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Aug
12 |
awarded | Nice Question |
Nov
12 |
awarded | Popular Question |
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 |
comment |
Liouville's theorem with your bare hands
Added a "first principles" proof :-) |
Dec
20 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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. |