the hypergeometric function, let $\left|q\right|<1$ and

$\Theta (q)=\sum _{n=-\infty }^{\infty }q^{n^{2}}$.

Then

$\Theta (q)^{2}={}_{2}F_{1}\left({\frac {1}{2}},{\frac {1}{2}},1,z\right)$

where

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

Similarly, let $\left|q\right|<1$ and

$\operatorname {H} (q)=1+240\sum _{n=1}^{\infty }\sigma _{3}(n)q^{n},$

with $\sigma _{3}$ being a divisor function. Then

$\operatorname {H} (q)={}_{2}F_{1}\left({\frac {1}{6}},{\frac {5}{6}},1,z\right)^{4}$

where

$q=\exp \left(-2\pi {\frac {{}_{2}F_{1}(1/6,5/6,1,1-z)}{{}_{2}F_{1}(1/6,5/6,1,z)}}\right),\quad z\in \mathbb {C} \setminus \{0,1\}.$

More formulas of this nature can be given, as explained by Ramanujan's theory of elliptic functions to alternative bases.

Miscellaneous

$\Gamma (s)\Gamma (1-s)={\frac {\pi }{\sin \pi s}}$ (Euler's reflection formula, see Gamma function)

$\pi ^{-s/2}\Gamma \left({\frac {s}{2}}\right)\zeta (s)=\pi ^{-(1-s)/2}\Gamma \left({\frac {1-s}{2}}\right)\zeta (1-s)$ (the functional equation of the Riemann zeta function)

$e^{-\zeta '(0)}={\sqrt {2\pi }}$

$e^{\zeta '(0,1/2)-\zeta '(0,1)}={\sqrt {\pi }}$ (where $\zeta (s,a)$ is the Hurwitz zeta function and the derivative is taken with respect to the first variable)

$\pi =\mathrm {B} (1/2,1/2)=\Gamma (1/2)^{2}$ (see also Beta function)

$\pi ={\frac {\Gamma (3/4)^{4}}{\operatorname {agm} (1,1/{\sqrt {2}})^{2}}}={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}$ (where agm is the arithmetic–geometric mean)

$\pi =\operatorname {agm} \left(\theta _{2}^{2}(1/e),\theta _{3}^{2}(1/e)\right)$ (where $\theta _{2}$ and $\theta _{3}$ are the Jacobi theta functions)

$\pi =-{\frac {\operatorname {K} (k)}{\operatorname {K} \left({\sqrt {1-k^{2}}}\right)}}\ln q,\quad k={\frac {\theta _{2}^{2}(q)}{\theta _{3}^{2}(q)}}$ (where $q\in (0,1)$ and $\operatorname {K} (k)$ is the complete elliptic integral of the first kind with modulus $k$; reflecting the nome-modulus inversion problem)

$\pi =-{\frac {\operatorname {agm} \left(1,{\sqrt {1-k'^{2}}}\right)}{\operatorname {agm} (1,k')}}\ln q,\quad k'={\frac {\theta _{4}^{2}(q)}{\theta _{3}^{2}(q)}}$ (where $q\in (0,1)$)

$\operatorname {agm} (1,{\sqrt {2}})={\frac {\pi }{\varpi }}$ (due to Gauss, $\varpi$ is the lemniscate constant)

$i\pi =\operatorname {Log} (-1)=\lim _{n\to \infty }n\left((-1)^{1/n}-1\right)$ (where $\operatorname {Log}$ is the principal value of the complex logarithm)

$1-{\frac {\pi ^{2}}{12}}=\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n{\bmod {k}})$ (where ${\textstyle n{\bmod {k}}}$ is the remainder upon division of n by k)

$\pi =\lim _{r\to \infty }{\frac {1}{r^{2}}}\sum _{x=-r}^{r}\;\sum _{y=-r}^{r}{\begin{cases}1&{\text{if }}{\sqrt {x^{2}+y^{2}}}\leq r\\0&{\text{if }}{\sqrt {x^{2}+y^{2}}}>r\end{cases}}$ (summing a circle's area)

$\pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}$ (Riemann sum to evaluate the area of the unit circle)

$\pi =\lim _{n\to \infty }{\frac {2^{4n}n!^{4}}{n(2n)!^{2}}}=\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}=\lim _{n\rightarrow \infty }{\frac {1}{n}}\left({\frac {(2n)!!}{(2n-1)!!}}\right)^{2}$ (by combining Stirling's approximation with Wallis product)

$\pi =\lim _{n\to \infty }{\frac {1}{n}}\ln {\frac {16}{\lambda (ni)}}$ (where $\lambda$ is the modular lambda function)

$\pi =\lim _{n\to \infty }{\frac {24}{\sqrt {n}}}\ln \left(2^{1/4}G_{n}\right)=\lim _{n\to \infty }{\frac {24}{\sqrt {n}}}\ln \left(2^{1/4}g_{n}\right)$ (where $G_{n}$ and $g_{n}$ are Ramanujan's class invariants)

See also

• List of mathematical identities
• Lists of mathematics topics
• List of trigonometric identities – Equalities that involve trigonometric functions
• List of topics related to π – Topics related to the mathematical constant
• List of representations of e

References

Other

• Tóth, László (2020), "Transcendental Infinite Products Associated with the +-1 Thue-Morse Sequence" (PDF), Journal of Integer Sequences, 23: 20.8.2, arXiv:2009.02025.

Further reading

• 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.