Analytic solutions to $f(5 x + x^3) = f(5 x) + f(3 x)$?

Question

I was wondering about real-analytic solutions to the equation

$$f(5 x + x^3) = f(5 x) + f(3 x)$$

If I plug in $x=0$ it follows $f(0)=0$. Also $f(-x) = -f(x)$.

My initial guess was a sum of arcsines but that seems a bad idea ? afterall the $x^3$ part is "inside" and not "outside " ( outside like f(x)^3 ...).

My second guess was a sum of cube roots, but that also seems like a bad idea ?

My third guess was to rewrite it as an integral but I got stuck trying.

My 4 th idea was to use a helping (Abel) equation : $ g(5x + x^3) = g(x) + 1 $ and express $f$ in terms of $g$.

We could also compute the taylor coefficients from the equations but Im not sure if that gives insight ?

Is $f$ a hypergeometric function ??

What are good asymptotics for $f$ ?

I have strong arguments that $f$ is not entire. unless ofcourse $f(x) = 0$. Notice if $a(x)$ is a solution then $C a(x)$ is also one for a real constant $C$.

And how about $C^{oo}$ solutions that are nowhere analytic ? Does that even make sense ?

Would it help to take the derivative on both sides ?

Is there a nonzero elementary solution ?

Is the solution unique up to a constant multiple ?

Is this related to fractals ? How about addition formula's ?

And how would it look on the complex plane ?

Answer 1

It doesn't quite look like you may find a solution which is real-analytic at zero. Otherwise you would have a power-series $f(x) = cx^n + O(x^{n+1})$ with non-zero $c$ which yields $c (5^n - 5^n - 3^n) = 0$. But you could, of course, adjust the constants a bit...

Answer 2

When there is an analytic solution $f(x)\not\equiv0$ there is an integer $r\geq0$ and an analytic $g$ with $$f(x)=x^r \>g(x),\quad g(0)\ne0\ .$$ Your functional equation then implies $$x^r(5+x^2)^r g(5x+x^3)\equiv (5x)^r g(5x)+(3x)^r g(3x)\ ,$$ so that after division by $x^r$ we obtain $$(5+x^2)^r g(5x+x^3)\equiv 5^r g(5x)+3^r g(3x)\ .$$ Putting $x=0$ here gives $$5^r g(0)=(5^r+3^r) g(0)\ ,$$ which is not compatible with $g(0)\ne0$.