most recent 30 from http://mathoverflow.net 2013-06-20T04:35:19Z http://mathoverflow.net/feeds/user/19642 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131477/connection-between-the-number-of-vertices-and-the-number-of-lattice-points-of-the/131482#131482 Stephen Sturgeon 2013-05-22T16:56:38Z 2013-05-22T16:56:38Z

<p>In response to your first question the number of vertices does not control the number of lattice points. Consider the polytopes $P=conv[(1,1),(1,-1),(-1,-1),(-1,1)]$, and $Q=conv[(0,1),(2,-1),(-2,-1)]$. They have the same number of lattice points but different number of vertices. </p> <p>In regard to your second question it is easier to determine the vertices since these just come down to finding solutions to the defining hyperplanes (since you are not defining your polytope as the convex hull of a set of vertices I am assuming you are starting with the hyperplanes). </p> <p>For your third question, there is clearly no bound as you could always take a scalar multiple of your polytope to get more lattice points.</p>

http://mathoverflow.net/questions/131454/local-complete-intersection/131457#131457 Stephen Sturgeon 2013-05-22T14:52:36Z 2013-05-22T14:52:36Z

<p>A ideal defining a complete intersection has a regular sequence as a generating set. Then a local complete intersection would be a quotient ring which has a regular sequence as a generating set after some localization.</p>

http://mathoverflow.net/questions/131334/upper-bound-of-a-series/131344#131344 Stephen Sturgeon 2013-05-21T14:06:38Z 2013-05-21T14:06:38Z

<p>Since $\displaystyle \frac{k^a}{(k+1)^a+(k+2)^a}&lt;\frac{k^a}{k^a+k^a}=\frac{1}{2}$ then $\displaystyle\sum_{k=1}^N\frac{k^a}{(k+1)^a+(k+2)^a}&lt;\sum_{k=1}^N\frac{1}{2}=\frac{N}{2}$.</p>

http://mathoverflow.net/questions/131238/real-root-of-a-cubic-equation/131242#131242 Stephen Sturgeon 2013-05-20T16:29:27Z 2013-05-20T16:29:27Z

<p>I am assuming you have some different example in mind since as Barry Cipra pointed out your objective is not true in your example (and any example that eventually has all positive coefficients). First of all every cubic has a real root by the intermediate value theorem. Hence if you want to prove the positivity I recommend you start with the general solution to the cubic. It is ugly, but you could perhaps derive positivity (if it is true) in your class of examples. </p>

http://mathoverflow.net/questions/130577/how-to-find-overlap-between-two-convex-hulls-along-with-the-overlap-area/130591#130591 Stephen Sturgeon 2013-05-14T14:55:10Z 2013-05-14T14:55:10Z

<p>This is partial answer. I assume you are dealing with convex polygons. Then you could put all your vertices in a matrix ($M_1$ for $P_1$, $M_2$ for $P_2$). Then for the convex hull to overlap you would need that $M_1 \bar{x}$=$M_2 \bar{y}$ has a solution, where $\bar{x}=(x_1,...,x_m)$ and $\bar{y}=(y_1,...,y_n)$, $x_i,y_i\geq0$ and $\sum_{i=1}^mx_i=1$, and $\sum_{i=1}^ny_i=1$. Then if you can perform the same row operations on both matrices such that one row becomes negative in one matrix and the corresponding row positive in the other matrix then you have shown that the polytopes do not overlap. This is equivalent to performing a change of basis to put one polytope on the positive side of a hyperplane and the other on the negative side of the same hyperplane. </p>

http://mathoverflow.net/questions/130483/an-example-of-a-ring-r-with-the-property-that-for-each-pann-rr-in-rm-min/130487#130487 Stephen Sturgeon 2013-05-13T13:46:14Z 2013-05-13T18:19:31Z

<p>I believe the graded ring $k[x,y]/(xy)$ satisfies this property. The minimal primes are just $(x)$, $(y)$ and the annihilators are just the other ideal. The dimension is 1. This is not however local.</p>

http://mathoverflow.net/questions/128519/secondary-polytope-simplicial Stephen Sturgeon 2013-04-23T18:34:11Z 2013-04-25T13:39:14Z

<p>Is the secondary polytope of a simplicial polytope necessarily simplicial?</p>

http://mathoverflow.net/questions/128519/secondary-polytope-simplicial/128715#128715 Stephen Sturgeon 2013-04-25T13:39:14Z 2013-04-25T13:39:14Z

<p>The secondary polytope of a simplicial polytope is not neccessarily either simplicial or simple. The secondary polytope of the cyclic 4-polytope with 8 vertices in given as an example in by Billera, Filliman, and Sturmfels in "Constructions and complexity of secondary polytopes" - Adv. Math. 83 (1990), no. 2, 155–179. It is clearly neither simplicial or simple.</p>

http://mathoverflow.net/questions/105977/reversing-the-carry-operation-in-multiplication/105980#105980 Stephen Sturgeon 2012-08-30T18:07:25Z 2012-08-30T18:07:25Z

<p>Your "inverse" operation is not clearly defined. You need some condition on the $b_k$ in order to ensure uniqueness. For instance, $1001$, $121$, $113$ all evaluate to the same value in base two. </p>

http://mathoverflow.net/questions/105976/polar-duality-and-1 Stephen Sturgeon 2012-08-30T17:43:58Z 2012-08-30T18:06:58Z

<p>We define the polar dual of a polytope $P$ as the set <code>$$\{x\in \mathbb{R}^n: x \cdot a\geq -1 \text{ for all } a\in P\}$$</code> Why do we require $-1$ instead of $-2$ or any other constant?</p>

http://mathoverflow.net/questions/82286/rational-binomial-identity Stephen Sturgeon 2011-11-30T16:21:10Z 2011-11-30T18:21:23Z

<p>Can anyone give a reference, a proof, or a reference that explains why Maple can evaluate this identity mathematically correctly:</p> <p>$$n-i-1=(d-1)\sum_{l=1}^{n-i-1}\frac{\binom{n-i-1}{l}}{\binom{n-i+d-3}{l}}$$</p>

http://mathoverflow.net/questions/131398/identity-for-number-of-monomials Stephen Sturgeon 2013-05-22T15:03:48Z 2013-05-22T15:03:48Z

Could you give a small example of $b(n,k)$ and the associated monomials in some small case?

http://mathoverflow.net/questions/131445/computational-ring-theory Stephen Sturgeon 2013-05-22T14:56:57Z 2013-05-22T14:56:57Z

Macaulay2 is also good software for commutative rings.

http://mathoverflow.net/questions/130727/is-there-a-name-for-this-graph Stephen Sturgeon 2013-05-15T15:05:21Z 2013-05-15T15:05:21Z

What is wrong with calling it the Hasse diagram of the inclusion poset?

http://mathoverflow.net/questions/130508/ideals-of-a-noetherian-ring-r-cohen-macaulay-as-r-modules Stephen Sturgeon 2013-05-13T18:24:00Z 2013-05-13T18:24:00Z

You will at least need to look outside of the polynomial ring. No ideals are Cohen-Macaulay as $k[x_1,...,x_n]$ modules by Boij-S\&quot;oderberg decomposition.

http://mathoverflow.net/questions/130483/an-example-of-a-ring-r-with-the-property-that-for-each-pann-rr-in-rm-min/130487#130487 Stephen Sturgeon 2013-05-13T18:18:21Z 2013-05-13T18:18:21Z

Consider $R=k[x,y]/(xy)$. Then R is just the vector space spanned by $(1,x,x^2,...,y,y^2,...)$ because any mixed terms of $xy$ will be eliminated. Thus we can calculate the annihilator by considering products of things in this basis. $x^i*y^j=0$ where as all other products just give a power of a variable. Hence the annihilators are proven. We note that $(x)$ and $(y)$ are the minimal primes. Then $(x) \subset (x,y) \subset R$ is a maximal chain of primes so the dimension is 1.

http://mathoverflow.net/questions/130480/can-you-find-any-uncountable-set Stephen Sturgeon 2013-05-13T13:33:14Z 2013-05-13T13:33:14Z

Your arguments are flawed in several places. You should talk to a mathematician in person to have all of it refuted and cleared up in your mind. I will just say that your continuum method is clearly flawed if you consider any real number with an infinite expansion. Then your &quot;mirror&quot; correspondence relates it to an infinite sequence of integers, but when you write out an infinite sequence of integers this is not an integer. Hence the &quot;mirror&quot; operation only produces an integer when you are dealing with a real number with finite decimal expansion. This doesn't even cover the rationals.

http://mathoverflow.net/questions/128730/a-question-about-graph-theory Stephen Sturgeon 2013-04-25T16:58:24Z 2013-04-25T16:58:24Z

Your question seems a bit unclear in terminology. Is this the same question: Given two graphs G and H on the vertex set {v_1,...,v_n} such that deg(v_i) in G is the same as deg(v_i) in H, and the vertices {v_{i_1},...,v_{i_k}} form a k-cycle in G if and only if they form a k-cycle in H, then is G isomorphic to H?

http://mathoverflow.net/questions/128519/secondary-polytope-simplicial/128559#128559 Stephen Sturgeon 2013-04-25T13:32:03Z 2013-04-25T13:32:03Z

I apologize for the confusion. The secondary polytope of an n-gon is the associahedron, which is simple not simplicial. By duality I can always study the dual simplicial polytope in this situation. My question should be rephrased as &quot;Is the secondary polytope always either simple or simplicial?&quot;