20060729, 10:50  #1 
Sep 2005
127 Posts 
Proved several years ago!

20060729, 12:54  #2 
Aug 2003
Snicker, AL
7·137 Posts 
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 
20060729, 14:01  #3 
Sep 2005
127 Posts 
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... J 
20060729, 18:17  #4 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Also, bearnol's "solution" contains lots of indecipherable notation...

20060729, 19:43  #5 
Sep 2005
127 Posts 
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. J 
20060729, 21:21  #6 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Is that supposed to be
or Looks like you're multiplying the expression by n^{2s1}, 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 
20060729, 22:11  #7  
Dec 2003
Hopefully Near M48
2×3×293 Posts 
Quote:
Besides, in general. For example, . Last fiddled with by jinydu on 20060729 at 22:12 

20060730, 07:43  #8  
Sep 2005
7F_{16} Posts 
Quote:


20060730, 07:48  #9  
Sep 2005
177_{8} Posts 
Quote:


20060730, 09:22  #10 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
> The latter
Where does the n^(2s1) term come from? > My proof only applies when zeta(s) = 0. Then zeta(s) DOES = zeta(1s) 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 20060730 at 10:09 Reason: 1/2, not 1 
20060730, 10:29  #11  
Sep 2005
7F_{16} Posts 
Quote:
1) It 'converts' the s term into 1s 2) because the real part of s is less than 1, the real part of 2s1 is also less than 1. Hence the summation must still sum to zero Quote:


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