

The Riemann Hypothesis - one of the outstanding unsolved problems in math - RiderOfGiraffes
http://benvitalis.wordpress.com/2011/02/08/the-riemann-hypothesis-a-journey-through-the-prime-num3ers/

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btilly
I can't ever read a claim that the Riemann Hypothesis is unsolved without
putting a small asterisk beside that claim. See
[http://www.lrb.co.uk/v26/n14/karl-sabbagh/the-strange-
case-o...](http://www.lrb.co.uk/v26/n14/karl-sabbagh/the-strange-case-of-
louis-de-branges) for why.

~~~
freyrs3
The proof has been checked and it contained errors. See:

[http://mathoverflow.net/questions/38049/what-exactly-has-
lou...](http://mathoverflow.net/questions/38049/what-exactly-has-louis-de-
branges-proved-about-the-riemann-hypothesis/38057#38057)

~~~
btilly
The cited paper that found holes is from before the proof discussed in the
article I pointed. The update points to a more recent paper that is behind a
paywall, so I can't find what it says.

~~~
thebooktocome
de Branges currently claims (as of September 2010) that he's fixed the problem
pointed out by Li. There's supposed to be a pre-print in the works, but I
haven't heard anything lately.

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gulbrandr
"We would like to see the prime numbers distributed as regularly as possible."

Why? I don't understand why mathematicians seek regularity in prime numbers so
much.

~~~
btilly
In this case it is statistical regularity. :-)

The primes act very much like they were a random set of numbers with n having
probability 1/log(n) of being prime. (Mathematicians always mean "natural log"
when they say log.) Except, of course, that this is obviously false. Because,
for example, there is only one even prime.

It is more accurate to say that the primes act very much like they were {2}
plus a random set of odd numbers with n having probability 2/log(n) of being
prime if it is odd. Except, of course, that this is obviously false. Because,
for example, there is only one prime that is divisible by 3.

One can continue refining this statement in the obvious way of special casing
more and more primes. Other than the obvious corrections that come from these
refinements, anything that you'd conjecture from this description is either
known to be true, or believed to be true. The number of primes below n. The
number of primes below n in an arithmetic sequence. The density of twin
primes. And so on, and so forth.

Now let's focus on the average of primes below n. (Mathematicians call that
number pi(n).) If the primes all were independently random, then you'd expect
that to be close to the sum, for i in 2 to n, of 1/log(i), which is
approximately the integral from 2 to n of 1/log(x). The usual name for that
integral is li(x). li(x) is approximately n/log(n). This estimate is known to
be true. That's called the prime number theorem.

One of many equivalent forms of the Riemann Hypothesis is that pi(n) - li(n)
is O(sqrt(n)). From the "random prime" hypothesis we can see why this is
reasonable. It turns out that for large n, (pi(n) - li(n))/sqrt(li(n)) should
follow a standard normal distribution. From which it is easy to show that the
Riemann Hypothesis should have 100% odds of being true.

~~~
btilly
When I woke up and thought about this problem again, I realized that I didn't
explain one detail. Why should (pi(n) - li(n))/sqrt(li(n)) follow a standard
normal distribution?

Here is why. pi(n) is the sum of a bunch of many independent random variables,
each of which looks like a flip of a biased coin. If you pick the right
version of the central limit theorem (see
<http://en.wikipedia.org/wiki/Lyapunov_condition> for the right version), it
is clearly going to be approximately normal.

Its average will be the sum of the averages, which is 1/log(2) + 1/(log(3) +
... + 1/log(n) which is going to be li(n) plus O(1). If n is large the li(n)
term dominates.

Its variance will be the sum of the variances of the individual random
variables. The variance of a coin with probability x of turning up 1 is
x(1-x). Therefore the sum of the variances is (1/log(2) - (1/log(2) __2) +
(1/log(3) - (1/log(3) __2) + ... + (1/log(n) - (1/log(n) __2) which is (modulo
a O(1) error) the integral from 2 to n of 1/log(x) - (1/log(x)) __2 which is
li(x) - O(n/log(n) __2), which is dominated by the li(x) term.

Therefore the average and variance of pi(n) are approximately li(x). So (pi(n)
- li(n))/sqrt(li(n)) should have approximately a standard normal distribution.

------
mckoss
And for the JS heads - a prime number sieve in JavaScript:

<http://wiki.pageforest.com/#js-patterns/prime-sieve>

~~~
harlowja
And for a classic.

<http://alpha61.com/primenumbershittingbear/>

