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Pi(x) function in number theory solved

  1. Jul 4, 2005 #1
    Here you are the best formula to calculate Pi(x) by means of a 4-dimensional integral:

    [tex]\pi(t)=\frac{1}{4\pi^2}\int_0^t\int_{c-i\infty}^{c+i\infty}\int_{d-i\infty}^{d+i\infty}\int_0^{\infty}dxdsdqdn\frac{n^{-s+2}x^{q-1}LnR(qn}}{R(4-s)}[/tex]

    Where R(s is the Riemann Zeta function.....
     
    Last edited: Jul 4, 2005
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  3. Jul 4, 2005 #2

    matt grime

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    please elaborate: in what sense that is the "best"? do you for instance mean that of all the 4 dimensional integrals it is the "best" or of all methods of evalutaing pi it is the "best", and in either case "best" means what?

    since you state "calculate pi" i presume you mean the fastest method. please prove that it is the fastest known method.

    it is undoubtedly an interesting relation, but how useful or original is it?
     
    Last edited: Jul 4, 2005
  4. Jul 6, 2005 #3
    I think is better becasuse always require the same number of operations,let,s use a random number generator to generate random number on the interval

    [tex](0,t)x(-\infty,\infty)x(0,\infty)x(-\infty,\infty)[/tex]

    where we have made the change of variables s=c+iu and q=d+iv

    then we generate N numbers (as t is differente each time the numbers will be different but always the same N) and compute the integral by Montecarlos,s Method the error wil be 1/N^(0,5) if we take N=100 poiints the error will be 0,1,but as we know Pi(x) is an integer we don,t need to have a lot of accuracy,the number of terms will be:

    N+Sum(1,N)
     
  5. Jul 6, 2005 #4
    It doesn't always require the same number of operations because you are integrating up to variable t, which can vary (hence the term "variable"). In fact, because you will need to integrate numerically it may take even more operations to get a good approximation than a summation would.
     
  6. Jul 6, 2005 #5

    Zurtex

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    What sort of accuracy and run time would this require to calculate something like Pi(10^100)?
     
  7. Jul 6, 2005 #6
    To Zurtex: i have no computer program to calculate montecarlo method so if they published my work (in fact if i were famous perhpas would win the field medal with this work) we would know the time required..

    To Keebs: i have taken the 4-dimensional volume to be [tex](0,t)x(-\infty,\infty)x(0,\infty)x(-\infty,\infty)[/tex] and in this 4 dfimensional volume we choose N points,yes the volume varies as t varies,so the random points will be different each time we want to calculate PI(t) but the number of points N is always the same.
     
  8. Jul 6, 2005 #7

    Hurkyl

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    That is necessarily false. For example, the time it takes to simply write the answer is unbounded.

    Let me restate:

    Suppose you say that you can compute pi(n) in N operations. I can demonstrate the existance of some number, m, such that it takes more than N operations to simply write pi(m), let alone compute it.
     
  9. Jul 6, 2005 #8
    But as t increases the volume increases, thus a computer will require more time to compute the integral.
     
  10. Jul 7, 2005 #9
    I don,t think there is a big problem with my integral,just use Montecarlo method to calculate it or simply use another method...

    Hurtkyl can you prove your statement?..(mathematically) and according to your post any method used to compute Pi(x) will require "unbounded" time.

    my method is also better as gives you the Pi(x) in an integral form,so you can apply the numerical method of calculating integrals to it..whereas the others require to calculate a series that if n is big to compute Pi(n) you will need to make at least a sum sum(1,n)
     
    Last edited: Jul 7, 2005
  11. Jul 7, 2005 #10

    matt grime

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    oh dear, and i was hopi9ng to give you the ebnefit of the doubt this time since i don't like to discourage people from investigating mathematics but for pity's sake egt a grip.

    hurkyl can prove his claim since there are an infinite number of primes, surely a fields' medal winner can see that? he is simply stating that there is no constant T such that pi(x) can be evaluated in time less then T for all x. which is what you are claiming can be done. proof. suppose that printinig a single digit takes t seconds then there are at most T/t printiable digits that may appear in pi(x) implying pi(x) is bounded, however pi(x) grows without bound.

    right, you have applied an inverse transform to a function. no biggee. numerical integration is exactly the same as summing a series so claiming your representation is better is a tad off the mark to say the least. until you can get this in perspective you will remain shunned by those mathematicians who have declined to give your thoughts time in the past. how can yuo honestly claim to have a better way of calculating pi(x) without actually knowng the first thing about complexity of algorithms? this is maths, you have to prove your claims especially ones that are apparenty nonsensical.

    you do not need a computer to estimate the order of the algorithm
     
  12. Jul 7, 2005 #11
    Thanks for your replies,according to you and Hurkyl then all methods involving calculating pI(x) require bigger time as x increase....

    My method is new (is the first method to calculate pi(x) involving an integral and not a sum) so i think is worth publishing,the problem perhaps is that i am not famous and can not compete with these snobish teachers of important universities publishing nonsenses and useless things , sorry i won,t comply anymore...

    another question let,s suppose that the integral equation for Pi(x) could be approximate by numerical methods by:

    [tex]\pi(a)K(a,s)+\sum_jK(s,x_j)\pi(x_j)=g(s) [/tex]

    then we could set a system of equation to calculate Pi(a) am i wrong?....
     
    Last edited: Jul 7, 2005
  13. Jul 7, 2005 #12

    shmoe

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    No it isn't-expressions of pi(x) in the form of an integral are not new at all. For computationally useful algorithms of this type, there's a paper by Lagarias and Odlyzko called Computing pi(x): an Analytic Method from the 80's (there are probably a few others out there as well, but afaik this is the first to provide a potentially useful method). An electronic roughish copy is on Odlyzko's website if you don't have access to the journal http://www.dtc.umn.edu/~odlyzko/ . Read it. Their integral is not only simpler, but you can see what a full on run-time analysis of something like this looks like as well as proofs of correctness. You can also find on his website some papers on approximating values of [tex]\zeta(s)[/tex] (for cripes sake, why do you still refuse to use the standard notation and wonder why people can't be bothered to take you seriously?), something that can not be done to an arbitrary accuracy in a constant amount of time (another roadblock in your claimed constant runtime).

    Get your head out of the clouds- Odlyzko has not only trumped what you claim to have done, but has a large body of work involving numeric techniques and other nifty results. Guess what? He doesn't have a fields medal either and he extremely well respected.
     
  14. Jul 7, 2005 #13

    Zurtex

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    I know nothing about number theory this advanced, but I've seen many people on this forum alone post Pi(x) represented by an integral.

    But at the end of the day an integral is just numerically calculated by a sum and the amount of run time and error on a computer needs to be calculated otherwise it's pointless just saying you have this fantastic integral that you've not even shown an example of how it works.
     
  15. Jul 7, 2005 #14
    Yes Shmoe i know i should change it to [tex] \zeta(s) [/tex] i will edit my equation...

    i checked teh webpage you gave me but i can not find the integral for PI(x) would you mind putting it on your reply?..thanks, i would be interesete d in watching the form of this integral representation of Pi(x)
     
  16. Jul 7, 2005 #15

    shmoe

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    Look under his "computational number theory" category of papers.
     
  17. Jul 7, 2005 #16
    Now my methdo would also be valid to calculate a(n) being this function n iff n is prime and 0 otherwise,now we have only 2 values and the expresion would be:

    [tex]a(t)=\frac{1}{4\pi^2}\int_{t-1}^t\int_{c-i\infty}^{c+i\infty}\int_{d-i\infty}^{d+i\infty}\int_0^{\infty}dxdsdqdn\frac{n ^{-s+2}x^{q-1}LnR(qn}}{\zeta(4-s)}[/tex]

    now the volume is alwsays the same and not depending on x so the number of operations required to calculate a(n) would be always the same..
     
  18. Jul 7, 2005 #17
    And to calculate an integral of Pi(n) you will need many more terms than sum(1,n) if you're going to use integrals, because calculating integrals numerically requires splitting up the curve into little tiny rectangles, finding the area of all of the rectangles, and then summing up all of the little areas.

    Haha, you know all of those summation formulas for pi(x), right? All of them can easily be transformed into an integral form.

    Nope, as t gets bigger you're going to require more operations to perform the multiplication and addition of the bigger values.
     
    Last edited: Jul 7, 2005
  19. Jul 7, 2005 #18

    shmoe

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    The higher the t, the larger the magnitude of the x term in the integrand, so you have to calculate zeta more accurately to get the integrand to a necessary precision, and this is not free. This is not going to let you calculate a(t) in constant time (wrt to t). (this is assuming that your expression is even correct)
     
  20. Jul 8, 2005 #19
    To shmoe,i have checked all the articles but found no integral representation fro Pi(x) they only speak about calculating Riemann,s Zeta function could you write the integral representation or tell me what,s the name of the paper where the integral appears?..thanks in advance shmoe....

    another question is the precision...we now that a(n) will be 0 or 1 so why we need such a high precision?... the same problem would happen with x/ln(x) so why this formula is accepted

    So the total time will be= time required to calculate Riemmann,s function with accuracy + tiem to calculate an integral with fixed volume (and as th evolume is fixed using montecarlo,s method with always the same points will be enough....
     
    Last edited: Jul 8, 2005
  21. Jul 8, 2005 #20

    Zurtex

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    Have you ever tried to approximate an integral by a numerical method? Have you ever tried to work out the error generated by a method of approximating an integral my numerical methods?

    Because I have and I can tell you that when you put it on to a computer you have to do a lot of working to reduce error and keep runtime to a minimum. Even integrating a fairly simple rational function required looking at areas where the error would be quite high and ways of reducing it.

    But your representation of Pi(x) is even worse, not only do you have the error of approximating the integral but you also have the error of approximating the functions and at least the Zeta function is a pain in the ass to approximate.

    Perhaps if you gave us an example of how easy you think this representation of Pi(x) is then that would help us understand what you think is so great about your method. Perhaps start on a low number like Pi(20) and show how it is the same run time of another relatively small number like Pi(10000000).
     
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