Solving the functional equation $f(x) = f(\frac{x}{\phi}) f(\frac{x}{\phi^2} - 1)$ I'm trying to find a function $f(x)$ such that the spacing between consecutive roots looks like the infinite Fibonacci word:
$$1, \phi^{-1}, 1, 1, \phi^{-1}, 1, \phi^{-1}, 1, 1, \phi^{-1}, 1, 1, \phi^{-1}, \ldots$$
If I'm not mistaken, any solution to the functional equation $f(x) = f(\frac{x}{\phi}) f(\frac{x}{\phi^2} - 1)$ must have roots at the points that I want.
And I simply have no idea where to go from here.
How can I solve this functional equation?
Update. I've found that this one similar problem has an easy solution. Change the denominators of $\phi$ and $\phi^2$ both to $2$, so that we have the equation $g(x) = g(\frac{x}{2}) g(\frac{x}{2} - 1)$. A change of variables gives us this equation:
$$g(2x) = g(x) g(x - 2)$$
Which differs from this double-angle formula only by scaling on the $x$-axis:
$$\sin 2 \theta = \sin \theta \sin (\theta + \frac{\pi}{2})$$
Thus, we have the easy solution $g(x) = \sin (-\pi \frac{x}{4})$. It's not obvious how to apply this solution to the original problem, however.
 A: Substituting $(1+ϕ)x$ for $x$ in your functional equation and transforming according to properties of $ϕ$ leads to the relation $f((1+ϕ)x)=f(ϕx)\cdot f(x-1)$. For $x=1-ϕ$ this implies $f(-ϕ)=f(-1)\cdot f(-ϕ)$, so we must have either $f(-ϕ)=0$ or $f(-1)=1$.
A: Let the golden mean be denoted $G$.
Assuming $f$ is meromorphic over the entire complex plane.
Also assuming $f$ is nonconstant.
If we let $x = 0$ we get $f(0) = f(0)*f(-1)$
Hence either $f(0)$ is 0 or $f(-1)$ is 0 or both.
Case A : $f(-1) = 0$
$f(-G)=f(-1)*f(-1/G-1) = 0$ hence $f(-G)=0$
Case B : $f(0) = 0$
$f(-G) = f(-1)f(-1/G -1) = f(-1)*f(-G)$
Hence $f(-1)$ is 1 or $f(-G) = 0$
(we can reuse the equation to get more and more values depending on previous ones )
--
if we substitute $x = x'* G^2$ we get
$f(G^2 x) = f(G x) f(x-1)$
differentiating
$G^2 f'(G^2 x) = G f'(x) f(x-1) + f(G x) f'(x-1)$
$\log(f(G^2) x) = \log(f(G x)) + \log(f(x-1))$
Lets call $log(f(x))$ := $T(x)$
if we assume $T(x)$ has a taylor series in a neigbourhood $x$ with radius $R \geq \max(G^2 |x|,|x|+1)$ then the taylor coefficients can be solved if we know the value of $T(x)$ (at $x$). The key is the expression of of the taylor series of $F(x-1)$ in terms of the taylor coefficients of $F(x)$ which is classic ( what $F$ does not matter here ).
If $T(0) = 0$ then the solution is unique. 
A: Let $a,b \geq 0$.
Consider the function 
$f_{a,b}(x)=\sin \left( \dfrac{\pi(x-b)}{a} \right)$.
The non-negative roots of $f_{a,b}(x)$ are: $b,b+a,b+2a,\ldots$.
Thus, to get a sequence of roots which can be decomposed into A.P.s, we can just multiply suitable functions of the above form.
For example, to get non-negative roots with gaps of $a,b,c,a,b,c,\ldots$, such as the sequence$0,a,a+b,a+b+c,2a+b+c,2a+2b+c,2a+2b+2c,\ldots$, the following function works:
$f_{a+b+c,0}(x)f_{a+b+c,a}(x)f_{a+b+c,a+b}(x)$
