this post was submitted on 26 Dec 2024

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Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating numbers should appear somewhere in Pi? (lemmy.dbzer0.com)

submitted 1 week ago* (last edited 1 week ago) by Melatonin@lemmy.dbzer0.com to c/asklemmy@lemmy.ml

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How about ANY FINITE SEQUENCE AT ALL?

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[–] putoelquelolea@lemmy.ml 3 points 1 week ago (1 children)

My guess would be that - depending on the number of digits you are looking for in the sequence - you could calculate the probability of finding any given group of those digits.

For example, there is a 100% probability of finding any group of two, three or four digits, but that probability decreases as you approach one hundred thousand digits.

Of course, the difficulty in proving this hypothesis rests on the computing power needed to prove it empirically and the number of digits of Pi available. That is, a million digits of Pi is a small number if you are looking for a ten thousand digit sequence

[–] Melatonin@lemmy.dbzer0.com 2 points 1 week ago (1 children)

But surely given infinity, there is no problem finding a number of ANY length. It's there, somewhere, eventually, given that nothing repeats, the number is NORMAL, as people have said, and infinite.

The probability is 100% for any number, no matter how large, isn't it?

Smart people?

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[–] LodeMike@lemmy.today 2 points 1 week ago (1 children)

Yes

[–] weker01@sh.itjust.works 3 points 1 week ago* (last edited 1 week ago) (1 children)

Can you prove this? Or link a proof?

[–] LodeMike@lemmy.today 2 points 1 week ago (5 children)

I don't know of one but the proof is simple. Let me try (badly) to make one up:

If it doesn't go into a loop of some kind, then it necessarily must include all finite strings (that's a theoretical compsci term).

Basically, take a string of any finite length, and then view pi in inrements of this length. Calculate it out to double the amount of substrings of length of your target string's interval you have [or intervals]. Check if your string one of those intervals. If not, do it again until it is, doubling how long you calculate each time.

Because pi is non-repeating, each doubling in intervals must necessarily include at least one new interval from all other previous ones. And because your target string length is finite, you have a finite upper limit to how many of these doublings you have to search. I think it's n in the length of your target string.

Someone please check my work I'm bad at these things, but that's the general idea. It's also wildly inefficient This doesn't work with Infinite strings because of diagnonalization.

[–] kogasa@programming.dev 4 points 1 week ago

Your first sentence asserts the claim to be proved. Actually it asserts something much stronger which is also false, as e.g. 0.101001000100001... is a non-repeating decimal which doesn't include "2". While pi is known to be irrational and transcendental, there is no known proof that it is normal or even disjunctive, and generally such proofs are hard to come by except for pathological numbers constructed specifically to be normal/disjunctive or not.

[–] weker01@sh.itjust.works 2 points 1 week ago (1 children)

Let me give another counterexample. Let x be the binary expansion of pi i.e. the infinite string representing pi in base 2.

Now you will not find 2 in this sequence by definition but it's still a non-repeating number.

Now one can validly say that we restricted our alphabet and we should look only for finite strings with digits that actually occure in the number. The answer is the string "23456789" concatenated with x.

[–] LodeMike@lemmy.today 1 points 1 week ago (1 children)

That's like saying your car is busted because it can't drive on a road made of broken glass.

[–] weker01@sh.itjust.works 1 points 1 week ago* (last edited 1 week ago)

That's mathematics. It do be like that sometimes. Counterexamples can be stupid but still valid.

It's on you to prove your claims.

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