ITT /sci/ solves a Millennium Prize Problem. How about... o

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>>8276641
Everything can be solved by IUT, we should use it.

>>8276641
Just write something stupid on 4chan. All the zeroes will come to you and expose themselves willingly!

>>8276641
I really doubt youll ever fin a zero elsewhere
>true believer

Lets focus on figuring out a proof

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>>8276845
Nah, it would be easier to just find a zero that invalidates the hypothesis.

Just one little counterexample disproves it immediately. Much easier than trying to figure out an entire goddamn proof.

>>8276858
like I said
>believer
>I will study my fucking brain off.. ill give it a shot
>only then we will find the truth

>>8276858
find a goddAMN ZERO i DARE YOU
SHATTRE my life

>>8276858
That would only be easier if the conjecture was false, if it was true then you can spend millennia trying to find one and come out plum handed.

>>8277216
let him try

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>>8276641

I actually did some work on the RH a little while back:
https://files.catbox.moe/1c46nx.pdf
Sorry if it's a bit hard to make sense of. The symbols are basically the ones Mathematica uses.

>>8276641
Why does he look ugly in this pic? I thought he was handsome

>>8278793
The fuck are you talking about? He looks as sexy as ever.

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>>8278805

>>8276641
>If some random Russian dude could figure out the Poincare conjecture

I did a little more work on it:
https://files.catbox.moe/6atcqt.pdf
Li's criterion says that proving [math]\lambda_n > 0[/math] for all [math] n \in \mathbb{N} [/math] is equivalent to proving the Riemann hypothesis.

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>set up product integral function [math] f(a)=\displaystyle \prod_{-\infty}^{\infty}\zeta(a+xi)^{dx} [/math]
>for some fixed a, if the zeta function has a zero along the line a+xi, then f(a) is going to be 0
>use real induction and prove that f(a) is nonzero and divergent on the set (0,1) \ {1/2}
>you have just proven the riemann hypothesis

[math] \displaystyle \prod_{-\infty}^{\infty}\zeta(a+xi)^{dx} = \exp(\int_{\mathbb{R}}\ln(\zeta(a+xi))dx [/math]
[math] \displaystyle \zeta(x) = \prod_{p\text{ prime}} \frac{1}{1-p^{-x}} [/math]
[math] \displaystyle \ln(\prod_{p\text{ prime}} \frac{1}{1-p^{-x}}) = \sum_{p \text{ prime}}\ln(\frac{1}{1-p^{-x}}) [/math]
[math] \displaystyle \sum_{p \text{ prime}}\ln(\frac{1}{1-p^{-x}}) = \sum_{p \text{ prime}}(\ln(1)-\ln(1-p^{-x})) = -\sum_{p \text{ prime}}\ln(1-p^{-x})[/math]
[math] \displaystyle \exp(\int_{\mathbb{R}}\ln(\zeta(a+xi))dx = \exp(-\sum_{p \text{ prime}}\int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx)[/math]
[math] \displaystyle \exp(-\sum_{p \text{ prime}}\int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx) = \prod_{p \text{ prime}}\exp(-\int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx) [/math]

it gets messier

[math] \displaystyle \int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx = \frac{i\text{Li}_2(p^{a+bi})}{\ln(p)}+b\ln(1-p^{-a-bi})-b\ln(1-p^{a+bi})+\frac{1}{2}ib^2\ln(p)+C [/math]

continued in next post

>>8279393

Although I have not examined its mathematical content, I wish you to know that I saw your [math] \LaTeX [/math] and found it unusually beatiful, at a glance.

Two issues which immediately pop out are the exponential "dx" in the beginning, and small parentheses throughout, which do not match the large operators.

Your post brings to mind two things:

1) What /is/ THE Riemann Zeta function, over the complexes, once all the brouhaha has been accounted for? A single, singular formula, which isn't just the early real motivation. This is something I've never actually checked.

2) Waht is the "Li" function? I can readily search this but I'm inviting you to add context, is the point.

>>8279393
you messed up your latex pretty bad
you forgot a parenthesis on line 5 and changed an x to a b on the last line
also, i'm hoping you just copy/pasted that integral over R, because that's a definite integral with a constnat of integration

>>8279422
[math] Li [/math] is usually the logarithmic integral function.

stare zero and let halves come

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>>8279592
>wants to solve perhaps THE greatest mathematical problem of the 20th century
>can't even into LaTeX right
Wonderful. Just, friggin, wonderful.

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Here's a few more hours worth of work on it:
https://files.catbox.moe/ri9i0v.pdf
Prove the [math]\lambda_n[/math] sequence is monotonically increasing, and by reference to the fact that [math]\lambda_1 > 0[/math], the result follows by ordinary induction.

Bump :^)

>>8282390
Instead of just mindlessly bumping, how about you help?

If we can determine some sort of closed-form answer for the incomplete Bell polynomial at the bottom, the we can remove the last derivative operator. Then we just take [math] \lim_{z \to 1} \lambda_n [/math] and we'll have a simple genral form for the terms in the series, which should make it much easier to analyze, seeing as doing the same for the pre-proceesed cases of [math]n = 1[/math] and [math]n = 2[/math] appear to result in a series expressed in terms of the various Stieltjes gammas.

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>>8279393
>real induction
>no discussion of the choice of logarithm
>no discussion of convergence whatsoever
>no words

>>8281121
You do know that proofs usually have words in them right ? At least explain what you're doing if you are expecting anyone to even glance at it

>>8283186
It's just scratch work, f a m. Though I agree his work would be more intelligible if he explained in words.

>>8283198
Still, he could at least explain why he is making these computations and why he thinks they are relevant. He seems to expect people to help. The least one can expect is to know what is happening. It's like asking someone to read disgusting, non-commented code

>>8283214
I explained why they're relevant in >>8281121 and >>8282423 and what I'm doing is attempting to put it into a form for which a growth rate can be determined.