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Prime number counting function with error O(x^e)

  1. Sep 25, 2005 #1
    i have discovered a formula for [tex]\pi(x^{a})[/tex] being Pi the prime number counting function in terms of a triple integral....but you wll say..this is already made what is this good for?..in fact if you knew [tex]\pi(x^{a})[/tex] with a total error O(x^d) by setting a=Ad and making A--->oo (infinite) the total error would be e=1/A O(x^e) with e the smallest positive number,or expanding in e powers to lineal order x^{e}=1+eln(x), so i have discovered a formula with the smallest possible error term...and this is completely new...:)
     
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  3. Sep 25, 2005 #2

    Hurkyl

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    There is no "smallest positive number", and the smallest possible error term is O(0).
     
  4. Sep 25, 2005 #3
    then i made a mistake in it sorry...i mean we can always choose and a for [tex]\pi(x^{a})[/tex] so the error term goes like this O(x^e) with e can be chosen to be the smallest number it ocurs to us (for example e=10^{-100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000} i know i,m exagerating but this is the sense of my proof,i have reduced the error term to O(x^e) with choosing a=dA and making A tends to infinite the error term would go like this O(x^e) with e an infinitesimal number ( i know this exist i have seen proofs with results a+e being e an infinitesimal number.

    -The formula for [tex]\pi(x^a)[/tex] is a triple integral..how would we calculate it?..the solution is to use Gauss-Hermite Quadrature formula so the integral becomes:

    [tex]\sum_{i,j,k}C_{i,j,k}W(x_i,x_j,x_k)F(x_i,x_j,x_k)[/tex]

    where the C.s are constants, W is a weight function in the form W(x,y,z)=exp(-x^{2}-y^{2}-z^{2}) and F is the integrand of the our triple integral....in fact is a complex integral so we first have to make the change of variable s=c+iu , q=d+iv (the i,j and k indexes are summed over the roots of Hermite Polynomials)...

    -Of course the smallest error term would be O(0) but with my method the error goes like O(x^e) for very small e it is almost like 1+eln(x),that is if we ignore the lineal term,the error would be constant for every x, including the error lineal term the error would go like ln(x) and as far as i know the smallest error term for prime counting function is O(x^{2/3}ln(x)) mine is smaller...
     
    Last edited: Sep 25, 2005
  5. Sep 25, 2005 #4

    Hurkyl

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    I'm suspicious, because my attempts to evaluate your previous integrals with Mathematica failed, returning with some astronomically huge value, which suggests that the integrals, in fact, do not converge.

    But those worries aside, what is the error term for this numerical integration method? Can you find out how many terms you must add in order to get a sufficiently good upper bound on the error term? How large do the numbers in the summands become?

    (And it would be nice to know how many bits of precision must be used to carry out the arithmetic)
     
  6. Sep 26, 2005 #5
    i can not calculate the error it will be in general O(x^d) with d=a+b+c
    a=number of time used to evaluate the integral.
    b=time used to write and evaluate the [tex]\zeta(s), ln\zeta(s) [/tex]
    c=time to write [tex]\pi(x^a)[/tex]
    the error term in doing the integration is calculated as the error term in Gauss Hermite quadrature formula,it will be in general the product of the errors in calculating each integral.....
     
  7. Sep 26, 2005 #6

    shmoe

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    This looks like nonsense to me. If you can find [tex]\pi(x^{a})[/tex] with a total error O(x^d) then your a and d are fixed. As this A changes, your error analyisis will no longer be correct. Before you tried to do some kind of change of variables to make the error look smaller, but actually did nothing at all, is this what you're doing again?

    then there's the whole "smallest positive number" problem...

    No you haven't, not in any number theory paper at least. You'll often see an error like [tex]O(x^{1+\epsilon})[/tex] and they say you can take any [tex]\epsilon>0[/tex]. They are not saying that [tex]\epsilon[/tex] is an "infintessimal number", just that the big O bound of that form holds for any possible fixed [tex]\epsilon>0[/tex] that you like, though the big O constant may depend on epsilon (sometimes they write [tex]O_\epsilon (x^{1+\epsilon})[/tex] to make this dependance explicit, but not usually).

    So you're saying the more time you spend on calculating the integral (presumably to get a more accurate value for it), the larger your error term? That sounds bad...
     
  8. Sep 26, 2005 #7

    HallsofIvy

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    ??? The time you take to do calculate something has nothing to do with "error" in calculating. Do you mean "how many times" you do something rather than "time used" or "time to"?
     
  9. Sep 26, 2005 #8
    But you have the problem of calculating a complex triple integral...in general the error of calculating a triple integral will be of the form O(x^d) with d a constant fixed for every integral i mean..to calculate a triple integral in the form:

    [tex]\int_{\Re}dxdydzF(t,x,y,z)[/tex] the error term will be like O(t^d) will depend only on the value t inside it...

    -In the case a and d were related we would have a=f(d) so we could choose a acurately so d is the smallest possible quantity so the error will be the smallest possible number it ocurs to us.
     
  10. Sep 26, 2005 #9

    shmoe

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    That's swell, but then a has changed as well, and your end result is that you've done nothing.

    Let's suppose you know [tex]f(x)=O(x^{3})[/tex] (for some function f). You then know that [tex]f(x^{1/3})=O(x)[/tex]. This doesn't mean you've found a smaller bound for your function f.
     
  11. Sep 27, 2005 #10
    the qeustion is that i have given a formula for [tex]\pi(x^a)[/tex] with error O(x^d) so let,s suppose we want to calculate [tex]\pi(10^{1000})[/tex] then set a=1000 and x=10 so the total error would be O(10^d) that is the advantage of my method.... if you want to judge the formula by yourselves it is:

    [tex]\pi(x^a)=\int_{c-i\infty}^{c+i\infty}\int_0^{\infty}\int_{d-i\infty}^{d+i\infty}dqdnds\frac{x^{q}n^{s}}{qs\zeta(s)}D_{n}(n^{-1}Ln\zeta(bnq) [/tex]

    where b=1/a this is the triple integral we must calculate if we make the
    change of variable c+iu=q d+iv=s the limits of integration turn real...
     
    Last edited: Sep 27, 2005
  12. Sep 27, 2005 #11

    Zurtex

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    :rofl:

    "d=a+b+c "

    You say a = 1000 here, now assuming b and c are both 0 or greater that means your error term is:

    O(10^1000) or more and as your calculation is of Pi(10^1000) that means your error term is even greater than the value of Pi(x) your supposed to be calculating :rofl:


    Or assuming you mean runtime, really hard to tell from your posts, such huge amounts of run time are totally useless.
     
    Last edited: Sep 27, 2005
  13. Sep 27, 2005 #12
    No zurtex sorry it was my fault..when is put d=a+b+c a is not the power of the [tex]\pi(x^a)[/tex] d is the error made in calculating the Riemann Zeta function and the triple integral......
     
  14. Sep 27, 2005 #13

    shmoe

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    So you're saying that you are free to choose a to be whatever you like, apparently not affecting d or the error term at all? This is as ridiculous as having an error term in the first place for what you're also claiming is an exact formula.

    Again you're missing the distinction between runtime and an error term. If you have some kind of exact algorithm, then there's no error term. What's interesting is how long it takes the algorithm to run (usually measured in number of some kind of operation, like the number of multiplications).

    Maybe you're claiming that for a fixed run time you get a certain error. If this is the case, then there is no way that your error is independant of the size of the argument, which is what you'd be claiming.

    Why do I get the feeling that this is going to be a complete repetition of your earlier threads on this?
     
  15. Sep 27, 2005 #14
    I have the "exact" formula for Pi(x^a) in the form:

    The following code was used to generate this LaTeX image:



    [tex]\pi(x^a)=\int_{c-i\infty}^{c+i\infty}\int_0^{\infty}\int_{d-i\infty}^{d+i\infty}dqdnds\frac{x^{q}n^{s}}{qs\zeta(s)}D_{n}(n^{-1}Ln\zeta(bnq)) [/tex]

    but unfortunately this integral can not be made exactly and also zeta can not be calculated exactly (only an approximation to it) so in general there will be a running time in the form O(x^d) (the bigger x is the most difficult is to calculate it) but we have calculated Pi(x^a) instead of Pi(x) so the true error term will be O(x^{d/a}) the running time needed to calculate the integral will depend on x....

    the main advantage of my method is that you calculate Pi(x^a) so you need to calculate a low x a put a big a for example a triple integral would require a running time O(x^d) if you want to calculate Pi(10^1000) put x=10 and a=1000 so the error would be O(10^d) i don,t understand why math referees can not understand this method is better than other analytic method to calculate Pi(x) known..¡¡i really don,t like mathematicians and their nasty rigour¡¡ :mad: :mad: :mad: (Hurkyl don,t get angry or ban me,but sometimes with they saying rigour and rigour they get on my nerves,in fact this method is new as it calculates Pi(x^a) instead of Pi(x).
     
    Last edited: Sep 27, 2005
  16. Sep 27, 2005 #15

    shmoe

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    Do you even know what you're claiming anymore? I sure don't, in one breath O(x^d) is the runtime, the next an error term. Fill in the blanks please:

    Using this triple integral, you can calculate pi(x) with error _____ in _____ time. No unspecified constants please, i.e. no "d"'s allowed, no exponent a, this is a question about pi(x).


    I'll say again, there is absolutely no way that your error term and run time are independant of the argument of the prime counting function. What will it take to get this nonsense kicked out of this forum?
     
  17. Sep 27, 2005 #16
    -as you can see the power a of [tex]\pi(x^a)[/tex] appears as b=1/a in the form:
    [tex]ln\zeta(bnq)[/tex] so the bigger b is the higher the error in calculating the integral will be but b=1/a so big b means small a,so the smallest error will get for a big (the bigger the a is the better,and the smaller error) so choosing a--->oo we will get a very small error
     
  18. Sep 27, 2005 #17

    shmoe

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    This is the same integral you had before right? But with a missing constant and your old [tex]e^{st}[/tex] term, and a mysterious subscript n in the [tex]D_n[/tex] notation? Where did this b in the log come from? Some sort of change of variables that you aren't going to bother mentioning?

    You're going to have something of the form [tex]x^{s}[/tex] in your integrand if you're trying to find pi(x), it comes from Perron's (this was your [tex]e^{st}[/tex] term, for some reason you wanted to use x=e^t). Maybe you're trying to hide this in your x^q term, but that doesn't seem dependant on a... No change of variables will let you escape the fact that as the argument of pi grows, so does the absolute value of this x^s term, either increasing your error or the time it takes to calculate the integral to a given precision.

    A simple reality check is in order. Do you really believe that you are capable of computing pi(x) to within a constant error in a constant amount of time (that is neither depend on x)? Honestly? Actually you seem to be making an even more ludicrous claim that as x grows (in your notation-a is growing but x is remaining fixed), the error is actually decreasing.
     
  19. Sep 27, 2005 #18

    Gokul43201

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    Why, oh why, aren't these threads in the IR section ?
     
  20. Sep 28, 2005 #19
    the b comes from the fact that:

    [tex]s\int_0^{\infty}dt\pi(e^{at})e^{-st}=sb\int_0^{\infty}du\pi(e^u)e^{-sbt}=sb\int_1^{\infty}dr\pi(r)r^{-sb-1}=bR(bs) [/tex]

    where R(bs) is equal to [tex]\sum_p{p^{-sb}}^[/tex] and the sum is extended over all primes,and the R(sb) function is related to [tex]ln\zeta(bns)[/tex] where b=1/a.
     
    Last edited: Sep 28, 2005
  21. Sep 28, 2005 #20

    CRGreathouse

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    Please calculate the following with your [tex]\pi(x^a)[/tex] formula:
    1. [tex]\pi(10^{22})[/tex] ([tex]x=10^{22},a=1[/tex])
    2. [tex]\pi((10^{11})^2)[/tex] ([tex]x=10^{11},a=2[/tex])
    3. [tex]\pi((10^2)^{11})[/tex] ([tex]x=10^2,a=11[/tex])
    4. [tex]\pi(10^{22})[/tex] ([tex]x=10,a=22[/tex])

    Thank you!
     
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