|
|
In article <e32geu$jak$1@xxxxxxxxxxxxxxxxxxxxxxxx>,
"Tsukamoto Tenma" <1@xxx> wrote:
>I have read somewhere that
>Pi/2=3/2*5/6*7/6*11/10*13/14*17/18*...
>where the nominators are prime numbers, with corresponding denominator its
>closest integer in the form 4n+2. I'm looking for a proof of this, any help
>appreciated.
First prove a similar infinite product where the numerators are the
odd prime numbers and the denominators are the closest integers that
are 0 mod 4:
3 5 7 11 13 17 19
pi/4 = - - - -- -- -- -- ... [1]
4 4 8 12 12 16 20
Define a multiplicative function a(n) on the integers by defining its
values on the primes. Let a(2) = 0. Let a(p) = 1 for primes which
are 1 mod 4 and a(p) = -1 for primes which are 3 mod 4.
The product in [1] is then equal to
--- 1
| | ----------
p 1 - a(p)/p
--- a(n)
= > ---- [2]
--- n
n
It can pretty easily be shown that a(2n) = 0 and a(2n+1) = (-1)^n.
Therefore, the sum in [2] is equal to the Gregory series for pi/4:
1 1 1 1 1
- - - + - - - + - - ... [3]
1 3 5 7 9
This proves [1]. The product you cite above is equal to
--- 1
| | ---------- [4]
p 1 + a(p)/p
The product of [2] and [4] is then
--- 1
| | ---------
p>2 1 - 1/p^2
3 --- 1
= - | | ---------
4 p 1 - 1/p^2
3 --- 1
= - > ---
4 --- n^2
n
= pi^2/8 [5]
Thus, [4] must be (pi^2/8)/(pi/4) = pi/2
Rob Johnson <rob@xxxxxxxxxxxxxx>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
|
|