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Inverse Floor Function.

by TheLegace
Tags: floor, function, inverse
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TheLegace
#1
Jul6-09, 10:19 AM
P: 27
1. The problem statement, all variables and given/known data



2. Relevant equations

find g^-1({-2,-1,0}).
These problems really throw me off, in my textbook there is a simlar problem but with x/2, and the inverse of the function yeilds a range of values. I would like to know if the same is true here.

3. The attempt at a solution

Since this is a floor function would the range be something like

n<x+2<n+1

Any help is appreciated.
Thank You.
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cipher42
#2
Jul6-09, 10:49 AM
P: 49
You can only define an true inverse function to a function that is one-to-one. Well, Either that or you have to narrow down your domain/range to a region where it is one-to-one. Think about the the simple function [tex]f(x)=x^2[/tex]. Since both f(-2) and f(2) are equal to 4, we cannot define an inverse function (call g(x)) since we would not know what number to assign to g(4) - should it be -2 or should it be 2? However, if we only look at positive values of x, then we are free to define [tex]g(x) = f^{-1}(x) = \sqrt{x}[/tex] since [tex]g(f(x))=\sqrt{f(x)}=\sqrt{x^2}=x[/tex] (note that the last equality holds because we are only interested in positive values of x - otherwise both -x and x would do).

The problem with defining an inverse function comes about when multiple values in the domain of the function get sent to the same value in the range. Sometimes, like with the above function, we can handle it by only looking at a portion of the graph. However, with your step function, you can see that this solution does not really work. Every single value between 0 and 1 gets sent to 0 by the step function! So defining an inverse function is out of the question.

However, there is another mathematical object, called a preimage, which happens to have the same notation as an inverse function, that is probably what this question is asking for. If we have a function, f, then the preimage of a, written [tex]f^{-1}(x)[/tex] is the set of all values, x, in the domain so that [tex]f(x)=a[/tex]. As an example, take what we did earlier with f(x) = x^2. Under this function, the preimage of 4 is the set {-2,2} since both f(-2) = 4 and f(2) = 4.

So in your problem, then want you to tell them all of the values of x that will give g(x)=-2, g(x)=-1, or g(x)=0. Lets look at the first one. What points will give you g(x)=-2. Well, since we're rounding down, as long as x is between -2 and -1, then I will have g(x)=-2. So the preimage of -2 for this function is all x in the range -2<x<1. So you were right that it will be a range. Now you just need to find the preimages of -1 and 0 under g and then combine them all together. Does that make sense?
HallsofIvy
#3
Jul6-09, 11:13 AM
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The very, very first time I had to present a proof in a graduate class, it involved f-1(A) for a set A- and proceeded through the entire proof assuming f had an inverse function! Very embarassing.

For set A, g-1(A) is defined as [itex]\left{ x| g(x)\in A}[/itex], the set of all x such that g(x) is in A. If g is not "onto", g-1(A) may be the empty set for some A, but it still exists. If g is not "one-to-one", g(A) may not be a singleton set even when A is. The notation does NOT imply that g must be invertible.

In particular, g-1({-2, -1, 0}) is the set of all x such that g(x)= -2 or g(x)= -1 or g(x)= 0. Since g is the least integer function, g maps all numbers between -2 and -1 into -2, all numbers from -1 to 0 into -1, and all numbers from 0 to 1 into 0.

TheLegace
#4
Jul6-09, 11:30 AM
P: 27
Inverse Floor Function.

Oh wow, I actually understand it now. Thank You Very Much.

I am thinking that when you find inverse of a function you get some sort of function, and maybe a distinct values, but I see that it is just the way things are mapped in each respective set, since the floor function yeilds an integer, the input/x values from the function will be real numbers from a respective range.

Thank You so Very Much.

TheLegace.
TheLegace
#5
Jul6-09, 01:11 PM
P: 27
Quote Quote by HallsofIvy View Post
The very, very first time I had to present a proof in a graduate class, it involved f-1(A) for a set A- and proceeded through the entire proof assuming f had an inverse function! Very embarassing.

For set A, g-1(A) is defined as [itex]\left{ x| g(x)\in A}[/itex], the set of all x such that g(x) is in A. If g is not "onto", g-1(A) may be the empty set for some A, but it still exists. If g is not "one-to-one", g(A) may not be a singleton set even when A is. The notation does NOT imply that g must be invertible.

In particular, g-1({-2, -1, 0}) is the set of all x such that g(x)= -2 or g(x)= -1 or g(x)= 0. Since g is the least integer function, g maps all numbers between -2 and -1 into -2, all numbers from -1 to 0 into -1, and all numbers from 0 to 1 into 0.
Hi, sorry one last question. For the ranges are you accounting for the floor(x+2), the +2 part of function, because, if g(x)=-2, then x+2=-2, the largest negative value that could be will be -4, and -3, so maybe -2<x+2<-1, then -4<x<-3, correct, beause the answer needs to be shown in that way, and example for this would be if g(-3.5)=floor(-1.5)= -2, so to yield values of -2, we would need all real numbers from -4 to -3, not including -3 of course, but including -4, I hope thats right.

Thank You.
TheLegace.


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