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Old 2006-07-29, 10:50   #1
bearnol
 
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Default Proved several years ago!

http://www.bearnol.pwp.blueyonder.co...th/riemann.htm
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Old 2006-07-29, 12:54   #2
Fusion_power
 
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Methinks bearnol's "solution" suffers from chicken or egg syndrome.

The last serious attempt at a Reimann proof was from de Branges about a year ago. There are several attempts recently published but most are deficient in one or another aspect. A google search for "Riemann Branges" will bring up some interesting reading.

Fusion
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Old 2006-07-29, 14:01   #3
bearnol
 
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My solution is correct and complete - think about it for more than 2 seconds and you'll see. There is some surprisingly clever logic hidden in those few lines :)
btw, it _also_ demonstrates pretty clearly - in rebuff to your assertion of chicken and egg - why the zeros of zeta occur in symmetric pairs about the real axis, a fact I was not even aware of until after I came up with the proof, but which I later (also) read to be true...
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Old 2006-07-29, 18:17   #4
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Also, bearnol's "solution" contains lots of indecipherable notation...
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Old 2006-07-29, 19:43   #5
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It uses, I believe, one of a number of fairly standard ways of expressing some mathematical notation (eg superscripts) in plain ASCII. I deliberately chose this format (ie plain text) (a number of years ago) to try and make the proof as accessible as possible to as wide a range of browsers as possible.
Since the proof's exposition (if not inspiration) is so simple, anyone knowing the form of the zeta function should be able to follow it very easily, regardless of exact coding anyway.
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Old 2006-07-29, 21:21   #6
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Is that supposed to be

\left( \sum_{n=1}^{\infty} n^{-s} \right) \cdot n^{2s-1}

or

\sum_{n=1}^{\infty} \left( n^{-s} \cdot n^{2s-1} \right)

Looks like you're multiplying the expression by n^{2s-1}, even though n is the variable of summation. Then you pull that multipler into the sum even though it's not constant.

Is that what you were trying to do?

Alex
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Old 2006-07-29, 22:11   #7
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Quote:
Originally Posted by jinydu
Also, bearnol's "solution" contains lots of indecipherable notation...
For example, "n**-s"

Besides, \zeta(s)\neq\zeta(1-s) in general. For example, \zeta(1)\neq\zeta(0).

Last fiddled with by jinydu on 2006-07-29 at 22:12
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Old 2006-07-30, 07:43   #8
bearnol
 
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Quote:
Originally Posted by akruppa
Is that supposed to be

\left( \sum_{n=1}^{\infty} n^{-s} \right) \cdot n^{2s-1}

or

\sum_{n=1}^{\infty} \left( n^{-s} \cdot n^{2s-1} \right)

Alex
The latter
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Old 2006-07-30, 07:48   #9
bearnol
 
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Quote:
Originally Posted by jinydu
Besides, \zeta(s)\neq\zeta(1-s) in general. For example, \zeta(1)\neq\zeta(0).
True, zeta(s) <> zeta(1-s) in general. My proof only applies when zeta(s) = 0. Then zeta(s) DOES = zeta(1-s)
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Old 2006-07-30, 09:22   #10
akruppa
 
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> The latter

Where does the n^(2s-1) term come from?

> My proof only applies when zeta(s) = 0. Then zeta(s) DOES = zeta(1-s)

The trivial zeros are not symmetric around s=1/2.

Alex

EDIT2: All posts but first moved to separate thread.

Last fiddled with by akruppa on 2006-07-30 at 10:09 Reason: 1/2, not 1
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Old 2006-07-30, 10:29   #11
bearnol
 
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Quote:
Originally Posted by akruppa
> The latter

Where does the n^(2s-1) term come from?
I _choose_ the 2s-1 term. It is chosen for two reasons:
1) It 'converts' the s term into 1-s
2) because the real part of s is less than 1, the real part of 2s-1 is also less than 1. Hence the summation must still sum to zero


Quote:
Originally Posted by akruppa
EDIT2: All posts but first moved to separate thread.
Thanks
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