מכפלת ואליס

במתמטיקה, מכפלת ואליס לחישוב פאי, הנקראת על שם ג'ון ואליס שגילה אותה בשנת 1655, היא הנוסחה הבאה: ∏ 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}}} בשנת 2015 הפיזיקאי קארל ריצ'רד האיגן (Carl Richard Hagen) הוכיח כי יש קשר בין מכפלת ואליס לחישוב מודל בוהר של אטום חמצן. 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)} נציב 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}}} נגדיר: I ( n ) = ∫ 0 π sin n ⁡ x d x {\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}xdx} נבצע אינטגרציה בחלקים: 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 xdx\\dv&=\sin xdx\\\Rightarrow v&=-\cos x\end{aligned}}} ⇒ I ( n ) = ∫ 0 π sin n ⁡ x d x = ∫ 0 π u d v = u v | x = 0 x = π − ∫ 0 π v d u = − sin n − 1 ⁡ x cos ⁡ x | x = 0 x = 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}xdx=\int _{0}^{\pi }udv=uv|_{x=0}^{x=\pi }-\int _{0}^{\pi }vdu\\{}&=-\sin ^{n-1}x\cos x|_{x=0}^{x=\pi }-\int _{0}^{\pi }-\cos x(n-1)\sin ^{n-2}x\cos xdx\\{}&=0-(n-1)\int _{0}^{\pi }-\cos ^{2}x\sin ^{n-2}xdx,n>1\\{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}xdx\\{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}xdx-(n-1)\int _{0}^{\pi }\sin ^{n}xdx\\{}&=(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}}} נשתמש במשוואה פונקציונלית הזו בדרך הבאה: 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 xdx=-\cos x|_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\I(2n)&=\int _{0}^{\pi }\sin ^{2n}xdx={\frac {2n-1}{2n}}I(2n-2)={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}I(2n-4)\\\end{aligned}}} נחזור על התהליך ונראה כי: = 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}xdx={\frac {2n}{2n+1}}I(2n-1)={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}I(2n-3)} נחזור על התהליך ונראה כי: = 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}}} , (מהתוצאות הנ"ל) ניתן לראות על פי כלל הסנדוויץ' כי: ⇒ 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 } מכפלת ואליס, באתר MathWorld (באנגלית)