Does the decimal expansion of Pi contain the decimal expansion of Pi?
and so on...
and so on....
Does the decimal expansion of Pi contain the decimal expansion of Pi?
and so on...
and so on....
Well, only once.
And no more than that.
If pi contained a subset of digits identical to its own decimal expansion, then the sequence of digits before the subset would be repeated within the subset; and the subset would contain a similar sequence of digits repeated within a further subset ... And so on. Which would mean that pi was a repeating decimal, which would mean it could be expressed as a ratio, which would mean it was a rational number. Which it isn't. Therefore pi cannot contain a subset of digits that repeat its own decimal expansion.
Grant Hutchison
Maybe beyond some point in the digits of pi, every other digit is a digit of pi, in order. Would that count as pi containing itself?
Good point. If you're willing to just encounter the digits, skipping over the digits that don't match rather than regularly skipping a fixed-size gap, then you will be able to re-find the digits of pi, just not consecutively.
Containment can have different levels. My own fetish number (down there in my signature, the Thue-Morse sequence), an infinite non-repeating sequence, does contain itself in a way. Regard the pattern of "01" and "10" pairs you see, or the "0110" and "1001" quads. Both (and infinitely more) appear in the same pattern. If you change every 01 to zero and every 10 to one, you get the same infinite sequence.
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...
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Oh, I neglected to mention the more obvious containment of the Thue-Morse (see signature) sequence within itself. It is exactly as Chuck wondered about pi. From Thue-Morse, take every other digit. You get Thue-Morse. Take every 4th digit. Take every 8th, or 16th. You get Thue-Morse again. Every 32nd, etc.
Again, this is a nonrepeating infinite sequence. And it's so nonrepeating that it completely avoids any triples, any repetition of one group three times: no 000, no 010101, no 101101101, etc. Way cooler than a number deserves to be.
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...
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Well, that's what I was talking about - an infinite subset, which in turn must contain its own identical infinite subset, which in turn contains another identical infinite subset, and so on. That would make it a repeating decimal. But pi is an irrational number, which means its non-repeating. Therefore it can't contain such nested identical infinite subsets.
Grant Hutchison
Putting what I've been saying on a mathematical footing:
If pi contains itself, then we could multiply it by 10^{n} (for some integer value of n, corresponding to the number of decimal places at which the contained copy starts) and recover pi, like this:
pi * 10^{n} = x + pi (where x is an integer that contains all the digits before the contained copy)
Rearranging gives us:
pi = x/(10^{n}-1)
Which is a ratio of two integers, which conflicts with our knowledge that pi is irrational.
Grant Hutchison
And of course if we take Strange's suggestion, that it shows up once, that would be the case where x and n are both zero, which are the only values for which grant's equation works.
Conserve energy. Commute with the Hamiltonian.
Last edited by wd40; 2017-Oct-01 at 03:36 PM.
I'm quite aware of that; it can't be represented by a finite expansion. Infinities behave strangely: if I number each one of the real numbers between 0 and 1 with a whole number, there would be a one-to-one correlation between 0<x<1 and all whole numbers. This can be repeated between any two arbitrary real numbers that aren't equal, e.g., 0.5<x<0.6.
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...
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Not repeating but terminating and in part repeating in bases of pi and its roots:
https://www.wolframalpha.com/input/?...e+(sqrt(pi,6))
There probably is a way to make the digits repeat, but the base would probably be unrepresentable using existing mathematical notations.
Is there a base n where pi = 1010?
Probably, but the n would be an irrational number unrepresentable in full but calculable to any number of given decimal points.
Corrections are welcome.
ETA Perhaps:
http://m.wolframalpha.com/input/?i=s...80%3Dn%5E3%2Bn
Last edited by a1call; 2017-Oct-02 at 04:57 AM.
I've always loved the fact that
e^(pi*i)=-1
I love irrational transcendental numbers too. Somewhere in pi there are a billion zeroes in a row.
And if you assign each triplet of digits as the ASCII codes for characters, pi somewhere contains this post, your post, in fact the entire forum, plus the DNA of everything living creature on the Earth, plus our entire history and every book ever written and every conversation ever engaged in. An infinite number of times.
clop
It should be impossible to copyright anything since the material previously existed in pi and is therefor not an original work.
Fortunately, that does not disqualify a copyright. If two people create an original expression of the same idea in the same way, idependently, they both may own it.
What could be disqualified is a non-creative copying of some part of pi. But that would have to be shown, not assumed.
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...
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This kind of stuff makes my head hurt...
We can't know for sure either way, I get that, I think... But would it not be sensible to assume that it must? Since that its non repeating and is infinite then should we assume that every possible sequence of digits is in there? Or is it because the possible sequence of digits is infinite also which gives it the ambiguity?
I think mathematicians believe it's likely that it does, but the last time I looked no-one had proved that it does. We don't even know if all digits 0 to 9 occur an infinite number of times in the decimal expansion of pi. It might just stop having nines at some point, for instance, leaving us with only a finite number of nines.
And it's trivially easy to come up with an infinite, non-repeating sequence that doesn't contain all finite sequences - 0110001111000001111110000000... for instance.
Grant Hutchison
It's easy to find counter examples. For instance, Liouville's constant is transcendental, irrational of course, infinite and non-repeating, but has only zeroes, and isolated single digits of 1, except for the first two digits.
ETA:Or is it because the possible sequence of digits is infinite also which gives it the ambiguity?
Similarly, Liouville's constant is .1100010000000000100000... where the nth 1 is at the n! position.
In mathematics, this property is called normality. A normal number is a real number in which every sequence of digits occurs with the same frequency as every other possible sequence of digits with the same length. It's possible to prove that "almost all" (that term has a specific mathematical meaning in this case) real numbers are normal, but it turns out to be tantalizingly hard to show that any specific real number is or is not normal. As grant points out, although most mathematicians suspect that pi is normal, it has not been proven to be so.
Conserve energy. Commute with the Hamiltonian.
The math involved in normal-numbers is way beyond my IQ, but with my naive interpretation, it should be possible to exist a not-normal number that contains any given finite sequence of digits, infinite times, just not with a uniform density. Same way a loaded Die will land 100 consecutive 1s, just much more rarely than it lands with 100 consecutive 6s.
There might be sequences which never occur in pi, but not being normal may or may not be relevant.
Corrections/insights are appreciated.
Last edited by a1call; 2017-Oct-04 at 12:56 AM.