List of formulae involving π

Template:Pi box The following is a list of significant formulae involving the mathematical constant [[pi|Template:Pi]]. Many of these formulae can be found in the article Pi, or the article [[Approximations of π|Approximations of Template:Pi]].

Euclidean geometry

$\pi =\frac{C}{d}=\frac{C}{2r^{2r+1}}$

where Template:Math is the circumference of a circle, Template:Math is the diameter, and Template:Math is the radius. More generally,

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

where Template:Math and Template:Math are, respectively, the perimeter and the width of any curve of constant width.

$A=\pi r^2$

where Template:Math is the area of a circle. More generally,

$A=\pi ab$

where Template:Math is the area enclosed by an ellipse with semi-major axis Template:Math and semi-minor axis Template:Math.

$C=\frac{2\pi }{\mathrm{agm}(a,b)}(a_1^2-\sum _{n=2}^{\mathrm{\infty }}{2}^{n-1}(a_n^2-b_n^2))$

where Template:Math is the circumference of an ellipse with semi-major axis Template:Math and semi-minor axis Template:Math and $a_n,b_n$ are the arithmetic and geometric iterations of $\mathrm{agm}(a,b)$, the arithmetic-geometric mean of Template:Math and Template:Math with the initial values $a_0=a$ and $b_0=b$.

$A=4\pi r^2$

where Template:Math is the area between the witch of Agnesi and its asymptotic line; Template:Math is the radius of the defining circle.

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

where Template:Math is the area of a squircle with minor radius Template:Math, $\mathrm{\Gamma }$ is the gamma function.

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

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

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

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

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

where Template:Math is the perimeter of the lemniscate of Bernoulli with focal distance Template:Math.

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

where Template:Math is the volume of a sphere and Template:Math is the radius.

$SA=4\pi r^2$

where Template:Math is the surface area of a sphere and Template:Math is the radius.

$HV=\frac{1}{2}{\pi }^2{r}^4$

where Template:Math is the hypervolume of a 3-sphere and Template:Math is the radius.

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

where Template:Math is the surface volume of a 3-sphere and Template:Math is the radius.

Regular convex polygons

Sum Template:Math of internal angles of a regular convex polygon with Template:Math sides:

$S=(n-2)\pi$

Area Template:Math of a regular convex polygon with Template:Math sides and side length Template:Math:

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

Inradius Template:Math of a regular convex polygon with Template:Math sides and side length Template:Math:

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

Circumradius Template:Math of a regular convex polygon with Template:Math sides and side length Template:Math:

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

Physics

• The cosmological constant:

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

• Heisenberg's uncertainty principle:

  $\mathrm{\Delta }x\mathrm{\Delta }p\ge \frac{h}{4\pi }$

• Einstein's field equation of general relativity:

  $R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R+\mathrm{\Lambda }{g}_{\mu \nu }=\frac{8\pi G}{c^4}{T}_{\mu \nu }$

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

  $F=\frac{|q_1{q}_2|}{4\pi {\epsilon }_0{r}^2}$

• 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 ${\theta }_0$ ($\mathrm{agm}$ is the arithmetic–geometric mean):

  $T=\frac{2\pi }{\mathrm{agm}(1,\cos ({\theta }_0/2))}\sqrt{\frac{L}{g}}$

• Period of a spring-mass system with spring constant $k$ and mass $m$:

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

• 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)

${\int }_0^2\sqrt{4-x^2}dx=\pi$ (integrating a quarter of a circle with a radius of two $x^2+y^2=4$ to obtain $4\pi /4$)

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{sech}xdx=\pi$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_t^{\mathrm{\infty }}{e}^{-1/2t^2-x^2+xt}dxdt={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_t^{\mathrm{\infty }}{e}^{-t^2-1/2x^2+xt}dxdt=\pi$

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

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

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

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$ (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).

${\int }_0^{\mathrm{\infty }}\ln (1+\frac{1}{x^2})dx=\pi$

${\int }_0^1\frac{x^4(1-x{)}^4}{1+x^2}dx=\frac{22}{7}-\pi$ (see also [[Proof that 22/7 exceeds π|Proof that 22/7 exceeds Template:Pi]]).

${\int }_0^1\frac{x^2(1+x{)}^4}{1+x^2}dx=\pi -\frac{17}{15}$

${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{x+1}dx=\frac{\pi }{\sin \pi \alpha },0<\alpha <1$

${\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{x(x+a)(x+b)}}=\frac{\pi }{\mathrm{agm}(\sqrt{a},\sqrt{b})}$ (where $\mathrm{agm}$ is the arithmetic–geometric mean; see also elliptic integral)

Note that with symmetric integrands $f(-x)=f(x)$, formulas of the form ${\int }_{-a}^{a}f(x)dx$ can also be translated to formulas $2{\int }_0^{a}f(x)dx$.

Efficient infinite series

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

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

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

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

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

The following are efficient for calculating arbitrary binary digits of Template:Pi:

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

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

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

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

Plouffe's series for calculating arbitrary decimal digits of Template:Pi:

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

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)

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

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

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

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

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

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

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

$\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{3n+1}-\frac{1}{3n+2})=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\cdots =\frac{\pi }{3\sqrt{3}}$

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

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

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

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

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

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

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

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

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

In general,

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^{2k+1}}=(-1{)}^k\frac{E_{2k}}{2(2k)!}{\left(\frac{\pi }{2}\right)}^{2k+1},k\in {\mathbb{N}}_0$

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

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

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

$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{(n^2+n)/2+1}\left|G_{((-1{)}^{n+1}+6n-3)/4}\right|=|G_1|+|G_2|-|G_4|-|G_5|+|G_7|+|G_8|-|G_{10}|-|G_{11}|+\cdots =\frac{\sqrt{3}}{\pi }$ (see Gregory coefficients)

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

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

$\sum _{n=1}^{\mathrm{\infty }}\frac{F_{2n}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}=\frac{4{\pi }^2}{25\sqrt{5}}$ (where $F_{2n}$ is the $2n$th Fibonacci number)

$\sum _{n=1}^{\mathrm{\infty }}\frac{L_{2n}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}=\frac{{\pi }^2}{5}$ (where $L_n$ is the $n$th Lucas number)

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

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

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

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

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

The last two formulas are special cases of

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

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

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

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

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

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

Machin-like formulae

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

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

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

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

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

$\frac{\pi }{4}=4\arctan \frac{1}{5}-\arctan \frac{1}{239}$ (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

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

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

$\frac{\sqrt{3}\pi }{6}=\left(\prod _{\genfrac{}{}{0ex}{}{p\equiv 1(\mathrm{mod}6)}{p\in \mathbb{P}}}\frac{p}{p-1}\right)\cdot \left(\prod _{\genfrac{}{}{0ex}{}{p\equiv 5(\mathrm{mod}6)}{p\in \mathbb{P}}}\frac{p}{p+1}\right)=\frac{5}{6}\cdot \frac{7}{6}\cdot \frac{11}{12}\cdot \frac{13}{12}\cdot \frac{17}{18}\cdots$

$\frac{2{\pi }^2}{25}=\left(\prod _{\genfrac{}{}{0ex}{}{p\equiv 1(\mathrm{mod}5)}{p\in \mathbb{P}}}\frac{p^2}{(p-1{)}^2}\right)\left(\prod _{\genfrac{}{}{0ex}{}{p\equiv 2,3(\mathrm{mod}5)}{p\in \mathbb{P}}}\frac{p^2}{p^2+1}\right)\left(\prod _{\genfrac{}{}{0ex}{}{p\equiv 4(\mathrm{mod}5)}{p\in \mathbb{P}}}\frac{p^2}{(p+1{)}^2}\right)=\frac{4}{5}\cdot \frac{9}{10}\cdot \frac{49}{50}\cdot \frac{121}{100}\cdot \frac{169}{170}\cdots$,

Where p is p also the prime number.

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

$\frac{\pi }{2}=\prod _{n=1}^{\mathrm{\infty }}{(1+\frac{1}{n})}^{(-1{)}^{n+1}}={(1+\frac{1}{1})}^{+1}{(1+\frac{1}{2})}^{-1}{(1+\frac{1}{3})}^{+1}\cdots$ (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:

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

where ${ϵ}_n=(-1{)}^{t_n}$ and $t_n$ is the Thue–Morse sequence Template:Harv.

One elegant infinite product formula:

$\pi =\prod _{n=1}^{\mathrm{\infty }}(1+\frac{(n+2)+(2n+1)+(2n-1)}{(n+2)\times (2n+1)\times (2n-1)})$

Arctangent formulas

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

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

where $a_k=\sqrt{2+a_{k-1}}$ such that $a_1=\sqrt{2}$.

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

where $F_k$ is the $k$th Fibonacci number.

$\frac{\pi }{4}=\arctan \frac{a}{b+c}+\arctan \frac{b}{a+c},$

For Pythagorean triple (a,b,c).

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

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 $z$:

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

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

Also

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

Suppose a lattice $\mathrm{\Omega }$ is generated by two periods ${\omega }_1,{\omega }_2$. We define the quasi-periods of this lattice by ${\eta }_1=\zeta (z+{\omega }_1;\mathrm{\Omega })-\zeta (z;\mathrm{\Omega })$ and ${\eta }_2=\zeta (z+{\omega }_2;\mathrm{\Omega })-\zeta (z;\mathrm{\Omega })$ 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

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

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

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

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

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

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

$\frac{2}{\pi }=\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{1+\frac{3}{1+\frac{5}{\ddots }}}}}}$ (The partial numerators consist of two interleaved sequences: the sequence of natural numbers and the sequence of odd numbers, alternating by index.)

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

Iterative algorithms

$a_0=1,a_{n+1}=(1+\frac{1}{2n+1})a_n,\pi =\underset{n\to \mathrm{\infty }}{lim}\frac{a_n^2}{n}$

$a_1=0,a_{n+1}=\sqrt{2+a_n},\pi =\underset{n\to \mathrm{\infty }}{lim}{2}^n\sqrt{2-a_n}$ (closely related to Viète's formula)

$\omega (i_n,i_{n-1},\dots ,i_1)=2+i_n\sqrt{2+i_{n-1}\sqrt{2+\cdots +i_1\sqrt{2}}}=\omega (b_n,b_{n-1},\dots ,b_1),i_k\in \{-1,1\},b_k=\{\begin{array}{cc}0& \text{if }{i}_k=1\\ 1& \text{if }{i}_k=-1\end{array},\pi =\underset{n\to \mathrm{\infty }}{lim}\frac{2^{n+1}}{2h+1}\sqrt{\omega \left(\underset{n-m}{\underbrace{10\dots 0}}{g}_{m,h+1}\right)}$ (where $g_{m,h+1}$ is the h+1-th entry of m-bit Gray code, $h\in \{0,1,\dots ,2^m-1\}$)

$\mathrm{\forall }k\in \mathbb{N},a_1=2^{-k},a_{n+1}=a_n+2^{-k}(1-\tan (2^{k-1}{a}_n)),\pi =2^{k+1}\underset{n\to \mathrm{\infty }}{lim}{a}_n$ (quadratic convergence)

$a_1=1,a_{n+1}=a_n+\sin a_n,\pi =\underset{n\to \mathrm{\infty }}{lim}{a}_n$ (cubic convergence)

$a_0=2\sqrt{3},b_0=3,a_{n+1}=\mathrm{hm}(a_n,b_n),b_{n+1}=\mathrm{gm}(a_{n+1},b_n),\pi =\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{b}_n$ (Archimedes' algorithm, see also harmonic mean and geometric mean)

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

Asymptotics

$(\genfrac{}{}{0ex}{}{2n}{n})\sim \frac{4^n}{\sqrt{\pi n}}$ (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)

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

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

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

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

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

Hypergeometric inversions

With ${}_2{F}_1$ being the hypergeometric function:

$\sum _{n=0}^{\mathrm{\infty }}{r}_2(n)q^n={}_2{F}_1(\frac{1}{2},\frac{1}{2},1,z)$

where

$q=\exp (-\pi \frac{{}_2{F}_1(1/2,1/2,1,1-z)}{{}_2{F}_1(1/2,1/2,1,z)}),z\in ℂ\setminus \{0,1\}$

and $r_2$ is the sum of two squares function.

Similarly,

$1+240\sum _{n=1}^{\mathrm{\infty }}{\sigma }_3(n)q^n={}_2{F}_1{(\frac{1}{6},\frac{5}{6},1,z)}^4$

where

$q=\exp (-2\pi \frac{{}_2{F}_1(1/6,5/6,1,1-z)}{{}_2{F}_1(1/6,5/6,1,z)}),z\in ℂ\setminus \{0,1\}$

and ${\sigma }_3$ is a divisor function.

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

Perhaps the most notable hypergeometric inversions are the following two examples, involving the Ramanujan tau function $\tau$ and the Fourier coefficients $\mathrm{j}$ of the J-invariant (Template:Oeis):

$\sum _{n=-1}^{\mathrm{\infty }}{\mathrm{j}}_n{q}^n=256\frac{(1-z+z^2{)}^3}{z^2(1-z{)}^2},$

$\sum _{n=1}^{\mathrm{\infty }}\tau (n)q^n=\frac{z^2(1-z{)}^2}{256}{}_2{F}_1{(\frac{1}{2},\frac{1}{2},1,z)}^{12}$

where in both cases

$q=\exp (-2\pi \frac{{}_2{F}_1(1/2,1/2,1,1-z)}{{}_2{F}_1(1/2,1/2,1,z)}),z\in ℂ\setminus \{0,1\}.$

Furthermore, by expanding the last expression as a power series in

$\frac{1}{2}\frac{1-(1-z{)}^{1/4}}{1+(1-z{)}^{1/4}}$

and setting $z=1/2$, we obtain a rapidly convergent series for $e^{-2\pi }$:

$e^{-2\pi }=w^2+4w^6+34w^{10}+360w^{14}+4239w^{18}+\cdots ,w=\frac{1}{2}\frac{2^{1/4}-1}{2^{1/4}+1}.$

Miscellaneous

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

$\pi =\sqrt{6\zeta (2)}$ (derived from Euler's solution to Basel problem, see Riemann zeta function)

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

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

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

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

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

$\pi =\mathrm{agm}({\theta }_2^2(1/e),{\theta }_3^2(1/e))$ (where ${\theta }_2$ and ${\theta }_3$ are the Jacobi theta functions)

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

$\mathrm{N}(2\varpi )=e^{2\pi },\mathrm{N}(\varpi )=e^{\pi /2}$ (where $\mathrm{N}$ is the Gauss N-function)

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

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

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

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

$\pi =\underset{n\to \mathrm{\infty }}{lim}\frac{2^{4n}n{!}^4}{n(2n){!}^2}=\underset{n\to \mathrm{\infty }}{lim}\frac{2^{4n}}{n{(\genfrac{}{}{0ex}{}{2n}{n})}^2}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2$ (by combining Stirling's approximation with Wallis product)

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

$\pi =\underset{n\to \mathrm{\infty }}{lim}\frac{24}{\sqrt{n}}\ln \left(2^{1/4}{G}_n\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{24}{\sqrt{n}}\ln \left(2^{1/4}{g}_n\right)$ (where $G_n$ and $g_n$ are Ramanujan's class invariants)

See also

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References

Other

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Further reading

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• Kazuya Kato, Nobushige Kurokawa, Saito Takeshi: Number Theory 1: Fermat's Dream. American Mathematical Society, Providence 1993, Template:Isbn.