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bio website stanford.edu/~rjlo
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Feb
27
awarded  Revival
Nov
18
comment Long gaps between primes
Following up on @GerhardPaseman's comment, the expected order of magnitude is $c \log^2 n$ (though proving this is hard!). See, for example, this paper of Granville: www.dms.umontreal.ca/~andrew/PDF/cramer.pdf. It discusses Cramer's conjecture, why it's probably wrong (at least in a precise form), and what the correct modification should be.
Nov
18
comment Long gaps between primes
The Wikipedia article en.wikipedia.org/wiki/Prime_gap on prime gaps covers this topic. In particular, C can be taken to be arbitrarily large.
Nov
3
comment What is known about the sum x^{n^2}/n?
I'm not an expert, but the Eichler integral can be defined, purely formally, as the operator that sends, with your notation, $x^n \mapsto x^n/n$ (recall that for modular forms $x=exp(2\pi i z)$, so this is integration $dz$). Your series is then a formal ``half-integral'' of the standard theta function. I've seen such things arise in talks, though I don't know anything about them myself (and I trust your ability to google as much as my own). I'd recommend searching for "half-derivative" rather than "half-integral", since the latter appears quite frequently with another meaning.
Oct
29
awarded  Informed
Oct
23
comment On extended Riemann Hypothesis and coefficients of Selberg Class L-functions
Even with an infinite sum, I think it's impossible if $d>1$. Let $G(s)=L(s,f)H(s)$. We need to balance three things: 1) The sum of $f*h$ being bounded, 2) $H(s)$ analytic in $\Re(s)>0$, maybe by asking for $S_h(X) \ll X^\epsilon$, and 3) The non-existence of zeros of $H(s)$ in $\Re(s)>0$. It's not even obvious to me that 1 and 2 are compatible once $d>1$: each is equivalent to a system of inequalities on $h(n)$. 1 has coefficients $S_f(X/n)$, whereas 2 has coefficients all one. Once $S_f(X)$ has size (as in $d>1$), these systems are on different scales, so a common solution is not obvious.
Oct
21
answered On extended Riemann Hypothesis and coefficients of Selberg Class L-functions
Apr
17
comment Off critical line zeros for half integer weight $L$-functions
Have you done any computations yourself? While I'm dubious that this should be true for almost any form, it's worth noting that $L(s,\theta_\chi)=L(2s-1/2,\chi)$ for a non-trivial Dirichlet character $\chi$, so RH presumably holds in this case. In general, though, the multiplicative structure of half-integral weight eigenforms is more complex, and I'd be very surprised if it were to hold if the form is orthogonal to the space of unary theta functions.
Nov
30
comment Is this extension of the Selberg class trivial?
You're absolutely right that there are issues with $L(s,f)^{1/2}$, which is why it's not actually something I want to consider. I brought it up mostly to clarify points 1 and 2. Maybe a prototype question would be this: Can $L(s,f)^{1/2}L(s,g)^{1/2}$ ever be sensibly continued to an entire function, where $L(s,f)$ and $L(s,g)$ are primitive elements of the Selberg class? It's known that each has zeros disjoint from the other, but maybe all the simple zeros coincide, or are there are no simple zeros, or... Conjecturally, this can't happen, but that's the sort of thing I'm imagining.
Nov
30
awarded  Commentator
Nov
30
comment Is this extension of the Selberg class trivial?
Right, this absolutely falls under the purview of pretentiousness; in fact, my question about more than square root cancellation can be viewed as a counterpart to Halasz's theorem which says that anything with large sums must come from (pretend to be) one of a natural set of examples. There, though, the natural examples are not Dirichlet characters. Instead, they are the additive characters, $n^{it}$. Indeed, non-principal Dirichlet characters don't have large partial sums - the partial sums are bounded! Thus, they are the natural examples for exceptional (more than square root) cancellation.
Nov
30
asked Is this extension of the Selberg class trivial?
Nov
24
awarded  Yearling
Aug
27
awarded  Yearling
Aug
27
awarded  Yearling
Aug
25
revised The behavior of a certain greedy algorithm for Erdős Discrepancy Problem
added 971 characters in body
Aug
25
revised The behavior of a certain greedy algorithm for Erdős Discrepancy Problem
added 550 characters in body
Aug
25
comment The behavior of a certain greedy algorithm for Erdős Discrepancy Problem
I will add data for the squarefree problem to my answer in a second, but let me just quickly give you the gist -- it's the same basic behavior, with $1/3$ appearing just as clearly.
Aug
25
comment The behavior of a certain greedy algorithm for Erdős Discrepancy Problem
His. And the ratio of $\log D/\log N$ I alluded to, not in graph form: 0.3010299957, 0.345687124, 0.3788254007, 0.3972102138, 0.3260186782, 0.3302216042, 0.3378116896, 0.3352640758, 0.3368385727, 0.3590944053, 0.336264762, 0.3602395359, 0.3305424391, 0.3367194215. Close enough to 1/3 that I suspect something is happening.
Aug
24
answered The behavior of a certain greedy algorithm for Erdős Discrepancy Problem