0

While there have been other posts regarding fractional derivatives, unfortunately I am unable to find a straight-forward answer as to what is the square-root-derivative of $\frac{dy}{dx}$, such that

$$\frac{dy}{dx} = \left( \frac{d^{\frac{1}{2}}}{d^{\frac{1}{2}}x}\right)\left( \frac{d^{\frac{1}{2}}}{d^{\frac{1}{2}}x}\right)y$$

Could anyone write down the direct formula for the square-root-derivative of $\frac{dy}{dx}$ please?

Gonçalo
  • 15,869
James
  • 922
  • 1
    Could you use the Riemann-Liouville formula? – Sean Roberson Jul 23 '24 at 20:30
  • @SeanRoberson thanks, i would be using it numerically, so any procedural way to get at the square-root-derivative will do (it need not have a nice/closed form analytical expression) Could you elaborate a little on how to use the Riemann Liouville formula numerically, please? – James Jul 23 '24 at 20:39
  • 1
    Without knowing specifics of$y,$ you can do this with Fourier transforms, but it is messy. – Thomas Andrews Jul 23 '24 at 20:41
  • 3
    There are multiple competing notions of what a half derivative is and each of them require different assumptions and domains to be valid. Of these, you will not get an easier formula in general than what you can find on Wikipedia. For numerical representations of periodic functions, a Fourier multiplier approach as Thomas mentions may work, but otherwise you will have to use one of the singular integral representations – whpowell96 Jul 23 '24 at 21:09
  • 2
    There is a tag for this ([tag:fractional-calculus]). There are many possible answers, as the Wikipedia article you linked says (including listing different answers and formulas). And there are already questions and answers about it: *Calculating Half of a derivative?, $\frac{d^{1/2}}{dx^{1/2}}$, what does it mean? and many more questions – The Art Of Repetition Jul 23 '24 at 22:01
  • @ThomasAndrews thank you, Fourier transforms would work fine, I think. Do you know whether common identities like $\nabla \times (\nabla \times \vec E) = \nabla (\nabla \cdot \vec E) - \nabla^2 \vec E$ will automatically remain true if we replace $\nabla$ with its fractional derivative $\nabla^{\frac{1}{2}}$? – James Jul 24 '24 at 02:50
  • @ThomasAndrews Plus a quick question about this method please: given a function $f$, we first take its Fourier transform $F$, then compute its derivative using $F^\frac{1}{2}=(2\pi i \bar x )^\frac{1}{2} F$. Is it a theorem that the inverse Fourier transform of $F^\frac{1}{2}$, i.e. $inv(F^\frac{1}{2}) = f^\frac{1}{2}$, will always be the fractional derivative of the original function $f$? – James Jul 24 '24 at 03:39
  • That is not a theorem, it is a definition of the half derivative for the class of functions where all the relevant transforms are well-defined – whpowell96 Jul 24 '24 at 05:18
  • 1
    Also, generally speaking, numerical approximation of fractional derivatives is a topic of active research. To my knowledge, all such methods require approximations of integral transforms like an FFT or specialized quadrature to approximate, e.g., Riemann-Liouville or Caputo fractional derivative. You may be better off searching on Google Scholar than this forum – whpowell96 Jul 24 '24 at 05:21
  • 1
    It is necessary to go back up to 1695 to find a letter of G.W.Leibniz to G.A.L’Hospital in which a fractional differential $d^{1/2}x$ is mentioned and qualified as “apparently paradoxical”. General-public essay : https://fr.scribd.com/doc/14686539/The-Fractional-Derivation-La-derivation-fractionnaire pp.7-12. – JJacquelin Jul 24 '24 at 06:33
  • @whpowell96 thank you. "it is a definition of the half derivative" had me worried, as i need the double application of $d^\frac{1}{2}d^\frac{1}{2}y$ to actually give the same answer as a single application of $dy$ using regular calculus. I suppose this must be so, since people seem fine with this Fourier transform method for obtaining $d^\frac{1}{2}$? – James Jul 24 '24 at 09:08

0 Answers0