The Riemann Hypothesis, explained(medium.com)
medium.com
The Riemann Hypothesis, explained
https://medium.com/@JorgenVeisdal/the-riemann-hypothesis-explained-fa01c1f75d3f
27 comments
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.
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.
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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...
> 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.
E.g. "Riemann hypothesis likely remains unsolved despite claimed proof" https://www.newscientist.com/article/2180504-riemann-hypothe...
> 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.
Very nice, can I link to it? I'm building a resource for [approaches to the Riemann Hypothesis](https://pnphard.pro/blems/riemann/)
How can the last graph be correct? Shouldn't it look something like this: http://mathworld.wolfram.com/RiemannPrimeCountingFunction.ht...
Because you define J(x) as https://cdn-images-1.medium.com/max/2000/1*EjlcuLa3Z_wnmYREs... so it shouldn't really approximate pi(x) that well.
Because you define J(x) as https://cdn-images-1.medium.com/max/2000/1*EjlcuLa3Z_wnmYREs... so it shouldn't really approximate pi(x) that well.
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.
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.
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.
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.
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.
> [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,
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.
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)) ...
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.
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
--- Zeta In Terms Of Vectors for RH
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
--- Zeta In Terms Of Vectors for RH
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.
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...
Anyone with a little more in-depth knowledge able to comment on this supposed proof?
Anyone with a little more in-depth knowledge able to comment on this supposed proof?
It doesn't really have a proof of anything in there - just some empirical observations and heavy hand-waving.
Youtube video on this subject that I found to be very interesting, even for a layman person like me: https://youtu.be/d6c6uIyieoo
A basic video on Prime gaps and related problems discussion intro video by Terry Tao https://youtu.be/pp06oGD4m00
Whew, theres a lot there. Thanks!
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-...
[2] https://mourafiq.com/2015/09/09/extensions-of-the-factorial-...
[3] https://mourafiq.com/2015/10/09/extensions-of-the-factorial-...