פורטל:מתמטיקה/נוסחה נבחרת/16

• מגוון נוסחאות ל-$\pi$
  • $$\frac{2}{\pi }=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdots$$
  • $$\pi =\frac{3}{4}\sqrt{3}+24(\frac{1}{12}-\sum _{n=2}^{\mathrm{\infty }}\frac{(2n-2)!}{(n-1){!}^2(2n-3)(2n+1)2^{4n-2}})$$
  • $\sum _{k=0}^{\mathrm{\infty }}\frac{k!}{(2k+1)!!}=\sum _{k=0}^{\mathrm{\infty }}\frac{2^{k}k{!}^2}{(2k+1)!}=\frac{\pi }{2}$
  • $\frac{1}{\pi }=12\sum _{q=0}^{\mathrm{\infty }}\frac{(-1{)}^q(6q)!(545140134q+13591409)}{(3q)!(q!{)}^3{\left(640320\right)}^{3q+3/2}}$
  • $\zeta (2):=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$
  • $$\frac{\pi }{4}=\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{2k+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$
  • $\frac{\pi }{4}=\left(\prod _{p\equiv 1(\mathrm{mod}4)}\frac{p}{p-1}\right)\cdot \left(\prod _{p\equiv 3(\mathrm{mod}4)}\frac{p}{p+1}\right)=\frac{3}{4}\cdot \frac{5}{4}\cdot \frac{7}{8}\cdot \frac{11}{12}\cdot \frac{13}{12}\cdots ,$
  • $\prod _{n=1}^{\mathrm{\infty }}\frac{4n^2}{4n^2-1}=\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \frac{8}{7}\cdot \frac{8}{9}\cdots =\frac{4}{3}\cdot \frac{16}{15}\cdot \frac{36}{35}\cdot \frac{64}{63}\cdots =\frac{\pi }{2}$ (see also Wallis product)
  • $\left(\frac{1}{2}\right)!:=\mathrm{\Gamma }\left(\frac{3}{2}\right):=\underset{0}{\overset{\mathrm{\infty }}{\int }}\sqrt{x}{e}^{-x}dx=\frac{\sqrt{\pi }}{2}$