User hans lundmark - MathOverflow

Answer by Hans Lundmark for Bertrand theorem - central forces

2011-01-08T10:16:37Z

Since you say central forces in the title, this looks to me like the standard Bertrand theorem described in Goldstein's Classical Mechanics (Section 3.6, proof in Appendix A). Or is there something nonstandard about your assumptions that I have missed?

Answer by Hans Lundmark for Nice Classes of Non-Closable Operators

2010-11-27T16:57:35Z

Here's another simple example of a non-closable operator, from Reed & Simon Vol. I, Section VIII.4, Example 4 (p. 252): Let $D(T)$ consist of those $\psi\in L^2(\mathbf{R})$ such that $\int|f\psi|<\infty$, where $f$ is a fixed function not in $L^2(\mathbf{R})$, and set $T\psi=(f,\psi)\psi_0$ for some fixed $\psi_0 \in L^2(\mathbf{R})$. Then $D(T)$ is dense in $L^2(\mathbf{R})$; however, as they show by a short computation, $D(T^*)$ consists of the vectors orthogonal to $\psi_0$, so it's not dense.

Answer by Hans Lundmark for Local linearization of ODE at singular point

2010-08-30T07:09:49Z

In three dimensions, Hartman gave the example $dx/dt=ax$, $dy/dt=(a-b)y+cxz$, $dz/dt=-bz$ where $a>b>0$ and $c \neq 0$. On the other hand, any $C^2$ planar flow is $C^1$ linearizable (another result by Hartman), so you will not find any polynomial examples in the plane. See Linearization via the Lie Derivative by Chicone & Swanson for references and more details.

(Strengthening the smoothness assumption will not give any more than $C^1$ linearizability even in the planar case, as shown by the $C^{\infty}$ example $dx/dt=-x$, $dy/dt=-2y+x^2$, which is easily solved explicitly: $x(t)=x_0 e^{-t}$, $y(t)=(y_0+x_0^2 t) e^{-2t}$. The solution curves are of the form $y=Ax^2-B x^2 \ln|x|$, and are therefore only $C^1$ at the origin, whereas the linearized system of course has smooth solution curves.)

Answer by Hans Lundmark for Are there other nice math books close to the style of Tristan Needham?

2010-07-15T14:38:12Z

Nonlinear dynamics and chaos by Steven Strogatz. Lots of pictures, intutive and clear explanations, interesting applications, great humor.

Answer by Hans Lundmark for What's your favorite equation, formula, identity or inequality?

2010-07-15T09:10:49Z

I have a soft spot for Heine's formula from the theory of orthogonal polynomials (since the proof is such a pretty calculation):

If $\mu$ is a measure with finite moments $\beta_k=\int x^k d\mu(x)$, then

$$\det(\beta_{i+j})_{i,j=0,\ldots,k-1} = \frac{1}{k!} \int \cdots \int \Delta(x_1,\ldots,x_k)^2 d\mu(x_1) \cdots d\mu(x_k)$$

where $\Delta$ is the Vandermonde determinant.

Answer by Hans Lundmark for Strengthening the Induction Hypothesis

2010-07-13T14:31:30Z

Here's a small example from the book Concrete Mathematics:

p. 78: Let $K_0=1$ and $K_{n+1}=1+\min(2 K_{\lfloor n/2 \rfloor}, 3 K_{\lfloor n/3 \rfloor})$ for $n\ge 0$. ("One of the authors of this book has modestly decided to call these the Knuth numbers.")

p. 97, exercise 3.25: Prove or disprove that the Knuth numbers satisfy $K_n \ge n$ for $n\ge 0$.

Induction fails when trying to prove $K_n \ge n$ directly (as explained in the text on p. 79), but works easily for the stronger statement $K_n > n$.

Answer by Hans Lundmark for Eigenvectors of a certain big upper triangular matrix

2010-05-30T10:16:12Z

Your matrix is totally nonnegative (i.e., all minors are nonnegative). This is because it can be factorized as the matrix of binomial coefficients (which is totally nonnegative by the Karlin–McGregor–Lindström–Gessel–Viennot lemma) times a diagonal matrix with positive entries $1/(2k)!!$ on the diagonal.

For an oscillatory matrix (i.e., a totally nonnegative matrix such that some power of it is totally positive), there is a theorem by Gantmacher & Krein which says that the eigenvalues are real and simple, and the eigenvector corresponding to the $k$th largest eigenvalue has $k-1$ sign changes. (Theorem 5.3 in Pinkus, Totally positive matrices.)

Unfortunately that doesn't apply here, since a power of an upper triangular matrix is upper triangular, so that some minors (below the diagonal) are always zero; hence your matrix is not oscillatory. But perhaps it is possible to use similar ideas to prove the sign changes in your case?

Answer by Hans Lundmark for Undergraduate Level Math Books

2010-05-25T15:32:11Z

For (applied) ODEs: Nonlinear dynamics and chaos by Steven Strogatz.

A very inspiring book! The explanations are crystal clear with lots of pictures. And it's funny too – the "Romeo and Juliet" illustration of 2-dimensional linear systems (Section 5.3) is a classic.

Answer by Hans Lundmark for Geodesics on a hyperbolic paraboloid

2010-05-24T07:57:14Z

I haven't worked out the details, but this should be doable in paraboloidal coordinates, in the same way that the geodesics on an ellipsoid were computed by Jacobi using his famous ellipsoidal coordinates (invented for that very purpose).

In the notation of the Wikipedia article, take the parameters to be $A=1$ and $B=-1$; then the coordinate surface $\mu=0$ is your paraboloid $2z=y^2-x^2$, and the geodesics on any coordinate surface should be possible to integrate (more or less) explicitly, although the details might be a bit messy. The idea is to rewrite the Hamiltonian for geodesic motion, $H=(p_x^2+p_y^2+p_z^2)/2$, in the new coordinates and apply the Hamiltion–Jacobi method (the Hamilton–Jacobi equation is separable in these coordinates).

I haven't got time to look for a good reference explaining Jacobi's work right now. I'll update the answer later if I find something.

Edit: Here's what I think is the most convenient, and least error-prone, way of setting up the equations for the geodesics. Change to paraboloidal coordinates in $R^3$; I'll call them $(u_1, u_2, u_3)$ instead of $(\lambda, \mu, \nu)$. Since this is an orthogonal coordinate system, the Euclidean metric tensor is diagonal, $ds^2=\sum_{k=1}^3 h_k^2 du_k^2$, where $h_1$, $h_2$, $h_3$ are the scale factors given in the Wikipedia article. The hyperbolic paraboloid that you are interested in is the coordinate surface $u_2=0$, which is a Riemannian manifold in itself, with coordinates $(u_1,u_3)$ and metric tensor given by $ds^2 = h_1^2 du_1^2 + h_3^2 du_3^2$ (where of course $u_2=0$ should be substituted into the expressions for the scale factors). On any Riemannian manifold, the geodesic equations are the canonical Hamiltonian equations given by the Hamiltonian function $H=\frac{1}{2} g^{ij} p_i p_j$, where $g^{ij}$ is the inverse metric tensor. In this case, we get $H(u_1,u_3,p_1,p_3) = \frac{1}{2} \left( \frac{p_1^2}{h_1(u_1,u_3)^2} + \frac{p_3^2}{h_3(u_1,u_3)^2} \right)$. So just feed this function to Mathematica, and numerically integrate the equations $$\dot{u}_1 = \partial H/\partial p_1,$$ $$\dot{u}_3 = \partial H/\partial p_3,$$ $$\dot{p}_1 = -\partial H/\partial u_1,$$ $$\dot{p}_3 = -\partial H/\partial u_3,$$ with suitable initial conditions. This will give you a geodesic emanating from a given point in a given direction. The result is in terms of paraboloidal coordinates, of course, but it is trivial to express it in terms of Cartesian coordinates (for plotting) using the formulas defining the change of variables. Finding a geodesic between two given points seems more complicated; perhaps use some "shooting" algorithm?

As I wrote above, it should be possible to integrate the equations by hand, but numerical integration seems to suffice for your purposes.

For the explicit integration of the geodesics on the ellipsoid, Jacobi's own lectures are probably as good a source as anything else (if you can read German). They are available at the Internet Archive. Elliptic coordinates are described in Lecture 26, the geodesics on the ellipsoid in Lecture 28 (p. 212).

Answer by Hans Lundmark for A plausible inequality.

2010-04-02T18:48:25Z

A similar (maybe slightly simpler) counterexample: $n=2$, $x=(2,0)$, $y=(0,2)$, arbitrary $a_1>a_2$.

Answer by Hans Lundmark for If the second derivative of a function on $\mathbb R^n$ is everywhere nondegenerate, does it follow that the first derivative is an injection?

2010-04-02T09:21:15Z

Regarding comment 3: If you restrict to polynomial functions $g$, then the question becomes a famous open problem (the Jacobian conjecture).

Comment by Hans Lundmark

2011-09-18T13:26:05Z

@suvrit: To the two different parenthesizations mentioned in the preceding sentence.

Comment by Hans Lundmark

2011-09-13T09:29:22Z

"Lindström", to be precise.

Comment by Hans Lundmark

2011-08-13T09:01:55Z

Your terminology seems a little nonstandard. $X_f$ is usually called the Hamiltonian vector field associated to $f$. When one speaks of "the Hamiltionian", this refers the the function $f$.

Comment by Hans Lundmark

2011-02-23T17:50:43Z

...and when I clicked it, I ended up at the Chern essay. (Perhaps not so surprising in hindsight, since both links point to the same URI.)

Comment by Hans Lundmark

2011-02-08T15:49:00Z

Broken link. Hopefully this works better: fr.wikipedia.org/wiki/…

Comment by Hans Lundmark

2011-02-04T08:50:29Z

@Dick: I think the new name was introduced for marketing reasons, and is mainly used by the followers of David Hestenes.

Comment by Hans Lundmark

2011-02-03T11:00:36Z

...and by the way, I wish I could refer you to the book that a colleague of mine is writing, but unfortunately it is not finished yet: mai.liu.se/~anaxe/GMA.html

Comment by Hans Lundmark

2011-02-03T10:57:35Z

@Dick: GA is really just another name for Clifford algebras, and there are determinants everywhere if you do coordinate calculations in a Clifford algebra. For example, the highest graded part of the Clifford product of two (homogeneous) multivectors is the exterior product of those multivectors, and exterior product is related to determinants in a way that you're probably familiar with.

Comment by Hans Lundmark

2011-01-24T22:04:47Z

@unknown: Bob Palais, "$\pi$ Is Wrong", Opinion column in Math Intelligencer, Vol. 23, No. 3, 2001.

Comment by Hans Lundmark

2011-01-09T15:06:25Z

Here's another reference that can be read on Google books: [Boccaletti & Pucacco, Theory of orbits](books.google.com/…).

Comment by Hans Lundmark

2010-12-08T07:58:36Z

@Victor: Here are two examples by Flaschka: scholar.google.com/… (although this might be considered "cheating", since the Phys Rev B paper was probably labelled "II" instead of "I" by accident).

Comment by Hans Lundmark

2010-11-30T09:47:28Z

Kenny, this site is not for general math questions, but for questions of interest to research mathematicians. Please read the FAQ (link at the top of the page) where you will find suggestions for sites that are more appropriate for your question.

Comment by Hans Lundmark

2010-11-16T09:23:12Z

If you are bothered by the existence question, identify these numbers with the real 2 by 2 matrices of the form [x,y ; y,x]. The real numbers then correspond to multiples xI of the identity matrix, whereas j corresponds to [0,1 ; 1,0] (not a matrix corresponding to a real number, but still its square is the identity matrix, which corresponds to the real number +1).

Comment by Hans Lundmark

2010-11-07T21:21:46Z

"Submitted on 19 Oct 2010"; that's pretty good timing! ;)

Comment by Hans Lundmark

2010-11-07T21:05:35Z

Gessel & Viennot give credit to the earlier results in their paper: "Arguments similar to the one of Lemma 5 have been used by Chaundy [3], Karlin and MacGregor [14], and Lindström [18]". The paper by Chaundy that they refer to is "The unrestricted plane partition" from 1932.