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This extension allows us to ask what happens when we take the "negative oneth" derivative
We would like to obtain the "left-cancellation" property of the derivative
To do this we are forced into the definition that the "negative oneth" derivative being the antiderivativepic.twitter.com/EPVXapmrpo
I got a question, maybe it's trivial: what kind of transformation does a half derivative generate? Some kind of translation+scaling?
I'm not sure.. I'm not an expert and don't know off the top of my head
unfortunately wolfram alpha doesn't quite get it:pic.twitter.com/eeHVxR53rR
Therefore no function equals its half-derivative :(
Fun: What about supersymmetry and the exponential expansion: exp(a^(1/2)D^(1/2))f(x)? Instead of the generator of space(time) translations via Taylor: exp(aD)f(x)=f(x+a), where D=d/d And susy transformations are "square roots" of spacetime translations since roughly Q^2~ {Q,Q}~P
Small note: Applying exp(a^(1/2)D^(1/2)) twice: exp(a^(1/2)D^(1/2))*exp(a^(1/2)D^(1/2))f(x) and using e.g. BCH formula, with a commuting a^(1/2)D^(1/2) operator [a^(1/2)D^(1/2), a^(1/2)D^(1/2)]=0 Retrieving: exp(aD)f(x)=f(x+a)? (a^(1/2)D^(1/2))^2=aD??
Better still compute the fractional derivative in Fourier space. Then you can generalize to all distributions.
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The fractional derivative of course returns the usual derivative when the derivative "powers" add up to an integer
