0
$\begingroup$

I have a function $f(a,b) = \frac{ab}{(\frac{a+b}{2})^2}$, and (to me) it has some cool properties (e.g $f(a,b) = f(b,a)$, $f(x,0) = 0$, $f(x, x) = 1$, $0 \leq f \leq 1$, etc.). Now I wanted to know the inverse function for this. It's a rational function, but I couldn't figure out how to do this. $$ y=\frac{xb}{(\frac{x+b}{2})^2}=\frac{4xb}{(x+b)^2} \\ x=\frac{4yb}{(y+b)^2} \implies x(y+b)^2=4yb \implies \frac{(y+b)^2}{y}=\frac{4b}{x} \\ \frac{y^2 + b^2}{y}=\frac{4b}{x}-2b \implies y+\frac{b^2}{y}=\frac{4b}{x}-2b $$ At this point, I have no idea where to go. I think that I'm missing out on some fundamental rule of solving this, but I don't have a clue about where to go. How do I simplify the $y+\frac{b^2}{y}$ ?
NOTE: The final function $f^{-1}(y,b)$ is such that $f^{-1}(f(x,b),b) = x$ for any $x$ and $b$ such that $x$ and $b$ are not both 0. One can be zero, but not both (since $f(0, 0)$ is undefined).

$\endgroup$
  • $\begingroup$ Can you write the condition for an inverse of $f$? Like, it is a function $g(a,b)$ such that ...? $\endgroup$ – Dietrich Burde Jan 25 at 19:45
  • $\begingroup$ @DietrichBurde Updated. $\endgroup$ – ARaspiK Jan 26 at 10:01
0
$\begingroup$

You can write $$y=\frac {4xb}{(x+b)^2}\\ x^2y+2bxy+-4bx+b^2y=0\\ x=\frac{4b-2by\pm\sqrt{(4b-2by)^2-4b^2y^2}}{2y}\\ x=\frac{2b-by\pm\sqrt{4b^2-4by}}{y}$$

$\endgroup$
  • $\begingroup$ Can you please detail how you went from $x^2y+2bxy+-4bx+b^2y=0$ to the next step? $\endgroup$ – ARaspiK Jan 26 at 7:31
  • $\begingroup$ I just plugged into the quadratic formula. We have a quadratic in $x$. $\endgroup$ – Ross Millikan Jan 26 at 15:16
  • $\begingroup$ Ohhhhh. Somehow I completely missed that! Thanks. $\endgroup$ – ARaspiK Jan 27 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.