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Jens
2010-Feb-25, 07:12 AM
I was thinking about irrational numbers, and got a funny idea. I know it is said that pi never repeats itself, which is fine. But sequences do repeat themselves. So for example, the sequence 99 appears in various places. The sequence 999 would also appear less frequently, but still appears. Same with the sequence 9999. So somewhere there will be a sequence of one millions 9's, though, right? But there is never an infinitely long sequence of 9's, right?

agingjb
2010-Feb-25, 10:03 AM
An infinitely long sequence of 9's (or any other digit, or an infinitely long repeated sequence) in a decimal expansion of a real number would imply that the number was rational. (An infinitely long sequence of digits would be all of the rest of the expansion from the point at which it started.)

It is assumed that every finite sequence of digits appears infinitely often in the expansion of pi - at unpredictably varying intervals - but I suspect that this is just an assumption that would be hard or impossible to prove.

Ivan Viehoff
2010-Feb-25, 02:07 PM
It is certainly possible for an irrational number to have a decimal expansion in which certain patterns fail to occur. There are uncountably many of them. For example
9.009990000999990000009999999...etc is an irrational number, and one in which sequences of a million nines will certainly occur, in fact very frequently from around about the 500 billion digit mark, if I did my sums right. But clearly it has many sequences that never occur, such as anything not involving a 0 or a 9, nor indeed most sequences involving only 0s and 9s, such as 909. And obviously there are uncountably many irrational numbers in which a sequence of a million nines occurs right at the beginning of their decimal expansion.

I recall reading something about the digits of pi being indistinguishable from a sequence of random numbers. I suspect there isn't a proof of it, but ratehr that people have taken sections of its decimal expansion and shown that they are indistinguishable from a sequence of random numbers. No doubt google will tell you, I can't be bothered just now.

As an exercise, you can calculate how many random digits from 0 and 9 you would have to generate to have a 99% probability of a sequence of a million 9s having occurred. I think it might be an amusingly large number. I somehow suspect one will find there are not remotely enough atoms in the visible universe to store all the numbers, even if you could store one number per atom. Since the number of alphabetic symbols in a Shakespeare play is probably of the order of a few hundred thousand, it's actually a very similar calculation as to how many monkeys you would need bashing typewriters at random to get a Shakespeare play. Even if the visible universe was filled with monkeys bashing typewriters, I think it would take a very long time.

hhEb09'1
2010-Feb-25, 02:16 PM
There's the Feynman point (http://en.wikipedia.org/wiki/Feynman_point). :)

A string of six 9s occurs surprisingly early in the expansion of pi, at the 762nd place. The odds of that happening are apparently less than one in a thousand.

Chuck
2010-Feb-25, 02:44 PM
That's to give engineers a good place to round it off.

agingjb
2010-Feb-25, 02:51 PM
Transcendental numbers can be constructed that have clearly a predictable ("non-random") sequence of digits (as Ivan Viehoff shows above). Can an irrational algebraic number be shown to have a sequence of digits that would clearly fail to be "random" (discounting any apparently oddities in the first "few" digits)?

Ivan Viehoff
2010-Feb-25, 06:11 PM
There's the Feynman point (http://en.wikipedia.org/wiki/Feynman_point). :)

A string of six 9s occurs surprisingly early in the expansion of pi, at the 762nd place. The odds of that happening are apparently less than one in a thousand.

In a random number sequence, the odds of 314159 occurring in the first 762 places is just as unlikely as 999999. Obviously some numbers had to be the first 762 numbers. There are so many things that can be called "unlikely coincidences" that probably some unlikely coincidence will be found.

Obviously the decimal expansion of pi is (at best) pseudo-random, because there is a method of generating those numbers. The usual pseudo-random number methods are cyclical, so if you know the cycle length and what has gone before, then you know what the last number in the cycle is. Modern electronic one-armed-bandits/fruit machines use pseudo-random sequences because it gives the house a guaranteed return on the cycle. (British gaming regulators are dubious about whether this should be legal.) With such machines, it really is true that if they haven't paid out much for some time, they do become more likely to have some pay outs, which is not the behaviour of true random numbers.

In a true random number sequence, you can choose a level of probability arbitrarily close to 1 (but not 1 itself) and calculate the number of digits you'd have to go to to get that probability of a specific sequence of numbers turning up. There's even a branch of number theory called probabilistic number theory that does things like that, and manages to show that certain propositions concerning numbers are so "overwhelmingly likely" as to "exclude" the possibility that they are false. I'm a bit dubious about this myself. In a pseudo-random sequence, it may be that it is impossible for a specific sequence of numbers ever to turn up, though if one was treating them as true random numbers then it would eventually occur with probability arbitrarily close to 1. I suspect we will never know whether somewhere in the expansion of pi there is a sequence of a million nines, unless by chance, like the six 9s, it turns up surprisingly early. Very, very early.

Ivan Viehoff
2010-Feb-25, 06:20 PM
Can an irrational algebraic number be shown to have a sequence of digits that would clearly fail to be "random" (discounting any apparently oddities in the first "few" digits)?
An algebraic number, for those unfamiliar with the term, is things like the square root of 2 and the cube root of 10. In particular, an algebraic number is the zero of a finite polynomial with integer (or equivalently rational) coefficients.

My guess is not, but I don't really know.

hhEb09'1
2010-Feb-25, 06:22 PM
In a random number sequence, the odds of 314159 occurring in the first 762 places is just as unlikely as 999999. Obviously some numbers had to be the first 762 numbers. They both occur in the first 768 digits. :)

With such machines, it really is true that if they haven't paid out much for some time, they do become more likely to have some pay outs, which is not the behaviour of true random numbers.The linear recursive sequences can be so long that the difference is probably insignificant. Or, it should be. Worse, though, would be if someone could identify the current position in the LRS, and know what the next results would be.