bio | website | |
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location | Stanford, California | |
age | ||
visits | member for | 3 years, 6 months |
seen | Mar 26 at 1:29 | |
stats | profile views | 580 |
Mar 6 |
answered | Does this 'alternating' Euler product converge for all $\Re(s) > 0$? |
Jun 25 |
awarded | Yearling |
Jun 25 |
awarded | Promoter |
Jun 15 |
comment |
A strange matrix equality
Is this an actual identity? If you take A to have zero trace and B to have nonzero trace then the above implies AAB=BAA |
Dec 20 |
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Liouville's theorem with your bare hands
Added a "first principles" proof :-) |
Dec 20 |
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Liouville's theorem with your bare hands
Can you prove Riemann's theorem "with your bare hands" or does it require Cauchy/Morera ? |
Dec 20 |
revised |
Liouville's theorem with your bare hands
added 535 characters in body |
Dec 20 |
answered | Liouville's theorem with your bare hands |
Dec 3 |
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There are n horses. At a time only k horse can run in the single race. How many minimum races are required to find the top m fastest horses?
Indeed, for k=2 you are looking at $m$ order statistics, which is linear in $n$. |
Dec 1 |
awarded | Enthusiast |
Nov 27 |
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Fundamental motivation for several complex variables
Thanks for showing this; it's a great proof that I hadn't seen before! However, isn't this just single variable complex analysis? |
Nov 27 |
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Solve equation with matrix variable
How closed form would you want it to be? Even in the $1\times1$ case of real variables, this reduces to finding roots of polynomials, for which there is no closed form past degree 5. |
Nov 12 |
awarded | Scholar |
Nov 12 |
accepted | Provable zero-free region for any entire function that analytically is similar to zeta(s) |
Nov 12 |
comment |
Provable zero-free region for any entire function that analytically is similar to zeta(s)
Thank you fedja for the solution, it is exactly what was requested. And thanks to everyone else who commented; it is great to see interest in this question and I hope this is just the beginning. What makes the above solution "easy" is that the zeros can escape to the left. This is closely tied to how we don't dictate growth rate to the left. This makes a big difference. For example, knowing the growth rate between 1.5 and 2, and between -0.5 and -1 allows you to prove that the gaps between zeros is at most $1/\log_3(T)$ even if the zeros have complex powers, so end up as singularities. |
Nov 12 |
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Uncertainty principle (really for Mellin, but never mind that!)
It's also known that if you have an entire function of a given exponential type, then if it is bounded on a wide enough sector then it has to be constant. |
Nov 12 |
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Uncertainty principle (really for Mellin, but never mind that!)
There are some nice consequences of Phragmen-Lindelof (and its proof) along these lines in Titchmarsh's Theory of Functions. For example there's the following result of Carlson. Suppose $f(z)$ is holomorphic in some sector of interior angle $\theta$, is exponentially bounded $|f(z)|\ltlt e^{k|z|}$ and exponentially decays on the boundary. $\exp$ gives an example for $\theta<\pi$ and the result is that if $\theta=\pi$ then $f(z)\equiv0$. In fact, having such a non-trivial function would give you a stronger Phragmen-Lindelof principle that could prove that the exponential function does not e |
Nov 11 |
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Provable zero-free region for any entire function that analytically is similar to zeta(s)
There's a type of argument one may pursue to move the zeros to a line. Suppose you have a nice function with given zeros on a line, nice in the sense of having a Hadamard Product. Then by taking the logarithmic derivative you can express the $f'(z)/f(z)$ as a sum of $\frac1{z-t}$ where $t$ ranges over the zeros. This is an integral of $\frac1{z-t}$ with respect to a discrete measure. But any function decaying to the right and $L^2$ on vertical lines can be expressed as such an integral with a continuous measure, by Cauchy integral formula. Pick a discrete measure that approximates this. |
Nov 11 |
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Provable zero-free region for any entire function that analytically is similar to zeta(s)
That is indeed an answer and fits all of my requirements. The one thing I should have perhaps added is that the function should be order 1 in the sense of growing like $exp(z^{1+\epsilon})$, so that it has a nice Hadamard Product and can be analyzed in terms of its zeros. Now I need to understand how we can make your function look, in particular the growth rate as a whole. The zeros of $F(z)$ are near the 0 axis but moving away at a rate of $\log T$. Regardless of the $y_n$ there will be infinitely many in horizontal strips, Lthough the exact choice of $y_n$ will effect the density |
Nov 10 |
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Provable zero-free region for any entire function that analytically is similar to zeta(s)
@Juan, right: having on the order of $T\log T$ zeros would be good. This is close to the statement of polynomial growth. It would be good to add that the function itself should be bounded as $|f(z)|\lt\lt|z|^{d|z|}Q^|z|$ for some $d$ and $Q$ |