User aravind - MathOverflow

Answer by Aravind for What would you want to see at the Museum of Mathematics?

2010-12-27T13:53:26Z

The trammel of Archimedes. See http://blog.makezine.com/archive/2010/06/my_10_favorite_mechanical_animation.html

Answer by Aravind for triple with large LCM

2010-12-22T08:49:50Z

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 $ap+r < bp+r < cp+r$, we have $c/a < 2$. The LCM is at least $\frac{a}{c}p^3$ (the case where a fraction of the numbers have r=0 is easy).

The main advantage seems to be the choice of many primes.

Two possibilities:

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.
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).

Answer by Aravind for Particular problem solved by solving a more general problem.

2010-12-14T11:12:17Z

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.

A problem on sums of arctangents of rationals

2010-12-03T05:24:14Z

Let $S$ be a set of rational numbers.

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 '$\{\tan^{-1}(x): x \in S\}$'.

In particular, let $\hat S$ be the closure of $S$ under both negation and the binary operation $p*q=\frac{(p+q)}{(1-pq)}$.

For any natural number $b \geq 2$, $S_b =\{ \frac{1}{b^k}: k \geq 1 \}$. Prove that $0 \notin \hat {S_b}$.

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 $b$).
$tan^{-1}(1/2) + tan^{-1}(1/3) = \pi/4$.

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?

Comment by Aravind

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>1. By the way, I am curious about how you obtained your example of numbers which multiply to an integer.

Comment by Aravind

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.