User the cheese stands alone - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:52:40Z http://mathoverflow.net/feeds/user/12722 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58211/possibility-of-an-elementary-differential-geometry-course Possibility of an Elementary Differential Geometry Course The Cheese Stands Alone 2011-03-11T23:43:42Z 2012-05-05T15:46:16Z <p>I have to admit I'm not sure if this is an appropriate question. It's related to research in math education, but not directly to math.</p> <p>I've found that in talking to professional physicists and engineers, most of them find some use for differential geometry nowadays. One theoretical physicist went as far as to say you could "do nothing serious without it." Yet at most schools (at least the few I've looked at) differential geometry is reserved for graduate students in math and advanced math undergraduates. No schools I looked at had an elementary differential geometry class in, say, a similar style as the calculus sequence. Some of the people I talked to also expressed a lot of difficulty in learning it for the first time on their own. I myself am taking an advanced graduate course in General Relativity, and a good portion of the difficulty of the students is in misuderstanding the fundamental concepts of differential geometry.</p> <p>To cover differential geometry rigorously, of course one needs quite a bit of advanced mathematics, including topology and analysis. But universities teach elementary calculus classes, most of which are not terribly rigorous, but are sufficient for the purposes of non-mathematicians. Linear algebra, multivariate calculus, and a bit of differential equations would (in my mind) be sufficient to teach a course for engineers. You might argue that one needs to know the theory of manifolds first, but I see this as analagous to studying calculus without really knowing the structure of $\mathbb{R}$.</p> <p>From my viewpoint, differential geometry is the logical extension of calculus. Based on it's huge (and growing) impact on applied disciplines, It seems logical to have a course in it for engineers and physicists, which I would put immediately after the final semester of calculus (assuming the students have also had linear algebra).</p> <p>So my question is this: Are there specific instances, either textbooks or courses at a university, of differential geometry classes taught with the intent of being useful for engineers and scientists, which assume only basic calculus knowledge and linear algebra? (Obviously, there are books like "Differential Geometry for Physicists," but I really mean something that would be used by mathematicians teaching such a course). If so, how successful have these courses/books been? If not, or if the attempts have been unsuccessful, is there any particular reason as to why it is not feasable/common?</p> http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Answer by The Cheese Stands Alone for Finding the degree of minimal polynomials The Cheese Stands Alone 2011-04-17T02:37:11Z 2011-04-17T18:48:24Z <p>No. Some conditions are needed on the $a_i$ and $p_i$. For instance, take n=2, $a_1 = a_2 = 2$, $p_1 = p_2 = 2$. Then $x = 2 \sqrt{2}$, which has minimal polynomial $x^2 - 8$. As an even simpler example, n=1, $a_1 = 2$, $p_1 = 4$, then $x$ is rational.</p> <p>For a less trivial example, take $a_1= 4$, $a_2 = 6$, $p_1=p_2=2$. Check that this has a polynomial of degree 12. In fact, this isn't really true at all. </p> <p>One can, however, prove that the degree of the minimal polynomial is at most $\prod a_n$, which is an easy exercise in field theory. Any graduate algebra textbook covering Galois theory will be more than sufficient to prove this; just remember the degree of the minimal polynomial is the same as the dimension of the extension field viewed as a vector space over the base field.</p> <p>EDIT:</p> <p>After much miscommunication on my part, we've reached the following results:</p> <p>Suppose $a_1,\ldots,a_n$ are pairwise relatively prime positive integers, $p_1, \ldots, p_n$ integers such that $\sqrt[a_i]{p_i}$ is of degree $a_i$ for each i. Then $\sqrt[a_1]{p_1} + \cdots + \sqrt[a_n]{p_n}$ is of degree $\displaystyle \prod_{i=1}^n a_i$.</p> <p>The condition that each $\sqrt[a_i]{p_i}$ is met (by Eisenstein Criterion) should there be a prime $q_i$ such that $q_i | p_i$ and $q_i^2 \not{|} p_i$ for each i.</p> http://mathoverflow.net/questions/59514/maximum-surface-area-among-convex-subsets-of-the-unit-sphere-of-a-given-volume Maximum surface area among convex subsets of the unit sphere of a given volume The Cheese Stands Alone 2011-03-25T04:04:09Z 2011-03-25T04:04:09Z <p>The following problem is listed in Steven Lay's "Convex Sets and Their Applications" (1982) as unsolved (paraphrased):</p> <p>Let $B$ be the unit ball in $\mathbb{R}^3$ and $0 &lt; V &lt; \pi$. Define $\mathcal{F}$ as the family of all convex subsets of $B$ with volume $V$. Find the member of $\mathcal{F}$ with maximum surface area.</p> <p>The conjectured answer (as of 1982) is $B \cap$ { $(x_1,x_2,x_3)\in \mathbb{R}^3 | |x_1| &lt; c$} for appropriately chosen $c$.</p> <p>Does anyone know if this problem has been solved, or if any progress has been made on it? I couldn't find any recent references in the literature to it.</p> http://mathoverflow.net/questions/58688/subspace-of-mathbbrn-spanned-by-the-image-of-convex-n-1-polyhedra-under Subspace of $\mathbb{R}^n$ spanned by the image of convex $(n-1)$-polyhedra under the face-counting map The Cheese Stands Alone 2011-03-16T21:35:24Z 2011-03-17T14:12:46Z <p>Fix $n \in \mathbb{N}$. A convex polyhedron $C$ in $\mathbb{R}^n$ is the convex hull of finitely many points with nonempty interior. For $H$ a supporting hyperplane, ie $C$ is contained in one of the two closed half-spaces bounded by $H$, we call $H \cap C$ a $j$-face of $C$, where $j$ is the affine dimension of $H \cap C$. By convention, $\varnothing$ is called a $-1$-face of $C$ and $C$ an $n$-face of itself.</p> <p>Define a function $F$ from the set of convex polyhedra to $\mathbb{R}^{n+2}$ by coordinates, so that $F(C) = (a^C_{-1}, ..., a^C_n)$, where $a^C_j$ is the number of $j$-faces of $C$ for $j=-1,...,n$. Let $W$ be the affine subspace of $\mathbb{R}^{n+2}$ generated by $\operatorname{im} F$.</p> <p>It's clear that $a^C_{-1}=1$ and $a^C_n=1$. Euler's formula $\displaystyle \sum_{j=-1}^n (-1)^j a^C_j = 0$ (which may be more familiar as the Euler characteristic $V+E-F=2$ in the case of $n=3$) is a third affine relation between the $a^C_j$'s. Hence, $\operatorname{dim}W \le n-1$. </p> <p>Is it always true for any n that $\operatorname{dim}W = n-1$? Put differently, for any $n$, are the three equations above the only affine relationships that must be satisfied by $a^C_j$'s for all convex polyhedra $C \subset \mathbb{R}^n$, or is there some $n$ in which there is another relation?</p> <p>I seem to recall an affirmative answer to this, but I can't remember how it was solved or where I found it.</p> http://mathoverflow.net/questions/58203/erdos-distance-problem-n12/58207#58207 Answer by The Cheese Stands Alone for Erdos distance problem n=12 The Cheese Stands Alone 2011-03-11T23:07:24Z 2011-03-11T23:07:24Z <p>I wish I could get images to work, but here goes my poor explanation:</p> <p>Take a regular hexagonal lattice with distance 1 between nearest neighbors, and choose a 15-point equilateral triangle in this lattice (15 is a triangular number). Remove the 3 verticies of the triangle. You'll be left with 12 points and 5 distinct distances.</p> <p>edit: just checked the OEIS reference, and it's available on google books. The picture you want is on page 200 at <a href="http://books.google.com/books?id=cT7TB20y3A8C&amp;printsec=frontcover&amp;dq=Research+Problems+in+Discrete+Geometry&amp;source=bl&amp;ots=amqJ7zFfB4&amp;sig=U99_5spjO8UIwbehycahkz6M2yg&amp;hl=en&amp;ei=Hql6TeyFKpDrrAHm7bzCBQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=5&amp;ved=0CEMQ6AEwBA#v=onepage&amp;q&amp;f=false" rel="nofollow">http://books.google.com/books?id=cT7TB20y3A8C&amp;printsec=frontcover&amp;dq=Research+Problems+in+Discrete+Geometry&amp;source=bl&amp;ots=amqJ7zFfB4&amp;sig=U99_5spjO8UIwbehycahkz6M2yg&amp;hl=en&amp;ei=Hql6TeyFKpDrrAHm7bzCBQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=5&amp;ved=0CEMQ6AEwBA#v=onepage&amp;q&amp;f=false</a></p> http://mathoverflow.net/questions/57820/is-there-a-mathematical-axiomatization-of-time-other-than-perhaps-entropy/57899#57899 Answer by The Cheese Stands Alone for Is there a mathematical axiomatization of time (other than, perhaps, entropy)? The Cheese Stands Alone 2011-03-09T01:46:31Z 2011-03-09T03:58:28Z <p>A bit tangential, but I'm surprised no one mentioned time in general relativity (Ken Knox discussed special relativity, but the case in general relativity is subtly different). This discusses relativity from a perspective closer to a physicist's, since it's a bit more elementary (and hence easier to understand) in my view. As discussed at the end, relativity in general is somewhat incompatible with statistical mechanics, at least under the standard approximations, so this is almost surely not what you're looking for, but it may be of use to someone.</p> <p>In GR, space-time is a 4-manifold which is endowed with a Lorentzian metric $g_{ab}$, which is a rank 2 covariant tensor. The scalar product of two vectors $V$ and $W$ is then $g_{ab}V^aW^b$, from which we can compute things as in special relativity (where the metric is $\eta_{ab}$, which is $0$ if $a \ne b$, -1 if $a=b$ and $x^a$ is a spacial coordinate (i.e. x,y,z), and 1 if $a=b$ and $x^a=t$ is the time coordinate).</p> <p>If a vector (field) V satisfies $g_{ab}V^aV^b >0$, we call it time-like, and similarly for trajectories based on their tangent vectors. These are the possible trajectories for particles with positive mass. Any time-like trajectory corresponds in the limit of low mass and low speed to a local frame of reference in which it is the 'time', and if the trajectory is a geodesic then the frame is inertial. Particles with 0 mass (i.e. light) have null trajectories, with $g_{ab}V^aV^b =0$. If $g_{ab}V^aV^b &lt;0$, the vector is space-like. Particles with no forces (other than gravity) acting on them travel on geodesics. The trajectory of a massive particle is called it's world line. The sign conventions here are often reversed, so care is advised. For a time-like path, we can compute its length by $ds^2 = g_{ab} x^a x^b$, where $x$ are your coordinates.</p> <p>For any observer, they observe themself as stationary, and traveling forward in time with unit speed. Time for that observer corresponds exactly to the length of their trajectory. That observer can even set up a local set of coordinates in which the metric is approximately $\eta_{ab}$, provided all masses are sufficiently far away. Locally, then, time behaves like in special relativity. The big difference between special and general relativity is that the latter has no inertial reference frames in general, so observers can only measure times in at their own location.</p> <p>To summarize, time is another coordinate in space-time, just like space, but it isn't universal (it's observer dependent), and the only real restrictions on it is that it must be the part of the metric that has positive signature (in the sense above).</p> <p>Since this was no-doubt useless in explaining the difference between time in special and general relativity, I recommend Hughston &amp; Tod's <em>An introduction to general relativity</em>. It's fairly light reading and has an introduction to special relativity, but it's rigorous enough for mathematicians reading it. A number of other books at a higher level are available, of which Hawking &amp; Ellis, Wald, and Misner, Thorne, &amp; Wheeler are all good references.</p> <p>However, if you're looking for a formulation in which the second law of thermodynamics is provable, or even where entropy is defined, relativity is the wrong place to look. Even in special relativity, concepts like thermal equilibrium depend on a particular reference frame, so entropy may be defined in one inertial frame but not another. There has historically been great debate over how temperature (which is the thermodynamic conjugate of entropy) should transform under Lorentz transformations, and it's still not totally resolved.</p> http://mathoverflow.net/questions/74475/notable-math-from-those-without-math-phds Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-09-04T02:28:44Z 2011-09-04T02:28:44Z Insofar as theoretical computer science is a part of mathematics, Chomsky should qualify. Despite being a linguist, his work helped shape modern theoretical computer science. http://mathoverflow.net/questions/70831/reconstructing-matrices Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-07-21T18:30:31Z 2011-07-21T18:30:31Z @Gerry My solution can be modified to accomodate this fairly easily, concatenating all the entries of A and B into A'(1,1) and setting B' to be a matrix with all entries -1. http://mathoverflow.net/questions/70831/reconstructing-matrices Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-07-20T18:22:15Z 2011-07-20T18:22:15Z To expand on Emil's comment a little bit, we need to know if you want to exclude similarity over $\mathbb{Q}$, $\mathbb{C}$, $\mathbb{F}_2$, or what. In any case, it isn't clear to me whether A' and B' must both contain only zeros and ones, or if they can contain integers. If the latter, an easy solution is to view the matrix A as encoding an integer via decimal expansion, and stick that in the (1,1) position of A', then choose the remaining diagonal entries to make sure A and A' have different traces. http://mathoverflow.net/questions/68272/is-it-possible-to-construct-a-finite-mathematical-universe Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-06-20T13:09:38Z 2011-06-20T13:09:38Z It seems to me that the &quot;conventional mathematics of infinity&quot; was started by Cantor much less than 400 years ago. Of course, there is a concept of infinity in calculus, but that's something totally different in my book. http://mathoverflow.net/questions/8056/what-are-good-non-english-languages-for-mathematicians-to-know/8070#8070 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-06-05T06:20:26Z 2011-06-05T06:20:26Z If you're going to choose a language like C which is really only useful for computational programs (and software design, but I assume most mathematicians don't do too much of that), you might as well say Fortran rather than C. It's still the language of choice in most hard sciences. http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-04-17T18:39:26Z 2011-04-17T18:39:26Z @Georges Elencwajg that's exactly what I meant. Sorry for the terrible miscommunication. I've added this to the answer, with all the hypotheses clearly stated. http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-04-17T18:37:38Z 2011-04-17T18:37:38Z In this case, the proof I had in mind is by induction on n on the stronger statement that if $a$ and $b$ are of degree $k$, $m$, for $m,n$ relatively prime, the degree of $a+b$ is $km$. I was under the impression this was a well-known result, but it may be incorrect. If it is true, then Eisenstein gives that $x^{a_i} - p_i$ is the minimal polynomial for $sqrt[a_i]{p_i}$, so $sqrt[a_i]{p_i}$ is of degree $a_i$. I suppose we need the much stronger condition that $gcd(a_i,a_j)=1$ for $i \not{=} j$ rather than just $gcd(a_1,\ldots,a_n)=1$ for this. I blame this obvious mistake on lack of sleep. http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-04-17T18:05:11Z 2011-04-17T18:05:11Z Perhaps it will be helpful to enunciate my claim fully. Suppose $p_i$ are integers, $a_i$ natural numbers, for $i=1,\ldots,n$, such that for each $p_i$ there is a prime $q_i$ such that $q_i|p_i$ and $q_i ^2 \not{|} p_i$, and additionally that $gcd(ai,\ldots ,an)=1$. Then I claim that $\sqrt[a_1]{p_1} + \cdots + \sqrt[a_n]{p_n}$ has degree $a_1 \cdots a_n$ over $\mathbb{Q}$. All the other claims I made, the &quot;proofs&quot; I thought I had were flawed except in the case n=2. Please disregard them. http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-04-17T06:57:13Z 2011-04-17T06:57:13Z Scratch what I said above; it's not true unless each $p_i^{1/a_i}$ has minimal polynomial of degree $a_i$. This holds in the case that some prime q divides $p_i$, but $q^2$ doesn't divide $p_i$, by the Eisenstein Criterion. With more advanced arguments, a little bit stonger statements can be made. It fails in the case where we take $\sqrt{4}$. Another similar case is that the kth root of unity $e^{i2 \pi /k}$ satisfies the polynomial $a^{k−1}+\cdots+x+1$, which is of degree k−1. But when $p_i^{1/a_i}$ is of degree $a_i$ over $\mathbb{Q}$, the above should hold. http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-04-17T03:00:45Z 2011-04-17T03:00:45Z There are a good number of cases where your statement will hold, but it's not true in general. If the $a_i$ and the $p_i$ aren't related in any obvious way, it's probably true for small values of n, but I'd still check. Wolfram alpha (www.wolframalpha.com) can compute the minimal polynomials in sufficiently small examples, and most computational algebra engines can do it for arbitrary numbers of the form you want. http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-04-17T02:58:40Z 2011-04-17T02:58:40Z I'm not totally sure what having $gcd(a_n,p_n)=1$ gives you (or conditions regarding pairs $(a_n,p_n)$), but if you have $gcd(a_1,...,a_n)=1$ then your statement holds. If $gcd(p_1, \ldots ,p_n)=1$ and $a_1= \cdots = a_n$, it should also hold. I'm not sure about the general case for $gcd(p_1, \ldots ,p_n)=1$, though I wouldn't be surprised if your statement held then. http://mathoverflow.net/questions/60257/approximate-search-space-on-a-5x5x5-cube-with-3-different-possible-classes Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-04-01T05:30:44Z 2011-04-01T05:30:44Z If all you need is an estimate, I'd run a Monte Carlo code. Something like: 1) Randomly generate a position 2) Check if it's a valid position or not 3) Check under what rotations, reflections the position is invariant. 4) Repeat Once you have a lot of trials (a computer can easily do a few billion), sum the reciprocals of the numbers from step 3 of those positions which were valid. Then estimate from this the probability that a randomly chosen configuration fits your criteria, and multiply by the number of positions (3^125), for a decent estimate. http://mathoverflow.net/questions/59824/use-of-traces-in-physics Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-03-29T00:54:33Z 2011-03-29T00:54:33Z I'll attempt to justify some of why traces are important, though not solely from a physics perspective. The trace of a matrix $M$ comes naturally (as does the determinant) from the characteristic polynomial $p(\lambda) = \det(\lambda I-M)$. Namely, for any algebraically closed field, it is the sum of the roots of $p(\lambda)$ counted with multiplicity (the determinant is the product). This is clearly coordinate-independent and a rather fundamental quantity. The usual definition lacks any intuition, but is more useful for generalizing to arbitrary matrix rings and for efficient computation. http://mathoverflow.net/questions/59806/place-of-analytic-geometry-in-modern-undergraduate-curriculum/59819#59819 Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-03-28T12:08:34Z 2011-03-28T12:08:34Z From my limited experience, these types of geometry courses are not generally taught for math majors. Typically, they are more directed towards math education majors or engineers who are trying to get a math minor. Non-euclidean geometries are certainly mentioned occasionally, but are not always explored with any depth. The reason for this is that geometry isn't totally necessary anymore. It would be difficult to do anything serious without, say, knowing what a group is, but geometry has become a niche topic. Graduate students typically learn some advanced geometry, but undergrads rarely do. http://mathoverflow.net/questions/59514/maximum-surface-area-among-convex-subsets-of-the-unit-sphere-of-a-given-volume Comment by The Cheese Stands Alone The Cheese Stands Alone 2011-03-27T19:40:38Z 2011-03-27T19:40:38Z I'm sure the solution is known for the 2D problem, but I can't find a source for the solution at the moment. I'm fairly sure the maximum achieved by the 2D analogue of the conjectured set, but not totally sure.