# Carlo Beenakker

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## Registered User

 Name Carlo Beenakker Member for 2 years Seen 4 hours ago Website Location Leiden, The Netherlands Age
physicist at Leiden University
 1d comment How can I randomly draw an ensemble of unit vectors that sum to zero?@Allen, wouldn't the rescaling to unit length spoil the sum to zero? 2d comment Derivation of Bessel functionsyes, $u\ll 1$ 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. 2d revised The relations between the Perelman’s entropy functional and notions of entropy from statistical mechanicsPerelman quote 2d revised The relations between the Perelman’s entropy functional and notions of entropy from statistical mechanicsrewording 2d answered The relations between the Perelman’s entropy functional and notions of entropy from statistical mechanics 2d comment The relations between the Perelman’s entropy functional and notions of entropy from statistical mechanicsit seems to be little more than a formal correspondence, judging from page 11 of arxiv.org/abs/math.DG/0211159 2d revised Derivation of Bessel functionsp_0 2d answered Derivation of Bessel functions 2d comment Derivation of Bessel functionsas 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. May17 comment how to proof this Stirling related equationsorry, I misunderstood the left and right hand side both as fractions, my mistake May17 comment Series expansion with remaining $log n$the series expansion is easiest if you first take the logarithm, and then you find directly a powerseries in $n^{-1}\ln n$, $$-\frac{k(k+1)}{2n}\ln n+2\ln(2kn)\left[\sum_{p=0}^{\infty}\frac{1}{p}k^p(1+k)^p(2kn)^{-p}(\ln n)^p \right]$$ May17 comment How to maximize the determinant of a matrix of the form VDV^H 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. May17 comment how to proof this Stirling related equationthe left hand side is infinite, so the inequality is obviously false. May17 comment How to maximize the determinant of a matrix of the form VDV^H since $Det A = (Det V^H V) \prod_n D_n$, this is trivial May16 revised computing Bernoulli numberszeta function May16 answered computing Bernoulli numbers May15 answered help on antiderivative of a vector function May15 comment help on antiderivative of a vector functionhmm, votes to close .... is it trivial? May15 accepted Principal value of integral May15 revised Eigenvalues of random Hamiltonian matricesadded reference May14 comment Principal value of integralI added several more intermediate steps; is it clear now? May14 revised Principal value of integralmath corrected May14 revised Principal value of integralexplanation May14 revised Principal value of integralderivation May14 comment Principal value of integralthe integral can be done as a contour integral in the complex plane, using the definition of principal value which I gave in a comment to your question above (so as the average of a contour closed in the upper and lower half of the complex plane, once picking up the pole at $x=i\epsilon$ and once at $x=-i\epsilon$; I will try to add this calculation to my answer later today. May14 comment Principal value of integral@Gerald: it is equivalent to the textbook definition ${\cal P}\int_a^b dx f(x)=\frac{1}{2} \lim_{\epsilon\rightarrow 0}\left(\int_{a-i\epsilon}^{b-i\epsilon} dx f(x)+ \int_{a+i\epsilon}^{b+i\epsilon} dx f(x)\right)$ May14 revised Principal value of integraltypo May14 answered Principal value of integral May14 comment Principal value of integralI would presume each pole at $x_n=n\pi$ is excluded in an interval $(x_n-\epsilon,x_n+\epsilon)$, and then the limit $\epsilon\rightarrow 0$ is taken. May13 comment Eigenvalues of random Hamiltonian matricesno, we checked numerically that the eigenvalue density vanishes linearly with the distance from either real or imaginary axis; moreover, there is a linear repulsion of pairs of eigenvalues on the real axis, as well as on the imaginary axis; in these respects the Hamiltonian ensemble behaves just like the Ginibre ensemble. May13 accepted Eigenvalues of random Hamiltonian matrices May13 comment Analytical continuation of electrostatic potentialssince your charge density solves the Poisson equation, your potential has the Fourier representation $V(\vec{r})\propto\int d\vec{k} \exp(i\vec{k}\cdot\vec{r})k^{-2}\rho_{C}(\vec{k})$. So you see that if the Fourier transformed charge density $\rho_C(\vec{k})$ decays only as a power law in $k$, then the potential must diverge exponentially for any imaginary $\vec{r}$. This seems unavoidable. May13 comment Eigenvalues of random Hamiltonian matrices@Federico: you're completely correct, I did not properly define the limit in which the density becomes uniform; I've corrected that now; the uniformity applies to the rescaled density; the nonuniformities in the rescaled density are of order $n^{1/2}$, so the factor $n^{-1}$ removes them in the largeâˆ’$n$ limit. May13 revised Eigenvalues of random Hamiltonian matricesrescaling May13 comment Eigenvalues of random Hamiltonian matricesthank you for asking: this is certainly not an artifact; in the Ginibre ensemble the eigenvalue density is known to vanish linearly on approaching the real axis; basically the depletion zone signals the accumulation of eigenvalues on the real axis, as if the eigenvalues are "attracted" towards the real axis, where they condense. It seems that the repulsion from the imaginary axis is stronger than linear, but this is something we have not yet checked. May13 revised Eigenvalues of random Hamiltonian matricestypo May13 answered Eigenvalues of random Hamiltonian matrices May11 comment A problem about Joint sine and cosine fourier transformhmm, there is no square root in the denominator? if there was a square root, then we could define $\alpha=\arccos(\lambda/\sqrt{\lambda^2+h^2})$ and your kernel would simply be $K(x,\lambda)=\sqrt{2/\pi}\cos(\lambda x-\alpha)$ May11 revised Lower bounds on derivative around zero set of a positive smooth function.small edit May11 revised Lower bounds on derivative around zero set of a positive smooth function.simpler May11 revised Lower bounds on derivative around zero set of a positive smooth function.one more step May11 answered Lower bounds on derivative around zero set of a positive smooth function. May9 comment Prove that the sum of a certain infinite series is 1can you give some background how this identity to unity arises? does that suggest a way to a proof? May8 accepted Probability distribution for two-state system that depends on residence time May8 revised Probability distribution for two-state system that depends on residence timeremove header May8 comment Probability distribution for two-state system that depends on residence time@ionlet: I expanded the answer, hopefully answering all your queries; is it clear now? May8 revised Probability distribution for two-state system that depends on residence timeexpanded May8 revised Probability distribution for two-state system that depends on residence timecorrected May7 answered Probability distribution for two-state system that depends on residence time May5 accepted Non-asymptotic results for bulk of random Wishart matrix