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Do we know that every possible finite sequence exists in pi?
by univacky 10y ago
Do we know that every possible finite sequence exists in pi?
- lern_too_spel 10y agoWe do not.
- falcolas 10y agoIIRC, since there are an infinite number of digits in Pi, and there are no infinitely repeating finite sequences in Pi, there are therefore an infinite number of finite sequences in Pi. Granted, the index might be larger to represent than the actual sequence you're indexing, but...
- ikeboy 10y agoThat doesn't imply that every finite sequence is in pi.
- singlow 10y agoI agree. There are an infinite number of finite sequences that are not my social security number and do not contain my social security number. An infinite list is still infinite though it lack any specific finite sequence.
- skykooler 10y ago"An infinite number of finite sequences" does not mean "all finite sequences". For example, the number 1.10100100010000100001... contains an infinite number of finite sequences, but does not contain, for example, any sequences with a 2, or even the sequence '10101'.
- aesthetics1 10y agoYou cannot bend the rules in this way and still equate it to the question at hand - you're introducing characters outside of base2 in your example. If you stuck within the realm of base2, '2' in its binary form certainly appears in the pattern. Not only that, but Pi is normal (all digits distributed evenly), and your sequence clearly is not. I understand what you are saying - indeed there are infinite finite sequences - but it just does not apply here.
- JamilD 10y agoYou can trivially modify their example to be 1.02003000400005... cycling through the digits and the argument still holds. As another commenter pointed out, pi is not proven to be normal, but is suspected to be.
- SomeStupidPoint 10y agoThey already showed 21 (in binary, 10101) doesn't appear anywhere, even as a binary pattern.
- evanb 10y agoThe example isn't base two. It's base 10. It just happens to be a number that only has 0s and 1s. Like eleven. Or 1000.
- TOMDM 10y agoTo phrase it another way, there are infinitely many sizes of Infinity. Just because you have one infinite sequence, does not mean it is the same size as another Infinity and as such may include or exclude elements of the other Infinity.
- monochromatic 10y agoPi is not known to be normal.
- Roujo 10y ago> there are therefore an infinite number of finite sequences in Pi Sure, but this doesn't answer the question: Do we know that every possible finite sequence is present? You can have the former without the latter being true. For example, a sequence of the form "1011011101111011111..."[0], while infinite, never contains the finite sequence "1234". [0] https://oeis.org/A094946 https://oeis.org/A094946
- ronreiter 10y agoNot sure that counts as proof.
- wruza 10y agohttps://en.wikipedia.org/wiki/Normal_number https://en.wikipedia.org/wiki/Normal_number Pi is not proven to be normal, but is suspected.
- inlineint 10y agoNote that normality of a number in a given base is a sufficient condition for existence of all finite sequences in its expansion in this base, but not necessary. It might turn out that Pi is not normal say in base 2, but all possible finite sequences of 0s and 1s occur in it, just with some of them being slightly more frequent than others of the same length.
- wruza 10y agoFrom that article, it seems that correct term is "disjunctive sequence".
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