Timeline for Prime race modulo $12$. When is the first sign change?
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13 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2017 at 18:18 | answer | added | Stopple | timeline score: 11 | |
| Aug 9, 2017 at 16:33 | comment | added | Peter Humphries | @reuns, yes, it is complicated; see Odlyzko and te Riele's paper for a similar proof. Basically you need to use many zeroes of these $L$-functions to overcome the constant $C$, which can be computationally quite demanding. | |
| Aug 9, 2017 at 15:17 | comment | added | reuns | @PeterHumphries Right tks, I found where is the problem (and what is your $C$) : $\displaystyle\sum_{p \le x} \chi(p) = \sum_{p^k \le x} \chi(p^k)-\frac{1}{2}\sum_{p^k \le x^{1/2}} \chi(p^k)^{\color{red}2}+ \mathcal{O}(x^{1/3})$ so $\sum_p \chi(p) p^{-s}$ has a singularity at $s=1/2$ if $\chi^2 = |\chi|$. You mentioned the structure of the imaginary zeros, is it complicated to see why (under GRH) $\pi(x;q,a)-\pi(x;q,b)$ changes of sign infinitely often in the OP's case ? | |
| Aug 9, 2017 at 14:40 | comment | added | Peter Humphries | @reuns, this is still missing the point. Try to make this rigorous and you will see that the issue is the second order term ($k = 2$ in your notation). Rubinstein and Sarnak show that under GRH $\frac{\sqrt{x}}{\log x} (\pi(x;q,a) - \pi(x;q,b)) = C + \sum_{\gamma} c_{\gamma} x^{i\gamma} + \cdots$ for certain constants $C, c_{\gamma}$, and the sum is over zeroes of certain Dirichlet $L$-functions. The problem is the term $C$, which you're neglecting. | |
| Aug 9, 2017 at 14:16 | comment | added | Peter Humphries | ...that's what I've been saying all along. In any case, it is certainly unknown unconditionally that for arbitrary $q$, $\pi(x;q,a) - \pi(x;q,b)$ changes sign infinitely often. Even under GRH, it could be the case that it does not change sign infinitely often (though this could only happen if the the imaginary ordinates of zeroes of certain Dirichlet $L$-functions were highly linearly dependent over $\mathbb{Q}$). | |
| Aug 9, 2017 at 14:04 | comment | added | Peter Humphries | @reuns, no, this is incorrect. Landau's theorem only shows that $\psi(x;q,a) - \psi(x;q,b)$ changes sign infinitely often, but this is not enough to conclude that $\pi(x;q,a) - \pi(x;q,b)$ does too. | |
| Aug 9, 2017 at 13:25 | comment | added | Gerry Myerson | Greg Martin discusses mod 12 races in arxiv.org/pdf/math/0010086.pdf (although he doesn't specifically consider the current question). | |
| Aug 9, 2017 at 13:25 | comment | added | Peter Humphries | In any case, the first sign change is probably unknown. The usual way to find sign changes is well known and is based on the work of Ingham. It is the same method that Odlyzko and te Riele use to disprove Mertens conjecture. | |
| Aug 9, 2017 at 13:23 | comment | added | Peter Humphries | That being said, Jason Sneed has apparently proven unconditionally that this does indeed change sign infinitely often (see math.uiuc.edu/~ford/wwwpapers/barriersIII.pdf). | |
| Aug 9, 2017 at 13:22 | comment | added | Peter Humphries | @reuns, there is an issue using partial summation to move from $\psi(x;q,a)$ to $\pi(x;q,a)$ that introduces an additional term that means that Landau's theorem may not work. As a toy model, see how Corollary 15.4 of Montgomery and Vaughan is only one-sided for $\pi(x)$ but two-sided for $\psi(x)$. | |
| Aug 9, 2017 at 13:15 | comment | added | Peter Humphries | @reuns, what you've written is not correct; look up Chebyshev's bias. | |
| Aug 9, 2017 at 13:14 | comment | added | reuns | The sign of $\pi(x;q,a)- \pi(x;q,b)$ changes infinitely often unconditionally, this is because otherwise $\sum_n \Lambda(n) n^{-s} (1_{n \equiv a \bmod q}-1_{n \equiv b \bmod q})$ would have a pole at $s= \sigma$ its abscissa of convergence. Finding the first "non-trivial pole" should give some hints on the first sign change. | |
| Aug 9, 2017 at 12:30 | history | asked | joro | CC BY-SA 3.0 |