Numerical coincidence involving the number 1663 - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:56:46Z http://mathoverflow.net/feeds/question/81368 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81368/numerical-coincidence-involving-the-number-1663 Numerical coincidence involving the number 1663 David Eppstein 2011-11-19T18:55:58Z 2011-11-19T23:43:03Z <p>The <a href="http://en.wikipedia.org/wiki/Mathematical_coincidence" rel="nofollow">numerical coincidence</a></p> <p>$\displaystyle \frac{1663e^2}{3} \approx 2^{12}$.</p> <p>showed up in a comment of <a href="http://cstheory.stackexchange.com/questions/9015/connecting-cells-by-line-and-column-permutations-in-a-finite-grid" rel="nofollow">this mostly-unrelated question</a>.</p> <p>Numerically, it's not a surprise that $e^2$ is close to a rational number whose numerator and denominator are in this range — similarly good approximations to most numbers can be obtained by truncating the continued fraction at the desired level of accuracy. What's much more of a surprise to me is the appearance of a 12th power in this expression. Is there a good explanation for having such a smooth number here? E.g. see the <a href="http://en.wikipedia.org/wiki/Ramanujan%27s_constant" rel="nofollow">j-invariant explanation</a> for Ramanujan's observation that</p> <p>$e^{\pi\sqrt{163}}$</p> <p>is close to an integer, or the Pisot number explanation for the fact that even powers of the golden ratio are close to integers.</p> http://mathoverflow.net/questions/81368/numerical-coincidence-involving-the-number-1663/81374#81374 Answer by Michael Lugo for Numerical coincidence involving the number 1663 Michael Lugo 2011-11-19T20:57:35Z 2011-11-19T20:57:35Z <p>Let's rewrite this as $12288 \approx 1663 e^2$. The coincidence, then, is that $1663e^2$ is close to an integer. In fact $1663e^2 \approx 12288.0002925$, or $1663e^2 - 12288 \approx 0.0002925$. That is, $1663e^2 - 12288 \approx 1/3448$.</p> <p>Since we'd be equally impressed if $1663e^2$ were just below an integer, that means that we're impressed because $1663e^2$ is in an interval of width about $2/3448$ or $1/1724$; call this $\epsilon$.</p> <p>But we'd expect $1663\epsilon$ integers $n \le 1663$ to have the property that $ne^2$ is within $\epsilon/2$ of an integer. And $1663\epsilon \approx 1$. So this is really not impressive.</p> <p>(Except that $12288$ has such a simple prime factorization.)</p> http://mathoverflow.net/questions/81368/numerical-coincidence-involving-the-number-1663/81375#81375 Answer by Aeryk for Numerical coincidence involving the number 1663 Aeryk 2011-11-19T21:17:20Z 2011-11-19T21:17:20Z <p>12288/1663 is the 7th convergent in the infinite continued fraction representation of $e^2$, so it naturally will be a very good rational approximation.</p> <p>For completeness, here's the list of the first 10 convergents (via Mathematica): {7, 15/2, 22/3, 37/5, 133/18, 2431/329, 12288/1663, 14719/1992, 27007/3655, 176761/23922}</p> http://mathoverflow.net/questions/81368/numerical-coincidence-involving-the-number-1663/81376#81376 Answer by Alan Haynes for Numerical coincidence involving the number 1663 Alan Haynes 2011-11-19T21:19:57Z 2011-11-19T21:19:57Z <p>The explanation is that $1663/12288$ is a convergent in the continued fraction expansion of $e^{-2}$. In other words, the continued fraction expansion of $e^{-2}$ starts out as $$[0;7,2,1,1,3,18,5,1,...].$$ If you truncate this expansion after the $5$ then you end up with the rational number $$[0;7,2,1,1,3,18,5]=\frac{1663}{12288}.$$ By basic properties of continued fractions it follows that $$\left|e^{-2}-\frac{1663}{12288}\right|\le\frac{1}{12288^2},$$ which explains the behaviour you are observing. Furthermore there is a theorem which says that all `best approximations' to a real number come from truncating the continued fraction expansion.</p>