On Fibonacci Functions with Period 𝑘

Abstract

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. 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.

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 lim_𝑥→∞𝑓(𝑥+1)/𝑓(𝑥)=(1+√5)/2.

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 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 Period 𝑘

Definition 1. Let 𝑘 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. Let 𝑓(𝑥)=𝑎^𝑥/𝑘 be a Fibonacci function with period 𝑘∈ℕ, where 𝑎>0. It follows that 𝑎^{(𝑥/𝑘)+2}=𝑎^{(𝑥/𝑘)+1}+𝑎^𝑥/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. Let 𝑓∶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. Let 𝑥∈ℝ. Since 𝑓(𝑥+2𝑘)=𝑓(𝑥+𝑘)+𝑓(𝑥), it follows that 𝑓^′(𝑥+2𝑘)=𝑓^′(𝑥+𝑘)+𝑓^′(𝑥).

Proposition 4. Let 𝑓∶ℝ→ℝ be a Fibonacci function with period 𝑘∈ℕ, and define 𝑔_𝑡(𝑥)=𝑓(𝑥+𝑡) for all 𝑥∈ℝ, where 𝑡∈ℝ. Then 𝑔_𝑡 is also a Fibonacci function with period 𝑘.

Proof. Let 𝑥∈ℝ. Then 𝑔_𝑡(𝑥+2𝑘)=𝑓(𝑥+2𝑘+𝑡)=𝑓(𝑥+𝑡+𝑘)+𝑓(𝑥+𝑡)=𝑔_𝑡(𝑥+𝑘)+𝑔_𝑡(𝑥).

Example 5. Let 𝑘∈ℕ and 𝑡∈ℝ. Define 𝑔_𝑡∶ℝ→ℝ by 𝑔_𝑡(𝑥)=((1+√5)/2)^{(𝑥+𝑡)/𝑘} for all 𝑥∈ℝ. Then 𝑔_𝑡 is a Fibonacci function with period 𝑘.

Theorem 6. Let 𝑓∶ℝ→ℝ be a Fibonacci function with period 𝑘∈ℕ, and let {𝐹_𝑛}_𝑛∈ℕ be a sequence of Fibonacci numbers with 𝐹₀=0, 𝐹₁=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. Let 𝑥∈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)=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 𝑓(()𝑘)(()𝑘)()𝑥+𝑛+2=𝑓𝑥+𝑛+1+𝑓𝑥+𝑛𝑘=𝐹_𝑛+1𝑓(𝑥+𝑘)+𝐹_𝑛𝑓(𝑥)+𝐹_𝑛𝑓(𝑥+𝑘)+𝐹_𝑛−1𝑓(𝑥)=𝐹_𝑛+1+𝐹_𝑛𝑓(𝑥+𝑘)+𝐹_𝑛+𝐹_𝑛−1𝑓(𝑥)=𝐹_𝑛+2𝑓(𝑥+𝑘)+𝐹_𝑛+1𝑓(𝑥). (1) This proof is completed.

3. Odd Fibonacci Functions with Period 𝑘

Definition 7. Let 𝑘 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. Let 𝑓(𝑥)=𝑎^𝑥/𝑘 be an odd Fibonacci function with period 𝑘∈ℕ, where 𝑎>0. It follows that 𝑎^{(𝑥/𝑘)+2}=−𝑎^{(𝑥/𝑘)+1}+𝑎^𝑥/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. Let 𝑓∶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. Let 𝑥∈ℝ. Since 𝑓(𝑥+2𝑘)=−𝑓(𝑥+𝑘)+𝑓(𝑥), it follows that 𝑓^′(𝑥+2𝑘)=−𝑓^′(𝑥+𝑘)+𝑓^′(𝑥).

Proposition 10. Let 𝑓∶ℝ→ℝ be an odd Fibonacci function with period 𝑘∈ℕ, and define 𝑔_𝑡(𝑥)=𝑓(𝑥+𝑡) for all 𝑥∈ℝ, where 𝑡∈ℝ. Then 𝑔_𝑡 is also an odd Fibonacci function with period 𝑘.

Proof. Let 𝑥∈ℝ. Then 𝑔_𝑡(𝑥+2𝑘)=𝑓(𝑥+2𝑘+𝑡)=−𝑓(𝑥+𝑡+𝑘)+𝑓(𝑥+𝑡)=−𝑔_t(𝑥+𝑘)+𝑔_𝑡(𝑥).

Example 11. Let 𝑘∈ℕ and 𝑡∈ℝ. Define 𝑔_𝑡∶ℝ→ℝ by 𝑔_𝑡(𝑥)=((−1+√5)/2)^{(𝑥+𝑡)/𝑘} for all 𝑥∈ℝ. Then 𝑔_𝑡 is an odd Fibonacci function with period 𝑘.

Theorem 12. Let 𝑓∶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. Let 𝑥∈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 𝑓(()𝑘)(()𝑘)()𝐹𝑥+𝑛+2=−𝑓𝑥+𝑛+1+𝑓𝑥+𝑛𝑘=−_−𝑛−1𝑓(𝑥+𝑘)+𝐹_−𝑛𝑓(𝑥)+𝐹_−𝑛𝑓(𝑥+𝑘)+𝐹_−𝑛+1𝑓(𝑥)=−𝐹_−𝑛−1+𝐹_−𝑛𝑓()+𝑥+𝑘−𝐹_−𝑛+𝐹_−𝑛+1𝑓(𝑥)=𝐹_−𝑛−2𝑓(𝑥+𝑘)+𝐹_−𝑛−1𝑓(𝑥). (2) This proof is completed.

4. 𝑓-Even Functions with Period 𝑘

Definition 13. Let 𝑘 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. Define 𝛼(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. Let 𝑘∈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 that 𝑔 is a Fibonacci function with period 𝑘. For any 𝑥∈ℝ, we have ()()𝑔()()𝑔()(𝑥)()𝑔()()𝑔(𝑥)()𝑔()(𝑥)𝑔(𝑥)=()+(𝑥).𝛼𝑔𝑥+2𝑘=𝛼𝑥+2𝑘𝑥+2𝑘=𝛼𝑥+𝑘𝑥+𝑘+𝑔=𝛼𝑥+𝑘𝑥+𝑘+𝛼𝑥+𝑘=𝛼𝑥+𝑘𝑥+𝑘+𝛼𝛼𝑔𝑥+𝑘𝛼𝑔 (3)
Hence, 𝛼𝑔 is a Fibonacci function with period 𝑘.
Next, we assume that 𝛼𝑔 is a Fibonacci function with period 𝑘. Let 𝑥∈ℝ. Then 𝛼()𝑔()()𝑔()=()=()+(𝑥)()𝑔()(𝑥)𝑔(𝑥)()𝑔()()𝑔(𝑥)()𝑔()(𝑥).𝑥+𝑘𝑥+2𝑘=𝛼𝑥+2𝑘𝑥+2𝑘𝛼𝑔𝑥+2𝑘𝛼𝑔𝑥+𝑘𝛼𝑔=𝛼𝑥+𝑘𝑥+𝑘+𝛼=𝛼𝑥+𝑘𝑥+𝑘+𝛼𝑥+𝑘=𝛼𝑥+𝑘𝑥+𝑘+𝑔 (4)
By the assumption of 𝛼, we obtain that 𝑔(𝑥+2𝑘)=𝑔(𝑥+𝑘)+𝑔(𝑥). Hence, 𝑔 is a Fibonacci function with period 𝑘.

Example 16. Let 𝑘∈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. Let 𝑘∈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 that 𝑔 is an odd Fibonacci function with period 𝑘. For any 𝑥∈ℝ, we have ()()𝑔()()()(𝑥)()𝑔()()𝑔(𝑥)()𝑔()(𝑥)𝑔(𝑥)()+(𝑥).𝛼𝑔𝑥+2𝑘=𝛼𝑥+2𝑘𝑥+2𝑘=𝛼𝑥+𝑘−𝑔𝑥+𝑘+𝑔=−𝛼𝑥+𝑘𝑥+𝑘+𝛼𝑥+𝑘=−𝛼𝑥+𝑘𝑥+𝑘+𝛼=−𝛼𝑔𝑥+𝑘𝛼𝑔 (5)
Hence, 𝛼𝑔 is an odd Fibonacci function with period 𝑘.
Next, we assume that 𝛼𝑔 is an odd Fibonacci function with period 𝑘. Let 𝑥∈ℝ. Then 𝛼()𝑔()()𝑔()=()()+(𝑥)()𝑔()(𝑥)𝑔(𝑥)()𝑔()()𝑔(𝑥)()()(𝑥).𝑥+𝑘𝑥+2𝑘=𝛼𝑥+2𝑘𝑥+2𝑘𝛼𝑔𝑥+2𝑘=−𝛼𝑔𝑥+𝑘𝛼𝑔=−𝛼𝑥+𝑘𝑥+𝑘+𝛼=−𝛼𝑥+𝑘𝑥+𝑘+𝛼𝑥+𝑘=𝛼𝑥+𝑘−𝑔𝑥+𝑘+𝑔 (6)
By the assumption of 𝛼, we obtain that 𝑔(𝑥+2𝑘)=−𝑔(𝑥+𝑘)+𝑔(𝑥). Hence, 𝑔 is an odd Fibonacci function with period 𝑘.

Example 18. Let 𝑘∈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. Let 𝑘 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. Define 𝛼(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. Let 𝑘∈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 that 𝑔 is a Fibonacci function with period 𝑘. For any 𝑥∈ℝ, we have ()()𝑔()()𝑔()(𝑥)()𝑔()()𝑔(𝑥)()𝑔()(𝑥)𝑔(𝑥)()+(𝑥).𝛼𝑔𝑥+2𝑘=𝛼𝑥+2𝑘𝑥+2𝑘=−𝛼𝑥+𝑘𝑥+𝑘+𝑔=−𝛼𝑥+𝑘𝑥+𝑘−𝛼𝑥+𝑘=−𝛼𝑥+𝑘𝑥+𝑘+𝛼=−𝛼𝑔𝑥+𝑘𝛼𝑔 (7)
Hence, 𝛼𝑔 is an odd Fibonacci function with period 𝑘.
Next, we assume that 𝛼𝑔 is an odd Fibonacci function with period 𝑘. Let 𝑥∈ℝ. Then 𝛼()𝑔()()𝑔()()−()+(𝑥)=()−(𝑥)()𝑔()(𝑥)𝑔(𝑥)()𝑔()()𝑔(𝑥)()𝑔()(𝑥).𝑥+𝑘𝑥+2𝑘=−𝛼𝑥+2𝑘𝑥+2𝑘=−𝛼𝑔𝑥+2𝑘=−𝛼𝑔𝑥+𝑘𝛼𝑔𝛼𝑔𝑥+𝑘𝛼𝑔=𝛼𝑥+𝑘𝑥+𝑘−𝛼=𝛼𝑥+𝑘𝑥+𝑘+𝛼𝑥+𝑘=𝛼𝑥+𝑘𝑥+𝑘+𝑔 (8)
By the assumption of 𝛼, we obtain that 𝑔(𝑥+2𝑘)=𝑔(𝑥+𝑘)+𝑔(𝑥). Hence, 𝑔 is a Fibonacci function with period 𝑘.

Example 22. Let 𝑘 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. Let 𝑘∈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 that 𝑔 is an odd Fibonacci function with period 𝑘. For any 𝑥∈ℝ, we have ()()𝑔()()()(𝑥)()𝑔()()𝑔(𝑥)()𝑔()(𝑥)𝑔(𝑥)=()+(𝑥).𝛼𝑔𝑥+2𝑘=𝛼𝑥+2𝑘𝑥+2𝑘=−𝛼𝑥+𝑘−𝑔𝑥+𝑘+𝑔=𝛼𝑥+𝑘𝑥+𝑘−𝛼𝑥+𝑘=𝛼𝑥+𝑘𝑥+𝑘+𝛼𝛼𝑔𝑥+𝑘𝛼𝑔 (9)
Hence, 𝛼𝑔 is a Fibonacci function with period 𝑘.
Next, we assume that 𝛼𝑔 is a Fibonacci function with period 𝑘. Let 𝑥∈ℝ. Then 𝛼()𝑔()()𝑔()()()+(𝑥)()−(𝑥)()𝑔()(𝑥)𝑔(𝑥)()𝑔()()𝑔(𝑥)()()(𝑥).𝑥+𝑘𝑥+2𝑘=−𝛼𝑥+2𝑘𝑥+2𝑘=−𝛼𝑔𝑥+2𝑘=−𝛼𝑔𝑥+𝑘𝛼𝑔=−𝛼𝑔𝑥+𝑘𝛼𝑔=−𝛼𝑥+𝑘𝑥+𝑘−𝛼=−𝛼𝑥+𝑘𝑥+𝑘+𝛼𝑥+𝑘=𝛼𝑥+𝑘−𝑔𝑥+𝑘+𝑔 (10)
By the assumption of 𝛼, we obtain that 𝑔(𝑥+2𝑘)=−𝑔(𝑥+𝑘)+𝑔(𝑥). Hence, 𝑔 is an odd Fibonacci function with period 𝑘.

Example 24. Let 𝑘 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. If 𝑓 is a Fibonacci function with period 𝑘∈ℕ, thenlim_𝑥→∞𝑓(𝑥+𝑘)𝑓(𝑥)=1+√52. (11)

Conjecture 26. If 𝑓 is an odd Fibonacci function with period 𝑘∈ℕ, thenlim_𝑥→∞𝑓(𝑥+𝑘)𝑓(𝑥)=−1−√52. (12)

Acknowledgments

The author would like to thank the referees for their useful comments and suggestions.