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  The fractional q-derivation

+ 0 like - 1 dislike

I define the fractional q-derivation:


$$=\frac{1}{((1-q)x)^{\alpha}}\sum_{n=0}^{\infty} (-1)^n [(1-q^{C_n^{\alpha}})/(1-q)]f(q^n x)q^{-n}$$

We have:

$$D_q^{\alpha} \circ D_q^{\beta} =D_q^{\alpha+\beta}$$

$$D_q^{\alpha} T=q^{\alpha} T D_q^{\alpha}$$

with $T(f)(x)=f(qx)$.

Have we algebraic properties of the fractional q-derivations?

asked Feb 23, 2022 in Mathematics by Antoine Balan (-80 points) [ revision history ]
edited Feb 23, 2022 by Antoine Balan

Is it a fractional derivative? What is fractional in it? There is no limit process like $\varepsilon\to 0$ in your first formula, so it is an expression rather than a derivative of any kind.

"Have we algebraic properties of the fractional q-derivations?"

What sort of question is this?

Most expressions have algebraic properties. The relations you provide for your $D$-operator are algebraic properties.

So what?

Are you asking for further algebraic properties? If the "fractional q-derivation" is indeed your definition, it would seem you are the person first called upon to work these out.

The formula I proposed is false but it is plausible and you can consult arxiv.org for the good definition of the fractional q-derivative.

As I remarked on a previous occasion, you maybe should put more care in your questions.

Furthermore, I suppose that the sources on arxiv.org to be consulted "for the good definition" contain also algebraic properties of the respective fractional q-derivative.

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