Comment on "Hybrid deformed algebra", by Andre A. Marinho, Francisco A. Brito [arXiv:1904.07843]

TL;DR

This paper critiques a previous claim by clarifying that the $-oscillator does not generate Fibonacci sequences in the limit $ ightarrow 0$, and discusses conditions under which Fibonacci sequences emerge from $q$-deformations.

Contribution

It corrects a misconception about the $$-oscillator's relation to Fibonacci sequences and clarifies the conditions for Fibonacci sequence emergence in $q$-deformed systems.

Findings

The $$-oscillator does not produce Fibonacci sequences as previously claimed.

Fibonacci sequences appear only when $q$-deformation is combined with $$-deformation.

The sequence reduces to standard integers in the limit $ ightarrow 0$.

Abstract

In this note, we show that the $\mu$-oscillator does not lead to the Fibonacci sequence as claimed in \cite{Marinho:2019zny} since $[n]^{\mu}=n$ in the limit $\mu \rightarrow 0$. Thus we obtain the sequence $[0]^{\mu}=0, 1, 2,...$. We only obtain the Fibonacci sequence when the $q$-deformation is associated to the $\mu$-deformation via the basis number $$\lim_{\mu\rightarrow 0} [n]^{\mu}_{q_1, q_2} = \frac{q_1^{2n}-q_2^{2n}}{q_1^2-q_2^2}.$$

Equations

$$\underset{\mu \to 0}{\lim }[n{]}_{q_1,q_2}^{\mu }=\frac{q_1^{2n}-q_2^{2n}}{q_1^2-q_2^2}.$$

$$[0]=0$$

$$[1]=1$$

$$[2]=\alpha$$

$$[3]={\alpha }^2+\beta$$

$$[n{]}^{\mu }=\frac{n}{1+n\mu }$$

$$[0{]}^{\mu }=0$$

$$[1{]}^{\mu }=\frac{1}{1+\mu }\ne 1.$$

$$\underset{\mu \to 0}{\lim }[1{]}^{\mu }=1.$$

$$\underset{\mu \to 0}{\lim }[n{]}^{\mu }=n$$

$$\underset{\mu \to 0}{\lim }[2]=2$$

Full text

Comment on [1]

Johar M. Ashfaque♠

♠*Max Planck Institute for Software Systems, Campus E1 5,

66123 Saarbrücken, Germany

In this note, we show that the $\mu$-oscillator does not lead to the Fibonacci sequence as claimed in [1] since $[n]^{\mu}=n$ in the limit $\mu\rightarrow 0$. Thus we obtain the sequence $[0]^{\mu}=0,1,2,...$. We only obtain the Fibonacci sequence when the $q$-deformation is associated to the $\mu$-deformation via the basic number

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1 Comment

In this note, we show that the $\mu$-oscillator does not necessarily lead to the Fibonacci sequence as claimed in [1] since $[n]^{\mu}=n$ in the limit $\mu\rightarrow 0$.

The $q$-oscillator Fibonacci basic number necessarily satisfies the Fibonacci relation since

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and setting $\alpha=\beta=1$ yields the Fibonacci sequence.

However, when considering the $\mu$-oscillator

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giving

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and

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It is only in the limit $\mu\rightarrow 0$ that

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Now, since

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we have

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