Invertible Functions

Date: 06/06/2003 at 13:48:18
From: Jason
Subject: Invertible Functions

I ran across a problem while inverting functions that stumped me and 
I was hoping you would be able to explain it to me.  

Consider the following function as a function from R -> R (Real) and 
say whether the function is invertible.

   h(x) = (sgn x)* sqrt( abs(x) )

where sgn is +1 if x is positive, -1 if x is negative, and 0 if x is 
0.

I understand the solution to be:

   h-1(t) = sgn(t) * t^2

but I don't understand how to get the solution from the original 
equation setup as:

   (sgn (h-1(t) ) * sqrt( abs( (h-1(t)) ) ) = t

I understand that to begin solving for an inversion you can set: 

   h( h-1(t) ) = t

(where h-1(t) is the inversion)

and then substituting:

   (sgn (h-1(t) ) * sqrt( abs( (h-1(t)) ) ) = t

but from this point I become stumped when finding the answer to be:

   h-1(t) = sgn(t) * t^2

Thank you so much for your assistance,
Jason Graves

Date: 06/06/2003 at 15:00:33
From: Doctor Peterson
Subject: Re: Invertible Functions

Hi, Jason.

Problems involving absolute values generally require the use of cases. 
Let's try that approach. You want to solve

  (sgn(y) * sqrt(abs(y)) = t

for y. First consider the case y>=0:

  (1 * sqrt(y)) = t

has the solution

  y = t^2

Second, in the case y<0, the problem becomes

  (-1 * sqrt(-y)) = t

and the solution is

  y = -(-t)^2 = -t^2

In each case, our y satisfies the conditions for the case, so there is 
no extra restriction needed. But we do need to think about how to tell 
from t which case we are in. To do that, note that from the original 
equation, the sign of t is always the same as the sign of y.

Now we can combine the solutions this way:

  y = t^2  if  t >= 0
     -t^2  if  t < 0

or,

  y = sgn(t) * t^2

And that is just the solution you were looking for.

What I just did is exactly what I would do less formally by merely 
looking at the graph of h. I would think of each part of it 
separately, noting that the top half is y = sqrt(x) and the bottom 
half is y = sqrt(-x); then I would invert each of those, to x = y^2 
and x = -y^2; then I would think about how to tell which function to 
use for a given y, namely that we multiply by the sign of y.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/