Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Which kinds of fields of mathematics do I have to know about in order to understand the Riemann hypothesis millenium prize problem?

share|improve this question
2  
Not an answer, but a reference: Riemann's Zeta Function by H. M. Edwards –  draks ... Apr 22 '12 at 22:28
10  
If you mean, 'in order to understand the terminology involved', then not too much. Basically the only definition needed is "the analytic continuation of a function". If you actually want to understand the theory behind it, the background is enormous... we're talking about a significant amount of number theory, algebra, analysis, geometry... and I'm not sure anyone actually fully understands the problem (hence it's unsolved!) –  Alastair Litterick Apr 22 '12 at 22:29
6  
Here is a rather simple statement equivalent to Riemann hypothesis. Let $\gamma$ be the Euler–Mascheroni constant, about 0.57, $\log$ be the natural logarithm, and $\sigma(n)$ be the sum of divisors of $n$ (for example, $\sigma(10)=1+2+5+10=18$). The Riemann hypothesis is equivalent to the statement that for any natural number $n\geq 5041$, $\sigma(n) < e^{\gamma} n \log (\log n)$. Source. However, this doesn't explain why is it important, what does it really mean etc. –  sdcvvc Apr 22 '12 at 23:03
2  
Although you did not ask for reference, I am putting 3 links of introductory books: 1) Stalking the Riemann Hypothesis; 2) Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics; 3) The Riemann Hypothesis: The Greatest Unsolved Problem in Mathematics. All these links will further point to further books. –  Sniper Clown Apr 22 '12 at 23:28
1  
Also The Riemann Hypothesis: A Resource For The Afficionado and Virtuoso Alike includes some modern surveys as well as original papers. –  Dan Brumleve Apr 22 '12 at 23:47
add comment

4 Answers

up vote 16 down vote accepted

Just to understand the$^\dagger$ statement of the problem, you would have to be familiar with complex analysis and analytic number theory. The $\zeta$ function itself is an analytic object from number theory and to understand its significance (just on the surface!) you would have to study it in these realms. Of course it is also a function on $\Bbb C$ after analytic continuation - attained using a functional equation - with a simple pole at $1$, and understanding what this means and how to manipulate the function deftly will mean studying complex analysis.

$^\dagger$I refer to the statement that $\zeta(s)$ has all nontrivial zeros on the critical line. There are actually a lot of equivalent statements that require very little knowledge of complex analysis (you'll still need to pick up a few definitions of arithmetic functions from analytic NT for many of them, these aren't too hard). You can find a lot of equivalences listed here for example.

Beyond that, to understand the modern approaches to RH and related or generalized conjectures and all of the theory there is surrounding this creature, you must go much further in algebraic number theory at the very least, and travel to many other worlds like modular forms, differential geometry, quantum theory and random matrices, etc. - basically at least a basic knowledge of most advanced subjects in analysis, algebra and geometry, and then especially deeply in pertinent areas.

share|improve this answer
    
There are equivalent statements which require no knowledge of complex analysis to understand. For example, "the Mertens function is $O(n^{\frac{1}{2} + \epsilon})$ for all $\epsilon \gt 0$". Perhaps even simpler is the same statement about the summatory Liouville function (which avoids making the distinction between squareful and squarefree numbers). –  Dan Brumleve Apr 22 '12 at 23:03
    
@DanBrumleve: That's a good point, I should have said that. There are numerous equivalent statements! By "the" statement of course I mean the one about $\zeta(s)$. –  anon Apr 22 '12 at 23:08
    
As with most things in math, you won't even begin to understand the hypothesis itself until you understand most of the equivalent statements and why they are equivalent. –  Neal Apr 22 '12 at 23:21
    
I quite like the recasting of the hypothesis as an eigenvalue problem myself... –  J. M. Apr 23 '12 at 0:52
add comment

Definitely, the number theory. You can find more connections to other fields in "The Millenium Problems" by Keith Devlin.

share|improve this answer
add comment

The Riemann Hypothesis is just a conjecture about the zeros of a function. Understanding this is simple.

However, understanding why this is important and what it means to number theory is far from simple.

share|improve this answer
5  
Understanding the meaning of zeros of a function may be simple, but understanding the definition of $\zeta$ outside of the region $Re(s) \gt 1$ has some prerequisites. –  Dan Brumleve Apr 22 '12 at 23:37
    
Yes but understanding what the RH is saying is a matter of understanding that "We have a function andwe think all of it's zeros line on a specific line". You do not really have to know what the function is to understand the content of the hypothesis. As I say though, understanding the actual function and why it is important etc is a different matter! –  fretty Apr 23 '12 at 10:16
add comment

I used to try to explain the complex analysis part. I have given up on that. See http://en.wikipedia.org/wiki/Prime-counting_function for details...

The short version is that the prime counting function $\pi(x)$ is approximated pretty well by $$ \frac{x}{\log x}. $$ It is known that a better approximation is given by $$ \mbox{li} \, (x). $$

The Riemann hypothesis is equivalent to a specific form of the statement that $ \mbox{li} \, (x) $ is a really good approximation, see http://en.wikipedia.org/wiki/Prime-counting_function#The_Riemann_hypothesis and TABLE

share|improve this answer
    
Another soundbite at the same level of precision: "parity is random". –  Dan Brumleve Apr 22 '12 at 23:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.