Ken Ono - The Riemann Hypothesis (March 14, 2018)

More details: www.simonsfoun...

[@bardamu9662]

Looks like the RH is the key to many instrumental locked mathematical problems. Truly fascinating and exposed by a brilliant lecturer. Thanks

[@johnstfleur3987]

THANK YOU PROFESSOR ONO.

[@vincentvandernoort1346]

Very interesting! A few times Ono refers forward to a second lecture he will give on 'Friday''. Is this second lecture also videotaped?

[@vincentvandernoort1346]

Ok, found it here: www.simonsfoundation.org/event/conference-on-number-theory-geometry-moonshine-and-strings-ii/ (among with other lectures from the conference). I don't know whether or not it is also here in KZbin

[@koenth2359]

​@@vincentvandernoort1346 Thanks. That was on march 16. ( Did he honour sqrt(10)-day in that video?)

[@robert-skibelo]

Really interesting lecture. Since precision is important in mathematics maybe it's worth mentioning that the slide title "Infinitude of primes aprés Euler" misspells "après" and it's unlikely that Euclid, an ancient Greek, wore medieval dress.

[@yosoylibre]

I hypothesize that since infinite series can be indeterminate like 1+2+... then it depends on how we treat them, that some results might seem contradictory. So if we order these integers like in some tests, it's very possible that we get this infinite sum to get a finite result that doesn't necessarily have to be what our intuition expects, like -1/12. So I agree with Srinivāsa Aiyangār Rāmānujan.

[@explorerc607]

Hi, Prof Ono, I am confused by the Euler product formula you presented. As far as I'm concerned, only s is more than 1, the formula makes sense. So beginning from 17:21 in the video, your proof said "if π (n)=k, then Euler's product gives ...." I think it is wrong becase s equals 1.

[@MichelJosephCardin]

What would be all the things that I would need to study and learn about; in order for me to understand everything that there is to understand; where it would then put me at that same place as everyone else that are trying to solve these mathematical/physical problems. I do understand that there is much to get through and where should I look to that there would be everything explained in ways that would be successful within every people's different schooling stages? Thank you.

[@carloingravalloba]

Yes, this one is rich of references 😮 and interesting. But having cited Jansen and Hardy however you not have talked about the works of Littlewood and Hardy to integrate your discussion about it ?

[@davidwilkie9551]

Mathematics for Mathematicians, to show how the question is assembled by default and reassembled using the approach of disproof, mechanical elimination of mismatched concepts.

[@cufflink44]

Though I have only a modest background in mathematics, I have discovered an error in Prof. Ono's lecture. It's Eratosthenes, not Erastothenes. 😄

[@MichelJosephCardin]

Oh my dear; I am so glad that I looked at it because I was gonna look for that error of yours and I have but grade 10 but with all 100's though but have zero ideas about any of the terminologies and was going to challenge myself at looking for that "mathematical numeric error" lol I would had been feeling quite stupid LOL Cheers. Maybe you found a dyslexic side of the Prof. LOL

[@brendawilliams8062]

Thankyou

[@ayadav4714]

I attempted this for two years. Proved two years are not enough to even understand it.

[@rns01111]

Fascinating

[@hoahuynh2734]

Euclid didn't specifically say the smallest number or prime was 2. His proof says DIFFERENTLY. A) More GENERAL: For any list of numbers, there are some number(s) which is(are) CO-PRIME(s) to the original list. B) More SPECIFIC: Any list of primes is INCOMPLETE. Examples: List {2, 3}; then 2*3 + 1 = 7; new prime not accounted yet. List {5, 7}; then 5*7 + 1 = 36 = 2*2*3*3; two new primes 2 and 3 are not accounted yet.

[@realimage3656]

私はリーマン予想の 1/2 という数値が発生する事に関して、 The Real number 0 and the imaginary number 0 are spatiotemporally offset by 1. という事が関連しているのではないだろうかと思っています。

[@wdobni]

somebody needs to come up with an integral that accurately predicts how many years it will be before the RH is proved....it took 300 years to prove fermat's last theorem...based on the summing the partial derivatives of the hermitian operators that motivate undergraduate mathematicians it is likely to be 9,300.447 years before RH is proved

[@hipotesisderiemannderivada5763]

You can find it zeta of function (1/2 + 14.13 ... i) = 0

[@tokajileo5928]

Polya was a Hungarian mathematitian an his name is mispelled. "ly' is one sound in Hungarian and it is spelled like "y" in "yellow"

[@vikraal6974]

Poya?

[@Maxyl56]

Is there a continuation of this video?

[@connorkearley7381]

i want it g

[@wallstreetoneil]

don't the negative even integers zeros come from the sin(pi*S/2) and the Gamma Function, being that you are minus'ing a negative, get evaluated at a positive Gamma(s) value?

[@powerdriller4124]

Reimann would have taken as a sufficient proof the 100 million zeros that computers have found on the x=0.5 line, and the exhaustive scannings in the 0

[@johnstfleur3987]

EXACT REASON.

[@diktakt1187]

41 23 Quadratic ramanuj

[@kenichimori8533]

We can then here be number.

[@kenichimori8533]

Riemann Surface is infinity integral 0∮0 pi0+1= 5/5/5/5/5/5 (Reintegral)

[@kenichimori8533]

Rieman hypothesis size is 0 bytes free.

[@kenichimori8533]

Index zero axiom = 04030201 = 8 + 02 - 01

[@sr71black]

Que pasa ahora con la fórmula de beimar

[@alexanderealley9992]

The answer to the Riemann Hypothesis is Infinity. Infinity times infinity equals infinity to the power of infinity. Infinity squared equals infinity to the power of infinity. If 2 is a prime then so is infinity. You are all welcome.

[@jerrari7323]

مهم

[@kenichimori8533]

Solution Rule is algebraic geometry.Pr

[@johnstfleur3987]

I AM THE PROOF.

[@johnstfleur3987]

THIS IS JESUS CHRIST.

[@johnstfleur3987]

8.

[@sunkhirous]

All roots of Zeta lie on zero line in form of S = 0+ i( 2pik -pi)/lnP^2^n , P is prime.

[@raphaelreichmannrolim25]

No, they don't... And the terms in Euler product give poles, not zeros. They are zeros of inverse Zeta. And its Euler product diverge to zero in Re(s)=

[@sunkhirous]

They’re not zeros of inverses Zeta .

[@sunkhirous]

The fact is if you sum Cosine of the roots you will get the so called roots of Zeta function .

[@KingHim00]

I solved Riemann's hypothesis that every zero does end up on the line and I have proof to show you in a way for you to check and see if every zero ends up on a critical line

[@SLAMgamer11]

you do?

[@KingHim00]

@@SLAMgamer11 yes go look on my channel

[@99bits46]

just take arbitrary 'z' of the form ( 1/2 + ix ) and plug it in Re(z) and do some tricks with algebra, you should have your Re(z) = 0. If you can't do it then the hypothesis is not true.

[@kenichimori8533]

ζ(n) = 1/12+1/13+1/14+1/15+1/16+1/17+1/18+1/19=20(n) Ten Zeta is Twelve Octave=20(n+1)=20(n+2)=0

[@johnstfleur3987]

"50%."

[@pramu_dithawickrama_tunga1672]

Reimann ekta conjetre 2k kiyal liyal tibada🤔 Mata anwanum magecidea ekta anuwanum ema wena ba😂u ecra chora na Prsma tiyan nikan ponya wagecbalbal encna😅u oktma klin hitwna. I ema chu dial 1k nami Wena ba🤔

[@middu4228]

Omg lol

[@kenichimori8533]

P = NP = 0110

[@kenichimori8533]

Modular function pr = Mod n = It's moon

[@johnstfleur3987]

I AM ZERO.

[@kenichimori8533]

Zeros Quantum approach. P = NP = 0001

[@kenichimori8533]

ln = 3n = f(e) euler function.

[@KingHim00]

I solved this just uploaded my notes go check it out.

[@williejohnson5172]

Solved it : kzbin.info/www/bejne/naPbhIGnasxpl6s

[@artificialresearching4437]

I might be wrong, but this could work: kzbin.info/www/bejne/hqe4o5WmqL2Ngrc P.S. A piece of advice: make video 1.5 faster, I speak very slowly)

[@kenichimori8533]

Riemann Geomerty 0 = ζ(s)^n=8=3+3

[@kenichimori8533]

P = NP = 3pik

[@pramu_dithawickrama_tunga1672]

Ema wen hari drlabi sp kind of lobe ekadi Mokda me wage adhskin adre klaoth Mama oyta me loek okM denum mata wetrma adre one🤔ema adre krla mata hukamth oananam sedi gamwak ektweuan mwene😂 1.mama oyata me butiak loek ona deyak deum adre kiyana absrrt de mata wetrk oen kiyal dena gen kela adre klaoth 😂koal uge or gamige amth hikwoth wdma wenwa😅evra adre kela ow Adre kradi mara awbidyak oem bro 🤣 Tyuselta hena jike. Me loke wadkrem kodma dwaum duwanwa.... 1k 1s me loek enen na adma niwam yanwa

[@arekkrolak6320]

"At least 41% of infinitely many..." and we are now entering the realm of BS :)

[@olbluelips]

Use your brain. There's nothing wrong with a percent of infinity. For example, 50% of infinitely many integers are divisible by 2. Around 33% by 3, etc.

[@kenichimori8533]

ζζζpi size 0 bytes.

[@bonniemassaro11]

This is so silly, the problem is the solution itself. It is a hypothesis until proven untrue. And it is based on a straight line. Therefore, untrue. No longer science. This is where Good ol' Riemann went wrong, there are no straight lines in physics, just a wobble with energy / eather in the middle, that is your 1/2 and the reason no-one can figure it out, the hypothesis includes an assumption. A straight line. And there is absolutely a pattern in the prime numbers, follow the space between 7s. Bet bitcoin gets nervous. Good trick all, way to keep a bunch of brainiacs busy trying to figure something out, that within itself is not based on truth. Beware of assumptions!!! Indeed.