Ïnverse functions and composition

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Discussion Overview

The discussion revolves around the conditions under which the composition of two functions and their inverses can be commutative, specifically examining the equation (f ∘ g⁻¹) = (f⁻¹ ∘ g). Participants explore the requirements for functions f and g to satisfy this relationship, focusing on the concept of involutions and their properties.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the conditions under which (f ∘ g⁻¹) = (f⁻¹ ∘ g) holds true.
  • Another participant notes that (f ∘ g)⁻¹ = (g⁻¹ ∘ f⁻¹) and attempts to derive implications from this relationship.
  • It is suggested that for (f ∘ g⁻¹) to equal (f⁻¹ ∘ g), one of the functions must be an involution, but the nature of both functions remains unclear.
  • Participants identify specific functions such as f(x) = a - x and f(x) = a/x as involutions, while questioning if these are the only forms of involutions for real functions.
  • A new function, f(x) = (1/(x-a)) + a, is introduced as another example of an involution, indicating the complexity of identifying all such functions.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the complete characterization of involutions and whether the identified functions are exhaustive. There is no consensus on the nature of f and g or the general conditions for the commutativity of the composition involving inverses.

Contextual Notes

The discussion highlights limitations in identifying all possible involutions and the complexity involved in proving properties related to function composition and inversion.

Who May Find This Useful

Readers interested in function theory, mathematical properties of inverses, and the study of involutions may find this discussion relevant.

mnb96
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Hi,
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.
 
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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?
 
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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?
 
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