@ParamanandSingh So we need moderators to help and enforce EOQS on askers as well, thanks in advance. I recall seeing quite a number of PSQs from this very user, so this is hardly the first time...
Maybe this actually proves the RH, maybe it doesn't (of which i'm yet to spot any flaw). In any case, it would be interesting to see why it is flawed.
Let: $p$ be a prime and $\mu$ be the M'obius function. Denote by $\Lambda$ the von-Mangoldt function and define $T(x):=\sum_{n\leq x} \log n, \the...
Do you see any real mistake in this proposed proof ?
Let: $p$ be a prime and $\mu$ be the M'obius function. Denote by $\Lambda$ the von-Mangoldt function and define $T(x):=\sum_{n\leq x} \log n, \theta(x)=\sum_{p\leq x} \log p, \psi(x)=\sum_{n\leq x} \Lambda(n)$. For $\Re(s)>1$, note that $-\frac...
I suspect sockpuppets because one of them also upvotes that crank's comments, including one that says nothing more than "whoever you are, it's pointless to waste my energy on the likes of you".
The roots $f(s)=0$ to the function:
$$f(s)=\frac{1}{\frac{1}{s^*}+\frac{1}{s}}+s^2-s$$
are all the complex numbers $s$ with real part:
$$\Re(s)=\frac{1}{2}$$
Mathematica:
(*start*)
Reduce[1/(1/s + 1/Conjugate[s]) + s^2 - s == 0, s]
(*end*)
Question
Does it make sense to multiply that function w...
@Peter Maybe - like I - they don't want to be associated with users who don't apply the same standards to themselves as they apply to other users, e.g. your recent dupe answer, which only got more upvotes when I mentioned it for DV/Del above. When this happens frequently with active users here it is very problematic.
This request is what I refer to - one of the most egregious duplications I've seen in recent times since the answerer could have triviall located the dupe (e.g. searching in common terms in their answer such as 317 and covering congruence)
^^ 371
e.g. the dupe target I gave is the 2nd match result in an MSE search on "371"
Probably the simplest dupe search I've ever encountered.
The roots $f(s)=0$ to the function:
$$f(s)=\frac{1}{\frac{1}{s^*}+\frac{1}{s}}+s^2-s$$
are all the complex numbers $s$ with real part:
$$\Re(s)=\frac{1}{2}$$
Mathematica:
(*start*)
Reduce[1/(1/s + 1/Conjugate[s]) + s^2 - s == 0, s]
(*end*)
Question
Does it make sense to multiply that function w...
