Timeline for Does the Euler product for $L(s,\chi_4)$ also converge in the right half of the critical strip?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Mar 6, 2017 at 19:37 | comment | added | André LeClair | The is is a very late reply, but in any case, last year Franca and I submitted a paper to arXiv. that provides an argument for why the EP should converge for Re s > 1/2 without assuming the RH (for non-principal characters). We provided a strategy to prove it, but could not prove one of the ingredients, which is that a particular series behaves like a random walk. So it may not be "hopeless" as KConrad wrote, though it does still seem difficult. The reason I think there is some hope is that random walks are rather generic. | |
| Jun 19, 2015 at 15:46 | comment | added | joro | For $p$ up to $10^9$ the finite product is $0.9798542...$ for $s=0.5001$. | |
| Jun 19, 2015 at 15:18 | comment | added | joro | @NoamD.Elkies This appears closely related to prime race modulo 4 (Chebyshev bias). Take $s=0.5001$. For $p$ up to $10^8$ the finite product is $0.960060883...$, while $L(s)=0.667719642658...$. This would mean there are significantly more primes congruent to $3$, but not too much to make the product zero... | |
| Jun 18, 2015 at 15:08 | comment | added | Noam D. Elkies | Good question. The answer is basically the pole at $s=1$. The connection with the $L$-function zeros is via the logarithmic derivative $L'/L$, which is singular when $L$ has either a zero or a pole. Dirichlet $L$-functions other than the Riemann zeta function have no poles, so only their zeros affect the analysis; but when you try to do the same for $\zeta$ you run into the $s=1$ pole before any zero shows up. | |
| Jun 18, 2015 at 13:41 | comment | added | joro | @NoamD.Elkies Why is the trivial character corresponding to zeta exception to convergence? | |
| Jun 18, 2015 at 8:57 | vote | accept | Agno | ||
| Jun 17, 2015 at 23:58 | history | edited | GH from MO |
edited tags
|
|
| Jun 17, 2015 at 23:57 | comment | added | GH from MO | Related question here: mathoverflow.net/questions/63714/… | |
| Jun 17, 2015 at 23:14 | comment | added | KConrad | To see the equivalence that Noam mentions, see Theorem 3.3 (and the two lemmas preceding it) in math.uconn.edu/~kconrad/articles/eulerprod.pdf, setting $d = 1$ and $\alpha_{p,1} = \chi(p)$. Note that for infinite products like an Euler product, the term "converges" means "converges and is not $0$." Since nobody has ever proved the $L$-function of a nontrivial Dirichlet character has no zeros in any vertical strip $1-\varepsilon < {\rm Re}(s) < 1$, it is basically hopeless at present to expect anyone to prove the Euler product converges at any $s$ with ${\rm Re}(s) < 1$. | |
| Jun 17, 2015 at 21:40 | comment | added | Noam D. Elkies | The Euler product for every Dirichlet series (for a nonprincipal character) converges at $s=1$, but that's more-or-less equivalent with the Prime Number Theorem in arithmetic progressions, and in particular Euler's heuristic argument for $\chi_4$ does not readily yield a rigorous proof for that $L$-function. Convergence for real part $> 1/2$ is equivalent to the Riemann hypothesis for the same $L$-function. (While I was editing this ABCDveve wrote similarly in his/her answer.) | |
| Jun 17, 2015 at 21:38 | answer | added | ABCDveve | timeline score: 8 | |
| Jun 17, 2015 at 21:07 | history | asked | Agno | CC BY-SA 3.0 |