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A trick to calculate Pi(a) for any a?

  1. Feb 19, 2005 #1
    Let be the integral equation satisfied by Pi(x)

    LnR(s)/s=Int(0,8)K(x,y)Pi(x) with K(x,s)=1/x(x^2-1) then we split the integral into 2 ones

    Int(0,8=Infinite)=Int(0,r)+Int(r,8) in the second one we make the change x=1/t, we will chose r so we have that after using Gaussian integration

    LnR(s)/s=Sum(j)K(xj,s)Pi(xj)+K(xk,s)Pi(a) we have chosen r so this happen

    then we take (sk) so we have the system of equations:

    LnR(sk)/sk=Sum(j)K(xj,sk)Pi(xj)+CkK(a,s)Pi(a) now from this system we have only to solve the value of Pi(a) and this is valid whatever a is

    that is whatever the integral is,using gaussian integration we can always obtain Int(a,b)f(x)=f(c)ck+sum(j)cjf(xj) only have to choose a d so

    Int(a,b)=Int(a,d)+Int(d,b) in the firs integral after making the change of variable x=(d-a)xk+(d+a)/2=c so d=2c-(1+xk)a/1+xk
    Last edited: Feb 19, 2005
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  3. Feb 19, 2005 #2

    matt grime

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    You realize that makes no sense? Seriously, I've read it 3 times and I can't understand a word of it as a piece of English. Learn Latex, please. Put in the missing steps. Explain clearly what you're trying to do. I mean, what is R? I can guess you mean the Riemann zeta function. Ln is log presumably. What are the indexing sets of the sums? Integers? Natural numbers, with or without zero? What does "solve the value of Pi(a)" mean? Tidy it up and maybe people can understand what you're getting at.

    nb I'm guessing you mean the prime counting function, by the way, for Pi(a).....
    Last edited: Feb 19, 2005
  4. Feb 19, 2005 #3
    Is not so hard to understand i only want to say i have discovered a numerical method to obtain pI(a) whatever a is by using Gaussian integration xk and xj are the roots of legendre Polinomials (as it happens in gaussian integration) only Matt you must apply Gaussian integration to the integral equation for Pi(x) is very simple
  5. Feb 19, 2005 #4

    matt grime

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    Jose, I know many things about the zeta function, and pi(x). And I'm pointing out that you are not helping yourself by continually making obfuscatory posts like this. Imagine someone wanders in and sees that post without knowing your history? What are they to make of it? Would you care to domonstrate how effective you "numerical method" is? By, say, doing something with it? For instance pi(x) satisfies




    and this lovely formula

    [tex] \pi(x) = -1 + \sum_{r=3}^{n} [(r-2)! - r \lfloor \frac{(r-2)!}{r} \rfloor ][/tex]

    Here is another Wolfram page that shows that using Mellin transforms in calculating pi(x) is well known


    here is something that shows how to use the zeroes of the zeta function to calculate pi(x)


    remember you wanted an integral where the zeroes of the zeta function were the poles?

    We are not saying what you have is wrong. But I'd ask you to re-evaluate your claims. Once you claimed to have solved pi(x), remember? And try and write it in a clearer format so that others might be able to decipher already quite tricky things (you are in this thread presuming that people inuitively know what pi, R, Ln, gaussian integration, legendre polynomials are without explicit reference to them, never mind not explicitly saying what the indexing sets of the sums are). And finally, what is the effectiveness of your algorithm, what is its cost?
  6. Feb 19, 2005 #5
    Hello thanks for the links,but they are even worse than my method (you need to know some primes) another way i am trying to do is by given the Kernel K(x,s)=1/x(x^s-1) get an operator L so LK(x,s)=d(x-s) where d is the delta function..where could i find this?..thanks. i have trie dby expanding the Kernel into a series of this kind K(s,x)=Sum(n)an(x)gn(s) where gn(s) is a set of orthonormal functions but i don,t know what to do to calculate L where L is the operator L=b0(x)+b1(x)D+b2(x)D^2
    Last edited: Feb 19, 2005
  7. Feb 19, 2005 #6


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    O.K eljose, using your method could you please compute Pi(a) for some number greater than 10^50 and show how it is computationally efficient.
    Last edited: Feb 19, 2005
  8. Feb 19, 2005 #7
    eljose - how much math education do you have?
  9. Feb 20, 2005 #8
    i,m a physicist my math education is this one received during the career...yes is not enoguh but what else can i do?..:)
  10. Feb 20, 2005 #9

    matt grime

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    Primes can be calculated, recursively if one must, using pi(x). If you look closely at the legendre method you'll see this isn't a serious problem. You're a little like James Harris who believed he had a better prime counting algorithm since he didnt' need to know the prime in advance, but "calculated them on the fly" each time he evaluated pi. Which is obviously a silly thing to do.

    This calculation of the primes costs something, obviosuly. But then you're attempting to numerically integrate things over infinite volumes at one point in one of your ideas, (or know where all the zeros of zeta are, plus the residues...).

    You need to get over the fact that what you have found is a simple relation gotten by merely applying some well known formulae to a function(*) and not actually a way to evaluate pi(a). PROOF: you've not managed to evaluate pi(a) at any point.

    The relation might be interesting, but you need to develop it further. A typical paper on something like this would be about 30 pages of densely written mathematics, typset in LaTeX, and with a bibliography, and an abstract. Then you might be able to get someone to submit it to Arxiv. But why? I mean, lots of us have many unpublished ideas kicking around.

    You also claimed your method is independent of x at one point in the pi(x)/x^4 thing. Not strictly true, since you need to bound the absolute error in the numerical integral by 10^{-4n} when x is of order 10^n. So larger x's require more accurate computation. This may indeed be more efficient than other known methods: prove if it is so, but I serioulsy doubt it. Ozlydsko has a table of efficiencies of various methods on his website. And there are tables of known primes up to huge sizes out there too.

    * I believe Wiener did this in print 50 years ago, and it is a very common technique in number theory. See for instance the proof that Selbergs L-functions of degree 2 can be characterised in simple invariancies of Hecke operators. (note, doing this from memory so may not be entirely accurate in the names)
  11. Feb 28, 2005 #10
    Here you are matt the math in latex,let be the integral equation for pi(x):


    we use an approximate formula of integration related to PI(a) where a is the point we want to calculate PI(x) function in we have


    now choosing several s_j we have a system we only have to solve this system to obtain pi(a) with K the kernel [tex]\frac{1}{x({x^s}-1)}[/tex]
    Last edited: Feb 28, 2005
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