- Anybody seen this before?

http://www.geocities.com/rze17/zeros.pdf

(home page http://www.geocities.com/rze17/math.html)

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths/

http://www.users.globalnet.co.uk/~perry/DIVMenu/

BrainBench MVP for HTML and JavaScript

http://www.brainbench.com - I'm looking at the proof for if I could find a flaw.

The only thing I have observed is that he took off the null roots of the polynomial, and I was wondering if the zeta function had any zero root or not (if they were infinite, then the proof is not valid).

I went to Mathworld and find this equality:

zeta(1-s) = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s)

I thought that "If s=1 then zeta(1-s) = zeta(0) = .... · cos(pi/2) = 0." but I have just realised that zeta(1) is the harmonic series! And then the value is in principle indetermined... can anyone say?

What happens if we quit a zero root? zeta(1-s)/s = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s) /s ... and then evaluate in s=1 again...

and I'd apply the L'Hôpital rule if I knew how to derivate zeta(1-s) - the s in the denominator is carried out -.

If it can have an infinity of zero roots, and then the proof is not rigorous.

Regards. Jose Brox.

----- Original Message -----

From: Jon Perry

To: Prime Numbers

Sent: Friday, February 28, 2003 5:31 PM

Subject: [PrimeNumbers] Infinite zeroes proof

Anybody seen this before?

http://www.geocities.com/rze17/zeros.pdf

(home page http://www.geocities.com/rze17/math.html)

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths/

http://www.users.globalnet.co.uk/~perry/DIVMenu/

BrainBench MVP for HTML and JavaScript

http://www.brainbench.com

Yahoo! Groups Sponsor

ADVERTISEMENT

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com

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[Non-text portions of this message have been removed] - Well, in the same page of Mathworld (http://mathworld.wolfram.com/RiemannZetaFunction.html)

we can see the graphic of zeta(x) and it isn't zero in x=0, so forget my previous message.

Jose.

----- Original Message -----

From: Jose Ramón Brox

To: Prime Numbers

Sent: Friday, February 28, 2003 6:30 PM

Subject: Re: [PrimeNumbers] Infinite zeroes proof

I'm looking at the proof for if I could find a flaw.

The only thing I have observed is that he took off the null roots of the polynomial, and I was wondering if the zeta function had any zero root or not (if they were infinite, then the proof is not valid).

I went to Mathworld and find this equality:

zeta(1-s) = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s)

I thought that "If s=1 then zeta(1-s) = zeta(0) = .... · cos(pi/2) = 0." but I have just realised that zeta(1) is the harmonic series! And then the value is in principle indetermined... can anyone say?

What happens if we quit a zero root? zeta(1-s)/s = 2· (2pi)^(-s) · cos(1/2 · s · pi) · gamma(s) · zeta(s) /s ... and then evaluate in s=1 again...

and I'd apply the L'Hôpital rule if I knew how to derivate zeta(1-s) - the s in the denominator is carried out -.

If it can have an infinity of zero roots, and then the proof is not rigorous.

Regards. Jose Brox.

----- Original Message -----

From: Jon Perry

To: Prime Numbers

Sent: Friday, February 28, 2003 5:31 PM

Subject: [PrimeNumbers] Infinite zeroes proof

Anybody seen this before?

http://www.geocities.com/rze17/zeros.pdf

(home page http://www.geocities.com/rze17/math.html)

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths/

http://www.users.globalnet.co.uk/~perry/DIVMenu/

BrainBench MVP for HTML and JavaScript

http://www.brainbench.com

Yahoo! Groups Sponsor

ADVERTISEMENT

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com

The Prime Pages : http://www.primepages.org/

Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.

[Non-text portions of this message have been removed]

Yahoo! Groups Sponsor

ADVERTISEMENT

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com

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[Non-text portions of this message have been removed] - I knew of the use of Euler-Maclaurin in

Sondow, Jonathan

The Riemann hypothesis, simple zeros and the asymptotic

convergence degree of improper Riemann sums.

Proc. Amer. Math. Soc. 126 (1998), no. 5, 1311--1314.

It is not clear to me what is added to that by

Nadim Restom web postings.

David - On Fri, 28 Feb 2003, [iso-8859-1] Jose Ramón Brox wrote:
> Well, in the same page of Mathworld (http://mathworld.wolfram.com/RiemannZetaFunction.html)

zeta(0) = -1/2.

> we can see the graphic of zeta(x) and it isn't zero in x=0, so forget my previous message.

lim_{s \to 1} cos(pi/2*s)*zeta(s) = -pi/2.