User aravind - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:54:28Z http://mathoverflow.net/feeds/user/10926 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50343/what-would-you-want-to-see-at-the-museum-of-mathematics/50492#50492 Answer by Aravind for What would you want to see at the Museum of Mathematics? Aravind 2010-12-27T13:53:26Z 2010-12-27T13:53:26Z <p>The trammel of Archimedes. See <a href="http://blog.makezine.com/archive/2010/06/my_10_favorite_mechanical_animation.html" rel="nofollow">http://blog.makezine.com/archive/2010/06/my_10_favorite_mechanical_animation.html</a></p> http://mathoverflow.net/questions/41698/triple-with-large-lcm/50140#50140 Answer by Aravind for triple with large LCM Aravind 2010-12-22T08:49:50Z 2010-12-23T07:12:22Z <p>Here is a rewriting of the proof of ${(\frac{n}{\log n})}^3$ lower bound which seems to allow some leeway to play with. Let $p$ be a prime around and less than $\frac{n}{3\log n}$. Then there must be $3\log n$ numbers in the same residue class mod $p$. For three of these, say $ap+r &lt; bp+r &lt; cp+r$, we have $c/a &lt; 2$. The LCM is at least $\frac{a}{c}p^3$ (the case where a fraction of the numbers have r=0 is easy).</p> <p>The main advantage seems to be the choice of many primes.</p> <p>Two possibilities:</p> <ol> <li><p>We want our prime to be around $\frac{n}{10}$ (say) rather than $\frac{n}{3\log n}$. but the problem is we cannot claim $>> \log n$ residues for such a prime. Is it feasible to prove that for one of the many primes we have in the range, there must be a residue which appears many times? Looks hard to me though.</p></li> <li><p>Instead of looking for the same residues, for a pair $ap+r,bp+s$, we can look to minimize $as-br$ since the gcd of this pair must divide this. So we can look at numbers $a/r$. Also we can set up relations -for example there must be three (many?) numbers $ap+r,bp+s,cp+t$ such that $\frac{a+1}{r}=\frac{b+1}{s}=\frac{c+1}{t}$ (mod p).</p></li> </ol> http://mathoverflow.net/questions/21214/particular-problem-solved-by-solving-a-more-general-problem/49375#49375 Answer by Aravind for Particular problem solved by solving a more general problem. Aravind 2010-12-14T11:12:17Z 2010-12-14T11:12:17Z <p>There are perhaps many examples found in inequalities because a given inequality with a mass of variables and functions can easily hide "Simple Inequality" of which it is a special case.</p> http://mathoverflow.net/questions/48142/a-problem-on-sums-of-arctangents-of-rationals A problem on sums of arctangents of rationals Aravind 2010-12-03T05:24:14Z 2010-12-06T18:38:43Z <p>Let $S$ be a set of rational numbers.</p> <p>For "special" sets $S$, we can ask if $\pi$ can be written as a $\mathbb{Q}$-linear or $\mathbb{Z}$-linear combination of elements from '$\{\tan^{-1}(x): x \in S\}$'.</p> <p>In particular, let $\hat S$ be the closure of $S$ under both negation and the binary operation $p*q=\frac{(p+q)}{(1-pq)}$.</p> <ol> <li><p>For any natural number $b \geq 2$, $S_b =\{ \frac{1}{b^k}: k \geq 1 \}$. Prove that $0 \notin \hat {S_b}$.</p> <p>Note: $1 \notin \hat S$ since any rational number $P/Q$ in $\hat {S_b}$ in non-reduced form satisfies $(P,Q) \in \{(0,1), (0,-1),(1,0),(-1,0)\}$ (mod $b$).</p></li> <li><p>$tan^{-1}(1/2) + tan^{-1}(1/3) = \pi/4$.</p> <p>Let $b_1,b_2 \geq 2$ be two natural numbers. For which pairs does 1 belong to the closure of $S_{b_1} \cup S_{b_2}$? For which pairs does zero belong to the closure? Are there pairs for which $\pi$ can be written as a $\mathbb{Q}$-linear combination of the arctangents of their negative powers but not as $\mathbb{Z}$-linear combinations?</p></li> </ol> http://mathoverflow.net/questions/48142/a-problem-on-sums-of-arctangents-of-rationals/48383#48383 Comment by Aravind Aravind 2010-12-06T08:54:10Z 2010-12-06T08:54:10Z Thanks for your answer, Paul. Looking at the problem in terms of multiplication of Gaussian integers does make it look clearer. Now, norm considerations can be used to rule out numbers like 1+i as the product of numbers in lZ+i, for an integer l&gt;1. By the way, I am curious about how you obtained your example of numbers which multiply to an integer. http://mathoverflow.net/questions/48142/a-problem-on-sums-of-arctangents-of-rationals Comment by Aravind Aravind 2010-12-03T09:34:57Z 2010-12-03T09:34:57Z This is not a homework problem. In the post, I define the closure operation AFTER considering the problem of expressing $\pi$ as a sum of arctans in order to reduce this to a (hopefully) simpler problem about rational numbers.