Infinite product for pi
Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series for π . Sn is the approximation after taking n terms. Each subsequent subplot magnifies the shaded area horizontally by 10 times. (click for detail) In mathematics , the Wallis product for π , published in 1656 by John Wallis ,[1] states that
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 ) ⋅ ( 8 7 ⋅ 8 9 ) ⋅ ⋯ {\displaystyle {\begin{aligned}{\frac {\pi }{2}}&=\prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)\\[6pt]&={\Big (}{\frac {2}{1}}\cdot {\frac {2}{3}}{\Big )}\cdot {\Big (}{\frac {4}{3}}\cdot {\frac {4}{5}}{\Big )}\cdot {\Big (}{\frac {6}{5}}\cdot {\frac {6}{7}}{\Big )}\cdot {\Big (}{\frac {8}{7}}\cdot {\frac {8}{9}}{\Big )}\cdot \;\cdots \\\end{aligned}}} Proof using integration [ edit ] Wallis derived this infinite product using interpolation, though his method is not regarded as rigorous. A modern derivation can be found by examining ∫ 0 π sin n x d x {\displaystyle \int _{0}^{\pi }\sin ^{n}x\,dx} for even and odd values of n {\displaystyle n} , and noting that for large n {\displaystyle n} , increasing n {\displaystyle n} by 1 results in a change that becomes ever smaller as n {\displaystyle n} increases. Let[2]
I ( n ) = ∫ 0 π sin n x d x . {\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}x\,dx.} (This is a form of Wallis' integrals .) Integrate by parts :
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 {\displaystyle {\begin{aligned}\Rightarrow I(n)&=\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=-\sin ^{n-1}x\cos x{\Biggl |}_{0}^{\pi }-\int _{0}^{\pi }(-\cos x)(n-1)\sin ^{n-2}x\cos x\,dx\\[6pt]{}&=0+(n-1)\int _{0}^{\pi }\cos ^{2}x\sin ^{n-2}x\,dx,\qquad n>1\\[6pt]{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}x\,dx\\[6pt]{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}x\,dx-(n-1)\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=(n-1)I(n-2)-(n-1)I(n)\\[6pt]{}&={\frac {n-1}{n}}I(n-2)\\[6pt]\Rightarrow {\frac {I(n)}{I(n-2)}}&={\frac {n-1}{n}}\\[6pt]\end{aligned}}} Now, we make two variable substitutions for convenience to obtain:
I ( 2 n ) = 2 n − 1 2 n I ( 2 n − 2 ) {\displaystyle I(2n)={\frac {2n-1}{2n}}I(2n-2)} I ( 2 n + 1 ) = 2 n 2 n + 1 I ( 2 n − 1 ) {\displaystyle I(2n+1)={\frac {2n}{2n+1}}I(2n-1)} We obtain values for I ( 0 ) {\displaystyle I(0)} and I ( 1 ) {\displaystyle I(1)} for later use.
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 {\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x{\Biggl |}_{0}^{\pi }=\pi \\[6pt]I(1)&=\int _{0}^{\pi }\sin x\,dx=-\cos x{\Biggl |}_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\[6pt]\end{aligned}}} Now, we calculate for even values I ( 2 n ) {\displaystyle I(2n)} by repeatedly applying the recurrence relation result from the integration by parts. Eventually, we end get down to I ( 0 ) {\displaystyle I(0)} , which we have calculated.
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 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)} = 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}}} Repeating the process for odd values I ( 2 n + 1 ) {\displaystyle I(2n+1)} ,
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)} = 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}}} We make the following observation, based on the fact that sin x ≤ 1 {\displaystyle \sin {x}\leq 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)} Dividing by I ( 2 n + 1 ) {\displaystyle 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}}} , where the equality comes from our recurrence relation. By the squeeze theorem ,
⇒ 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 } See the main page on Gaussian integral .
Proof using Euler's infinite product for the sine function [ edit ] While the proof above is typically featured in modern calculus textbooks, the Wallis product is, in retrospect, an easy corollary of the later Euler infinite product for the sine function .
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)} Let x = π 2 {\displaystyle x={\frac {\pi }{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)\\[6pt]\Rightarrow {\frac {\pi }{2}}&=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\[6pt]&=\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}}} [1] Relation to Stirling's approximation [ edit ] Stirling's approximation for the factorial function n ! {\displaystyle n!} asserts that
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].} Consider now the finite approximations to the Wallis product, obtained by taking the first k {\displaystyle k} terms in the product
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}},} where p k {\displaystyle p_{k}} can be written as
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}}}\\[6pt]&={1 \over {2k+1}}\cdot {{2^{4k}\,(k!)^{4}} \over {[(2k)!]^{2}}}.\end{aligned}}} Substituting Stirling's approximation in this expression (both for k ! {\displaystyle k!} and ( 2 k ) ! {\displaystyle (2k)!} ) one can deduce (after a short calculation) that p k {\displaystyle p_{k}} converges to π 2 {\displaystyle {\frac {\pi }{2}}} as k → ∞ {\displaystyle k\rightarrow \infty } .
Derivative of the Riemann zeta function at zero [ edit ] The Riemann zeta function and the Dirichlet eta function can be defined:[1]
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\\[6pt]\eta (s)&=(1-2^{1-s})\zeta (s)\\[6pt]&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0\end{aligned}}} Applying an Euler transform to the latter series, the following is obtained:
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\\[6pt]\Rightarrow \eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\[6pt]&=-{\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]\\[6pt]&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\[6pt]&=-{\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)\\[6pt]&={\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)\\[6pt]&={\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}}} See also [ edit ] External links [ edit ]