Wokring in q-calculus where everything is defined on the set $$R_q =\{\pm q^k,k \in \mathbb{Z} \} \cup \{ 0\}$$ In which they define the q-derivative (or q-difference operator) as $$D_qf(x)=\frac{f(x)−f(qx)}{(1−q)x}, \quad x\neq 0$$ $$D_qf(x)=f'(0), \quad x=0 $$ My question is, what is meant by q-differentialbe here as the only singularity is at $x=0$ which is excluded. Then every function should be q-differentiable or not?
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q-differentiability isn't meaningfully defined on its own. It's only when, for example: we take the limit as $q \rightarrow 1$, that we say it is 'differentiable' – Mako Nov 14 '23 at 12:04
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Also note that: $$D_qf(x)=f'(x), \quad x=0 $$ doesn't make much sense. Rather it should be $$\lim_{q \rightarrow 1}D_qf(x) = f'(x)$$ – Mako Nov 14 '23 at 12:08
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1Sorry, I mean that $D_qf(x)=f'(0)$ at $x=0$. – Irfan Ali Nov 16 '23 at 13:32