User drvitek - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:56:06Z http://mathoverflow.net/feeds/user/8345 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59840/surprising-and-useful-physical-intuition-for-mathematical-objects/60023#60023 Answer by drvitek for Surprising and Useful Physical Intuition for Mathematical Objects drvitek 2011-03-29T22:59:57Z 2011-03-29T23:56:51Z <p>I don't know if you've heard of this before, but there is an extremely elegant physical proof of the existence and properties of the Fermat point. It even illustrates the degeneracy that starts to occur for largest angle at least $2\pi/3$.</p> <p>Consider your desired triangle as a flat, sturdy, sheet. Make very small notches at the vertices. Now take three ideal ropes and at one end of each fix identical balls. Join all three free ends. Place the rope-ball configuration on the top of the triangle, and slide each of the ropes through the notches. Let the balls hang off the triangle.</p> <p>Now, allow the system to assume its minimal potential energy. The intersection of the ropes will move to minimize the sum of the distances to the vertices, as potential energy is linear in height. So the intersection will move to the Fermat point.</p> <p>But here is the tricky part. As the intersection stabilizes, we know that the forces on it must be zero. The forces all have the same magnitude, so if they are all zero then they must have equal angles between them. So the lines from the Fermat point <em>X</em> to the vertices of triangle <em>ABC</em> satisfy $\angle AXB = \angle BXC = \angle CXA = 2\pi/3$.</p> <p>Of course, this suggests that in a suitably obtuse triangle the intersection will fall right through one of the notches, and indeed for angles greater than $2\pi/3$ the Fermat point is at the largest angle.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://cs.smith.edu/~orourke/MathOverflow/FermatPoint.jpg" alt="FermatPoint"></p> http://mathoverflow.net/questions/43791/hanging-a-ball-with-string/43808#43808 Answer by drvitek for Hanging a ball with string drvitek 2010-10-27T14:56:34Z 2010-10-30T05:59:38Z <p>There is another stable solution of total length $3\pi + h + \epsilon$ for any $\epsilon > 0$. We take a circle of constant latitude $\delta &lt; 0$ (sufficiently small) and then connect this circle to the north pole via two diametrically opposed strings. This is then clearly stable. Furthermore this is equivalent to your solution (plus Scott Carnahan's epsilon-modification).</p> <p>However, there is in fact a stable solution of total string-length $2\pi+h+\epsilon$. We simply take the almost-equatorial circle in the last solution and drag it to the south pole, so that we have a small circle there along with two diametrically opposed support strings connecting it to the north pole. This solution is (barely) stable, although physically it is not exactly easy to implement (a very small but non-negligible disturbance will suffice to remove the ball).</p> <p>The reason for the two answers: this was my thought process in action.</p> <p><strong>EDIT</strong>: Please disregard the above; both of these solutions are unstable.</p> <p>In fact Scott's example is part of a general class of solutions of total length $3\pi+h+\epsilon.$ Take any two diametrically opposite points, and draw a small spherical triangle containing one of the points. Connect all six possible edges between the four points by geodesics, and finally rotate the arrangement so that one of the strings passes through the north pole.</p> <p>Here is a short proof that any stable configuration must have at least four points where three or more strings meet. If there are three or less points, there is a hemisphere $H$ which contains all of the points. Take the complement $H^C$ of this hemisphere; we can remove any strings in $H^C$ because they cannot be geodesics. As $H^C$ doesn't contain the north pole, the sphere can fall out.</p> <p>Note that if any strings are not geodesics between junctions, we can ignore the strings.</p> <p>Here is a short proof (that is not quite rigorous) that could provide a lower bound. Suppose there is not a loop (that is, a set of points connected by geodesics) that is completely contained in the southern hemisphere. Then we may drag any points in the southern hemisphere into the northern hemisphere. (This statement is the part I can't make completely rigorous.) So the sphere must be unstable. Now if we have a loop in the southern hemisphere, there must be at least two strings meeting at the north pole, as otherwise we could simply slide all of the string off one side of the sphere.</p> <p>So we must have a loop in the southern hemisphere and (not necessarily direct) connections from at least two of these points to the north pole. I can't figure out how to work a good lower bound from here.</p> http://mathoverflow.net/questions/44211/product-of-sine/44220#44220 Answer by drvitek for Product of sine drvitek 2010-10-30T05:22:55Z 2010-10-30T05:22:55Z <p>From Andreescu and Andrica, <em>Complex Numbers from A to Z</em> p. 48.</p> <p>$$\prod_{1 \le k \le n} \sin{\frac{(2k-1)\pi}{2n}} = \frac{1}{2^{n-1}}.$$</p> <p>Ibid., p. 50.</p> <p>"The following identities hold:</p> <p>a) $$\prod_{1\le k \le n-1; \gcd{(k,n)} = 1}\sin{\frac{k\pi}{n}} = \frac{1}{2^{\phi(n)}}$$ whenever $n$ is not a power of a prime.</p> <p>b) $$\prod_{1\le k \le n-1; \gcd{(k,n)} = 1}\cos{\frac{k\pi}{n}} = \frac{(-1)^{\frac{\phi(n)}{2}}}{2^{\phi(n)}}$$ for all odd positive integers $n$.</p> <p>The first one completely answers your question; the next two are just for fun!</p> http://mathoverflow.net/questions/42043/is-there-order-to-the-number-of-groups-of-different-orders/42748#42748 Answer by drvitek for Is there order to the number of groups of different orders? drvitek 2010-10-19T07:34:33Z 2010-10-19T07:34:33Z <p>The most general statement that I have seen is due to Holder and gives the following result: if $n$ is squarefree, and if $f_p(n)$ is the number of primes $q$ with $q \mid n$ and $p \mid q-1$, then the number of groups of order $n$ is given by $$g(n) = \sum_{d\mid n}\prod_{p\mid d,\;d > 1} \frac{p^{f_p(n/d)}-1}{p-1}.$$</p> <p>This formula is nic ein that it is exact; of course, the asymptotics here are very hard to extract, and given that there tend to be more groups of non-squarefree order than of squarefree order it is not terribly useful in asymptotic estimates.</p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/42525#42525 Answer by drvitek for Awfully sophisticated proof for simple facts drvitek 2010-10-17T17:34:44Z 2010-10-17T17:34:44Z <p>A number of high school contest problems in number theory reduce to Mihailescu's theorem. (The only perfect powers with a difference of 1 are 8 and 9.)</p> http://mathoverflow.net/questions/42466/algebraic-geometry/42470#42470 Answer by drvitek for Algebraic geometry drvitek 2010-10-17T07:36:00Z 2010-10-17T07:36:00Z <p>Hartshorne. (It even has <a href="http://en.wikipedia.org/wiki/Hartshorne%2527s_Algebraic_Geometry" rel="nofollow">its own Wikipedia page</a>!)</p> http://mathoverflow.net/questions/42332/on-a-special-kind-of-graph-connectig-n-point-to-n-points/42357#42357 Answer by drvitek for On a special kind of graph connectig n point to n points. drvitek 2010-10-16T02:03:53Z 2010-10-16T02:03:53Z <p>Ross's answer is more helpful in terms of structure, but I will point out that your graph is also an example of a <em>circulant graph</em>. Informally, a circulant graph is defined by parameters $n$ and $v$, with $v$ a vector in <code>$\mathbb{N}^m\;(m&lt;n)$</code>. One takes the empty graph on $n$ vertices labeled $1, 2, \cdots, n$, and joins vertex $i$ to vertices $i + v_1, i+v_2, \cdots, i+v_m$ (reduced modulo $n$). So for example the complete graph on three vertices has $n = 3$ and $v = (1,2)$.</p> <p>Your case is the circulant graph on $2n$ vertices with vector $v = (3,5,\cdots,2n-1)$. (There are lots of other vectors $v$ which give isomorphic graphs, but this one is the easiest to see why it works.)</p> http://mathoverflow.net/questions/41473/bounding-the-roots-of-the-sum-of-two-polynomials/41477#41477 Answer by drvitek for Bounding the roots of the sum of two polynomials drvitek 2010-10-08T01:33:27Z 2010-10-08T01:33:27Z <p>You could test if $\min{|p_1(t)|} > \max{|p_2(t)|}$ (or the reverse) over the interval $[t_{min},t_{max}]$, but this looks to be harder than computing $p_3$. This is an odd question...</p> http://mathoverflow.net/questions/40412/if-you-could-redesign-a-high-school-mathematics-curriculum-from-the-ground-up-wh/40430#40430 Answer by drvitek for If you could redesign a high school mathematics curriculum from the ground up, what would you include? drvitek 2010-09-29T06:02:29Z 2010-09-29T06:02:29Z <p>I was fortunate enough to go to a high school that had a course called Further Math (if any of you have heard of the IB curriculum, this class covered most of the topics on the IB Further Math exam) where we did basic propositional logic, Peano arithmetic, basic group theory (up to normal subgroups) and set theory up to Cantor-Schroeder-Bernstein, plus some statistics. I loved this class, although the only new thing I saw all year was CSB, and I think students might like this more than something such as calculus that they only really work with in physics class.</p> <p>I would say that every high school should start students with a general logic class - introduce the mathematical formalisms (up to truth tables) and then work with real-world examples of disingenuous statements or the like. I think that students very much like learning about how adults lie to them. :P</p> <p>Beyond that, as much as I hate to say it, no student should leave high school in an ideal world without knowing probability and statistics (up to the basic properties of the normal distribution). It has become such a powerful tool in our times, and yet it is so easy to lie to people with statistics: take Twain's quote, "There are three kinds of lies: lies, damned lies, and statistics."</p> <p>I guess the idea here is that most high school students will be distracted enough by learning about how adults are lying to them not to notice that they're also being taught how to think critically and how to pursue truth - and that's what mathematics is really about, right?</p> <p>Edit: just saw Andy's comment.</p> http://mathoverflow.net/questions/38435/testing-permutations-to-see-if-they-generate-s-n Testing permutations to see if they generate $S_n$ drvitek 2010-09-12T02:26:00Z 2010-09-12T14:38:54Z <p>Alright, so <a href="http://mathoverflow.net/questions/38026/generating-n-cycles" rel="nofollow">a similar question was recently asked</a> about the theoretical bound for generating certain permutations in polynomial time. I had been thinking about a related problem in algorithms (with applications to a specific problem in graph theory - namely, discrete moves of sets of points among the vertices of a graph) and H A Helfgott's question inspired me to ask here.</p> <p>Suppose I have some "black box" that spits out permutations $\rho_i \in S_n$. I know the following things about the permutations it spits out:</p> <ul> <li>$\rho_i$ is of cycle type $(k_i,1,1,\cdots,1)$.</li> <li>This "black box" is fast in $n$ (linear in $n$ or so, maybe plus a few log terms).</li> <li>If I run this black box long enough, it will spit out all of the $k$-cycles in some subgroup $H \subseteq S_n$. I don't know what $H$ is <em>a priori</em>, although I can tell you (based on other constraints of the general problem) if $H \subseteq A_n$.</li> </ul> <p>Let $G \subseteq S_n$ be the group generated by the $\rho_i$. (Note that $G$ may not in fact be either $H$ or $S_n$.) </p> <p>I'd like to test if $A_n \subseteq G$.</p> <ol> <li>Is there a computationally efficient test to see if the $\rho_i$ act primitively on $[1,n]$? I want to say that if they act transitively and if the $k_i$ do not all share some nontrivial factor, they act primitively, but I am not sure of this.</li> <li>Assuming that the answer to (1) is yes, I can guarantee that the natural action of $G$ on $[1,n]$ is transitive and primitive. Does this guarantee that $G = A_n$? If not, what computationally non-intensive criterion do I need to add to guarantee that $G = A_n$?</li> </ol> <p>Note: right now my algorithm for solving this problem is somewhere in that scary, scary territory beyond $O(n!)$ (yeah, <em>that's</em> how I'm testing to see if the darn thing is the alternating group), so any polynomial-time algorithm here would be super-awesome.</p> http://mathoverflow.net/questions/37498/sequences-of-evenly-distributed-points-in-a-product-of-intervals/37514#37514 Answer by drvitek for Sequences of evenly-distributed points in a product of intervals drvitek 2010-09-02T16:31:26Z 2010-09-02T16:48:13Z <p>One way to interpret this result is that it comes from the periodicity of the continued fraction expansion of $\phi = 1 + \frac{1}{1+\frac{1}{\cdots}}$ in the sense that it has no "better-than-expected" rational convergents, whereas for example with $\pi = (3;7,15,1,292,\cdots)$ we may stop at the 292 to get a good approximation (355/113 I believe).</p> <p>So one may look at numbers of the form $x_n = (n;n,n,n,\cdots)$, which satisfy $x_n^2 -nx_n - 1 = 0$, or $$x_n = \frac{n+\sqrt{n^2+4}}{2}.$$ So a few good sequences may be for example $\left\{nx_2\right\}$ where $x_2 = 1+\sqrt{2}$, the so-called "silver ratio", or the same for $x_3 = (3+\sqrt{13})/2.$</p> <p>EDIT: These are in some cases pretty good approximations; one way to measure the "well-distribution" of such a sequence is to take the fractional parts $\{\lfloor nx_n \rfloor: n = 1, \cdots, M\}$, sort them, compute the maximum difference between consecutive terms, and multiply this by $M$ to get some number in the range $[1,M)$. This can be accomplished in one line in Mathematica as follows:</p> <pre><code>WellDistribution[x_,M_]:= Max[Differences[Sort[Table[N[FractionalPart[x*m]], {m, 1, M}]]]]*M; </code></pre> <p>Some interesting things happen with this when we vary $n$; perhaps I'll make a new post out of it.</p> http://mathoverflow.net/questions/36995/asymptotic-growth-of-a-certain-integer-sequence/37103#37103 Answer by drvitek for Asymptotic growth of a certain integer sequence drvitek 2010-08-30T00:14:40Z 2010-08-30T14:21:15Z <p>This is building off of the work of Richard Borherds' answer and the comment I made there. We can provide a way to compute an upper bound which would be much better than the current exponential bound in $n$, but relies on enumerating solutions to certain Diophantine equations.</p> <p>We are going to try and enumerate the <em>distinct</em> values of the expression $$e_11^n+e_22^n+\cdots+e_kk^n,$$ where $e_i = \pm 1$ for $i \in [1,n]$. Let $S$ be the set of all values attained by this expression. We have obviously $|S| \le 2^k$ as there are $2^k$ ways to choose the signs, but we know that two different choices for the $e_i$ may give the same numeric value. An illustrative example occurs in case $(k,n) = (5,2)$, where we have $$+1^2+2^2+3^2+4^2-5^2 = +1^2+2^2-3^2-4^2+5^2 = 5.$$ If we have two choices for the $e_i$ - for now, call the two lists of coefficients $a_i$ and $b_i$ that coincide, we may form sets $A = \{i \in [1,n]: a_i = +1, b_i = -1\}$ and $B = \{i \in [1,n]: a_i = -1, b_i = +1 \}$. Then we know that $A \cap B = \emptyset$, and more importantly, $\sum_{a\in A}a^n = \sum_{b \in B}b^n$.</p> <p>Now define a function $s(k,n,a,b)$, with $a \ge b$ and $a+b \le k$, to count the number of pairs of sets $(A,B)$ satisfying the following conditions:</p> <ul> <li>$A \cup B \subseteq [1,k]$;</li> <li>$A \cap B = \emptyset$;</li> <li>$|A| = a$ and $|B| = b$;</li> <li>if $a = b$, then $\max{A} &lt; \max{B}$;</li> <li>$\sum_{a\in A}a^n = \sum_{b \in B}b^n$.</li> </ul> <p>Then we may write $$|S| \ge 2^k - \sum_{m=3}^k2^{k-m-1}\sum_{1 \le j \le m/2}s(k,n,m-j,j).$$ Indeed, we see that if some set of choices of $e_i$ give the same numerical value, all but one will be removed by this inclusion-exclusion counting. We may simplify this a bit further by writing $$f(k,n) = \sum_{m=3}^k2^{-m-1}\sum_{1 \le j \le m/2}s(k,n,m-j,j)$$ as our equation then becomes $|S| \ge 2^k(1-f(k,n))$. Now once we have established suitable bounds on $f(k,n)$, we may proceed as Richard did, noting that we simply solve for the least valid value of $k$ in the inequality $$2^k(1-f(k,n)) \ge 2\sum_{i=1}^ki^n.$$</p> <p>Where does Fermat's Last Theorem come in? Well, in case $n \ge 3$, we need only sum from $m = 4$ to $k$ - there are no solutions for $m = 3$, as a solution for $m = 3$ takes the form $x^n + y^n = z^n$! So we lose what is (possibly?) the largest contributing term to $f(k)$.</p> <p>EDIT: I was off by a factor of two. Darn.</p> http://mathoverflow.net/questions/35468/widely-accepted-mathematical-results-that-were-later-shown-wrong/35470#35470 Answer by drvitek for Widely accepted mathematical results that were later shown wrong? drvitek 2010-08-13T10:23:12Z 2010-08-13T10:40:55Z <p>Kempe's "proof" of the four-color theorem springs to mind. Wikipedia says that Kempe published it in 1879 and it was proven to be incorrect by Heawood in 1890. As I recall, the flaw in the original argument was as follows: Kempe defined a structure on a planar graph called a Kempe chain, and argued that certain of these chains could not intersect. There was a subtle flaw in this argument (which I can't seem to find a decent explanation of) and it failed for certain large graphs - the chains can in fact intersect. Heawood provided a 25-node example of intersecting chains; the smallest counterexamples are the Fritsch and Soifer graphs on 9 nodes.</p> <p>Edit: I didn't address the reknown of Kempe's proof. Wikipedia says that it was "widely acclaimed" (interesting coincidence of wording) while <a href="http://www.ams.org/notices/199807/thomas.pdf" rel="nofollow">Thomas 1998</a> provides an excellent history but says little on this matter. I don't know if this could be truly considered "widely acclaimed" based on an uncited Wikipedia entry.</p> http://mathoverflow.net/questions/35455/does-subgroup-structure-of-a-finite-group-characterize-isomorphism-type/35460#35460 Answer by drvitek for Does subgroup structure of a finite group characterize isomorphism type? drvitek 2010-08-13T09:39:26Z 2010-08-13T09:39:26Z <p>The modular group of order 16 and the group C8 x C2 have the same subgroup lattice. Does this provide a counterexample to what you are trying to prove?</p> <p>Reference: <a href="http://www.opensourcemath.org/gap/small_groups.html" rel="nofollow">http://www.opensourcemath.org/gap/small_groups.html</a></p> http://mathoverflow.net/questions/34581/conjectureit-can-be-to-partition-prime-sequence-to-two-parts-of-equal-or-consec/35458#35458 Answer by drvitek for Conjecture:It can be to Partition prime sequence to two parts of equal (or consecutive) sum drvitek 2010-08-13T09:22:45Z 2010-08-13T09:22:45Z <p>Scott Carnahan had an interesting idea; let's formalize it into an actual solution. We will show that, given $n \ge 2$ a positive integer, $p_1, \cdots, p_n$ the first <em>n</em> primes, we have some $e_1, \cdots, e_n$ with $e_i = \pm 1$ such that $|e_1p_1 + e_2p_2 + \cdots + e_np_n| \le 1$. (Note that we may further stipulate that $e_n = 1$.) A simple parity argument from here suffices to prove the conjecture.</p> <p>We will prove this by induction on $n$. The cases $2 \le n \le 6$ are trivial to verify, and were provided already by a-boy. We now fix $n \ge 7$.</p> <p>We first need some asymptotics in the form of the Bertrand-Chebyshev theorem; we use the formulation that for $m > 1$ there is a prime between $m$ and $2m$.</p> <p>Write $S_k = e_np_n + e_{n-1}p_{n-1} + \cdots + e_{n-k+1}p_{n-k+1}$, and let $M(k)$ be the minimum of $|S_k|$ over all tuples $(e_n, e_{n-1}, \cdots, e_{n-k+1})$. We stipulated earlier that $e_n = 1$, so we have $M(1) = p_n$. Two facts that will be useful to us in the future are that (1) $M(k+1) \le |M(k)-p_{n-k}|$ and (2) if $|a| \le |b|$, then $\min{\{|a+b|,|a-b|\}} \le |b|$.</p> <p>We claim that $M(k) \le p_{n-k+1}$ for $k = 1, 2, \cdots, n-2$. We prove this by induction on $k$. The claim for $k = 1$ is trivial. Now if $M(k) \le p_{n-k}$, then we are done, as $M(k+1) \le \min{\{|M(k)+p_{n-k}|,|M(k)-p_{n-k}|\}} \le p_{n-k}$ by fact (2).</p> <p>Now suppose $p_{n-k} &lt; M(k) \le p_{n-k+1}$. Write $2m+1 = p_{n-k+1} \ge p_3 = 5$, so that $m > 1$. In this case we know that $m &lt; p_{n-k} &lt; M(k) \le 2m$. But then $M(k+1) \le M(k) - p_{n-k} \le 2m-(m+1) = (m-1) &lt; p_{n-k}$ as desired.</p> <p>The fact that $M(k) \le p_{n-k+1}$ is eminently useful. </p> <p>Indeed, we may use it to dispatch of the even case immediately. Set $k = n-6$. Then we have $M(n-6) \le 17$. As all sums considered in $M(n-6)$ are sums of an even number of odd terms, we in fact have $M(n-6) \le 16$ and even. Now we simply note that all odd numbers between -15 and 15 are realizable as sums and differences of the first 6 primes, which is left as an easy computational exercise.</p> <p>In the odd case, we consider $k = n-5$. Then $M(n-5) \le 13$. For the same parity reasons as above, we have in fact $M(n-5) \le 12$. And again, we note that all even numbers between -12 and 12 are realizable as sums and differences of the first 5 primes - another easy computational exercise.</p> <p>The limits of $n-6$ and $n-5$ are the best possible for our small-case analysis.</p> <p>If we were to establish an algorithm for this, we could just do the greedy algorithm on choosing $e_n$, then $e_{n-1}$, and so on, each time choosing $e_k$ so as to minimize $S_{k+1}$ (or randomly if $S_k = 0$). Our claim that $M(k) \le p_{n-k}$ will continue to be satisfied by the greedy algorithm, as the proof of the claim does not involve changing prior $e_i$. Thus our greedy-algorithm mimium modulus must satisfy the same inequality, and we continue until we are at $n-6$ or $n-5$, then finish as in our nonconstructive proof.</p> http://mathoverflow.net/questions/35150/sum-equals-product/35152#35152 Answer by drvitek for Sum Equals Product drvitek 2010-08-10T17:54:18Z 2010-08-10T18:13:29Z <p>Let the positive, unequal integers be $a_1 &lt; a_2 &lt; \cdots &lt; a_k$ with $a_1+a_2+\cdots+a_k = a_1a_2\cdots a_k = n$. Obviously if $k = 1$ all positive integers work; suppose from now on that $k > 1$. Note that $a_k \ge k$. Then we have $a_1 + a_2 + \cdots + a_k &lt; ka_k$, while $a_1a_2\cdots a_k \ge (k-1)!a_k$. So we must have $ka_k > (k-1)!a_k$, which means that $k > (k-1)!$. This means $k \le 3$. If $k = 2$, then we have $a_1 + a_2 = a_1a_2$, or $\frac{1}{a_1}+\frac{1}{a_2} = 1$. This can easily be seen to have only the solution $(2,2)$, which doesn't satisfy our hypothesis that the $a_i$ be unequal.</p> <p>The case $k = 3$ is then the only tricky case. If $a_1 > 1$, then we have $a_1+a_2+a_3 &lt; 3a_3$ and $a_1a_2a_3 \ge 6a_3$. So $a_1 = 1$. We then must find solutions to $1 + a_2 + a_3 = a_2a_3$. This can be rewritten as $(a_2-1)(a_3-1) = 2$, which as $a_2 &lt; a_3$ are integers immediately gives $a_2 = 2, a_3 = 3$.</p> <p>In summary, all possible solutions are $\{n\}$ for all positive integers $n$ and $\{1,2,3\}$.</p> <p>History: this is definitely a classic problem; see for example <a href="http://amc.maa.org/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2006.pdf%20%222006%20USA%20Mathematical%20Olympiad%20problem%20#4%22" rel="nofollow">2006 USA Mathematical Olympiad problem #4</a> (pdf, problem is on second page), which is your problem with several restrictions removed.</p> http://mathoverflow.net/questions/89720/how-to-mentor-an-exceptional-high-school-student/89722#89722 Comment by drvitek drvitek 2012-02-28T17:45:39Z 2012-02-28T17:45:39Z As a college student and fellow RSI alum, I gotta say it does wonders for a kid who hasn't seen research mathematics up close and personal. http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples/27546#27546 Comment by drvitek drvitek 2011-05-05T01:43:06Z 2011-05-05T01:43:06Z @Qiaochu: Your argument is nice, but irrelevant to this minimal counterexample: 271441 is 521 squared. http://mathoverflow.net/questions/8846/proofs-without-words/31419#31419 Comment by drvitek drvitek 2011-05-04T00:13:11Z 2011-05-04T00:13:11Z Indeed, there is a proof with only eleven words: rearrangement and arithmetic-geometric inequalities. (Details are left to the reader.) http://mathoverflow.net/questions/2147/most-helpful-math-resources-on-the-web/2445#2445 Comment by drvitek drvitek 2011-05-03T15:49:06Z 2011-05-03T15:49:06Z Simon: try <a href="http://www.trillia.com/online-math/index.html" rel="nofollow">trillia.com/online-math/index.html</a> for a modified version that somebody got. http://mathoverflow.net/questions/59840/surprising-and-useful-physical-intuition-for-mathematical-objects/60023#60023 Comment by drvitek drvitek 2011-03-30T01:09:53Z 2011-03-30T01:09:53Z Haha, and I just wrote that you always provide beautiful illustrations. You've done it again! http://mathoverflow.net/questions/59840/surprising-and-useful-physical-intuition-for-mathematical-objects/59893#59893 Comment by drvitek drvitek 2011-03-29T22:49:59Z 2011-03-29T22:49:59Z I've noticed this but haven't pointed it out; your posts have the most beautiful visuals here. Sometimes pretty pictures are quite nice! http://mathoverflow.net/questions/54769/is-there-evidence-whether-undergraduate-math-courses-improve-problem-solving Comment by drvitek drvitek 2011-02-08T17:15:29Z 2011-02-08T17:15:29Z Anna: I just saw your post as I submitted my comment. Several more things: that's a mission statement for the mathematics major, not for the department as a whole. Still, at least in reference to Duke, you may want to look at the 149S course, which is undergraduate problem-solving and has an enrollment of about 20 each year, not all of which end up being math majors. It's also taught by upperclassmen with some guidance from a professor, which is slightly unusual but works rather well. http://mathoverflow.net/questions/54769/is-there-evidence-whether-undergraduate-math-courses-improve-problem-solving Comment by drvitek drvitek 2011-02-08T17:10:54Z 2011-02-08T17:10:54Z Anna: I would be surprised if there was any high-quality research on this, simply because &quot;problem-solving&quot; as a general skill -- as opposed to mathematical problem-solving, which has some overlap, but is definitely distinct -- is very hard to define or measure across-the-board. You might try specific situations or something like Raven's Progressive Matrices -- which I would suspect would correlate decently with mathematical training -- but you're running into a definition problem, especially as &quot;problem-solving&quot; is nowadays being used as a sort of catch-all term in education-speak. http://mathoverflow.net/questions/54536/a-reverse-hadamard-inequality/54555#54555 Comment by drvitek drvitek 2011-02-06T21:25:26Z 2011-02-06T21:25:26Z Your formula doesn't look right; all the $W_j$ are equal. http://mathoverflow.net/questions/27324/what-are-some-naturally-occurring-high-degree-polynomials/27325#27325 Comment by drvitek drvitek 2011-02-06T16:00:05Z 2011-02-06T16:00:05Z This answer is not so bad. The cyclotomic polynomial of degree 105 is the first one with a coefficient not in {-1,0,1}. http://mathoverflow.net/questions/50343/what-would-you-want-to-see-at-the-museum-of-mathematics/50439#50439 Comment by drvitek drvitek 2011-01-03T22:57:13Z 2011-01-03T22:57:13Z I question your choice of &quot;intuitive&quot; for Cantor's proof (adopting for a moment the perspective of a layperson) although beautiful is more than apt! http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/49037#49037 Comment by drvitek drvitek 2010-12-17T13:21:24Z 2010-12-17T13:21:24Z I think that Ax-Grothendieck can be lumped with this in a sort of &quot;unexpected model-theoretic arguments&quot; category. http://mathoverflow.net/questions/44211/product-of-sine/44253#44253 Comment by drvitek drvitek 2010-11-01T01:55:07Z 2010-11-01T01:55:07Z @Ace: Thank you for taking the time to check my answer, find that it was off by one, and do the hard work in extending it! Have an upvote! http://mathoverflow.net/questions/44211/product-of-sine/44220#44220 Comment by drvitek drvitek 2010-11-01T01:52:48Z 2010-11-01T01:52:48Z @Vagabond: Gah! A fencepost error! (To be honest, though, it was late, I knew where some semi-relevant formulas were, and I regurgitated them.) http://mathoverflow.net/questions/43791/hanging-a-ball-with-string/43808#43808 Comment by drvitek drvitek 2010-10-30T05:39:12Z 2010-10-30T05:39:12Z (Edit: added solution that actually works, distinct from Scott's.)