The first part of this series surveyed previous attempts at contacting aliens. The second part proposed a base-neutral notation system for encoding messages to aliens.In the previous post in this series, we used continued fractions to represent any real number as a unique sequence of whole numbers. Such sequences are ideal for sending real numbers to aliens via radio signals. Now we have to choose which numbers to send them, out of an infinite choice of candidates.
I propose sending them two numbers. The first of these is the topic of this post: Khinchin’s Constant, KK equals 2.685452001065306445… in base ten, or [2,1,2,5,1,1,2,1,1,3,10,…] as as the sequence corresponding to its continued fraction..
What is so special about K? It is one of the very few numbers capable of giving the driest of mathematical texts exclamatory hiccups. Mathworld prefaces its introduction to K with “Amazingly, …“. The bible of mathematical constants, the stolidly named Mathematical Constants, irrupts with “Here occurs one of the most astonishing facts in mathematics.”
And yet K is virtually unknown to a wider audience. Pi, e, i, the golden mean and the square root of two are all well ensconced in high school maths curricula, though not K.
To explain why I think K would make a excellent number to send to aliens, it will help to first derive it. This is easy to do, because we’ve already done all the hard work exploring continued fractions in the previous post. Pick any random real number — you know, one that in base ten would look something like 14.7631809156… with additional random digits continuing off to the right ad nauseam. Then, represent this number as a continued fraction to find the unique sequence of whole numbers that corresponds to it, just as we’ve done before.
Now consider the first n terms of this sequence — that’s n whole numbers, starting with 14,1,3,4,… in our example. To find the geometric mean of this group of n numbers, we multiply them together and then take the nth rootTo find the arithmetic mean of n numbers, you add them up and divide the sum by n. To find the geometric mean of n numbers, you multiply them and then take the nth root.. What Aleksandr Yakovlevich Khinchin proved in 1934 is that as you make n larger and larger, the geometric mean of the first n terms of this sequence converges on our constant K, 2.685452001065306445…, regardless of the number we picked.
That, to mathematicians, was an utterly unexpected result. There are two reasons why, I think. First, most of the numbers we use every day correspond to sequences whose geometric means evidently do not converge on K. All rational numbers correspond to finite sequences, and therefore cannot possibly lead to KThe rational number 1.23, for example, corresponds exactly to [1,4,2,1,6,1]. (To be tedious but thorough: What if the number lies between 0 and 1? Divide the number into 1 first to get the same sequence without a zero as the first term. This works because there is a unique, one-to-one correspondence between a number and that number divided into 1. Or else just ignore the zero and start from the second term.). Nor can any irrational number that is not transcendental, because its sequence always obeys a pattern: The Golden Mean, for example, corresponds to the sequence [1,1,1,1,1,1,1,1,1,…], whose geometric mean is obviously 1. The sequence corresponding to the square root of 8, [2,1,4,1,4,1,4,1,4,…] converges on 2, not K.
Second is that by inspection, we can easily construct an infinite number of infinite sequences (all of them having a unique correspondence to a real number) that clearly do not converge on K. [1,2,3,4,5,6,7,8,…]’s geometric mean will grow to infinity. [100,100,100,100,…]’s geometric mean is 100. [101,101,101,101,…]’s geometric mean is 101, and so on.
So how can Khinchin’s proof hold? It can because there are innumerably more real numbers that do obey Khinchin than do not. And if I asked you to choose one real number at random (as I did), the probability that you’d pick one that does not obey Khinchin is zero. Zilch. GuaranteedIf you really want to know more about why that is so, the answer involves countable vs. uncountable sets and Georg Cantor..
A couple of things about K, then:
, because we’ve crunched enough numbers and had a look, but we don’t know for sure. A few other useful mathematical constants seem to as well. Still, most of our workaday numbers do not converge on K, no doubt for the same reason that these numbers caught our eye in the first place — they concern themselves with ordered systems.All this is very interesting. But the real clincher as to why we should beam K to aliens — the thing about this number that takes it to a whole new level, as it were, is this: It would appear that Khinchin’s Constant obeys itself. The geometric mean of the first n terms of K, [2,1,2,5,1,1,2,1,1,3,10,…], also converges on K, for as far as we’ve looked. Khinchin’s Constant appears to be autological.
Take a look for yourself. Convinced?
What does this imply? It implies that the sequence of whole numbers that describes K, [2,1,2,5,1,1,2,1,1,3,10,…], is simultaneously described by KHofstadter illustrates the notion with this Escher print:
He goes on to posit that such systems are the basic building blocks for self-awareness in far more complex systems, such as ourselves.. It implies that K is infused with that essential quality of self-loopiness, of continuous folding back on itself, that Douglas Hofstadter identifies in Gödel, Escher, Bach as being at the core of all self-referential systems.
What K embodies, then, is the seemingly paradoxical ability to describe the properties of the system that produces it. It’s akin to what happens when a mind contemplates the laws of physics that govern the mind. Our aliens, which for our purposes here are really just stand-ins for complex self-aware systems, would not escape noticing this analogy — and not just because we’re giving them a massive hint by sending K as [2,1,2,5,1,1,2,1,1,3,10,…].
By including K in our message, then, we are broadcasting that we consider self-referential systems to be special — a prerequisite for the kind of complexity that underpins self-awareness, which in turn allows for the understanding of messages from outer space.
The second number I propose to send will take a completely different tack.
Isn’t this a bit on the esoteric side, though? Pi is after all a lot more useful and well known, in the sense that it permeates any science that makes use of maths. Khinchin’s constant was completely unknown to me until I read this, and it is still very closely related to what I do on a full time basis (ergodic theory)… Anyway, thanks for pointing it out – it is an interesting number.
To be sure, it’s esoteric, but I wasn’t agitating for greater recognition of K in curricula or anything like that:-)
The aim of this entire exercise is really to see how far we can strip assumptions from meaning and if there is any meaning left at the end of it. I found it interesting that if an ability to detect isomorphism is the one minimum thing all self-aware systems have in common (Hofstadter’s thesis), then the “meaning” of K is something we might be able to share without making further assumptions.
I’m assuming systems that are not self-aware could never get it into their “heads” to send or listen for messsages from outer space, because otherwise I don’t see how we can stop ourselves from deciding pulsars aren’t aliens trying to talk to us.
Stefan
Am I correct in assuming you believe numbers to be of more relevance to any alien life than our idiosyncratic spoken/written languages?
Aren’t you assuming that alien life has some concept of numeracy then? Is that a safe assumption?
Stefan!
You don’t seem to be the only one to spend time thinkling about how to tell the universe that we are here.
http://www.nyteknik.se/pub/ipsart.asp?art_id=40198
Thanks, Janne. Here a the paper about the idea, in English.
Having given this some more consideration, here’s another five cents: There’s nothing really “amazing” about K itself. Almost all real numbers have the property that the number k appears with a certain probability P(k) in the continued fraction expansion, in the long run. In fact P(k) = ln(1 + 1/(k(k+2)) )/(ln (2)). So these numbers have many different means in common, and as long as you make sure the particular average is among “almost all real numbers” as mentioned above, this average is going to have the same property as K – it is fixed under the particular averaging procedure.
I think you might end up with the aliens simply believing that we consider the geometric mean in general as the one important thing.
Damn. That’s been bothering me too. When I first tried to visualize the process by which the terms of a continued fraction are obtained for an arbitrary real number — whereby the fractional part of the previous operation is divided into one, the integer part of the result becomes the next term and the fractional part of the result is used in the next iteration — I assumed that there was something important to the relationship between successive fractional parts, because this is the case with rational numbers. The aha moment came when it struck me that for most real numbers, these fractional parts have to cycle around an infinite number of different real numbers between 0 and 1 by the time our continued fraction contains an infinite number of terms; if they do not, then we get a finite cycle of terms, and the number can no longer be transcendental (and is therefore “countable”). All you have to do then is find the probability of each term occuring, as you do, and then calculate the geometric mean, which has to be the same for all these numbers.
At the same time, the probability calculation that you come up with (Wolfram has it too, on p.914) is surprising to me, because if the distribution of the fractional parts were spread out evenly between 0 and 1, as I expected it would be, this would be akin to replacing the iterated fractional parts with a random number between 0 and 1 — in which case there’d be a 50% chance of getting a number between .5 and 1, and hence 50% of the terms in a continued fraction would have to be 1, not 41.50% as it turns out to be. That part still has me stumped, but that’s my fault for being ignorant of ergodic theory.
So I agree with you that the first part of the above post, which relates the “amazing”s and “astonishing”s of real mathematicians, may have turned out to be a little histrionic (but hey, I’m the amateur), because the mechanism by which K is produced is not really all that abstruse. At the same time, the second part of the post — about the autological qualities of K — I think that still holds, though K is clearly not the only number for which this is the case. Mathworld identifies a whole class of such means, for example, though K appears to me to be the simplest mean to calculate.
One thing I don’t understand is why nobody bothers to calculate the arithmetic mean. Because it doesn’t converge?
I’d say you’re a lot more insightful than ignorant:) The fractional parts are not evenly distributed between 0 and 1 – they have a density 1/(ln(2)*(1+x)). In terms of the transformation T(x) that maps x (between 0 and 1) to “the next fractional part”, this density is invariant – which is what allows us to say things about typical statistics of cft’s.
You can compare it with base-10 expansions. If T(x) is the map sending a number between 0 and 1 to 10*x (mod 1) – i.e. just throwing away the first digit in the decimal expansion – then T has an invariant density which is just our ordinary uniform one. This is what tells us that for almost all real numbers, the digits in the decimal expansion occur with equal probability. (And repeating this in other bases, taking the intersection of “almost all” a countable number of times tells us that almost all numbers have uniformly distributed decimal expansions in _any_ base. No actual such numbers are known, of course…)
You’re right that the arithmetic mean doesn’t converge – the probabilities P(k) go like 1/k^2 when k is large, and the arithmetic mean is given by sum[k*P(k)] which is divergent.
Is the next number going to be as interesting? 😉