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On the blending of real/complex analysis with number theory. The study involves distribution of prime numbers and other problems and helps giving asymptotic estimates to these.
7
votes
Precise relation between prime number theorem and zero-free region
Your heuristic is wrong: $G(x)=x\exp{(-a\sqrt{\log{x}}})$ follows from $f=\frac{c}{\log{(|t|+3)}}$ for some fixed real $c>0$.
I really don't want to tell you the answer, because this is a great exerc …
7
votes
Accepted
Contour integration of $\zeta(s)\zeta(2s)$
This case, at least, you can do by hand; if $c(n)$ is the $n$th coefficient of $\zeta(s)\zeta(2s)$, then
$\sum_{n\leq X}c(n)=\sum_{nm^2\leq X}1=\sum_{m}\sum_{n\leq X/m^2}1 = \sum_{m < \sqrt{X}}(m^{-2 …
4
votes
Accepted
Cos[Im[zetazero(n)]Log(prime)] spans a countable dense set in [-1,1]?
For any fixed real $\alpha$, the fractional parts of the numbers $\alpha \gamma$, where $\beta+i\gamma$ runs over all zeros of $\zeta(s)$ in the critical strip with $0<\gamma < T$, become uniformly d …
2
votes
What should be learned in an introductory analytic number theory course?
This may be a stretch, but what about the Burgess bound for character sums? The Riemann hypothesis over finite fields is a ubiquitous tool in number theory now, and this seems as a good an introducti …
13
votes
Accepted
Equivalent forms of the Grand Riemann Hypothesis
Well, suppose pi is a cuspidal automorphic representation of GL(n)/Q. This has the structure of a tensor product, indexed by primes p, of representations pi_p of the groups GLn(Qp). The Satake isomo …
6
votes
Examples of asymptotic formulas with optimal error term
If $r_k(n)$ is the number of representations of $n$ as a sum of $k$ squares, and $k\geq 5$, then the estimate
$\sum_{n \leq X}r_k(n)=C_k X^{\frac{k}{2}}+O(X^{\frac{k}{2}-1})$
is known, and is best p …
12
votes
Accepted
Sato-Tate measure for GL(3) Automorphic forms
(2017-11-26 edit by j.c.: earlier versions of this answer consisted of David Hansen's screenshot of the following, with the text "Here is a screenshot of a semi-answer which froze my computer when I h …
12
votes
Accepted
Primes of the form a^2+1
Hi Franz,
Unfortunately I doubt this Euler product has very good behavior. If you believe the Hardy-Littlewood conjectures, then $\sum_{n\leq X}\Lambda(n^2+1) \sim cX$ where $c=\prod_{p>2}(1-\chi_{ …
9
votes
What information do the roots of the generating function of the nontrivial zeroes of the Rie...
I would be willing to wager vast sums of money that this function has the unit circle as a boundary of essential singularities. As $x\to 1$ it diverges like $u(x) \sim c (1-x)^{-2}\log{(1-x)}$.
Edit …
3
votes
Achieving consecutive integers as norms from a quadratic field
Something small, but maybe useful, which no one seems to have pointed out: as $p\to\infty$, $(\mathbb{Z}/p\mathbb{Z})^{\times}$ contains arbitrarily long strings of consecutive quadratic residues. In …
3
votes
Reference for the expected number of prime factors of n larger than n^alpha is -log alpha
I believe you can extract this from a paper of Andrew Granville, "Prime divisors are Poisson distributed". There is an electronic copy of this on his website.
0
votes
Very strong multiplicity one for Hecke eigenforms
I believe the following statement is a folklore conjecture: Let pi1 and pi2 be a pair of cuspidal automorphic representations of GL2 whose local constituents agree at a set of places of positive Diric …
0
votes
Very strong multiplicity one for Hecke eigenforms
Sorry, I meant to say that furthermore one of the pi's should be genuine in Shahidi's sense, or in otherwords not twist-equivalent to itself. Automorphic inductions of grossencharakters, as in Kevin' …
10
votes
Accepted
PNT for general zeta functions, Applications of.
Hi Anweshi,
Since Emerton answered your third grey-boxed question very nicely, let me try at the first two. Suppose $L(s,f)$ is one of the L-functions that you listed (including the first two, which …
29
votes
If the Riemann Hypothesis fails, must it fail infinitely often?
To elaborate on my comment, I really do think this won't be resolved until RH itself is resolved, or at least until new tools are brought onto the scene, because it's so far from what we know about th …