most recent 30 from http://mathoverflow.net 2013-05-19T17:54:30Z http://mathoverflow.net/feeds/user/6542 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65768/a-mass-spring-model-for-hair-simulation/80899#80899 unknown (google) 2011-11-14T15:32:24Z 2011-11-14T15:32:24Z

<p>Notice that the only unknown quantities in your force equation are the updated velocities $v_i^{n+1}$, and that force depends linearly on these. Everything else is either known from the last time step, e.g. $x_i^n$, or is a simulation parameter/material constant.</p> <p>Therefore, after substituting for $\mathbf{a}$ in your first equation, you will end up with a set of <em>linear</em> equations in the unknowns $v_i^{n+1/2}$; in particular the system will be of the form $$(\mathbf{I} - c\Delta t^2\ \mathbf{d}\mathbf{d}^T)\mathbf{v}^{n+1/2} = \mathbf{b}$$ for some scalar $c$ and right-hand side $\mathbf{b}$. You can readily solve this system without Newton's method. In particular, for sufficiently small time steps $\Delta t$ the matrix is positive-definite, in which case I recommend using Conjugate Gradients to solve the system iteratively without ever even needing to form the dense matrix $\mathbf{d}\mathbf{d}^T.$</p>

http://mathoverflow.net/questions/74841/an-example-of-a-beautiful-proof-that-would-be-accessible-at-the-high-school-level/76575#76575 unknown (google) 2011-09-27T22:58:39Z 2011-09-27T22:58:39Z

<p>Minkowsky's Theorem (every convex region in the plane of area greater than 4 that's symmetric about the origin contains a lattice point other than (0,0)) is not at all obvious (are you sure you can't squeeze a sufficiently large "blob of irrational slope" in there?) but has a beautiful, simple, and surprising geometric proof.</p>

http://mathoverflow.net/questions/52178/finding-an-element-of-a-vector-subspace-contained-in-the-first-orthant unknown (google) 2011-01-15T18:49:24Z 2011-01-15T23:06:25Z

<p>Given a matrix $M$, I want to find a nontrivial vector in the kernel of $M$ that also lies in the first orthant, if such a vector exists. That is, I want to simultaneously solve</p> <p>$$Mx = 0$$ $$x \geq 0$$ $$x \neq 0$$</p> <p>I'm having trouble phrasing this problem in a way that can be efficiently solved numerically. One approach I've tried is to solve</p> <p>$$\min_x \|Mx\|^2\quad s.t. \quad x\geq 0, x_i = 1$$</p> <p>using nonnegative least squares for every $i$, and looking for solutions whose minimum is 0. If the minimum is positive for every $i$, the original problem had no solution. Unfortunately, in addition to being inefficient (I have to do $\dim x$ solves), standard least squares packages are having great difficulty converging for this approach.</p> <p>Is there a better way to solve this problem?</p>

http://mathoverflow.net/questions/74841/an-example-of-a-beautiful-proof-that-would-be-accessible-at-the-high-school-level/76575#76575 2011-11-20T11:14:46Z 2011-11-20T11:14:46Z

Draw the region $R$ in the plane. Cut the plane into 2x2 squares by cutting along the lines $x=2i$ and $y=2j$ for all integers $i,j$; each square contains some part of $R$ (possible none.) Stack the squares on top of each other. Since $R$ has area greater than 4, there exist two squares whose parts of $R$ overlap. Write down what this means algebraically, apply symmetry and convexity, and construct the nontrivial lattice point.