4 ms·
A bit rusty on the math, but taking a quick gander at the Wikipedia page (https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Newman_constant https://en.wikipedia.o
by vincentchu 9y ago
A bit rusty on the math, but taking a quick gander at the Wikipedia page (https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Newman_constant https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Newman_const...):
"In brief, the Riemann hypothesis is equivalent to the conjecture that Λ ≤ 0."
This result seems to imply that the De Bruijn-Newman constant (Λ) is non-negative, i.e., >=0. Thus, were one able to prove that Λ=0, then one would have also shown the Riemann Conjecture to be true? Similarly, if one were to show Λ to be not equal to 0, then the Riemann Conjecture would be false.
Again, just a quick reading. Sure others who are better/smarter can chime in!
Update: Tao expands on this point in the bottom of his blog post.
- CogitoCogito 9y agoYes so if you want to prove the Riemann hypothesis one possibility would be to similarly show some sort of contradiction if Λ > 0. This is sort of the opposite approach that Rogers and Tao have appeared to use. Of course maybe that side is much much harder (it is equivalent to the Riemann hypothesis after all).
- vincentchu 9y agoI have essentially no real knowledge about this topic, but I suppose some progress has been made by tightening the bounds for this constant? Can somebody with more info chime in--- is this a huge, groundbreaking amount of progress?
- CogitoCogito 9y agoI would presume that entirely different methods would be needed to prove that the constant can't be positive. There are many types analytical proofs in math that break down to "negative", "zero" and "positive" cases and often a couple of those are relatively easy whereas others are extremely difficult. The Calabi conjecture is an example. So really it's hard to say. You might even say this proof made things harder. It guarantees that you have to prove the constant is 0 if you want to prove the conjecture. Before you might have hoped prove negativity.
- cookingrobot 9y agoRight, near the bottom of the paper he writes “...this result does not make the Riemann hypothesis any easier to prove, in fact it confirms the delicate nature of that hypothesis”
- Someone 9y agoReading https://en.wikipedia.org/wiki/De_Bruijn–Newman_constant https://en.wikipedia.org/wiki/De_Bruijn–Newman_constant, it moves the bound from −1.1×10−12 (probably an approximation) to zero. So, not huge in absolute terms, but on the other hand, huge, as it removes all remaining negative numbers.
- tzahola 9y agoHmm. So the Riemann hypothesis requires Λ to be <= 0. Now they’ve showed that Λ >= 0. Since the intersection of the two ranges is non-measurable, probabilistically the Riemann hypothesis must be false! :^)
- nimish 9y ago{0} is measurable (It's a borel set) with the standard lebesgue measure and has measure 0. Of course, with some other measure it might not be measurable, but that would be a weird measure.
- RBerenguel 9y agoBeen thinking for a while how such a measure would be defined, but my measure theory abilities are a bit rusty (heck, unused for 10 years already, time flies). I'd bet it's impossible (in \mbb{R})... and sadly I'll now keep thinking about it until I remember how to prove for sure it or fall asleep, thanks :D
- deleted 9y ago[deleted]
- vostok 9y agoThe Lebesgue measure is not defined for all subsets of R. It is defined on the Lebesgue sigma algebra which is larger than the Borel sigma algebra. You could instead choose a sigma algebra that does not contain {0}. Trivially, you could choose the standard Lebesgue measure restricted to {{}, R}.
- RBerenguel 9y agoOh well, I was thinking of a measure where 0 was in some Borel set of it, otherwise it loses a bit the fun of it
- vostok 9y ago