
Is e Normal? - lisper
http://www.mathpages.com/home/kmath519/kmath519.htm
======
pdpi
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.

~~~
Retr0spectrum
Are there any cases where a number is normal in one base but not another?

~~~
markild
If you look at the definition on Wikipedia[1], you can see that a normal
number, per definition, is base agnostic.

[1]:
[https://en.wikipedia.org/wiki/Normal_number](https://en.wikipedia.org/wiki/Normal_number)

~~~
danbruc
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.

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ecesena
This doesn't prove that the number in the edited example isn't simply normal
in a base b for some b.

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amadvance
That number is rational, and all rational numbers are not normal.

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Fargren
But they can be simply normal. It's a much weaker property.

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ghshephard
I'm guessing based on
[https://en.wikipedia.org/wiki/Transcendental_number](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?

~~~
danbruc
The source looks like this so I guess it is a font issue on your side, I see
the proper symbol for pi.

    
    
      <span style="font-family:Symbol">p</span>

~~~
markild
One can argue that its more a typographic issue than a font issue.

The author probably did see a "pi", but should have been using an embedded
font, or even better, the unicode "GREEK SMALL LETTER PI", π.

~~~
jerf
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.

~~~
robinhouston
There is always U+1D70B MATHEMATICAL ITALIC SMALL PI, but it isn’t available
in very many fonts.

~~~
jerf
Unicode: The solution to, and the cause of, all life's font problems.

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yoodenvranx
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

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Grue3
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.

~~~
CarolineW
sqrt(2) has the CF [1;2,2,2,2,2,...] and yet it appears to be normal.
Similarly sqrt(5) has the CF [2;4,4,4,4,4,...]

It's generally expected that non-integer square roots are normal, and yet they
_always_ have repeating CFs.

All that combines to mean that I would be surprised if _e_ turns out not to be
normal.

~~~
irremediable
Yeah, great counterexample. I was thinking, "but continued fractions... that
rings a bell..." \-- couldn't figure out what seemed off about it, though.

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Jabbles
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.

[1]
[http://www.numberworld.org/digits/E/](http://www.numberworld.org/digits/E/)

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LeifCarrotson
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...

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sirsar
> _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.

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catpolice
"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.

~~~
arsenide
Here's a (slighty) less hand-wavey proof.

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.

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placebo
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?

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amelius
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!

~~~
Houshalter
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.

~~~
brianpgordon
Some sequences of bits are copyrighted, or otherwise illegal to distribute:

[https://en.wikipedia.org/wiki/Illegal_number](https://en.wikipedia.org/wiki/Illegal_number)

[https://en.wikipedia.org/wiki/Illegal_prime](https://en.wikipedia.org/wiki/Illegal_prime)

~~~
Houshalter
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.

