It's worth noting that the article slightly confuses the definition of normal number: A number isn't "normal with respect to a given base". It's either normal, or it's not. The definition is in terms of a given base, but it must hold true for all bases.
I think the "confusion" is justified. It's a good thing(tm) to start with the concept of "normal with respect to a base" and then move on to say that "normal" or "absolutely normal" means "normal for every base (greater than 1)."
Wikipedia does this, adding the extra term "simply normal" for the case where we are talking about relative to a single base:
We say that x is simply normal in base b if
the sequence S(x,b) is simply normal, and
that x is normal in base b if the sequence
S(x,b) is normal. The number x is called a
normal number (or sometimes an absolutely
normal number) if it is normal in base b
for every integer b greater than 1.
For the purposes of explanation, especially to non-technical audiences, I think the approach taken is fully justified.
I think the question was whether it is possible for a number to be normal in one base but not in another. Intuitively I would say that randomness should transcend the base and therefore a number that is normal in one base should be normal in any other base and therefore absolutely normal. But if that were the case the distinction between normal in a specific base and absolutely normal would be unnecessary. So I don't know but would also be interested in an example in case one exists and is known.
EDIT: It is easily possible for simply normal numbers, i.e. if you only consider single digit frequencies but not frequencies of digit pairs, triples and so on. 0.(0123456789) is simply normal in base 10 because every digit occurs with frequency one tenth but it is not normal in base 10 because only ten out of one hundred digit pairs occur. But it is also not simply normal in base 10¹⁰ because it then consists only of the single digit representing 0123456789 repeated indefinitely.
I am not sure if I get your point. The number is simply normal in base 10 but is not simply normal in base 10¹⁰. And that is what Retr0spectrum asked with simply normal replaced with just normal, i.e. for a number that is normal in at least one base but not normal in at least one other base.
I'm guessing based on https://en.wikipedia.org/wiki/Transcendental_number that what the author in-consistently refers to as "p" (though, later there is a reference to "e or pi" - is actually pi, or π - I don't think I've ever seen pi referred to as "p" - anybody care to hazard a guess as to why the author did so?
My only problem with the Unicode GREEK SMALL LETTER PI is that it consistently renders on the systems I've used as a square with the bottom portion cut off. This must be intentional, because I just zoomed in to this page to an absurd degree and it consistently renders that way even at 36+ pts. It makes sense to me that in day-to-day life with real Greek the simpler rendering is superior, it doesn't match the traditional mathematical rendering very well.
If you go to this[1] page and click "View All", you'll see that it's "Verdana" or "Tahoma", two pretty common web fonts, that are likely to be the culprit.
Please note the postscript at the bottom ofthe article:
> Postscript: In November 2013 I received an email from Dan Corson informing me that, using digits of e computed by a program called “y-cruncher”, he had checked the number of CPSs up to 100 million digits, and found that it does finally approach the expected value, although even by this point it has not quite ever reached the expected value
It wouldn't be surprising that e is not normal. After all, it's continuous fraction is regular ([2; 1, 2, 1, 1, 4, ... 1, 1, 2k, ...]), while for a true random number there wouldn't be any pattern to it.
There used to be five hundred billion digits of e available here[1], which would remove this reason to think something strange was happening, not that it would prove it.
Because the digits are (in terms of compression) random, and take at least 3.3 bits each, 500,000,000,000 digits is over 200 gigabytes of data! Downloading that would take a long time and cost a lot for both parties. Generating it again would only take 12.8 days...
> Downloading of digits is no longer available due to the massive bandwidth requirements. Your best bet is to directly contact one of the record holders and see if they still have a copy of the digits.
Seems like a case for the high-bandwidth high-latency hard-drive-by-mail method.
"Is it conceivable that different irrational numbers have different "frequencies", such that high-frequency numbers exhibit their expected averages very quickly, whereas low-frequency numbers may require billions of digits to exhibit the expected averages?"
Yes, and obviously so. Take a (base 10) normal irrational and divide it by 10^1000000000000. The first trillion or so digits of the result will be 0, but the result is still irrational (because duh) and (base 10) normal because the frequencies of the digits have to converge in the infinite limit and giving one digit a large finite head start won't change its overall proportion in the limit. QED.
Note that for a base 10 normal irrational by the definition of "convergence" in this case means || (number of digits of value a) / (number of digits of value b) ||_\infty < ε (sup norm over all a, b from 0 to 9, a ≠ b) for some ε > 0 such that for some fixed n (the number of digits in the decimal expansion, halted), we have that whenever N > n this inequality holds.
In the case where there are a fixed number m of leading "a"s (whatever digit a is), we have ||(m + dig_a)/(dig_b)||_\infty = ||m/dig_b + dig_a/dig_b||_\infty -- we only need this to be less than SOME fixed ε (not necessarily the same ε as above) for every N > n for some n we choose.
By assumption ||dig_a / dig_b||_\infty < ε whenever n > N. Choose n' > n such that dig_b satisfies m/dig_b < ε
(Note: this is possible as dig_b increases as n' grows: since m is constant and if dig_b were to be constant for n > N' then ||dig_a/dig_b||_\infty would be larger than ε for some n).
Then we have ||m/dig_b||_\infty < ε and ||dig_a/dig_b||_\infty < ε so ||m/dig_b + dig_a/dig_b||_\infty < 2ε.
Thus it converges -- satisfying the same definition as above chooseing n = n' and ε = 2ε.
I just thought of what might be a silly question: In a situation where we can't prove that a probability distribution converges to some theoretical assumed limit, can we assume it even converges to anything? All we have is circumstantial evidence. Or who is to say that somewhere along the line it doesn't converge to something else? Even if tested to the billionth digit, it's still practically zero compared to testing to the billionth busy beaver number etc. etc. I guess this is more of a philosophical question - given no mathematical proof, why is any sample size of N that is tested not better in any sense than a sample of value N=1?
I guess "e" is not normal, because otherwise any bitstring would occur somewhere in its binary representation, and given that some bitstrings are copyrighted, this would present a legal problem!
Bits are not copyrighted. This is just such a weird misunderstanding of information theory.
If bits were copyrighted, you could just transform them into different bits. Change a single bit and then it's no longer copyrighted. Or xor them all. Or only every other bit. Or xor the whole thing with a different file, like a one time pad.
You could also change the content itself. Say it's a video. Shift all the pixels to the right. Or tint it slightly red. Or play it on a monitor and record it with a camera, then invert the resulting video.
There are an infinite number of possible transforms that result an infinite number of possible bits. What matters is information, not bits. They are not the same thing! Information is the bits, and the method you use to transform them together.
Often it's proposed that you can compress a file by storing the index where it occurs in pi. The problem is the index where your file starts will be bigger than the actual file! So you no longer need the file, but you still need this other set of bits that represents the index. The bits have changed, but the information has not.
If you don't have the index, then the whole procedure is entirely useless. Yes, pi theoretically contains season 5 of breaking bad - and an infinite number of variations and derivative works of it. But you will never ever find it unless you already know where to look.
And no computer could possibly calculate pi to that many digits in millions of years, so the whole thing is impractical anyway.
Yes but it's the information that they contain that's illegal, not the bits themselves. You could xor them all and they would still be illegal to distribute. Likewise no one would care if you spread them but didn't give any context to what their use was.
Anyway those examples are trade secrets and other legal issues, they weren't copyrighted.
