F(f(x))=x*a.

[tommy1729]

Mphlee answer is correct and nice.

I wonder if you yourself considered the one periodic function we always mention to you.

Im gonna play abit myself:

let c^2 = a.

Then f(x) = c x is a solution.

Assume f(x + g(x)) is another solution.

Then f( f(x + g(x)) + g(f(x + g(x))) )  = a x.

so

c ( c x + c g(x) + g( c x + c g(x) )) = ax + a g(x) + c g( c x + c g(x) )

so 

a g(x) + c g( c x + c g(x) ) = 0

now this equation might not help much unless we redefine with the 1-periodic function.

However that equation implies that g(x) cannot be a nonlinear polynomial.

Now lets return to f(f(x)) = a x.

clearly f(x) cannot be a polynomial because the degree of f(f(x)) would be 2 times the degree of f(x) , yet a x is degree 1.

thus g(x) is not a polyomial is trivial.

Let f(x) be entire.

if f(x) has a single fixpoint then it must be 0 since f(f(x)) = ax has fixpoint 0.

if f(x) has another fixpoint then f(f(x)) =/= ax everywhere.


So f(x) entire must be of the form x^t exp(T(x))

but iterations of exp(T(x)) never return to polynomials.

So f(x) is not entire.

**

Let f(x) be analytic near its fixpoint 0.

then f(x) is approximated by a polynomial near its fixpoint 0.

so f(f(x)) must be degree 2 Q where Q is the degree of f(x).

this implies Q is infinite or does not exist since f(f(x)) = a x has degree 1.

Or at least suggests ... this is not formal yet but a taylor idea.
with error terms this might work however.

So it seems there are no closed form solutions and Mphlee answer is the most general case.

just my 50 cent

have you discovered anything ?

what did you try ?


regards

tommy1729