38,025 reputation
13167256
bio website math.ucla.edu/~tao
location Los Angeles
age 40
visits member for 5 years, 10 months
seen 11 hours ago
Professor of Mathematics at UCLA

2d
comment $\frac{1}{2}<\sigma<1$, is $f(n) = \Bigl| \,1+ \frac{1}{2^{\sigma + i n}} + \cdots + \frac{1}{n^{\sigma + i n}} \Bigr|$ from $O(\log n)$?
The growth of zeta (after deleting the pole at s=1) on the critical line is related to the growth on other lines by Lindelof's theorem: en.wikipedia.org/wiki/Lindel%C3%B6f%27s_theorem . Note also that Montgomery's paper cited above also applies to lines to the right of the critical line.
Aug
27
comment Why should we care about “higher infinities” outside of set theory?
Perhaps an analogy will help. Your question seems to me to be similar to "why should we care about $0.999\dots = 1$ outside of analysis?". On the one hand, it would be rare to have a situation outside of analysis in which one could start with the assumption $0.999\dots \neq 1$ and accidentally end up with a more blatant contradiction. On the other hand, the effort required to purge $0.999\dots = 1$ from being a consequence of the foundations of mathematics outside of analysis, would greatly reduce the ability to efficiently work on those areas for very little gain.
Aug
27
comment Why should we care about “higher infinities” outside of set theory?
I think it actually quite hard to fence off the Power Set Axiom (and a fortiori, the existence of different uncountable cardinalities) from the rest of mathematics. For instance, in analysis, establishing the Hahn-Banach theorem or the existence of the Stone-Cech compactification requires pretty much all of the axioms of set theory. To me, it's not so much that the multiplicity of uncountable cardinals comes up directly all that much in mathematics, but rather that they are inevitable byproducts of the very useful set theory axioms that we rely quite heavily in mathematics.
Aug
27
comment Why should we care about “higher infinities” outside of set theory?
Given that Cantor's theorem is more or less immediate from the Power Set Axiom together with relatively uncontroversial axioms, perhaps the real question here is "why should we admit the Power Set Axiom in mathematics outside of set theory?"
Aug
26
awarded  Notable Question
Aug
26
comment $\frac{1}{2}<\sigma<1$, is $f(n) = \Bigl| \,1+ \frac{1}{2^{\sigma + i n}} + \cdots + \frac{1}{n^{\sigma + i n}} \Bigr|$ from $O(\log n)$?
However, mean value theorems will show that the desired bound is true for "most" values of n. See e.g. Titchmarsh's book on the zeta function.
Aug
26
comment $\frac{1}{2}<\sigma<1$, is $f(n) = \Bigl| \,1+ \frac{1}{2^{\sigma + i n}} + \cdots + \frac{1}{n^{\sigma + i n}} \Bigr|$ from $O(\log n)$?
Actually, from looking at Montgomery's paper ams.org/mathscinet-getitem?mr=460255 it seems that the bound $\zeta(\sigma+it) = O(\log t)$ cannot hold for all sufficiently large $t$; presumably the argument can extend to the case when $t$ is restricted to be integer.
Aug
26
comment $\frac{1}{2}<\sigma<1$, is $f(n) = \Bigl| \,1+ \frac{1}{2^{\sigma + i n}} + \cdots + \frac{1}{n^{\sigma + i n}} \Bigr|$ from $O(\log n)$?
This is equivalent asking for the bound $\zeta(\sigma+it) = O( \log t )$ for integer values of $t$, which is basically a variant of the Lindelof hypothesis.
Aug
23
comment Bounded gaps between primes in arithmetic progressions
The stated result also appears in this paper of Benatar: arxiv.org/abs/1305.0348 , who also studies prime gaps in other sets of positive relative density. Other relevant literature includes the work of Thorner arxiv.org/abs/1401.6677 who studies Chebotarev sets (more general than arithmetic progressions) and Maynard arxiv.org/abs/1405.2593 (who studies dense clusters of primes in well-distributed subsets). Finally there is older work of Freiberg arxiv.org/abs/1110.6624 that uses the older GPY technology to obtain a related result.
Aug
22
comment Is there a version of the Titchmarsh Convolution theorem to find singular support?
... as is done for instance in this 1977 paper of Bengel: link.springer.com/article/10.1007%2FBF01362427
Aug
22
comment Is there a version of the Titchmarsh Convolution theorem to find singular support?
Gah, you're right. One needs the additional observation that the convolution of two distributions that are one-dimensionally smooth on non-parallel line segments will be two-dimensionally smooth except on the boundary of the supporting parallelogram, and upon taking another derivative one gets smooth measures on the one-dimensional boundary edges plus some zero-dimensional stuff on the vertices. This should eventually let one arrive at the same 1-skeleton that you have in your answer. Probably the "right" way to proceed is via wave front sets, though...
Aug
22
comment Matrix equation $XAXBXC=I$
By making the change of variables $X = A^* \tilde X$ one may reduce without loss of generality to the case $A=I$, or equivalently to the equation $BXC = X^{-2}$.
Aug
22
comment Is there a version of the Titchmarsh Convolution theorem to find singular support?
One can also use the theory of distributions. The third (distributional) derivative $\nabla^3(f*f*f)$ of $f$ is equal to $\nabla f * \nabla f * \nabla f$ (distributing indices appropriately). $\nabla f$ is supported (again, as a distribution) on the boundary of the hexagon, which is compact, hence $\nabla f * \nabla f * \nabla f$ is supported on the Minkowski sum of three copies of this boundary, and so these are the only places where $f*f*f$ can be singular.
Aug
22
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Oh, good point, so it's still a 3D flow and so potentially rather complicated, unless one can use the asymptotic regime to simplify things by focusing on the top order terms. This approach, by the way, should also help explain the cardiod shape you observed for the other zeroes.
Aug
21
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
We only need to check the cases $n > 6000$ so one can also hope to do some asymptotic analysis, focusing on the terms with the largest powers of $n$.
Aug
21
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Differentiating the identity $x^2 f' = (1-x^2) f$ a total of $n$ times gives the recurrence $\phi_{n+1} = (x^2 - 2nx - 1) \phi_n + nx (-(n-1)x-2) \phi_{n-1} - n(n-1) x^2 \phi_{n-2}$ (I think). This can be combined with the identity $\phi_{n+1} = (x^2-1) \phi_n - x^2 \phi'_n$ to get a third order linear ODE for $\phi_n$, so the curve $(\phi_n(x), \phi'_n(x), \phi''_n(x))$ follows some explicit vector field in ${\bf R}^3$, which can be projected down to ${\bf RP}^2$. Perhaps the topology of this vector field is giving the claim?
Aug
15
comment What is the $q$-analog of $\Gamma(z)\Gamma(1-z)=\frac\pi{\sin(\pi z)}$?
This is a very different use of the letter "q", but the "q-aspect" version of the identity $\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin \pi z}$ in analytic number theory is the identity $\tau(\chi) \tau(\overline{\chi}) = q \chi(-1)$ for Gauss sums $\tau(\chi) = \sum_{n \in {\bf Z}/q{\bf Z}} \chi(n) e(n/q)$ of Dirichlet characters of conductor $q$. (Gauss sums show up in the functional equation for Dirichlet L-functions in much the same way that Gamma factors show up for the Riemann zeta function.)
Aug
6
awarded  Good Answer
Aug
5
comment Are limits decidable? Should definitions be decidable?
As for a more computable notion of a limit, you may be interested in the concept of metastability, which I discuss at terrytao.wordpress.com/2007/05/23/…
Aug
5
comment Are limits decidable? Should definitions be decidable?
However, one can informally interpret "existential" axioms (such as the axiom of choice or the least upper bound axiom) from a "computational" perspective as providing various "oracles" that extend a base computational model. I discuss this (rather non-rigorously) at terrytao.wordpress.com/2010/03/19/…