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Riemann Hypothesis proof

Anonymous Fri 17 Apr 03:53:36 2020 No.11573322 View Reply Original Report

Quoted By: >>11573404 >>11573577 >>11574177

Riemann Hypothesis proof. The divergent sums are analytically continued, as Riemann did. Please peer review it.

Anonymous

Anonymous Fri 17 Apr 2020 04:12:18 No.11573379 Report

Quoted By: >>11574135

Shouldn't there be pairs of poles where the trivial zeros are? If this proof doesn't acknowledge their existence, there's something wrong with it.

Anonymous

Anonymous Fri 17 Apr 2020 04:23:53 No.11573404 Report

Quoted By: >>11574140

>>11573322
You have no idea what analytic continuation is. It's not good practice to use terms you can't define.

Anonymous

Anonymous Fri 17 Apr 2020 05:03:16 No.11573474 Report

Quoted By: >>11574169

Dafuq actually convolution means? This is trash

Anonymous

Anonymous Fri 17 Apr 2020 05:51:02 No.11573577 Report

Quoted By: >>11574161

>>11573322
You lost me at "the inverse laplace transform will diverge if there are poles in the right half plane".
https://en.wikipedia.org/wiki/Inverse_Laplace_transform
If you use the mellin inversion theorem then you can choose the real part of the vertical line to be to the right of all poles.
Convergence of your final sum does not say anything specific about what the real part of the vertical line must have been.
The only thing you can conclude is that the real parts of the poles are bounded above by some finite real number.

Anonymous

Anonymous Fri 17 Apr 2020 13:34:29 No.11574135 Report

Quoted By:

>>11573379

As far as I'm aware the trivial zeros are the negative natural numbers. They they come in pairs. Since they are on the left half of the plane they are convergent

Anonymous

Anonymous Fri 17 Apr 2020 13:35:39 No.11574140 Report

Quoted By: >>11576213

>>11573404

But I do. I read and comprehend riemanns original paper. Tell me where I miss applied the term

Anonymous

Anonymous Fri 17 Apr 2020 13:41:13 No.11574161 Report

Quoted By: >>11575862

>>11573577

Inverse Laplace transforms are of the form

1/(s-a) => e^at.

So a pole with a positive real part will have an inverse transform that can't be integrated over.

I'm aware people like to associate the zeta function with a Mellin transform, but for my purpose the Laplace is better.

Anonymous

Anonymous Fri 17 Apr 2020 13:44:03 No.11574169 Report

Quoted By:

>>11573474
>Dafuq actually convolution means? This is trash

Convolution is a well defined operation and many integral transforms have convolution properties. A convolution is where the resulting functions value is equal to the area under one function multiplied by the product if both input functions and one is flipped on the horizonal axis and shifted by an amount equal to the dependant variable

Anonymous

Anonymous Fri 17 Apr 2020 13:46:15 No.11574177 Report

Quoted By: >>11574178

>>11573322
This is not mathematics.

Anonymous

Anonymous Fri 17 Apr 2020 13:47:17 No.11574178 Report

Quoted By:

>>11574177

This is not an argument

Anonymous

Anonymous Fri 17 Apr 2020 23:55:13 No.11575862 Report

Quoted By: >>11576173

>>11574161
Okay. I see what you mean.
The real problem is you need to show the convergence of the dirichlet series for 1/zeta(1/2).
You used a sleight of hand by using the analytically continued value of 1/zeta at 1/2 as a bogus proof of the convergence of the dirichlet series for 1/zeta at 1/2.
The convergence of that dirichlet series needs a direct proof which would require some method to show the limit of the partial sums exists and is finite.

Anonymous

Anonymous Sat 18 Apr 2020 01:47:55 No.11576173 Report

Quoted By: >>11576195 >>11576212

>>11575862

I use the convergence properties of the zeta function to evaluate the dirichlet series at 1/2

Anonymous

Anonymous Sat 18 Apr 2020 01:54:35 No.11576195 Report

Quoted By: >>11576338

>>11576173
Nonmathematical gibberish.

Anonymous

Anonymous Sat 18 Apr 2020 02:00:10 No.11576212 Report

Quoted By: >>11576344

>>11576173
In mathematics you have to write all the details down if you're a beginner. You go too fast and you try to go over the complexity by thinking in these abstract loose wishy washy ways but when you write all the details down and you understand perfectly what you've done, thats when you're doing mathematics.
Do that for a few years, then try again.
Right now you're a schizo poster.

Anonymous

Anonymous Sat 18 Apr 2020 02:00:29 No.11576213 Report

Quoted By: >>11576353

>>11574140
Analytic continuation does NOT mean you're magically allowed to simply use the usual $\prod \frac{1}{1-p^{-s}}$ formula everywhere, even where it doesn't make any sense. The analytic continuation has a totally different formula once you move into other regions of the plane. If the old formula could be used everywhere, then there would be no point in analytic continuation at all.
The analytic continuation of zeta does not satisfy that product formula anymore.

Anonymous

Anonymous Sat 18 Apr 2020 02:41:35 No.11576338 Report

Quoted By: >>11576399

>>11576195
>Nonmathematical gibberish

Let's roll that back
Why did Riemann do?
Dirichlet series:
Interesting properties with natural and prime numbers but not convergent under 2

Zeta function:
No primes number properties but converges and has zeros

Riemann plays a shell game here and claims the zeros of the zeta function say something about the primes in the dirichlet series.

If I'm spouting non mathematical gibberish then the there is no relevance to proving or disproving the RH since he is doing the same thing.

Anonymous

Anonymous Sat 18 Apr 2020 02:42:39 No.11576344 Report

Quoted By:

>>11576212

What detail did I leave out. I'll elaborate if you want

Anonymous

Anonymous Sat 18 Apr 2020 02:45:00 No.11576353 Report

Quoted By:

>>11576213

I think you misunderstand . Analytical continuation is exactly using different expressions with different favorable properties to make conclusions are functions they collectively seem to describe. Don't blame me for applying something implicit in the hypothesis itself

Anonymous

Anonymous Sat 18 Apr 2020 03:01:45 No.11576399 Report

Quoted By: >>11576518

>>11576338
>Riemann plays a shell game
No, everything Riemann did is based on a rigorously proven analytic continuation. Everything you did is based on a handwave.

Anonymous

Anonymous Sat 18 Apr 2020 03:08:25 No.11576422 Report

Quoted By: >>11576443

Tooker is this you? Did you finally give up the infinity-hat nonsense and change your persona?

Anonymous

Anonymous Sat 18 Apr 2020 03:14:21 No.11576443 Report

Quoted By: >>11576452 >>11576467

>>11576422
Nah, Tooker knows Latex. This is a miniTooker.

Anonymous

Anonymous Sat 18 Apr 2020 03:18:34 No.11576452 Report

Quoted By:

>>11576443
Tooker 2: Electric boogaloo

Anonymous

Anonymous Sat 18 Apr 2020 03:25:44 No.11576467 Report

Quoted By: >>11576493

>>11576443
Trying to argue with other cranks makes you realize just how much Tooker spoils us.
You can't argue with these guys at all, they just post a bunch of typo-riddled word salad they blasted out of their ass into Word, and can't understand anything a word you say to them. Tooker might be wrong, but he's lucid enough to have fun debating with.

Anonymous

Anonymous Sat 18 Apr 2020 03:33:42 No.11576493 Report

Quoted By: >>11576502

>>11576467
Doesn't tooker have a bachelor's in math? Then he'd be more qualified than 50% of this board in a lot of things, he's still wrong though. OP's probably some chump with no life who thought he could solve the RH in 15 lines, probably a highschooler.

Anonymous

Anonymous Sat 18 Apr 2020 03:37:33 No.11576502 Report

Quoted By:

>>11576493
I thought it was physics, but I could be wrong. Regardless, he was a graduate student before he snapped, so he has a legitimate bachelor's in some kind of math-heavy field.

Anonymous

Anonymous Sat 18 Apr 2020 03:42:34 No.11576518 Report

Quoted By:

>>11576399

I am using the same analytic continuation. There's no need for me to reinvent the wheel here. Anyways, analytic continuation isn't the controversy. Where the zeros of the zeta function is

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