Inverse of power-2 rational function












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).






quadratics inverse-function rational-functions







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  • $begingroup$
    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












$begingroup$


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).






quadratics inverse-function rational-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    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












0








0


1



$begingroup$


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).






quadratics inverse-function rational-functions







share|cite|improve this question











$endgroup$




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).






quadratics inverse-function rational-functions




quadratics inverse-function rational-functions






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share|cite|improve this question













share|cite|improve this question




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edited Jan 26 at 6:50







ARaspiK

















asked Jan 25 at 19:40









ARaspiKARaspiK

11




11












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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






$begingroup$
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




$begingroup$
@DietrichBurde Updated.
$endgroup$
– ARaspiK
Jan 26 at 10:01










1 Answer
1






active

oldest

votes


















0












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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0












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 25 at 20:23









Ross MillikanRoss Millikan

299k24200374




299k24200374












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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




$begingroup$
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




$begingroup$
Ohhhhh. Somehow I completely missed that! Thanks.
$endgroup$
– ARaspiK
Jan 27 at 10:06


















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