Timeline for Geometric interpretation of the half-derivative?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Dec 23, 2022 at 15:04 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Feb 10, 2021 at 6:00 | answer | added | Tom Copeland | timeline score: 8 | |
| Dec 24, 2018 at 13:39 | comment | added | Tom Copeland | In terms of Cauchy-like contour integrals, see math.stackexchange.com/questions/125343/… and math.stackexchange.com/questions/1537/… | |
| Feb 27, 2016 at 21:05 | comment | added | Tom Copeland | See archive.org/stream/theoryoflinearop033341mbp#page/n39/mode/2up | |
| Feb 12, 2015 at 0:43 | comment | added | Tom Copeland | You can relate locality vs. non-locality to simple poles vs. branch cuts in the appropriate complex-contour/convolution integral reps for fractional differintegro ops of this type. The regular derivatives can also be evaluated with a nonlocal Cauchy contour integral, as well, with no singularity in the gamma fct. | |
| Mar 16, 2014 at 4:22 | comment | added | Tom Copeland | The Fourier trf. really has very little to do with this operator, hand waving to the contrary. Similarly, $D^{-1}_{FT}g(x)=\int_{-\infty }^{x}g(u)du$, for Fourier integral ops from the Fourier convolution thm, convolving the Heaviside step fct $H(x)$ with $g(x)$, whereas $D^{-1}_{LPT}g(x)=H(x)\int_{0 }^{x}g(u)du$, for Laplace trf ops from the LPT conv. thm. Note that $FT(H(x))= 1/(i 2 \pi f) + \delta(f)/2$, whereas $LPT(H(x))= 1/p$ | |
| Jan 6, 2014 at 1:20 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Include fig found by AlexR.
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| Jan 5, 2014 at 2:45 | vote | accept | Joseph O'Rourke | ||
| Jan 4, 2014 at 23:56 | comment | added | Alex R. | For what it's worth there is a nice geometric interpretation of a fractional integral in this book, page 52, figure 5.1. | |
| Jan 4, 2014 at 16:54 | comment | added | Terry Tao | A correction to my previous comment: the first order operator $d/dx$ is indefinite on the real line (the spectral variable $\xi$ can be either positive real or negative real), and so it is not natural to consider fractional powers of this operator (one has to arbitrarily choose a branch cut for $\xi^\alpha$). But if one is working on the half-line instead, then the spectral variable $\xi$ now naturally lives on the upper half-plane (Fourier-Laplace transform) and one now has a canonical interpretation of $\xi^\alpha$. So fractional powers of $d/dx$ are reasonable in half-line settings. | |
| Jan 4, 2014 at 12:35 | comment | added | Joseph O'Rourke | @RicardoAndrade: Thanks; followed your tagging suggestions. | |
| Jan 4, 2014 at 12:34 | history | edited | Joseph O'Rourke |
edited tags
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| Jan 4, 2014 at 6:00 | comment | added | Qiaochu Yuan | One obstruction to an obvious geometric interpretation is that the ordinary derivative can be regarded as generalizing to, for example, the exterior derivative, but I'm not aware of a generalization of the half-derivative this broad without extra structure (fractional powers of the Laplacian require a Riemannian metric, for example). | |
| Jan 4, 2014 at 5:25 | comment | added | Ricardo Andrade | Would it perhaps be appropriate to replace the deprecated tag 'geometry' with the tag 'geometric-intuition'? Also, would the top-level tag 'ca.analysis-and-odes' be adequate? | |
| Jan 4, 2014 at 4:07 | answer | added | Andrey Rekalo | timeline score: 52 | |
| Jan 4, 2014 at 2:49 | comment | added | Terry Tao | Also, fractional derivatives depend continuously on the exponent $\alpha$ (at least in the distributional topology), as can be seen on the Fourier side. The non-locality disappears as $\alpha$ approaches a natural number due to denominators such as $\Gamma(-\alpha)$ that appear in the formulae for the kernel away from the origin (which is something like $\frac{1}{\Gamma(-\alpha)} |x-y|^{-1-\alpha}$ in one dimension). | |
| Jan 4, 2014 at 2:46 | comment | added | Terry Tao | I haven't found fractional powers of the first derivative $d/dx$, which is indefinite, to be of much use. Fractional powers $\Delta^\alpha$ of the (analyst's) Laplacian $\Delta = -\sum_{i=1}^n \frac{\partial}{\partial x_i^2}$, on the other hand, are much more useful (note that the Laplacian is positive definite, in contrast to the first order operator.) The square root of the Laplacian has a natural geometric interpretation as the Dirichlet-to-Neumann operator for the upper half-plane. | |
| Jan 4, 2014 at 2:44 | comment | added | Ryan Reich | I can't answer your question, but it reminds me of this one, in whose answer locality of the derivative is obtained from the Leibniz rule. | |
| Jan 4, 2014 at 2:43 | comment | added | Tomas | Well, my view of fractional derivatives is from it's kernel (by using Fourier transform), though this has nothing to do with geometric interpretation. For integral derivatives,like $\frac{d^k}{dx^k}$, the Schwartz kernel is $\delta_0^{(k)}$, which measures only on a point, however, when considering the fractional derivative,like $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$ on $\mathbb{R}$, the respective kernel is $|x-y|^{-1-\frac{1}{2}}$, which is nonlocal anymore. | |
| Jan 4, 2014 at 2:32 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Typo.
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| Jan 4, 2014 at 2:27 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |