- Ok, so a quick question for anyone who'd like to throw in an answer.
Is the Riemann Hypothesis true? Please, no references to 'proofs' that
haven't made it into respected journals yet...
I'm asking for opinion, rather that proof. Who thinks it's true? Who
thinks it isn't? Who thinks it don't matter too much?
This is only a little fun, of course...
- Yes, it's true. But I'll keep my proof to myself for now, hehe.
Actually, I know it has to do with the observation that solutions to
the Zeta function in the complex plane are either trivially on a line
or else symmetric around a line, but that's about all I know. It's
beyond me.... But I figure that nature should be symmetric...
--- In firstname.lastname@example.org, "Andrew Swallow"
> Ok, so a quick question for anyone who'd like to throw in an answer.
> Is the Riemann Hypothesis true? Please, no references to 'proofs'
> haven't made it into respected journals yet...Who
> I'm asking for opinion, rather that proof. Who thinks it's true?
> thinks it isn't? Who thinks it don't matter too much?
> This is only a little fun, of course...
> Personally, I think it is true (unbiased opinion). There is a section ofWhat do you mean 'trivial'?
> mathematics connected to RH that says that any zero off the critical line
> effectively controls the prime distribution. Hence most of the zeroes on the
> critical line are trivial, which has been proved otherwise!
- On Mon, Aug 04, 2003 at 07:42:34PM +0100, Jon Perry wrote:
> It was a joke! The idea is still true - each non-critical zero 'pulls' allI still don't get you. But I presume you're trying to say that zeros off
> the other zeroes drastically more than zeroes on the line. While it is
> possible for there to be an inifinte such number of nc zeroes, this kind of
> preposterism has long by held at arm's length.
> Hence if there is a zero of the critical line, the zeroes on the line can be
> considered trivial - VERY unlikely, or there is a 'leaning tower of nc
> zeroes', equally unlikely.
the critical line have a much larger effect than those on the line. But
of course it would depend on how many such zeros there were, hence the
idea of zero-density results. I presume that's what you mean?
> And if b is not 1/2, then some of the x^(1-b) are smaller than others, e.g.Hmm ok, I guess that's one way of looking at it. But like I said, it all
> we may have x^(1/4), which produces smaller values than x^(1/2) or indeed
> x^(3/4) - it's symmetric root. And this translates to a larger term than
> expected - especially as we tend x to infinity.
> As all the nc roots provide such a component, the prime distribution is
> therefore 'pulled' by these values - thus we have a leaning tower effect.
depends on estimating how many non-critical zeros there are. As long as
there aren't too many, the tower wouldn't budge.
> No, because even just one affects all the zeroes after it, and it's effectDepends what kind of prime information you're looking for really. You're
> would amplify with x.
right in the sense that the strength of the error term in the PNT is
equivalent to the size of the known zero free region for the zeta
function. So just one zero really can upset the PNT. But I don't see
why one zero can affect all the others though.
- Well ok, I think we've boiled it down to explaining the obvious by now.
Yes, a non-critical zero makes a larger contribution. It's not
necessarily the end of the world though, it all depends what you're
trying to prove.
On Wed, Aug 06, 2003 at 09:58:29AM +0100, Jon Perry wrote:
> The fundamental stumbling block in Riemanns's pi(x) estimate is the presence
> of the
> terms. Riemann claims that these oscillate in sign, and therefore
> effectively cancel each other out, and do not need to be included in the
> final pi(x) sum. One of his reasons for the claim is the Riemann Hypothesis.
> But if a root does not lie on the critical line, then a contributing factor
> is x^(1/4) say, not x^(1/2), which dominates the sum, and prevents the
> cancelling out process, because as x grows, this one term grows quicker than
> the others.