Riemann Hypothesis -- New Euler Zeta function and its Complex Roots-Euler complex product

[pridoni54]

please ignore my last response.

thanks

[pridoni54]

Present product formula with the NOTE: The factors are all the prime and their even power. NECESSARILY CONSISTS the factor (1+(1/81)^S)

(1+(1/2)^S)(1+(1/3)^S)(1+(1/4)­^S)(1+(1/5)^S)(1+(1/7)^S)(1+(1­/9)^S)...(1+(1/79)^S)(1+(1/81)­^S)...

It is MISTAKE. WHY ? because the factor (1+(1/81)^S) NEVER APPEARED in the product formula as the term (1/81)^s paired with the term (1/9)^s.

Please understand that I don't try to argue with you, but try with you to find the way how to develop the method.

[pridoni54]

Thanks for your response. 81=3^4 is the square square of prime 3 but it isn't the factor of the formula. Also 625=5^4 is the square square of prime 5 but it isn't the factor of the formula and so on......

Please check your FORMULATION NOTE:The factors are all the prime and their even power.

Regards

[sunkhirous]

@pridoni54 To understand these product better ,multiply them by Euler prime product and is = 1 . Also to note that Andrew Odlyzko made the same mistake on 64 in his email to me in 2010 .

[pridoni54]

Please see related work

Grave Concerns On Riemann Hypothesis - Prime Numbers

PS.......64=2^6

[sunkhirous]

@pridoni54 The roots are primes, square of primes,square square of primes,square square square of primes and so on... so the 64 factor will no appear in the product .

[pridoni54]

Hi,

Subject of my question: IS IT CORRECT your posted product formula?

The formula consist the following factors :(1+(1/4)^S); (1+(1/9)^S); (1+(1/16)^S); ( 1+(1/25)^S);

And doesn't consist the following factors: (1+(1/64)^S); (1+(1/81)^S); (1+(1/625)^S)

On another hand 64 =2^8; 81=3^4; 625=5^4 the prime numbers even power.

Please advise.

Thanks

[nicodougy]

Men stop wasting your time with this video. This thing is just nonsense. It is worst than nonsense actually since it is claiming that other derivations (which have sense them!) are wrong. Ok a sum of cosines will never converge!!!!! To converge you need your limit to go to zero...cos(nx) don't....

[juik104]

zeta zeros have real part zero?? lmfaoo what are you smokin??

[phicomingatya]

Wake up from your dream world, man!

[sunkhirous]

A few months after this utube, The equation was acknowledged and accepted by many universities and mathematicians and by top 10 universities in the world.

[TheSkysYourLimit]

your multiplicative formula posted is wrong, any person with an iq of 5 can prove this easily. not only is it wrong, but the correct version (which is well known by orangutans) doesn't tell you anything about the functional equation or give any insight into the distribution of zeros that lie within the critical strip

"...*where* going the wrong direction..." ? (incorrect use of the word, so i certainly hope you are 5 years old or English isn't your first language)

[boobah1067]

@TheSkysYourLimit Sorry, I am not familiar with these things as much as you people. What do you say is wrong with his multiplicative formula? (And yes, English is not his first language, as I find obvious enough not to comment on, since his errors are typical of Hungarians and other foreign-language speakers, but not of bad native speakers.) Thanks.

[sunkhirous]

Any one can easily prove the Equation by multiplying it to Euler product and get answer = 1

[phicomingatya]

@sunkhirous You still don't see your error??? The euler product has only primes. Yours got 4 and 16. Duh! Your equation is false.

[boobah1067]

@phicomingatya Oops, it was you who said there was a 16 and then you also said his equation has none. So I misspoke in the other post, saying it was someone else. But anyway, how does his not give 16? I don't know why it doesn't. And second, his formula gives not only primes yet Euler's does, but could the two combine in Riemann's fomula?

[phicomingatya]

@boobah1067 You got a point here. Now that I look at sunkhirous' formula again, you are actually right. 16 is in it.

Though if you want to use a formula like this, the factors should be all the primes and there 2^n-th power instead of their even power. Else you get 64^-s (64 is an even power of 2) and 4^-s * 16^-s (which is also 64^-s).

Then you have to check if the formula is convergent for Re(s) < 1. One way to do that is by checking if the log of the formula is convergent.

[pridoni54]

@sunkhirous

Hi Sunkhirous

Very interesting and original effort.

Please check your multiplicative formula posted for S=2. It is not equal to ((pi)^2)/8.

Each of the member of your formulas is greater than 1.

The product of the first few members is more than ((pi)^2)/8.

friendly advice: Don't stop, keep going, but first find mistake....and before make conclusion always carefully check result.

v/r Pridon

[sunkhirous]

@pridoni54 For S>1 use the zeta product on line 4 for Basel problem =pi^2/6 or 1.644.... and equation A is for complex plane <=1

[phicomingatya]

Dude, there is no 1/16^s in your equation. Your equation is wrong. Even for Re(s) > 1 it is wrong, big time!

[boobah1067]

@phicomingatya Why is there no 16 in his? Another commenter, named phicomingatya, made a comment and said (maybe incorrectly) that BECAUSE his equation got 4 and 16, it was wrong. Now you are saying his equation is wrong because it doesn't show 16 for Re(s). Please explain.

[phicomingatya]

@boobah1067 The formula sunkhirous is using for the Riemann Sum, only works for s with a real part that is above 1. So it is actually useless to prove or disprove Riemann Hypothesis to begin with. Because the critical line is at real part of 0.5.

But also his formula does not even represent the Riemann Sum. Because the Riemann Sum (for Re(s)>1) is the sum of ALL numbers to the negative s-power. Expanding his infinite multiplication will not get all numbers summed.

[boobah1067]

@phicomingatya But in the Bernoulli system (and seemingly the Fibonnaci system) 1 is AT half as well as being a numeric point at 0 and 2. So 1/2 (1 at 1/2 -- one of two) is the first modular 0 in Sunkhirous' system, no? It is the first point where 1 is expressible in a system where 2, 4, etc. are the next 1 each time. Yes?

[phicomingatya]

@boobah1067 I don't know what you are talking about.

Why are you asking these questions? And what is your background in mathematics? Seems like you are randomly naming some non-existing mathematic systems and also without any relationship to the Riemann hypothesis.

Anyway, have you by any chance read my other reply to your post?

[boobah1067]

@phicomingatya Hi. I want to understand what his equation is doing and how he can claim it's been accepted. If it has, then it has merits and the "non-existing" system you mention must be simply recognized. If it works, it works for a reason. Now, Sunkhirous has said he doesn't understand what 1/2 is doing but that he has a 0-point modulus. If so, perhaps it relates to the Bernoulli system: 1 @ half. --- Re. other post, which one did u mean? I replied to both in 1 message, I thot. Cheers.

[sunkhirous]

I've explained everything in my other pages and the paradox where mathematicians where going the wrong direction also Riemann functional equation do have real part 1/2 .The purpose of this utube is 1-Mathematicians and physicists use it 2 - that all roots of zeta are shown and have real parts zero ,One purpose of these roots is to evaluate Prime numbers.

[sunkhirous]

@sunkhirous On Riemann functional equation, I meant Riemann Xi function have real part 1/2.

[boobah1067]

@sunkhirous What is the difference between the Xi function (with 1/2) and the Zeta function (you have re-envisioned to have real part 0)? I mean, what is the Xi function -- since I can't find anything to explain it simply. Thanks.

[sunkhirous]

@boobah1067 Π􏱃s􏱄(s−1)π−s ζ(s) = ξ(t), or ξ(t)=1−(tt+1)􏱋 ψ(x)x−3 cos(1tlogx)dx is the XI function from 1859 Riemann papers .

[boobah1067]

@sunkhirous Hi. I was just thinking about you today. See, I know the look of the function, but I don't know it in plainer terms: what is its purpose?

[boobah1067]

@sunkhirous And by the way, some symbols didn't display. Could you write those out in English? ... Plus if you don't mind, let me know the overall point of the function: what does it do? What does it mean? I think about numbers but only understand some of the equation parts.

[nicodougy]

I don't mean to be rude or anything like this. I am just saying that you need to check what you are doing before saying things like "there are no zero with Re(s)=1/2". That is just wrong...point! Just take the expression of Zeta (the real one for Re(s)<1), take the reported value of the first zero (you can find it on internet), plug it and you will find zero. Now your expression with Re(s)=0 is a sum of a bunch of cosine, there is simply no way it can converge!

[boobah1067]

@nicodougy I'm not as familiar w/ this material. Pls tell me what you mean there is no convergence and the cosines don't match his equation? Seems to me that if he's right, the 0s he found are modular zeros, so if u are matching them up with the 1st zero of the usual Riemann function, then his are a kind of p-adic system. No? If that's right, then yes, yr matches would not be numeric (they'd show as divergences), but would match in absolute sense with ever-changing consistent scales. Am I right?

[sunkhirous]

I thought it would be useful to mention that I've emailed over 100s of known mathematicians and physicists around world and I received 5% reply and support (also some invalid objections that shows they did not understand the use of Euler Zeta related to computing Prime numbers) also I see on arxiv some trying to use the result by going around the issue and use Laplacian determanants ,Weight functions and other way of expressing Riemann Hypothesis and adding more confusion...!

[nicodougy]

OK, think whatever you want, Hardy prove there are infinite many zeros on the critical line but to your point of view it is a common mistake! There is a guy that is looking for zeros since long already, he is quite famous and believe me he is looking also out of the line 1/2 because his goal is to disproof he Riemann Hypothesis. Just an advice, read a bit what over people (that are actually not that stupid) did before...and again your series does NOT converge for your s=i*whatever

[sunkhirous]

A common mistake is that mathematicians believe billions Zeta zeros have been proven to be on line 1/2 which the method is based on the hight T ,1-Assuming that all real part zeros are on line1/2 2- Then they count all imaginary part zeros to the hight T .

[nicodougy]

If it was just a joke, then forgive my last post. But there are funnier jokes than this one then!

[nicodougy]

Man, I am sorry to have to tell ou it's wrong, but you know the greatest (Euler, Riemann, Gauss, Hardy, Ramanudjan,...) have worked on that, so if it was that easy they would have found long time ago! Especially you should know that it is KNOWN there are zeros for Re=1/2 and actually it is proven there are infinity of them! The problem is you do not even start with the good expression, your expression is only valid for Re(s)>1...no doubt they rejected your Arxiv, if it is wrong at line1!

[sunkhirous]

I will soon have video on Riemann Hypothesis from A to Z and how to use the roots,and maybe on Prime Twins and how to find them.