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I was looking for a geometrical interpretations of fractional derivatives and fractional integrals.

I would be glad to see any kind of intuitive and preferably visual interpretation of the objects of fractional calculus.

Can anyone recommend the source or share personal opinion on the topic?

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  • $\begingroup$ In this thread I answer a very similar question, pointing out how they emerge in stochastic dynamics. $\endgroup$ – Nikolaj-K Apr 26 '15 at 7:51
  • $\begingroup$ @NikolajK, your answer looks depressingly complicated. Isn't there a simpler way of depicting concepts of fractional calculus? $\endgroup$ – Vlad Apr 26 '15 at 7:58
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    $\begingroup$ There are various interprations, depending on the domain of application considered in physics. For example, in electrotechnics :fr.scribd.com/doc/14686539/… . Concerning pure geometrical interpretation, see : arxiv.org/pdf/math/0110241.pdf $\endgroup$ – JJacquelin Apr 26 '15 at 8:33
  • $\begingroup$ @JJacquelin Thank you, I'll have to go through it and try to extract a short and self-contained example or illustration $\endgroup$ – Vlad Apr 26 '15 at 8:46
  • $\begingroup$ @JJacquelin that "geometric" interpretation is nothing more than its area projected with a gamma function. Most can parse that from the formula. Plus there's not really a way to construct that without using non geometric methods. $\endgroup$ – Zach466920 Apr 26 '15 at 13:48
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Please see the following link

www.scielo.org.mx/pdf/rmf/v60n1/v60n1a6.pdf

about "A physical interpretation of fractional calculus" and,

vectron.mathem.pub.ro/dgds/v15/D15-ta.pdf

entitled "The geometric and physical interpretation of fractional order."

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$$\left(\frac{d}{dx}\right)^{1/2} x^a = \frac{\Gamma(a+1)}{\Gamma(a+1/2)} x^{a-1/2}$$ and $$\left(\frac{d}{dx}\right)^{1/2} e^{ax} = \sqrt{a} e^{ax}$$ so it's look cool, but unfortunately it's not local : the distribution filter to get $\left(\frac{d}{dx}\right)^{1/2} f$ from $f$ is not point-wise as $\delta'(x)$ :

$$f'(x) = [f \ast \delta'](x)$$

$$\left(\frac{d}{dx}\right)^{1/2}f(x) = [f \ast d_{1/2}](x)$$ (where $\ast$ is the convolution) but $d_{1/2}(x)$ is not even supported on a finite interval. it means that from the Taylor series or the Fourier series of a function it's easy to calculate the $1/2$th derivatives, but from a simple function such as $1/\sin(x)$ it becomes much more difficult, and even impracticable.

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