In de wiskunde is het Wallis-product , dat in 1655 werd geconstrueerd door John Wallis , een voorstelling van het getal
π
/
2
{\displaystyle \pi /2}
in de vorm van een oneindig product :
∏
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 }{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+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}}\ldots ={\frac {\pi }{2}}}
Bewijs met Eulers oneindige productformule voor de sinusfunctie[1]
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De productformule voor de sinus luidt:
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)}
Met
x
=
1
2
π
{\displaystyle x={\tfrac {1}{2}}\pi }
volgt dan:
2
π
=
∏
n
=
1
∞
(
1
−
1
4
n
2
)
{\displaystyle {\frac {2}{\pi }}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)}
,
dus
π
2
=
∏
n
=
1
∞
(
4
n
2
4
n
2
−
1
)
=
∏
n
=
1
∞
(
2
n
)
(
2
n
)
(
2
n
−
1
)
(
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 }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\&{}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+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}}\ldots \end{aligned}}}
Bewijs met een integraal[2]
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Definieer
I
(
n
)
=
∫
0
π
sin
n
(
x
)
d
x
.
{\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}(x)dx.}
Pas voor
n
≥
2
{\displaystyle n\geq 2}
partiële integratie toe, zodat
I
(
n
)
=
∫
0
π
sin
n
−
1
(
x
)
d
(
−
cos
(
x
)
)
=
−
cos
(
x
)
sin
n
−
1
(
x
)
|
x
=
0
π
+
∫
0
π
cos
(
x
)
d
(
sin
n
−
1
x
)
=
0
+
(
n
−
1
)
∫
0
π
sin
n
−
2
(
x
)
cos
2
(
x
)
d
x
=
(
n
−
1
)
∫
0
π
sin
n
−
2
(
x
)
(
1
−
sin
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
)
,
{\displaystyle {\begin{aligned}I(n)&=&\int _{0}^{\pi }\sin ^{n-1}(x)\,\mathrm {d} (-\cos(x))\\&=&\left.-\cos(x)\sin ^{n-1}(x)\right|_{x=0}^{\pi }+\int _{0}^{\pi }\cos(x)\,\mathrm {d} (\sin ^{n-1}x)\\&=&0+(n-1)\int _{0}^{\pi }\sin ^{n-2}(x)\cos ^{2}(x)\,\mathrm {d} x\\&=&(n-1)\int _{0}^{\pi }\sin ^{n-2}(x)(1-\sin ^{2}(x))\,\mathrm {d} x\\&=&(n-1)\int _{0}^{\pi }\sin ^{n-2}(x)\,\mathrm {d} x-(n-1)\int _{0}^{\pi }\sin ^{n}(x)\,\mathrm {d} x\\&=&(n-1)I(n-2)-(n-1)I(n),\end{aligned}}}
ofwel
I
(
n
)
=
n
−
1
n
⋅
I
(
n
−
2
)
{\displaystyle I(n)={\frac {n-1}{n}}\cdot I(n-2)}
Herhaalde toepassing hiervan voor
n
=
2
k
{\displaystyle n=2k}
levert
I
(
2
k
)
=
2
k
−
1
2
k
⋅
2
k
−
3
2
k
−
2
⋅
…
⋅
3
4
⋅
1
2
⋅
I
(
0
)
{\displaystyle I(2k)={\frac {2k-1}{2k}}\cdot {\frac {2k-3}{2k-2}}\cdot \ldots \cdot {\frac {3}{4}}\cdot {\frac {1}{2}}\cdot I(0)}
en voor
n
=
2
k
+
1
{\displaystyle n=2k+1}
volgt
I
(
2
k
+
1
)
=
2
k
2
k
+
1
⋅
2
k
−
2
2
k
−
1
⋅
…
⋅
4
5
⋅
2
3
⋅
I
(
1
)
{\displaystyle I(2k+1)={\frac {2k}{2k+1}}\cdot {\frac {2k-2}{2k-1}}\cdot \ldots \cdot {\frac {4}{5}}\cdot {\frac {2}{3}}\cdot I(1)}
Samen geven deze twee vergelijkingen
I
(
2
k
+
1
)
I
(
2
k
)
⋅
I
(
0
)
I
(
1
)
=
2
k
2
k
+
1
⋅
2
k
2
k
−
1
⋅
…
⋅
4
5
⋅
4
3
⋅
2
3
⋅
2
1
{\displaystyle {\frac {I(2k+1)}{I(2k)}}\cdot {\frac {I(0)}{I(1)}}={\frac {2k}{2k+1}}\cdot {\frac {2k}{2k-1}}\cdot \ldots \cdot {\frac {4}{5}}\cdot {\frac {4}{3}}\cdot {\frac {2}{3}}\cdot {\frac {2}{1}}}
Er geldt
I
(
0
)
=
∫
0
π
d
x
=
π
{\displaystyle I(0)=\int _{0}^{\pi }\,\mathrm {d} x=\pi }
en
I
(
1
)
=
∫
0
π
sin
(
x
)
d
x
=
−
cos
(
x
)
|
x
=
0
π
=
2
{\displaystyle I(1)=\int _{0}^{\pi }\sin(x)\,\mathrm {d} x=-\cos(x){\Big |}_{x=0}^{\pi }=2}
Omdat
I
(
n
)
{\displaystyle I(n)}
een dalende rij is, geldt
2
k
2
k
+
1
=
I
(
2
k
+
1
)
I
(
2
k
−
1
)
≤
I
(
2
k
+
1
)
I
(
2
k
)
≤
I
(
2
k
+
1
)
I
(
2
k
+
1
)
=
1
{\displaystyle {\frac {2k}{2k+1}}={\frac {I(2k+1)}{I(2k-1)}}\leq {\frac {I(2k+1)}{I(2k)}}\leq {\frac {I(2k+1)}{I(2k+1)}}=1}
Als
k
→
∞
{\displaystyle k\to \infty }
gaat de linkerkant naar 1, dus wegens de insluitstelling volgt
lim
k
→
∞
I
(
2
k
+
1
)
I
(
2
k
)
=
1
{\displaystyle \lim _{k\to \infty }{\frac {I(2k+1)}{I(2k)}}=1}
De bovenstaande formule kan dus worden herschreven tot
π
2
=
lim
k
→
∞
I
(
2
k
+
1
)
I
(
2
k
)
I
(
0
)
I
(
1
)
=
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
6
5
⋅
6
7
⋅
…
{\displaystyle {\frac {\pi }{2}}=\lim _{k\to \infty }{\frac {I(2k+1)}{I(2k)}}{\frac {I(0)}{I(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 \ldots }
Relatie met de formule van Stirling
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Externe verwijzingen
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