Invertible FunctionsDate: 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/ |
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