What Are the Implications of the Riemann Hypothesis for Prime Numbers?

[Borogoves]

let \zeta(z)=\sum_{n \in \mathbb{N}} n^{-z} ~ {{a+ib}}>1

then, \zeta(z)=0 iff z=-2n where n is a natural number.

pi(x)=\int_0^\infty\frac{dx}{\xS[x+1]} gamma(x+)

where S[x+1]= \sum_{n \in \mathbb{N}} n^-{x+1}

I have discovered that pi(x)=\int_a^b\frac{dx}/logx = 1/log b+ 2/log b + 3!/logb +...

furthermore, pi(x)-\int_^x\frac{dx}+1/2 \int_^x(1/2)\frac{dx} (logx)^-1 ~ (x^1/3) /logx

is this already kmown ?

 

[matt grime]

line one you need to explain: what does {{a+ib}}~>1 mean?

line 2 is false, since your expression for zeta does not exist at the points you claim are roots.

there are zeroes of zeta (with the proper definition) at other points than those you claim.

 

[shmoe]

latex problems have rendered your post mostly incomprehensible. While you still can't view latex in preview mode, you can edit it after you post. I suggest you do so. While you are at it, you might want to give some justifications for whatever it is you are claiming.

 

[Borogoves]

line one you need to explain: what does {{a+ib}}~>1 mean?

line 2 is false, since your expression for zeta does not exist at the points you claim are roots.

there are zeroes of zeta (with the proper definition) at other points than those you claim.

latex has not displayed as intended.

It is a definition, it cannot be false.

{{a+ib}}>1 where the former is the norm.

a(bc)=/= (ab)c and a(ab)=a^b (ab)b=ab^2

I'm claiming that the integral is bound by ~ (x^1/3) /logx where x is large, say >10^10

1/log b+ 2/log b + 3!/logb +... is of continued fraction form

 

[eljose]

i think i have caught up with his reasoning he wants to prove RH by writing that:

\int_{0}^{\infty}dx[x]/x^{s+1}=\zeta(s)

and then for s=1/2 somehow he obtains a bound for the integral, where s=a+ib

 

[matt grime]

The expression you give is not a valid expression for the zeta function for Re(z)<1. You have not got the correct definition of the function. I suggest you take one second to look at the series and think. What happens if you put z=-2 in?

 

[Borogoves]

I was refferring to the real zeroes. stop the niggling remarks

 

[matt grime]

Niggling remarks? You asserted that *the only zeroes* of zeta(z) are at negative even integers. That is false, and to compound the error the expression you give for zeta(z) isn't even defined at those points. Those are not niggling remarks. You did not say you were referring to only the real zeroe, and if you don't have the correct definition of the zeta function what makes you think the rest of your argument is valid?

Your latex is still incorrect.

 

[Borogoves]

I define zeta to be:

1/s-1 \-s \int_{0}^{\1}dx\h(x)x^s-1



Evidently, somebody is attempting to edit my posts.

 

[Hurkyl]

Borogoves:

You don't seem to care about your work enough to write it correctly.

Of what you've written, the stuff I can make out seems to be patently false. (e.g. that first equation with integral in your opening post... the left hand side is a function, and the right hand side is a constant!)

Furthermore, you don't seem to be paying any heed to the criticisms and advice you've been offered.


If things don't quickly take a turn for the better, I'm going to close this thread.

 

[Borogoves]

I have mailed my results to a few professors ,none of whom have responded.
applying mellins transform to get:
\zeta (s) = \frac{1}{s-1} - <br /> s \int_0^1 h(x) x^{s-1} \, dx
\phi : \kappa \longrightarrow\sigma

where h(x) is the gauss kuzmin wirsing operator

\int_0^(pi/2)\ 1\fract 1\ sqrt{sin\theta - 1 }{2!cos\theta-1}... \dx ~ 1/ log 3! + 1/ log 5! + 1/ log 7! in continued fractions.

is this known ?

 

[Borogoves]

please try to correct the latex, I cannot get it to display my results.

 

[Hurkyl]

Spaces are helpful. Use them liberally. And don't use a backslash (\) unless you're actually trying to invoke a LaTeX command, like the command "\int" to get it to produce an integral, or the command "\frac" to produce a fraction.

I've edited your first LaTeX image in #11 to be what I think you want, as a demonstration. I also put some weird spacing in the code so you can see that you don't have to worry about it appearing in the output.

(You may need to reload the page to see the new image)

And, of course, you can click on the image or edit your post to see what the new LaTeX code looks like.

 

[eljose]

Borogoves..you can try to submit your work to "clatech math archive" or something similar they seem to accept math and physics papers...

By the way could anybody write the Gauss-Kuzmin operator h(x)?..thanks.

 

[shmoe]

{{a+ib}}>1 where the former is the norm.

The dirichlet series \sum_{n=1}^\infty n^{-s} converges when the real part of s is greater than 1, not when it's norm is greater than 1. real part of s greater than 1 implies the norm is greater than 1, but the converse is not true. You may consider this a "niggling remark", but mistakes like this make people stop reading and not bother responding to you.

I have mailed my results to a few professors ,none of whom have responded.
applying mellins transform to get:
\zeta (s) = \frac{1}{s-1} - <br /> s \int_0^1 h(x) x^{s-1} \, dx
\phi : \kappa \longrightarrow\sigma

where h(x) is the gauss kuzmin wirsing operator

You mean h(x) is the Guass map here: h(x)=1/x-[1/x] where [] denotes the floor. The gauss-kuzmin-wirsing operator is the frobenius-perron operator of this map (or the transfer operator, or other names), though it's relation to zeta is similar.

I don't know where "phi", "kappa", or "sigma" came from though.I suggest you take a look at https://www.physicsforums.com/showthread.php?t=8997 for some latex info. Take the time to present whatever it is you are doing in a legible manner. The less effort it takes to decipher your posts, the better the responses you'll get. A quick latex tip, {} are used for grouping things like

\sum_{n=0}^{blah\ blah\ blah}\sqrt{\frac{2}{blah\ blah}}

 

[eljose]

the method by borogoves is somehow described in the web:

http://en.wikipedia.org/wiki/Gauss-Kuzmin-Wirsing_operator

\zeta(s)=(s/s-1)-s\int_{2}^{1}dxG[x^{s-1}]

then i don,t know how he managed to get an asymptotic¿? approach to this integral..the idea is not bad but perhaps it could be mor useful to describe a Gauss-Kuzmin operator so:

\sum_{p}p^{-s}=(s/s-1)-s\int_{0}^{1}dxG_{p}[x^{-s+1}]