(no subject)

From: Alain Verghote (alainverghote_at_yahoo.fr)
Date: 12/02/04 

Date: Thu, 2 Dec 2004 18:38:37 +0000 (UTC)

On 27 Nov 2004, alain Verghote wrote:
>Dear friends,
>
>I think it interesting to show you. g:R->R x>1 .
>You will catch 'the knack' with 1/p p steps are needed
>to loop onto the next integer.
>So I choose g(x)=c*(x-1)!*(x-1+1/p)!*(x-1+2/p)!....(x-1+(p-1)/p)!
> g(x+1/p)= c*(x-1+1/p)!*(x-1+2/p)!....(x-1+(p-1)/p)!*x!
>I adopt the writing convention (x-1)! instead of gamma(x).
>Try solving this way : h(y)*h(y+1)=1/y ,
>courage,
>
>Alain.

Here is the solution to h(y)*h(y+1)=1/y , (1)
>>From terms of form (y/a-b)!.
With our conventions we try h(y)=c*(y/2-1)!/(y/2-1/2)!
 we adjust coeff c to obtain (1) ;we propose :
      h(y)=sqrt(2)/2*y/2-1)!/(y/2-1/2)!

 Thanks a lot,Alain

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