Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating numbers should appear somewhere in Pi?
Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating numbers should appear somewhere in Pi?
How about ANY FINITE SEQUENCE AT ALL?
It's almost sure to be the case, but nobody has managed to prove it yet.
Simply being infinite and non-repeating doesn't guarantee that all finite sequences will appear. For example, you could have an infinite non-repeating number that doesn't have any 9s in it. But, as far as numbers go, exceptions like that are very rare, and in almost all (infinite, non-repeating) numbers you'll have all finite sequences appearing.
No, the fact that a number is infinite and non-repeating doesn't mean that and since in order to disprove something you need only one example here it is: 0.1101001000100001000001... this is a number that goes 1 and then x times 0 with x incrementing. It is infinite and non-repeating, yet doesn't contain a single 2.
A number for which that is true is called a normal number. It’s proven that almost all real numbers are normal, but it’s very difficult to prove that any particular number is normal. It hasn’t yet been proved that π is normal, though it’s generally assumed to be.
The jury is out on whether every finite sequence of digits is contained in pi.
However, there are a multitude of real numbers that contain every finite sequence of digits when written in base 10. Here's one, which is defined by concatenating the digits of every non-negative integer in increasing order. It looks like this:
0 . 0 1 2 3 4 5 6 7 8 9 10 11 12 ...This is what allows pifs to work!
It has not been proven either way but if pi is proven to be normal then yes. https://en.m.wikipedia.org/wiki/Normal_number
no. it merely being infinitely non-repeating is insufficient to say that it contains any particular finite string.
for instance, write out pi in base 2, and reinterpret as base 10.
11.0010010000111111011010101000100010000101...it is infinitely non-repeating, but nowhere will you find a 2.
i've often heard it said that pi, in particular, does contain any finite sequence of digits, but i haven't seen a proof of that myself, and if it did exist, it would have to depend on more than its irrationality.
Yes.
And if you're thinking of a compression algorithm, nope, pigeonhole principle.
I’m going to say yes to both versions of your question. Infinity is still infinitely bigger than any expressible finite number. Plenty of room for local anomalies like long repeats and other apparent patterns.
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