3 ms·
I really wonder if you have studied basic analytic number theory, because that bound is actually well-known and is equivalent to the PNT. Anyway, Montgomery-Vau
by Quant10 4y ago
I really wonder if you have studied basic analytic number theory, because that bound is actually well-known and is equivalent to the PNT. Anyway, Montgomery-Vaughan explicitly and clearly states in equation 6.12 that there exists some constant c>0 such that
psi(x) = x + O(x exp(-c√log x))
uniformly for x \geq 2. Since psi(x)=0 for x in [1, 2), this uniformity trivially extends to x \geq 1.
Also, arXiv moderators don't review papers, they are too busy for that. Submissions claiming to solve famous open problems are classified basing more on author reputation/submission history. I personally know a few moderators in math and physics.
- mjreacher 4y agoYes you would be right I haven't taken a class on analytic number theory. Perhaps excuse my lack of knowledge on the subject but why would psi(x)=0 for x in [1, 2)? Λ(2) is not 0. You are right that arXiv moderators don't review papers in the sense of peer review but they still check over them and I haven't heard of an occasion where something that was originally placed in General Mathematics or General Physics was later moved to another category. Safe to say if there was anything substantial in the proof we would have heard of something from the experts in number theory by now.
- Quant10 4y agoDude, x in [1, 2) means 1 \leq x < 2. Anyway, your first comment would make one think that you are an expert in analytic number theory who found some crucial flaw in the proof. But since that's clearly not the case, i won't comment any further on this thread.
- deleted 4y ago[deleted]