
The Riemann Hypothesis, explained - jorgevei
https://medium.com/@JorgenVeisdal/the-riemann-hypothesis-explained-fa01c1f75d3f
======
mmq
Very nice article.

I always wanted to extend my tutorial[1][2][3] with a 4th part, to make the
connection between Euler Gamma function and Riemann Zeta Function.

I will link to your article, it's really well written.

[1] [https://mourafiq.com/2015/08/30/extensions-of-the-
factorial-...](https://mourafiq.com/2015/08/30/extensions-of-the-factorial-
testing-the-convergenge-of-series.html)

[2] [https://mourafiq.com/2015/09/09/extensions-of-the-
factorial-...](https://mourafiq.com/2015/09/09/extensions-of-the-factorial-
stirling-approximation.html)

[3] [https://mourafiq.com/2015/10/09/extensions-of-the-
factorial-...](https://mourafiq.com/2015/10/09/extensions-of-the-factorial-
euler-gamma-function.html)

------
yaks_hairbrush
This is a really nice explanation -- it flowed really well, and I liked the
sieve approach to proving the product-over-primes formula (the other standard
way relies on a clever trick, and being clever is fun-but-not-good-for-
exposition).

I did get hung up on the complex graph showing the first few non-trivial
zeros. It's a very busy graph! I ended up realizing that you graphed the
contours Re(zeta(s))=0 and Im(zeta(s))=0.

------
gdne
If you like this topic and didn’t read all the way to the bottom, the author’s
recommendation of “Prime Obsession” by John Derbyshire is spot on. It’s a
fantastic book.

~~~
TheCondor
That is a great read, I was also going to recommend it.

------
carapace
From the title I expected something about Michael Atiyah.

E.g. "Riemann hypothesis likely remains unsolved despite claimed proof"
[https://www.newscientist.com/article/2180504-riemann-
hypothe...](https://www.newscientist.com/article/2180504-riemann-hypothesis-
likely-remains-unsolved-despite-claimed-proof)

> 24 September 2018

> Michael Atiyah claims to have found a proof for the Riemann hypothesis

> One of the most famous unsolved problems in mathematics likely remains
> unsolved. At a hotly-anticipated talk at the Heidelberg Laureate Forum
> today, retired mathematician Michael Atiyah delivered what he claimed was a
> proof of the Riemann hypothesis, a challenge that has eluded his peers for
> nearly 160 years.

------
rfurmani
Very nice, can I link to it? I'm building a resource for [approaches to the
Riemann
Hypothesis]([https://pnphard.pro/blems/riemann/](https://pnphard.pro/blems/riemann/))

~~~
dvh
Article cannot be read if your read more than 3 articles on medium

------
petters
How can the last graph be correct? Shouldn't it look something like this:
[http://mathworld.wolfram.com/RiemannPrimeCountingFunction.ht...](http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html)

Because you define J(x) as [https://cdn-
images-1.medium.com/max/2000/1*EjlcuLa3Z_wnmYREs...](https://cdn-
images-1.medium.com/max/2000/1*EjlcuLa3Z_wnmYREs-SnJQ.png) so it shouldn't
really approximate pi(x) that well.

~~~
gjm11
I think they're mislabelled and misdescribed a bit, and what those graphs
actually show is what you get if you (1) compute an approximation to J using
only a limited number of Riemann zeros, and then (2) use the inversion formula
to compute what pi(x) would be if #1 had given you the exact value of J. So
the y-axis shouldn't be labelled "J(x)", it should be labelled something more
like "pi(x)".

Caution: I haven't actually done the computation and checked.

[EDITED to add:] Now I have, or at least one of them, for the "first 35 roots"
one. I took "first 35 roots" to mean "first 35 with positive imaginary part,
plus their complex conjugates". I got a graph that was much wigglier, and a
much closer match to pi(x) than the one in the article. So then I thought
maybe it was meant to be 35 roots in total -- though you really do want to
take those conjugates in pairs, so the odd number is strange. Anyway, I tried
with the first 17 pairs: still much too wiggly. With the first _five_ pairs of
zeros, I get a good (but not perfect) match for the graph in the article.

------
graycat
It is easy to take prime number 2 (from the famous treatise _A Short Table of
Even Primes_ ) and "divide it into" 6\. And it is easy to take 6 and factor it
into primes 2 and 3.

Where the OP wrote

> Those numbers you can’t divide into other numbers, except when you divide
> them by themselves or 1?

would have read better with

"Those numbers you can't factor into ..."

I.e., it is easy to take 2 and "divide it into 6" but can't factor 2 into a
product of other numbers except itself and 1.

~~~
asdf1234tx
It's the very same thing that struck me right away. The English is off, even
though I understood perfectly well what the author intended to communicate.
Then it clicked, he's Norwegian, and most assuredly English is a second
language.

------
omazurov
_> [The Euler Product Formula] states that the sum of the zeta function is
equal to the product of the reciprocal of one minus the reciprocal of primes
to the power s. This astonishing connection laid the foundation for modern
prime number theory..._

Here is the Euler formula for middle school.

We start with an infinite product of power series for all primes:

(1 + 2 + 2^2 + ... + 2^k + ...) *

(1 + 3 + 3^2 + ... + 3^k + ...) * ...

(1 + p + p^2 + ... + p^k + ...) * ...

Let's open the parentheses in an orderly manner without running into infinity
too prematurely (i.e. taking only a finite number of non-unit terms in each
product):

1 + 2 + 3 + 2^2 + 5 + 2 * 3 + 7 + 2^3 + 3^2 + 2 * 5 + 11 + 2^2 * 3 + ...

Now, if we apply any completely multiplicative function f(ab) = f(a)f(b) to
the terms we get

(1 + f(2) + f(2^2) + ... + f(2^k) + ...) *

(1 + f(3) + f(3^2) + ... + f(3^k) + ...) * ...

(1 + f(p) + f(p^2) + ... + f(p^k) + ...) * ... =

1 + f(2) + f(3) + f(4) + f(5) + f(6) + ...

When f(n) = n^(-s) we get the zeta function on the right and an infinite
product of converging sums (s > 1) on the left.

The converging sums S = 1 + p^(-s) + p^(-2s) + ... are of course 1/(1 -
p^(-s)) as S = 1 + S * p^(-s).

Voilà,

    
    
      1 + 2^(-s) + 3^(-s) + ... = 1/(1 - 2^(-s)) * 1/(1 - 3^(-s)) * ... * 1/(1 - p^(-s)) ...
    

In case you think the Möbius formula is totally beyond the middle school
curriculum ("primitive roots" and stuff):

SUM m(n) * n^(-s) = (1 - 2^(-s))(1 - 3^(-s))(1 - 5^(-s))...(1 - p^(-s))...

(just open the parentheses on the right, that's what middle-schoolers do).

And what happens if we multiply our power series S by 1 - p^(-s)? They cancel
each other, so the whole product is 1.

The "astonishing connection" of the Euler formula turns out to be the
fundamental theorem of arithmetic in disguise.

------
zetaITOvectors
It may boil down to the way we view the problem: whether to focus on the zeta
values or the structure of the function.

The true mystery can be the viewed from a vector angle, i.e. view the zeta
value as $\zeta(c) = V(c) \bullet V(0)$, where $V(x) = (1^{-x}, 2^{-x},
3^{-x}, ...)$ for x a complex variable.

A more thorough treatment is @ [https://read.barnesandnoble.com/book/egg-n-
our-face-3#1](https://read.barnesandnoble.com/book/egg-n-our-face-3#1)

\--- Zeta In Terms Of Vectors for RH

------
mayankkaizen
Book 'Prime Obsession' by John Derbyshire is also very good detailing the
history and basic math behind Riemann Hypothesis. It is very accessible, well
written and captivating.

------
aldoushuxley001
I remember hearing of a purported proof of the Riemann Hypothesis by George
Spencer Brown: [http://empslocal.ex.ac.uk/people/staff/mrwatkin//zeta/GSB-
RH...](http://empslocal.ex.ac.uk/people/staff/mrwatkin//zeta/GSB-RHproof.pdf)

Anyone with a little more in-depth knowledge able to comment on this supposed
proof?

~~~
vlasev
It doesn't really have a proof of anything in there - just some empirical
observations and heavy hand-waving.

------
rakic
[https://i.kym-
cdn.com/photos/images/original/000/572/078/d6d...](https://i.kym-
cdn.com/photos/images/original/000/572/078/d6d.jpg)

------
juliend2
Youtube video on this subject that I found to be very interesting, even for a
layman person like me:
[https://youtu.be/d6c6uIyieoo](https://youtu.be/d6c6uIyieoo)

------
thepoet
A basic video on Prime gaps and related problems discussion intro video by
Terry Tao [https://youtu.be/pp06oGD4m00](https://youtu.be/pp06oGD4m00)

------
asdffdsa321
Whew, theres a lot there. Thanks!

------
dest
Really nice article. Thanks for sharing.

~~~
jorgevei
Thanks!

------
Yhippa
Paywall'd.

~~~
TrinaryWorksToo
Incognito it

