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Ïnverse functions and composition

  1. Feb 26, 2009 #1
    given two functions f and g, is there any known condition under which the following is valid:

    [tex](f \circ g^{-1}) = (f^{-1} \circ g)[/tex]

    Basically I have to find out the requirements for [tex]f[/tex] and [tex]g[/tex] for which composition is commutative in respect to inversion.
  2. jcsd
  3. Feb 26, 2009 #2
    attempt to a solution

    I probably did a step forward but I'm still stuck.
    We know that: [tex](f \circ g)^{-1}=(g^{-1} \circ f^{-1})[/tex].

    This obviously implies that the following must be true:
    [tex](f \circ g^{-1})=(f \circ g^{-1})^{-1}=(g \circ f^{-1})[/tex]

    If consider only the body in the parentheses [tex]h = f \circ g[/tex], we have the condition:

    [tex]h = h^{-1}[/tex]

    OK! Now, what is the family of functions that fulfills the property [tex]f = f^{-1}[/tex]?
    I already found [tex]f(x)=a-x[/tex] and [tex]f(x)=a/x[/tex]
    Are there others?
    Last edited: Feb 26, 2009
  4. Feb 26, 2009 #3
    I just found out that a function such that [tex]f=f^{-1}[/tex] is called involution.
    I also sketched out a proof which proves that the composition of two involutions is not an involution.

    This would partly answer my question, implying that in order to satisfy [tex](f \circ g^{-1}) = (f^{-1} \circ g)[/tex] only one of the two function can be an involution; but still nothing useful is known about the nature of f and g

    And still, what remains unanswered is: are [tex]f(x)=a-x[/tex] and [tex]f(x)=\frac{a}{x}[/tex] the only existing involutions for real functions?
    Last edited: Feb 27, 2009
  5. Feb 26, 2009 #4
    Last edited: Feb 27, 2009
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