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Question on inverse function

  1. Sep 23, 2006 #1
    there have been a thread going on about worst/best notations. as i said there, i was confused with [tex]f^{-1}(x)[/tex] when i first came across it in high school. i thought [tex]f^{-1}(x)[/tex] is the same as [tex]{1}\over{f(x)}[/tex]

    but now i am wondering, is there any function for which
    [tex]f^{-1}(x)[/tex] = [tex]{1}\over{f(x)}[/tex]?
     
    Last edited: Sep 23, 2006
  2. jcsd
  3. Sep 23, 2006 #2

    Hurkyl

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    If you just play with that equation, I think it's easy enough to see how to find such functions.

    i.e. if f(a) = b, then what do you know about f^-1(b)? An about f(b)?

    also, if you pick an ordering on things, to see that f cannot be a continuous function.
     
  4. Sep 23, 2006 #3
    if f(a) = b, then f^-1(b) = a.
    and f(b) = f(f(a))
    but where is it going?

    i am afraid i didn't get it.
     
  5. Sep 23, 2006 #4

    Hurkyl

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    You can do better than that. (And once you see it, keep going for a while)
     
    Last edited: Sep 23, 2006
  6. Sep 23, 2006 #5
    f(b)=f(f(f^-1(b)))
    perhaps this what hurkyl meant.
     
  7. Sep 23, 2006 #6

    Hurkyl

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    Nope. He has the lovely equation

    [tex]f^{-1}(x) = \frac{1}{f(x)}[/tex]

    which I expect him to use sometime while doing this problem.
     
  8. Sep 23, 2006 #7
    ok, if f^-1(b) = 1/f(b), then
    a = 1/f(f(a))
    f(f(a)) = 1/a
    f(a) = f^-1(1/a)

    but i am still in the dark.
     
  9. Sep 23, 2006 #8

    Hurkyl

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    With the work you've done there, you should be able to tell me the values of

    f(b)

    and

    f^(-1) (1/a)

    directly, rather than leave them in terms of other values of f.
     
  10. Sep 23, 2006 #9
    in my last post, i got
    f(a) = f^-1(1/a)
    then f(b) = f(f(a)) = f(f^-1(1/a)) = 1/a = 1/f^-1(b)

    now we have f(b) = 1/f^-1(b)

    but, how can i tell the values of f(b) and f^-1(1/a) directly, rather than leave them in terms of other values of f?
     
  11. Sep 23, 2006 #10

    Hurkyl

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    Argh, I have to go; I was hoping to keep giving you hints so you could figure out all the important steps yourself. I'll write out what I had been planning to say with answers... but please try to answer the questions yourself before reading my answer, and moving onto the next question! (I'll put the answers in white -- highlight for easier reading)


    (From here on, I'm going to set g = f^-1, to make it easier to write)

    There is a loose end I missed before -- we have to assume the domain of our function does not contain zero. Do you see why?

    --------------

    f(0) = a
    g(a) = 0
    g(a) = 1/f(a)
    1/f(a) = 0 <==== there is no possible value of f(a) that works here

    --------------

    So, for the rest of this post, let's assume that we do not require 0 to be in the domain of f.



    We know f(a) = b and g(b) = a. I've already said you can tell me the values of f(b) and g(1/a). What are they?

    ----------

    f(b) = 1/a
    g(1/a) = b

    ----------


    Are there any other values of f or g that you can compute?

    ----------

    yes: we can compute both f and g at the points a, b, 1/a, 1/b.
    ----------



    What are those values?

    ----------

    f(a) = b
    f(b) = 1/a
    f(1/a) = 1/b
    f(1/b) = a
    g(a) = 1/b
    g(b) = a
    g(1/a) = b
    g(1/b) = 1/a
    ----------


    Do you think your equation g(x) = 1/f(x) can be used to relate the values we just computed to anything else?

    ----------

    No. I don't have a rigorous proof of this, but it seems clear to be just from playing with the equation.
    ----------


    So, we know that if we pick the value b for f(a), then this forces upon us a particular choice of f(b), f(1/a), and f(1/b), and nothing else.

    Play with some possible orderings of a and b, such as:

    1/b < 1/a < a < b

    and plot the values of f at those points. If f is continuous, can it possibly be invertible?



    Can you tell me the possible values of f(1) and f(-1)? (Hint: we've already done the work: set a = 1, and see what that tells us)

    ----------

    f(a) = b, so f(1) = b
    f(1/a) = 1/b, so f(1) = 1/b
    So, b = 1/b, and thus b = 1, or b = -1.
    The same is true if we set a = -1.
    So, we can either choose f(1) = 1 and f(-1) = -1, or we can choose f(1) = -1 and f(-1) = 1

    ----------


    Now, what about the rest of the values of f? All we have to do is to organize the rest of the numbers into groups of four, of the form: {a, b, 1/a, 1/b}, then define f(a) = b, et cetera. It seems clear to me that there is no problem in doing this organization, and thus we've constructed an f satisfying your equation.
     
  12. Sep 23, 2006 #11
    thanks. i also have to go now. i will sit with your hints tomorrow and try to solve it myself. wish me luck :)
     
    Last edited: Sep 23, 2006
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