The following is a list of significant formulae involving the mathematical constant .
Many of these formulae can be found in the article Pi, or the article Approximations of .
Euclidean geometry
\pi = \frac Cd
where  is the circumference of a circle,  is the diameter.
More generally,
\pi=\frac{s}{w}
where  and  are, respectively, the perimeter and the width of any curve of constant width.
A = \pi r^2
where  is the area of a circle and  is the radius.
V = {4 \over 3}\pi r^3
where  is the volume of a sphere and  is the radius.
SA = 4\pi r^2
where  is the surface area of a sphere and  is the radius.
H = {1 \over 2}\pi^2 r^4
where  is the hypervolume of a 3-sphere and  is the radius.
SV = 2\pi^2 r^3
where  is the surface volume of a 3-sphere and  is the radius.
Physics
The cosmological constant:
\Lambda = {{8\pi G} \over {3c^2}} \rho
Heisenberg's uncertainty principle:
\Delta x\, \Delta p \ge \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_1q_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
Period of a simple pendulum with small amplitude:
T \approx 2\pi \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^2EI}{L^2}
Formulae yielding {{pi}}
Integrals
2 \int_{-1}^1 \sqrt{1-x^2}\,dx = \pi  (integrating two halves y(x)=\sqrt{r^2-x^2} to obtain the area of a circle of radius r=1)
\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 (integral form of arctan over its entire domain, giving the period of tan).
\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=\piA000796 - OEIS
\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 ).
Note that with symmetric integrands f(-x)=f(x), formulas of the form \int_{-a}^af(x)\,dx can also be translated to formulas 2\int_{0}^af(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)
12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + 3/2}}=\frac 1 \pi  (see Chudnovsky algorithm)
\frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}=\frac 1 \pi  (see Srinivasa Ramanujan, Ramanujan–Sato series)
The following are efficient for calculating arbitrary binary digits of :
\sum_{k=0}^\infty \frac{(-1)^k}{4^k}\left(\frac{2}{4k+1}+\frac{2}{4k+2}+\frac{1}{4k+3}\right)=\pi page 126
\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)
\frac{1}{2^6} \sum_{n=0}^{\infty} \frac{{(-1)}^n}{2^{10n}} \left( - \frac{2^5}{4n+1} - \frac{1}{4n+3} + \frac{2^8}{10n+1} - \frac{2^6}{10n+3} - \frac{2^2}{10n+5} - \frac{2^2}{10n+7} + \frac{1}{10n+9} \right)=\pi
Plouffe's series for calculating arbitrary decimal digits of :
\sum_{k=1}^\infty k\frac{2^kk!^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 B2n is a Bernoulli number.
\sum_{n=1}^\infty \frac{3^n - 1}{4^n}\, \zeta(n+1) = \piWeisstein, Eric W. "Pi Formulas", MathWorld
\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}{3^n(2n+1)}=1-\frac{1}{3\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}{\sqrt{12}} (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 \frac1{(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)^{n}}{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)^{n}}{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)^{n}}{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}
\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) (page 647)
\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\sqrt5} (where F_n is the n-th Fibonacci number)
\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 \varepsilon (n) is the number of prime factors of the form p\equiv 1\,(\mathrm{mod}\,4) of n; Euler, 1748)Carl B. Boyer, A History of Mathematics, Chapter 21., pp.
488–489
Some formulas relating  and harmonic numbers are given here.
Machin-like formulae
\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}
\frac{\pi}{2} = \sum_{n=0}^\infty \arctan\frac{1}{F_{2n+1}} = \arctan\frac{1}{1} + \arctan\frac{1}{2} + \arctan\frac{1}{5} + \arctan\frac{1}{13} + \cdots
where F_n is the n-th Fibonacci number.
Infinite series
Some infinite series involving π are:<br/>
where (x)_n  is the Pochhammer symbol for the rising factorial.
See also Ramanujan–Sato series.
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 ,
\prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1} = \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 = \frac{4}{3} \cdot \frac{16}{15} \cdot \frac{36}{35} \cdot \frac{64}{63} \cdots = \frac{\pi}{2}  (see also Wallis product)
Viète's formula:
\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots = \frac2\pi
A double infinite product formula involving the Thue-Morse sequence:
\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} = \frac{\pi}{2},
where \epsilon_n = (-1)^{t_n} and t_n is the Thue-Morse sequence .
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} .
\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.
Continued fractions
\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}}}}}
\pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \cfrac{7^2}{2 + \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 third 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}}{2 h+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\} )
a_1=1,\, a_{n+1}=a_n+\sin a_n,\, \pi =\lim_{n\to\infty}a_n (cubic convergence) page 49
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.
Miscellaneous
n!
\sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n (Stirling's approximation)
e^{i \pi} +1 = 0 (Euler's identity)
\sum_{k=1}^{n} \varphi (k) \sim \frac{3n^2}{\pi^2} (see Euler's totient function)
\sum_{k=1}^{n} \frac {\varphi (k)} {k} \sim \frac{6n}{\pi^2} (see Euler's totient function)
\pi =\Beta (1/2,1/2)=\Gamma (1/2)^2 (see also Beta function and Gamma 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 page 225)
\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) page 41
\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)
\lim_{n\rightarrow \infty}\frac{1}{n^2} \sum_{k=1}^n (n\bmod k) = 1-\frac{\pi^2}{12} (where  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} \le 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 \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 Stirling's approximation)
\pi=\lim_{n\to\infty}\frac{1}{n}\ln\frac{16}{\lambda (ni)} (the numbers \lambda (ni) are positive and algebraic for every n\in\mathbb{N}; \lambda is the modular lambda function)
See also
References
.
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, .
