User thierry zell - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:14:17Z http://mathoverflow.net/feeds/user/8212 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110641/proving-that-a-complicated-function-is-eventually-concave/110681#110681 Answer by Thierry Zell for Proving that a complicated function is eventually concave Thierry Zell 2012-10-25T15:26:18Z 2012-10-25T15:26:18Z <p>Your function $f(\gamma)$, and its second derivative, are definable in the Pfaffian closure of $\mathbb{R}_{\exp}$. In particular, since this Pfafian closure is an o-minimal structure, this means that the set $\lbrace \gamma \mid f''(\gamma)=0\rbrace$ has only finitely many connected components, which might be enough for your purpose.</p> <p>More details: let $$\phi(x,y)= e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}\log\left(\frac{\sum_{x\in\mathcal{X}}p_{x|x'}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}{\sum_{x\in\mathcal{X}}p_{x}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}\right),$$ and let $$\Phi(x,y)=\int_0^y \phi(x,t)dt.$$</p> <p>This function $\phi$ is certainly definable in the exponential field, and the function $\Phi$ is definable in its Pfaffian closure. Now, $$\int_{-\infty}^\infty \phi(x,y)dy = \lim_{A \to \infty} \Phi(x,A) - \lim_{B\to -\infty} \Phi(x,B).$$ Since these limits are first-order definable, these integrals, and thus your function $f(\gamma)$, are definable in the Pfaffian closure of $\mathbb{R}_{\exp}$.</p> http://mathoverflow.net/questions/109705/real-functions-with-finitely-many-zeroes/110647#110647 Answer by Thierry Zell for Real functions with finitely many zeroes Thierry Zell 2012-10-25T11:15:21Z 2012-10-25T11:15:21Z <p>Emil's answer is the definitive one, but I thought I would add some details. Wilkie's result about $\mathbb{R}_{\exp}$ that he mentions relies in part on Khovanskii's theory of <em>fewnomials</em>. </p> <p>In a way, Wilkie's theorem is overkill for your purpose, especially if you're interested in elementary function, since Wilkie's result deals with the multitude of definable functions in the expansion that are definable but hard to describe succinctly. </p> <p>On the other hand, Khovanskii's original result is much more hands on (though in no way constructive), relying on three purely elementary ingredients: perturbation, Rolle's theorem, and the Bezout inequality. So if you need at all to "look under the hood" and see <strong>why</strong> such a result may be true, you may want to take a look at Khovanskii's book. The beginning is rather accessible and contains a detailed proof of what you need.</p> http://mathoverflow.net/questions/56813/applications-of-the-notion-of-of-gromov-hausdorff-distance/110097#110097 Answer by Thierry Zell for Applications of the notion of of Gromov-Hausdorff distance Thierry Zell 2012-10-19T14:58:20Z 2012-10-19T14:58:20Z <p>Another graph-theoretic application.</p> <p>Given an undirected finite graph $G$, the <a href="http://en.wikipedia.org/wiki/Colin_de_Verdi%25C3%25A8re_graph_invariant" rel="nofollow">Colin de Verdière graph invariant</a> $\mu(G)$ is defined as the maximum multiplicity of the second smallest eigenvalue of a matrix $M\in O_G$, provided that this eigenvalue is structurally stable. Here, $O_G$ is an interesting (for $G$) subset of the symmetric matrices called the Schrödinger-like operators on $G$.</p> <p>It turns out that this invariant is very nice in the following sense: if $H$ is a minor of $G$, then the invariant verifies $\mu(H) \leq\mu(G)$. A trick used to prove this result is to put weights on the edges of $G$ to turn it into a metric space (with the obvious distance of smallest sum of weight in a path). Then, any minor $H$ can be obtained as a Gromov-Hausdorff limit of $G$ with suitable weights (contracted edges see their weight go to zero and deleted edges have their weight go to infinity), and conversely, any sequence of weights on the edges of $G$ that converges gives a minor.</p> <p>Now that I wrote it down, it does not seem like such a powerful remark, but, as I recall, this shift in point of view is (at the very least) very convenient to prove the result.</p> http://mathoverflow.net/questions/109238/arbitrary-union-of-meagre-open-sets/109303#109303 Answer by Thierry Zell for Arbitrary union of meagre open sets Thierry Zell 2012-10-10T15:17:00Z 2012-10-10T15:17:00Z <p>I'll post this answer CW, because I don't have time to work out the details. Hopefully, someone can fill these in or shoot down the strategy.</p> <p>On wikipedia, there is a cute remark about characterizing meagre sets using a <em>Banach-Mazur game</em>. Basically, you have two players who take turns to build a nested sequence of open sets $O_n$. If $U=\cap_{n=1}^\infty O_n$ is the resulting intersection, one of the players aims to have $U \cap X =\varnothing$ and the other player aims to have a point from $X$ in $U$.</p> <p>Then, $X$ is meagre iff the player who wants the empty intersection has a winning strategy.</p> <p>Couldn't this characterization be used in this problem? As I mentioned, I haven't been able to make the details work right...</p> http://mathoverflow.net/questions/108109/cantor-sets-inside-cantor-sets Cantor Sets Inside Cantor Sets Thierry Zell 2012-09-25T22:02:31Z 2012-09-26T13:02:08Z <p>(Or: "I heard you liked Cantor Sets...")</p> <p>I'm working on a student project, and the following construction came up very naturally: If $C$ is the usual Cantor Set, build a countable union of copies of this set as follows: start from $C$, and add a one-third scale copy of $C$ in the interval $[1/3,2/3]$, add three one-ninth scale copies of $C$ in the intervals $[1/9, 2/9]$, $[4/9,5/9]$ and $[7/9, 8/9]$, etc.</p> <p><strong>To Clarify:</strong> basically, every time you remove an interval in the classical construction, in this version, you do not remove the whole interval but you replace it with an appropriately scaled copy of the Cantor set. But you also add Cantor sets in the "holes of the holes", so to speak. So that's why in the second stage, there is a copy of $C$ in $[4/9,5/9]$, that fills in the middle third gap that appears in the copy of $C$ that was added at the first stage.</p> <p>I guess the best way to see it is that at step $n$, you add to the set previously constructed copies of $C$ scaled at $3^{-n}$ in <strong>every</strong> empty interval where such a copy will fit.</p> <p>I have my reasons for wanting to look at this construction, but that got me wondering: it looks so natural that it may very well come up in several contexts.</p> <p>So. Anyone knows where this construction first appeared? Is it especially notable? Does it illustrate any especially interesting property? I would hate to miss something good about it.</p> http://mathoverflow.net/questions/36025/explicit-computations-using-the-haar-measure Explicit computations using the Haar measure Thierry Zell 2010-08-18T21:09:43Z 2012-07-12T17:14:52Z <p>This question is somewhat related to my previous one on <a href="http://mathoverflow.net/questions/35655/measure-on-real-grassmannians" rel="nofollow">Grassmanians.</a> The few times I've encountered the Haar measure in the course of my mathematical education, it's always been used in a very theoretical setting: in the right setting, it exists, it is unique (if the setting is <em>really</em> nice), and you can integrate against it to define new objects that will have nice properties because the measure itself does.</p> <p>So my question is: how practical is it to compute with? I'm talking about very concrete examples here, e.g. "G=O(4), I integrate f(M)=[some explicit function with a matrix input] and the answer I get is 42". From conversations, I got the feeling that the construction of the Haar measure allows you <em>in principle</em> to write such a computation explicitly. I'm concerned with the tractability of the computation itself. Examples would be great.</p> http://mathoverflow.net/questions/94184/does-your-dissertation-matter-for-industry-research-jobs/94243#94243 Answer by Thierry Zell for Does your dissertation matter for industry research jobs? Thierry Zell 2012-04-16T19:38:51Z 2012-04-16T19:38:51Z <p>My usual gripe: strictly speaking, the answer is "it depends" and anyone who says otherwise is making assumptions that are unwarranted given how little we know about Mr A (Dr A?) and his situation. The <strong>type of industry</strong> matters, of course (some have a more established history of employing mathematicians than others), as well as the <strong>country</strong> that Mr A lives in. So as much as I can appreciate the encouraging words that we can give to Mr A, I want to insist that any answer must by definition have a <strong>localized domain of validity</strong> that may or may not overlap Mr A's situation.</p> <p>So I don't know if the topic of Mr A's dissertation will matter, but it seems to me that <strong>it shouldn't matter to Mr A:</strong> according to the question, Mr A has essentially wrapped up the dissertation and so, to paraphrase Rumsfeld, will have to look for a job with the topic he has rather than the topic he wishes he had.</p> <p>So if I can offer Mr A some advice, it would be rather to <strong>do his homework</strong> about the type of industry he wants to join. The fact that the dissertation topic was relevant to the industry may or may not matter, but what <strong>will</strong> matter for sure is your ability to make the case that <strong>you (not your dissertation)</strong> have something to offer to your prospective employer. I think scaaahu made a good point in pointing out that some employers/colleagues may have misconceptions about what mathematicians do. But more generally, you should research what the industry does and in what way you can be of help.</p> <p>(For the record: I know a few mathematicians who have successful careers in various industries. As far as I can tell, all of them were very diligent in preparing for the industry in question, and they targeted one specific type of activity.)</p> http://mathoverflow.net/questions/90169/combination-with-repetition-with-limit/90170#90170 Answer by Thierry Zell for Combination with repetition with limit Thierry Zell 2012-03-04T01:34:03Z 2012-03-04T01:34:03Z <p>I didn't get the chance to work out the precise formula, but maybe this will help. It will give you a shorter answer anyway. In a nutshell: linearity of expectation is your friend.</p> <p>Let $X$ be the number of empty boxes, and let $X_i$ be the random variable which is $0$ if box #$i$ is not empty and $1$ otherwise, so that $X=X_1 + \cdots +X_n.$ Then, $$\mathbb{E}(X)=\mathbb{E}(X_1)+\cdots+\mathbb{E}(X_n)=n\mathbb{E}(X_1).$$ Now, $\mathbb{E}(X_1)=\mathbb{P}(X_1=1)$ which is easy to compute since it's just the probability of distributing the balls so that box #1 is empty, which is a classical application of inclusion-exclusion and the combination with repetition formula. </p> <p>This way, you will get something much simpler. There is still an alternating sum, and there is no way around that, but there only one level of summation rather than the two nested levels in your formulas.</p> <p>I hope this helps!</p> http://mathoverflow.net/questions/89706/sperner-lemma-applications/89709#89709 Answer by Thierry Zell for Sperner Lemma Applications Thierry Zell 2012-02-27T23:14:20Z 2012-02-27T23:14:20Z <p>Francis Su wrote a paper called <em>Rental Harmony: Sperner's Lemma in Fair Division</em> that, as the name indicates, uses Sperner's lemma to solve some fair division problems. It the 2001 Merten Hasse award winning paper, and as such can be found free of charge <a href="http://mathdl.maa.org/mathDL/22/?pa=content&amp;sa=viewDocument&amp;nodeId=2706" rel="nofollow">here</a></p> http://mathoverflow.net/questions/85730/a-trick-or-a-general-technique-probabilistic-method/85874#85874 Answer by Thierry Zell for A trick or a general technique? (Probabilistic Method) Thierry Zell 2012-01-17T01:22:27Z 2012-01-17T01:22:27Z <p>Well, this is going to sound a bit silly, but your main tool, the fact that $$E[P−QE[P]/E[Q]]=0,$$ follows trivially from the linearity of expectation, and to me, the linearity is where the magic happens.</p> <p>The linearity of expectation may not sound like a big deal, it's a fairly easy fact in probability theory, but I'm always surprised by its non-obvious consequences for combinatorics in general and the probabilistic method in particular. My favorite application of linearity is that the expectation of the number $X$ of fixed points for a permutation of $n$ elements taken uniformly at random is exactly 1. It's a one-line proof using linearity versus highly non-trivial derangement juggling if you just write it out explicitly: $$E[X]=\sum_{k=0}^n k\binom{n}{k} d_{n-k} \frac{1}{n!};$$ where $d_{n-k}$ is the number of derangements on $n-k$ elements, computed using the inclusion-exclusion formula. (BTW, it <strong>is</strong> possible to show that this formula yields~1, it's just not all that easy.)</p> http://mathoverflow.net/questions/82465/resubmitting-a-paper/82479#82479 Answer by Thierry Zell for Resubmitting a paper Thierry Zell 2011-12-02T16:15:48Z 2011-12-02T16:15:48Z <p>Research mathematics is a small world. It is rather likely that the people who are currently editing and refereeing your paper will play a significant role in the development of your career in the next 5 years or even more. They might not take too kindly to the withdrawal, especially if they think that the move was not really warranted (i.e., if they don't think that your result is really much too good for the journal you submitted). So think to yourself: is <strong>this</strong> a risk worth taking? Is your result that badly served by the current journal?</p> <p>As for the impact factor, have you really tried to find out what people think about those? Deans and other paper pushers might put a lot of store by them, and so it's definitely something to keep in mind in the long run, but your colleagues' opinion of a journal is not based on a somewhat artificial numerical construct, so for your immediate future, the actual name and reputation of the journal is much more important than a somewhat arbitrary numerical score. (It is a fact that the math journals with the highest impact factor are <strong>not necessarily</strong> the most prestigious ones.)</p> http://mathoverflow.net/questions/80135/degree-of-a-real-algebraic-variety-and-regular-morphisms/80140#80140 Answer by Thierry Zell for Degree of a real algebraic variety and regular morphisms Thierry Zell 2011-11-05T17:45:43Z 2011-11-05T17:45:43Z <p>I would encourage you to read <strong>Algorithms in Real Algebraic Geometry</strong> by Saugata Basu, Richard Pollack, Marie-Françoise Roy, which contains all the state of the art results about effective results in real algebraic geometry. It is a free download from <a href="http://perso.univ-rennes1.fr/marie-francoise.roy/bpr-ed2-posted2.html" rel="nofollow">http://perso.univ-rennes1.fr/marie-francoise.roy/bpr-ed2-posted2.html</a> Projections are an instance of quantifier elimination, a procedure whose complexity is not entirely understood, but definitely very bad, even for a single existential quantifier.</p> <p>What you will find is that things are a lot more tricky than you realize right now. In particular, there is a definite failure of Bézout-like theorems over the reals, and fact which clearly appears in Fulton's book. </p> <p>Your definition dramatically underestimates the value of the degree. Here is an example derived from Fulton's book. Take $$f(x,y)= \prod_{i=1}^d (x-i)^2+\prod_{j=1}^d(y-j)^2.$$ Then, $V(f)$ has degree $\leq 2d$ by your definition, but it is made of $d^2$ isolated points. From a geometrc point of view, this is something whose degree should probably be at least $d^2$. The same example works in more variables, of course.</p> http://mathoverflow.net/questions/80056/using-slides-in-math-classroom/80106#80106 Answer by Thierry Zell for Using slides in math classroom Thierry Zell 2011-11-05T03:03:22Z 2011-11-05T03:03:22Z <p>I've already given my opinion, and this is more of a remark: how the pros and cons are weighed between blackboard and slides should be influenced by a whole collection of classroom factors, and the first one among them should probably be class size.</p> <p>This is a rather obvious remark, but I thought it was worth pointing out; Jaap Eldering's answer brought it to the forefront for me, because he mentioned doing examples on slides to avoid making mistakes, and my first reaction was: "making mistakes in class is good!". </p> <p>And then it occurred to me that I can use mistakes in the classroom fairly effectively because I only teach small classes. In a big classroom, I would simply not be able to receive instant feedback efficiently enough to do this as well, and I would not be comfortable trying.</p> <p>In a very large lecture hall, the blackboard will often lose a lot of its advantages given how large you have to write.</p> http://mathoverflow.net/questions/80056/using-slides-in-math-classroom/80058#80058 Answer by Thierry Zell for Using slides in math classroom Thierry Zell 2011-11-04T15:43:55Z 2011-11-04T15:43:55Z <p>I think you already touched on the two main points: pretty pictures are so much better than anything done on a chalkboard is the pro, but you cannot decently unwind any argument on slides. </p> <p>I've used them intensively, I do it a lot less now. (Here's a con you did forget about: they take a <strong>lot</strong> of time to prepare, even when you're only revising them.) If the room lends itself well to it, the hybrid method is best: use the slides only when they beat the board. Rooms that have a screen in the corner, rather than in front of the board, are best for this.</p> <p>Also, it seems that it's easier to fall asleep to slides than to a lecture, so be aware of that. Make sure that the room is never too dark (the quality of the screen material can be critical here too: good screens should be readable in full light). And switching your routine, never showing slides for too long, helps keeping the students awake.</p> http://mathoverflow.net/questions/79811/how-many-0-1-solutions-would-this-system-of-underdetermined-linear-equations-hav/79838#79838 Answer by Thierry Zell for How many 0, 1 solutions would this system of underdetermined linear equations have? Thierry Zell 2011-11-02T14:33:42Z 2011-11-02T14:33:42Z <p>Your problem is an instance of what is (also) known as <a href="http://en.wikipedia.org/wiki/Binary_integer_programming#Integer_unknowns" rel="nofollow">Binary Integer Programming.</a> As noted in the other answers, the decision problem is NP-complete and the counting problem is #P-hard. </p> <p>I do know that there has been some work on finding solutions, and there are libraries available that are pretty efficient. (I've played around with <a href="http://sourceforge.net/projects/lpsolve/" rel="nofollow">lp_solve</a>). Despite the progresses made, my experience with this suggests that computations are prohibitively long on non-trivial instances (unlike some NP-hard problems which have decent practical algorithms). So I'm rather pessimistic about finding <strong>all</strong> solutions in practice, at least with an out-of-the-box algorithm.</p> <p>You might be able to do better by exploiting the structure of your specific problems, but I wouldn't know where to start. I'd be interested in suggestions about this myself.</p> http://mathoverflow.net/questions/79742/motivating-algebra-and-analysis-for-average-undergraduates/79783#79783 Answer by Thierry Zell for Motivating Algebra and Analysis for Average Undergraduates Thierry Zell 2011-11-02T00:02:25Z 2011-11-02T00:02:25Z <p>I have yet to teach such a course, but I would motivate abstract algebra based on plane geometry, especially group theory. It certainly worked for me to some extent, though I suspect that's because I saw a lot more plane geometry in high school than our students ever do. Various flavors of plane transformations, dihedral groups, regular polyhedra and crystalline structures, you can even go into combinatorics with Polya enumeration$\dots$ I think a geometric picture would be valuable even after moving away from geometric problems.</p> <p>As for analysis, I would go for the big theorems, which seems where the difference between analysis and calculus lies. For instance, after doing a bunch of limits of integrals depending on an integer parameter, wouldn't it be nice to know that limits and integrals can be interchanged when the convergence is uniform? The big theorems can be presented not for the sake of abstraction, but because they make boring computational tasks easier.</p> <p>I don't know if this answers your question, though I'm putting it out there anyway; this is almost too obvious, you probably wanted more details or a different tack altogether.</p> http://mathoverflow.net/questions/79044/what-items-must-appear-on-a-mathematicians-cv/79052#79052 Answer by Thierry Zell for What items MUST appear on a mathematician's CV? Thierry Zell 2011-10-25T05:40:42Z 2011-10-25T05:40:42Z <h2>An obvious answer</h2> <p>One cannot insist enough (and I am surprised to see this has been hinted at already, but not stated quite this boldly) on the fact that different positions have different expectations, e.g. many non-academic employers seem to expect a single-page résumé.</p> <p><strong>Even when looking for an academic position, candidates routinely keep more than one version of their CV.</strong> In order to tailor to the specifics of the various positions (more or less emphasis on research, post-doc vs. tenure-track). Even if you put the exact same items in all your CVs, the <strong>order</strong> in which they are presented, <strong>which items are emphasized and detailed</strong> is a good way prove that you understand the expectations of the position, and make sure that the relevant items are easily found by the reader.</p> <p>I cannot be more specific since this question yet again commits the sin of being a non-geographically specific career question: needless to say that localization also plays a role.</p> http://mathoverflow.net/questions/78988/when-does-mathscinet-review-a-paper/78989#78989 Answer by Thierry Zell for When does mathscinet review a paper? Thierry Zell 2011-10-24T15:53:14Z 2011-10-24T15:53:14Z <p>There are many things that the Mathematical Reviews does not review. Off the top of my head, </p> <ol> <li>Non-mathematical items that may appear in mathematical publications;</li> <li>Many proceedings are not refereed, or if they are, the individual articles are not always refereed. </li> </ol> <p>As to why MathSciNet would maintain a list of these items without reviewing them, I'm assuming that it's for the sake of completeness. </p> http://mathoverflow.net/questions/77229/dimension-of-a-real-affine-variety/78711#78711 Answer by Thierry Zell for dimension of a real affine variety Thierry Zell 2011-10-20T21:59:41Z 2011-10-20T21:59:41Z <p>Definition 2 and 3 are equivalent. What you want here is the notion of <em>Cylindrical Algebraic Decomposition</em> (see e.g. Bochnak, Coste and Roy's book). The $d$ you're looking for in both cases is the dimension of the largest cell in the decomposition, and it is a routine result that this dimension does not depend on the decomposition itself.</p> <p>As for Definition 1, I am not sure I understand exactly what you're saying. You're using a total-degree monomial ordering, then taking the zero-set of the leading terms of the variety's ideal... I'll come back to this if I remember.</p> http://mathoverflow.net/questions/77911/level-of-detail-on-a-phd-application/77998#77998 Answer by Thierry Zell for Level of detail on a Phd application Thierry Zell 2011-10-13T04:35:17Z 2011-10-13T04:35:17Z <p>If it's really a 100-word summary, I don't really see how there can be any level of detail. But ultimately, the best thing to do and the one that will get you the most reliable answer is not to ask Math Overflow users, but would be to contact someone in Math in the university of your choice and ask them. Professors usually try to be helpful with their future recruits.</p> http://mathoverflow.net/questions/77694/implicit-function-theorem-for-algebraic-sets/77698#77698 Answer by Thierry Zell for implicit function theorem for algebraic sets Thierry Zell 2011-10-10T14:57:54Z 2011-10-10T14:57:54Z <p>This is a well-known theorem. I imagine it has been routinely used for many years, so tracking down a historical reference would be hard. The following argument will work for any $f$ which is definable over the real field, i.e. whose <em>graph</em> is a semi-algebraic set.</p> <p>The graph $\Gamma \subset A\times B$ of your mapping has dimension $\dim(A)$, since the restriction to $\Gamma$ of the projection $\pi_A: A\times B \to A$ is 1-to-1. Now do a cylindrical algebraic decomposition of $A\times B$ adapted to $\Gamma$ and the projection $\pi_B$. Then, any cell in $f(B)$ is the image of a cell of $\Gamma$. In particular, this is true for all the $\dim(A)$-dimensional cells that appear in $\Gamma$; if $C$ is such a cell, any point $b$ in the image $\pi_B(C)$ verifies $\dim(f^{-1}(b)) \geq \dim(A)-\dim(\pi_B(C)) \geq \dim(A)-\dim(B)$.</p> <p>So not only can you find such a $b$, but there should be quite a few of them.</p> http://mathoverflow.net/questions/75922/is-beauty-at-the-high-school-level-even-possible/75935#75935 Answer by Thierry Zell for Is beauty at the high school level even possible? Thierry Zell 2011-09-20T11:33:23Z 2011-09-20T11:33:23Z <p>I have not read's Rota's paper, so I am not sure what point he is trying to make. Though I have glanced at the abstract, which begins with: </p> <blockquote> It has been observed that whereas painters and musicians are likely to be embarrassed by references to the beauty in their work, mathematicians instead like to engage in discussions of the beauty of mathematics. Professional artists are more likely to stress the technical rather than the aesthetic aspects of their work. </blockquote> <p>Maybe this statement is explained and supported in the main text, but, at first glance, this difference in attitude does not strike me as obvious (I'm pretty sure I've heard more discussions about technical aspects from mathematicians than from artists).</p> <hr> <p>To come back to your question, the fact that the untrained and the trained experience beauty differently is by no means specific to mathematics: the trained benefit from having a rich context that dramatically alters their appreciation of the aesthetic aspects of a work; I cannot think of a discipline where this would fail to be true.</p> <p>What does it mean for high school math? Well, just because I've lost some of my enthusiasm for Euclidean geometry since high school, because I've discovered more beautiful things still, it does not mean that the <strong>nature</strong> of the aesthetic emotions I feel for Euclidean geometry is any different from the emotions I feel for other beautiful mathematics. And if the intensity has somewhat waned in the case of Euclidean geometry (and I'm not even sure it has), it was certainly very powerful back in high school.</p> <p>So, basically: <strong>who cares if the beauty that one shows in high school math class does not meet the standards of beauty of professional mathematicians?</strong> What if it makes these folks go "meh"? The teacher is not talking to them, but to the students! So, as long as the students experience the same kind of feeling of mathematical beauty, the work is done: they know there is more to this, that mathematical beauty is out there.</p> <hr> <p>To finish, a couple of tangential remarks:</p> <ul> <li>I'm not sure I buy the whole premise anyway. To me, some of the most beautiful ideas in math are commonly seen at the high-school level: they are beautiful because they are so simple yet so powerful (coordinates, change of variables,...). Though, on the other hand, I'm not sure it is possible to appreciate the beauty of these ideas without the benefit of hindsight.</li> <li>I have purposefully stayed away from the question: <em>What proportion of students can one reach with that?</em>, in part because I have no idea how to answer the question, and also because this discussion would be so system- and country-dependent that MO is not the place for it. But I think this should be part of your introspection.</li> </ul> http://mathoverflow.net/questions/75698/examples-of-seemingly-elementary-problems-that-are-hard-to-solve/75700#75700 Answer by Thierry Zell for Examples of seemingly elementary problems that are hard to solve? Thierry Zell 2011-09-17T21:14:31Z 2011-09-17T21:14:31Z <p>The <a href="http://en.wikipedia.org/wiki/Jacobian_conjecture" rel="nofollow">Jacobian conjecture</a>.</p> http://mathoverflow.net/questions/75299/intersections-of-real-algebraic-sets-a-bezout-type-question/75316#75316 Answer by Thierry Zell for intersections of real algebraic sets (a bezout-type question) Thierry Zell 2011-09-13T14:40:04Z 2011-09-13T20:28:47Z <p>A personal preference, undoubtedly, but I find that a differential geometry point of view can really help with the questions you're asking.</p> <p><strong>Dimension:</strong> Any semialgebraic set can be decomposed (non-uniquely) as a union of smooth manifolds (stratification), and you can always define the dimension as the maximal dimension that occurs in your decomposition. It does not depend of the decomposition. Of course, there are other ways of defining dimension, especially for varieties. But note that your set can have smooth points of varying dimensions (e.g. <a href="http://en.wikipedia.org/wiki/Whitney_umbrella" rel="nofollow">Whitney's umbrella</a> or Cartan's umbrella).</p> <p><strong>Bézout-type results:</strong> There are many in real algebraic geometry, all made the more complicated by the fact that real roots don't always exist, of course. In your case, the result would follow from the fact that there are effective bounds on the number of connected components of intersections of algebraic varieties (and other semi-algebraic sets) in terms of the defining equations. The best bounds come from the <em>critical point method</em>, which is an efficient algorithm to produce one point per connected component, based on finding critical points of fairly simple Morse functions (e.g. generic projections or distance from a generic point). The bounds come from the Bézout inequality applied to polynomial systems defining those critical points. So a couple of derivatives of your polynomials will show up, and having a million points is complete overkill in your case, provided that you can guarantee that the intersection must be 0-dimensional when $P$ is not contained in $Q$ <strike> (which should be the case here).</strike> (oops! I was thinking of the wrong dimension!)</p> <p><strong>Added later:</strong> See section 12.6 of the book by <a href="http://perso.univ-rennes1.fr/marie-francoise.roy/bpr-ed2-posted2.html" rel="nofollow">Basu-Pollack-Roy</a> for an exposition of the critical method. It is downloadable free of charge at the link.</p> http://mathoverflow.net/questions/74841/an-example-of-a-beautiful-proof-that-would-be-accessible-at-the-high-school-level/74969#74969 Answer by Thierry Zell for An example of a beautiful proof that would be accessible at the high school level? Thierry Zell 2011-09-09T03:43:56Z 2011-09-09T03:43:56Z <p>Sperner's lemma (in dimension 2 to keep it visual). The proof in Francis Su's <a href="http://mathdl.maa.org/mathDL/22/?pa=content&amp;sa=viewDocument&amp;nodeId=2706" rel="nofollow">Monthly paper</a>, <em>Rental harmony: Sperner's lemma in fair division</em> is especially easy to visualize. Theris a non-empty content, you can have students ponder the role of the hypotheses. And fair division applications allow to motivate it via concrete applications.</p> http://mathoverflow.net/questions/74415/is-the-preimage-of-the-closure-the-closure-of-the-preimage-under-a-quotient-map/74466#74466 Answer by Thierry Zell for Is the preimage of the closure the closure of the preimage under a quotient map? Thierry Zell 2011-09-03T22:29:20Z 2011-09-03T22:29:20Z <p>The answer seems to be yes in the case of interest to the OP. Nevertheless, the answer to the question as stated is <strong>no</strong>, and this is the chance for me to pull out my favorite example of badly behaved quotient map.</p> <p>Let $\mathcal{S} = [0,1]^{\mathbb{N}}$, and define for all $\mathbf{x} \in \mathcal{S}$ the support of $\mathbf{x}$ to be $\mathrm{supp}(\mathbf{x})=\bigcup_{i\in \mathbb{N}} \mathbf{x}_i .$ Let $\mathcal{X} \subset [0,1] \times \mathcal{S}$ be the set defined by $$\mathcal{X} =\lbrace (x,\mathbf{x}) \mid x\in \mathrm{supp}(\mathbf{x})\rbrace.$$</p> <blockquote> <p><strong>Fact 1:</strong> $\mathcal{X}$ is a metric space.</p> </blockquote> <p>We can define a distance $d$ by: $d[(x,\mathbf{x}), (y,\mathbf{y})]:=1$ if $\mathbf{x}\neq \mathbf{y}$, and $d[(x,\mathbf{x}),(y,\mathbf{x})]:=|x-y|.$</p> <blockquote> <p><strong>Fact 2:</strong> The map $q:\mathcal{X} \to [0,1]$ defined by $q(x,\mathbf{x})=x$ is a quotient map.</p> </blockquote> <p>The map $q$ satisfies $|q(x,\mathbf{x})-q(y,\mathbf{y})| \leq d[(x,\mathbf{x}), (y,\mathbf{y})]$, so $q$ must be continuous. So it's enough to check that for any $F \subseteq [0,1]$, $q^{-1}(F)$ closed implies that $F$ is closed. Consider a convergent sequence $(x_n)$ in $F$ whose limit is $\ell$: it can be lifted to the sequence $(x_n, \mathbf{x})\in q^{-1}(F)$ (where the second factor is constant). This sequence converges too, so if we assume that $q^{-1}(F)$ is closed, the limit $(\ell, \mathbf{x})$ must be an element of $q^{-1}(F)$, and thus $\ell$ is in $F$; it follows that $F$ is closed.</p> <blockquote> <p><strong>Fact 3:</strong> The map $q$ fails to verify $q^{-1}(\overline{S})=\overline{q^{-1}(S)}$ for all $S$.</p> </blockquote> <p>Consider $S=(0,1)$. Then $q^{-1}(\overline{S})=\mathcal{X}$, but $\overline{q^{-1}(S)}$ does not contain the element that corresponds to the constant sequence 1.</p> http://mathoverflow.net/questions/74295/interesting-mathematical-topics-arising-from-biology/74309#74309 Answer by Thierry Zell for Interesting mathematical topics arising from Biology Thierry Zell 2011-09-01T23:09:09Z 2011-09-01T23:09:09Z <p>A very thorough introduction to some now classical topics can be found in <a href="http://en.wikipedia.org/wiki/James_D._Murray" rel="nofollow">James D. Murray's</a> now two-volume book published by Springer. Expect lots of ODE's and PDE's in that one. </p> <p>As far as more exotic math is concerned, a complete overview would be difficult: it seems people throw everything they have and see what works. I've seen some interesting talks involving combinatorics, others involving algebraic geometry.</p> http://mathoverflow.net/questions/72220/plagiarism-in-the-community/72317#72317 Answer by Thierry Zell for Plagiarism in the community Thierry Zell 2011-08-07T23:03:20Z 2011-08-07T23:03:20Z <p>Ask yourself: do you want to be part of a community, or do you want to do math by yourself in the woods?</p> <p>There is no wrong answer: doing either is a fine and potentially rewarding lifestyle, but giving up on the community means giving up on both the positives and negatives that it brings, and most answers here make clear that the positives outweigh the negatives.</p> http://mathoverflow.net/questions/14574/your-favorite-surprising-connections-in-mathematics/71325#71325 Answer by Thierry Zell for Your favorite surprising connections in Mathematics Thierry Zell 2011-07-26T15:53:49Z 2011-07-26T15:53:49Z <p>Here is one of my favorite, that I learned from A. G. Khovanskii: let $f$ be a univariate rational function with real coefficients. Then, you can think of $f$ as inducing a continuous self-map of $\mathbb{RP}^1 \cong S^1$, in particular, it has a <em>topological degree</em>, say $[f]$, and if $f$ happens to be a polynomial, it is obvious that $[f]=0$ if $\deg(f)$ is even, and that $[f]=\pm 1$ if $\deg(f)$ is odd (depending on the sign of the main coefficient).</p> <p>If the decomposition of $f$ in continued fraction is $$f=P_0+\frac{1}{P_1+\frac{1}{P_2+\ddots}}$$ Then one can prove easily that $[f]$ is the (finite) sum: $[f]=\sum_{i \geq 0} (-1)^i[P_i]$. (Khovanskii himself taught this to high-schoolers in Moscow.)</p> <p>The interesting connection for me follows: for any real polynomial $P$, the topological degree of the fraction $P'/P$ is clearly the (negative of the) number of real roots of $P$. Thus, the computation formula above applied to $[P'/P]$ allows us to recover <a href="http://en.wikipedia.org/wiki/Sturm%27s_theorem" rel="nofollow">Sturm's theorem</a>. </p> <p>I don't know if it really qualifies as a new proof of the theorem, but it's definitely a different point of view on that proof.</p> http://mathoverflow.net/questions/42286/link-repository-of-international-dissertations Link Repository of International Dissertations Thierry Zell 2010-10-15T13:40:37Z 2011-07-11T17:27:43Z <p>This question (cry for help?) grew out of Colin Tan's question: <a href="http://mathoverflow.net/questions/42146/does-anyone-have-a-copy-of-schmids-effective-work-on-hilbert-17th" rel="nofollow">does anyone have a copy of schmid’s effective work on hilbert 17th?</a> which was a request for a copy of an Habilitationsschrift.</p> <p>We've all been in this situation: somebody works on a dissertation of some kind, obtains interesting results, but either can't be bothered to publish because they're leaving the profession, or do publish the essentials, but have to omit some important details because of space considerations.</p> <p>Inter-library loan can be an option in these cases, but it's very likely to take a <em>very long time</em> if it works at all. In the US, it appears that most universities require you to sign away your reproduction right to an independent company which will keep your dissertation on microfilm (correct me if I'm wrong, this is CW after all!). So in that case, you <em>can</em> track down a copy of the dissertation in question (if you're willing to pay for it) -- that is, if you can manage to track down the reproduction service in question.</p> <p>Some countries, on the other hand, feel the need to maintain an online repository of the student work done in their universities. Here again, it involves a non-trivial slog through various library links before finding the right place (and may demand a better command of the language than for math reading). Hence, the question, which is more of a suggestion:</p> <blockquote> <p>How about we gather a list of links to various dissertation repositories / reproduction services right here?</p> </blockquote> <p><strong>Suggestion:</strong> Since no answer is "better" than any other, and to improve readability, it would be best to keep this to a <strong>single</strong> list answer that can be edited with the various contributions. Thanks in advance!</p> http://mathoverflow.net/questions/121116/to-what-extent-should-a-pure-mathematician-care-about-the-meaning-of-things Comment by Thierry Zell Thierry Zell 2013-02-07T22:03:46Z 2013-02-07T22:03:46Z @quid: in addition to the whole &quot;work&quot; and &quot;mean&quot;, I was very disturbed by the &quot;should&quot; as well... The only thing mathematicians &quot;should&quot; do is mathematics. Let's leave the <b>how</b> to do it to each individual... http://mathoverflow.net/questions/116024/fourier-application-in-biology Comment by Thierry Zell Thierry Zell 2012-12-10T21:53:21Z 2012-12-10T21:53:21Z Why? . http://mathoverflow.net/questions/115972/apparently-simple-probability Comment by Thierry Zell Thierry Zell 2012-12-10T13:37:31Z 2012-12-10T13:37:31Z If your events are not independent, there is no way this works, is there? For $x$ small enough, take $B_1=B_3$ totally disjoint from $B_2$. Then, $P(A_2)=2x$ which is larger than the bound of $x^2+x$ that you have. http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate Comment by Thierry Zell Thierry Zell 2012-12-04T16:30:26Z 2012-12-04T16:30:26Z I don't think this is, at least as stated, a good question for MO (and apparently I am not the only one, because I did not go as far as to downvote). Also, I would challenge some of your assumptions. The notation $\frac{df}{dx}$ is used very frequently in the books I'm reading, for instance. Also, does $\frac{d}{dx}$ denote a single operator? As Poincare remarked: &quot;Mathematics is the art of giving the same name to different things&quot;. http://mathoverflow.net/questions/112231/is-notices-of-the-american-mathematical-society-available-before-1995 Comment by Thierry Zell Thierry Zell 2012-11-12T22:42:12Z 2012-11-12T22:42:12Z The earliest Notices entry I was able to get from MathSciNet was 1983. I'm not quite sure why. http://mathoverflow.net/questions/110641/proving-that-a-complicated-function-is-eventually-concave/110681#110681 Comment by Thierry Zell Thierry Zell 2012-11-01T22:28:32Z 2012-11-01T22:28:32Z I'm sorry that I didn't have the time to come back to this. I have been in email contact with the OP. Emil raises some good objections, and I am not sure how to fix them yet or when I'll have the time to tackle this. Sorry about that! http://mathoverflow.net/questions/109705/real-functions-with-finitely-many-zeroes/110647#110647 Comment by Thierry Zell Thierry Zell 2012-10-25T14:11:33Z 2012-10-25T14:11:33Z The paper should cover the same things. I would still advise you to try for the book though because I would assume it has more context and is more reader-friendly. From where I am, most of the book is available on Google books and Chapter I is probably the most useful for you. http://mathoverflow.net/questions/109279/when-should-i-publish-my-results Comment by Thierry Zell Thierry Zell 2012-10-10T15:04:54Z 2012-10-10T15:04:54Z Unless you have reasons to believe that the referee is not very representative of your shared academic community, why not trust the referee's advice? http://mathoverflow.net/questions/108433/for-which-millennium-problems-does-undecidable-true Comment by Thierry Zell Thierry Zell 2012-09-30T21:12:29Z 2012-09-30T21:12:29Z @Denis: You might be thinking of a result from Erdos that I learned in proofs from THE BOOK: suppose $\lbrace f_\alpha \rbrace$ is a family of holomorphic functions such that when you evaluate the family at some fixed $z\in \mathbb{C}$, you can only get countably many values. Then, $\neg CH$ implies that such a family must itself be countable, but there are such families of continuum size if $CH$ holds. There is a link to the original paper in this MO answer: <a href="http://mathoverflow.net/questions/1924/what-are-some-reasonable-sounding-statements-that-are-independent-of-zfc/1958#1958" rel="nofollow" title="what are some reasonable sounding statements that are independent of zfc">mathoverflow.net/questions/1924/&hellip;</a> http://mathoverflow.net/questions/108209/why-do-we-need-algorithms-and-why-is-non-convex-optimization-difficult Comment by Thierry Zell Thierry Zell 2012-09-27T15:13:21Z 2012-09-27T15:13:21Z Nitpick/clarification of Qiaochu's first remark: the quadratic formula is irrelevant to saying that $x=\sqrt{2}$ is the positive solution to $x^2=2$, of course. That's how the square root function is defined; what the quadratic formula does is solve more general equations in terms of square roots. Tying this in with the original question: the quadratic formula does <b>not</b> give &quot;boom, instant solution&quot;, because you need an algorithm to compute the square root. And if you say: &quot;I can leave the square root as is&quot;, then you're comparing apples and oranges... Sorry, but very naive question IMHO. http://mathoverflow.net/questions/108109/cantor-sets-inside-cantor-sets/108148#108148 Comment by Thierry Zell Thierry Zell 2012-09-26T16:13:01Z 2012-09-26T16:13:01Z Thanks Pietro! I remained deliberately vague about the nature of that project in my question, because I didn't want to steer answers in any direction, but we were precisely looking at such examples to construct null sets. I love the invariant superset interpretation. I doubt I would have come up with it on my own. http://mathoverflow.net/questions/108109/cantor-sets-inside-cantor-sets Comment by Thierry Zell Thierry Zell 2012-09-26T06:39:53Z 2012-09-26T06:39:53Z @R W: I'm not sure that there is anything especially notable about it. But if there is, I'd like to know more about it. http://mathoverflow.net/questions/108109/cantor-sets-inside-cantor-sets/108115#108115 Comment by Thierry Zell Thierry Zell 2012-09-26T04:45:40Z 2012-09-26T04:45:40Z Thanks for your detailed answer, and the enlightening pictures. Very cool stuff. http://mathoverflow.net/questions/108109/cantor-sets-inside-cantor-sets/108121#108121 Comment by Thierry Zell Thierry Zell 2012-09-26T04:33:21Z 2012-09-26T04:33:21Z Gerald Edgar pointed out that my original description was somewhat ambiguous. I think that Andreas managed to read my intent, because his characterization fits the picture I have. I have tried to fix the original post, I hope it's clearer now. http://mathoverflow.net/questions/108109/cantor-sets-inside-cantor-sets Comment by Thierry Zell Thierry Zell 2012-09-26T04:30:47Z 2012-09-26T04:30:47Z @Gerald: Indeed, I was struggling to phrase it, so much so that it was still not what I meant even after my first edit. Now, I think we have it.