Riemann Hypothesis proof

[Anonymous No.11565675]

Please review this Riemann Hypothesis proof.

[Anonymous No.11565684]

Needs to be a little more crunchy.

[Anonymous No.11565742]

>>11565684

What do you mean?

[Anonymous No.11565979]

Just so you'll stop posting this, I'll bite.

The summation you gave at the beginning is nonconvergent for s <= 1, it's not the analytically continued version of the function. The prime definition and inverse definition =/= the analytically continued definition. Furthermore, I have no idea what you mean by "squarefree" and "actually convolution". Also, you're just doing a change of coordinates, just simply inputting 1/2 into the function doesn't allow you to say anything about the function, you're just shifting the convergent set forward by 1/2 on the complex plane, you're just saying "take the input s and add 1/2 to it before putting it through the function". There are a lot more problems with you're "proof" which I won't get into, but if you don't believe me, submit it, but I guarantee it's not going to be accepted.

[Anonymous No.11566038]

Not bad, just add some infinity hat, that will make it

[Anonymous No.11566040]

>>11565675
The Riemann zeta function already has a definition using the Mellin transform.

[Anonymous No.11566098]

>>11565979
>The summation you gave at the beginning is nonconvergent for s <= 1

It's analytically continued. That's the first line.

[Anonymous No.11566134]

>>11566098
You know nothing about complex analysis. The function, read my lips, the function you gave, multiplying over the primes like that, is nonconvergent for s < 1. Just simply saying that you're taking the analytic continuation of the function means nothing, you never gave an expression for the nonconvergent domain, I could say that in the domain s < 1 the function equals chicken, and it'd be equally valid.

[Anonymous No.11566164]

>>11566134
>Just simply saying that you're taking the analytic continuation of the function means nothing

I disagree. When Riemann first introduced the zeta function, it was a series that didn't converge for half the plane. The them analytically continued the function to extend it to the left hand plane. I'm just doing the same thing .

I know the convergent issue you are concerned about. Like riemann I am analytically continuing the function in the divergent region

[Anonymous No.11566172]

>>11565675
Thank you for posting it here

[Anonymous No.11566180]

>>11566172
Nice samefag OP

[Anonymous No.11566183]

>>11566164
Tell me how you're analytically continuing it then.

[Anonymous No.11566186]

>>11566164
is this an elaborate Tooker satire?

[Anonymous No.11566194]

>>11566180

But I'm OP and I didn't post that.

[Anonymous No.11566199]

>>11566183
>Tell me how you're analytically continuing it then.

The exact same way riemann does. My proof is true if the zeta function itself is true. If you're just gonna say nuh uh it diverges , then the zeta function itself is just a pile of bullshit.

[Anonymous No.11566201]

>>11566186
>is this an elaborate Tooker satire?

Read the proof and find out.

[Anonymous No.11566203]

>>11566194
Poster count doesn't lie.

[Anonymous No.11566209]

>>11566201
>Read the proof and find out.
It's written in MSWord instead of Latex so it's automatically wrong and would cause cancer if read.

[Anonymous No.11566247]

>>11565675
Op is a faggot lel

[Anonymous No.11566260]

>>11566180
>>11566194
>>11566203
that was me, you can't count, and I long for the day when someone proves the riemann hypothesis and publishes on /sci/

[Anonymous No.11566266]

>>11566199
>My proof is true if the zeta function itself is true

[Anonymous No.11566466]

>>11566199
>Zeta function is true
Eat shit fuck off and get RAPED if we're being honest. Post details or get fucked straight into hell.

[Anonymous No.11566573]

>>11565675
>wants a person reviewing proof
>finds bullshit
>gets mad when being called out for bullshit

cope harder

[Anonymous No.11566583]

>not even written in TeX
11 on the 1 to 10 crackpot scale

[Anonymous No.11566638]

Error on second line.
p^s = Exp[(Log[p])]^s = Exp[s*Log[p]] != Exp[-s+Log[p]]

[Anonymous No.11567148]

>>11566638

I reviewed this and don't see it. It's 1/p^s any way. It's never just p^s perhaps that's your errror

[Anonymous No.11567166]

>>11566573

I'm not mad. I fully anticipated the "nuh uh it doesn't converge" argument. That's why the first line says I take analytic continuation for granted.

Let's make this perfectly. The riemann hypothesis and zeta function. Riemann creates a function that has all these very interesting properties that require the function. To be zero where it actually diverges. Riemanns answer to that is he analytically continued his sum.

If Riemann can do that then so can i. This is probably why no proof is accepted snyway. Proof are being held to a higher standard than the hypothesis itself. Let's explore

You don't take the analytic continuation pill: ok fine. My feelings aren't hurt. My proof is just no sense, but so is the meaningfulness of the riemann hypothesis, since it's divergent

You swallow the analytic continuation pill. Ok, fine. Still I don't really care, but then the zeta function means something very profound for prime numbers, and my proof is legitimate, since it takes advantage of those properties

[Anonymous No.11567180]

>>11567166
no one cares

[Anonymous No.11567193]

>>11566266
>>My proof is true if the zeta function itself is true

This isn't exactly true. It's true if you believe it has any relevance to prime numbers and converges on the critical line, since it actually diverges there. Riemann got around this by invoking analytic continuation. So am I.

[Anonymous No.11567198]

>>11566466

I give exact details. I continue the divergent series in the exact same riemann does. Where is the ambiguity here?

[Anonymous No.11568196]

Bumping myself for any real criticisms