Timeline for What does Mellin inversion "really mean"?
Current License: CC BY-SA 3.0
18 events
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| Aug 9 at 6:42 | comment | added | Dr Potato | This is the most super-duplicate question I've seen in this forum so far, and there are good reasons why nobody mark them all as the same. This is because yet we want to see things from several angles. If it indeed may be solved with Fourier analysis, it could be done the other way around. I could say: Fourier is in time (additive) shifts and Mellin in scale shifts (multiplicative) And still I ask: So what is the Inverse Mellin transform anyways? | |
| S Apr 21, 2018 at 22:34 | history | suggested | kjetil b halvorsen | CC BY-SA 3.0 |
Removed unnecessary thanks
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| Apr 21, 2018 at 20:21 | review | Suggested edits | |||
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 19, 2013 at 17:11 | answer | added | Michael Rubinstein | timeline score: 9 | |
| Feb 13, 2012 at 12:21 | answer | added | user19475 | timeline score: 4 | |
| Nov 13, 2011 at 4:25 | answer | added | Phil Isett | timeline score: 18 | |
| Nov 13, 2011 at 1:07 | answer | added | Frank Thorne | timeline score: 24 | |
| Nov 4, 2011 at 0:40 | answer | added | paul garrett | timeline score: 29 | |
| Nov 3, 2011 at 20:33 | history | edited | mathphysicist |
intuition tag added
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| Nov 3, 2011 at 14:52 | history | edited | Frank Thorne | CC BY-SA 3.0 |
edited body
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| Nov 3, 2011 at 13:38 | answer | added | Tom Copeland | timeline score: 12 | |
| Nov 3, 2011 at 1:33 | comment | added | Matt Young | I think of the Mellin transform as the equivalent gadget to the Fourier transform, but for the group of positive reals instead of the real line. The change of variables going from one transform to the other is either an exponential or log (depending on which way it's going) which corresponds to the usual isomorphism between the two groups. The function $t \rightarrow t^s$ is a character on the positive reals which is analogous to the function $x \rightarrow e^{2 \pi i x y}$ which is a character on the real line. The Mellin transform is natural for studying multiplicative functions. | |
| Nov 3, 2011 at 0:50 | comment | added | Peter Humphries | I agree with Greg here that Montgomery and Vaughan is a good pointer on why Mellin inversion works. But personally, the only justification I need is changing variables so that it becomes the same as Laplace inversion. This change of variables shows why one integral is over the positive reals while the other is a line integral in the complex plane. Of course, this all depends on how willing you are to believe why Laplace inversion is morally true, or Fourier inversion, for that matter. | |
| Nov 3, 2011 at 0:06 | comment | added | Aaron Bergman | Shouldn't this be something like Pontryagin duality for the multiplicative group of positive reals (maybe with some analytic continuation thrown in?). | |
| Nov 2, 2011 at 23:44 | comment | added | Gerry Myerson | Morally, why is the Fourier inversion formula true? Morally, why is the Laplace inversion formula true? | |
| Nov 2, 2011 at 23:15 | comment | added | Greg Martin | You could also take a look at section 5.1 of Montgomery and Vaughan's Multiplicative Number Theory. It has several specific examples of Mellin transform pairs, together with brief remarks on how the formulas are proved. I believe that the basic method really is the same as Perron's formula: if $y$ is small then you drag the contour integral to the right and get a contribution of $0$, while if $y$ is large then you drag the contour to the left and pick up the residue of the integrand at $s=0$. | |
| Nov 2, 2011 at 21:55 | history | asked | Frank Thorne | CC BY-SA 3.0 |