
The Erdos discrepancy problem - robinhouston
http://arxiv.org/abs/1509.05363
======
kittenfluff
My interpretation in layman's terms of what this paper has proved -

Take an infinite sequence of +1 and -1, for example

    
    
      1, 1, -1, 1, -1, -1, -1, 1, 1, -1, ...
    

You get an evenly spaced subsequence by starting on the n'th element, and
picking every n'th element after that (and stopping after finitely many
terms). So for example, we could pick every 2nd element of this sequence and
get

    
    
      1                (stopping after 1 term)
      1, 1             (stopping after 2 terms)
      1, 1, -1         (stopping after 3 terms)
      1, 1, -1, 1      (stopping after 4 terms)
      1, 1, -1, 1, -1  (stopping after 5 terms)
    

The discrepancy of an evenly spaced subsequence is obtained by adding together
all the members of the subsequence and taking the absolute value. So the
discrepancies of the sequences above are 1, 2, 1, 2, and 1.

The challenge is to find an evenly-spaced subsequence with as large a
discrepancy as possible. For example, in a given sequence, can you find an
evenly-spaced subsequence with a discrepancy of 10? of 100? of a million?

The paper has (apparently) proved that for _any_ sequence, there is no upper
limit to the discrepancies of evenly spaced subsequences, i.e. no matter how
large the discrepancy of a subsequence you have found, there is always one
larger.

The amazing thing about this result, to me, is the fact that it holds for
_any_ sequence of +1 and -1. Even if you try to engineer a sequence whose
subsequences all have very small discrepancy, in some sense "there isn't
enough room". You are always doomed to come up with a sequence containing
evenly-spaced subsequences of arbitrarily large discrepancy.

~~~
jsprogrammer
It seems it would come down to an under-sampling problem. You should be able
to weave through the sequence any way you want, but if you weave through at a
frequency different (ie. lower) than the frequency needed to sample the small
discrepancy, you will begin to measure artifacts of the "original signal" (ie.
your small discrepancy sequences).

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myle
It looks like another success story for the polymath project. Maybe the way
research is done at least in some fields is time to change?

~~~
privong
> It looks like another success story for the polymath project. Maybe the way
> research is done at least in some fields is time to change?

[https://en.wikipedia.org/wiki/Polymath_Project](https://en.wikipedia.org/wiki/Polymath_Project)
for those who aren't aware of the polymath project.

Regarding your second sentence, could you elaborate? Research in many fields
is already highly collaborative (e.g., particle physics), so I'm not sure what
you are saying might/should change.

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rjmunro
Good explanation of the problem here:
[https://www.youtube.com/watch?v=pFHsrCNtJu4](https://www.youtube.com/watch?v=pFHsrCNtJu4)

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jacobolus
For anyone wondering what this problem is, see
[http://michaelnielsen.org/polymath1/index.php?title=The_Erdő...](http://michaelnielsen.org/polymath1/index.php?title=The_Erdős_discrepancy_problem)

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jballanc
Very interesting result, and it feels like this could be more important than
it might seem at first glance. If this result could be further generalized for
sampling any continuously integrable function, it could have interesting
implications for Maxwell's Demon.

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sampo
> f: N → {−1,+1} taking values in {−1,+1}

Is there a need to mention the {−1,+1} twice?

~~~
arsenide
No, but does it hurt? It emphasizes the main focus of the problem

~~~
re501
Well since it's so obviously redundant you start to wonder that maybe it isn't
redundant and maybe you're wrong and not getting something when in-fact you're
right and it is redundant. To me it more annoying than emphasizing.

When you want to emphasize you should make it clear that you're emphasizing.

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mrcactu5
it was a conjecture only a few days ago
[https://terrytao.wordpress.com/2015/09/11/the-erdos-
discrepa...](https://terrytao.wordpress.com/2015/09/11/the-erdos-discrepancy-
problem-via-the-elliott-conjecture/)

