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This is the first episode of the RH Saga. We embark on a journey into the world of L-functions, by introducing the Riemann Hypothesis and the dream of a new geometry over the "field with one element".The aim of RH Saga Season 1 is to map the landscape of L-functions, as a foundation for future in-depth exploration of some of the most immortal math problems of all time. This video is part of a PeakMath course. Join the journey at https://www.peakmath.org/---Chapters:00:00 - Introduction to the RH Saga01:50 - Introduction to Episode 1: The Dream03:17 - Chapter 1: Intro to F107:58 - Summary of Chapter 1 09:26 - Chapter 2: Recap of RH18:46 - Chapter 3: Proof of RH?21:40 - Summary of Chapter 3Links:1. "Numbers as Functions" - Yuri Maninhttps://arxiv.org/pdf/1312.5160.pdf2. "Riemann's Hypothesis" - Brian Conreyhttps://tinyurl.com/conrey903. SageMathCellhttps://sagecell.sagemath.org/4. SageMathhttps://www.sagemath.org/5. SageMathCell permalink for Riemann spectrum codehttps://tinyurl.com/e1-spectrum 6. SageMathCell permalink for cosine waves codehttps://tinyurl.com/e1-cosineErrata:Around 6:44: Just to clarify - the "fractional Gamma value" \Gamma(p/q) may not be a period in itself, but \Gamma(p/q)^q is certainly a period.
This is the third episode of the RH Saga. We continue our journey by considering the zeros of L-functions beyond the Riemann zeta function. The location of these zeros is the subject of the generalized Riemann hypothesis. The story of these zeros also give a first indication of the general relationship between prime numbers and L-functions.The overall aim of RH Saga Season 1 is to map the landscape of L-functions, as a foundation for future in-depth exploration of some of the most immortal math problems of all time. Chapters:00:00 - Intro02:00 - Review of examples05:04 - Analytic continuation12:39 - Zeros in the critical strip16:44 - Cosine waves21:35 - Final remarks---Links:1. LMFDBhttps://www.lmfdb.org/2. "The Music of the Primes" on Amazon UK:https://www.amazon.co.uk/Music-Primes...---Errata:The integral written at approx. 11:00 has a slight mistake. Instead of the number 2, the variable s should be used inside the sine function in the integrand. Correction displayed on screen at 11:20.---
Riemann HypothesisThe prime number theorem determines the average distribution of the primes. The Riemann hypothesis tells us about the deviation from the average. Formulated in Riemann?s 1859 paper, it asserts that all the ?non-obvious? zeros of the zeta function are complex numbers with real part 1/2.Some numbers have the special property that they cannot be expressed as the product of two smaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called prime numbers, and they play an important role, both in pure mathematics and its applications. The distribution of such prime numbers among all natural numbers does not follow any regular pattern. However, the German mathematician G.F.B. Riemann (1826 ? 1866) observed that the frequency of prime numbers is very closely related to the behavior of an elaborate function ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ? called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation ζ(s) = 0 lie on a certain vertical straight line.This has been checked for the first 10,000,000,000,000 solutions. A proof that it is true for every interesting solution would shed light on many of the mysteries surrounding the distribution of prime numbers.
All this leads to several basic questions.Is there a theory in the global case, playing the same role as cohomology doesfor Zeta functions of varieties over a field of positive characteristic? Is there ananalogue of a Frobenius automorphism in the classical case? Is there a generalindex theorem by which one can prove the classical Riemann hypothesis? We arehere in the realm of conjectures and speculation. In the adelic setting propoundedby Tate and Weil, the papers [Conn], [Den], [Hara] offer glimpses of a possible setupfor these basic problems.On the other hand, there are L-functions, such as those attached to Maasswaveforms, which do not seem to originate from geometry and for which we stillexpect a Riemann hypothesis to be valid. For them, we do not have algebraic andgeometric models to guide our thinking, and entirely new ideas may be needed tostudy these intriguing objects.Problems of the Millennium: the Riemann HypothesisE. Bombieri
https://www.claymath.org/millennium-problems/rules/Rules for the Millennium Prize ProblemsThe revised rules for the Millennium Prize Problems were adopted by the Board of Directors of the Clay Mathematics Institute on 26 September, 2018. Please read this document carefully before contacting CMI about a proposed solution. In particular, please note that:CMI does not accept direct submission of proposed solutions.The document is a complete statement of the rules and procedures: CMI will not offer any futher guidance or advice.Before CMI will consider a proposed solution, all three of the following conditions must be satisfied: (i) the proposed solution must be published in a Qualifying Outlet (see ?6), and (ii) at least two years must have passed since publication, and (iii) the proposed solution must have received general acceptance in the global mathematics community
Here are some interesting results which might be useful in solving the problem.Quotehttps://en.wikipedia.org/wiki/Riemann_zeta_function#Riemann's_functional_equationThis zeta function satisfies the functional equationwhere Γ(s) is the gamma function. This is an equality of meromorphic functions valid on the whole complex plane. The equation relates values of the Riemann zeta function at the points s and 1 − s, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer s = −2n, known as the trivial zeros of ζ(s). When s is an even positive integer, the product sin(πs/2)Γ(1 − s) on the right is non-zero because Γ(1 − s) has a simple pole, which cancels the simple zero of the sine factor.Quotehttps://en.wikipedia.org/wiki/Riemann_zeta_function#Other_resultsThe fact thatfor all complex s ≠ 1 implies that the zeros of the Riemann zeta function are symmetric about the real axis. Combining this symmetry with the functional equation, furthermore, one sees that the non-trivial zeros are symmetric about the critical line Re(s) = ½It is also known that no zeros lie on a line with real part 1.Quotehttps://en.wikipedia.org/wiki/Riemann_hypothesis#Zeros_on_the_critical_lineHardy (1914) and Hardy & Littlewood (1921) showed there are infinitely many zeros on the critical line, by considering moments of certain functions related to the zeta function. Selberg (1942) proved that at least a (small) positive proportion of zeros lie on the line. Levinson (1974) improved this to one-third of the zeros by relating the zeros of the zeta function to those of its derivative, and Conrey (1989) improved this further to two-fifths. In 2020, this estimate was extended to five-twelfths by Pratt, Robles, Zaharescu and Zeindler[22] by considering extended mollifiers that can accommodate higher order derivatives of the zeta function and their associated Kloosterman sums.Quotehttps://mathworld.wolfram.com/RiemannHypothesis.html It is known that the zeros are symmetrically placed about the line I(s)=0. This follows from the fact that, for all complex numbers s,1. s and the complex conjugate s* are symmetrically placed about this line.2. From the definition (1), the Riemann zeta function satisfies zeta(s*)=zeta(s)*, so that if s is a zero, so is s*, since then zeta(s*)=zeta(s)*=0*=0.It is also known that the nontrivial zeros are symmetrically placed about the critical line R(s)=1/2, a result which follows from the functional equation and the symmetry about the line I(s)=0. For if s is a nontrivial zero, then 1-s is also a zero (by the functional equation), and then 1-s* is another zero. But s and 1-s* are symmetrically placed about the line R(s)=1/2, since 1-(x+iy)*=(1-x)+iy, and if x=1/2+x', then 1-x=1/2-x'.From above results, it can be inferred that for any point in the critical strip, ζ(s*)=ζ(1-s) if and only if s is in critical line, where s*=complex conjugate of s.
https://en.wikipedia.org/wiki/Riemann_zeta_function#Riemann's_functional_equationThis zeta function satisfies the functional equationwhere Γ(s) is the gamma function. This is an equality of meromorphic functions valid on the whole complex plane. The equation relates values of the Riemann zeta function at the points s and 1 − s, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer s = −2n, known as the trivial zeros of ζ(s). When s is an even positive integer, the product sin(πs/2)Γ(1 − s) on the right is non-zero because Γ(1 − s) has a simple pole, which cancels the simple zero of the sine factor.
https://en.wikipedia.org/wiki/Riemann_zeta_function#Other_resultsThe fact thatfor all complex s ≠ 1 implies that the zeros of the Riemann zeta function are symmetric about the real axis. Combining this symmetry with the functional equation, furthermore, one sees that the non-trivial zeros are symmetric about the critical line Re(s) = ½It is also known that no zeros lie on a line with real part 1.
https://en.wikipedia.org/wiki/Riemann_hypothesis#Zeros_on_the_critical_lineHardy (1914) and Hardy & Littlewood (1921) showed there are infinitely many zeros on the critical line, by considering moments of certain functions related to the zeta function. Selberg (1942) proved that at least a (small) positive proportion of zeros lie on the line. Levinson (1974) improved this to one-third of the zeros by relating the zeros of the zeta function to those of its derivative, and Conrey (1989) improved this further to two-fifths. In 2020, this estimate was extended to five-twelfths by Pratt, Robles, Zaharescu and Zeindler[22] by considering extended mollifiers that can accommodate higher order derivatives of the zeta function and their associated Kloosterman sums.
https://mathworld.wolfram.com/RiemannHypothesis.html It is known that the zeros are symmetrically placed about the line I(s)=0. This follows from the fact that, for all complex numbers s,1. s and the complex conjugate s* are symmetrically placed about this line.2. From the definition (1), the Riemann zeta function satisfies zeta(s*)=zeta(s)*, so that if s is a zero, so is s*, since then zeta(s*)=zeta(s)*=0*=0.It is also known that the nontrivial zeros are symmetrically placed about the critical line R(s)=1/2, a result which follows from the functional equation and the symmetry about the line I(s)=0. For if s is a nontrivial zero, then 1-s is also a zero (by the functional equation), and then 1-s* is another zero. But s and 1-s* are symmetrically placed about the line R(s)=1/2, since 1-(x+iy)*=(1-x)+iy, and if x=1/2+x', then 1-x=1/2-x'.
For every nontrivial zero of Zeta function, ζ(s) =0, Re(Y(s)) = - ~ (negative infinity)Im(Y(s) = undefined (due to switching between two different, discontinuous values)The plot of Y function suggests that those conditions can only be satisfied where Re(s) = 1/2, as Riemann predicted.
Quote from: hamdani yusuf on 01/07/2022 05:59:51At this point, it should be obvious that direct attack on the problem is impossible.But it doesn't stop us from trying anyway. We can start from the functional equation.Quote from: hamdani yusuf on 29/06/2022 17:17:50Here are some interesting results which might be useful in solving the problem.Quotehttps://en.wikipedia.org/wiki/Riemann_zeta_function#Riemann's_functional_equationThis zeta function satisfies the functional equationwhere Γ(s) is the gamma function. This is an equality of meromorphic functions valid on the whole complex plane. The equation relates values of the Riemann zeta function at the points s and 1 − s, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer s = −2n, known as the trivial zeros of ζ(s). When s is an even positive integer, the product sin(πs/2)Γ(1 − s) on the right is non-zero because Γ(1 − s) has a simple pole, which cancels the simple zero of the sine factor.When Riemann zeta function produces 0 result, ζ(s) = 0, at least one of these terms is 01) 2s = 0 → s = -∞2) πs-1 = 0 → s = -∞3) sin(πs/2).Γ(1-s) = 0 → s ∈ {negative even numbers}4) ζ(1-s) = 0 = ζ(s) Point #3 gives trivial zeros, while point #4 gives non-trivial zeros. when 1-s=s → 1=2s → s=1/2 But ζ(1/2) <> 0
At this point, it should be obvious that direct attack on the problem is impossible.
Here are some interesting results which might be useful in solving the problem.Quotehttps://en.wikipedia.org/wiki/Riemann_zeta_function#Riemann's_functional_equationThis zeta function satisfies the functional equationwhere Γ(s) is the gamma function. This is an equality of meromorphic functions valid on the whole complex plane. The equation relates values of the Riemann zeta function at the points s and 1 − s, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer s = −2n, known as the trivial zeros of ζ(s). When s is an even positive integer, the product sin(πs/2)Γ(1 − s) on the right is non-zero because Γ(1 − s) has a simple pole, which cancels the simple zero of the sine factor.
Quote from: hamdani yusuf on 29/06/2022 17:17:50From above results, it can be inferred that for any point in the critical strip,ζ(s*)=ζ(1-s) if and only if s is in critical line,where s*=complex conjugate of s.Quote from: hamdani yusuf on 13/07/2022 03:51:564) ζ(1-s) = 0 = ζ(s)Quote from: hamdani yusuf on 29/06/2022 17:17:50The fact thatζ(s) = (ζ(s*))*But we only care about the case where ζ(s) = 0. 0* = 0 → ζ(s) = ζ(s*) = ζ(1-s) = 0when s* = 1-s → s+s* = 1 → Re(s)+Im(s).i + Re(s)-Im(s).i = 1 → 2.Re(s) = 1 → Re(s) = 1/2
From above results, it can be inferred that for any point in the critical strip,ζ(s*)=ζ(1-s) if and only if s is in critical line,where s*=complex conjugate of s.
4) ζ(1-s) = 0 = ζ(s)
The fact thatζ(s) = (ζ(s*))*
To be clear, I don't dispute the validity that all trivial zeros are on the real line.
Y function convincingly shows the symmetry of Riemann's Zeta function around the critical line. It also shows the increasing stability for higher imaginary part. But somehow some of us aren't convinced that all non-trivial zeros of Riemann's Zeta function are on the critical line.
We get a nice full wave when the imaginary part is exactly 2*pi
Y function guarantees that at the critical line, its output is always negative infinity, because on that line, s=1-s*. Consequently,Y(s) = Ln (ζ(s)-ζ(s)) = Ln(0) = -~To prove Riemann's hypothesis, a further step is required. It needs to show that ζ(s)<>ζ(1-s*) when s<>1-s*.This is where the backslash function can help.B(s) = Ln (ζ(s) / ζ(1-s*)) = Ln(ζ(s)) - Ln(ζ(1-s*))