Em matemática , o produto de Wallis para π , expresso em 1655 por John Wallis, estabelece que
∏ n = 1 ∞ ( 2 n 2 n − 1 ⋅ 2 n 2 n + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ = π 2 {\displaystyle \prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\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}}} Comparação da convergência do produto de Wallis (asteriscos em roxo) e diversas séries infinitas históricas para π. Sn é a aproximação após tomar n termos. Cada subsequente subplotagem magnifica a área sombreada horizontalmente por 10 vezes. (clicar para detalhe) Dedução Wallis deduziu este produto infinito como ele é apresentado atualmente em livros de cálculo, examinando ∫ 0 π sin n x d x {\displaystyle \textstyle \int _{0}^{\pi }\sin ^{n}x\,dx} n , e notando que para grandes n , aumentando n por 1 resulta em uma mudança que torna-se menor quando n aumenta. Como o cálculo infinitesimal moderno ainda não existia, e a análise matemática naquela época era inadequada para discutir as questões de convergência, esta era uma pesquisa difícil bem como também uma tentativa.
O produto de Wallis é, em retrospecto, um corolário fácil da posterior fórmula de Euler para a função senoidal. Em 2015 os pesquisadores Carl Richard Hagen e Tamar Friedmann, em uma descoberta surpresa, encontraram a mesma fórmula nos cálculos da mecânica quântica dos níveis de energia de um átomo de hidrogênio .[ 1] [ 2] [ 3] [ 4] [ 5]
Prova usando o produto infinito de Euler para a função seno[ 6] sin x x = ∏ n = 1 ∞ ( 1 − x 2 n 2 π 2 ) {\displaystyle {\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right)} Seja x = π 2
⇒ 2 π = ∏ n = 1 ∞ ( 1 − 1 4 n 2 ) ⇒ π 2 = ∏ n = 1 ∞ ( 4 n 2 4 n 2 − 1 ) = ∏ n = 1 ∞ ( 2 n 2 n − 1 ⋅ 2 n 2 n + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋯ {\displaystyle {\begin{aligned}\Rightarrow {\frac {2}{\pi }}&=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)\\\Rightarrow {\frac {\pi }{2}}&=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\&=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdots \end{aligned}}} Prova usando integração[ 7] Seja:
I ( n ) = ∫ 0 π sin n x d x {\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}x\,dx} (uma forma da integral de Wallis). Integração por partes:
u = sin n − 1 x ⇒ d u = ( n − 1 ) sin n − 2 x cos x d x d v = sin x d x ⇒ v = − cos x {\displaystyle {\begin{aligned}u&=\sin ^{n-1}x\\\Rightarrow du&=(n-1)\sin ^{n-2}x\cos x\,dx\\dv&=\sin x\,dx\\\Rightarrow v&=-\cos x\end{aligned}}} ⇒ I ( n ) = ∫ 0 π sin n x d x = − sin n − 1 x cos x | 0 π − ∫ 0 π ( − cos x ) ( n − 1 ) sin n − 2 x cos x d x = 0 + ( n − 1 ) ∫ 0 π cos 2 x sin n − 2 x d x , n > 1 = ( n − 1 ) ∫ 0 π ( 1 − sin 2 x ) sin n − 2 x d x = ( n − 1 ) ∫ 0 π sin n − 2 x d x − ( n − 1 ) ∫ 0 π sin n x d x = ( n − 1 ) I ( n − 2 ) − ( n − 1 ) I ( n ) = n − 1 n I ( n − 2 ) ⇒ I ( n ) I ( n − 2 ) = n − 1 n ⇒ I ( 2 n − 1 ) I ( 2 n + 1 ) = 2 n + 1 2 n {\displaystyle {\begin{aligned}\Rightarrow I(n)&=\int _{0}^{\pi }\sin ^{n}x\,dx\\{}&=-\sin ^{n-1}x\cos x|_{0}^{\pi }-\int _{0}^{\pi }(-\cos x)(n-1)\sin ^{n-2}x\cos x\,dx\\{}&=0+(n-1)\int _{0}^{\pi }\cos ^{2}x\sin ^{n-2}x\,dx,\qquad n>1\\{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}x\,dx\\{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}x\,dx-(n-1)\int _{0}^{\pi }\sin ^{n}x\,dx\\{}&=(n-1)I(n-2)-(n-1)I(n)\\{}&={\frac {n-1}{n}}I(n-2)\\\Rightarrow {\frac {I(n)}{I(n-2)}}&={\frac {n-1}{n}}\\\Rightarrow {\frac {I(2n-1)}{I(2n+1)}}&={\frac {2n+1}{2n}}\end{aligned}}} Este resultado será usado abaixo:
I ( 0 ) = ∫ 0 π d x = x | 0 π = π I ( 1 ) = ∫ 0 π sin x d x = − cos x | 0 π = ( − cos π ) − ( − cos 0 ) = − ( − 1 ) − ( − 1 ) = 2 I ( 2 n ) = ∫ 0 π sin 2 n x d x = 2 n − 1 2 n I ( 2 n − 2 ) = 2 n − 1 2 n ⋅ 2 n − 3 2 n − 2 I ( 2 n − 4 ) {\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x|_{0}^{\pi }=\pi \\I(1)&=\int _{0}^{\pi }\sin x\,dx=-\cos x|_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\I(2n)&=\int _{0}^{\pi }\sin ^{2n}x\,dx={\frac {2n-1}{2n}}I(2n-2)={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}I(2n-4)\end{aligned}}} Repetindo o processo,
= 2 n − 1 2 n ⋅ 2 n − 3 2 n − 2 ⋅ 2 n − 5 2 n − 4 ⋅ ⋯ ⋅ 5 6 ⋅ 3 4 ⋅ 1 2 I ( 0 ) = π ∏ k = 1 n 2 k − 1 2 k {\displaystyle ={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}\cdot {\frac {2n-5}{2n-4}}\cdot \cdots \cdot {\frac {5}{6}}\cdot {\frac {3}{4}}\cdot {\frac {1}{2}}I(0)=\pi \prod _{k=1}^{n}{\frac {2k-1}{2k}}} I ( 2 n + 1 ) = ∫ 0 π sin 2 n + 1 x d x = 2 n 2 n + 1 I ( 2 n − 1 ) = 2 n 2 n + 1 ⋅ 2 n − 2 2 n − 1 I ( 2 n − 3 ) {\displaystyle I(2n+1)=\int _{0}^{\pi }\sin ^{2n+1}x\,dx={\frac {2n}{2n+1}}I(2n-1)={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}I(2n-3)} Repetindo o processo,
= 2 n 2 n + 1 ⋅ 2 n − 2 2 n − 1 ⋅ 2 n − 4 2 n − 3 ⋅ ⋯ ⋅ 6 7 ⋅ 4 5 ⋅ 2 3 I ( 1 ) = 2 ∏ k = 1 n 2 k 2 k + 1 {\displaystyle ={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n-4}{2n-3}}\cdot \cdots \cdot {\frac {6}{7}}\cdot {\frac {4}{5}}\cdot {\frac {2}{3}}I(1)=2\prod _{k=1}^{n}{\frac {2k}{2k+1}}} sin 2 n + 1 x ≤ sin 2 n x ≤ sin 2 n − 1 x , 0 ≤ x ≤ π {\displaystyle \sin ^{2n+1}x\leq \sin ^{2n}x\leq \sin ^{2n-1}x,0\leq x\leq \pi } ⇒ I ( 2 n + 1 ) ≤ I ( 2 n ) ≤ I ( 2 n − 1 ) {\displaystyle \Rightarrow I(2n+1)\leq I(2n)\leq I(2n-1)} ⇒ 1 ≤ I ( 2 n ) I ( 2 n + 1 ) ≤ I ( 2 n − 1 ) I ( 2 n + 1 ) = 2 n + 1 2 n {\displaystyle \Rightarrow 1\leq {\frac {I(2n)}{I(2n+1)}}\leq {\frac {I(2n-1)}{I(2n+1)}}={\frac {2n+1}{2n}}} Pelo teorema do confronto,
⇒ lim n → ∞ I ( 2 n ) I ( 2 n + 1 ) = 1 {\displaystyle \Rightarrow \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}=1} lim n → ∞ I ( 2 n ) I ( 2 n + 1 ) = π 2 lim n → ∞ ∏ k = 1 n ( 2 k − 1 2 k ⋅ 2 k + 1 2 k ) = 1 {\displaystyle \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}={\frac {\pi }{2}}\lim _{n\rightarrow \infty }\prod _{k=1}^{n}\left({\frac {2k-1}{2k}}\cdot {\frac {2k+1}{2k}}\right)=1} ⇒ π 2 = ∏ k = 1 ∞ ( 2 k 2 k − 1 ⋅ 2 k 2 k + 1 ) = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ ⋯ {\displaystyle \Rightarrow {\frac {\pi }{2}}=\prod _{k=1}^{\infty }\left({\frac {2k}{2k-1}}\cdot {\frac {2k}{2k+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \cdots } Relação com a aproximação de Stirling A fórmula de Stirling para n ! estabelece que
n ! = 2 π n ( n e ) n [ 1 + O ( 1 n ) ] {\displaystyle n!={\sqrt {2\pi n}}{\left({\frac {n}{e}}\right)}^{n}\left[1+O\left({\frac {1}{n}}\right)\right]} Considere agora as aproximações finitas para o produto de Wallis, obtidas tomando os primeiros k termos no produto:
p k = ∏ n = 1 k 2 n 2 n − 1 2 n 2 n + 1 {\displaystyle p_{k}=\prod _{n=1}^{k}{\frac {2n}{2n-1}}{\frac {2n}{2n+1}}} pk pode ser expresso como
p k = 1 2 k + 1 ∏ n = 1 k ( 2 n ) 4 [ ( 2 n ) ( 2 n − 1 ) ] 2 = 1 2 k + 1 ⋅ 2 4 k ( k ! ) 4 [ ( 2 k ) ! ] 2 {\displaystyle {\begin{aligned}p_{k}&={1 \over {2k+1}}\prod _{n=1}^{k}{\frac {(2n)^{4}}{[(2n)(2n-1)]^{2}}}\\&={1 \over {2k+1}}\cdot {{2^{4k}\,(k!)^{4}} \over {[(2k)!]^{2}}}\end{aligned}}} Substituindo a aproximação de Stirling nesta expressão (para k ! e (2k )!) pode-se deduzir (após curto cálculo) que pk converge para π ⁄2 k → ∞.
ζ'(0)[ 6] A função zeta de Riemann e a função eta de Dirichlet podem ser difinidas:
ζ ( s ) = ∑ n = 1 ∞ 1 n s , ℜ ( s ) > 1 η ( s ) = ( 1 − 2 1 − s ) ζ ( s ) = ∑ n = 1 ∞ ( − 1 ) n − 1 n s , ℜ ( s ) > 0 {\displaystyle {\begin{aligned}\zeta (s)&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\Re (s)>1\\\eta (s)&=(1-2^{1-s})\zeta (s)\\&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0\end{aligned}}} Aplicando uma transformação de Euler à última série, obtém-se o seguinte
η ( s ) = 1 2 + 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 [ 1 n s − 1 ( n + 1 ) s ] , ℜ ( s ) > − 1 ⇒ η ′ ( s ) = ( 1 − 2 1 − s ) ζ ′ ( s ) + 2 1 − s ( ln 2 ) ζ ( s ) = − 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 [ ln n n s − ln ( n + 1 ) ( n + 1 ) s ] , ℜ ( s ) > − 1 {\displaystyle {\begin{aligned}\eta (s)&={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {1}{n^{s}}}-{\frac {1}{(n+1)^{s}}}\right],\Re (s)>-1\\\Rightarrow \eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {\ln n}{n^{s}}}-{\frac {\ln(n+1)}{(n+1)^{s}}}\right],\Re (s)>-1\end{aligned}}} ⇒ η ′ ( 0 ) = − ζ ′ ( 0 ) − ln 2 = − 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 [ ln n − ln ( n + 1 ) ] = − 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 ln n n + 1 = − 1 2 ( ln 1 2 − ln 2 3 + ln 3 4 − ln 4 5 + ln 5 6 − ⋯ ) = 1 2 ( ln 2 1 + ln 2 3 + ln 4 3 + ln 4 5 + ln 6 5 + ⋯ ) = 1 2 ln ( 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ ⋯ ) = 1 2 ln π 2 ⇒ ζ ′ ( 0 ) = − 1 2 ln ( 2 π ) {\displaystyle {\begin{aligned}\Rightarrow \eta '(0)&=-\zeta '(0)-\ln 2=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[\ln n-\ln(n+1)\right]\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\&=-{\frac {1}{2}}\left(\ln {\frac {1}{2}}-\ln {\frac {2}{3}}+\ln {\frac {3}{4}}-\ln {\frac {4}{5}}+\ln {\frac {5}{6}}-\cdots \right)\\&={\frac {1}{2}}\left(\ln {\frac {2}{1}}+\ln {\frac {2}{3}}+\ln {\frac {4}{3}}+\ln {\frac {4}{5}}+\ln {\frac {6}{5}}+\cdots \right)\\&={\frac {1}{2}}\ln \left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \cdots \right)={\frac {1}{2}}\ln {\frac {\pi }{2}}\\\Rightarrow \zeta '(0)&=-{\frac {1}{2}}\ln \left(2\pi \right)\end{aligned}}} Referências ↑ Friedmann, Tamar; Hagen, C. R. (2015). «Quantum mechanical derivation of the Wallis formula for π». Journal of Mathematical Physics . 56 : 112101. Bibcode:2015JMP....56k2101F. arXiv:1510.07813 . doi:10.1063/1.4930800 ↑ "Discovery of classic pi formula a 'cunning piece of magic'." University of Rochester (November 10, 2015 ↑ "New derivation of pi links quantum physics and pure math." American Institute of Physics (November 10, 2015) ↑ "New derivation of pi links quantum physics and pure math." Phys.org (November 10, 2015) ↑ "Revealing the hidden connection between pi and Bohr's hydrogen model." Physics World (November 17, 2015) ↑ a b «Wallis Formula» ↑ «Integrating Powers and Product of Sines and Cosines: Challenging Problems» Ligações externas Hazewinkel, Michiel, ed. (2001), «Wallis formula», Enciclopédia de Matemática , ISBN 978-1-55608-010-4 (em inglês), Springer «Why does this product equal π/2? A new proof of the Wallis formula for π.». 3Blue1Brown. 20 de abril de 2018 – via YouTube 
![{\displaystyle n!={\sqrt {2\pi n}}{\left({\frac {n}{e}}\right)}^{n}\left[1+O\left({\frac {1}{n}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b5b277240d28bdacc697262a5e35f7f569e21be)
![{\displaystyle {\begin{aligned}p_{k}&={1 \over {2k+1}}\prod _{n=1}^{k}{\frac {(2n)^{4}}{[(2n)(2n-1)]^{2}}}\\&={1 \over {2k+1}}\cdot {{2^{4k}\,(k!)^{4}} \over {[(2k)!]^{2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cecacd3ce92c09a5f7d065de7e058b353e41f8a8)
![{\displaystyle {\begin{aligned}\eta (s)&={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {1}{n^{s}}}-{\frac {1}{(n+1)^{s}}}\right],\Re (s)>-1\\\Rightarrow \eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {\ln n}{n^{s}}}-{\frac {\ln(n+1)}{(n+1)^{s}}}\right],\Re (s)>-1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a563692b798d4345d47a14d2127d3f5051972380)
![{\displaystyle {\begin{aligned}\Rightarrow \eta '(0)&=-\zeta '(0)-\ln 2=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[\ln n-\ln(n+1)\right]\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\&=-{\frac {1}{2}}\left(\ln {\frac {1}{2}}-\ln {\frac {2}{3}}+\ln {\frac {3}{4}}-\ln {\frac {4}{5}}+\ln {\frac {5}{6}}-\cdots \right)\\&={\frac {1}{2}}\left(\ln {\frac {2}{1}}+\ln {\frac {2}{3}}+\ln {\frac {4}{3}}+\ln {\frac {4}{5}}+\ln {\frac {6}{5}}+\cdots \right)\\&={\frac {1}{2}}\ln \left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \cdots \right)={\frac {1}{2}}\ln {\frac {\pi }{2}}\\\Rightarrow \zeta '(0)&=-{\frac {1}{2}}\ln \left(2\pi \right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a89cf6dfdcc7c51b4fc4c1b42c88236611c7b72)
. doi:10.1063/1.4930800 
