Re: " How do you solve : f^-1(x- f(y)) = sqrt(1+x - y^2) ?"

From: A N Niel (anniel_at_nym.alias.net.invalid)
Date: 07/12/04 

Date: Mon, 12 Jul 2004 08:11:36 -0400

In article <Ken.Pledger-5C9DD4.15222812072004@bats.mcs.vuw.ac.nz>, Ken
Pledger <Ken.Pledger@mcs.vuw.ac.nz> wrote:

> In article <200407111445.i6BEj1o08585@proapp.mathforum.org>,
> alainverghote@yahoo.fr (Alain Verghote) wrote:
>
> > Dear site-friends,
> >
> > Please do tell me if you know a solving method for
> > such an equation; f is a real and bijective function .
> > f^-1 the inverse.
> >
> > Thanks, Alain.
>
>
>
> If the equation holds for every x and y, then in particular it
> holds when x = y^2.
>
>

so...
  f^-1(y^2- f(y)) = sqrt(1+ y^2 - y^2)
  y^2 - f(y) = f(1)
  y^2-f(1) = f(y)

and for y=1 we get

  1-f(1) = f(1)
  f(1) = 1/2

so the solution must be

  f(y) = y^2 - 1/2

Well, it's not bijective, is it? Proceeding anyway...

  f^{-1}(x) = (1/2) sqrt(4x+2)

Let's try it...

  f^{-1}(x-f(y)) = (1/2) sqrt(4(x-y^2+1/2)+2) = sqrt(1+x-y^2),

so it works.