Comparació de la convergència del producte de Wallis (asteriscs liles) i diverses sèries infinites per π. Sn és l'aproximació després de prendre n termes. Cada subgràfica amplia la precisió de la imatge en un factor de 10. En matemàtiques , el producte de Wallis és una expressió que s'utilitza per representar el valor de π que va ser descoberta pel matemàtic anglès John Wallis el 1655 i que estableix queː[ 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
⋯ ⋯ -->
=
π 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}}}
Abans de res, s'ha de considerar que les arrels de sin(x)/x són ±nπ, on n = 1, 2, 3.... Llavors, es pot expressar el sinus com un producte infinit de factors lineals d'arrelsː
sin
-->
(
x
)
x
=
k
(
1
− − -->
x
π π -->
)
(
1
+
x
π π -->
)
(
1
− − -->
x
2
π π -->
)
(
1
+
x
2
π π -->
)
(
1
− − -->
x
3
π π -->
)
(
1
+
x
3
π π -->
)
{\displaystyle {\frac {\sin(x)}{x}}=k\left(1-{\frac {x}{\pi }}\right)\left(1+{\frac {x}{\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)}
k és una constant.
Per trobar la constant k , es pren el límit en ambdós costatsː
lim
x
→ → -->
0
sin
-->
(
x
)
x
=
lim
x
→ → -->
0
(
k
(
1
− − -->
x
π π -->
)
(
1
+
x
π π -->
)
(
1
− − -->
x
2
π π -->
)
(
1
+
x
2
π π -->
)
(
1
− − -->
x
3
π π -->
)
(
1
+
x
3
π π -->
)
⋯ ⋯ -->
)
=
k
{\displaystyle \lim _{x\to 0}{\frac {\sin(x)}{x}}=\lim _{x\to 0}\left(k\left(1-{\frac {x}{\pi }}\right)\left(1+{\frac {x}{\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)\cdots \right)=k}
lim
x
→ → -->
0
sin
-->
(
x
)
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\sin(x)}{x}}=1}
sin
-->
(
x
)
x
=
(
1
− − -->
x
π π -->
)
(
1
+
x
π π -->
)
(
1
− − -->
x
2
π π -->
)
(
1
+
x
2
π π -->
)
(
1
− − -->
x
3
π π -->
)
(
1
+
x
3
π π -->
)
⋯ ⋯ -->
{\displaystyle {\frac {\sin(x)}{x}}=\left(1-{\frac {x}{\pi }}\right)\left(1+{\frac {x}{\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)\cdots }
sin
-->
(
x
)
x
=
(
1
− − -->
x
2
π π -->
2
)
(
1
− − -->
x
2
4
π π -->
2
)
(
1
− − -->
x
2
9
π π -->
2
)
⋯ ⋯ -->
{\displaystyle {\frac {\sin(x)}{x}}=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots }
1
π π -->
/
2
=
(
1
− − -->
1
2
2
)
(
1
− − -->
1
4
2
)
(
1
− − -->
1
6
2
)
⋯ ⋯ -->
=
∏ ∏ -->
n
=
1
∞ ∞ -->
(
1
− − -->
1
4
n
2
)
{\displaystyle {\frac {1}{\pi /2}}=\left(1-{\frac {1}{2^{2}}}\right)\left(1-{\frac {1}{4^{2}}}\right)\left(1-{\frac {1}{6^{2}}}\right)\cdots =\prod _{n=1}^{\infty }(1-{\frac {1}{4n^{2}}})}
π π -->
2
=
∏ ∏ -->
n
=
1
∞ ∞ -->
(
4
n
2
4
n
2
− − -->
1
)
{\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }({\frac {4n^{2}}{4n^{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 =\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 }
