User carlo beenakker - MathOverflow

Answer by Carlo Beenakker for Derivative of a determinant of a matrix field

2013-05-24T17:48:40Z

your identity follows simply by using $\log({\rm det}\; A)= {\rm tr}\; (\log A)$, so

$$\frac{\partial}{\partial x_i}{\rm det}\;A= \frac{\partial}{\partial x_i} \exp({\rm tr}\;\log A)= ({\rm det}\;A) \frac{\partial}{\partial x_i}{\rm tr}\;\log A= ({\rm det}\;A)\;{\rm tr}\;\left(A^{-1}\frac{\partial}{\partial x_i}A\right)$$

this identity is known as Jacobi's formula.

Answer by Carlo Beenakker for Matrix norms / eigenvalues / singular values / another thing

2013-05-22T21:24:55Z

this is called the numerical radius of a matrix

Answer by Carlo Beenakker for Did Oresme know the zeroth power?

2013-05-22T15:35:42Z

According to Florian Cajori, "A history of mathematical notations", page 100, it was Nicolas Chuquet in 1484 who introduced the exponential notation in "Le triparty en la science des nombres", including positive, negative and zero powers. This is more than a century after Oresme, who interpreted roots as fractional powers, but did not have the notion of raising to the zeroth (or negative) power.

The explicit notation which Chuquet used was $.12.^2$ for what we would write as $12x^2$. So the exponent is written without its base. He multiplies $.12.^0$ by $.10.^2$ to obtain $.120.^2$, correctly evaluating $x^0$ times $x^2$ as $x^2$, hence interpreting $x^0$ as unity. Negative powers are indicated by a letter $m$ (moins), as in $12.^{1.\tilde{m}.}$ for $12x^{-1}$. Algebraic multiplication, involving the product of the coefficients and the sum of the exponents, is a familiar process with Chuquet. As Cajori writes:

Chuquet elaborates the exponential notation to a completeness apparently never before dreamed of. On this subject Chuquet was about one hundred and fifty years ahead of his time; had his work been written at the time when it was written, it would, no doubt, have greatly accelerated the progress of algebra. As it was, his name was known to few mathematicians of his time.

Answer by Carlo Beenakker for Exact solutions to nonlinear Klein-Gordon equation

2013-05-22T12:53:52Z

regularization of the Klein-Gordon equation proceeds in the same way as for the Schrödinger equation; you restrict $x$ to the interval $(0,L)$ and impose periodic boundary conditions $\phi(0,t)=\phi(L,t)$; this quantizes the wave vector $p=p_n(t)$, $n\in\mathbb{Z}$ --- for the linear Schrödinger equation the quantization would be time independent, now it depends on time; in the limit $L\rightarrow\infty$ you would ignore the quantization, calculate the energy $E_L$ by integrating $x$ from $0$ to $L$, and then obtain a finite limit $L^{-1}E_L$; this just means that the energy density per unit length is finite, even though the entire energy content of the infinite system is infinite (as it should be).

Answer by Carlo Beenakker for The relations between the Perelman's entropy functional and notions of entropy from statistical mechanics

2013-05-18T18:55:13Z

Perelman himself wrote about his entropy formula for the Ricci flow that "The interplay of statistical physics and (pseudo)-riemannian geometry occurs in the subject of Black Hole Thermodynamics, developed by Hawking et al. Unfortunately, this subject is beyond my understanding at the moment."

Subsequently, this connection has been explored in some detail in two papers by Samuel and Chowdury: "Geometric flows and black hole entropy" and "Energy, entropy and the Ricci flow". For a discussion of both these papers, see chapter 6 of "On the Emergence Theme of Physics" by Robert Carroll.

Perelman has given a gradient formulation for the Ricci flow, introducing an "entropy function" which increases monotonically along the flow. We pursue a thermodynamic analogy and apply Ricci flow ideas to general relativity. We investigate whether Perelman's entropy is related to (Bekenstein-Hawking) geometric entropy as familiar from black hole thermodynamics. From a study of the fixed points of the flow we conclude that Perelman entropy is not connected to geometric entropy. However, we notice that there is a very similar flow which does appear to be connected to geometric entropy. The new flow may find applications in black hole physics suggesting for instance, new approaches to the Penrose inequality.

Answer by Carlo Beenakker for Derivation of Bessel functions

2013-05-18T14:56:36Z

I'll make an attempt at providing the steps you are seeking to go "from Euler equation to Bessel function".

You start from the Euler equation, describing conservation of momentum,

$$\rho\frac{\partial \vec{u}}{\partial t}+\rho\vec{u}\cdot\nabla\vec{u}=-\nabla p$$

and the continuity equation, describing conservation of mass,

$$\frac{\partial\rho}{\partial t}=-\nabla\cdot(\rho \vec{u})$$

These are nonlinear equations, to make them tractable you'll want to linearize them, both in the velocity $\vec{u}$ and in the deviations $\delta\rho=\rho-\rho_0$ of the density from the uniform density $\rho_0$. This approximation throws away lots of interesting physics (shock waves, turbulence,...), but without it no simple solution exists.

The linearized equations read

$$\rho_0\frac{\partial \vec{u}}{\partial t}=-\nabla p$$

$$\frac{\partial\delta\rho}{\partial t}=-\rho_0\nabla\cdot\vec{u}$$

We may also assume a linear relation $p=p_0+C^2\delta\rho$ between the pressure $p$ and the density variations. (This is a socalled adiabatic equation of state, the coefficient $C^2$ must be positive for mechanical stability.) We define $\xi=\delta\rho/\rho_0$, take the divergence of the first equation and the time derivative of the second equation,

$$\nabla\cdot\frac{\partial \vec{u}}{\partial t}=-C^2\nabla^2 \xi$$

$$\frac{\partial^2\xi}{\partial t^2}=-\frac{\partial}{\partial t}\nabla\cdot\vec{u}$$

Finally, we substitute the first equation into the second one, exchanging the order of differentiation with respect to time and space, to arrive at a wave equation for $\xi$,

$$\frac{\partial^2\xi}{\partial t^2}=C^2\nabla^2 \xi$$

The quantity $C>0$ represents the speed of sound.

We seek a solution of this equation that is a harmonic function of time, so it oscillates with frequency $\omega$. Rather than working with sines or cosines, it is more convenient to use a complex notation, writing

$$\xi(\vec{r},t)={\rm Re}\;e^{-i\omega t}f(\vec{r})$$

The complex function $f$ satisfies the Poisson equation,

$$C^2\nabla^2 f=-\omega^2 f$$

Let's seek a solution with cylindrical symmetry, so $f(R)$ depends only on the radial coordinate $R=\sqrt{x^2+y^2}$. The Poisson equation in cylindrical coordinates takes the form

$$\frac{d^2}{dR^2}f(R)+\frac{1}{R}\frac{d}{dR}f(R)=-(\omega/C)^2f$$

The solution is a Bessel function

$$f(R)={\rm constant}\times J_0(\omega R/C)$$

The full solution thus becomes

$$\delta\rho/\rho_0=A\cos(\omega t+B)J_0(\omega R/C)$$

where $A$ and $B$ are arbitrary coeffients.

And we're done :)

Answer by Carlo Beenakker for computing Bernoulli numbers

2013-05-16T17:34:21Z

these are socalled "scaled" Bernoulli numbers; an efficient algorithm is discussed by Brent & Harvey, arXiv:1108.0286, Eq. 8.

alternatively, you can use any method that computes the Riemann zeta function (say via the Euler product), because of the identity

$$\frac{B_{2n}}{(2n)!}=(-1)^{n+1}2\zeta(2n)(2\pi)^{-2n}$$

(note that $B_{2n+1}=0$ for $n\geq 1$)

Answer by Carlo Beenakker for help on antiderivative of a vector function

2013-05-15T20:59:05Z

I'm sure I'll be quickly corrected if I'm wrong, but I don't think that a scalar function $g(\vec{x})$ exists, such that $\partial/\partial x_i g(\vec{x})=S_{ij}x_j/||\vec{x}||$ (summation over repeated indices implied). A tensorial function does exist:

$$\frac{\partial}{\partial x_{j}}S_{ij}||\vec{x}||=S_{ij}x_j/||\vec{x}||.$$

Answer by Carlo Beenakker for Eigenvalues of random Hamiltonian matrices

2013-05-13T08:23:22Z

In the course of a physics project in my group, I have had an opportunity to learn more about the eigenvalue statistics of Hamiltonian matrices. (Our physics problem actually involved skew-Hamiltonian matrices, so I made a small detour, joined by Jonathan Edge & Jan Dahlhaus.)

The ensemble is the one you suggested: $2n\times 2n$ real matrices $H$ with Hamiltonian symmetry and normally distributed elements. It is convenient to rescale the eigenvalues $\varepsilon_k$ of $H$ by a factor $\sqrt{2n}$, and separate the real and imaginary parts:

$(2n)^{-1/2}\varepsilon_k=x_k+iy_k$.

The eigenvalue density in the complex plane $x+iy$ consists of three parts: a two-dimensional density $\rho_{c}(x,y)$ of the complex eigenvalues, a one-dimensional density $\rho_{r}(x)$ of the real eigenvalues and another one-dimensional density $\rho_{i}(y)$ of the imaginary eigenvalues.

Based on numerical experiments, I can offer three conjectures:

1) For large $n$, the rescaled complex eigenvalues $x_k+iy_k$ uniformly cover a disc of unit radius,

$\lim_{n\rightarrow\infty}n^{-1}\rho_{c}(x,y)=2/\pi$ for $x^2+y^2<1$.

2) For large $n$, the rescaled real eigenvalues $x_k$ uniformly cover the interval $-1<x<1$ , with density

$\lim_{n\rightarrow\infty}n^{-1/2}\rho_{r}(x)=1/\sqrt{\pi}$ .

Therefore the expectation value of the number $n_{r}$ of real eigenvalues satisfies $\lim_{n\rightarrow\infty}n^{-1/2}E[n_{r}]=2/\sqrt{\pi}$.

3) Also the rescaled imaginary eigenvalues $iy_{k}$ have a uniform density in the large-$n$ limit, in the interval $-1<y<1$ , but this density is less than the density of the real eigenvalues. The expectation value of the number $n_{i}$ of imaginary eigenvalues satisfies $\lim_{n\rightarrow\infty}n^{-1/2}E[n_{i}]={\rm constant}\approx 0.72$ .

Conjectures 1 and 2 were proven by Edelman and collaborators in the absence of Hamiltonian symmetry, so when all $(2n)^{2}$ real matrix elements of $H$ are chosen from independent normal distributions. [This is known as the (real) Ginibre ensemble.] Our numerics suggests that, for large matrices, the Hamiltonian symmetry only affects the (rescaled) eigenvalue distribution within a distance of order $n^{-1/2}$ from the imaginary axis.

By way of illustration, I include a plot of the eigenvalues $\varepsilon$ of $200$ real matrices of size $100\times 100$ (so $n=50$), with normally distributed matrix elements, both with the Hamiltonian symmetry (left) and without (right). These eigenvalues are shown without rescaling, so they cover a disc of radius $\sqrt{2n}=10$.

Answer by Carlo Beenakker for Principal value of integral

2013-05-14T14:04:31Z

formula 3.749.2 from Gradshteyn & Ryzhik gives:

$$\int_0^{\infty}\frac{1-x\;{\rm cotan}x}{x^2+\epsilon^2}dx=\frac{\pi}{2\epsilon}-\frac{\pi}{e^{2\epsilon}-1}\quad{\rm for}\quad \epsilon>0.$$

taking the limit $\epsilon\downarrow 0$ gives your $\pi/2$; G&R do not explicitly say that their formula is a principal value integral, but it's the only sensible way to avoid the poles of the cotangent at $\pi,2\pi,...$; note that there is no singularity at $x=0$, so the limit $\epsilon\downarrow 0$ gives no ambiguity.

Here's the derivation by contour integration, as promised. The integral over $1/(x^2+\epsilon^2)$ is elementary, so I only do the one involving the cotangent:

$${\cal P}\int_{0}^{\infty}dx\frac{x\;{\rm cotan}x}{x^2+\epsilon^2}= \frac{1}{4}\lim_{\delta\rightarrow 0}\left(\int_{-\infty+i\delta}^{\infty+i\delta}dx\frac{x\;{\rm cotan}x}{x^2+\epsilon^2}+ \int_{-\infty-i\delta}^{\infty-i\delta}dx\frac{x\;{\rm cotan}x}{x^2+\epsilon^2}\right)$$

$$=\frac{1}{4}\lim_{\delta\rightarrow 0}\left(\int_{-\infty+i\delta}^{\infty+i\delta}\frac{dx}{\sin x}\frac{x\;e^{ix}}{x^2+\epsilon^2}+ \int_{-\infty-i\delta}^{\infty-i\delta}\frac{dx}{\sin x}\frac{x\;e^{-ix}}{x^2+\epsilon^2}\right)$$

$$=\frac{1}{4}\left(\int_{C_+}\frac{dz}{z-i\epsilon}\frac{z\;e^{iz}}{(z+i\epsilon)\sin z}+ \int_{C_-}\frac{dz}{z+i\epsilon}\frac{z\;e^{-iz}}{(z-i\epsilon)\sin z}\right)$$

$$=\frac{1}{4}\times 2\pi i\times\left(\lim_{z\rightarrow i\epsilon}\frac{z\;e^{iz}}{(z+i\epsilon)\sin z}- \lim_{z\rightarrow -i\epsilon}\frac{z\;e^{-iz}}{(z-i\epsilon)\sin z}\right)$$

$$=\frac{1}{4}\times 2\pi i\times 2\times \frac{1}{\sin i\epsilon}\frac{i\epsilon\;e^{-\epsilon}}{2i\epsilon}$$

$$=\frac{\pi}{e^{2\epsilon}-1}$$

In the first equality I inserted the definition of principal value; in the second equality I used that $xe^{\pm ix}/\sin x = x\;{\rm cotan}x\pm ix$ and the second term vanishes upon integration because it is an odd function of $x$; in the third and following equalities I have closed the contour in the upper half of the complex plane for the first integral (contour $C_+$, picking up the pole at $z=i\epsilon$), and in the lower half of the complex plane for the second integral (contour $C_-$, pole at $z=-i\epsilon$). And so I arrive at the answer from Gradshteyn & Ryzhik, confirming that theirs was indeed a principal value result.

Answer by Carlo Beenakker for Lower bounds on derivative around zero set of a positive smooth function.

2013-05-11T12:58:07Z

Taylor expansion in $x$ of $f$ around $x=0$:

$f(x,y)=x^{n} g(y) +$ terms of order $x^{n+1}$, with $g(y)$ not identically zero.

The requirement that $f(0,y)=0$ and $f(x,y)>0$ for all $y$ and for all $x\neq 0$ implies that $g(y)\geq 0$ for all $y$ and that $n=2p\geq 2$ is a positive even integer. Choose a $y_0$ where $g(y_0)=c>0$.

Set $y=y_0$ and make a Taylor expansion in $x$ of $f_y=\partial f/\partial y$ around $x=0$:

$f_y(x,y_0)=x^{m} c' + {\rm order}(x^{m+1})$ with $m\geq n$.

(The power $m$ may be larger than $n$ if $f_y=0$ at $y=y_0$.) We thus conclude that

$\lim_{x\rightarrow 0}x^{-2p}\left[f(x,y_0)+\int_0^x f_y(s,y_0)ds\right]=c>0$,

which amounts to your statement.

Answer by Carlo Beenakker for Probability distribution for two-state system that depends on residence time

2013-05-07T21:53:58Z

first simple case: $p_+=p_-\equiv p_0$ and $\kappa_+=\kappa_-\equiv \kappa$; then all you need to know is the time $\delta t$ since the last switching event, which has an exponential distribution, hence:

$$p(x,t)=(1-e^{-\kappa t})^{-1}\int_0^t d\delta t\; \kappa e^{-\kappa \delta t}p_0(x,\delta t)$$

now the general case; you'll need to distinguish even from odd number of switching events, and find the distribution $P_{\rm even}(\delta t)$ of the time $\delta t$ since the last switching event, given that there have been an even number of switches in a time $t$, and similarly for an odd number; the a priori probability that there have been an even or odd number of switches is just given by the Poisson distribution (summing over $n$ even or $n$ odd). The integral over $P_{\rm even}(\delta t)p_{+}(x,\delta t)$ and $P_{\rm odd}(\delta t)p_{-}(x,\delta t)$ then gives the full answer.

a bit more explicit, still assuming $\kappa_+=\kappa_-\equiv \kappa$ for simplicity; it is convenient to set the first switching event at time $0$, so that the number of switching events $m=n+1$ in a time $t>0$ is $\geq 1$; the probability $P_{m,t}(\delta t)$ that there have been $m=n+1\geq 1$ switching events in a time $t$, while the last switching event was a time $\delta t\in[0,t)$ ago is given by a slight modification of the Poisson distribution,

$$ P_{m,t}(\delta t)=\frac{1}{(m-1)!}[\kappa(t-\delta t)]^{m-1}\kappa e^{-\kappa t}.$$

summing over all $m=1,2,3,\ldots$ we recover the exponential probability $\sum_{m}P_{m,t}(\delta t)=\kappa\exp(-\kappa\delta t)$ we had before, but now we have to distinguish between even $m$ (= odd $n$) and odd $m$ (= even $n$):

$$P_{\rm even}(\delta t)=\sum_{m=1,3,5}^{\infty}P_{m,t}(\delta t)=\cosh[\kappa(t-\delta t)]\kappa e^{-\kappa t}$$

$$P_{\rm odd}(\delta t)=\sum_{m=2,4,6}^{\infty}P_{m,t}(\delta t)=\sinh[\kappa(t-\delta t)]\kappa e^{-\kappa t}$$

and we're done:

$$p(x,t)=(1-e^{-\kappa t})^{-1}\int_0^t d\delta t\; [P_{\rm even}(\delta t)p_+(x,\delta t) +P_{\rm odd}(\delta t)p_-(x,\delta t)]$$

$$\quad\quad=(1-e^{-\kappa t})^{-1}\int_0^t d\delta t\; \kappa e^{-\kappa t}\left[\cosh[\kappa(t-\delta t)]p_+(x,\delta t) +\sinh[\kappa(t-\delta t)]p_-(x,\delta t)\right]$$

if we take $p_+=p_-\equiv p_0$ we recover the earlier result.

Answer by Carlo Beenakker for Non-asymptotic results for bulk of random Wishart matrix

2013-05-04T22:12:57Z

You ask for the average of the singular values of the Wishart matrix. I'm pretty sure there is no closed form expression valid for any $n$. If instead you would ask for the average of the square of the singular value, then the answer is very simple, this is just unity independent of $n$.

More generally, if $X_{n}$ has dimension $p\times n$, with $p\leq n$, and $w_k$ is an eigenvalue of $W_n=n^{-1}X_nX_n^T$, then

$$E\left(p^{-1}\sum_{k=1}^{p}w_k\right)=1,$$

regardless of the value of $p$ or $n$. This follows directly from equation 17.8.2 in Mehta's book on Random-Matrix Theory, applied to the Wishart-Laguerre probability distribution of the $w_k$'s.

Similar closed form expressions exist for the integer moments $E(p^{-1}\sum_k w_k^m)$, see this paper by Livan and Vivo. The singular values are a fractional moment ($m=1/2$). In principle these are given by an integral over Laguerre polynomials, but I do not think this integral can be carried out in closed form.

Answer by Carlo Beenakker for Does the paper "On the cobordism ring $\Omega_*$ and a complex analogue II" exist?

2013-05-03T15:07:13Z

Here's what John Milnor writes about this, in his collected works:

The projected Part II of this paper was never written. In fact I am chagrined to discover that I have never published any details about some of the announced results which were intended to appear in it. I was very grateful when Thom gave me permission to reprint his 1959 Bourbaki lecture, "Travaux de Milnor sur le cobordism", which gives a better account of this work than anything which I have published.

(page 249 of Part III of the Collected Papers of John Milnor, AMS 2007).

The paper by René Thom which Milnor mentions can be accessed here.

Answer by Carlo Beenakker for Quantum algorithms for dummies

2013-05-03T06:32:41Z

the classic reference for a computer scientist wanting to get up to speed on quantum algorithms is:

Quantum Computer Science by David Mermin.

In the 1990's it was realized that quantum physics has some spectacular applications in computer science. This book is a concise introduction to quantum computation, developing the basic elements of this new branch of computational theory without assuming any background in physics. It begins with an introduction to the quantum theory from a computer-science perspective. It illustrates the quantum-computational approach with several elementary examples of quantum speed-up, before moving to the major applications: Shor's factoring algorithm, Grover's search algorithm, and quantum error correction. The book is intended primarily for computer scientists who know nothing about quantum theory, but will also be of interest to physicists who want to learn the theory of quantum computation, and philosophers of science interested in quantum foundational issues.

Answer by Carlo Beenakker for A simple and good reference about solitons

2013-05-02T19:38:45Z

For a mathematically inclined introduction:

Glimpses of Soliton Theory: The Algebra and Geometry of Nonlinear PDEs

For real-world applications:

Waves Called Solitons: Concepts and Experiments

I would think these two books cover most of the terrain. If you're looking for a "math-free" intro, or want to refer someone to such an intro:

The Versatile Soliton.

Answer by Carlo Beenakker for Distributions of Time Derivatives of Stochastic Processes

2013-05-01T18:06:25Z

I would think the answer is "no" to both questions.

Let me abbreviate $v(t)=du/dt$, so $u(t)=u(0)+\int_0^t v(t')dt'$. Now ask for the expectation of $u^2(t)$. You'll need to know how $v(t')$ and $v(t'')$ are correlated for any $t',t''$ in the interval $(0,t)$. That information is not given.

Similarly, in the other direction, to find the distribution of the velocity $v(t)$ you need to know how the position $u(t)$ at nearby times is correlated, and that information is not given.

Answer by Carlo Beenakker for Exact sampling from 2D Ising model where coupling is constant?

2013-04-29T21:27:25Z

The entropic sampling technique can efficiently sample quite large lattices, several thousands spins. It has been applied to your Hamiltonian, in a uniform magnetic field ($b_i$ independent of $i$), in several works:

Systematic enumeration of configuration classes for entropic sampling of Ising models, Bruno Jeferson Lourenço, Ronald Dickman

Complete high-precision entropic sampling, Ronald Dickman, A. G. Cunha-Netto

Entropic sampling dynamics of the globally-coupled kinetic Ising model, Beom Jun Kim, M.Y. Choi

Answer by Carlo Beenakker for Expected edit distance

2013-04-27T18:08:54Z

The only rigorous bound I am aware of is due to Gonzalo Navarro*

$$c\geq 1-{\rm e}/\sqrt{\sigma},$$

for an alphabet of $\sigma$ characters. Obviously, for the binary string ($\sigma=2$) this bound is ineffective. Navarro also mentions a large-$\sigma$ conjecture $c=1-1/\sqrt{\sigma}$, which for the binary string would give $c=0.2929$, quite close to your numerical finding.

*G. Navarro, A guided tour to approximate string matching (2001)

Answer by Carlo Beenakker for On solution of a discrete-time equation

2013-04-28T17:53:47Z

you'll want to solve this equation iteratively, considering $S(k)$ as known and $S(k+1)$ as unknown; for $F(k)$ invertible you then have a Sylvester equation, of the form $F^{-1}(k)S(k+1)-S(k+1)F^{T}(k)=C(k)$, which has a unique solution iff $F^{-1}(k)$ and $F(k)$ have no common eigenvalue. The Wikipedia page gives algorithms to solve this equation, implemented in several software packages.

Answer by Carlo Beenakker for The discrete theory of compressible fluids dynamics

2013-04-27T18:33:17Z

There's unpublished work by Gay-Balmaz and Pavlov, Variational Discretization of Compressible Fluids, described here, with an instructive summary of the difficulties involved in extending the discrete theory from incompressible to compressible fluids.

Answer by Carlo Beenakker for What is flexible about flexible algebras?

2013-04-24T08:55:13Z

I browsed through the 1948 paper by Albert where he introduced the notion of a flexible algebra. It is presented as a property that gives a certain flexibility to how algebraic operations are to be carried out. Like all good names, it is so natural it hardly needs an explanation.

Answer by Carlo Beenakker for Expectation of the trace of an inverse of a random matrix

2013-04-23T20:25:09Z

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; including for convenience a factor $1/N$, the quantity you seek is

$$N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r\in(0,1]$ (Marchenko-Pastur distribution). I find in this way the answer

$$\lim_{N,M\rightarrow\infty}N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=(2r\gamma)^{-1}\left(-1-\gamma\sqrt{ab}+\sqrt{(1+a\gamma)(1+b\gamma)}\right)$$

with $a=(1+\sqrt r)^2$ and $b=(1-\sqrt r)^2$. As a check, you can take the limit $\gamma\rightarrow 0$ of this expression and obtain $1$, as it should.

Answer by Carlo Beenakker for What are the best known lower and upper bounds for the second Chebyshev function $\psi(x)$

2013-04-23T13:27:59Z

The most recent results on bounds for $\psi(x)$ are from this year:

Sharper estimates for Chebyshev's functions $\vartheta$ and $ψ$, February 2013.

In this article we present some improved results for Chebyshev's functions $\vartheta$ and $\psi$ using the new zero-free region obtained by H. Kadiri and the first $10^{13}$ zeros of the Riemann zeta function on the critical line calculated by Xavier Gourdon. The methods in the proofs are similar to those of the Rosser-Shoenfeld papers on this subject.

Answer by Carlo Beenakker for Who first observed that Conjugate Gradient for Symmetric Positive Definite linear systems is a Krylov method?

2013-04-23T11:18:25Z

Krylov Subspace Methods for Solving Large Unsymmetric Linear Systems, Y. Saad, Mathematics of Computation 37, 105-126 (1981).

The purpose of the present paper is to generalize the conjugate gradient method regarded as a projection process onto the Krylov subspace K„. We shall say of a method realizing such a process that it belongs to the class of Krylov subspace methods. It will be seen that these methods can be efficient for solving large nonsymmetric systems.

Answer by Carlo Beenakker for Young transform reference

2013-04-21T09:58:44Z

Your transform is a logarithmic variation on the Young-Fenchel transform, which has an extensive literature, for example:

• On the Young-Fenchel transform for convex functions

• Variational Principles of Continuum Mechanics (chapter 5 on Young-Fenchel transformations)

More generally, one can define the Fenchel-Moreau transform,

$$(\mathscr F_{\phi}\;g)(y) = -\inf_{x}\; [g(x)-\phi(x,y)], $$

with respect to a coupling function $\phi(x,y)$. The Young-Fenchel transform corresponds to a bilinear $\phi$. Choosing $\phi(x,y)=\log(\sum_{n}x_n y_n)$ and $g(x)=\log f(x)$ gives essentially your transform.

Answer by Carlo Beenakker for Simulating random sequential adsorption in reverse

2013-04-18T13:10:44Z

Some realizations of surface density $P$ have a sufficiently regular arrangement of discs that there is no space left to reach the larger density $Q$, without overlapping discs. For these realizations the occurrence of Process 2 can therefore be excluded. I would think that this is the only conclusion you can reach with certainty, but it is reasonable to assume that if you see a highly irregular arrangement of discs, with closely packed regions separated by large voids, then this was produced by Process 2 and not directly by Process 1.

So indeed, these two procedures are not statistically equivalent.

Answer by Carlo Beenakker for Albanese dual to the Picard scheme

2013-04-17T11:54:45Z

The original reference is: On Picard Varieties, Wei-Liang Chow, American Journal of Mathematics 74, 895-909 (1952), with references to earlier work by Weil and Igusa.

Answer by Carlo Beenakker for What is the correspondence between combinatorial problems and the location of the zeroes of polynomials called?

2013-04-16T21:04:48Z

Rota himself called this correspondence "the critical problem". You can find the full quote in Michael Lugo's blog.

As explained by Garrett Birkhoff (in his book on Lattice Theory), the critical problem consists in locating the zeros of the characteristic polynomial of a geometric lattice, as defined by the Möbius function. Many extremal combinatorial problems, notably the classical problem of graph-coloring, can be re-stated as critical problems.

Combinatorial Geometries is a classic reference.

Other combinatorially defined polynomials (with further references) can be found in Lugo's blog mentioned above.

Answer by Carlo Beenakker for Does the derivative of log have a Dirac delta term?

2013-04-15T09:04:26Z

In integral form, this amounts to the Sokhotski-Plemelj theorem:

$\lim_{\epsilon\rightarrow 0^{+}}\int_{-\infty}^{\infty}dx f(x) \frac{d}{dx}\log (x+i\epsilon)=-i\pi f(0)+{\cal P}\int_{-\infty}^{\infty}dx f(x)\frac{1}{x}$.

The symbol ${\cal P}$ indicates that the Cauchy principal value of the integral is to be taken. Formalization then amounts to stating the conditions on $f$ so that the principal value integral exists.

The logarithm for $x<0$ is defined as $\log x= \log(-x) + i\pi$. You can avoid the delta function by including the absolute value signs in the logarithm:

$\int_{-\infty}^{\infty}dx f(x) \frac{d}{dx}\log|x|={\cal P}\int_{-\infty}^{\infty}dx f(x)\frac{1}{x}$.

Comment by Carlo Beenakker

2013-05-25T08:07:26Z

same question as: mathoverflow.net/questions/130336/…

Comment by Carlo Beenakker

2013-05-24T21:21:34Z

no unique solution without initial condition....

Comment by Carlo Beenakker

2013-05-24T20:12:17Z

@R S: "intuition" ? at least in the way my mind works, this one line proof is as "intuitive" as it gets...

Comment by Carlo Beenakker

2013-05-24T14:07:17Z

the "quantum collapse" is completely unrelated to any notion of "number of states", it is a statement about the nature of the state before and after the measurement of a certain Hermitian operator: from a superposition of eigenstates before the measurement to an eigenstate after the measurement.

Comment by Carlo Beenakker

2013-05-23T17:59:50Z

you're asking for an explanation of Perelman's intuition to relate the Ricci flow to thermodynamics; many asked that question, an insightful analysis has been given here: arxiv.org/abs/1303.5193

Comment by Carlo Beenakker

2013-05-22T23:30:50Z

you might as well take $a=0$, since this is just an additive constant for each eigenvalue

Comment by Carlo Beenakker

2013-05-19T21:30:56Z

@Allen, wouldn't the rescaling to unit length spoil the sum to zero?

Comment by Carlo Beenakker

2013-05-19T02:46:18Z

yes, $u\ll 1$ means that the quadratic (convective) terms are neglected; and indeed, the Bessel equation, which has Bessel functions as solutions, is a linear equation, so it cannot include the convective term; see my full answer below.

Comment by Carlo Beenakker

2013-05-18T15:25:05Z

it seems to be little more than a formal correspondence, judging from page 11 of arxiv.org/abs/math.DG/0211159

Comment by Carlo Beenakker

2013-05-18T09:22:49Z

as far as I can see, the solution you have written down for the density $\rho$ neglects the nonlinear term $u\cdot\nabla u$, so any effect of convection is neglected and it only applies in the limit of small velocities $u$; in that limit, indeed, the differential equation for $\rho$ is just the Poisson equation, which in cylindrical coordinates has the Bessel function solution; all the complications of hydrodynamics (shock waves, turbulence, etc.) come from the nonlinear convective term $u\cdot\nabla u$ which is neglected; once you add that term, there is no closed form solution.

Comment by Carlo Beenakker

2013-05-17T13:33:12Z

sorry, I misunderstood the left and right hand side both as fractions, my mistake

Comment by Carlo Beenakker

2013-05-17T12:49:09Z

utdallas.edu/~herve/Abdi-SVD2007-pretty.pdf

Comment by Carlo Beenakker

2013-05-17T12:47:14Z

the series expansion is easiest if you first take the logarithm, and then you find directly a powerseries in $n^{-1}\ln n$, $$-\frac{k(k+1)}{2n}\ln n+2\ln(2kn)\left[\sum_{p=0}^{\infty}\frac{1}{p}k^p(1+k)^p(2kn)^{-p}(\ln n)^p \right]$$

Comment by Carlo Beenakker

2013-05-17T10:44:44Z

thank you, Federico, I stand corrected; the correct formula is $${\rm Det}A=\sum_{S}|{\rm Det}V_{S}|^2\prod_{k\in S}D_{kk}$$ where $S$ is a subset of $M$ indices out of $1,2,...2M$ and $V_S$ is an $M\times M$ matrix constructed from $V$ by deleting the $M$ columns that are not in $S$. it would seem that to maximize this is in general not trivial.

Comment by Carlo Beenakker

2013-05-17T10:27:03Z

the left hand side is infinite, so the inequality is obviously false.