5 rev 2020.11.11.37991, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. 3 site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. To learn more, see our tips on writing great answers. Sign up, Existing user? After Sir Michael Atiyah’s presentation of a claimed proof of the Riemann Hypothesis earlier this week at the Heidelberg Laureate Forum, we’ve shared some of the immediate discussion in the aftermath, and now here’s a round-up of … "This [criterion] is just one of many equivalent formulations of the Riemann hypothesis," Thompson said. A standard method is proposed to prove strictly that the Riemann Zeta function equation has no non-trivial zeros. Here's the Riemann hypothesis again: The real part of every non-trivial zero of the Riemann zeta function is 1/2. How do I match both capital and small letters using regex in bash? But the truth is that nobody knows for sure. First, we wish to extend to the strip . You will receive a verification email shortly. Its validity has become one of the most famous open questions in mathematics. clearly $F_K(x)$ is well-defined for any $x > 1$ (it is a finite product), and $\lim_{K \to \infty} F_K(x)$ exists too because the logarithm of the infinite product is $-\sum_p \ln(1-p^{-x})$ which is absolutely convergent since $\ln(1-p^{-x}) \sim -p^{-x}$ and that $\sum_p p^{-x} < \sum_{n=1}^\infty n^{-x}$ which is (absolutely) convergent. That little claim might not sound very important. This was the method by which Euler originally discovered the formula. $$\prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}} \prod_{p \in P}^{p \le A}p^{-v_ps} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}} (\prod_{p \in P}^{p \le A}p^{-v_p})^{-s}$$. Similarly for the right-hand side the infinite coproduct of reals greater than one does not guarantee divergence, e.g., Instead, the denominator may be written in terms of the primorial numerator so that divergence is clear. "So often in number theory, what ends up happening is if you assume the Riemann hypothesis [is true], you're then able to prove all kinds of other results," Lola Thompson, a number theorist at Oberlin College in Ohio, who wasn't involved in this latest research, said. /Filter /FlateDecode finally, $\zeta(x)- F_K(x) = \sum_{n=1}^\infty |a_n(K)-1| n^{-x} > 0$, it is absolutely convergent, it is decreasing in $K$, and it clearly $\to 0$ when $K \to \infty$ since every term $\to 0$. All the other zeros are "non-trivial" and complex numbers. The fact that this trick works, she said, convinces many mathematicians that the Riemann hypothesis must be true. valid for (Guillera and Sondow 2005). \zeta(x)(\frac{1}{2^x})=\frac{1}{2^x}+\frac{1}{4^x}+... \\ We proved a large chunk of this criterion.". >> $$\lim_{K \to \infty} F_K(x) = \prod_p \frac{1}{1-p^{-x}} = \zeta(x) \qquad\qquad (\forall \ x > 1)$$. /Length 1950 p The positive natural numbers are given by, Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ( ��j,I�����ۏ^�XA_f��"����Xa�m��f#4y�)��T/Trƫ�3+�$(���1��Hڦ"��� That's a pretty abstract mathematical statement, having to do with what numbers you can put into a particular mathematical function to make that function equal zero. In other words, there are a lot of other ideas that, like this criterion, would prove that the Riemann hypothesis is true if they themselves were proven. A standard method is proposed to prove strictly that the Riemann Zeta function equation has no non-trivial zeros. I'm wondered wether you can prove this gap with the same kind of argumentation or wether you will need a new base for the complete proof. @daniels_pa : I wrote it, by induction. How do I show$\prod_{p}(1-p^2)^{-1}= \frac{\pi ^2}{6}$? However, Proof. Here's the Biden-Harris plan to beat COVID-19, Woman sheds coronavirus for 70 days without symptoms. But there's another kind of number: imaginary numbers. The Riemann zeta function involves what mathematicians call "complex numbers." "What we have done in our paper," said Ken Ono, a number theorist at Emory University and co-author of the new proof, "is we revisited a very technical criterion which is equivalent to the Riemann hypothesis … and we proved a large part of it. New York, Use MathJax to format equations. Let's break that down according to how Thompson and Ono explained it. , the sieved right-hand side approaches 1, which follows immediately from the convergence of the Dirichlet series for There is no need to be rude :) Sometimes we are all tired and short with time. s ) Why does the same UTM northing give different values when converted to latitude? Stay up to date on the coronavirus outbreak by signing up to our newsletter today. This convergence was proven by Euler! ζ(2)=112+122+132+⋯=A1+A2+A3+⋯=∑An=π26\zeta \left( 2 \right) =\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\dots ={ A }_{ 1 }+{ A }_{ 2 }+{ A }_{ 3 }+\dots =\sum { { A }_{ n } } =\frac { { \pi }^{ 2 } }{ 6 } ζ(2)=121​+221​+321​+⋯=A1​+A2​+A3​+⋯=∑An​=6π2​. (by the fundamental theorem of arithmetic). ) How do you cook more successfully in a different kitchen? Leonhard Euler proved the Euler product formula for the Riemann zeta function in his thesis Variae observationes circa series infinitas (Various Observations about Infinite Series), published by St Petersburg Academy in 1737. Would the Millennium Falcon have been carried along on the hyperspace jump if it stayed attached to the Star Destroyer? Those ellipses mean the series in the function keeps going on like that, forever. Why do you need this extra step? $$\prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks}$$, Product over a sum is the sum over the cartesian products of the products. However this proof isn't a 'rigorous proof' as my professor says. The Riemann hypothesis states that when the Riemann zeta function crosses zero (except for those zeros between -10 and 0), the real part of the complex number has to equal to 1/2. $$\lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-s}$$,$s$may be complex,$ s = u + it \$ Ask specific questions about the challenge or the steps in somebody's explanation. It can be seen that the right side is being sieved. In math, a function is a relationship between different mathematical quantities. For each natural number nnn and each real number kkk greater than 222 it is obviously true, that: There are no comments in this discussion. Explanations are more than just a solution — they should $$\sum_{v \in V } g(f(v)) = \sum_{w \in \{f(v) : v \in V \} } g(w)$$ He thus proved the equivalence of both formulas (4) and (3). If the proof turns up along this track, then that will likely mean Ono and his colleagues have developed an important underlying framework for solving the Riemann hypothesis. Infinite sums can have different values depending on order. A real number can be anything from minus 3, to zero, to 4.9234, pi, or 1 billion. Thank you for signing up to Live Science. The Riemann zeta function is one of the most Euler's important and fascinating functions in mathematics. Ono and his colleagues, in a paper published May 21 in the journal Proceedings of the Natural Academy of Sciences (PNAS), proved that in many, many cases, the criterion is true. explain the steps and thinking strategies that you used to obtain the solution. 2 And in doing so, they have reopened an old avenue that might eventually lead to an answer to the old question: Is the Riemann hypothesis correct? -D*ef̳�6gѣ���Y�D'#. Here's what it looks like.