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Inverse Functions

  1. Sep 24, 2003 #1
    What is the general solution to the equations f(g(x))=h(x) and g(f(x))=h(x), or how do you find the particular soltuion to the equations given a function f (given any f what is the general form of h(x) such that g(x) exists)?

    My thoughts on the topic:

    Specific example:
    Suppose we have f(x)=x+1 and h(x)=x. Then g(x)=x-1.
    Suppose instead h(x)=x+2. Then,
    f(g(x))=g(x)+1=x+2
    g(x)=x+1
    g(f(x))=x+1+1=x+2 so it checks.
    Now suppose h(x)=3x
    f(g(x))=g(x)+1=3x
    g(x)=3x-1
    g(f(x))=3(x+1)-1=3x+2 doesn't check.

    More general case:
    Suppose now f(x)=x2
    f(g(x))=[g(x)]2=h(x)
    g(x)=sqrt(h(x))
    g(f(x))=[squ]h(x2)=h(x)
    So g(x) exists for a given h(x) if
    h(x2)=[h(x)]2
    I don't think I can solve this any further. I feel resigned to the fact that I must check to see that this holds for some specific h(x) rather than finding the general solution.
    Certainly, h(x)=xk works. Any further thoughts?

    Edit:
    Oh by the way, h(x)=x of course works for all f provided that f has an inverse. So any general form should be able to reduce to h(x)=x.
     
    Last edited: Sep 24, 2003
  2. jcsd
  3. Sep 24, 2003 #2

    Hurkyl

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    Try making the substitution x = f(z)
     
  4. Sep 24, 2003 #3
    On the verge of tears... can't make substitution work... must change major....no more hints....
     
  5. Sep 24, 2003 #4

    Hurkyl

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    Hrm, while that substitution is interesting, it doesn't seem to help as much as I had hoped...

    (it only lets me prove that f, g, and h all commute)


    Ok, here's the next suggestion! See if you can apply some interesting operations and substitutions to solve each of the initial two equations formally for g-1(x). I get:

    g-1(x) = h-1(f(x)) = f(h-1(x))

    and this solution, if it exists, checks out.

    With the information given, I doubt you can make a much stronger statement than this.
     
  6. Sep 26, 2003 #5

    HallsofIvy

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    You are asking "given any f what is the general form of h(x) such that g(x) exists? " (such that f(g(x))= h(x)).

    It doesn't depend of h so much as f: If f has an inverse, then

    f-1(f(g(x))= f-1(h(x)) so
    g(x)= f-1(h(x)).
     
  7. Sep 26, 2003 #6
    Your answer doesn't check out. I've come up with several expressions that did something similar. The tricky part is that you have to utilize both f(g(x))=g(f(x)) and f(g(x))=h(x). In your answer you don't utilize the first one.
    Choose f(x)=x3, h(x)=2x
    f-1(x)=x1/3
    g(x)=(2x)1/3
    but g(f(x))=x*21/3
     
  8. Sep 26, 2003 #7

    Hurkyl

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    (false therefore true) is a true implication.

    Note that Ivy's conclusion began with "If f has an inverse".
     
  9. Sep 26, 2003 #8
    But f(x)=x3 does have an inverse. In fact I purposely chose this one rahter than
    f(x)=x2 for that exact reason.

    Say we have f(x)=ex and h(x)=2x. Then g(x)=ln(2x) but g(f(x))=ln(2ex)=ln2+x.

    Anyway the point is you can reach Ivy's conclusion from just the fact that f(g(x))=h(x). Simply apply the inverse of f on both sides. But this is true regardless of the fact that g(f(x))=f(g(x)). A solution should be dependent on this fact.
     
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