Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,720 comments
1,470 users with positive rep
818 active unimported users
More ...

  The fractional q-derivation

+ 0 like - 1 dislike
514 views

I define the fractional q-derivation:

$$D_q^{\alpha}(f)(x)=[(f(qx)-f(x))/((q-1)x)]^{\alpha}=$$

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

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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...