Integral equation of second kind and prime number counting function

[eljose79]

in this postcrpit i would like to say i have fund a second order integral equation (fredholm type) for the prime number counting function in particular for Pi(2^t)/2^2t function being Pi(t) the prime number counting function,teh equation is like this is we call Pi(2^t)/2^2t=g(t) then we have

g(s)+LnR(5s)/5s=Int(1,Infinite)K(s,t)g(t)dt for a certain symmetrical Kernel K(s,t) see the paper adjoint this Kernel is:

K(s,t)=nexp(-n^2(t-s)^2)+2^2(t+s)/2^5st-1

R(s) is Riemman,s Zeta function
with this calculating the prime number counting function is obvious,...just solve the integral equation by some approximate method to get Pi(2^2t)/2^2t

my results give an expresion for Pi(t) as

Pi(t)=Sum(n)a(n)fn(log2(t))

being fn(t) a set of orthonormal function...

yes,i treid to submit to journals but the snobbish referees (which only want famous names in the papers) did not give me a chance.

 

[matt grime]

of course they'll run a mile if you say things like 'can be approximated' we already know how to do that thanks, and you're not actually giving an analytic solution which is what you're implying you've got.

 

[eljose79]

yes you are right we know Pi(x) for big x,also exact formulae for PI(x) exist but you need to know some primes or are imposible to calculate for big x,if you want to take a look at my work is at:

es.briefcase.yahoo.com

Yahoo ID:(user name) josegarc20042005
contraseña:(password) eljose

is inside the folder "Mis doucmentos" (my documents) with the name pifunction.pdf hope it is interesting for you

in this work i present a new approach but solving a second kind fredholm equation by using orthonormal series

 

[matt grime]

So now you're just saying you've found another approximation for Pi(x), so is yours better than all the others, does it genuinely involve new mathematics, is it of interest from the research community, have you found someone to communicate it to the journal on your behalf, what did the rejection letter actually say, why aren't you prepared to just submit it as your thesis for PhD which i recall your doing, there's no pressure to publish unimportant results in pure, this isn't stats so why push it, why not accept it might not actually be very good rather than without corroboration insult the journal, if the previous standard of presentation was anything to go by it won't meet the standards required for publication. Do you know that hasn't been thought of before? Because it doesn't appear as though your working in a new area with a new technique (the ideas involved appeared in the 30s didn't they?)

 

[eljose79]

-Yes you,re right i am not using new math,but with old mathematics i have got an expression (in a Fourier series) for Pi(x), you don,t need toknow any prime or use a very complicate expression for gettting Pi(x), yo don,t need to know even number theory only you need to solve a simple second kind fredholm equation,tell me where i can find similar examples and i will agree you,i am the first in solving prime number counting function by means of Fourier series,that,s completely new (if i am wrong tell me a Fourier series for pI(x) please)

-I submitted my manuscritp to several math teachers at my university,but i did not have a response at all...this is what i call to be an snob,i offered myself to do Phd but i got no reply what opinion should i have about these people?.

I claim to have solve the Pi(x) function for any x this is also new by mean of a series
at least gausss,Newton and others were given an chance to be known, i suppose that if my work was made by ...(put a famous mathematician name here) it would have appeared in press,television,and other mass media telling us how inteligent this person is... but when you are not famous you have nothing to do.

 

[matt grime]

pi(x) doesn't have a Fourier series, it is defined on the whole of R and not periodic.
have you proved your 'series' is even asymptotic to pi, and is a better approximation than li(x)?

any paper submitted to a journal must conform to the house typesetting standard of the journal. try writing it out in free hand in your first language and then talking to the mathematicians at university rather than sending them a pdf like that.

 

[eljose79]

pi(2^t)/2^2t would have a series of orthonormal function in (1,infinite).

I,m not a mathematician (don,t have a degree in math) but a physicist,perhaps it sounds funny but real, so i do not know any math teacher in the branch of number theory,pehaps i will send them a manuscript in my mother language (spanish) by letter explaining my ideas ,because i do not know them.


Anothe curious formula for the mobius function can be derived as two inverse mellin transform.

M-^1(1/R(4-s))
---------------=moebius function
M-^1(R(4-s) where R(s) is the Riemann Zeta function.

 

[matt grime]

So, if you've "solved pi(x)", why don't you simple tell us what pi(10,000,000,000,000,000,000) is? or have you in fact just reduced this to another equally hard question? For instance, i can calculate pi(x) for any given x without knowing a single prime number using a recursively defined function such as Legendre's or Meissels.

And whilst you may have a series solution, it is not a Fourier series, which is a ver specific type of series and not just any old series whose terms are orthogonal wrt to an arbitrary inner product.

as to one of your earlier requests i suggest you look up kolmogorovs three primes proof to see where series expansions are used in number theory - they are very common, and shouldn't be beyond your abilities to find (for chrissake the zeta function!)

 

[eljose79]

Well i think the merit or succes of my method is that i have reduced the problem of calculating Pi(x) to the problem of solving a second order Integral Fredholm equation with a given Kernel K(s,t) there are not tricks here,you don,t need to know any prime or any strange function, only solve an integral equation by any means...is this is imposible then i will have to admit i,ve failed.

I can not calculate Pi(100000000000000) or any value by myself, because i only know how to solve an integral equation by means of series.

 

[matt grime]

oh, so you haven't actually done anything other than create something possibly harder to evaluate? and despite not knowing how to calculate anything you felt that you could claim you'd solved something? that's a large claim to make without any evidence. no wonder you've been dismissed by those mathematicians.

as i keep pointing out this kind of method is not new, and there are plenty of different ways of calculating pi(x) without needing to know the prime numbers before hand. they are all of limited computational use. moreover you've not shown how many terms of your series it's necessary to calculate in order to know pi(x) for any x.

you could start by tidying up the presentation of that article and putting doing it in latex for instance so that we can make a better judgement of it.

 

[eljose79]

My method can calculate Pi(x) for any x by means of a series expansion in orthonormal functions, i also give the form of coefficients (an integral involving LnR(t) being R(t) the Riemann Zeta function i avoids the problem of computation,..the question is ¿do you know how to solve an integral Fredholm equation of second kind? is the question is "yes" you will agree with me in the utility of the method..why do you think my method can not be useful?

If you don,t like this method i have found another integral representation

Pi(x)=1/2pi.iInt(c-i8,c+i8)dq1/(4-q)x^4-qInt(0,8)dnn^2M^-1(R(4-s))LnR(4n-nq)
this comes from the representation for the moebius function as

Moebius=M^-1(1/R(4-s))
--------------- M^-1 is the Mellin inverse transform
M-^1(R(4-s)

 

[matt grime]

i think your method is not useful simply because you have been completely unable to use it yourself. There are lots of "prime counting functions", well, strictly speaking there is only one, but there are different implementations of it, and they all benefit from the fact that they can be computed, something you have singly failed to do. Thus we can say there is no untility in your theory since there is no utility in it. Prove me wrong, use it to calculate pi(x) for some (large) x's and we might believe you've got a computationally efficient algorithm.

 

[eljose79]

I think my method is good to express Pi(2^t)/2^2t in a series of orthonormal functions,do you think is it useless?... in fact we have

Pi(2^2t)/2^2t+CLnR(5t)2^2t/t=Sum(l)Al,mrlFm(t) where Al,m are the Fourier coefficients of expressing the Kernel K(s,t) in a double Fourier series, Fm(t) is a set of orthonormal functions and rl can be obtained by solving the linear system.

Bm+rm=Sum(l)Al,mrl =system of linear equation.

from these we could calculate rl and Pi(2^2t)/2^2t

Another question..do you think my version of moebius function in terms of

moebius(x)= M-^1(1/R(4-s))/M-^1(R(4-s)) is true?..if so i claim to have solved pi(x) as an integral.

 

[matt grime]

That is a complete misuse of "solved": you have found an expansion (allegedly) in terms of some other sets of functions (I don't recall you posting here the formlae for the set of orthonormal functions). If you're equation is any good, then use it, go on, demonstrate that you can calculate pi(10,000,000,000) in a reasonalbe number of steps with some accuracy. A simple challenge.

So, to summarize, again:

1. that is not what solved means
2. demonstrate its utility
3. write down the basis of this orthonormal system
4. write down the estimate of the error in truncating the sum of orthonormal functions after N terms.

 

[eljose79]

You,re right Matt perhaps i have not solved anything, i only gave an Integral equation of Fredholm type (second Kind) but i do not know how to solve it or to obtain the error,perhaps i made a mistake but i think my work could still be interesting as i have reduced the calculation of Pi(2^t)/2^2t to solve an Integral equation by any means,there are lots and lots of method to solve this integral equation, now i am working in another method to obtain Pi(x) using dirichlet series inversion formula .

 

[matt grime]

You ought to find out if you've even made the problem any easier, and the other work that's been done on this. Use Google wisely.

 

[matt grime]

here's something else to consider.

you've done this expansion, right? now in order to work out coefficients in an expansion like this, you need to do an integral, right? so what is the integrand. reading back over your pdf it isn't clear, mainly because of the fact yo'uve not doen this in a suitable latex type program. it appears that you've magically managed to omit the prime counting function itself from the integrand (you apper to only have log(R(s)) where R is teh Riemann Zeta Function, though it is hard to tell: you need to improvee the presentation.

you can post latex here in this forum. so can you clarify what the integrand is, and where the g(s) has vanished to.