Spisak formula koje sadrže π

Ovaj članak je spisak značajnih formula koje sadrže matematičku konstantu π.

Klasična geometrija

$O=2\pi r=\pi d,\,$$O=2\pi r=\pi d,\,$

gdje je O obim kruga, r je radijus, a d je dijametar.

$P=\pi r^{2},\,$$P=\pi r^{2},\,$

gdje je P površina kruga, a r je radijus.

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

gdje je V zapremina sfere, a r je radijus.

$P=4\pi r^{2}\,$$P=4\pi r^{2}\,$

gdje je P površina sfere, a r je radijus.

S=2rπ*h

gdje je S površina kalote(isječka), r radijus, h visina.

Analiza

Integrali

$\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx={\frac {\pi }{2}}$$\int _{{-1}}^{1}{\sqrt {1-x^{2}}}\,dx={\frac {\pi }{2}}$ (pogledajte π)

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

$\int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi$$\int _{{-\infty }}^{\infty }{\frac {dx}{1+x^{2}}}=\pi$ (integral oblika arctan preko svog cijelog domena, date period tan).

$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}$$\int _{{-\infty }}^{{\infty }}e^{{-x^{2}}}\,dx={\sqrt {\pi }}$ (pogledajte normalna raspodjela).

$\oint {\frac {dz}{z}}=2\pi i$$\oint {\frac {dz}{z}}=2\pi i$ (kada se put integracije omota kednom u smijeru suprotnom od kazaljke na satu oko 0. Pogledajte Cauchyjeva integralna formula)

$\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}.$$\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}.$

$\int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi$$\int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi$ (pogledajte dokaz da 22 podijeljeno sa 7 prelazi vrijednost π]]).

Valjani beskonačni redovi

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

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

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

${\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$${\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$

Slijedeć formule su pogodne za računanje proizvoljnih binarnih decimala broja π:

$\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$$\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$ (pogledajte Bailey-Borwein-Plouffeova formula)

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

Ostali beskonačni redovi

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}$$\sum _{{n=0}}^{{\infty }}{\frac {(-1)^{{n}}}{2n+1}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}$ (pogledajte Leibnizova formula za pi)

$\zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}$$\zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}$ (pogledajte Baselov problem i zeta funkcija)

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

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

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

Machinove formule

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

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

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

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

${\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}$${\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}}=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 }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}$

Beskonačni proizvodi

$\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 {\pi }{2}}$$\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 {\pi }{2}}$ (pogledajte Wallisov proizvod)

Vietaova 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 }}$${\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 }$

Tri neprekidna razlomka

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

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

$\pi ={\cfrac {4}{1+{\cfrac {1}{2+{\cfrac {9}{2+{\cfrac {25}{2+{\cfrac {49}{\ddots }}}}}}}}}}\,$$\pi ={\cfrac {4}{1+{\cfrac {1}{2+{\cfrac {9}{2+{\cfrac {25}{2+{\cfrac {49}{\ddots }}}}}}}}}}\,$

Za više informacija o trećem identitetu, pogledajte Eulerova formula za neprekidni razlomak.

(Također pogledajte članke neprekidni razlomak i uopćeni neprekidni razlomak.)

Razno

$n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}$$n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}$ (Stirlingova aproksimacija)

$e^{i\pi }+1=0\;$$e^{{i\pi }}+1=0\;$ (Eulerov identitet)

$\sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}$$\sum _{{k=1}}^{{n}}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}$ (pogledajte Eulerova fi funkcija)

$\sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}$$\sum _{{k=1}}^{{n}}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}$ (pogledajte Eulerova fi funkcija)

$\Gamma \left({1 \over 2}\right)={\sqrt {\pi }}$$\Gamma \left({1 \over 2}\right)={\sqrt {\pi }}$ (pogledajte gama funkcija)

$\pi ={\frac {\Gamma \left({1/4}\right)^{4/3}\mathrm {agm} (1,{\sqrt {2}})^{2/3}}{2}}$$\pi ={\frac {\Gamma \left({1/4}\right)^{{4/3}}{\mathrm {agm}}(1,{\sqrt {2}})^{{2/3}}}{2}}$ (gdje je ags aritmetičko-geometrijska sredina)

Fizika

• Kosmnološka konstanta:

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

• Heisenbergov princip neodređenosti:

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

• Einsteinova jednačina polja opće relativnosti:

$R_{ik}-{g_{ik}R \over 2}+\Lambda g_{ik}={8\pi G \over c^{4}}T_{ik}$$R_{{ik}}-{g_{{ik}}R \over 2}+\Lambda g_{{ik}}={8\pi G \over c^{4}}T_{{ik}}$

• Coulombov zakon za električnu silu:

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

• Magnetna permeabilnost za slobodni prostor:

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

Također pogledajte

• Pi
• Izračunavanje π
• Spisak tema vezanih za π

Reference

• Peter Borwein, The Amazing Number Pi