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

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

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

π = L w {\displaystyle \pi ={\frac {L}{w}}}

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

A = π r 2 {\displaystyle A=\pi r^{2}}

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

A = π a b {\displaystyle A=\pi ab}

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

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

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

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

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

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

where A is the area of a squircle with minor radius r, Γ {\displaystyle \Gamma } is the gamma function.

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

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

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

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

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

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

V = 4 3 π r 3 {\displaystyle V={4 \over 3}\pi r^{3}}

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

S A = 4 π r 2 {\displaystyle SA=4\pi r^{2}}

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

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

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

S V = 2 π 2 r 3 {\displaystyle 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 ) π {\displaystyle S=(n-2)\pi }

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

A = n s 2 4 cot π n {\displaystyle A={\frac {ns^{2}}{4}}\cot {\frac {\pi }{n}}}

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

r = s 2 cot π n {\displaystyle r={\frac {s}{2}}\cot {\frac {\pi }{n}}}

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

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

Physics

Λ = 8 π G 3 c 2 ρ {\displaystyle \Lambda ={{8\pi G} \over {3c^{2}}}\rho }
Δ x Δ p h 4 π {\displaystyle \Delta x\,\Delta p\geq {\frac {h}{4\pi }}}
R μ ν 1 2 g μ ν R + Λ g μ ν = 8 π G c 4 T μ ν {\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }}
F = | q 1 q 2 | 4 π ε 0 r 2 {\displaystyle F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}}
  • Magnetic permeability of free space:[note 1]
μ 0 4 π 10 7 N / A 2 {\displaystyle \mu _{0}\approx 4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}}
  • Approximate period of a simple pendulum with small amplitude:
T 2 π L g {\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}}
  • Exact period of a simple pendulum with amplitude θ 0 {\displaystyle \theta _{0}} ( agm {\displaystyle \operatorname {agm} } is the arithmetic–geometric mean):
T = 2 π agm ( 1 , cos ( θ 0 / 2 ) ) L g {\displaystyle T={\frac {2\pi }{\operatorname {agm} (1,\cos(\theta _{0}/2))}}{\sqrt {\frac {L}{g}}}}
R 3 T 2 = G M 4 π 2 {\displaystyle {\frac {R^{3}}{T^{2}}}={\frac {GM}{4\pi ^{2}}}}
F = π 2 E I L 2 {\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}}

A puzzle involving "colliding billiard balls":

b N π {\displaystyle \lfloor {b^{N}\pi }\rfloor }

is the number of collisions made (in ideal conditions, perfectly elastic with no friction) by an object of mass m initially at rest between a fixed wall and another object of mass b2Nm, when struck by the other object.[1] (This gives the digits of π in base b up to N digits past the radix point.)

Formulae yielding π

Integrals

2 1 1 1 x 2 d x = π {\displaystyle 2\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx=\pi } (integrating two halves y ( x ) = 1 x 2 {\displaystyle y(x)={\sqrt {1-x^{2}}}} to obtain the area of the unit circle)
sech x d x = π {\displaystyle \int _{-\infty }^{\infty }\operatorname {sech} x\,dx=\pi }
t e 1 / 2 t 2 x 2 + x t d x d t = t e t 2 1 / 2 x 2 + x t d x d t = π {\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 }
1 1 d x 1 x 2 = π {\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi }
d x 1 + x 2 = π {\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi } [2][note 2] (see also Cauchy distribution)
sin x x d x = π {\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi } (see Dirichlet integral)
e x 2 d x = π {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}} (see Gaussian integral).
d z z = 2 π i {\displaystyle \oint {\frac {dz}{z}}=2\pi i} (when the path of integration winds once counterclockwise around 0. See also Cauchy's integral formula).
0 ln ( 1 + 1 x 2 ) d x = π {\displaystyle \int _{0}^{\infty }\ln \left(1+{\frac {1}{x^{2}}}\right)\,dx=\pi } [3]
sin x x d x = π {\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi }
0 1 x 4 ( 1 x ) 4 1 + x 2 d x = 22 7 π {\displaystyle \int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi } (see also Proof that 22/7 exceeds π).
0 x α 1 x + 1 d x = π sin π α , 0 < α < 1 {\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{x+1}}\,dx={\frac {\pi }{\sin \pi \alpha }},\quad 0<\alpha <1}
0 d x x ( x + a ) ( x + b ) = π agm ( a , b ) {\displaystyle \int _{0}^{\infty }{\frac {dx}{\sqrt {x(x+a)(x+b)}}}={\frac {\pi }{\operatorname {agm} ({\sqrt {a}},{\sqrt {b}})}}} (where agm {\displaystyle \operatorname {agm} } is the arithmetic–geometric mean;[4] see also elliptic integral)

Note that with symmetric integrands f ( x ) = f ( x ) {\displaystyle f(-x)=f(x)} , formulas of the form a a f ( x ) d x {\textstyle \int _{-a}^{a}f(x)\,dx} can also be translated to formulas 2 0 a f ( x ) d x {\textstyle 2\int _{0}^{a}f(x)\,dx} .

Efficient infinite series

k = 0 k ! ( 2 k + 1 ) ! ! = k = 0 2 k k ! 2 ( 2 k + 1 ) ! = π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}} (see also Double factorial)
k = 0 k ! ( 2 k ) ! ( 25 k 3 ) ( 3 k ) ! 2 k = π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {k!\,(2k)!\,(25k-3)}{(3k)!\,2^{k}}}={\frac {\pi }{2}}}
k = 0 ( 1 ) k ( 6 k ) ! ( 13591409 + 545140134 k ) ( 3 k ) ! ( k ! ) 3 640320 3 k = 4270934400 10005 π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}}={\frac {4270934400}{{\sqrt {10005}}\pi }}} (see Chudnovsky algorithm)
k = 0 ( 4 k ) ! ( 1103 + 26390 k ) ( k ! ) 4 396 4 k = 9801 2 2 π {\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {9801}{2{\sqrt {2}}\pi }}} (see Srinivasa Ramanujan, Ramanujan–Sato series)

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

k = 0 ( 1 ) k 4 k ( 2 4 k + 1 + 2 4 k + 2 + 1 4 k + 3 ) = π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{4^{k}}}\left({\frac {2}{4k+1}}+{\frac {2}{4k+2}}+{\frac {1}{4k+3}}\right)=\pi } [5]
k = 0 1 16 k ( 4 8 k + 1 2 8 k + 4 1 8 k + 5 1 8 k + 6 ) = π {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)=\pi } (see Bailey–Borwein–Plouffe formula)
k = 0 1 16 k ( 8 8 k + 2 + 4 8 k + 3 + 4 8 k + 4 1 8 k + 7 ) = 2 π {\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 }
k = 0 ( 1 ) k 2 10 k ( 2 5 4 k + 1 1 4 k + 3 + 2 8 10 k + 1 2 6 10 k + 3 2 2 10 k + 5 2 2 10 k + 7 + 1 10 k + 9 ) = 2 6 π {\displaystyle \sum _{k=0}^{\infty }{\frac {{(-1)}^{k}}{2^{10k}}}\left(-{\frac {2^{5}}{4k+1}}-{\frac {1}{4k+3}}+{\frac {2^{8}}{10k+1}}-{\frac {2^{6}}{10k+3}}-{\frac {2^{2}}{10k+5}}-{\frac {2^{2}}{10k+7}}+{\frac {1}{10k+9}}\right)=2^{6}\pi }

Plouffe's series for calculating arbitrary decimal digits of π:[6]

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

Other infinite series

ζ ( 2 ) = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + = π 2 6 {\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}} (see also Basel problem and Riemann zeta function)
ζ ( 4 ) = 1 1 4 + 1 2 4 + 1 3 4 + 1 4 4 + = π 4 90 {\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}}
ζ ( 2 n ) = k = 1 1 k 2 n = 1 1 2 n + 1 2 2 n + 1 3 2 n + 1 4 2 n + = ( 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} , where B2n is a Bernoulli number.
n = 1 3 n 1 4 n ζ ( n + 1 ) = π {\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi } [7]
n = 2 2 ( 3 / 2 ) n 3 n ( ζ ( n ) 1 ) = ln π {\displaystyle \sum _{n=2}^{\infty }{\frac {2(3/2)^{n}-3}{n}}(\zeta (n)-1)=\ln \pi }
n = 1 ζ ( 2 n ) x 2 n n = ln π x sin π x , 0 < | x | < 1 {\displaystyle \sum _{n=1}^{\infty }\zeta (2n){\frac {x^{2n}}{n}}=\ln {\frac {\pi x}{\sin \pi x}},\quad 0<|x|<1}
n = 0 ( 1 ) n 2 n + 1 = 1 1 3 + 1 5 1 7 + 1 9 = arctan 1 = π 4 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}} (see Leibniz formula for pi)
n = 0 ( 1 ) ( n 2 n ) / 2 2 n + 1 = 1 + 1 3 1 5 1 7 + 1 9 + 1 11 = π 2 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{(n^{2}-n)/2}}{2n+1}}=1+{\frac {1}{3}}-{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}-\cdots ={\frac {\pi }{2{\sqrt {2}}}}} (Newton, Second Letter to Oldenburg, 1676)[8]
n = 0 ( 1 ) n 3 n ( 2 n + 1 ) = 1 1 3 1 3 + 1 3 2 5 1 3 3 7 + 1 3 4 9 = 3 arctan 1 3 = π 2 3 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}(2n+1)}}=1-{\frac {1}{3^{1}\cdot 3}}+{\frac {1}{3^{2}\cdot 5}}-{\frac {1}{3^{3}\cdot 7}}+{\frac {1}{3^{4}\cdot 9}}-\cdots ={\sqrt {3}}\arctan {\frac {1}{\sqrt {3}}}={\frac {\pi }{2{\sqrt {3}}}}} (Madhava series)
n = 1 ( 1 ) n + 1 n 2 = 1 1 2 1 2 2 + 1 3 2 1 4 2 + = π 2 12 {\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}}}
n = 1 1 ( 2 n ) 2 = 1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 + = π 2 24 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}}
n = 0 ( 1 2 n + 1 ) 2 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + = π 2 8 {\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}}}
n = 0 ( ( 1 ) n 2 n + 1 ) 3 = 1 1 3 1 3 3 + 1 5 3 1 7 3 + = π 3 32 {\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}}}
n = 0 ( 1 2 n + 1 ) 4 = 1 1 4 + 1 3 4 + 1 5 4 + 1 7 4 + = π 4 96 {\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}}}
n = 0 ( ( 1 ) n 2 n + 1 ) 5 = 1 1 5 1 3 5 + 1 5 5 1 7 5 + = 5 π 5 1536 {\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}}}
n = 0 ( 1 2 n + 1 ) 6 = 1 1 6 + 1 3 6 + 1 5 6 + 1 7 6 + = π 6 960 {\displaystyle \sum _{n=0}^{\infty }\left({\frac {1}{2n+1}}\right)^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}}

In general,

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

where E 2 k {\displaystyle E_{2k}} is the 2 k {\displaystyle 2k} th Euler number.[9]

n = 0 ( 1 2 n ) ( 1 ) n 2 n + 1 = 1 1 6 1 40 = π 4 {\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}}}
n = 0 1 ( 4 n + 1 ) ( 4 n + 3 ) = 1 1 3 + 1 5 7 + 1 9 11 + = π 8 {\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}}}
n = 1 ( 1 ) ( n 2 + n ) / 2 + 1 | G ( ( 1 ) n + 1 + 6 n 3 ) / 4 | = | G 1 | + | G 2 | | G 4 | | G 5 | + | G 7 | + | G 8 | | G 10 | | G 11 | + = 3 π {\displaystyle \sum _{n=1}^{\infty }(-1)^{(n^{2}+n)/2+1}\left|G_{\left((-1)^{n+1}+6n-3\right)/4}\right|=|G_{1}|+|G_{2}|-|G_{4}|-|G_{5}|+|G_{7}|+|G_{8}|-|G_{10}|-|G_{11}|+\cdots ={\frac {\sqrt {3}}{\pi }}} (see Gregory coefficients)
n = 0 ( 1 / 2 ) n 2 2 n n ! 2 n = 0 n ( 1 / 2 ) n 2 2 n n ! 2 = 1 π {\displaystyle \sum _{n=0}^{\infty }{\frac {(1/2)_{n}^{2}}{2^{n}n!^{2}}}\sum _{n=0}^{\infty }{\frac {n(1/2)_{n}^{2}}{2^{n}n!^{2}}}={\frac {1}{\pi }}} (where ( x ) n {\displaystyle (x)_{n}} is the rising factorial)[10]
n = 1 ( 1 ) n + 1 n ( n + 1 ) ( 2 n + 1 ) = π 3 {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(2n+1)}}=\pi -3} (Nilakantha series)
n = 1 F 2 n n 2 ( 2 n n ) = 4 π 2 25 5 {\displaystyle \sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}} (where F n {\displaystyle F_{n}} is the n-th Fibonacci number)
n = 1 σ ( n ) e 2 π n = 1 24 1 8 π {\displaystyle \sum _{n=1}^{\infty }\sigma (n)e^{-2\pi n}={\frac {1}{24}}-{\frac {1}{8\pi }}} (where σ {\displaystyle \sigma } is the sum-of-divisors function)
π = n = 1 ( 1 ) ϵ ( n ) n = 1 + 1 2 + 1 3 + 1 4 1 5 + 1 6 + 1 7 + 1 8 + 1 9 1 10 + 1 11 + 1 12 1 13 + {\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 }   (where ϵ ( n ) {\displaystyle \epsilon (n)} is the number of prime factors of the form p 1 ( m o d 4 ) {\displaystyle p\equiv 1\,(\mathrm {mod} \,4)} of n {\displaystyle n} )[11][12]
π 2 = n = 1 ( 1 ) ε ( n ) n = 1 + 1 2 1 3 + 1 4 + 1 5 1 6 1 7 + 1 8 + 1 9 + {\displaystyle {\frac {\pi }{2}}=\sum _{n=1}^{\infty }{\frac {(-1)^{\varepsilon (n)}}{n}}=1+{\frac {1}{2}}-{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}-{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}+\cdots }   (where ε ( n ) {\displaystyle \varepsilon (n)} is the number of prime factors of the form p 3 ( m o d 4 ) {\displaystyle p\equiv 3\,(\mathrm {mod} \,4)} of n {\displaystyle n} )[13]
π = n = ( 1 ) n n + 1 / 2 {\displaystyle \pi =\sum _{n=-\infty }^{\infty }{\frac {(-1)^{n}}{n+1/2}}}
π 2 = n = 1 ( n + 1 / 2 ) 2 {\displaystyle \pi ^{2}=\sum _{n=-\infty }^{\infty }{\frac {1}{(n+1/2)^{2}}}} [14]

The last two formulas are special cases of

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

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

Some formulas relating π and harmonic numbers are given here. Further infinite series involving π are:[15]

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

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

Machin-like formulae

π 4 = arctan 1 {\displaystyle {\frac {\pi }{4}}=\arctan 1}
π 4 = arctan 1 2 + arctan 1 3 {\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}}
π 4 = 2 arctan 1 2 arctan 1 7 {\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}}
π 4 = 2 arctan 1 3 + arctan 1 7 {\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}}
π 4 = 4 arctan 1 5 arctan 1 239 {\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}} (the original Machin's formula)
π 4 = 5 arctan 1 7 + 2 arctan 3 79 {\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}}
π 4 = 6 arctan 1 8 + 2 arctan 1 57 + arctan 1 239 {\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}}
π 4 = 12 arctan 1 49 + 32 arctan 1 57 5 arctan 1 239 + 12 arctan 1 110443 {\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}
π 4 = 44 arctan 1 57 + 7 arctan 1 239 12 arctan 1 682 + 24 arctan 1 12943 {\displaystyle {\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

π 4 = ( p 1 ( mod 4 ) p p 1 ) ( p 3 ( mod 4 ) p p + 1 ) = 3 4 5 4 7 8 11 12 13 12 , {\displaystyle {\frac {\pi }{4}}=\left(\prod _{p\equiv 1{\pmod {4}}}{\frac {p}{p-1}}\right)\cdot \left(\prod _{p\equiv 3{\pmod {4}}}{\frac {p}{p+1}}\right)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,} (Euler)
where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
3 π 6 = ( p 1 ( mod 6 ) p P p p 1 ) ( p 5 ( mod 6 ) p P p p + 1 ) = 5 6 7 6 11 12 13 12 17 18 , {\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 ,}
π 2 = n = 1 ( 2 n ) ( 2 n ) ( 2 n 1 ) ( 2 n + 1 ) = 2 1 2 3 4 3 4 5 6 5 6 7 8 7 8 9 {\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+1)}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots } (see also Wallis product)
π 2 = n = 1 ( 1 + 1 n ) ( 1 ) n + 1 = ( 1 + 1 1 ) + 1 ( 1 + 1 2 ) 1 ( 1 + 1 3 ) + 1 {\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{n}}\right)^{(-1)^{n+1}}=\left(1+{\frac {1}{1}}\right)^{+1}\left(1+{\frac {1}{2}}\right)^{-1}\left(1+{\frac {1}{3}}\right)^{+1}\cdots } (another form of Wallis product)

Viète's formula:

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

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

π 2 = m 1 n 1 ( ( 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 ) ) ϵ n , {\displaystyle {\frac {\pi }{2}}=\prod _{m\geq 1}\prod _{n\geq 1}\left({\frac {(4m^{2}+n-2)(4m^{2}+2n-1)^{2}}{4(2m^{2}+n-1)(4m^{2}+n-1)(2m^{2}+n)}}\right)^{\epsilon _{n}},}
where ϵ n = ( 1 ) t n {\displaystyle \epsilon _{n}=(-1)^{t_{n}}} and t n {\displaystyle t_{n}} is the Thue–Morse sequence (Tóth 2020).

Arctangent formulas

π 2 k + 1 = arctan 2 a k 1 a k , k 2 {\displaystyle {\frac {\pi }{2^{k+1}}}=\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},\qquad \qquad k\geq 2}
π 4 = k 2 arctan 2 a k 1 a k , {\displaystyle {\frac {\pi }{4}}=\sum _{k\geq 2}\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},}

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

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

where F k {\displaystyle F_{k}} is the k-th Fibonacci number.

π = arctan a + arctan b + arctan c {\displaystyle \pi =\arctan a+\arctan b+\arctan c}

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

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

Complex functions

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

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

e z R z π Z {\displaystyle e^{z}\in \mathbb {R} \leftrightarrow \Im z\in \pi \mathbb {Z} }
e z = 1 z 2 π i Z {\displaystyle e^{z}=1\leftrightarrow z\in 2\pi i\mathbb {Z} } [16]

Also

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

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

η 1 ω 2 η 2 ω 1 = 2 π i . {\displaystyle \eta _{1}\omega _{2}-\eta _{2}\omega _{1}=2\pi i.}

Continued fractions

4 π = 1 + 1 2 2 + 3 2 2 + 5 2 2 + 7 2 2 + {\displaystyle {\frac {4}{\pi }}=1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}} [17]
ϖ 2 π = 2 + 1 2 4 + 3 2 4 + 5 2 4 + 7 2 4 + {\displaystyle {\frac {\varpi ^{2}}{\pi }}={2+{\cfrac {1^{2}}{4+{\cfrac {3^{2}}{4+{\cfrac {5^{2}}{4+{\cfrac {7^{2}}{4+\ddots \,}}}}}}}}}\quad } (Ramanujan, ϖ {\displaystyle \varpi } is the lemniscate constant)[18]
π = 3 + 1 2 6 + 3 2 6 + 5 2 6 + 7 2 6 + {\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}} [17]
π = 4 1 + 1 2 3 + 2 2 5 + 3 2 7 + 4 2 9 + {\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}}
2 π = 6 + 2 2 12 + 6 2 12 + 10 2 12 + 14 2 12 + 18 2 12 + {\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 }}}}}}}}}}}}

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 = ( 1 + 1 2 n + 1 ) a n , π = lim n a n 2 n {\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}}}
a 1 = 0 , a n + 1 = 2 + a n , π = lim n 2 n 2 a n {\displaystyle a_{1}=0,\,a_{n+1}={\sqrt {2+a_{n}}},\,\pi =\lim _{n\to \infty }2^{n}{\sqrt {2-a_{n}}}} (closely related to Viète's formula)
ω ( i n , i n 1 , , i 1 ) = 2 + i n 2 + i n 1 2 + + i 1 2 = ω ( b n , b n 1 , , b 1 ) , i k { 1 , 1 } , b k = { 0 if  i k = 1 1 if  i k = 1 , π = lim n 2 n + 1 2 h + 1 ω ( 10 0 n m g m , h + 1 ) {\displaystyle \omega (i_{n},i_{n-1},\dots ,i_{1})=2+i_{n}{\sqrt {2+i_{n-1}{\sqrt {2+\cdots +i_{1}{\sqrt {2}}}}}}=\omega (b_{n},b_{n-1},\dots ,b_{1}),\,i_{k}\in \{-1,1\},\,b_{k}={\begin{cases}0&{\text{if }}i_{k}=1\\1&{\text{if }}i_{k}=-1\end{cases}},\,\pi ={\displaystyle \lim _{n\rightarrow \infty }{\frac {2^{n+1}}{2h+1}}{\sqrt {\omega \left(\underbrace {10\ldots 0} _{n-m}g_{m,h+1}\right)}}}} (where g m , h + 1 {\displaystyle g_{m,h+1}} is the h+1-th entry of m-bit Gray code, h { 0 , 1 , , 2 m 1 } {\displaystyle h\in \left\{0,1,\ldots ,2^{m}-1\right\}} )[19]
k N , a 1 = 2 k , a n + 1 = a n + 2 k ( 1 tan ( 2 k 1 a n ) ) , π = 2 k + 1 lim n a n {\displaystyle \forall k\in \mathbb {N} ,\,a_{1}=2^{-k},\,a_{n+1}=a_{n}+2^{-k}(1-\tan(2^{k-1}a_{n})),\,\pi =2^{k+1}\lim _{n\to \infty }a_{n}} (quadratic convergence)[20]
a 1 = 1 , a n + 1 = a n + sin a n , π = lim n a n {\displaystyle a_{1}=1,\,a_{n+1}=a_{n}+\sin a_{n},\,\pi =\lim _{n\to \infty }a_{n}} (cubic convergence)[21]
a 0 = 2 3 , b 0 = 3 , a n + 1 = hm ( a n , b n ) , b n + 1 = gm ( a n + 1 , b n ) , π = lim n a n = lim n b n {\displaystyle a_{0}=2{\sqrt {3}},\,b_{0}=3,\,a_{n+1}=\operatorname {hm} (a_{n},b_{n}),\,b_{n+1}=\operatorname {gm} (a_{n+1},b_{n}),\,\pi =\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}} (Archimedes' algorithm, see also harmonic mean and geometric mean)[22]

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

Asymptotics

( 2 n n ) 4 n π n {\displaystyle {\binom {2n}{n}}\sim {\frac {4^{n}}{\sqrt {\pi n}}}} (asymptotic growth rate of the central binomial coefficients)
C n 4 n π n 3 {\displaystyle C_{n}\sim {\frac {4^{n}}{\sqrt {\pi n^{3}}}}} (asymptotic growth rate of the Catalan numbers)
n ! 2 π n ( n e ) n {\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}} (Stirling's approximation)
k = 1 n φ ( k ) 3 n 2 π 2 {\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}} (where φ {\displaystyle \varphi } is Euler's totient function)
k = 1 n φ ( k ) k 6 n π 2 {\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}}

Hypergeometric inversions

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

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

Then

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

where

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

Similarly, let | q | < 1 {\displaystyle \left|q\right|<1} and

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

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

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

where

q = exp ( 2 π 2 F 1 ( 1 / 6 , 5 / 6 , 1 , 1 z ) 2 F 1 ( 1 / 6 , 5 / 6 , 1 , z ) ) , z C { 0 , 1 } . {\displaystyle 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

Γ ( s ) Γ ( 1 s ) = π sin π s {\displaystyle \Gamma (s)\Gamma (1-s)={\frac {\pi }{\sin \pi s}}} (Euler's reflection formula, see Gamma function)
π s / 2 Γ ( s 2 ) ζ ( s ) = π ( 1 s ) / 2 Γ ( 1 s 2 ) ζ ( 1 s ) {\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)} (the functional equation of the Riemann zeta function)
e ζ ( 0 ) = 2 π {\displaystyle e^{-\zeta '(0)}={\sqrt {2\pi }}}
e ζ ( 0 , 1 / 2 ) ζ ( 0 , 1 ) = π {\displaystyle e^{\zeta '(0,1/2)-\zeta '(0,1)}={\sqrt {\pi }}} (where ζ ( s , a ) {\displaystyle \zeta (s,a)} is the Hurwitz zeta function and the derivative is taken with respect to the first variable)
π = B ( 1 / 2 , 1 / 2 ) = Γ ( 1 / 2 ) 2 {\displaystyle \pi =\mathrm {B} (1/2,1/2)=\Gamma (1/2)^{2}} (see also Beta function)
π = Γ ( 3 / 4 ) 4 agm ( 1 , 1 / 2 ) 2 = Γ ( 1 / 4 ) 4 / 3 agm ( 1 , 2 ) 2 / 3 2 {\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}}} (where agm is the arithmetic–geometric mean)
π = agm ( θ 2 2 ( 1 / e ) , θ 3 2 ( 1 / e ) ) {\displaystyle \pi =\operatorname {agm} \left(\theta _{2}^{2}(1/e),\theta _{3}^{2}(1/e)\right)} (where θ 2 {\displaystyle \theta _{2}} and θ 3 {\displaystyle \theta _{3}} are the Jacobi theta functions[23])
π = K ( k ) K ( 1 k 2 ) ln q , k = θ 2 2 ( q ) θ 3 2 ( q ) {\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)}}} (where q ( 0 , 1 ) {\displaystyle q\in (0,1)}  and K ( k ) {\displaystyle \operatorname {K} (k)} is the complete elliptic integral of the first kind with modulus k {\displaystyle k} ; reflecting the nome-modulus inversion problem)[24]
π = agm ( 1 , 1 k 2 ) agm ( 1 , k ) ln q , k = θ 4 2 ( q ) θ 3 2 ( q ) {\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)}}} (where q ( 0 , 1 ) {\displaystyle q\in (0,1)} )[24]
agm ( 1 , 2 ) = π ϖ {\displaystyle \operatorname {agm} (1,{\sqrt {2}})={\frac {\pi }{\varpi }}} (due to Gauss,[25] ϖ {\displaystyle \varpi } is the lemniscate constant)
i π = Log ( 1 ) = lim n n ( ( 1 ) 1 / n 1 ) {\displaystyle i\pi =\operatorname {Log} (-1)=\lim _{n\to \infty }n\left((-1)^{1/n}-1\right)} (where Log {\displaystyle \operatorname {Log} } is the principal value of the complex logarithm)[note 3]
1 π 2 12 = lim n 1 n 2 k = 1 n ( n mod k ) {\displaystyle 1-{\frac {\pi ^{2}}{12}}=\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n{\bmod {k}})} (where n mod k {\textstyle n{\bmod {k}}} is the remainder upon division of n by k)
π = lim r 1 r 2 x = r r y = r r { 1 if  x 2 + y 2 r 0 if  x 2 + y 2 > r {\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}}}\leq r\\0&{\text{if }}{\sqrt {x^{2}+y^{2}}}>r\end{cases}}} (summing a circle's area)
π = lim n 4 n 2 k = 1 n n 2 k 2 {\displaystyle \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)
π = lim n 2 4 n n ! 4 n ( 2 n ) ! 2 = lim n 2 4 n n ( 2 n n ) 2 = lim n 1 n ( ( 2 n ) ! ! ( 2 n 1 ) ! ! ) 2 {\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}} (by combining Stirling's approximation with Wallis product)
π = lim n 1 n ln 16 λ ( n i ) {\displaystyle \pi =\lim _{n\to \infty }{\frac {1}{n}}\ln {\frac {16}{\lambda (ni)}}} (where λ {\displaystyle \lambda } is the modular lambda function)[26][note 4]
π = lim n 24 n ln ( 2 1 / 4 G n ) = lim n 24 n ln ( 2 1 / 4 g n ) {\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)} (where G n {\displaystyle G_{n}} and g n {\displaystyle g_{n}} are Ramanujan's class invariants)[27][note 5]

See also

References

Notes

  1. ^ The relation μ 0 = 4 π 10 7 N / A 2 {\displaystyle \mu _{0}=4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}} was valid until the 2019 redefinition of the SI base units.
  2. ^ (integral form of arctan over its entire domain, giving the period of tan)
  3. ^ The n {\displaystyle n} th root with the smallest positive principal argument is chosen.
  4. ^ When n Q + {\displaystyle n\in \mathbb {Q} ^{+}} , this gives algebraic approximations to Gelfond's constant e π {\displaystyle e^{\pi }} .
  5. ^ When n Q + {\displaystyle {\sqrt {n}}\in \mathbb {Q} ^{+}} , this gives algebraic approximations to Gelfond's constant e π {\displaystyle e^{\pi }} .

Other

  1. ^ Galperin, G. (2003). "Playing pool with π (the number π from a billiard point of view)" (PDF). Regular and Chaotic Dynamics. 8 (4): 375–394. doi:10.1070/RD2003v008n04ABEH000252.
  2. ^ Rudin, Walter (1987). Real and Complex Analysis (Third ed.). McGraw-Hill Book Company. ISBN 0-07-100276-6. p. 4
  3. ^ A000796 – OEIS
  4. ^ Carson, B. C. (2010), "Elliptic Integrals", in Olver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.), NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN 978-0-521-19225-5, MR 2723248.
  5. ^ Arndt, Jörg; Haenel, Christoph (2001). π Unleashed. Springer-Verlag Berlin Heidelberg. ISBN 978-3-540-66572-4. page 126
  6. ^ Gourdon, Xavier. "Computation of the n-th decimal digit of π with low memory" (PDF). Numbers, constants and computation. p. 1.
  7. ^ Weisstein, Eric W. "Pi Formulas", MathWorld
  8. ^ Chrystal, G. (1900). Algebra, an Elementary Text-book: Part II. p. 335.
  9. ^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi. American Mathematical Society. ISBN 0-8218-3246-8. p. 112
  10. ^ Cooper, Shaun (2017). Ramanujan's Theta Functions (First ed.). Springer. ISBN 978-3-319-56171-4. (page 647)
  11. ^ Euler, Leonhard (1748). Introductio in analysin infinitorum (in Latin). Vol. 1. p. 245
  12. ^ Carl B. Boyer, A History of Mathematics, Chapter 21., pp. 488–489
  13. ^ Euler, Leonhard (1748). Introductio in analysin infinitorum (in Latin). Vol. 1. p. 244
  14. ^ Wästlund, Johan. "Summing inverse squares by euclidean geometry" (PDF). The paper gives the formula with a minus sign instead, but these results are equivalent.
  15. ^ Simon Plouffe / David Bailey. "The world of Pi". Pi314.net. Retrieved 2011-01-29.
    "Collection of series for π". Numbers.computation.free.fr. Retrieved 2011-01-29.
  16. ^ Rudin, Walter (1987). Real and Complex Analysis (Third ed.). McGraw-Hill Book Company. ISBN 0-07-100276-6. p. 3
  17. ^ a b Loya, Paul (2017). Amazing and Aesthetic Aspects of Analysis. Springer. p. 589. ISBN 978-1-4939-6793-3.
  18. ^ Perron, Oskar (1957). Die Lehre von den Kettenbrüchen: Band II (in German) (Third ed.). B. G. Teubner. p. 36, eq. 24
  19. ^ Vellucci, Pierluigi; Bersani, Alberto Maria (2019-12-01). "$$\pi $$-Formulas and Gray code". Ricerche di Matematica. 68 (2): 551–569. arXiv:1606.09597. doi:10.1007/s11587-018-0426-4. ISSN 1827-3491. S2CID 119578297.
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  22. ^ Eymard, Pierre; Lafon, Jean-Pierre (2004). The Number Pi. American Mathematical Society. ISBN 0-8218-3246-8. p. 2
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  • 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.