Timeline for Radon transform range theorem and radial functions
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2022 at 22:07 | vote | accept | phaedo | ||
| Jan 29, 2022 at 21:10 | comment | added | Greg O. | I think that's essentially correct. But, actually, there is a cleaner way to characterize the $H_n$ using spherical harmonics: $H_n$ is the space of harmonic polynomials up to degree $n$, where a harmonic polynomial is any polynomial whose Laplacian vanishes. See Theorem 1.1.3. of Bai and Xu's "Approximation Theory and Harmonic Analysis on Spheres and Balls": link.springer.com/content/pdf/10.1007/978-1-4614-6660-4.pdf | |
| Jan 25, 2022 at 20:54 | comment | added | phaedo | Ok I think your $H_n$ can probably be characterized as all homogeneous polynomials of degree $n-2k, k\in\{0, 1, \cdot, [n/2]$ where $[\cdot]$ denotes the floor function. Or am I missing something else? | |
| Jan 25, 2022 at 19:51 | comment | added | Greg O. | That's not true, though. For example, $H_1$ as I've defined it does not contain constant functions, since there is no linear combination of $\sin(\theta)$ and $\cos(\theta)$ that gives a constant. Likewise,$H_2$ does not contain $\sin(\theta)$ or $\cos(\theta)$, since there is no linear combination of $\sin^2(\theta)$, $\cos(\theta)\sin(\theta)$, $\cos^2(\theta)$ to give you those functions, etc. | |
| Jan 25, 2022 at 13:28 | comment | added | phaedo | Then I think it would be less ambiguous to say that $P_n$ is a polynomial of degree <= n | |
| Jan 23, 2022 at 17:11 | history | edited | Greg O. | CC BY-SA 4.0 |
added answer to Q2
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| Jan 23, 2022 at 16:22 | comment | added | Greg O. | To be more clear, the proof of the moment conditions amounts to the identity: $\int Rf(w,t) t^n dt = \int_{\mathbb{R}^d} f(x) \langle w,x\rangle^n dx$ for all $w = (w_1,...,w_n) \in \mathbb{S}^{d-1}$. The right-hand side may be formally expanded into a homogeneous polynomial of degree $n$ in the variable $w$, and this is what is meant by a "degree n" spherical polynomial. | |
| Jan 23, 2022 at 16:13 | comment | added | Greg O. | I don't think there is a uniquely defined degree in this case. But I don't think that matters in defining the moment conditions. Simply take $H_n$ to be the space of all functions realizable as a homogeneous polynomial of degree $n$ evaluated on the sphere. Then the moment conditions say that $P_n$ must belong to $H_n$. | |
| Jan 21, 2022 at 11:25 | comment | added | phaedo | Thanks for your answer. I realized this later. However: the degree of the zero polynomial is $-\infty$, not 1. In addition, if Pn is considered as a "spherical" homogeneous polynomial subject to trigonometric identity simplifications, how would you define its degree? | |
| S Jan 19, 2022 at 23:45 | review | First answers | |||
| Jan 19, 2022 at 23:56 | |||||
| S Jan 19, 2022 at 23:45 | history | answered | Greg O. | CC BY-SA 4.0 |