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Inverse functions and functions

  1. Aug 24, 2010 #1

    Char. Limit

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    1. The problem statement, all variables and given/known data

    This isn't really a homework question, just a curiousity, but here goes.

    Is it possible to find every function that is its own inverse? That is, can we find all functions that fit the definition

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

    ?

    I can name two... f(x) = x and f(x) = x-1. Are there others? If not, can that fact be proven?
     
  2. jcsd
  3. Aug 24, 2010 #2

    hunt_mat

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    Inverse functions can be thought of as a reflection about the line y=x, so start drawing
     
  4. Aug 24, 2010 #3

    Char. Limit

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    So, in other words, in order for a function to be its own inverse, it must be symmetrical about the line y=x. How can we prove this symmetry of a function?
     
  5. Aug 24, 2010 #4

    LCKurtz

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    It happens when y = f(x) and x = f(y). :tongue2:
     
  6. Aug 24, 2010 #5

    Char. Limit

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    Considering that that just leads to a circular definition, I guess I can conclude that there is no known way to find all functions that fit this formula, or to prove that the set of the functions you have that fit the formula is complete?
     
  7. Aug 24, 2010 #6

    LCKurtz

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    Complete in what metric? Note that that set of functions isn't even closed under addition:

    f(x) = x and g(x) = 1-x are both self-inverse but their sum is 1 which isn't even invertible.
     
  8. Aug 24, 2010 #7

    Char. Limit

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    Oh I don't mean closed... I simply meant that given a set of functions that are self-inverse, it is likely not possible to prove that the set of functions contains EVERY function that is self-inverse.

    Now I have four functions...

    f(x)=a-x for constant a
    f(x)=x
    f(x)=1/x^n for n=2m-1 for positive integer m

    And that's it really.
     
  9. Aug 24, 2010 #8

    Hurkyl

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    :confused: This doesn't make any sense.

    Depends what you mean by "find". I thought the solution given was satisfactory. *shrug*

    :confused: I suppose certain presentations of a set may not have an obvious proof, or may even be undecidable. However, in this very thread, we have not only presented a putative complete set, but have even sketched the proof that it is complete.

    There is also the trivial presentation of a set that contains all self-inverse functions: let S be the set of all functions that are self-inverse. The proof that this is complete is also trivial. :wink:


    Er, composing that with itself gives you
    f(f(x)) = xn2
     
    Last edited: Aug 24, 2010
  10. Aug 24, 2010 #9

    vela

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    This also works:

    [tex]f(x) = \left\{ \begin{array}{ll} -2x+3 & \textrm{if } x<1 \\ (3-x)/2 & \textrm{if } x \ge 1 \end{array} \right.[/tex]
     
  11. Aug 24, 2010 #10

    LCKurtz

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    Ahhh, you didn't mean complete in the technical sense..

    Perhaps you didn't grok hunt_mat's original reply to you in post #2. Start drawing. There are lots of them.
     
  12. Aug 24, 2010 #11

    LCKurtz

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    One additional thought for you. Think about a function defined as a set of ordered pairs. Then you could use this definition for a criterion:

    A function f is its own inverse if and only if for every (a,b) ε f, (b,a) ε f.
     
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