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I have heard about a generalization of the calculus named 'quantized calculus'. In this calculus the derivative is defined as

$$ df= [F,f]=Ff-fF $$

Here $ F(g(x))= \frac{i}{\pi}\int_{-\infty}^{\infty}dt \frac{g(t)-g(s)}{t-s}$. In any case if this is the 'quantized' derivative , how can one defined a 'quantized integral'? How can one recover the usual definition of derivative from this $ \frac{d}{dx} $?

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    I can tell $F$ is an operator, but I have no idea what $[F,f]$ means nor what $H$ is. In any case, this smells like physics...2012-04-24
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    $ [A,B] $ is the commutator between the 2 operator.. or similar , i think $ f$ is the element of an algebra.2012-04-24
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    I assume by the letter $H$ you actually meant $F$ then right? Would I be correct in also assuming $f$ is a function and $df$ as an operator?2012-04-24
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    ah of course i meant $F$ instead of H :) ... and $ df$ should be an operator ..2012-04-24
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    I don't know anything at all about quantized calculus, but most of the top hits Google shows me for the phrase state that the analogue of the classical integral in this setting is the "Dixmier trace" (whatever that is).2012-04-24
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    Hey, just as a side note... You may want to look up things called deravations. Let $R$ be a ring, then $\rho:R\times R\rightarrow R$. $\rho$ is a derivation if: 1. For every $a, b \in R$ we have that $\rho(ab)=a\rho(b)+\rho(a)b$ and $\rho(a+b)=\rho(a)+\rho(b)$2013-12-20

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