bio | website | |
---|---|---|
location | France | |
age | 33 | |
visits | member for | 3 years, 10 months |
seen | 4 hours ago | |
stats | profile views | 3,086 |
I'm a former student in physics fond of number theory, especially Hilbert's 8th problem, further generalizations of the Riemann Hypothesis and almost everything related to prime numbers and L Functions. I'm also interested in Galois Theory even though I still don't know much about it.
Jan 29 |
comment |
Is there a hidden symmetry in the prime numbers distribution?
Even though my not-that-smart phone has a display issue, you may have misunderstood something. There is absolutely no possibility for an even integer $m$ to fulfill $r_{0}(m)=12$, since the only even prime number is $2$. Maybe you meant $spr(m)=12$? |
Jan 28 |
asked | Is there a hidden symmetry in the prime numbers distribution? |
Jan 25 |
answered | Shortest/Most elegant proof for $L(1,\chi)\neq 0$ |
Jan 24 |
accepted | Rankin-Selberg convolution and product of degrees |
Jan 24 |
comment |
Rankin-Selberg convolution and product of degrees
I totally agree with GH from MO. |
Jan 24 |
asked | Rankin-Selberg convolution and product of degrees |
Jan 22 |
accepted | Has this formula about prime gaps already been conjectured and/or proven? |
Jan 22 |
revised |
Has this formula about prime gaps already been conjectured and/or proven?
edited title |
Jan 22 |
asked | Has this formula about prime gaps already been conjectured and/or proven? |
Jan 22 |
answered | japanese/chinese for mathematicians? |
Jan 20 |
comment |
Under what conditions $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ for $n\in\mathbf{N}$ implies isometry?
Ok, that makes sense. I'm not able to answer your question but I found it rather useful to get it clarified. |
Jan 19 |
comment |
Under what conditions $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ for $n\in\mathbf{N}$ implies isometry?
This question is interesting but I think it lacks a quantifier. Do you require your equality to hold for all $n\in\mathbb{N}$ or just for some $n$? |
Jan 13 |
comment |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
Yes. But as $T$ increases, the width of each rectangle gets smaller, getting "infinitesimal" as $T$ tends to $\infty$ (because we don't know a priori the possible abscissae of non trivial zeroes of $F$ so that we can't really ignore any value of $x\in [0,1/2]$ without a good reason to do so). |
Jan 13 |
comment |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
If I understand well, I took the limit as $T$ tends to $\infty$ too early? Would I be better off considering first a simple function depending on $T$, which may be increasing on $[0,1/2]$ for $T$ large enough, and considering the limit as $T$ tends to $\infty$ at the very last step? |
Jan 13 |
comment |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
I'm afraid you're right. Kind of feel naughty surreal numbers lurking behind me...Is there a way to prevent the whole ship from sinking? |
Jan 13 |
revised |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
added a missing dollar sign |
Jan 13 |
revised |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
changed the definition of the star product used here |
Jan 13 |
revised |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
defined the precise notion of convolution used here |
Jan 13 |
revised |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
slightly changed the definition of $\delta_{F}$ |
Jan 13 |
revised |
Does the proportion of non trivial zeroes of given real part increase on [0,1/2] for all L-functions?
added another question |