most recent 30 from http://mathoverflow.net 2013-05-18T17:51:02Z http://mathoverflow.net/feeds/question/106279 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106279/palindromic-continued-fraction Jack Huizenga 2012-09-03T22:03:33Z 2012-09-03T22:37:19Z

<p>Here's what I hope is a final question outside of my area that I need to understand a problem about stable vector bundles on $\mathbb{P}^2$. Thank you everybody for your help so far!</p> <p>Suppose I have a rational number $0&lt;\alpha&lt;1$ with a palindromic continued fraction expansion, i.e. </p> <p>$$\alpha = [0; a_1,\ldots,a_k],$$ where $a_i = a_{k+1-i}$, so that the sequence $a_1,\ldots,a_k$ is a palindrome. I believe from working with several examples that the last two convergents $p_k/q_k$, $p_{k-1}/q_{k-1}$ of this continued fraction satisfy $p_k = q_{k-1}$. That is, the denominator of the penultimate convergent is the numerator of $\alpha$.</p> <p>I assume this is well known, and if so a reference would be great! Thanks!</p>

http://mathoverflow.net/questions/106279/palindromic-continued-fraction/106283#106283 Gerry Myerson 2012-09-03T22:34:27Z 2012-09-03T22:34:27Z

<p>See Edward B Burger, A tail of two palindromes, American Mathematical Monthly 112, April 2005, pages 311 to 321, but especially page 317, Lemma 1 and discussion thereof. </p>

http://mathoverflow.net/questions/106279/palindromic-continued-fraction/106284#106284 D. Savitt 2012-09-03T22:37:19Z 2012-09-03T22:37:19Z

<p>It's straightforward to prove by induction on $k$ that $[a_k; a_{k-1},\ldots,a_1] = q_k/q_{k-1}$. (Let the left-hand side be $r_k$, observe that $r_{k+1} = a_{k+1} + 1/r_k$, and then note that the sequence $q_k/q_{k-1}$ satisfies the same recurrence.) Then $[0; a_k,\ldots,a_1] = q_{k-1}/q_k$; deduce for your palindromic $\alpha$ that $\alpha = q_{k-1}/q_k$. But of course $\alpha = p_k/q_k$ as well, and so we conclude.</p>