most recent 30 from http://mathoverflow.net 2013-05-25T19:23:16Z http://mathoverflow.net/feeds/question/48142 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48142/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 paul Monsky 2010-12-05T18:44:48Z 2010-12-05T18:44:48Z

<p>This is not an answer but an attempt to interpret your question in terms of the arithmetic of the field K=Q(i), where it seems much more natural.</p> <p>The image in (Reals)/((pi)(integers)) of the set of u for which tan(u) is a rational, r/s, or "is 1/0" is an additive subgroup, G. Now u-->s+ir is a homomorphism from G to the multiplicative group L*/Q* where L is the field Q(i), and you're really studying L*/Q*. Your first question is basically this: If b>1 is an integer, are the elements (b^n)+i of L*/Q*, n=1,2,3..., multiplicatively independent? This looks very hard to me.</p> <p>If instead of say the powers of 2, you look at the positive even integers, then there are relations. For example (2+i)^6, (8+i)^4, (12+i)^(-2) and (70+i)^2 multiply to an integer, and pi is 6 arc tan (1/2)+ 4 arc tan (1/8)- 2 arc tan (1/12) + 2 arc tan (1/70). A more tractable problem might be to show that for any l there are such relations expressing pi as a Z-linear combo of arc tan (1/k) with l dividing each k.</p>

http://mathoverflow.net/questions/48142/a-problem-on-sums-of-arctangents-of-rationals/48466#48466 Franz Lemmermeyer 2010-12-06T18:38:43Z 2010-12-06T18:38:43Z

<p>In addition to Paul's comment let me point out that the structure of the underlying group was investigated in </p> <ul> <li>Sam Northshield, Associativity of the secant method, Amer. Math. Mon. 109 (2002), 246-257,</li> </ul> <p>and that the relation with the arithmetic of ${\mathbb Z}[i]$ was observed by </p> <ul> <li>John Todd, A problem on arc tangent relations, Amer. Math. Mon. 56 (1949), 517-528</li> </ul> <p>The associativity of the secant method is, by the way, a consequence of the usual geometric interpretation of the group law on conics.</p>