Numerical coincidence involving the number 1663 - MathOverflow

Numerical coincidence involving the number 1663

2011-11-19T18:55:58Z

$\displaystyle \frac{1663e^2}{3} \approx 2^{12}$.

showed up in a comment of this mostly-unrelated question.

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 j-invariant explanation for Ramanujan's observation that

$e^{\pi\sqrt{163}}$

is close to an integer, or the Pisot number explanation for the fact that even powers of the golden ratio are close to integers.

Answer by Michael Lugo for Numerical coincidence involving the number 1663

2011-11-19T20:57:35Z

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, $1663e^2 - 12288 \approx 1/3448$.

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 $\epsilon$.

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.

(Except that $12288$ has such a simple prime factorization.)

Answer by Aeryk for Numerical coincidence involving the number 1663

2011-11-19T21:17:20Z

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.

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}

Answer by Alan Haynes for Numerical coincidence involving the number 1663

2011-11-19T21:19:57Z

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.