In this section, we obtain several results on Fibonacci functions using the periodicity.

**Proposition 3.1** *Let* g(x) *and* f(x) *be Fibonacci functions with* g(x)=a(x)f(x) *for some* a(x). *Then*

\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+1)}{a(x)}=1.

*Proof* If g(x) and f(x) are Fibonacci functions, then by Corollary 2.12 we obtain

\begin{array}{rcl}\frac{1+\sqrt{5}}{2}& =& \underset{x\to \mathrm{\infty}}{lim}\frac{g(x+1)}{g(x)}\\ =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+1)f(x+1)}{a(x)f(x)}\\ =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+1)}{a(x)}\underset{x\to \mathrm{\infty}}{lim}\frac{f(x+1)}{f(x)}\\ =& \frac{1+\sqrt{5}}{2}\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+1)}{a(x)}.\end{array}

This proves the proposition. □

**Corollary 3.2** *Let* g(x) *and* f(x) *be Fibonacci functions with* g(x)=a(x)f(x) *for some* a(x). *Then*

\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+k)}{a(x)}=1

*for all natural numbers* *k*.

*Proof* It follows from the following equation:

\begin{array}{rcl}\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+k)}{a(x)}& =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+k)a(x+k-1)\cdots a(x+1)}{a(x+k-1)a(x+k-2)\cdots a(x)}\\ =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+1)f(x+1)}{a(x)f(x)}\\ =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+k)}{a(x+k-1)}\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+k-1)}{a(x+k-2)}\cdots \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+1)}{a(x)}\\ =& 1.\end{array}

□

**Corollary 3.3** *Let* g(x) *and* f(x) *be Fibonacci functions with* g(x)=a(x)f(x) *for some* a(x). *If* y>0, *then*

\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y)}{a(x)}=\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y)}{a(x+y-\lfloor y\rfloor )}.

*Proof* It follows from Corollary 3.2 that

\begin{array}{rcl}\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y)}{a(x)}& =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y)a(x+y-\lfloor y\rfloor )}{a(x+y-\lfloor y\rfloor )a(x)}\\ =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y)}{a(x+y-\lfloor y\rfloor )}\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y-\lfloor y\rfloor )}{a(x)}\\ =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y-\lfloor y\rfloor +\lfloor y\rfloor )}{a(x+y-\lfloor y\rfloor )}\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y-\lfloor y\rfloor )}{a(x)}\\ =& \underset{x\to \mathrm{\infty}}{lim}\frac{a(x+y)}{a(x+y-\lfloor y\rfloor )}.\end{array}

□

A map t(x) is said to be *ultimately periodic of period* p>0 if

\underset{x\to \mathrm{\infty}}{lim}\frac{t(x+p)}{t(x)}=1.

Note that a(x) discussed in Proposition 3.1 is ultimately periodic of period 1.

**Example 3.4** Let t(x):=mx+b. If m\ne 0, then {lim}_{x\to \mathrm{\infty}}\frac{t(x+p)}{t(x)}={lim}_{x\to \mathrm{\infty}}\frac{m(x+p)+b}{mx+b}=1, showing that t(x) is ultimately periodic of period *p* for all p>0.

Using Example 3.4, we obtain the following example.

**Example 3.5** If t(x):={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\cdots +{a}_{0}, then t(x) is ultimately periodic of period *p* for all p>0.

**Example 3.6** If t(x):=sinx, then \frac{sin(x+p)}{sinx}=\frac{sinxcosp+cosxsinp}{sinx}=cosp+sinpcotx. It follows that

\underset{x\to \mathrm{\infty}}{lim}\frac{sin(x+p)}{sinx}=cosp+sinp\underset{x\to \mathrm{\infty}}{lim}cotx.

Since {lim}_{x\to \mathrm{\infty}}cotx does not exist, t(x) is not ultimately periodic of period p>0 unless sinp=0 and cosp=1, *i.e.*, p=2k\pi for any integer k>0.

**Proposition 3.7** *If* a(x) *and* b(x) *are ultimately periodic of period* p>0, *then* \alpha a(x)+\beta b(x) *is also ultimately periodic of period* p>0 *for all* \alpha ,\beta >0.

*Proof* Since a(x) and b(x) are ultimately periodic of period p>0, there exist {\u03f5}_{1}(x),{\u03f5}_{2}(x)>0 such that \frac{a(x+p)}{a(x)}=1+{\u03f5}_{1}(x) and \frac{b(x+p)}{b(x)}=1+{\u03f5}_{2}(x) where {\u03f5}_{i}(x)\to 0 (i=1,2). We know that \frac{1+{\u03f5}_{1}(x)}{1+{\u03f5}_{2}(x)}=1+\u03f5(x) for some \u03f5(x). In fact, \u03f5(x)=\frac{{\u03f5}_{2}(x)-{\u03f5}_{1}(x)}{1+{\u03f5}_{1}(x)}\to 0. This shows that

\begin{array}{rcl}\frac{\alpha a(x+p)+\beta b(x+p)}{\alpha a(x)+\beta b(x)}& =& \frac{1+\frac{\beta b(x+p)}{\alpha a(x+p)}}{1+\frac{\beta b(x)}{\alpha a(x)}}\frac{\alpha a(x+p)}{\alpha a(x)}\\ =& \frac{1+\frac{\beta}{\alpha}\frac{(1+{\u03f5}_{2}(x))b(x)}{(1+{\u03f5}_{1}(x))a(x)}}{1+\frac{\beta}{\alpha}\frac{b(x)}{a(x)}}\frac{a(x+p)}{a(x)}\\ =& \frac{1+\frac{\beta}{\alpha}(1+\u03f5(x))\frac{b(x)}{a(x)}}{1+\frac{\beta}{\alpha}\frac{b(x)}{a(x)}}\frac{a(x+p)}{a(x)}\\ \to & \frac{a(x+p)}{a(x)}\\ \to & 1.\end{array}

This proves the proposition. □

**Proposition 3.8** *If* a(x) *and* b(x) *are ultimately periodic of period* p>0, *then* a(x)b(x) *is also ultimately periodic of period* p>0.

*Proof* It follows from the following equation:

\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+p)b(x+p)}{a(x)b(x)}=\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+p)}{a(x)}\underset{x\to \mathrm{\infty}}{lim}\frac{b(x+p)}{b(x)}=1.

□

**Theorem 3.9** *The collection* {U}_{p} *of all functions which are ultimately periodic of period* p>0 *forms an algebra*.

*Proof* It follows immediately from Propositions 3.7 and 3.8. □

**Proposition 3.10** *If* a(x)\in {U}_{p} *and* a(x)\ne 0 *for all* x\in [\lambda ,\mathrm{\infty}) *for some* *λ*, *then* \frac{1}{a(x)}\in {U}_{p}.

*Proof* It follows from the following equation:

\underset{x\to \mathrm{\infty}}{lim}\frac{\frac{1}{a(x+p)}}{\frac{1}{a(x)}}=\underset{x\to \mathrm{\infty}}{lim}\frac{a(x)}{a(x+p)}=1.

□

**Proposition 3.11** *If* a(x)\in {U}_{p}, *then* \frac{1}{a(x)}\in {U}_{kp} *for all natural numbers* *k*.

*Proof* The proof is similar to that of Corollary 3.2. □

A map *f* defined on the set of all real numbers **R** is said to be *periodic of period* p>0 if f(x+p)=f(x) for all x\in \mathbf{R}. It is obvious that every map of period of periodic 1 is ultimately periodic of period 1.

**Proposition 3.12** *Let* f(x) *be a Fibonacci function and let* a(x) *be periodic of period* 1. *If* g(x):=a(x)f(x), *then* g(x) *is a Fibonacci function*.

*Proof* Given x\in \mathbf{R}, since a(x) is periodic of period 1, we have

\begin{array}{rcl}g(x+2)& =& a(x+2)f(x+2)\\ =& a(x+2)f(x+1)+a(x)f(x)\\ =& a(x+1)f(x+1)+a(x)f(x)\\ =& g(x+1)+g(x).\end{array}

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We ask the following question: Are there a Fibonacci function f(x) and a function a(x) which is ultimately periodic of period 1 but not periodic of period 1 such that g(x)=a(x)f(x) is also a Fibonacci function?

**Theorem 3.13** *Let* f(x), g(x) *be Fibonacci functions with* g(x)=a(x)f(x). *If* a(x+1)\ne a(x) *for all* x\in \mathbf{R}, *then*

\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+2)-a(x)}{a(x+2)-a(x+1)}=-\frac{1+\sqrt{5}}{2}.

*Proof* Since a(x+2)\ne a(x) for all x\in \mathbf{R}, we have

\begin{array}{rcl}a(x+2)[f(x+1)+f(x)]& =& a(x+2)f(x+2)\\ =& g(x+2)\\ =& a(x+1)f(x+1)+a(x)f(x).\end{array}

It follows that [a(x+2)-a(x+1)]f(x+1)=-[a(x+2)-a(x)]f(x), which implies

\frac{f(x+1)}{f(x)}=-\frac{a(x+2)-a(x)}{a(x+2)-a(x+1)}.

By Corollary 2.12, we obtain

\underset{x\to \mathrm{\infty}}{lim}\frac{a(x+2)-a(x)}{a(x+2)-a(x+1)}=-\underset{x\to \mathrm{\infty}}{lim}\frac{f(x+1)}{f(x)}=-\frac{1+\sqrt{5}}{2},

proving the theorem. □

If we let a(x):={x}^{2}, then {lim}_{x\to \mathrm{\infty}}\frac{{(x+2)}^{2}-{x}^{2}}{{(x+2)}^{2}-{(x+2)}^{2}}={lim}_{x\to \mathrm{\infty}}\frac{4x+4}{2x+3}=2\ne -\frac{1+\sqrt{5}}{2}. This shows that g(x)={x}^{2}f(x) cannot be a Fibonacci function for any Fibonacci function f(x).

Note that a(x) cannot be an increasing function on [\lambda ,\mathrm{\infty}) for some \lambda \in \mathbf{R}. In fact, we suppose that there is an \lambda \in \mathbf{R} such that a(x)<a(x+1) for all x\ge \lambda. Then a(x+2)-a(x+1)<a(x+2)-a(x). It follows that 1\le {lim}_{x\to \mathrm{\infty}}\frac{a(x+2)-a(x)}{a(x+2)-a(x+1)}=-\frac{1+\sqrt{5}}{2}, a contradiction.

Given \lambda \in \mathbf{R}, if we let \mathrm{\Phi}:=\frac{1+\sqrt{5}}{2} and we let

\frac{a(x+2)-a(x)}{a(x+2)-a(x+1)}=-\mathrm{\Phi},

(2)

then a(x+2)-a(x)=\mathrm{\Phi}a(x+1)-\mathrm{\Phi}a(x+2). It follows that

\begin{array}{rl}a(x+2)& =\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}a(x+1)+\frac{1}{1+\mathrm{\Phi}}a(x)\\ =\frac{1}{\mathrm{\Phi}}a(x+1)+\frac{1}{{\mathrm{\Phi}}^{2}}a(x).\end{array}

(3)

**Theorem 3.14** *Let* f(x) *be a Fibonacci function and let* a(x) *be a map with condition* (2). *If* g(x):=a(x)f(x) *is a Fibonacci function for all* x\ge \lambda, *then*

*for all* x\ge \lambda.

*Proof* Let a(x) be a function satisfying the condition (2). Since f(x) is a Fibonacci function, we have the following:

\begin{array}{rcl}g(x+2)& =& a(x+2)f(x+2)\\ =& [\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}a(x+1)+\frac{1}{1+\mathrm{\Phi}}a(x)][f(x+1)+f(x)]\\ =& \frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}a(x+1)f(x+1)+\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}a(x+1)f(x)\\ +\frac{1}{1+\mathrm{\Phi}}a(x)f(x+1)+\frac{1}{1+\mathrm{\Phi}}a(x)f(x)\\ =& g(x+1)+g(x)-\frac{1}{1+\mathrm{\Phi}}g(x+1)+\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}a(x+1)f(x)\\ +\frac{1}{1+\mathrm{\Phi}}a(x)f(x+1)-\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}g(x).\end{array}

Since g(x) is also a Fibonacci function, we obtain

\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}a(x+1)f(x)+\frac{1}{1+\mathrm{\Phi}}a(x)f(x+1)=\frac{1}{1+\mathrm{\Phi}}g(x+1)+\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}g(x).

It follows that

\frac{1}{1+\mathrm{\Phi}}[a(x)f(x+1)-g(x+1)]=\frac{\mathrm{\Phi}}{1+\mathrm{\Phi}}[g(x)-a(x+1)f(x)].

This shows that

\begin{array}{rcl}\mathrm{\Phi}& =& \frac{a(x)f(x+1)-g(x+1)}{g(x)-a(x+1)f(x)}\\ =& \frac{a(x)f(x+1)-a(x+1)f(x+1)}{a(x)f(x)-a(x+1)f(x)}\\ =& \frac{f(x+1)}{f(x)}\end{array}

for all x\ge \lambda. □