Re: Are there any Professional Physicists here that can solve this Physics Equation?

From: Perfectly Innocent (perfectlyInnocent_at_as-if.com)
Date: 06/07/04 

Date: 7 Jun 2004 06:11:30 -0700

Sam Wormley <swormley1@mchsi.com> wrote in message news:<40C3B3A2.7D14991E@mchsi.com>...
> Perfectly Innocent wrote:
> >
> [gone]
>
> Best look at http://mathworld.wolfram.com/topics/Algebra.html and
> review the nomenclature and operation.

I defined the nomenclature and operation of the question with great
precision. I'll add this clarification for you.

When x and y are in A, then x^y = (x+y)/(1+xy).
When x is in A and Z is in S, then x^Z is in S and is currently
unknown.

Find x^Z given that (for all x, y and Z):

x^(y^Z) = (x^y)^Z

0^Z=Z
x^Z > Z if x>0
x^Z < Z if x<0

0 is an element of A.
x and y are elements of A.
Z is an element of S.

perfectlyInnocent@as-if.com (Perfectly Innocent) wrote in message news:<c45b45b3.0405280044.22bedc0d@posting.google.com>...
>
> Let A={real numbers x such that |x|<1}
> Let S={real numbers Z such that 1 < Z < infinity}
>
> Define ^ on A by the rule x^y = (x+y)/(1+xy).
> It follows that (A, ^) is an Abelian group:
>
> x^y=y^x
> x^0=x
> x^(-x)=0
> x^(y^z)=(x^y)^z
>
> I'm looking for a function from AxS->S (also written ^) such that, for
> any x, y, in A and any Z in S:
>
> x^(y^Z) = (x^y)^Z
>
> 0^Z=Z
> x^Z > Z if x>0
> x^Z < Z if x<0

> Eugene Shubert
> http://www.everythingimportant.org

Relevant Pages