# How do we classify all possible extensions of the Fibonacci recursion to the complex plane?

Take the straight forward Fibonacci equation

$$F_0 = F_1 = 1$$ $$F_{n-2} + F_{n-1} = F_n$$

Let's consider a holomorphic function $F: \mathbb{C} \to \mathbb{C}$ such that

$$F(z)\Big{|}_{\mathbb{N}} = F_n$$ $$F(z-2) + F(z-1) = F(z)$$

Let's call such $F$, those that satisfy the Fibonacci equation in the complex plane. It is very easy to produce such functions. Taking the Binet identity

$$F_n = \frac{\phi^{n} - \psi^n}{\phi - \psi}$$

where

$$\phi = \frac{1 + \sqrt{5}}{2}$$ $$\psi = \frac{1 - \sqrt{5}}{2}$$

It follows for any $k,j \in \mathbb{Z}$, using the standard branch of $\phi^z$ and $\psi^z$ that the functions

$$F_{jk}(z) = \frac{e^{2\pi i j z}\phi^z -e^{2 \pi i k z}\psi^z}{\phi - \psi}$$

are a solution of the Fibonacci equation in the complex plane. These can't be all solutions though. Namely if $F$ and $G$ are solutions where we've simply chosen different $k$ and $j$ for each one, then

$$\frac{F}{2} + \frac{G}{2}$$

is equally a solution, which corresponds to no function from our list. This additionally implies that the infinite sum

$$\mathcal{F} = \sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk}F_{jk}(z)$$

is a solution to the Fibonacci equation in the complex plane if it converges everywhere and

$$\sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk} = 1$$

Are these all of the solutions?

• Wow -1 really fast, is there something obvious I'm missing here? – user78249 Jul 18 '17 at 9:02

## 1 Answer

Yes, these are all entire solutions. This follows from a general theorem which says that linear combinations of exponential solutions of such equations (linear homogeneous with constant coefficients) are dense in the set of all solutions, see, for example Gelfond, Calculus of finite differences, Chap V, sect. 7.

• Jesus! That is very interesting. You always seem to have straightforward answers to my questions and the references you give me make me slap my head and go "gotta read more." Thanks a bunch! I'll check out that book. – user78249 Jul 22 '17 at 5:02