Llista de fórmules amb π

A continuació es mostra una llista de fórmules que tenen a veure amb la constant matemàtica π.

${\displaystyle \pi =L/d\!}$

on L és la longitud d'una circumferència de diàmetre d.

${\displaystyle A=\pi r^{2}\!}$

on A és l'àrea d'un cercle de radi r.

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

on V és el volum d'una esfera de radi r.

${\displaystyle S=4\pi r^{2}\!}$

on S és la superfície exterior d'una esfera de radi r.

• La constant cosmològica:

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

• El principi d'incertesa de Heisenberg:

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

• Les equacions de camp d'Einstein de la relativitat general:

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

• La llei de Coulomb de la força elèctrica:

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

• Permeabilitat magnètica en el buit:

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

• Període d'un pèndol simple d'amplitud petita:

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

• La fórmula del vinclament:

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

${\displaystyle \int \limits _{-\infty }^{\infty }{\text{sech}}(x)dx=\pi \!}$

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

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

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

${\displaystyle \int \limits _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi \!}$ (forma integral de l'arctangent al llarg de tot el seu domini).

${\displaystyle \int \limits _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}\!}$ (veure Integral de Gauß).

${\displaystyle \oint {\frac {dz}{z}}=2\pi i\!}$ (Vegeu també fórmula de la integral de Cauchy)

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

${\displaystyle \int \limits _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi \!}$

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

${\displaystyle 12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {1}{\pi }}\!}$

${\displaystyle {\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {1}{\pi }}\!}$ (veure Srinivasa Ramanujan)

${\displaystyle {\frac {\sqrt {3}}{6^{5}}}\sum _{k=0}^{\infty }{\frac {((4k)!)^{2}(6k)!}{9^{k+1}(12k)!(2k)!}}\left({\frac {127169}{12k+1}}-{\frac {1070}{12k+5}}-{\frac {131}{12k+7}}+{\frac {2}{12k+11}}\right)=\pi \!}$^[1]

Les següents identitats són útils per calcular dígits binaris arbitraris de π:

${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)=\pi \!}$

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

${\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}\!}$ (vegeu també el problema de Basilea i la funció zeta de Riemann)

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

${\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}\!}$, on B_2n és un nombre de Bernoulli.

${\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi \!}$^[2]

${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{1}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}\!}$ (sèrie de Leibniz)

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

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

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

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

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

${\displaystyle \pi ={1}+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots \!}$ (Euler, 1748)

Després dels dos primers termes, els signes venen determinats de la següent manera: si el denominador és un nombre primer de la forma 4m - 1, el signe és positiu; si el denominador és un nombre primer de la forma 4m + 1, es signe és negatiu; per nombres compostos,el signe és igual al producte dels signes dels factors.^[3]

${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}\!}$ (la fórmula original de Machin)

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

${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}\!}$ (d'Euler)

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}\!}$ (de Hermann)

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}\!}$ (de Hutton o de Vega^[4])

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

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

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

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

on ${\displaystyle F_{n}}$ és l'enèssim nombre de Fibonacci.

Algunes sèries infinites relacionades amb pi són:^[5]

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

on

${\displaystyle (x)_{n}\!}$

és el símbol de Pochhammer del factorial decreixent.

${\displaystyle {\frac {\pi }{4}}={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{16}}\cdot {\frac {19}{20}}\cdot {\frac {23}{24}}\cdot {\frac {29}{28}}\cdot {\frac {31}{32}}\cdots \!}$ (Euler)

on els numeradors són els nombres primers senars; i cada denominador és el múltiple de 4 més proper al numerador.

${\displaystyle \prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {4}{3}}\cdot {\frac {16}{15}}\cdot {\frac {36}{35}}\cdot {\frac {64}{63}}\cdots ={\frac {\pi }{2}}\!}$

Fórmula de Vieète:

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

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

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

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

(vegeu també fracció contínua)

${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\!}$ (aproximació de Stirling)

${\displaystyle e^{i\pi }+1=0\!}$ (Identitat d'Euler)

${\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}\!}$ (veure Funció φ d'Euler)

${\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}\!}$ (veure Funció φ d'Euler)

${\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}\!}$ (veure trambé funció Gamma)

${\displaystyle \pi ={\frac {\Gamma \left({1/4}\right)^{4/3}\mathrm {agm} (1,{\sqrt {2}})^{2/3}}{2}}\!}$ (on agm és la Mitjana aritmètico-geomètrica)

${\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n\;{\bmod {\;}}k)=1-{\frac {\pi ^{2}}{12}}\!}$ (on mod és la funció mòdul, que dona el residu de la divisió de n entre k)

${\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}}$ (sumatori de Riemann per avaluar l'àrea d'un cercle unitat)

${\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}}$ (a través de l'aproximació de Stirling)