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

[Catullus]

Does anyone know of a nonlinear function , such that , for some a that does not equal 1? If a equals 1, then could equal . But what about if a does not equal 1?

[MphLee]

Let \(R\) be a ring and \(a\in R\). Consider any solutions \(\chi: R\to R\) of the f.eq \(\chi(ax)=1+\chi(x)\). This is an abel function of multiplication.

The set of all these solutions, denote it with \({\mathcal A}_a\), is a very big set. Note that for every \(r\in R\) we have the inclusion \({\mathcal A}_a +r\subseteq {\mathcal A}_a\) and, let \(\Theta\) be the set of solutions \(\theta: R\to R\) of the f.eq. \(\theta(ax)=\theta(x)\), then \({\mathcal A}_a+\Theta\subseteq {\mathcal A}_a\). This means we have many solutions.

Let \(h\in R\) be a root of the polynomial \(p\in R[x]\) where \(p(x)=2x-1\). Assume that \(\chi\in {\mathcal A}_a\) has a right inverse \(\phi\) (a retraction), i.e. \(\chi\) is surjective/ \(\chi\phi={\rm id}_R\), then define \(f(x):=\phi(\chi(x)+h)\). It satisfies

\[\begin{align}
f^2(x)& =\phi(     \chi(         \phi(\chi(x)+h)           )+h)&&&\\
         &  =\phi( \chi(x)+2h) &&&\\
&  =\phi( \chi(x)+1) &&&\\
& =ax &&&
\end{align}\]

In short: let \({\mathcal A}_a^{\rm surj}\) the set of surjective abel functions of multiplication by \(a\), and \(V(2x-1)\) be the set of roots of the previous polynomial, let \(\mathcal F_a\) be set of solutions of the f.eq \(f^2(x)=ax\) in \(R\), then we have a map
\[{\rm pt}\times_{R^R}({\mathcal A}_a^{\rm surj}\times R^R)\times V(2x-1) \to   \mathcal F_a\]

Sending triples \((\chi,\phi,h)\), where \(\chi\phi={\rm id}_R\) to functions \(f(x):=\phi(\chi(x)+h)\). A you can see the size of \(\mathcal F_a\) can be bounded from below by studying if this assignment is injective.

[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