Inverse Functions

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:

Hurkyl

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Try making the substitution x = f(z)
 
On the verge of tears... can't make substitution work... must change major....no more hints....
 

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.
 

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)).
 
Originally posted by HallsofIvy
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)).
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
 

Hurkyl

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(false therefore true) is a true implication.

Note that Ivy's conclusion began with "If f has an inverse".
 
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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