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Jan 23 at 15:37 comment added dohmatob And this is indeed the case for $p=2$; can't get real $D(1)$ if $b \gt 1/2$.
Jan 23 at 15:32 comment added dohmatob @Mark You're right. I was a bit sloppy there, and feared this would come back to hunt me. My hope is that $b \gt 1/2$ would break something down the line even in the complex / rigorous argument. Will fix asap.
Jan 23 at 14:45 comment added Mark Schultz-Wu I don't understand your comment "this is only solvable if $\dots$". Do you mean over $\mathbb{R}$? Because for $p\geq 3$ one expects $\widehat{Q}(1), \widehat{Q}(2),\dots$ to be $\mathbb{C}$-valued in general, so reasoning via solvability over $\mathbb{R}$ seems like it may be specific to the case of $p = 2$ where $\widehat{Q}(0), \widehat{Q}(1)\in\mathbb{R}$.
Jan 23 at 14:18 answer added R W timeline score: 1
Jan 23 at 11:25 history edited dohmatob CC BY-SA 4.0
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Jan 23 at 11:25 comment added dohmatob That was a typo, fixed.
Jan 23 at 8:56 comment added kodlu I think the Fourier transform does not map $Z_p$ to $Z_p$ but $Z_p$ to the complex numbers.
Jan 22 at 16:10 comment added dohmatob Ah, i think that might be it ! Good catch. You mean $\mathrm{Re} (\hat Q(k)) \ge 0$, or is $\hat Q(k)$ always guaranteed to be real ?
Jan 22 at 9:38 comment added JP McCarthy Isn't it therefore as simple as $\widehat{Q}(k)\geq 0$?
Jan 22 at 4:38 history edited dohmatob CC BY-SA 4.0
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Jan 22 at 1:08 history edited dohmatob CC BY-SA 4.0
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Jan 22 at 1:06 comment added dohmatob Fourier analysis indeed recovers the "manual" solution I obtained earlier using basic algebraic manipulation. On think that felt a bit unnatural was working with $D$ and $Q$ as if the where just any functions on $\mathbb Z_p$ (i.e not probability) distributions, and then plugging the probability constraints at the end. I was wondering if there is a standard nice way to do Fourier analysis of **probability functions on finite discrete structure like $\mathbb Z_p$.
Jan 22 at 1:00 history edited dohmatob CC BY-SA 4.0
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Jan 22 at 0:46 history edited dohmatob CC BY-SA 4.0
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Jan 22 at 0:43 comment added dohmatob The $(1+k)^{-\beta}$ thing: What I mean is that $Q(0) = c$, $Q(1) = c2^{-\beta}$, $Q(p-1) = cp^{-\beta}$, with $c$ chosen so that $\sum_k Q(k) = 1$.
Jan 22 at 0:41 comment added dohmatob OK, I read you. I've made an small update in the question which does a calculation based on your insightful comment. Thanks again.
Jan 22 at 0:40 history edited dohmatob CC BY-SA 4.0
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Jan 22 at 0:35 comment added paul garrett @dohmatob, $\mathbb Z/p$ is "self-dual", so general ideas of harmonic/Fourier analysis still apply... and "Fourier transform/series/etc" still converts convolution to pointwise multiplication. The meaning of the $(1+k)^{\beta}$ is very unclear, though...
Jan 22 at 0:20 comment added dohmatob @paulgarrett $\mathbb Z_p := \mathbb Z / p \mathbb Z$ is simply the integers modulo $p$. I guess your remark still holds about Fourier, that's $\widehat D^2 = \widehat Q$, right ?
Jan 21 at 23:57 comment added paul garrett @RW, thanks, I'm glad it was not just me wondering...
Jan 21 at 23:56 comment added R W @paul garrett - Judging by the "illustrative example", $\mathbb Z_p$ is the order $p$ cyclic group. Anyway, formula $Q(k) \propto (k+1)^{-\beta}$ from the question is not compatible with either interpretation.
Jan 21 at 22:49 comment added paul garrett Just to be clear, $\mathbb Z_p$ is the $p$-adic integers? In general, since this is a compact topological group, its dual is discrete, and on the Fourier expansion side, convolution becomes pointwise multiplication... ?!?
Jan 21 at 18:57 history edited dohmatob CC BY-SA 4.0
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Jan 21 at 12:04 history edited dohmatob CC BY-SA 4.0
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Jan 20 at 23:37 history edited dohmatob
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Jan 20 at 23:27 history asked dohmatob CC BY-SA 4.0