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Homework Help: Inverse of a multivalued function

  1. Jan 24, 2007 #1
    Hi, just a quick question concerning the invertibility of multivalued functions. Specifically, I am looking at

    y=x^2

    as a simple example. So for an inverse to exist we have to restrict the domain to (0,+infty) right (doesnt matter which branch Im taking, so Ill take this one.)

    Now my problem is the point 0, because the point (0) is actually a multiset. For

    y=x^2

    there is a double zero. So formally speaking the mapping of the origin is (0,0) -> (0). Is this correct? Because if it is, then this implies that the point y=0 has no inverse, regardless of our choice of branch, because there is no bijection at this point.

    That is my query. The point (0) is a branch point isnt it? So Im thinking that the inverse doesnt exist here. Is this correct? I've looked online and many sources say that if you restrict the domain to [0,+infty) you get an acceptable inverse, but I think it should be (0,+infty).

    Any help that anyone could give would be excellent. Im not being pedantic, this is important for what I am doing. Also sorry if some of my maths definitions are ambiguous; Im a physicist not a mathematician!

    Thank you,

    Stephen
     
  2. jcsd
  3. Jan 24, 2007 #2

    HallsofIvy

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    I think you have it backwards. For any y> 0 there exist two different values of x such that x2= y. That's why you have to select a "branch" in order to have an inverse. But there is only one x such that x2= 0. In this sense, 0 is "better behaved" than other values of x. It's not at all clear to me what you mean by (0,0)-> 0. Would you mean by that notation (-2,2)-> 4?

    Again, that's backwards. 0 is better behaved than other points which is why we can choose either [0 +infty) or (-infty, 0].

     
  4. Jan 24, 2007 #3
    yes sorry. By my horrible notation I meant (for example) (-2,2) -> 4. So if we take (0,+infty) then we have only (2) -> 4 (a bijection) from the domain.

    But if we keep x=0, we still have (0,0) -> (0) don't we? Its still a double zero, so there is no bijective map from x=0 to its image y=0. I don't think I've got it the wrong way round have I?

    My think my point is that for a quadratic there are always exactly two solutions. We have removed all of the negative solutions by choosing [0,+infty) as the domain, but there are still two solutions corresponding to y=0, which are x=(0,0).

    Ste
     
  5. Jan 24, 2007 #4
    Stating that there are exactly two solutions is a nice way to make the fundamental theorem of algebra work; the degree of the polynomial is equal to the number of roots (real or complex.)

    However, is x=0 different from x=0?

    Think of a function as a machine that turns one number into another number. In the case of f(x) = x^2, that machine would turn an input of -2 into an output of 4, and an input of 2 into an output of 4.

    An inverse function is a math machine that undoes what the original machine does. So, if I take the output of the function machine and put it into the inverse function machine, I should be back to what I started with. However, as you realize, such a machine can't exist if there were two *different* values of x that gave the same value of y. If I were to put 4 into the inverse machine, it wouldn't know whether the original input in the function machine was a 2 or a -2. That is why you restrict the original domain of the function (range of the inverse function.) However, in the case of putting 0 into the inverse function, you seem to be claiming that it could have come from two *different* x-values. 0 and 0. Are you sure they're different?

    I like your rigorous way of thinking though. Sometimes I warn students in other courses that they're over-thinking a problem. That's not necessarily a bad thing to do. How about if we call it a repeated root. The noun-marker "a" meaning one value.
     
  6. Jan 24, 2007 #5

    AlephZero

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    Drpizza is right. I think you are getting confused by terminology.

    A function is just a relation between two sets (i.e. a rule for converting an element in one set to an element in the other set). Functions in that sense don't have roots: equations have roots.

    Any polynomial P(x) has a unique factorization into linear factors (the fundamental theorem of algebra). If n of the factors are identical, then the equation P(x) = 0 can be said to have n repeated roots. Note the the roots belong to the equation P(x)=0, not to the function P(x).

    If you have any differentiable function f(x) (not necessarily a polynomial) and there is a value of x=X where f(X) = 0 and f'(X) = 0, then one can say the equation f(x) = 0 has a repeated root, in the sense that a power series expansion of f(x) would be of the form f(x) = (x-X)^2.g(x) for some function g(x).

    But, multiple roots have nothing to do with the existence of the inverse of a function. Consider f(x) = x^3. That has an inverse defined for all real numbers exactly the same as the function f(x) = x, even though the equation x^3 = 0 has (in the above sense) a triple root.

    The function x^2 is invertible on the domain x > 0 and also on the domain x >= 0. There is only one element of the set of real numbers called "zero", not two.
     
    Last edited: Jan 24, 2007
  7. Jan 24, 2007 #6
    hhhmm,

    ok, thanks for the responses. It turns out that my particular function is not invertible for an entirely different reason. However thank you all.

    I was confused because I started with the domain (-infty,+infty), and 'removed' (-infty,0), i.e. an open interval at 0. So I was thinking of the two zeros as x=epsilon, x=-epsilon, and taking epsilon -> 0. I thought that since the two zero solutions arose from the limits of two different objects in the domain, +epsilon and -epsilon, it might mean that they are distinct. But I guess I was wrong (right?).

    One last question concerning the matter then. Is x= -epsilon in the domain in the limit epsilon -> 0? I think it is since we are removing (-infty,0) as opposed to (-infty,0].

    Thank you,

    Ste
     
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