List of formulae involving π

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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

[math]\displaystyle{ \pi = \frac Cd = \frac C{2r} }[/math]

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

[math]\displaystyle{ \pi=\frac{L}{w} }[/math]

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

[math]\displaystyle{ A = \pi r^2 }[/math]

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

[math]\displaystyle{ A = \pi ab }[/math]

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

[math]\displaystyle{ A=4\pi r^2 }[/math]

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

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

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

[math]\displaystyle{ A=(k+1)(k+2)\pi r^2 }[/math]

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

[math]\displaystyle{ A=\frac{(-1)^k+3}{8}\pi a^2 }[/math]

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

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

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

[math]\displaystyle{ V = {4 \over 3}\pi r^3 }[/math]

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

[math]\displaystyle{ SA = 4\pi r^2 }[/math]

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

[math]\displaystyle{ H = {1 \over 2}\pi^2 r^4 }[/math]

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

[math]\displaystyle{ SV = 2\pi^2 r^3 }[/math]

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:

[math]\displaystyle{ S=(n-2)\pi }[/math]

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

[math]\displaystyle{ A=\frac{ns^2}{4}\cot\frac{\pi}{n} }[/math]

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

[math]\displaystyle{ r=\frac{s}{2}\cot\frac{\pi}{n} }[/math]

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

[math]\displaystyle{ R=\frac{s}{2}\csc\frac{\pi}{n} }[/math]

Physics

• The cosmological constant:

[math]\displaystyle{ \Lambda = {{8\pi G} \over {3c^2}} \rho }[/math]

• Heisenberg's uncertainty principle:

[math]\displaystyle{ \Delta x\, \Delta p \ge \frac h {4\pi} }[/math]

• Einstein's field equation of general relativity:

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

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

[math]\displaystyle{ F = \frac{|q_1q_2|}{4 \pi \varepsilon_0 r^2} }[/math]

• Magnetic permeability of free space:

[math]\displaystyle{ \mu_0 \approx 4 \pi \cdot 10^{-7}\,\mathrm{N}/\mathrm{A}^2 }[/math]

• Approximate period of a simple pendulum with small amplitude:

[math]\displaystyle{ T \approx 2\pi \sqrt\frac L g }[/math]

• Exact period of a simple pendulum with amplitude [math]\displaystyle{ \theta_0 }[/math] ([math]\displaystyle{ \operatorname{agm} }[/math] is the arithmetic–geometric mean):

[math]\displaystyle{ T=\frac{2\pi}{\operatorname{agm}(1,\cos (\theta_0/2))}\sqrt{\frac{L}{g}} }[/math]

• Kepler's third law of planetary motion:

[math]\displaystyle{ \frac{R^3}{T^2} = \frac{GM}{4\pi^2} }[/math]

• The buckling formula:

[math]\displaystyle{ F =\frac{\pi^2EI}{L^2} }[/math]

A puzzle involving "colliding billiard balls":

[math]\displaystyle{ \lfloor{b^N\pi}\rfloor }[/math]

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

[math]\displaystyle{ 2 \int_{-1}^1 \sqrt{1-x^2}\,dx = \pi }[/math] (integrating two halves [math]\displaystyle{ y(x)=\sqrt{1-x^2} }[/math] to obtain the area of the unit circle)

[math]\displaystyle{ \int_{-\infty}^\infty \operatorname{sech}x \, dx = \pi }[/math]

[math]\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 }[/math]

[math]\displaystyle{ \int_{-1}^1\frac{dx}{\sqrt{1-x^2}} = \pi }[/math]

[math]\displaystyle{ \int_{-\infty}^\infty\frac{dx}{1+x^2} = \pi }[/math] (see also Cauchy distribution)

[math]\displaystyle{ \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi} }[/math] (see Gaussian integral).

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

[math]\displaystyle{ \int_0^\infty \ln\left(1+\frac{1}{x^2}\right)\, dx=\pi }[/math]

[math]\displaystyle{ \int_{-\infty}^\infty \frac{\sin x} x \,dx=\pi }[/math]

[math]\displaystyle{ \int_0^1 {x^4(1-x)^4 \over 1+x^2}\,dx = {22 \over 7} - \pi }[/math] (see also Proof that 22/7 exceeds π).

[math]\displaystyle{ \int_0^\infty \frac{x^{\alpha-1}}{x+1}\, dx=\frac{\pi}{\sin \pi\alpha},\quad 0\lt \alpha\lt 1 }[/math]

[math]\displaystyle{ \int_0^\infty \frac{dx}{\sqrt{x(x+a)(x+b)}}=\frac{\pi}{\operatorname{agm}(\sqrt{a},\sqrt{b})} }[/math] (where [math]\displaystyle{ \operatorname{agm} }[/math] is the arithmetic–geometric mean; see also elliptic integral)

Note that with symmetric integrands [math]\displaystyle{ f(-x)=f(x) }[/math], formulas of the form [math]\displaystyle{ \int_{-a}^af(x)\,dx }[/math] can also be translated to formulas [math]\displaystyle{ 2\int_{0}^af(x)\,dx }[/math].

Efficient infinite series

[math]\displaystyle{ \sum_{k=0}^\infty \frac{k!}{(2k+1)!!} = \sum_{k=0}^\infty\frac{2^k k!^2}{(2k+1)!} = \frac \pi 2 }[/math] (see also Double factorial)

[math]\displaystyle{ \sum_{k=0}^\infty \frac{k!\,(2k)!\,(25k-3)}{(3k)!\,2^{k}}=\frac{\pi}{2} }[/math]

[math]\displaystyle{ \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k}}=\frac{4270934400}{\sqrt{10005}\pi} }[/math] (see Chudnovsky algorithm)

[math]\displaystyle{ \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}=\frac{9801}{2\sqrt{2}\pi} }[/math] (see Srinivasa Ramanujan, Ramanujan–Sato series)

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

[math]\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 }[/math]

[math]\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 }[/math] (see Bailey–Borwein–Plouffe formula)

[math]\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 }[/math]

[math]\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 }[/math]

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

[math]\displaystyle{ \sum_{k=1}^\infty k\frac{2^kk!^2}{(2k)!}=\pi +3 }[/math]

Other infinite series

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

[math]\displaystyle{ \zeta(4)= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots = \frac{\pi^4}{90} }[/math]

[math]\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)!} }[/math] , where B_2n is a Bernoulli number.

[math]\displaystyle{ \sum_{n=1}^\infty \frac{3^n - 1}{4^n}\, \zeta(n+1) = \pi }[/math]

[math]\displaystyle{ \sum_{n=2}^\infty \frac{2(3/2)^n-3}{n}(\zeta (n)-1)=\ln \pi }[/math]

[math]\displaystyle{ \sum_{n=1}^\infty \zeta (2n)\frac{x^{2n}}{n}=\ln\frac{\pi x}{\sin \pi x},\quad 0\lt |x|\lt 1 }[/math]

[math]\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} }[/math] (see Leibniz formula for pi)

[math]\displaystyle{ \sum_{n=0}^\infty \frac{(-1)^{(n^2-n)/2}}{2n+1}=1+\frac13-\frac15-\frac17+\frac19+\frac{1}{11}-\cdots=\frac{\pi}{2\sqrt{2}} }[/math] (Newton, Second Letter to Oldenburg, 1676)

[math]\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}} }[/math] (Madhava series)

[math]\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} }[/math]

[math]\displaystyle{ \sum_{n=1}^\infty \frac1{(2n)^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + \cdots = \frac{\pi^2}{24} }[/math]

[math]\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} }[/math]

[math]\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} }[/math]

[math]\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} }[/math]

[math]\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} }[/math]

[math]\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} }[/math]

In general,

[math]\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 }[/math]

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

[math]\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} }[/math]

[math]\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} }[/math]

[math]\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} }[/math] (see Gregory coefficients)

[math]\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} }[/math] (where [math]\displaystyle{ (x)_n }[/math] is the rising factorial)

[math]\displaystyle{ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(2n+1)}=\pi -3 }[/math] (Nilakantha series)

[math]\displaystyle{ \sum_{n=1}^\infty \frac{F_{2n}}{n^2 \binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5} }[/math] (where [math]\displaystyle{ F_n }[/math] is the n-th Fibonacci number)

[math]\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 }[/math]   (where [math]\displaystyle{ \epsilon (n) }[/math] is the number of prime factors of the form [math]\displaystyle{ p\equiv 1\,(\mathrm{mod}\,4) }[/math] of [math]\displaystyle{ n }[/math])

[math]\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 }[/math]   (where [math]\displaystyle{ \varepsilon (n) }[/math] is the number of prime factors of the form [math]\displaystyle{ p\equiv 3\, (\mathrm{mod}\, 4) }[/math] of [math]\displaystyle{ n }[/math])

[math]\displaystyle{ \pi=\sum_{n=-\infty}^\infty \frac{(-1)^n}{n+1/2} }[/math]

[math]\displaystyle{ \pi^2=\sum_{n=-\infty}^\infty \frac{1}{(n+1/2)^2} }[/math]

The last two formulas are special cases of

[math]\displaystyle{ \begin{align}\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{align} }[/math]

which generate infinitely many analogous formulas for [math]\displaystyle{ \pi }[/math] when [math]\displaystyle{ x\in\mathbb{Q}\setminus\mathbb{Z}. }[/math]

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

[math]\displaystyle{ \pi=\frac{1}{Z} }[/math]	[math]\displaystyle{ Z=\sum_{n=0}^{\infty } \frac{((2n)!)^3(42n+5)} {(n!)^6{16}^{3n+1}} }[/math]
[math]\displaystyle{ \pi=\frac{4}{Z} }[/math]	[math]\displaystyle{ Z=\sum_{n=0}^{\infty } \frac{(-1)^n(4n)!(21460n+1123)} {(n!)^4{441}^{2n+1}{2}^{10n+1}} }[/math]
[math]\displaystyle{ \pi=\frac{4}{Z} }[/math]	[math]\displaystyle{ Z=\sum_{n=0}^{\infty } \frac{(6n+1)\left ( \frac{1}{2} \right )^3_n} {{4^n}(n!)^3} }[/math]
[math]\displaystyle{ \pi=\frac{32}{Z} }[/math]	[math]\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 )^3_n} {{64^n}(n!)^3} }[/math]
[math]\displaystyle{ \pi=\frac{27}{4Z} }[/math]	[math]\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} }[/math]
[math]\displaystyle{ \pi=\frac{15\sqrt{3}}{2Z} }[/math]	[math]\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} }[/math]
[math]\displaystyle{ \pi=\frac{85\sqrt{85}}{18\sqrt{3}Z} }[/math]	[math]\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} }[/math]
[math]\displaystyle{ \pi=\frac{5\sqrt{5}}{2\sqrt{3}Z} }[/math]	[math]\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} }[/math]
[math]\displaystyle{ \pi=\frac{2\sqrt{3}}{Z} }[/math]	[math]\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}} }[/math]
[math]\displaystyle{ \pi=\frac{\sqrt{3}}{9Z} }[/math]	[math]\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}} }[/math]
[math]\displaystyle{ \pi=\frac{2\sqrt{11}}{11Z} }[/math]	[math]\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}} }[/math]
[math]\displaystyle{ \pi=\frac{\sqrt{2}}{4Z} }[/math]	[math]\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}} }[/math]
[math]\displaystyle{ \pi=\frac{4\sqrt{5}}{5Z} }[/math]	[math]\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!)^35^n{72}^{2n+1}} }[/math]
[math]\displaystyle{ \pi=\frac{4\sqrt{3}}{3Z} }[/math]	[math]\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}} }[/math]
[math]\displaystyle{ \pi=\frac{4}{Z} }[/math]	[math]\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}} }[/math]
[math]\displaystyle{ \pi=\frac{72}{Z} }[/math]	[math]\displaystyle{ Z=\sum_{n=0}^\infty \frac{(-1)^n(4n)!(260n+23)}{(n!)^44^{4n}18^{2n}} }[/math]
[math]\displaystyle{ \pi=\frac{3528}{Z} }[/math]	[math]\displaystyle{ Z=\sum_{n=0}^\infty \frac{(-1)^n(4n)!(21460n+1123)}{(n!)^44^{4n}882^{2n}} }[/math]

where [math]\displaystyle{ (x)_n }[/math] is the Pochhammer symbol for the rising factorial. See also Ramanujan–Sato series.

Machin-like formulae

[math]\displaystyle{ \frac{\pi}{4} = \arctan 1 }[/math]

[math]\displaystyle{ \frac{\pi}{4} = \arctan\frac{1}{2} + \arctan\frac{1}{3} }[/math]

[math]\displaystyle{ \frac{\pi}{4} = 2 \arctan\frac{1}{2} - \arctan\frac{1}{7} }[/math]

[math]\displaystyle{ \frac{\pi}{4} = 2 \arctan\frac{1}{3} + \arctan\frac{1}{7} }[/math]

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

[math]\displaystyle{ \frac{\pi}{4} = 5 \arctan\frac{1}{7} + 2 \arctan\frac{3}{79} }[/math]

[math]\displaystyle{ \frac{\pi}{4} = 6 \arctan\frac{1}{8} + 2 \arctan\frac{1}{57} + \arctan\frac{1}{239} }[/math]

[math]\displaystyle{ \frac{\pi}{4} = 12 \arctan\frac{1}{49} + 32 \arctan\frac{1}{57} - 5 \arctan\frac{1}{239} + 12 \arctan\frac{1}{110443} }[/math]

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

Infinite products

[math]\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, }[/math] (Euler)

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

[math]\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 , }[/math]

[math]\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 }[/math] (see also Wallis product)

[math]\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 }[/math] (another form of Wallis product)

Viète's formula:

[math]\displaystyle{ \frac{2}{\pi}=\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots }[/math]

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

[math]\displaystyle{ \frac{\pi}{2}=\prod_{m\geq1} \prod_{n\geq1} \left( \frac{(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)} \right) ^{\epsilon_n}, }[/math]

where [math]\displaystyle{ \epsilon_n = (-1)^{t_n} }[/math] and [math]\displaystyle{ t_n }[/math] is the Thue–Morse sequence (Tóth 2020).

Arctangent formulas

[math]\displaystyle{ \frac{\pi}{2^{k+1}}=\arctan \frac{\sqrt{2-a_{k-1}}}{a_k}, \qquad\qquad k\geq 2 }[/math]

[math]\displaystyle{ \frac{\pi}{4}=\sum_{k\geq 2}\arctan \frac{\sqrt{2-a_{k-1}}}{a_k}, }[/math]

where [math]\displaystyle{ a_k=\sqrt{2+a_{k-1}} }[/math] such that [math]\displaystyle{ a_1=\sqrt{2} }[/math].

[math]\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 }[/math]

where [math]\displaystyle{ F_k }[/math] is the k-th Fibonacci number.

[math]\displaystyle{ \pi =\arctan a+\arctan b+\arctan c }[/math]

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

[math]\displaystyle{ \pi =\arctan 1+\arctan 2+\arctan 3. }[/math]

Complex exponential formulas

[math]\displaystyle{ e^{i \pi} +1 = 0 }[/math] (Euler's identity)

The following equivalences are true for any complex [math]\displaystyle{ z }[/math]:

[math]\displaystyle{ e^z\in\mathbb{R}\leftrightarrow\Im z\in\pi\mathbb{Z} }[/math]

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

Also

[math]\displaystyle{ \frac{1}{e^z-1}=\lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{z-2\pi i n}-\frac{1}{2},\quad z\in\mathbb{C}. }[/math]

Continued fractions

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

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

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

[math]\displaystyle{ \pi = \cfrac{4}{1 + \cfrac{1^2}{3 + \cfrac{2^2}{5 + \cfrac{3^2}{7 + \cfrac{4^2}{9 + \ddots}}}}} }[/math]

[math]\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}}}}}} }[/math]

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

(See also Continued fraction and Generalized continued fraction.)

Iterative algorithms

[math]\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} }[/math]

[math]\displaystyle{ a_1=0,\, a_{n+1}=\sqrt{2+a_n},\, \pi =\lim_{n\to\infty} 2^n\sqrt{2-a_n} }[/math] (closely related to Viète's formula)

[math]\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}}{2 h+1} \sqrt{\omega\left(\underbrace{10 \ldots 0}_{n-m} g_{m, h+1}\right)}} }[/math] (where [math]\displaystyle{ g_{m, h+1} }[/math] is the h+1-th entry of m-bit Gray code, [math]\displaystyle{ h \in \left\{0,1, \ldots, 2^{m}-1\right\} }[/math])

[math]\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 }[/math] (quadratic convergence)

[math]\displaystyle{ a_1=1,\, a_{n+1}=a_n+\sin a_n,\, \pi =\lim_{n\to\infty}a_n }[/math] (cubic convergence)

[math]\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 }[/math] (Archimedes' algorithm, see also harmonic mean and geometric mean)

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

Asymptotics

[math]\displaystyle{ \binom{2n}{n}\sim \frac{4^{n}}{\sqrt{\pi n}} }[/math] (asymptotic growth rate of the central binomial coefficients)

[math]\displaystyle{ C_n\sim \frac{4^{n}}{\sqrt{\pi n^3}} }[/math] (asymptotic growth rate of the Catalan numbers)

[math]\displaystyle{ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n }[/math] (Stirling's approximation)

[math]\displaystyle{ \sum_{k=1}^{n} \varphi (k) \sim \frac{3n^2}{\pi^2} }[/math] (where [math]\displaystyle{ \varphi }[/math] is Euler's totient function)

[math]\displaystyle{ \sum_{k=1}^{n} \frac {\varphi (k)} {k} \sim \frac{6n}{\pi^2} }[/math]

Miscellaneous

[math]\displaystyle{ \Gamma (s)\Gamma (1-s)=\frac{\pi}{\sin \pi s} }[/math] (Euler's reflection formula, see Gamma function)

[math]\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) }[/math] (the functional equation of the Riemann zeta function)

[math]\displaystyle{ e^{-\zeta'(0)}=\sqrt{2\pi} }[/math]

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

[math]\displaystyle{ \pi =\Beta (1/2,1/2)=\Gamma (1/2)^2 }[/math] (see also Beta function)

[math]\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} }[/math] (where agm is the arithmetic–geometric mean)

[math]\displaystyle{ \pi = \operatorname{agm}\left(\theta_2^2(1/e),\theta_3^2(1/e)\right) }[/math] (where [math]\displaystyle{ \theta_2 }[/math] and [math]\displaystyle{ \theta_3 }[/math] are the Jacobi theta functions)

[math]\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)} }[/math] (where [math]\displaystyle{ q\in (0,1) }[/math] and [math]\displaystyle{ \operatorname{K}(k) }[/math] is the complete elliptic integral of the first kind with modulus [math]\displaystyle{ k }[/math]; reflecting the nome-modulus inversion problem)

[math]\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)} }[/math] (where [math]\displaystyle{ q\in (0,1) }[/math])

[math]\displaystyle{ \operatorname{agm}(1,\sqrt{2})=\frac{\pi}{\varpi} }[/math] (due to Gauss, [math]\displaystyle{ \varpi }[/math] is the lemniscate constant)

[math]\displaystyle{ i\pi=\operatorname{Log}(-1)=\lim_{n\to\infty}n\left((-1)^{1/n}-1\right) }[/math] (where [math]\displaystyle{ \operatorname{Log} }[/math] is the principal value of the complex logarithm)

[math]\displaystyle{ 1-\frac{\pi^2}{12}=\lim_{n\rightarrow \infty}\frac{1}{n^2} \sum_{k=1}^n (n\bmod k) }[/math] (where [math]\displaystyle{ n\bmod k }[/math] is the remainder upon division of n by k)

[math]\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} \le r \\ 0 & \text{if } \sqrt{x^2+y^2} \gt r \end{cases} }[/math] (summing a circle's area)

[math]\displaystyle{ \pi = \lim_{n \rightarrow \infty} \frac{4}{n^2} \sum_{k=1}^n \sqrt{n^2 - k^2} }[/math] (Riemann sum to evaluate the area of the unit circle)

[math]\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 }[/math] (by combining Stirling's approximation with Wallis product)

[math]\displaystyle{ \pi=\lim_{n\to\infty}\frac{1}{n}\ln\frac{16}{\lambda (ni)} }[/math] (where [math]\displaystyle{ \lambda }[/math] is the modular lambda function)

[math]\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) }[/math] (where [math]\displaystyle{ G_n }[/math] and [math]\displaystyle{ g_n }[/math] are Ramanujan's class invariants)

See also

• List of mathematical identities
• Lists of mathematics topics – None
• List of trigonometric identities – Equalities that involve trigonometric functions
• List of topics related to π – Wikipedia list article
• List of representations of e

References

Other

• Tóth, László (2020), "Transcendental Infinite Products Associated with the +-1 Thue-Morse Sequence", Journal of Integer Sequences 23: 20.8.2, https://cs.uwaterloo.ca/journals/JIS/VOL23/Toth/toth46.pdf .

Further reading

• Peter Borwein, The Amazing Number Pi
• Kazuya Kato, Nobushige Kurokawa, Saito Takeshi: Number Theory 1: Fermat's Dream. American Mathematical Society, Providence 1993, ISBN 0-8218-0863-X.