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Riemann Hypothesis and Primes

by philosophking
Tags: hypothesis, primes, riemann
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matt grime
#37
Aug9-05, 03:53 PM
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respond to a proof of what? I can't understand a word of what you're saying but that's not down to the mathematics.
Jonny_trigonometry
#38
Aug9-05, 05:04 PM
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Quote Quote by shmoe
I just noticed this:



It was Littlewood who did this. There's an upper bound for where the first sign change is guaranteed to happen. I can't recall what the current best upper bound is, but it's still well out of computational reach (the first was by Skewes and was something like 10^34 digits, this has been greatly improved though).
ya, I think I read somewhere that it was about 10^10^10
shmoe
#39
Aug9-05, 06:41 PM
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Quote Quote by eljose
[tex]\chi(1/2-is)=-1 [/tex] so the modulus of the function [tex]\chi(1/2-is) [/tex] would be equal to 1.
So? I've already explained that [tex]|\chi(s)|=1[/tex] does not imply s is on the critical line.

Quote Quote by eljose
so you seem to be very clever to find "gaps" and fails in my works now i have a proof for you that is to proof that the set of functional equation...

[tex]k^{2}\chi(1/2+b-it)=\chi(1/2-b-it) [/tex]

[tex] \chi(1/2-b+it)\chi^{*}(1/2-b+it)=1 [/tex]

[tex]k^2=[\chi(1/2-b+it) [/tex] with []=modulus of the complex number

have a fixed b as it solution....and more complicate to prove that the only solutions to it for b and k are k=b+1=1....
I'm not sure I understand what you're saying, but with t=0, k=1, and [tex]b=1/2-\alpha[/tex] where I defined [tex]\alpha[/tex] in http://www.physicsforums.com/showthread.php?t=84445 we have:

[tex](1)^2\chi(1/2+b)=1=\chi(1/2-b)[/tex]

[tex] \chi(1/2-b)\overline{\chi(1/2-b)}=1[/tex]

and

[tex](1)^2=|\chi(1/2-b)|[/tex]

So this looks like a solution that doesn't have b=0.

Or do you want a k and b that hold for all real values of t? Honestly it's hard to understand what you're saying, your post is really confusing (though bravo for finally using the standard symbol [tex]\chi[/tex]). If this is indeed what you're after your second equation is saying [tex]|\chi(1/2-b+it)|=1[/tex] for all real t, so just use an approximation like

[tex]|\chi(\sigma+it)|\sim\left(\frac{t}{2\pi}\right)^{1/2-\sigma}[/tex]

as [tex]t\rightarrow\infty[/tex], valid in any fixed vertical strip (this follows from Stirlings). This asymptotic implies that the modulus can't always be 1 unless [tex]\sigma=1/2[/tex], i.e. our only possibility is b=0. When b=0 and t is real, [tex]|\chi(1/2+it)|=1[/tex] follows from the definition of [tex]\chi[/tex]. k=-1 or +1 follows from your first or last equation.

Quote Quote by eljose
i am waiting for your respones oh masterminds of mathematics....
Tell you what, if you'd prefer me to never read or respond to your posts again just ask nicely and you've got it.
shmoe
#40
Aug9-05, 06:51 PM
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Quote Quote by Jonny_trigonometry
ya, I think I read somewhere that it was about 10^10^10
Here's the mathworld link:

http://mathworld.wolfram.com/SkewesNumber.html

truns out I was off. By a lot. Skewes original bound was in the 10^(10^34) digit range, but is now a "mere" 317 digits.

The largest value of pi(x) that's been computed exactly is in the range 10^22 if this table is any indication:

http://mathworld.wolfram.com/PrimeCountingFunction.html

so it may be some time before we see an explicit crossing
eljose
#41
Aug10-05, 06:19 AM
P: 501
sorry i didn,t want to offend you i was only looking solutions to the functional equation that holds for a series of values of t (t does not include 0) b and t are both real so the equations are satisfied:( b and t are real and k>0 and 1/2>b>-1/2 so 1/2-b and 1/2+b can take the values in the interval (0,1)

[tex]k^{2}\chi(1/2+b+it)=\chi(1/2-b+it) [/tex]

[tex] \chi(1/2-b-it)\chi^*(1/2+b-it)=1 [/tex]

[tex] k=[\chi(1/2-b-it)] [/tex]

with [] meaning the modulus of the complex number,to prove my assumption would be equivalent to prove that having f the real function satisfying:

[tex] f(a+ib)f^*(c+ib)=1 [/tex] with f satisfying [tex]f(a+ib)=f^*(a-ib) [/tex] then we have that a=c

Edit:operating with the second equation i get the condition:

[tex]\Ganma(1/2-b-it)\Ganma(1/2+b+it)Cos(\ipit-\pib/2)=\pi [/tex]

but don,t know if this will be enough to prove that b=0
eljose
#42
Aug10-05, 09:40 AM
P: 501
well i have made some advances in proving riemann hypothesis the final key to the jigsaw is to prove that the equality:

[tex]\int_0^{\infty}\int_0^{\infty}dxdy(x/y)^{-1/2}(x/y)^{-b}(x/y)^{-it}e^{-(x+y)}=\int_0^{\infty}\int_0^{\infty}dxdy(x/y)^{-1/2}(x/y)^{b/2}(x/y)^{-it}e^{-(x+y)} [/tex] holds only for b=0 that,s all for the moment :)

sorry i made a mistake so this conclusion wouldn,t be valid the only conclusion i have got is that the set of functional equations:

[tex]\chi(1/2+b-it)\chi(1/2-b+it)=1 [/tex]

[tex]k^{2}\chi(1/2-b+it)=\chi(1/2+b+it) [/tex]

you can check that if b=0 then k=1 (t is different from zero) the non trivial roots of the function [tex]\zeta(1/2+is) [/tex] can be written as s=t+ib. if exist a solution of the functional equations above with b non-zero then Riemann hypothesis would be false.....

also you can check that both equation are invariant with the symmetries:

[tex] (k,b,t)\rightarrow(k,b,-t) [/tex] and [tex](1/k,b,t)\rightarrow(k,-b,-t)[/tex]
Hurkyl
#43
Aug10-05, 04:24 PM
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You do realize that those iterated integrals are separable, right? They can be written as a function of x alone times a function of y alone.
eljose
#44
Aug12-05, 06:24 AM
P: 501
we also have that -1/2<b<-1/2 and the same for -b......
mathwonk
#45
Aug12-05, 05:33 PM
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dear philosophking, in the midst of the rampant lunacy being indulged here, in post 32 I have actually tried to answer to some extent your original question, based on my reading of riemanns original paper and in the context of his works in general.

does it mean anything to you?
marteinson
#46
Aug17-05, 10:26 PM
P: 30
I have an intuition about a method of attempting to solve the Goldbach conjecture. Does any more knowledgeable person here think the following would help simplify the problem?

Instead of attempting to prove that p + q = 2n showing that all even numbers can be expressed as the sum of two primes, would it be feasible to prove each of the following three subdivided problems?

That the three series of even numbers

A) 6n

B) 2+6n

C) 4+6n

are each composed of even numbers which can be written as the sum of two primes?

I see it as likely that the following facts may help: 1 + 6n is a series composed of primes (exactly half of them) along with certain composite integers which are the product of a pair of primes, while 5 + 6n is also a series similarly composed of the other half of the primes along with composites which are the product of a pair of primes.

Furthermore, adding two primes on the series 1 + 6n always gives a sum which is a member of 2+6n, AND adding any pair of primes on the series 5 + 6n always gives an even number on 4 + 6n, whereas adding one prime from 1+6n with another from the series 5+6n always gives a sum that is a member of 6n itself. This highly restricted pattern of series membership appears to reveal a tendency which could be exploited in an attempted proof.

Finally, adding the three series A B and C above would of course give all the even numbers, so proving that each of the three respects the Goldbach rule would prove the whole conjecture.
shmoe
#47
Aug18-05, 01:21 AM
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I'm no expert on the goldbach problem, but I guarantee that those who are working on it are well aware of modular arithmetic. It's very standard to consider breaking problems up into cases of congruence classes, indeed the goldbach problem is already talking about even integers.

"I see it as likely that the following facts may help: 1 + 6n is a series composed of primes (exactly half of them) along with certain composite integers which are the product of a pair of primes, while 5 + 6n is also a series similarly composed of the other half of the primes along with composites which are the product of a pair of primes."

I'm not sure what you mean by the part in bold above. What 'certain composites' are you refering to? Just the ones that are products of two primes? There are lots of composites in these sequences that are the products of many primes as I'm sure you know.

I wouldn't use the term "exactly half of them", they have the same asymptotic density, but you don't 'usually' expect a perfect split in the respective prime counting functions like the word 'exactly' suggests (see also the "Chebyshev bias" or "primes race")


Also, why not start a new thread for this? Maybe it could be split off.


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