User emilio pisanty - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:03:36Z http://mathoverflow.net/feeds/user/20729 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130484/analytical-continuation-of-electrostatic-potentials Analytical continuation of electrostatic potentials Emilio Pisanty 2013-05-13T13:38:55Z 2013-05-13T13:38:55Z <p>I'm having some trouble figuring out the properties with respect to analytical continuation of functions defined using an integral kernel. More particularly, I am working with the electrostatic potential $$V(x,y,z)=\int \frac{\rho_C(x',y',z')dx'dy'dz'}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}$$ where the charge density $\rho_C$ describes a localized system. </p> <p>(Scroll down for my specific question.)</p> <p>My initial expectation is that, if $\rho_C$ decays fast enough and contains no singular, delta-function-like points, then the potential $V$ will be a real analytic function of its variables $x,y,z$, inheriting that behaviour from the kernel. Because of this, I would expect $V$ to extend to an analytical function, probably on the whole complex plane. I additionally expect this analytical continuation to contain either branch cuts or singularities, or to tend to infinity for large complex arguments in some direction, because of Liouville's Theorem.</p> <p>In my particular case, $\rho_C$ is a molecular transition charge, which means that my knowledge about it must be numerically obtained through quantum chemistry methods, and the integral must be done numerically. I expect the real, physical $\rho_C$ to decay as $e^{-r}$ at large distances. On the other hand, the quantum chemical code approximates the charge density as a superposition of gaussians, $e^{-r^2}$.</p> <p>Both of these simple cases I can solve for analytically for real coordinates, using their spherical symmetry and Gauss's law. The results are deceptively simple: for an exponential charge density the potential goes as $$V_\text{exp}\propto e^{-r}\left(1+\frac{2}{r}\right)-\frac{2}{r},$$ while for a gaussian distribution $$V_\text{gauss}\propto \frac{\text{erf}(r)}{r}.$$ On the real line, both potentials look essentially identical. However, their behaviour when extended to complex coordinates is wildly different:</p> <ul> <li>On one hand, $\text{erf}$ is an odd, entire function, so $V_\text{gauss}$ is a function of $r^2=x^2+y^2+z^2$ and has therefore no branch cuts anywhere in any variable. In contrast, $V_\text{exp}$ is a function of the radial distance $r=\sqrt{x^2+y^2+z^2}$, so if one keeps $x,y$ real then $V_\text{exp}$ as a function of $z$ has branch cuts starting at $\pm i\sqrt{x^2+y^2}$, and its Riemann surface has two sheets.</li> <li>On top of that, $V_\text{exp}$ is bounded except for the neighbourhood of the branch cut, while $V_\text{gauss}$ blows up as $e^{+|z|^2}$ for large imaginary $z$.</li> </ul> <p>While these differences weird me out slightly, I can accept them as representing the very different origins of the two potentials. The difference in behaviour for $\text{Re}(x^2+y^2+z^2)&lt;0$ is worrisome, as it means the quantum chemical code cannot be trusted there, but I can accept that as inevitable as long as I have warning flags that indicate the code is out of its validity region.</p> <p>(The above is mostly background, but if I have made conceptual errors or I am working on misconceptions I would dearly like to know. My precise question follows)</p> <hr> <p>What really worries me, though, is the last point on the list: $V_\text{gauss}$ blows up as $e^{+|z|^2}$ for large imaginary $z$. This means that for $x=y=0$ and $z=i\zeta$, the integral $$V_\text{gauss}(0,0,i\zeta)=\int \frac{e^{-(x^2+y^2+z^2)}dx\ dy\ dz}{\sqrt{x^2+y^2+(z-i\zeta)^2}}$$ (dropping the primes) must go as $e^{+\zeta^2}$. Most importantly, I cannot see how to get this behaviour from a numerical integration procedure, which is necessary to get the right behaviour for a real molecule.</p> <p>For large $\zeta$, the integrand is bounded except for an integrable ring singularity at $x^2+y^2=\zeta^2$, $z=0$, but the charge density goes <em>down</em> as a gaussian on that singularity. I do not see a way of getting a super-exponential growth from that integral that does not involve shifting integration contours and thus evaluating $\rho_C$ on complex coordinates (which does not sit well with the real quantum chemical data on my molecule). On the other hand, Liouville's theorem makes me very wary of solutions to that integral that appear to return bounded entire functions for $V_\text {gauss}$.</p> <p>Can someone help me see a way out of this conundrum, or point me to appropriate references where this or similar issues are treated in some detail?</p> http://mathoverflow.net/questions/127612/a-2-f-1-equivalent-of-the-tricomi-u-function A ${}_2 F_1$ equivalent of the Tricomi $U$ function? Emilio Pisanty 2013-04-15T12:30:13Z 2013-04-15T12:30:13Z <p>The confluent hypergeometric function <code>${}_1F_1(a;b;z)$</code> has a natural partner in the Tricomi function <code>$U(a,b,z)$</code>, which provides a second, linearly independent solution to the confluent hypergeometric equation. This is not strictly necessary as one can usually find a second <code>${}_1F_1$</code> function with appropriate parameters as a second solution, but the Tricomi function seems to be a more natural choice.</p> <p>For the gaussian hypergeometric equation, a similar situation holds in that one can find appropriate parameters for two gauss hypergeometric functions <code>${}_2 F_1(a,b;c;z)$</code> that will provide a general solution. However, there seems to be no natural partner for <code>${}_2F_1$</code> analogous to $U$. Is this the case, or am I missing something? If this is indeed the case because of some fundamental reason, why is it?</p> <p>In case this is all too general, more precisely: is there a (canonical?) solution $f(a,b;c;z)$ to the <a href="http://en.wikipedia.org/wiki/Hypergeometric_function#The_hypergeometric_differential_equation" rel="nofollow">hypergeometric differential equation</a> that limits as $$\lim_{b\rightarrow\infty} f(a,b;c;\frac z b)=U(a,c,z)?$$</p> http://mathoverflow.net/questions/124356/on-a-hypergeometric-type-integral On a hypergeometric-type integral Emilio Pisanty 2013-03-12T22:46:09Z 2013-03-13T12:16:34Z <p>I'm having a bit of trouble with the integral $$\int_0^1 e^{-\frac{z^2}{2}u}\frac{u^{m-\frac{1}{2}} (1-u)^{n/2} }{ \left(1+ \left(s^2-1\right)u\right)^{m+\frac{n}{2}+1}} du.$$ (Here $m$ and $n$ are nonnegative integers, $s>0$ and I can assume that $z\in\mathbb{R}$.) For $s=0$ it reduces to <a href="http://dlmf.nist.gov/13.4.E1" rel="nofollow">the standard integral representation</a> of the confluent hypergeometric function ${}_1F_1$, which is great. I would like to extend it to the case $s>0$, which then looks very much like many integrals in Gradshteyn, Erdélyi, Prudnikov and the DLMF, but I can't quite match it to anything. </p> <p>The reason I ask here is that this is reducible to the form $$\int_0^1 u^{a-1} (1-u)^{b-a-1}{}_0F_0(-;-;-\zeta u)\times{}_1F_0(a';-;s'u),$$ which is the Euler type of integral that's apparently so common, but I can't find a general enough form for this.</p> <p>I know this is a bit of a long shot but I'm hoping someone with more intuition for hypergeometric functions will be able to point me in the right direction.</p> http://mathoverflow.net/questions/122629/the-paley-wiener-theorem-and-exponential-decay The Paley-Wiener theorem and exponential decay. Emilio Pisanty 2013-02-22T14:18:01Z 2013-02-23T04:52:08Z <p>Consider a function whose Fourier transform is supported on a half-ray: $$A(t)=\int_0^\infty \omega(E) e^{-iEt}d E,$$ where I can suppose $\omega(E)\geq 0$ and any suitable regularity conditions on $\omega(E)$ in the limit $E\rightarrow\infty$. I am interested in results constraining the rate of decay of $A(t)$ in the limit $t\rightarrow\infty$. <strong>Specifically, I would like to rule out the asymptote $\mathbf{|A(t)|\sim e^{-\Gamma t}}$.</strong></p> <p>The reference I have [1] uses the original 1934 Paley-Wiener theorem [2], which states that under these (or similar) assumptions for $A$ the integral $$\int_{-\infty}^\infty \frac{\left|\ln|A(t)|\right|}{1+t^2}dt&lt;\infty$$ must converge. This is strong enough to rule out the asymptote.</p> <p>However, I have looked up the proof in Paley and Wiener and I find it far too technical and non-self-contained for me to follow with any ease; it also has an air of old mathematics that has probably been replaced with cleaner arguments by now. I do get some of the intuition behind the appearance of the $1/(1+t^2)$ factor. (Namely, a unitary transform of the upper half-plane $z$ space into the unit circle in $\zeta=i\frac{z+1}{z-1}$, where the measure transforms as $\frac{d\zeta}{\zeta}\approx\frac{dz}{1+z^2}$.) I still don't find, however, any intuition into how the can't-be-too-fast decay of $A$ correlates with the support of $\omega$, or at least no intuition that can be turned into a rigorous argument.</p> <p>I am looking for references or arguments that prove in a clearer fashion that exponential decay of $A$ is impossible with such a Fourier domain, and particularly for ones that have clear intuition behind them that can be turned into a solid argument, even if the rigorous details are fiddly.</p> <ol> <li>L. Fonda, G. C. Ghirardi and A. Rimini. Decay theory of unstable quantum systems. <a href="http://dx.doi.org/10.1088/0034-4885/41/4/003" rel="nofollow"><em>Rep. Prog. Phys.</em> <strong>41</strong>, pp. 587-631 (1978)</a>. Page 592.</li> <li>R. Paley and N. Wiener. <em>Fourier Transforms in the Complex Domain</em> (Providence, Rhode Island: American Mathematical Society, 1934). Theorem XII, p16.</li> </ol> http://mathoverflow.net/questions/87695/electrical-resistance-across-the-diagonal-of-a-square-metal-plate/87707#87707 Answer by Emilio Pisanty for Electrical resistance across the diagonal of a square metal plate Emilio Pisanty 2012-02-06T19:35:23Z 2012-12-17T19:15:51Z <p>I don't know if you can have something as fancy as Thévenin's theorem in a general Riemannian manifold, but as a physicist I'd say you'd be better off looking at generalizations of the conductivity, $\sigma=1/\rho$, where $\rho$ is the resistivity and can be measured as $\rho=R A/L$ for a (homogeneous) wire of cross-section $A$, length $L$, and resistance $R$. Then you can formulate a local version of Ohm's law as $$\mathbf{j}=\sigma\mathbf{E},$$ where $\mathbf{j}$ is the current density, related to the current $I$ flowing through a surface $\Sigma$ as $I=\int_\Sigma \mathbf{j}\cdot\textrm{d}\mathbf{A}$, and $\mathbf{E}$ is the electric field, related to the potential difference $V(\mathbf{x})-V(\mathbf{y})$ between arbitrary points $\mathbf{x},\mathbf{y}$ as the line integral $\Delta V=\int_{\mathbf{x}}^{\mathbf{y}} \mathbf{E}\cdot\textrm{d}\mathbf{r}$ (independent of the integration path).</p> <p>As for the second question, you'd have to solve Laplace's equation $\nabla^2 V=0$ on a rectangular-box domain $[0,L]\times[0,L]\times[0,h]$, with appropriate boundary conditions - I'm not that sure there, but I'd set $V=0$ on $(0,0,z)$ and $V=V_0$ on $(L,L,z)$ to begin with; if that does not determine a unique solution then Neumann conditions on the rest of the boundary should do it. You then find the electric field $\mathbf{E}=-\nabla V$ to get the current density and integrate across an appropriate spanning surface, say the other diagonal plane.</p> <p>I'll try and flesh this out and fill this in when I have time. </p> <p>EDIT J. Cserti's paper Application of the lattice Green's function for calculating the resistance of an infinite networks of resistors, <em>Am. J. Phys.</em> <strong>68</strong> no. 10, pp 896 (2000), <a href="http://dx.doi.org/10.1119/1.1285881" rel="nofollow">doi:10.1119/1.1285881</a>, <a href="http://arxiv.org/abs/cond-mat/9909120" rel="nofollow">arXiv:cond-mat/9909120</a>, solves the discretized problem of an infinite network of resistors on a square grid (including as a special case <a href="http://xkcd.com/356/" rel="nofollow">the Nerd Sniping xkcd problem</a>). In the continuum limit that yields a solution to this problem.</p> http://mathoverflow.net/questions/111910/dertivative-of-a-special-function-with-respect-to-order/111920#111920 Answer by Emilio Pisanty for Dertivative of a Special Function with respect to Order Emilio Pisanty 2012-11-09T17:51:54Z 2012-11-09T17:51:54Z <p>For large $x$, if $a>0$, $I_m$ behaves asymptotically like $I_m(ax)\approx e^{ax}/\sqrt{ax}$. Therefore for large $x$ the integrand will look like $x^{m-1/2}e^{-(x-a)^2/2}$. This dies off fast enough that the improper integral converges uniformly in $m$ and you can differentiate inside the integral.</p> http://mathoverflow.net/questions/94517/is-zero-a-hydrogen-eigenvalue Is zero a hydrogen eigenvalue? Emilio Pisanty 2012-04-19T11:45:28Z 2012-10-02T15:52:21Z <p>This question has been bugging me for some time.</p> <p>Take the hamiltonian for the hydrogen atom: $$\hat{H}=-\frac{1}{2}\nabla^2-\frac{1}{r},$$ acting on (a domain contained in) $L^2(\mathbb{R}^3)$. It is standard fact that this is an unbounded operator which has a countable infinity of eigenvalues, all of which are negative and which accumulate around 0, and has a continuous spectrum on the whole of $(0,\infty)$. Physically, the former are bound states which correspond to elliptic Keplerian orbits in the classical problem, and the latter are unbound states and correspond to hyperbolic orbits. I also know that the spectrum of all unbounded operators is a closed set, so that 0 is definitely in $\sigma\left(\hat{H}\right)$.</p> <p>My question is then: to what part of the spectrum does 0 belong to (i.e. point, continuous, residual)? What are the corresponding eigenfunctions? What kind of degeneracy does it have? (I would expect it to admit a common eigenvector with any $l,m$ angular momentum numbers, but I'm far from sure.) How do the eigenfunctions correspond to the nearby bound and unbound states?</p> http://mathoverflow.net/questions/94517/is-zero-a-hydrogen-eigenvalue/108647#108647 Answer by Emilio Pisanty for Is zero a hydrogen eigenvalue? Emilio Pisanty 2012-10-02T15:52:21Z 2012-10-02T15:52:21Z <p>I looked in Anatoly's references and Quantum mechanics for Mathematicians by Leon A. Takhtajan does have the calculation of the continuum wavefunctions though it does not do the $k=0$ case.</p> <p>The eigenfunction $f_l$ at energy $E=\frac{1}{2}k^2$ and angular momentum $l$ must satisfy the eigenvalue equation $$f_l''+\left(\frac{2}{r}-\frac{l(l+1)}{r^2}+k^2\right)f_l=0.$$ After the obligatory asymptotics factorization of $f_l(r)=r^{l+1}e^{ikr}F_l(r)$, this equation reads $$F_l''+\left(\frac{2(l+1)}{r}-2ik\right)F_l'+\left(\frac{2}{r}-\frac{2ik(l+1)}{r}\right)F_l=0,$$ and Takhtajan gives the solution as a confluent hypergeometric function, $F_l(r)={}_1F_1\left(l+1+\frac{i}{k};2(l+1);2ikr\right)$, under $F_l(0)=1$.</p> <p>From this the $k=0$ case can be recovered as a limit in the same spirit as the $_2F1\rightarrow_1F_1$ confluence by letting the length $\lambda=\frac{1}{k}$ go to infinity. Thus at zero energy, $$F_l(r) =\lim_{\lambda\rightarrow\infty} {}_1F_1\left(l+1+i\lambda;2(l+1);\frac{-2r}{i\lambda}\right) =_0F_1\left(;2(l+1);-2r\right) =\frac{(2l+1)!}{2^\frac{2l+1}{2}}r^{-\frac{2l+1}{2}}J_{2l+1}(\sqrt{8r}).$$ (This still obeys $F_l(0)=1$). Alternatively, the Bessel function solution can be obtained directly by the appropriate transformations or by plugging the $k=0$ equation into Mathematica.</p> <p>In the asymptotic regime, $r\gg 2l+1$, one then gets $$f_l(r) =\frac{(2l+1)!}{2^l\sqrt{2\pi \sqrt{2}}}r^{\frac{1}{4}} \cos\left(\sqrt{8r}-(2l+1)\frac{\pi}{2}-\frac{\pi}{4}\right)$$ if my maths is right. However, in real cases this can only happen if $2l+1\ll r\ll \lambda$.</p> http://mathoverflow.net/questions/107159/pochhammer-symbol-of-a-differential-and-hypergeometric-polynomials Pochhammer symbol of a differential, and hypergeometric polynomials Emilio Pisanty 2012-09-14T09:36:06Z 2012-09-19T02:23:40Z <p>I have a minor result which I'm sure has come up somewhere before but I can't seem to find it.</p> <p>Consider a confluent hypergeometric function of the form $$\newcommand{\ff}{{}_1F_1} \ff(b+k;b;z)\textrm{, for }k\in\mathbb{N}.$$ Numerical tests suggest that this is always a polynomial of degree $k$ multiplied by an exponential. One can prove this in a dull fashion by using $\ff(b;b;z)=e^z$ and then applying recurrence relations, but I found a cleaner way using the series definition,</p> <p>$$\ff(b+k;b;z) =\sum_{n=0}^\infty\frac{(b+k)_n}{(b)_n}\frac{z^n}{n!}$$</p> <p>where $(b)_k=b(b+1)\cdots(b+k-1)$ is the Pochhammer symbol. By exploiting the identities $$\frac{(b+n)_k}{(b)_k}=\frac{\Gamma(b+k+n)\Gamma(b)}{\Gamma(b+k)\Gamma(b+n)}=\frac{(b+n)_k}{(b)_k} \textrm{ and }nz^n=z\frac{d}{dz}z^n,$$ one can easily prove that $$\ff(b+k;b;z)=\frac{\left(b+z\frac{d}{dz}\right)_k}{(b)_k}e^z,$$</p> <p>by being somewhat liberal with the meaning of the Pochhammer symbol. This is clearly the desired polynomial-times-exponential, and provides an explicit expression for the polynomial that looks kind of like a Rodrigues formula.</p> <p>Even better, if you put this together with <a href="http://dlmf.nist.gov/13.2.E39" rel="nofollow">Kummer's first transformation</a>, $$\ff\left(a;b;z\right)=e^{z}\ff\left(b-a;b;-z\right),$$ and the <a href="http://dlmf.nist.gov/18.11.E2" rel="nofollow">expression for Laguerre polynomials in terms of hypergeometric functions</a>, $L^{(\alpha)}<em>{n}\left(x\right)=\frac{\left(\alpha+1\right)</em>{n}}{n!}\ff\left(-n,\alpha+1,x\right)$, you get an analogous result for Laguerre polynomials,</p> <p>$$L^{(b-1)}_{k}\left(x\right)=\frac{1+k/b}{k!}e^z\left(b+z\frac{d}{dz}\right)_ke^{-z}.$$</p> <p>Are these results familiar to anyone? Do they fit inside a larger framework? They are not the best thing since sliced bread but they do have a nice simplicity to them, and particularly I would like to cite the appropriate reference if they have appeared before.</p> http://mathoverflow.net/questions/104924/analytic-continuation-of-instantaneous-eigenstates-of-a-time-dependent-hamiltonia Analytic continuation of instantaneous eigenstates of a time-dependent hamiltonian Emilio Pisanty 2012-08-17T16:17:57Z 2012-08-17T19:10:12Z <p>We are considering the instantaneous eigenstates of an analytically time-dependent hamiltonian and I would like to know how legitimate it is to extend them to the complex plane.</p> <p>Specifically, our hamiltonian describes an atom or a molecule in an external laser pulse. Thus it includes kinetic energy terms, all Coulomb potentials, and a laser interaction, and it can be written $$H(t)=H_0+\sum_{i=1}^N\mathbf{F}(t)\cdot\mathbf{r}_i.$$ Here $H_0$ can be assumed to be as well-behaved as necessary as long as one preserves the basic structure, but in general it's outside our capacity for exact solutions. $\mathbf{F}(t)$ is a vector-valued analytic (and if necessary entire) function of the time $t$ which is real for real times. We define the instantaneous eigenstates to obey the relation $$H(t)|n(t)\rangle=E_n(t)|n(t)\rangle$$ in Dirac notation.</p> <p><strong>My question:</strong> Since $\mathbf{F}(t)$ can be extended to complex times, can the eigenstates $|n(t)\rangle$ be extended as well?</p> <p>My specific worry is the following. One can always write, in this setting, $$H(t)=T+V_1(t)+iV_2(t),$$ where $T$ is the kinetic energy and all three of $T$, $V_1$ and $V_2$ are hermitian and well-behaved. However, for nonzero $V_2$ it appears that $H(t)$ is no longer normal: $$[H,H^\dagger]=[T+V_1+iV_2,T+V_1-iV_2]=2i[V_2,T],$$ which is in general nonzero since $T$ contains derivatives. It would appear then that formally none of the $|n(t)\rangle$ exist outside $t\in\mathbb{R}$. However, I would find such a strict cutoff somewhat strange - i.e., what if $V_2$ is nonzero but can somehow be neglected w.r.t. the other terms? surely there is some kind of a gray area around the real axis of some sort. Further, the existence and analyticity of the eigenstates for complex times, and in particular matrix elements involving them, are crucial to our results and physical intuition seems to indicate they should exist.</p> <p>For completeness, we are considering in particular matrix elements of the form $$\langle m(t_1)|U(t_1,t_2)A(t_2)U(t_2,t_3)|n(t_3\rangle,$$ where $A(t)$ is analytic in $t$ and hermitian for real times, $U(t,t')$ is the propagator and obeys $$i\frac{\partial}{\partial t}U(t,t')=H(t)U(t,t'),\quad U(t,t)=\textrm{id},$$ and all three of the times involved, $t_1, t_2$ and $t_3$, may be complex.</p> <p>Any insights will be deeply appreciated.</p> http://mathoverflow.net/questions/100364/a-good-reference-to-grok-hypergeometric-functions A good reference to grok hypergeometric functions? Emilio Pisanty 2012-06-22T14:52:15Z 2012-07-10T16:33:10Z <p>When I was introduced during my degree to special functions, I made friends with a number of nice functions - Laguerre, Legendre, Hermite, Bessel, and whatnot - but I made only a passing acquaintance with the hypergeometric and confluent hypergeometric functions, mostly limited to looking them up in Arfken and recoiling in horror at the dryness of the material and the lack of physical content in the calculations.</p> <p>I know, of course, that this lack of physical content is also accompanied by an astounding generality. After a while I did get the core of the idea, which I believe is "explore all special functions whose series coefficients are rational functions of $n$", and I do find it appealing, but I've not had the energy nor the motivation to follow that up and see what's interesting about the thing.</p> <p>However, it appears that the long-delayed moment is here and some pretty hairy integrals (think $\int_0^\infty x^k e^{-\alpha x^2}J_m(\beta x)dx$) have pushed some ugly "${}_1 F_1$" symbols onto my page. So my question is, then: <strong>what's a good introduction to hypergeometric and confluent hypergeometric functions?</strong> I'd like one where I can get an intuitive understanding of what to expect from them in different circumstances, what nice properties they have, and generally why it really <em>is</em> worth it to deal with them instead of their more specific cases like Laguerre, Legendre, Hermite, Bessel, etc.</p> http://mathoverflow.net/questions/58554/preferably-rare-audio-video-recordings-of-famous-mathematicians/101690#101690 Answer by Emilio Pisanty for (Preferably rare) Audio/Video recordings of famous mathematicians? Emilio Pisanty 2012-07-08T17:07:21Z 2012-07-08T17:07:21Z <p>While he's not exactly a mathematician, there is on Youtube a wealth of Richard Feynman lectures and interviews. My personal favourite is part of the "The Character of Physical Law" Messenger Lectures at Cornell, titled <a href="http://www.youtube.com/watch?v=-Km7-6-J81k&amp;feature=related" rel="nofollow">the distinction of past and future</a>. (Though, really, all the Messenger lectures are amazing.)</p> http://mathoverflow.net/questions/91232/about-a-christoffel-darboux-type-sum About a Christoffel-Darboux-type sum Emilio Pisanty 2012-03-15T00:10:29Z 2012-06-25T00:22:12Z <p>Hi! I've been using the Christoffel-Darboux identity for the Hermite polynomials, <code>$$\sum_{k=0}^n\frac{H_k(x)H_k(y)}{2^k k!}=\frac{1}{2^n n!}\frac{H_{n+1}(x)H_n(y)-H_n(x)H_{n+1}(y)}{x-y},$$</code> for some time, and it's been quite helpful. I would like to extend this to a sum of the form <code>$$\sum_{k=0}^n\frac{i^kH_k(x)H_k(y)}{2^k k!}$$</code> (or if possible an arbitrary phase $e^{ik\theta}$ replacing $i^k$), but I came up empty when looking for references. Can anyone point me in the right direction? or is this a lost cause? Cheers!</p> <p>EDIT, to provide some motivation: in <a href="http://arxiv.org/abs/1109.5724" rel="nofollow">this paper</a> we explore eigenstates of the position quadrature in a truncated quantum harmonic oscillator number basis. The sum I'm asking about is related to the momentum eigenstates in the truncated space, or equivalently to the Fourier transform of the position eigenstates.</p> http://mathoverflow.net/questions/99549/how-does-my-radio-work/100249#100249 Answer by Emilio Pisanty for How Does My Radio Work? Emilio Pisanty 2012-06-21T14:28:29Z 2012-06-21T14:54:24Z <p>Your expression for FM transmissions is not quite right - it's missing the radio frequency! The simple model that captures the essentials of what FM station $k$ is sending you is the function $$B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right),$$ where $\gamma_k\psi_k$ never gets close to $\omega_k$ (so you're only modulating the frequency and not completely disrupting it). If the interesting signal $\psi_k$ is a pure note at frequency $\omega$, then the spectrum of the actual radio signal can be found in terms of Bessel functions and consists of sidebands separated from the carrier by spacing $\omega$. (The number of sidebands is controlled by how large $\gamma_k$ is.)</p> <p>The real radio signal your device is getting, then, is $$F(t)=\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^m B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right).$$ Because of the modulation, none of the stations' radio signals are single peaks; instead they are spread over a bandwidth roughly given by the frequency content of the audio signals they encode. (For comparison, human hearing can detect 16 Hz to roughly 20,000 Hz, AM frequencies are medium-wave radio at 520 kHz to 1,610 kHz, and FM stations run at 87.5 to 108 MHz. Thus in reality the peaks are quite narrow!)</p> <p>To detect a signal, your device uses a combination of antennas, loops of wire, parallel plates, and the like, which contrive to give to the decoding device (the one that takes a radio signal and gives you an audio one) a voltage $f$ that's controlled by a damped harmonic oscillator equation of the form $$\frac{d^2}{dt^2}f-2\gamma\frac{d}{dt}f+\omega_0^2f=F,$$ where the resonance frequency $\omega_0$ is controlled by a knob on the device. The spectral response of this dynamical system is routinely evaluated in college ODE courses, and comes out as a Lorentzian bell-shaped curve centred at $\omega_0$ and of width $\gamma$. Choose $\gamma$ to match the spectral width of the typical radio station, and you've got a fantastic filter!</p> <p>EDIT: After doing some looking up, I find that the $\psi_k$ here is not exactly the audio signal the station is trying to encode, but rather something like its average over the interval $[0,t]$, so it is equivalent to it up to simple mathematical operations performed at the decoder.</p> http://mathoverflow.net/questions/43538/wonderful-applications-of-the-vandermonde-determinant/100243#100243 Answer by Emilio Pisanty for Wonderful applications of the Vandermonde determinant Emilio Pisanty 2012-06-21T13:35:27Z 2012-06-21T13:35:27Z <p>The Discrete Fourier Transform, which sends a vector $x=\left(x_j\right)_{j=0}^{N-1}$ to $y=\mathrm{DFT}(x)$ such that $$y_k=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i \times jk/N}x_j$$ has a matrix representation $$\mathrm{DFT}_{jk}=e^{2\pi i \times jk/N}=\left(e^{2\pi i /N}\right)^{j\times k},$$ which is in fact a <em>doubly</em> Vandermonde matrix: both it and its transpose are Vandermonde matrices. With this you can use the Vandermonde determinant to prove that $\mathrm{DFT}$ is nonsingular, and if you prove using other means that it is unitary (rather easy) then you will get, I think, a nontrivial expression for 1 as a product of differences of roots of unity.</p> http://mathoverflow.net/questions/98989/unicity-of-a-vector-space-frames-dual-frame Unicity of a vector space frame's dual frame Emilio Pisanty 2012-06-06T20:54:10Z 2012-06-06T20:54:10Z <p><a href="http://en.wikipedia.org/wiki/Frame_of_a_vector_space" rel="nofollow">The Wikipedia page on vector space frames</a> gives a construction to find a dual frame for a given frame. Specifically, given a set of vectors ${ e_k }$ in a Hilbert space $\mathcal{H}$ such that for all $v\in\mathcal{H}$ and some $A \leq B&lt;\infty$ we have $$A||v||^2\leq\sum_k|\langle e_k|v\rangle|^2\leq B ||v||^2,$$ the construction gives (among other things) a set of linear functionals $\phi_k:\mathcal{H}\rightarrow\mathbb{C}$ such that for all $v\in\mathcal{H}$ $$v=\sum_k e_k \cdot\phi_k(v).$$</p> <p>My question is, how unique are these functionals? (or alternatively, their dual images in $\mathcal{H}$.) The construction gives a natural way to find these dual images, which depends on the inverse of the map $S(v)=\sum_k\langle v|e_k\rangle e_k$ and thus on the inner product used, but that doesn't mean the corresponding functionals will depend on the inner product.</p> <p>If I have, as in one standard example, three noncollinear vectors in a two-dimensional space, then I would naively expect to have "one real degree of freedom" in choosing the coordinates. How does this show up? If there are other sets of functionals, what do their dual images in $\mathcal{H}$ look like? Is the one picked by $S$ special? if so, how? Does any of this depend on the (in)finiteness of the space dimension?</p> http://mathoverflow.net/questions/20551/sources-for-bibtex-entries/96237#96237 Answer by Emilio Pisanty for Sources for Bibtex entries Emilio Pisanty 2012-05-07T16:47:55Z 2012-05-07T16:47:55Z <p>Since it's not been mentioned, I'll chime in with <a href="http://www.mendeley.com/" rel="nofollow">Mendeley</a>, which provides BibTeX citations for papers on the site (minus the linebreaks, unfortunately). They also have a freeware reference manager.</p> http://mathoverflow.net/questions/95632/maximum-singular-value-of-a-random-1-1-matrix/95666#95666 Answer by Emilio Pisanty for Maximum Singular Value of a random +1/-1 matrix Emilio Pisanty 2012-05-01T12:56:01Z 2012-05-02T16:20:22Z <p>For what it's worth, <a href="http://quicc.net/other_files/svdrandom.cdf" rel="nofollow">here</a>'s some numerical data in Mathematica CDF format.</p> <p>Edit: I've been playing around some more and this <img src="http://quicc.net/other_files/randomsvdhistogram.png" alt="histogram"> is a histogram for 1,000,000 tries with $m=9$, $n=5$ for the five different singular values (each in a different colour). I'm intrigued by the peaks - @Alex, were you expecting that? There is also a significant portion of matrices with one zero singular value, but I am unsure whether it is due to numerical artifacts.</p> http://mathoverflow.net/questions/92798/conditions-ensuring-an-order-betweenthe-smallest-eigenvalues-of-two-positive-def/95535#95535 Answer by Emilio Pisanty for Conditions ensuring an order betweenthe smallest eigenvalues of two positive definite Jacobi matrices Emilio Pisanty 2012-04-30T00:10:31Z 2012-04-30T00:10:31Z <p>Felix is correct in the small $b$ limit. In the very large $b$ limit, on the other hand, the spectra of $J$ and $L$ will coincide, so one needs some more finesse. The following is a rough, approximate attempt at making some intuition.</p> <p>Consider the case where the first diagonal entry of $J$ is $a_1&lt;0$ with $|a_1|\ll |b|^2$, and all other diagonal entries in $J$ and $L$ are zero. Then you can expand $\det(tI-J)$ along the first row to get $$\det(tI-J)=b^{2n} \left(U_n(t/2)-\frac{a_1 }{b^2} U_{n-1}(t/2)\right),$$ where $U_n$ is the $n$th Chebyshev polynomial of the second kind. For small $a_1$, then, the question is which way does the second term "push" the eigenvalue? I don't have a proof but it's <a href="http://en.wikipedia.org/wiki/File%3aChebyshev_Polynomials_of_the_2nd_Kind_%28n%3D0-5,_x%3D%28-1,1%29%29.svg" rel="nofollow">graphically clear</a> that the signs of $U_n$ and $U_{n-1}$ coincide just <em>inside</em> of the former's leftmost zero, which means that a small, negative $a_1$ will push $J$'s minimum eigenvalue left, in accord with your conjecture.</p> <p>This can be extended to $L$ having a nonzero element in its diagonal, which will push its minimum eigenvalue either left or right but no more than $J$'s, at least at first order.</p> <p>Since your result is (or appears to be) true both for big and small $b$, I should expect it to be true everywhere, or to have some pretty interesting mathematics in the middle.</p> http://mathoverflow.net/questions/94086/partitions-of-an-interval Partitions of an interval Emilio Pisanty 2012-04-15T02:58:14Z 2012-04-15T03:14:41Z <p><a href="http://mathoverflow.net/questions/94038/is-f-a-polynomial-provided-that-it-is-partially-smooth" rel="nofollow">This question</a> asks about properties of functions which are "piecewise" polynomials. I would like to ask a specific question about the meaning of "piecewise" there.</p> <p>Specifically, consider "partitions" of an open interval of the form $(c,d)=\overline{ \bigcup_{n=1 }^\infty (a_n, b_n)}$, i.e. into a countable infinity of pairwise disjoint open intervals which gives a set dense in $(c,d)$. As I elaborated on a comment of Aaron Tikuisis, this permits ugly constructions like $$(-1,1)=(-1,0)\cup\bigcup_{k=1}^\infty \left(\frac{1}{k+1},\frac{1}{k}\right)$$ or even $$(0,1)=\bigcup_{k=1}^\infty \bigcup_{j=1}^\infty \left(\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j+1},\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j}\right),$$ for which a countable infinity of subintervals (those with $j=1$) have no next neighbour to the right. </p> <p>My question is, how far can this process be pushed? Can one give a decomposition $(c,d)=\overline{ \bigcup_{n=1 }^\infty (a_n, b_n)}$ where none of the $(a_n,b_n)$ have a next neighbour to the right? or even to both sides? If this is impossible, then can one meaningfully characterize how many of the subintervals can have this property? Has this appeared in the literature?</p> http://mathoverflow.net/questions/91161/eigenvectors-and-eigenvalues-of-tridiagonal-matrix/91229#91229 Answer by Emilio Pisanty for Eigenvectors and eigenvalues of Tridiagonal matrix Emilio Pisanty 2012-03-14T22:39:40Z 2012-03-16T19:26:55Z <p>Another way to look at this problem, from the ground up, is to expand the characteristic polynomial of your first matrix, $\mathcal{T}_n(p,q)$, along the last row and then along the last column, from which you'll quickly get <code>$$\det(\mathcal{T}_n(p,q)-\lambda)=-\lambda \det(\mathcal{T}_{n-1}(p,q)-\lambda)-pq \det(\mathcal{T}_{n-2}(p,q)-\lambda).$$</code> You can then compare this with the recurrence relations for the standard orthogonal polynomials; it is then rather easy to match it to that for the <a href="http://en.wikipedia.org/wiki/Chebyshev_polynomials" rel="nofollow">Chebyshev polynomials of the first kind</a>, <code>$$T_{n+1}(x)=2x T_n(x)-T_{n-1}(x),$$</code> with $T_0(x)=1$, $T_1(x)=x$,which should make it clear that the identification is <code>$$T_n(\lambda)=\frac{1}{2(\sqrt{pq})^n}\det\left(\mathcal{T}_n(p,q)+2\sqrt{pq}\lambda\right),$$</code> or the equivalent $\det(\mathcal{T}_n(p,q)-\lambda)=2(\sqrt{pq})^n T_n\left(\frac{-\lambda}{2\sqrt{pq}}\right)$. Since the Chebyshev polynomials are given by $T_n(\cos(\theta))=\cos(n\theta)$, this gives all the eigenvalues. The reason this happens is that (Denis' symmetric version of) your matrix is the Jacobi matrix for the Chebyshev polynomials. You can exploit this to get the eigenvectors in terms of lower order polynomials $T_m$, $m\leq n$, evaluated at the eigenvalue; the construction is in </p> <p>Gautschi, Walter. Orthogonal Polynomials, Computation and Approximation. Numerical Mathematics and Scientic Computation, Oxford University Press, 2004..</p> http://mathoverflow.net/questions/90221/why-are-lacunary-series-so-badly-behaved Why are lacunary series so badly behaved? Emilio Pisanty 2012-03-04T20:00:30Z 2012-03-05T18:53:08Z <p>Hi!</p> <p>I just came across the <a href="http://en.wikipedia.org/wiki/Ostrowski-Hadamard_gap_theorem" rel="nofollow">Ostroski-Hadamard gap theorem</a>, and while I can understand the proofs as well as the <a href="http://en.wikipedia.org/wiki/Lacunary_function#A_simple_example" rel="nofollow">principle</a> that the series $\sum_{n=0}^\infty z^{2^n}$ ought to have a singularity at every $2^n$-th root of unity for every $n$, I feel I'm missing some intuition into what exactly is going on. </p> <p>Specifically, there is certainly the intuition that the faster a power series' coefficients decrease, the larger the radius of convergence will be - say, comparing the geometric series with the exponential power series. When contrasted with lacunary series, this seems to fail: the coefficients seem to be increasingly "smaller", at least in an average sense, but the function becomes terribly ill-behaved. (One could try and argue that in the Cesàro sense the coefficients do tend to zero: if $\sum_{n=0}^\infty z^{2^n}=\sum_{k=0}^\infty a_k z^k$, then $\frac{1}{n}\sum_{k=0}^n a_k\approx\frac{\lfloor\log_2(n)\rfloor}{n}\rightarrow0$ as $n\rightarrow\infty$. On the other hand, the power series $\sum_{k=0}^\infty \frac{z^k}{k}$, while having the same radius of convergence, can easily, if non-uniquely, be analytically extended to the whole complex plane; I'd expect the same of any series of the form $\sum_{k=0}^\infty \frac{\log(k)}{k}z^k$.)</p> <p>Can anyone share some insight?</p> http://mathoverflow.net/questions/89496/subspace-where-an-operator-is-positive Subspace where an operator is positive Emilio Pisanty 2012-02-25T16:40:23Z 2012-02-25T16:40:23Z <p>Given a self-adjoint operator $\hat{T}$ on a Hilbert space $\mathcal{H}$, and assuming it has a basis of eigenvectors <code>$\{\phi_n\}$</code> such that $\hat{T}\phi_n=\lambda_n\phi_n$, one can consider the subspace <code>$$V=\textrm{span}\{\phi_n:\lambda_n&gt;0\},$$</code> inside of which $\hat{T}$ is positive definite. My question is, does there exist a way of defining this subspace invariantly in terms of $\hat{T}$ without making reference to its eigenvectors?</p> <p>(Some motivation: given the hydrogen-atom hamiltonian in quantum mechanics, $\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 -\frac{e^2}{r}$, the subspace where $\hat{H}&lt;0$ is the linear span of all bound states, which correspond to elliptical orbits in the analogous classical problem.)</p> http://mathoverflow.net/questions/130484/analytical-continuation-of-electrostatic-potentials Comment by Emilio Pisanty Emilio Pisanty 2013-05-13T17:50:55Z 2013-05-13T17:50:55Z @IgorKhavkine, does that mean I can expect to find points where Cauchy-Riemann is violated? If so, where should I look? http://mathoverflow.net/questions/1624/is-8-the-largest-cube-in-fibonacci-sequence/1629#1629 Comment by Emilio Pisanty Emilio Pisanty 2013-03-19T01:39:41Z 2013-03-19T01:39:41Z The link works but most of the definite articles are missing. http://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/1393#1393 Comment by Emilio Pisanty Emilio Pisanty 2013-03-18T13:16:13Z 2013-03-18T13:16:13Z ... and, similarly, you can translate between Fourier and Taylor series by seeing $\sum_{n=0}^\infty c_n e^{in\theta}$ as $\sum_{n=0}^\infty c_n z^n$ for $z=e^{i\theta}$. http://mathoverflow.net/questions/124356/on-a-hypergeometric-type-integral Comment by Emilio Pisanty Emilio Pisanty 2013-03-13T13:27:40Z 2013-03-13T13:27:40Z Yes, it's fixed now. http://mathoverflow.net/questions/117415/old-books-still-used/117493#117493 Comment by Emilio Pisanty Emilio Pisanty 2012-12-31T01:50:46Z 2012-12-31T01:50:46Z I'm unsure how widely Orthogonal Polynomials is used in the special functions community, but as a physicist it has saved me from quite a few tight spots that no other book could. Pretty hard to find and use, though. http://mathoverflow.net/questions/81960/the-dzhanibekov-effect-an-exercise-in-mechanics-or-fiction-explain-mathemati Comment by Emilio Pisanty Emilio Pisanty 2012-12-11T18:07:42Z 2012-12-11T18:07:42Z Related on physics.SE: Why does a cuboid spin stably around two axes but not the third? <a href="http://physics.stackexchange.com/q/34364/8563" rel="nofollow">physics.stackexchange.com/q/34364/8563</a> http://mathoverflow.net/questions/115125/iterating-random-matrix-operations Comment by Emilio Pisanty Emilio Pisanty 2012-12-02T00:30:33Z 2012-12-02T00:30:33Z I can't think of a better motivation for that problem. http://mathoverflow.net/questions/107159/pochhammer-symbol-of-a-differential-and-hypergeometric-polynomials Comment by Emilio Pisanty Emilio Pisanty 2012-09-15T17:03:42Z 2012-09-15T17:03:42Z @TomCopeland so it is. Let me check the details. http://mathoverflow.net/questions/33469/closed-form-for-an-orthogonal-polynomial-integral/33613#33613 Comment by Emilio Pisanty Emilio Pisanty 2012-09-14T08:55:07Z 2012-09-14T08:55:07Z &quot;The identities I needed were well-hidden in Abramowitz and Stegun and Gradshteyn and Ryzhik&quot; - as is so often the case. http://mathoverflow.net/questions/86864/when-can-a-family-of-polynomials-get-a-weight-function-to-be-made-orthogonal/102718#102718 Comment by Emilio Pisanty Emilio Pisanty 2012-07-20T15:57:21Z 2012-07-20T15:57:21Z Who is &quot;they?&quot; Formulating this as a question instead of an answer might give you better chances of a response, though it would be helpful if you flesh out your problem more. What, specifically, are you trying to do? What paper are you following? What specific problems are you having? http://mathoverflow.net/questions/99506/blackbox-theorems/99533#99533 Comment by Emilio Pisanty Emilio Pisanty 2012-06-28T11:56:35Z 2012-06-28T11:56:35Z @EvanJenkins: do you have a reference for the non-AC proof? When studying Haar measure construction I found a lot of texts doing only the compact case, and one text with an AC proof of the locally compact case. http://mathoverflow.net/questions/13638/which-popular-games-are-the-most-mathematical/13650#13650 Comment by Emilio Pisanty Emilio Pisanty 2012-06-28T11:43:54Z 2012-06-28T11:43:54Z Tic-tac-toe can in fact be mapped: <a href="http://xkcd.com/832/" rel="nofollow">xkcd.com/832</a> http://mathoverflow.net/questions/100824/existence-of-power-series-expansion-for-implicitly-defined-function Comment by Emilio Pisanty Emilio Pisanty 2012-06-28T01:09:48Z 2012-06-28T01:09:48Z Specifically, take $F[g(x)]=g(x)^3$ and $y(x)=x$, so they obey your regularity and invertibility conditions. The unique solution is $g(x)=x^{1/3}$, all of whose Taylor expansions around whatever point you choose have a finite radius of convergence, due to a branch point at $x=0$. http://mathoverflow.net/questions/100824/existence-of-power-series-expansion-for-implicitly-defined-function/100827#100827 Comment by Emilio Pisanty Emilio Pisanty 2012-06-28T01:06:18Z 2012-06-28T01:06:18Z If $F$ and $y$ are given and analytic, as in the OP's hypotheses, then your argument simply says that $y'(w)\neq0$ implies $g'(w)\neq0$, which is a condition on the unknown $g$ and thus probably acceptable. http://mathoverflow.net/questions/91232/about-a-christoffel-darboux-type-sum/100477#100477 Comment by Emilio Pisanty Emilio Pisanty 2012-06-25T00:29:52Z 2012-06-25T00:29:52Z (The asymptotic formulas they use, by the way, are not particularly satisfactory, since they do not account for the increasing spacing of the zeros near the edges. Better ones are in Szeg&#246;.)