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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)=x

f(g(x))=[g(x)]

g(x)=sqrt(h(x))

g(f(x))=[squ]h(x

So g(x) exists for a given h(x) if

h(x

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)=x

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.

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)=x

^{2}f(g(x))=[g(x)]

^{2}=h(x)g(x)=sqrt(h(x))

g(f(x))=[squ]h(x

^{2})=h(x)So g(x) exists for a given h(x) if

h(x

^{2})=[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)=x

^{k}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.

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