Zerrenda:Π duten formulak

Hona hemen π konstante matematikoarekin zerikusia duten formulen zerrenda.

Geometria klasikoa

$\pi =L/d$

non L d diametroko zirkunferentzia baten luzera baita.

$A=\pi r^2$

non A r erradioko zirkulu baten azalera baita.

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

non V baita r erradioko esfera baten bolumena.

$S=4\pi r^2$

non S r erradioko esfera baten kanpo-azalera baita.

Fisikoa

• Konstante kosmologikoa:

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

• Heisenbergen ziurgabetasun-printzipioa:

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

• Erlatibitate orokorraren Einsteinen eremu-ekuazioak:

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

• Indar elektrikoaren Culomb-en legea:

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

• Iragazkortasun magnetikoa hutsean:

${\mu }_0=4\pi \cdot 10^{-7}\mathrm{N}/{\mathrm{A}}^{\mathrm{2}}$

• Anplitude txikiko pendulu sinple baten periodoa:

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

• Gilborduraren formula:

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

Identitateak

Integralak

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\text{sech}(x)dx=\pi$

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

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

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

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{1+x^2}=\pi$ (arku tangente baten forma integrala bere eremu osoan).

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\sqrt{\pi }$ (ikus Gauß-en Integrala).

$\oint \frac{dz}{z}=2\pi i$ (Ikus, halaber, Cauchyren formula integral)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{x}dx=\pi$

$\underset{0}{\overset{1}{\int }}\frac{x^4(1-x{)}^4}{1+x^2}dx=\frac{22}{7}-\pi$

Serie infinitu eraginkorrak

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

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

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

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

Identitate hauek baliagarriak dira π-ren digitu bitar arbitrarioak kalkulatzeko:

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

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

Beste serie infinitu batzuk

$\zeta (2)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\frac{{\pi }^2}{6}$ (ikus, halaber, Basileako arazoa eta Riemann-en zeta funtzioa)

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

$\zeta (2n)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{2n}}=\frac{1}{1^{2n}}+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\frac{1}{4^{2n}}+\cdots =(-1{)}^{n+1}\frac{B_{2n}(2\pi {)}^{2n}}{2(2n)!}$ non B2n Bernoulliren zenbaki bat baita.

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

$\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{(-1{)}^n}{2n+1}\right)}^1=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots =\arctan 1=\frac{\pi }{4}$ (Leibnizko seriea)

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

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

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

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

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

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

Lehenengo bi terminoen ondoan, zeinuak honela zehazten dira: izendatzailea 4m - 1 formako zenbaki lehena bada, zeinua positiboa da; izendatzailea 4m + 1 formako zenbaki lehena bada, negatiboa da; zenbaki konposatuen bidez, zeinua faktoreen zeinuen biderkaduraren berdina da.

Machin-en formulak

$\frac{\pi }{4}=4\arctan \frac{1}{5}-\arctan \frac{1}{239}$ (Machinen jatorrizko formula)

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

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

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

$\frac{\pi }{4}=2\arctan \frac{1}{3}+\arctan \frac{1}{7}$ (Huttonena edo Vegarena)

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

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

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

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

non $F_n$ baita enegarren Fibonacci-ren zenbakia.

Serie infinitu batzuk

Hona hemen pi-rekin lotutako serie infinitu batzuk:

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

non

$(x{)}_n$

beheranzko faktorialeko Pochhammerren ikurra da.

Produktu infinituak

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

non zenbakitzaileak zenbaki lehenak baitira; eta izendatzaile bakoitza zenbakitzailetik hurbilen dagoen 4ren multiploa baita.

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

Vieèteko formula:

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

Frakzio jarraituak

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

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

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

(ikus, halaber, zatiki jarraitua)

Denetarik

$n!\sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$ (Stirling-en hurbilketa)

$e^{i\pi }+1=0$ (Eulerren identitatea)

$\sum _{k=1}^n\phi (k)\sim \frac{3n^2}{{\pi }^2}$ (ikus Eulerren Φ funtzioa)

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

$\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$ (ikus Gama funtzioaren trambea)

$\pi =\frac{\mathrm{\Gamma }{\left(1/4\right)}^{4/3}\mathrm{a}\mathrm{g}\mathrm{m}(1,\sqrt{2}{)}^{2/3}}{2}$ (non agm baita aritmetika ertaina-geometrikoa)

$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n^2}\sum _{k=1}^n(nmodk)=1-\frac{{\pi }^2}{12}$ (non mod baita modulu-funtzioa, n-ren eta k-ren arteko zatiketaren hondarra ematen duena)

(unitate zirkulu baten azalera ebaluatzeko Riemann-en batukaria)$\pi =\underset{n\to \mathrm{\infty }}{\lim }\frac{4}{n^2}\sum _{k=1}^n\sqrt{n^2-k^2}$

$\pi =\underset{n\to \mathrm{\infty }}{\lim }\frac{2^{4n}}{n{(\genfrac{}{}{0ex}{}{2n}{n})}^2}$ (Stirling-en hurbilketaren bidez)