Infinite product for pi
Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series forπ .Sn is the approximation after takingn terms. Each subsequent subplot magnifies the shaded area horizontally by 10 times.(click for detail) TheWallis product is theinfinite product representation ofπ
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}}} It was published in 1656 byJohn Wallis .[ 1]
Proof using integration [ edit ] Wallis derived thisinfinite 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} n {\displaystyle n} n {\displaystyle n} n {\displaystyle n} n {\displaystyle n} [ 2]
I ( n ) = ∫ 0 π sin n x d x . {\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}x\,dx.} (This is a form ofWallis' 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 forI ( 0 ) {\displaystyle I(0)} I ( 1 ) {\displaystyle I(1)}
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 valuesI ( 2 n ) {\displaystyle I(2n)} recurrence relation result from the integration by parts. Eventually, we end get down toI ( 0 ) {\displaystyle I(0)}
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 valuesI ( 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 thatsin 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 byI ( 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}}} By thesqueeze 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 onGaussian integral .
[ edit ] While the proof above is typically featured in modern calculus textbooks, the Wallis product is, in retrospect, an easy corollary of the laterEuler infinite product for thesine 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)} Letx = 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] [ edit ] Stirling's approximation for the factorial functionn ! {\displaystyle n!}
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 firstk {\displaystyle k}
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}},} wherep k {\displaystyle p_{k}}
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 fork ! {\displaystyle k!} ( 2 k ) ! {\displaystyle (2k)!} p k {\displaystyle p_{k}} π 2 {\displaystyle {\frac {\pi }{2}}} k → ∞ {\displaystyle k\rightarrow \infty }
Derivative of the Riemann zeta function at zero [ edit ] TheRiemann zeta function and theDirichlet 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}}} 
