5
$\begingroup$

The numerical coincidence

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

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ Powers of the golden ratio are close to the set $\sqrt{5}\mathbb Z$. $\endgroup$
    – André Henriques
    Commented Nov 19, 2011 at 19:45
  • $\begingroup$ Link fixed; thanks. And of course it's only the even powers of the golden ratio that are close to integers, for the reason you state. $\endgroup$
    – David Eppstein
    Commented Nov 19, 2011 at 19:58
  • 2
    $\begingroup$ $|329e^2-2431|$ is only about twice $|1663e^2-3\cdot 2^{12}|$. But there are bound to be some reasonably small numbers that give numerical coincidences of this or any similar flavor. I think any answer should address whether or not a given numerical coincidence is unlikely given the size of the numbers involved (taking as a model one in which numbers are selected uniformly at random from some range). $\endgroup$
    – Steve Huntsman
    Commented Nov 19, 2011 at 20:26

4 Answers 4

Reset to default
12
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • 7
    $\begingroup$ To me, the fact of being a convergent is a restatement of, rather than an explanation for, being a close approximation. And it doesn't say anything about why 12288 is so smooth. $\endgroup$
    – David Eppstein
    Commented Nov 19, 2011 at 21:27
  • 2
    $\begingroup$ Well of course, the two statements are equivalent- maybe I should have assumed that you already knew that. When you say that $12288$ is smooth, well it is also not very large. $\endgroup$
    – Alan Haynes
    Commented Nov 19, 2011 at 21:49
  • 2
    $\begingroup$ After all, some denominator of some rational approximation of some irrational number like $e^3$ or $\pi^\pi$ should be smooth. Wasn't $12288$ so smooth, the subject of this question would have been different. $\endgroup$
    – Pietro Majer
    Commented Nov 20, 2011 at 9:19
6
$\begingroup$

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}

Improve this answer
$\endgroup$
1
  • $\begingroup$ Similarly to $e^{\pi\sqrt{163}} \approx 640320^3+744$, the ninth convergent suggests $3655e^2 \approx 30^3+7$ $\endgroup$
    – Jaume Oliver Lafont
    Commented Jun 1, 2017 at 3:16
4
$\begingroup$

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.)

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ What I'm impressed by is not so much that $e^2$ is close to the ratio of two four-five digit integers, but that one of them is extremely smooth. $\endgroup$
    – David Eppstein
    Commented Nov 19, 2011 at 21:09
  • 4
    $\begingroup$ I think putting 1663 in the title has led people in the wrong direction. I get the feeling that what you wanted to ask was whether there's some reason why $2^{12}e^{-2}$ is so close to being in ${\bf Z}[1/3]$ (it's 554.33332, to 5 decimals). $\endgroup$
    – Gerry Myerson
    Commented Nov 19, 2011 at 23:00
  • 1
    $\begingroup$ Thanks — I rewrote the question in an attempt to aim more at what I want in an answer. $\endgroup$
    – David Eppstein
    Commented Nov 19, 2011 at 23:43
3
$\begingroup$

It can be related to the simpler $$\frac{13e^2}{3}\approx 2^5$$

because $$ \frac{3}{e^2} \approx \frac{1663}{2^{12}} = \frac{13·2^7-1}{2^{12}}= \frac{13}{2^5}-\frac{1}{2^{12}}$$

Further decomposition into alternating sign bits is $$\frac{1663}{2^{12}} = \frac{1}{2}-\frac{1}{2^3}+\frac{1}{2^5}-\frac{1}{2^{12}}$$

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.