In honor of pi day (3/14)…

If pi has an infinite number of digits in random sequence, then pi must contain all sequences.

It is commonly said that “pi does not repeat” and while generally true (pi does not predictably repeat) I think this assertion is clearly not strictly correct – there are an infinite number of repeated sequences in pi (e.g. “123123” occurs in the first million digits) as subsets of the sets of numbers in pi.

Therefore, I postulate that there exists at least one decimal position N in pi that is followed by the exact sequence of digits 1 to N-1. Namely, pi is “3.14159…314159..”. In regular expression syntax this question would actually just be phrased as /^([0-9]+){2}/, omitting the period after 3, naturally.

It should be provable that not only such an N exists but there is an infinite set of such repeating points. I would guess that that first N would be very large, maybe larger than the number of digits of pi yet computed (in the trillions), but if there is one there are almost assuredly an infinite number of other repeating points. This set would start with an almost improbably large number and I would suppose the numbers would very quickly get ridiculously larger.

There would be different such sets for different transcendentals, so perhaps we can discuss the creation of a function “D” that defines the infinite set of these “repeating points” for any transcendental. D(π) , D(e), D(φ), ~~D(√2)~~, etc.

The digits of pi aren’t really in random sequence — they form a pattern, namely that they’re the digits of pi! I don’t know much about what that implies. But let’s think about a truly random digit sequence, that is, an infinite sequence X1, X2, … where each number is chosen independently and uniformly from {0, …, 9}.

It’s true that in such a sequence, each finite sequence of digits appears with probability 1. But that doesn’t imply that there’s a position N that’s followed by the exact sequence of digits X1, …, X(N-1). The difficulty is that the probability of generating the sequence X1, …, X(N-1) decreases with N. For instance, the probability that position 2 has the desired property is 1/10, since it just needs to repeat one digit. But the probability that position 3 does the trick is only 1/100. And so on…the probability that position N does the trick is 1/10^(N-1).

So what’s the probability that a repeat occurs anywhere? Clearly it’s no more than P(repeat at 2) + P(repeat at 3) + P(repeat at 4) + …. That’s 1/10 + 1/100 + 1/1000 + …, which is a geometric series summing to (1/10) / (1 – 1/10) = 1/9. So the probability of a repeat like this occurring is no more than 1/9.

Cool question, though!

Brian,

Fantastically straightforward bit of probabilities there, thank you.

Do you think it could be proven at some point that N does not exist? (Proving N would be best done by finding it, obviously.) If we picked nine random transcendental numbers, it seems we’d have a good shot of finding one with such an N.

Follow-on question – could we then define the set of transcendental numbers that did have such a repeating point N?

Zoomm (swipes hand over head). Way over my head, but awesome postulate.

Yeah, if we generated a bunch of transcendental numbers, probably some of them would have repeats. The most common would be where the second digit is the same as the first. That would be about 10 times as common as repeating the first two digits.

As to proving that a number doesn’t contain one of these repeats…that seems hard. I wouldn’t know where to start.