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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 $f(x) = f(x / \phi) f(x / \phi^2 - 1)$ 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 $g(x) = g(x/2) g(x/2 - 1)$. 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:

$$\sin 2 \theta = \sin \theta \sin (\theta + \pi/2)$$

Thus, we have the easy solution $g(x) = \sin (-\pi x / 4)$. It's not obvious how to apply this solution to the original problem, however.

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    *Winther* has shown, that the equation has the simple solution $e^{a(x+1)}$. I understand that you want to greate the Fibonacci word in the form of $g(0):=1$, $g(1):=\phi^{-1}$ and $g(n):=g(n-1),g(n-2)$. It would be very kind of you if you could tell me what type of relation between f and g you expect. (Or is any of the answers below the answer to your question ?)2016-09-08
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    @user90369 Who are you asking? Me or someone else?2016-09-08
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    It's a question for you ... if it's o.k. .2016-09-08
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    Equivalently $f(x) = f(0)\prod_{n=2}^\infty f(\frac{x}{\phi^n}-1)$2017-01-20

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