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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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  • $\begingroup$ 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 ?) $\endgroup$ – user90369 Sep 8 '16 at 17:34
  • $\begingroup$ @user90369 Who are you asking? Me or someone else? $\endgroup$ – Tanner Swett Sep 8 '16 at 18:32
  • $\begingroup$ It's a question for you ... if it's o.k. . $\endgroup$ – user90369 Sep 8 '16 at 18:53
  • $\begingroup$ Equivalently $f(x) = f(0)\prod_{n=2}^\infty f(\frac{x}{\phi^n}-1)$ $\endgroup$ – user335907 Jan 20 '17 at 5:05
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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 $f(-1)=1$.

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  • $\begingroup$ That answer is either wrong or contains a typo. $\endgroup$ – mick Sep 9 '12 at 18:39
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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 )

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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.

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  • $\begingroup$ I dont know what type of pseudoperiodic the OP is talking about and i dont know about a closed form solution , if that is even possible. i could not help thinking about automorphic functions and moebius transforms but i dont know if that could help here. Maybe there are better answers possible. As for the roots , i consider it unlikely to be true for non-constant meromorhpic f(x). I would like to see some plots of this function on the complex plane. I intended to post about solutions with branches in the complex plane but i was uncertain that what i was about to say would be correct. I wonder. $\endgroup$ – mick Sep 7 '12 at 14:58
  • $\begingroup$ As for being "almost periodic", I think I was mistaken there; on second thought, I wouldn't expect $f$ to be almost periodic. As for the roots, I thought there was a theorem stating that if $S$ is a discrete, closed set of complex numbers, then there is a meromorphic function on $\mathbb{C}$ whose roots are exactly $S$. $\endgroup$ – Tanner Swett Sep 7 '12 at 15:26
  • $\begingroup$ But that theorem does not include an additional functional equation. Anyways did you compute those taylor coefficents or made a plot ? $\endgroup$ – mick Sep 7 '12 at 17:52
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    $\begingroup$ Hey who downvoted without leaving a message or answer ?? $\endgroup$ – mick Sep 7 '12 at 18:42
  • $\begingroup$ I dare him to give a much better answer :) $\endgroup$ – mick Sep 8 '12 at 19:02
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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:

$f_{a+b+c,0}(x)f_{a+b+c,a}(x)f_{a+b+c,a+b}(x)$

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  • $\begingroup$ That's great if my desired set of roots is a union of periodic sets. But the root set that I gave isn't a union of periodic sets, is it? Can this method be modified to produce the set of roots that I'm looking for? $\endgroup$ – Tanner Swett Sep 8 '16 at 18:40
  • $\begingroup$ Ah! I misread the Fibonacci word. $\endgroup$ – Aravind Sep 9 '16 at 6:58

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