Wallisin kaavat
Wallisin kaavat ovat menetelmiä, joilla voidaan laskea piin likiarvoja mielivaltaisen tarkasti. Kaavat on johtanut englantilainen matemaatikko John Wallis [ 1]
(1)
∏
n
=
1
∞
2
n
2
n
−
1
⋅
2
n
2
n
+
1
=
lim
n
→
∞
(
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
6
5
⋅
6
7
⋅
…
⋅
2
n
2
n
−
1
⋅
2
n
2
n
+
1
)
=
π
2
{\displaystyle \prod _{n=1}^{\infty }{\frac {2n}{2n-1}\cdot {\frac {2n}{2n+1}=\lim _{n\to \infty }\left({\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 \cdot {\frac {2n}{2n-1}\cdot {\frac {2n}{2n+1}\right)={\frac {\pi }{2}
(2)
lim
n
→
∞
(
n
!
)
2
⋅
2
2
n
(
2
n
!
)
n
=
π
{\displaystyle \lim _{n\to \infty }{\frac {(n!)^{2}\cdot 2^{2n}{(2n!){\sqrt {n}={\sqrt {\pi }
Kaavojen todistus
Wallisin kaavat pystytään todistamaan osittaisintegroinnin avulla.
Merkitään jokaiselle
n
∈
N
{\displaystyle n\in \mathbb {N} }
I
n
=
∫
0
π
2
sin
n
x
d
x
{\displaystyle I_{n}=\int _{0}^{\frac {\pi }{2}\sin ^{n}x\,{\text{d}x}
a
n
=
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
6
5
⋅
6
7
⋅
…
⋅
2
n
2
n
−
1
⋅
2
n
2
n
+
1
{\displaystyle a_{n}={\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 \cdot {\frac {2n}{2n-1}\cdot {\frac {2n}{2n+1}
b
n
=
(
n
!
)
2
⋅
2
2
n
(
2
n
!
)
n
{\displaystyle b_{n}={\frac {(n!)^{2}\cdot 2^{2n}{(2n!){\sqrt {n}
Tällöin
I
0
=
π
2
{\displaystyle I_{0}={\frac {\pi }{2}
I
1
=
1
{\displaystyle I_{1}=1}
Jos
n
≥
2
{\displaystyle \scriptstyle n\geq 2}
I
n
=
−
cos
π
2
sin
n
−
1
π
2
+
cos
0
sin
n
−
1
0
−
∫
0
π
2
−
cos
x
(
n
−
1
)
sin
n
−
2
x
⋅
cos
x
d
x
{\displaystyle I_{n}=-\cos {\frac {\pi }{2}\sin ^{n-1}{\frac {\pi }{2}+\cos 0\sin ^{n-1}0-\int _{0}^{\frac {\pi }{2}-\cos x(n-1)\sin ^{n-2}x\cdot \cos x\,{\text{d}x}
=
(
n
−
1
)
∫
0
π
2
cos
2
x
sin
n
−
2
x
d
x
=
(
n
−
1
)
∫
0
π
2
(
1
−
sin
2
x
)
sin
n
−
2
x
d
x
{\displaystyle =(n-1)\int _{0}^{\frac {\pi }{2}\cos ^{2}x\sin ^{n-2}x\,{\text{d}x=(n-1)\int _{0}^{\frac {\pi }{2}(1-\sin ^{2}x)\sin ^{n-2}x\,{\text{d}x}
=
(
n
−
1
)
(
I
n
−
2
−
I
n
)
{\displaystyle =(n-1)(I_{n-2}-I_{n})}
Siispä saadaan rekursiivinen kaava
I
n
{\displaystyle \scriptstyle I_{n}
I
n
=
n
−
1
n
⋅
I
n
−
2
{\displaystyle I_{n}={\frac {n-1}{n}\cdot I_{n-2}
Tämän avulla nähdään, että
I
2
n
=
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}={\frac {2n-1}{2n}\cdot I_{2n-2}={\frac {2n-1}{2n}\cdot {\frac {2n-3}{2n-2}\cdot I_{2n-4}
=
…
=
1
2
⋅
3
4
⋅
…
⋅
2
n
−
3
2
n
−
2
⋅
2
n
−
1
2
n
⋅
I
0
{\displaystyle =\ldots ={\frac {1}{2}\cdot {\frac {3}{4}\cdot \ldots \cdot {\frac {2n-3}{2n-2}\cdot {\frac {2n-1}{2n}\cdot I_{0}
I
2
n
+
1
=
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}={\frac {2n}{2n+1}\cdot I_{2n-1}={\frac {2n}{2n+1}\cdot {\frac {2n-2}{2n-1}\cdot I_{2n-3}
=
…
=
2
3
⋅
4
5
⋅
…
⋅
2
n
−
2
2
n
−
1
⋅
2
n
2
n
+
1
⋅
I
1
{\displaystyle =\ldots ={\frac {2}{3}\cdot {\frac {4}{5}\cdot \ldots \cdot {\frac {2n-2}{2n-1}\cdot {\frac {2n}{2n+1}\cdot I_{1}
Näin ollen
I
2
n
+
1
I
2
n
=
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
…
⋅
2
n
−
2
2
n
−
3
⋅
2
n
−
2
2
n
−
1
⋅
2
n
2
n
−
1
⋅
2
n
2
n
+
1
⋅
I
1
I
0
=
a
n
⋅
2
π
{\displaystyle {\frac {I_{2n+1}{I_{2n}={\frac {2}{1}\cdot {\frac {2}{3}\cdot {\frac {4}{3}\cdot {\frac {4}{5}\cdot \ldots \cdot {\frac {2n-2}{2n-3}\cdot {\frac {2n-2}{2n-1}\cdot {\frac {2n}{2n-1}\cdot {\frac {2n}{2n+1}\cdot {\frac {I_{1}{I_{0}=a_{n}\cdot {\frac {2}{\pi }
a
n
=
I
2
n
+
1
I
2
n
⋅
π
2
{\displaystyle a_{n}={\frac {I_{2n+1}{I_{2n}\cdot {\frac {\pi }{2}
Koska
sin
2
n
+
2
x
≤
sin
2
n
+
1
x
≤
sin
2
n
x
{\displaystyle \sin ^{2n+2}x\leq \sin ^{2n+1}x\leq \sin ^{2n}x}
x
∈
[
0
,
π
2
]
{\displaystyle x\in \left[0,{\frac {\pi }{2}\right]}
I
2
n
+
2
≤
I
2
n
+
1
≤
I
2
n
{\displaystyle I_{2n+2}\leq I_{2n+1}\leq I_{2n}
1
≥
I
2
n
+
1
I
2
n
≥
I
2
n
+
2
I
2
n
=
2
n
+
1
2
n
+
2
I
2
n
I
2
n
=
2
n
+
1
2
n
+
2
→
1
{\displaystyle 1\geq {\frac {I_{2n+1}{I_{2n}\geq {\frac {I_{2n+2}{I_{2n}={\frac {\frac {2n+1}{2n+2}I_{2n}{I_{2n}={\frac {2n+1}{2n+2}\to 1}
n
→
∞
{\displaystyle n\to \infty }
lim
n
→
∞
a
n
=
lim
n
→
∞
I
2
n
+
1
I
2
n
⋅
π
2
=
π
2
{\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\frac {I_{2n+1}{I_{2n}\cdot {\frac {\pi }{2}={\frac {\pi }{2}
◻
{\displaystyle \Box }
Koska
b
n
+
1
2
=
b
n
2
⋅
(
2
n
+
2
)
2
(
2
n
+
1
)
2
⋅
n
n
+
1
{\displaystyle b_{n+1}^{2}=b_{n}^{2}\cdot {\frac {(2n+2)^{2}{(2n+1)^{2}\cdot {\frac {n}{n+1}
a
n
=
a
n
+
1
⋅
(
2
n
+
1
)
(
2
n
+
3
)
(
2
n
+
2
)
2
{\displaystyle a_{n}=a_{n+1}\cdot {\frac {(2n+1)(2n+3)}{(2n+2)^{2}
induktiotodistuksella nähdään helposti, että
b
n
2
=
2
n
+
1
n
⋅
a
n
{\displaystyle b_{n}^{2}={\frac {2n+1}{n}\cdot a_{n}
n . Siten väite (2) seuraa väitteestä (1).
◻
{\displaystyle \Box }
↑ Lehtinen, Matti: Osittaisintegroinnin ihmeitä :
Wallisin ja Stirlingin kaavat [1]
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![{\displaystyle x\in \left[0,{\frac {\pi }{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56d27bfd717fd45e923abd4ad1de37d9169632c9)
