Given the edit of the OP, it's supposed to be some kind of counting function.
If he is it's a terrible abuse of notation. Using symbols in a way counter to normal mathematical convention renders ones work illegible.
Fair enough, I'm a nobody ...
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I need money, endorsements, or a professional affiliation.
I have been trying to get someone/anyone to look at my proof since 2006.
I don't have the correct 'political' affiliations mathwise.
What I have right now, is a promise of a peer review ... time will tell.
They are professional mathematicians and may help me clarify the proof for publication.
I've had people claim my work as their own before, I want the formula somewhere public (I've posted in more than one location) to have a 'paper' trail as it were.
Also, I like Civilization 8)
[note] Do you think feedback from <99%ile would be more helpful? 8)
If he is it's a terrible abuse of notation. Using symbols in a way counter to normal mathematical convention renders ones work illegible.
Unless I misunderstand something, the last line is false. Because you are taking the absolute of every number in the sum, the function can only be zero, when the i^4*y term is -1 for all x. For that to happen, y must be an odd multiple of 1/2. As 2 is prime, it cancels out the 1/2 in the y term. So for all Px, except for 2, the y term is not an odd multiple of 1/2 and thus the sum is not zero.
If you really do have a proof of this, post it. Don't just post definitions of functions and other mathematical facts that are somehow related to the twin prime conjecture, without actually presenting a proof. Then maybe somebody who knows a significant amount about pure math (there are a few on this forum; I'm certainly not one) will be able to evaluate it, and maybe help you with the problems in notation that Perfection noted.
Also, you need to reply to uppi's post, in which he finds a possible flaw in what you have posted.
To be frank, what you've posted so far rings a few alarms that will cause people to
think you're a "crank". Posting old standardized test scores saying you tested at the 99th percentile in math in some test does not help your case: most people who would score at the 99th percentile in a high school standardized test are still unable to come up with new mathematical research. It comes across as saying that, since you're pretty good at math, your proof must be right, which is obviously not true. Posting non-standard mathematical notation and not giving us your full proof rings more alarms.
One other problem comes up when you say that you have been trying to get people to look at your proof since 2006, and have been ignored because of a lack of funding and official credentials. This isn't a huge barrier in math: you can post this on a number of free sites, such as arXiv, which while not peer-reviewed still lets your work get out into the mathematical community while not allowing anybody to steal your idea and post it as their own. You need to do this if you haven't already. If arXiv doesn't take it, at least post it on forums frequented by real mathematicians, not gaming forums. They'll be able to give you better analysis than we can.
On a related note, the wiki article on this conjecture is interesting. It appears that some recent work really is zeroing in on a real proof. There may be a rigorous proof before too long.
i^(4I) = 1
i^(4(x/n)) = 1 when n is a divisor of x
|1+1| = 2
2/2 = 1
Twin Pairs = 6x+-1
Odd times Odd = Odd
Odd +- Even = Odd
All the odd terms are indivisible by P[].
P = 2 because P[0]/2 = 1 the numerator becomes Odd, r+-1 ... indivisible
P = 3 because r+-1 ... indivisible
P >=5 because r+-4,+-2 ... indivisible
Therefore the sum is 0, and it is why this formula lies at the center of the proof.
Example:
2*3*5((2I +- 1)/2)+-3+-1=
I=0 ... (2I+1)11,13,17,19,(2I-1)-11,-13,-17,-19 ... not divisible by 2,3, or 5
I=1 ... (2I+1)49,47,43,41,(2I-1)11,13,17,19 ... not divisible by 2,3, or 5
The function has a floor operator. Each term in the sum is either 0 or 1 (as long as the exponentiation is defined in terms of complex numbers)*.
|(a+bi)| = (a^2 + b^2)^1/2,
0 <= |(0+i)^x + 1| <= 2, x is a positive rational number
Consider floor (|(0+i)^x + 1|/2), x is a positive rational number
*Which may be more of a problem than I initially thought when I originally made this post.
Why? You are claiming that Infinity+4 (or whatever you mean) is indivisible by 5. If that statement was obvious, we would have solved the twin prime problem long ago.P >=5 because r+-4,+-2 ... indivisible
In that case, you could post on a non-standard site like viXra, which exists to publish papers that can't get into arXiv. It will be harder to get knowledgeable people to read it, but you could at least write your full proof out online in a way that will clearly mark the proof as yours, and then go around to various math-related forums asking for opinions. You should definitely consider using some sort of free pre-print site, especially if the peer review doesn't pan out (e.g. if it gets outright rejected without helpful comments). This way we could actually see your proof, which is necessary to make any real evaluation of your work, and you'll have timestamp evidence that the idea is yours in the event the same ideas mysteriously pop up in somebody else's paper later on.arXiv requires endorsement and/or a professional affiliation, others have asked for thousands to publish or even review. Some look down their nose and call me an idiot without even looking at the proof.
Still, peer review is finally coming 8)
[note] Clever or stupid, either way, I will be published or know to never publish, and that review is all I've been asking.
I have been trying to get someone/anyone to look at my proof since 2006.
- Avoid the +/- notation. I have no idea what you mean by that. That notation has no
place in a proof.
...
Why? You are claiming that Infinity+4 (or whatever you mean) is indivisible by 5. If that statement was obvious, we would have solved the twin prime problem long ago.
In any case, if you are worried that someone will steal the proof, posting these formulas will accomplish very little. Formulas are easily reformulated (especially when you use strange notation), so if anyone steals it he could (and as the notation needs work actually has to) reformulate the formulas in a way they're not recognizable. In a proof it is the argument that counts, not the formulas.
In that case, you could post on a non-standard site like viXra, which exists to publish papers that can't get into arXiv. It will be harder to get knowledgeable people to read it, but you could at least write your full proof out online in a way that will clearly mark the proof as yours, and then go around to various math-related forums asking for opinions. You should definitely consider using some sort of free pre-print site, especially if the peer review doesn't pan out (e.g. if it gets outright rejected without helpful comments). This way we could actually see your proof, which is necessary to make any real evaluation of your work, and you'll have timestamp evidence that the idea is yours in the event the same ideas mysteriously pop up in somebody else's paper later on.
You should post a formal proof then, so that it can be criticized.
I haven't been following the thread though, maybe you've addressed the lack of a proof already.
edit: I have read back a bit and see that this is a bit of a mess. Sorry, I went through university doing lots of formal mathematical proofs, and they all adhered to certain.. standards. It's not meant to be personal, I'm just comparing this proof to those standards because I've been trained to.
Example: x+-3+-1 represents the set (x+4),(x+2),(x-2),(x-4)
... the indivisibility is true for each element of the two twin pairs
The residue for all Primes >= 5 is +4,+2,-2, and -4, therefore any Prime >= 5, in the set of Primes, cannot divide the four elements of the two twin pairs.
Mike Tepper said:I'm having trouble with understanding what the guy wrote down (and from the other comments in the thread it looks like I'm not the only one.) Even if what he says does imply the twin prime conjecture, I suspect that there is some sort of logical issue in the inner workings. Like dividing infinity by some number is a number or somewhere whatever he says implies 0=1. And it's funny but if you let 0=1 you can prove just about anything.
In the last year or so there actually has been a lot of progress on the conjecture and while it isn't completely solved, what we do know now is head and shoulders above what has been known for a long time.
What we know now is there are infinitely many prime pairs with a gap that's smaller than 600. This of course sounds a lot bigger than 2, but it was first proved for pairs with a gap less than 7 x 10^7 last year and before that, there was nothing. (Well not nothing, there was evidence but no proof that there was infinitely many pairs with some finitely bounded gap.)
The other thing is if he really does have a well written out proof he can post it to arxiv.org, which is a server for preprint (unpublished) math and physics papers. Though I they do moderate and turn down anything they think can be easily be discredited.
The other thing is if he really does have a well written out proof he can post it to arxiv.org, which is a server for preprint (unpublished) math and physics papers. Though I they do moderate and turn down anything they think can be easily be discredited.