User ralph furmaniak - MathOverflow

Answer by Ralph Furmaniak for Liouville's theorem with your bare hands

2012-12-20T19:53:44Z

If your definition of holomorphic is that there is an absolutely convergent power series $\sum_{n=0}^{\infty} a_n z^n$ then Parseval's theorem implies that the mean square of the function on the circle of radius $R$ is $\sum_{n=0}^{\infty} |a_n^2| R^{2n}$ and it follows that $a_n=0$ for all $n>0$.

Edit (new approach)

Without loss of generality let's assume that $f(0)=0$ and $f'(0)=0$. Consider the function $f(x)/x$, continuous on all of $\mathbb{C}$. It is analytic on the punctured plane and converges to 0 on its boundary, hence by the maximum principle must be 0.

You can drop the $f'(0)=0$ condition above if you assume instead that $f$ is twice differentiable at 0, so then $f(x)/x$ is holomorphic on all of $\mathbb{C}$ and decreases to 0, so must be identically 0.

This does presuppose the maximum principle.

Provable zero-free region for any entire function that analytically is similar to zeta(s)

2012-11-03T00:02:57Z

Is there an entire function $f:\mathbb C\rightarrow\mathbb C$ such that for some $\delta>0$:

$f(z)$ is bounded when $\Re z>1+\delta$
$f(z)$ is unbounded when $\Re z=1$
$f(z)$ grows polynomially in vertical strips, ie for all $\sigma$ there is $C_\sigma$ so that $|f(\sigma+i t)|\ll|t|^{C_\sigma}$
$f(z)$ does not vanish when $\Re z>\frac12$ (provably!).

Conjecturally there is a very rich family: $L$-functions, but (4) is unproven.

If you drop (2), $1+e^{-z}$ works, or $\zeta(1/2+i t)$, or Wang zeta functions

If you drop (3), the Selberg zeta functions works, or $\exp(L(s))$

Edit: Note that $\zeta(s)$ is not entire so you can instead look at $\zeta(s)(s-1)/(s+2)$ or change the question to allow finitely many singularities, where the bounds are taken away from the singularities.

Answer by Ralph Furmaniak for A question that arises in trying to make mathematically precise a well known informal statement about analytic functions

2012-11-10T01:01:43Z

As mentioned, if you know that your points come from sampling a holomorphic function then you can use limits of finite differences to compute all of the derivatives at the limit point. Or do it step by step, compute $f(0)=\lim_{z\rightarrow0}f(z)$, then replace all points $(z,y)$ with $(z,(y-f(0))/z)$ and continue.

However, without assuming you're starting with a holomorphic function there is very little you can say using limits. For example you can take $f(z)=\exp(-z^2)$ and take $x_i$ converging to 0 from the right, and you'd get the power series expansion at $0$: $f(z)=0+0z+0z^2+\cdots$. Even worse, you could add a tiny random error to each sample so as to not effect any of the limits, but still mess up the function.

Answer by Ralph Furmaniak for Uncertainty principle (really for Mellin, but never mind that!)

2012-11-09T00:57:01Z

How about $$\frac{e^{-e^x}}{1+\epsilon x^2}$$ If you compute the Fourier transform you can shift the contour to height $\pm\pi/2$ to get an $e^{-|t|}$ times something decaying to 1, by Riemann-Lebesgue lemma

Edit.

Or you can look at a shift of your original example: $f(x+\log A)$ to get something on the lines of $$e^{-A e^x} e^{-B x^2} e^{C x}$$

Properties of a matrix-valued generalization of the $\Gamma$ function

2011-10-20T07:41:08Z

I am interested in the following function (Mellin transform of matrix exponential): $$\int_0^{\infty} x^{s-1} e^{-A-Bx}d x$$ Where $x$ and $s$ are scalars, but $A$ and $B$ are matrices with $B\succ 0$.

When $A$ and $B$ commute then it is easy to compute: simultaneously diagonalizing gives gamma functions on the diagonal. Are there any other cases in which you can say something about this function? Has it been studied? Is there an analytic continuation?

Answer by Ralph Furmaniak for Positive polynomials

2011-06-16T14:20:27Z

As mentioned, $P(x,y,z)$ being positive for positive x,y,z is the same as $P(x^2,y^2,z^2)$ always being positive, so we can just consider the case of positive polynomials. The single-variable case is a lot easier since a polynomial being non-negative is equivalent to it being a sum of squares, and there are good algorithms for decomposing a polynomial (even multivariate) as a sum of squares. There's an overview of this at: http://junction.stanford.edu/~lall/data/engr210b_0405/sum_of_squares_2004_11_07_01.pdf

In the multivariable case the positivstellensatz (also see: Hilbert's Seventeenth Problem http://en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem) tells us that the polynomial isn't necessarily a sum of squares, but that if you multiply it by some appropriate square you can get a sum of squares. These polynomials are sometimes called (semi-)definite polynomials.

We have no real good way of recognizing definite polynomials, which is to be expected since this problem is NP-complete (you can encode SAT in terms of arithmetic on variables being 0 or 1)

Why is Mellin-inverse of Gamma periodic?

2011-05-31T21:52:24Z

Specific Case

The periodicity is obvious from computation: $$\cal{M}^{-1}\{\Gamma\}(x) := \frac{1}{2\pi i}\int_{c}\Gamma(s)x^{-s}d s=e^{-x}$$ However, is there a way to see directly from the integral that it should be periodic in $x$ with period $2\pi i$?

One thing you can read off of the integral is that it satisfies a differential equation, although I don't see why this would give periodicity.

One way to tackle this could be by using a $\beta$-integral to prove multiplicativity of the function

Generalization

The above function naturally lives on $\mathbb{C}/(2\pi i\mathbb{Z})$.

What about $\cal{M}^{-1}\{\prod_{i=1}^n\Gamma(\lambda_i s+\mu_i)\}$ for some $\lambda_i>0,\mu_i\in\mathbb{C}$. In certain cases this function seems to naturally live on a quotient of a Siegel space or on a symmetric space. Is there some general result about this?

Answer by Ralph Furmaniak for Optimal tax Rate

2011-05-31T20:13:31Z

For person $i$ with income $I_i$ let $w_i$ be 1 or -1 depending on whether said person lives in $A$ or $B$. Then the condition that the person is living in the right place is a linear constraint on $w_i$ and the other $w$. To make this tractable we can allow $w_i$ to be in the interval $[-1,1]$, since the optium will still be at the endpoints (you can also interpret this is a probability that a person with tht income woul dbe in $A$ or $B$. This gives a linear system for equilibrium. Unfortunately when $M>0$ there will not be a single equilibrium (as an extreme, when $M$ is sufficiently large any state is an equilibrium). If $M=0$ you get a unique equilibrium (or if you somehow pick a canonical equilibrium) and you could try to do a numerical search on $T$ (ternary search?) Another way is to pick a target median income, form a linear constraint that the median is at least this target, and do a binary search to find the best feasible median.

A discussion of the applicability of this to economics is beyond the scope of this post :)

Answer by Ralph Furmaniak for Feasible space of SDP

2011-05-31T17:54:25Z

In this case isn't the convex hull simply a simplex? In this case you are looking for SDPs that just reduce to linear programs.

As an aside, curved boundaries are in a way most natural for SDPs: for example it is known that any convex algebraic region whose boundary is everywhere either strictly convex or positively curved is SD representible (see Semidefinite Representation of Convex Sets by Helton and Nie) (nb: the dimension of this representation is astronomical)

Answer by Ralph Furmaniak for Approximate functional equation for Dirichlet eta, does any exist?

2011-05-31T17:43:31Z

First of all, if you want just a single sum up to T, then just like with the zeta function you have an approximation: $$\eta(s)\sim\sum_{n=1}^{T/2\pi}\frac{(-1)^{n-1}}{n^s}$$

It turns out additionally that the full approximate functional equation (aka Riemann-Siegel formula) holds for really anything where you have a functional equation. You can see, for example, "The approximate functional equation for a class of zeta-functions" by Chandrasekharan and Narasimhan.

In this case, you can write the functional equation as relating $1-2^{-s}+3^{-s}-\cdots$ and $0.5^{-s}-1.5^{-s}+2.5^{-s}-\cdots$ (this form does not have an annoying $1-2^s$ term in it) so this latter term is what you find in the final formula: $$\eta(s) \sim \sum_{n\le x}\frac{(-1)^{n-1}}{n^s} - \chi(s)\sum_{n\le y}\frac1{(n-1/2)^s}$$

Also, any of the methods for proving the formula for $\zeta$ carry over naturally to $\eta$ (sometimes more naturally), so let me know if you had a particular one in mind.

Comment by Ralph Furmaniak

2013-06-15T22:08:46Z

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

Comment by Ralph Furmaniak

2012-12-20T20:25:07Z

Added a "first principles" proof :-)

Comment by Ralph Furmaniak

2012-12-20T20:24:31Z

Can you prove Riemann's theorem "with your bare hands" or does it require Cauchy/Morera ?

Comment by Ralph Furmaniak

2012-12-03T00:19:55Z

Indeed, for k=2 you are looking at $m$ order statistics, which is linear in $n$.

Comment by Ralph Furmaniak

2012-11-27T19:35:30Z

Thanks for showing this; it's a great proof that I hadn't seen before! However, isn't this just single variable complex analysis?

Comment by Ralph Furmaniak

2012-11-27T01:43:07Z

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.

Comment by Ralph Furmaniak

2012-11-12T08:21:06Z

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.

Comment by Ralph Furmaniak

2012-11-12T01:37:22Z

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.

Comment by Ralph Furmaniak

2012-11-12T01:36:06Z

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

Comment by Ralph Furmaniak

2012-11-11T23:39:41Z

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.

Comment by Ralph Furmaniak

2012-11-11T03:35:42Z

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

Comment by Ralph Furmaniak

2012-11-10T21:39:51Z

@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 $Q$

Comment by Ralph Furmaniak

2012-11-10T20:03:51Z

$e^{-e^{C x}}$ ends up hurting one of the bounds, but I think with $e^{-A e^x}$ you help the first super-exponentially without hurting the second at all, and when you add in the $e^{-x^2}$ you end up beating the second bound by something like $e^{-(\log t)^2}$

Comment by Ralph Furmaniak

2012-11-10T19:18:58Z

There's plenty more hypotheses that can be added, but I'm very interested in even a cheating solution, especially since the bounty expires in 2 days and there's been no ideas. For later, one may want $f(\sigma+i t)$ to have a tight min and max of $1-\sigma$ and $(1-\sigma)^{-1}$ respectively when $\sigma>1$. You may want large values on the 1-line to have positive density. You may want the mean square to exist on vertical lines between 1/2 and 1

Comment by Ralph Furmaniak

2012-11-09T18:48:32Z

You can also replace the $1+x^2$ with any function growing in the strip, the best example being along the lines of $e^{A e^x}$. In fact, if you translate your original example you get something decaying much more quickly, without changing the modulus of the Fourier transform: $$e^{-A e^x}e^{B x}$$