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Mar 16, 2007, 7:33:32 AM3/16/07

to

http://arxiv.org/abs/math.NT/0703367

An unpublished paper by Tribikram Pati, in which he claims to

have disproved the Riemann Hypothesis. I don't have the number

theory background to review it, but I gather the author is a

respected mathematician, possibly retired, from Allahabad, India.

I wondered if any of you could give opinions on it.

Reinhold

Mar 16, 2007, 9:45:02 AM3/16/07

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Hi

interesting, i thought everyone was expecting the opposite.

Did he also include a list of the complex zero's not on the critical

line?

Gerry

Mar 16, 2007, 10:35:24 AM3/16/07

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Will, at leaast this does not look like an immediate nonsense.

I'll probably take a look at it if I have time.

Mar 16, 2007, 6:43:54 PM3/16/07

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Reinhold Burger wrote:

Wait until his paper is vetted by the professionals.

Bob Kolker

Mar 16, 2007, 6:43:56 PM3/16/07

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One would assume that would settle the issue, but this paper does not do

that as far as I could tell by glancing at it. The author constructs a

chain of conclusions from the assumption (used essentially throughout)

that the hypothesis is true, ending in something that is supposed to be

obviously false. No construction of other zeros; it fits the word

"disproof" in the title.

It will be interesting to see what experts have to say about this.

--

David L. Johnson

"Business!" cried the Ghost. "Mankind was my business. The common

welfare was my business; charity, mercy, forbearance, and

benevolence, were, all, my business. The dealings of my trade were but

a drop of water in the comprehensive ocean of my business!" --Dickens,

Mar 26, 2007, 1:00:17 PM3/26/07

to

On 16 Mrz., 13:33, Reinhold Burger <rfbur...@cs.uwaterloo.ca> wrote:

> http://arxiv.org/abs/math.NT/0703367

>

> An unpublished paper by Tribikram Pati, in which he claims to

> have disproved theRiemannHypothesis. I don't have the number> http://arxiv.org/abs/math.NT/0703367

>

> An unpublished paper by Tribikram Pati, in which he claims to

> theory background to review it, but I gather the author is a

> respected mathematician, possibly retired, from Allahabad, India.

>

> I wondered if any of you could give opinions on it.

>

> Reinhold

on p.12, m stands for the multiplicity of a zero of the zeta function,

so

typically m=1. But then the step from the third to the fourth

displayed

formula reads

something + A < something,

for a positive constant A. Similarly, I do not understand how the

positive

constant B disappears a few lines later. I did not attempt to

investigate,

though, whether this is a major problem or not.

Stefan Wehmeier

ste...@math.upb.de

Mar 26, 2007, 5:00:20 PM3/26/07

to

In article <eu8u71$25l$1...@news.ks.uiuc.edu>,

stefanw <ste...@math.upb.de> wrote:

>But then the step from the third to the fourth displayed

>formula reads

>something + A < something,

>for a positive constant A. Similarly, I do not understand how the

>positive >constant B disappears a few lines later.

stefanw <ste...@math.upb.de> wrote:

>But then the step from the third to the fourth displayed

>formula reads

>something + A < something,

>for a positive constant A. Similarly, I do not understand how the

>positive >constant B disappears a few lines later.

He explains a couple of pages back that "A" stands for some unspecified

positive constant, which might not be the same constant each time it

appears, even in consecutive lines of a series of inequalities. Sort

of like big-O notation.

--

Tim Chow tchow-at-alum-dot-mit-dot-edu

The range of our projectiles---even ... the artillery---however great, will

never exceed four of those miles of which as many thousand separate us from

the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences

Mar 27, 2007, 12:30:03 PM3/27/07

to

On 26 Mrz., 23:00, t...@lsa.umich.edu wrote:

> In article <eu8u71$25...@news.ks.uiuc.edu>,

>

> stefanw <stef...@math.upb.de> wrote:

> >But then the step from the third to the fourth displayed

> >formula reads

> >something + A < something,

> >for a positive constant A. Similarly, I do not understand how the

> >positive >constant B disappears a few lines later.

>

> He explains a couple of pages back that "A" stands for some unspecified

> positive constant, which might not be the same constant each time it

> appears, even in consecutive lines of a series of inequalities. Sort

> of like big-O notation.

> In article <eu8u71$25...@news.ks.uiuc.edu>,

>

> stefanw <stef...@math.upb.de> wrote:

> >But then the step from the third to the fourth displayed

> >formula reads

> >something + A < something,

> >for a positive constant A. Similarly, I do not understand how the

> >positive >constant B disappears a few lines later.

>

> He explains a couple of pages back that "A" stands for some unspecified

> positive constant, which might not be the same constant each time it

> appears, even in consecutive lines of a series of inequalities. Sort

> of like big-O notation.

But then the final contradiction of the kind f(t) < f(t) (for some f)

poses no problem. If I am allowed to add some constant (not the same

on both sides of the inequality), the contradiction vanishes.

Stefan Wehmeier

ste...@math.upb.de

Mar 28, 2007, 11:00:12 AM3/28/07

to

In article <eubgqb$1vk$1...@news.ks.uiuc.edu>,

stefanw <ste...@math.upb.de> wrote:

>But then the final contradiction of the kind f(t) < f(t) (for some f)

>poses no problem. If I am allowed to add some constant (not the same

>on both sides of the inequality), the contradiction vanishes.

stefanw <ste...@math.upb.de> wrote:

>But then the final contradiction of the kind f(t) < f(t) (for some f)

>poses no problem. If I am allowed to add some constant (not the same

>on both sides of the inequality), the contradiction vanishes.

True enough. But perhaps this issue could be fixed easily by insisting

that A means A, and just changing A to A' at a few points in the paper.

This wouldn't be any sloppier than many other papers I've seen.

Mar 30, 2007, 10:00:22 AM3/30/07

to

Unfortunately, Louis de Branges has a proof of the Riemann Hypothesis

on his home page....

on his home page....

Gary McGuire

Mar 30, 2007, 11:00:06 AM3/30/07

to

Maybe.

May 19, 2007, 5:15:00 PM5/19/07

to

I see a logical error in the paper, namely, a vicious circle when

delta

is defined via delta. It is in equations (18) and (20).

If we substitute $\hat\sigma$ into (20), we get inequality of type

$A^*

< f(n,t)$ (the left "<" sign is never used in the sequel). Here t is

bound with delta via relation $t=t_0+\delta$, $t_0$ being a zero of

zeta

between n-1 and n+1, and we may consider $t_0$ uniquely determined by

n.

When choosing $A^*$ numbers t and delta are not yet defined, delta is

introduced later as $\delta=A^*/\log(n+1)$. Thus we are choosing

delta

as satisfying inequality of type $\delta<f(delta)$. In order to

define

it well, we should eliminate dependence on delta in f. It is easy to

do

for several factors using estimates $n-1<t_0+\delta<n+1$; but then

there

remains $|\zeta(1/2+it_0+i\delta)|^{\tilde A}$ which we cannot

estimate

by n only. Thus existence of such delta needs proof; and the rest of

the

paper shows this delta cannot exist. So here is the contradiction and

not in the RH.

May 23, 2007, 11:30:01 AM5/23/07

to

Your results looks fine. But the main definition of delta is not

"introduced later" - after (18), I think you point on (36) and on that

the author give attantion to (11) and (20) - but on (10) and (11). So

the main idea of the author was to show, that we need that delta to

proof RH, but there is no delta, so we can not proof RH after all (it

is not a disproof but a 'not proveable' - statement).

May 28, 2007, 12:30:03 PM5/28/07

to

On May 23, 7:30 pm, lange_kl...@hotmail.com wrote:

> Your results looks fine. But the main definition of delta is not

> "introduced later" - after (18), I think you point on (36) and on that

> the author give attantion to (11) and (20) - but on (10) and (11).

Delta is $A^*/log n$, so it is defined by A*, i.e. in (20). In (11) it

is only

announced: "... where A* is a ... number to be further specified".

I don't mean (36) because it is deduced correctly from (20) and below.

> the main idea of the author was to show, that we need that delta to

> proof RH, but there is no delta, so we can not proof RH after all (it

> is not a disproof but a 'not proveable' - statement).

No, not so. He wants to show that if we proved RH then there would

exist such a delta (what would lead to a contradiction).

May 30, 2007, 9:30:03 AM5/30/07

to

Yes, Pati by himself didn't uses the strategy showing that RH is not

proveable under the axioms of standard analysis. But indirectly - his

"disproof" shows - that the only way for standard analysis handle the

RH may lead to the conclusion, that I give before.

But you are right: He himself claims a contradicition found on RH.

My point was:

This main idea, I talked about, was unintended given by the author's

paper. But its path opened for that.

May 30, 2007, 10:30:00 AM5/30/07

to

In article <f3ju8r$cmf$1...@news.ks.uiuc.edu>, <lange...@hotmail.com> wrote:

>Yes, Pati by himself didn't uses the strategy showing that RH is not

>proveable under the axioms of standard analysis. But indirectly - his

>"disproof" shows - that the only way for standard analysis handle the

>RH may lead to the conclusion, that I give before.

>Yes, Pati by himself didn't uses the strategy showing that RH is not

>proveable under the axioms of standard analysis. But indirectly - his

>"disproof" shows - that the only way for standard analysis handle the

>RH may lead to the conclusion, that I give before.

Let me see if I understand. You claim that the author wanted to show

something like:

1. RH implies the existence of delta.

2. In fact, delta does not exist.

3. Therefore, RH is false.

However, the author did not show this. But are you claiming that the

author *did* in fact show, unintentionally, that

1'. If RH were provable using standard axioms, then delta would exist.

2'. In fact, delta does not exist.

3'. Therefore, RH is not provable using standard axioms?

It would be remarkable for a paper that makes no explicit mention of

axioms or related concepts from mathematical logic to show 1'-2'-3'

without showing 1-2-3. Without looking at the paper in detail, I would

suspect that Kuznetsova's interpretation is correct, that the author

has shown neither 1 nor 1'.

Jun 11, 2007, 10:00:08 AM6/11/07

to

Bernd Krötz is showing, that step by step Pati's paper is correct, but

the logical strategy behind them failed. With or without RH the

contradiction is there:

the logical strategy behind them failed. With or without RH the

contradiction is there:

http://guests.mpim-bonn.mpg.de/kroetz/RH.pdf

So, and that is my point:

I mean that this error of Patis strategy leads us to the idea, that RH

is not proveable/ or disproveable in boundarys of standard analysis

axioms.

Jun 12, 2007, 3:10:53 PM6/12/07

to

> Bernd Krötz is showing, that step by step Pati's paper is correct, but

> the logical strategy behind them failed.

> the logical strategy behind them failed.

I have read the text at the link. He says the same what I did.

Calculations are correct but it doesn't mean "Pati's paper is

correct".

If I choose a number x such that 0 < x < sin x and carry lots of

correct calculations with it -- how do you think, maybe I'll get

even more striking contradiction?

> I mean that this error of Patis strategy leads us to the idea, that RH

> is not proveable/ or disproveable in boundarys of standard analysis

> axioms.

It has nothing to do with axioms. It's a mere logical error of one

definite mathematician. This happens from time to time... Erratum

humanum est.

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