# On Fibonacci Functions with Periodk

Authors: Banyat Sroysang
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

A functionf:ℝ→ℝ is said to be a Fibonacci function if f(x+2)=f(x+1)+f(x) for all x∈ℝ. In 2012, some properties on the Fibonacci functions were presented. In this paper, for any positive integer k, a function f:ℝ→ℝ is said to be a Fibonacci function with period k if f(x+2k)=f(x+k)+f(x) for all x∈ℝ; we present some properties on the Fibonacci functions with period k.


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## Body

## 1. Introduction

Presently, there are many research articles about Fibonacci numbers (see [1]). Fibonacci numbers  are also involved in the golden ratio (see [2]). In 2008, Kim and Neggers [3] studied Fibonacci means. In 2009, Jung [4] studied Hyers-Ulam stability of Fibonacci functional equation. In 2010, Han et al. [5] studied a Fibonacci norm of positive integers. In 2012, Han et al. [6] studied Fibonacci sequences in groupoids. Moreover, they [7] gave some properties on Fibonacci functions; a function f:ℝ→ℝ is said to be a Fibonacci function if f(x+2)=f(x+1)+f(x), for all x∈ℝ, using the concept of f-even and f-odd functions. They also showed that if f is a Fibonacci function, then limx→∞f(x+1)/f(x)=(1+5)/2.In this paper, for any positive integerk, a function f:ℝ→ℝ is said to be a Fibonacci function with period k if f(x+2k)=f(x+k)+f(x) for all x∈ℝ; we present some properties on the Fibonacci functions with period k using the concept of f-even and f-odd functions with period k. Moreover, we also present some properties on the odd Fibonacci functions with period k.


## 2. Fibonacci Functions with Periodk

Definition 1.
Letk be a positive integer. A function f:ℝ→ℝ is said to be a Fibonacci function with period k if f(x+2k)=f(x+k)+f(x) for all x∈ℝ.Example 2.
Letf(x)=ax/k be a Fibonacci function with period k∈ℕ, where a>0. It follows that a(x/k)+2=a(x/k)+1+ax/k for all x∈ℝ, so a2=a+1. Then a=(1+5)/2. Thus, f(x)=((1+5)/2)x/k for all x∈ℝ.Proposition 3.
Letf:ℝ→ℝ be a Fibonacci function with period k∈ℕ. Assume that f is differentiable. Then f′ is also a Fibonacci function with period k.Proof.
Letx∈ℝ. Since f(x+2k)=f(x+k)+f(x), it follows that f′(x+2k)=f′(x+k)+f′(x).Proposition 4.
Letf:ℝ→ℝ be a Fibonacci function with period k∈ℕ, and define gt(x)=f(x+t) for all x∈ℝ, where t∈ℝ. Then gt is also a Fibonacci function with period k.Proof.
Letx∈ℝ. Then gt(x+2k)=f(x+2k+t)=f(x+t+k)+f(x+t)=gt(x+k)+gt(x).Example 5.
Letk∈ℕ and t∈ℝ. Define gt:ℝ→ℝ by gt(x)=((1+5)/2)(x+t)/k for all x∈ℝ. Then gt is a Fibonacci function with period k.Theorem 6.
Letf:ℝ→ℝ be a Fibonacci function with period k∈ℕ, and let {Fn}n∈ℕ be a sequence of Fibonacci numbers with F0=0, F1=1, and Fn+1=Fn+Fn-1 for all n∈ℕ. Then, for any n∈ℕ and x∈ℝ, f(x+nk)=Fnf(x+k)+Fn-1f(x).Proof.
Letx∈ℝ. We note that f(x+k)=F1f(x+k)+F0f(x) and f(x+2k)=F2f(x+k)+F1f(x). Now, we assume that f(x+nk)=Fnf(x+k)+Fn-1f(x) and f(x+(n+1)k)=Fn+1f(x+k)+Fnf(x), where n∈ℕ. Then
(1)f(x+(n+2)k)=f(x+(n+1)k)+f(x+nk)=Fn+1f(x+k)+Fnf(x)+Fnf(x+k)+Fn-1f(x)=(Fn+1+Fn)f(x+k)+(Fn+Fn-1)f(x)=Fn+2f(x+k)+Fn+1f(x).
This proof is completed.


## 3. Odd Fibonacci Functions with Periodk

Definition 7.
Letk be a positive integer. A function f:ℝ→ℝ is said to be an odd Fibonacci function with period k if f(x+2k)=-f(x+k)+f(x) for all x∈ℝ.Example 8.
Letf(x)=ax/k be an odd Fibonacci function with period k∈ℕ, where a>0. It follows that a(x/k)+2=-a(x/k)+1+ax/k for all x∈ℝ, so a2=-a+1. Then a=(-1+5)/2. Thus, f(x)=((-1+5)/2)x/k for all x∈ℝ.Proposition 9.
Letf:ℝ→ℝ be an odd Fibonacci function with period k∈ℕ. Assume that f is differentiable. Then f′ is also an odd Fibonacci function with period k.Proof.
Letx∈ℝ. Since f(x+2k)=-f(x+k)+f(x), it follows that f′(x+2k)=-f′(x+k)+f′(x).Proposition 10.
Letf:ℝ→ℝ be an odd Fibonacci function with period k∈ℕ, and define gt(x)=f(x+t) for all x∈ℝ, where t∈ℝ. Then gt is also an odd Fibonacci function with period k.Proof.
Letx∈ℝ. Then gt(x+2k)=f(x+2k+t)=-f(x+t+k)+f(x+t)=-gt(x+k)+gt(x).Example 11.
Letk∈ℕ and t∈ℝ. Define gt:ℝ→ℝ by gt(x)=((-1+5)/2)(x+t)/k for all x∈ℝ. Then gt is an odd Fibonacci function with period k.Theorem 12.
Letf:ℝ→ℝ be an odd Fibonacci function with period k∈ℕ, and let {F-n}n∈ℕ be a sequence of Fibonucci numbers with F0=0, F-1=1, and F-n-1=-F-n+F-n+1 for all n∈ℕ. Then, for any n∈ℕ and x∈ℝ, f(x+nk)=F-nf(x+k)+F-n+1f(x).Proof.
Letx∈ℝ. We note that f(x+k)=F-1f(x+k)+F0f(x) and f(x+2k)=F-2f(x+k)+F-1f(x). Now, we assume that f(x+nk)=F-nf(x+k)+F-n+1f(x) and f(x+(n+1)k)=F-n-1f(x+k)+F-nf(x), where n∈ℕ. Then
(2)f(x+(n+2)k)=-f(x+(n+1)k)+f(x+nk)=-(F-n-1f(x+k)+F-nf(x))+F-nf(x+k)+F-n+1f(x)=(-F-n-1+F-n)f(x+k)+(-F-n+F-n+1)f(x)=F-n-2f(x+k)+F-n-1f(x).
This proof is completed.


## 4.f-Even Functions with Period k

Definition 13.
Letk be a positive integer and let α:ℝ→ℝ be such that if αh=0, where h:ℝ→ℝ is continuous, then h=0. The function α is said to be an f-even function with period k if α(x+k)=α(x) for all x∈ℝ.Example 14.
Defineα(x)=x-⌊x⌋ for all x∈ℝ. Let h:ℝ→ℝ be a continuous function such that αh=0. For any x∉ℤ, we have α(x)≠0, so h(x)=0. Since ℝ∖ℤ is dense in ℝ and h is continuous, it follows that h=0. Let k∈ℕ and x∈ℝ. Then α(x+k)=x+k-⌊x+k⌋=x+k-⌊x⌋-k=x-⌊x⌋=α(x). Hence, α is an f-even function with period k.Theorem 15.
Letk∈ℕ and α:ℝ→ℝ be an f-even function with period k and let g:ℝ→ℝ be a continuous function. Then g is a Fibonacci function with period k if and only if αg is a Fibonacci function with period k.Proof.
First, we assume thatg is a Fibonacci function with period k. For any x∈ℝ, we have
(3)(αg)(x+2k)=α(x+2k)g(x+2k)=α(x+k)(g(x+k)+g(x))=α(x+k)g(x+k)+α(x+k)g(x)=α(x+k)g(x+k)+α(x)g(x)=(αg)(x+k)+(αg)(x).
Hence,αg is a Fibonacci function with period k.
Next, we assume thatαg is a Fibonacci function with period k. Let x∈ℝ. Then
(4)α(x+k)g(x+2k)=α(x+2k)g(x+2k)=(αg)(x+2k)=(αg)(x+k)+(αg)(x)=α(x+k)g(x+k)+α(x)g(x)=α(x+k)g(x+k)+α(x+k)g(x)=α(x+k)(g(x+k)+g(x)).
By the assumption ofα, we obtain that g(x+2k)=g(x+k)+g(x). Hence, g is a Fibonacci function with period k.Example 16.
Letk∈ℕ. Define α(x)=x-⌊x⌋ and g(x)=((1+5)/2)x/k for all x∈ℝ. For all x∈ℝ, we have αg(x)=(x-⌊x⌋)((1+5)/2)x/k. We recall that α is an f-even function with period k, and g is a Fibonacci function with period k. Hence, αg is a Fibonacci function with period k.Theorem 17.
Letk∈ℕ and α:ℝ→ℝ be an f-even function with period k and let g:ℝ→ℝ be a continuous function. Then g is an odd Fibonacci function with period k if and only if αg is an odd Fibonacci function with period k.Proof.
First, we assume thatg is an odd Fibonacci function with period k. For any x∈ℝ, we have
(5)(αg)(x+2k)=α(x+2k)g(x+2k)=α(x+k)(-g(x+k)+g(x))=-α(x+k)g(x+k)+α(x+k)g(x)=-α(x+k)g(x+k)+α(x)g(x)=-(αg)(x+k)+(αg)(x).
Hence,αg is an odd Fibonacci function with period k.
Next, we assume thatαg is an odd Fibonacci function with period k. Let x∈ℝ. Then
(6)α(x+k)g(x+2k)=α(x+2k)g(x+2k)=(αg)(x+2k)=-(αg)(x+k)+(αg)(x)=-α(x+k)g(x+k)+α(x)g(x)=-α(x+k)g(x+k)+α(x+k)g(x)=α(x+k)(-g(x+k)+g(x)).
By the assumption ofα, we obtain that g(x+2k)=-g(x+k)+g(x). Hence, g is an odd Fibonacci function with period k.Example 18.
Letk∈ℕ. Define α(x)=x-⌊x⌋ and g(x)=((-1+5)/2)x/k for all x∈ℝ. For all x∈ℝ, we have αg(x)=(x-⌊x⌋)((-1+5)/2)x/k. We recall that α is an f-even function with period k and g is an odd Fibonacci function with period k. Hence, αg is an odd Fibonacci function with period k.


## 5.f-Odd Functions with Period k

Definition 19.
Letk be a positive integer and let α:ℝ→ℝ be such that if αh=0 where h:ℝ→ℝ is continuous, then h=0. The function α is said to be an f-odd function with period k if α(x+k)=-α(x) for all x∈ℝ.Example 20.
Defineα(x)=sin(πx) for all x∈ℝ. Let h:ℝ→ℝ be a continuous function such that αh=0. For any x∉πℤ, we have α(x)≠0, so h(x)=0. Since ℝ∖πℤ is dense in ℝ and h is continuous, it follows that h=0. Let k be a positive odd integer and x∈ℝ. Then α(x+k)=sin(π(x+k))=sin(πx+πk)=-sin(πx)=-α(x). Hence, α is an f-even function with period k.Theorem 21.
Letk∈ℕ and α:ℝ→ℝ be an f-odd function with period k and let g:ℝ→ℝ be a continuous function. Then g is a Fibonacci function with period k if and only if αg is an odd Fibonacci function with period k.Proof.
First, we assume thatg is a Fibonacci function with period k. For any x∈ℝ, we have
(7)(αg)(x+2k)=α(x+2k)g(x+2k)=-α(x+k)(g(x+k)+g(x))=-α(x+k)g(x+k)-α(x+k)g(x)=-α(x+k)g(x+k)+α(x)g(x)=-(αg)(x+k)+(αg)(x).
Hence,αg is an odd Fibonacci function with period k.
Next, we assume thatαg is an odd Fibonacci function with period k. Let x∈ℝ. Then
(8)α(x+k)g(x+2k)=-α(x+2k)g(x+2k)=-(αg)(x+2k)=-(-(αg)(x+k)+(αg)(x))=(αg)(x+k)-(αg)(x)=α(x+k)g(x+k)-α(x)g(x)=α(x+k)g(x+k)+α(x+k)g(x)=α(x+k)(g(x+k)+g(x)).
By the assumption ofα, we obtain that g(x+2k)=g(x+k)+g(x). Hence, g is a Fibonacci function with period k.Example 22.
Letk be a positive odd integer. Define α(x)=sin(πx) and g(x)=((1+5)/2)x/k for all x∈ℝ. We have αg(x)=(sin(πx))((1+5)/2)x/k for all x∈ℝ. We recall that α is an f-odd function with period k and g is a Fibonacci function with period k. Hence, αg is an odd Fibonacci function with period k.Theorem 23.
Letk∈ℕ and α:ℝ→ℝ be an f-odd function with period k and let g:ℝ→ℝ be a continuous function. Then g is an odd Fibonacci function with period k if and only if αg is a Fibonacci function with period k.Proof.
First, we assume thatg is an odd Fibonacci function with period k. For any x∈ℝ, we have
(9)(αg)(x+2k)=α(x+2k)g(x+2k)=-α(x+k)(-g(x+k)+g(x))=α(x+k)g(x+k)-α(x+k)g(x)=α(x+k)g(x+k)+α(x)g(x)=(αg)(x+k)+(αg)(x).
Hence,αg is a Fibonacci function with period k.
Next, we assume thatαg is a Fibonacci function with period k. Let x∈ℝ. Then
(10)α(x+k)g(x+2k)=-α(x+2k)g(x+2k)=-(αg)(x+2k)=-((αg)(x+k)+(αg)(x))=-(αg)(x+k)-(αg)(x)=-α(x+k)g(x+k)-α(x)g(x)=-α(x+k)g(x+k)+α(x+k)g(x)=α(x+k)(-g(x+k)+g(x)).
By the assumption ofα, we obtain that g(x+2k)=-g(x+k)+g(x). Hence, g is an odd Fibonacci function with period k.Example 24.
Letk be a positive odd integer. Define α(x)=sin(πx) and g(x)=((-1+5)/2)x/k for all x∈ℝ. We have αg(x)=(sin(πx))((-1+5)/2)x/k for all x∈ℝ. We recall that α is an f-odd function with period k and g is an odd Fibonacci function with period k. Hence, αg is a Fibonacci function with period k.


## 6. Open Problems

Conjecture 25.
Iff is a Fibonacci function with period k∈ℕ, then(11)limx→∞f(x+k)f(x)=1+52.Conjecture 26.
Iff is an odd Fibonacci function with period k∈ℕ, then(12)limx→∞f(x+k)f(x)=-1-52.



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