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April 9th, 2017, 11:25 AM   #11
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Conclusion : Better to keep my interesting readings for me.
To @Mashke,

The article is not finished and that is not my claim by the authors claim.
The proof that you did not even read the article
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April 9th, 2017, 11:28 AM   #12
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Quote:
Originally Posted by mobel View Post
Conclusion : Better to keep my interesting readings for me.
To @Mashke,

The article is not finished and that is not my claim by the authors claim.
The proof that you did not even read the article
Did I need to? Since it doesn't prove RH, what is the point of reading it? For one thing I'm not a student of RH and couldn't tell a good argument from a bad one. I do happen to be a student of bad science journalism. Maybe I should read it on that basis.

You're the one who just admitted to knowingly posting a false claim in order to get people to read your posts. You can't fault me for calling that out.

Quote:
Originally Posted by mobel View Post
Conclusion : Better to keep my interesting readings for me.
Better conclusion: Post what's actually known, not a mis-statement for clicks. For example: "Interesting connection between physics and RH." Nobody will bust your chops for that. It's pretty clear that the physicists don't actually know any number theory. You're right, I got that from Woit and not from the original paper. When I first heard about that paper (prior to your post about it) I said to myself, "Bullpucky." And I didn't bother to read the paper.

Last edited by Maschke; April 9th, 2017 at 11:38 AM.
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April 9th, 2017, 11:38 AM   #13
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The next time I will keep any interesting article for me.
By bye
Have good day.
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April 14th, 2017, 01:17 AM   #14
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I think there are more seriuos attack to RH:

- One is realize that Riemann build his function starting from the easy (always unwritten...) definition of a prime number:

be:

$z= n!/n^2$ the function told us (from 5 to infty) if a number is a prime, or not

We can see that $z$ has the "same" behaviour of RH zeros:

It gives integers (equal to trivial zeros) when $n$ is a NON prime, so Rest from the Subtraction:

$z- \lfloor {z}\rfloor =0$


Gives a Rational when $n$ is a prime.

$z- \lfloor {z}\rfloor >0$

So our rest here are the digits following (or not) the integers.

Is clear it's ture by construction, or just because it's the most simple formula that fits for all the primes (except 2) and for all non primes (except 4).

Rh function works in the same way taking just $(logx/x)$ as reference and we can prove by the transfinite inductions that we can couple one by one of my simple zeros / non zeros with one trivial / non trivial zeros.

The only point is to prove we don't make mistakes in the computation of the next RH zeros and that we can correctly computate it as n+1.

There already is a proof that is only based on the Induction on the RH zeros I've no time to check, but I'm quite sure (still if fully right in the computation of the next zero) has to be supported by this Set/ Ordinal Number concerning to vanish the doubt that the induction fail (or can fail).

It will lead to more deep concept is well clear to engeneer, but not to mathematician:

- If there is a variation on the medium behavior of an event, that there must be a variable that cause it.

So it means that the number of freedom degree of a physical or a math problem imply the number of free behaviour.

I found the same evidence on FLT and Beal, but I'm not able to say you more...

As already written several times once we find the equation is true:

\[ \sum_{x=1}^{\pi1}{(2X-1)} + \sum_{X=\pi1+1}^{\pi2}{(2X-1)} = \sum_{x=1}^{\pi2}{(2X-1)} = (\pi2)^2 \]

and we ask if modifing 2 parameters:

\[ \sum_{x=1}^{A}{(2X-1)} + \sum_{X=1}^{B}{(2X-1)} =? \sum_{x=1}^{C}{(2X-1)} \]

1) the Lower limit of the second sum from ${X=\pi1+1}$ to $1$

2) the Upper limit of the second sum from ${\pi2}= C$ to $B=C-K$

where K is an unknown free parameter, we can again have a solution...

What we know is that moving the limits we are making a linear transformation of the problem: if one fells also the other has to fells too.

Now looking into the general equation if the variation of the limits happen on a Sum of linear terms (has happen for n=2) we can find several "equilibrium" points since we move on two linear changing of the values: Limits vs Terms of the Sum and their parallell,

while if the terms are NON linear, as happen to the general Term of the Sum for powers, that is $M_n (X^n-(X-1)^n)$ for n>2

We, maximum, will find just one equilibrium point for each variation of the limits since we loose proportionality, so we loose the linear scaling property.

Last edited by complicatemodulus; April 14th, 2017 at 01:20 AM.
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April 14th, 2017, 01:52 AM   #15
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Quote:
Originally Posted by mobel View Post
Conclusion : Better to keep my interesting readings for me.
To @Mashke,

The article is not finished and that is not my claim by the authors claim.
The proof that you did not even read the article
A link to the paper and an overview appeared in various places... I first saw it at Quanta magazine.

Also, there is no need to falsely 'advertise' your post since most people who frequent the forum will likely read it anyway. Hell i don't know about everyone else here but i just sit around all day hitting the 'new posts' button...
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