A Steep towards a formula for Pi(x)

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Let be Pi(x) the prime number counting function,we can write the integral relating Zeta function and Pi(x) function as:

[tex]\lnH(s)=\int_{-\infty}^{\infty}dt\frac{\pi(e^{e^t})e^{t-s})}{e^{f(t-s)}-1} [/tex] with the function:

[tex]H(s)=\zeta(e^{-s}) [/tex] and [tex]f(t-s)=e^{t-s} [/tex]

so the strategy to obtain the formula would be obtain a L differential operator of the form:

[tex]L=a_{0}(x)+a_{1}(x)D+a_{2}(x)D^{2} [/tex] D means d/dx

so we would have that [tex]LG(x-s)=\delta(x-s) [/tex] [tex]G(x-s)=\frac{e^{t-s}}{e^{f(t-s)}-1} [/tex] so applying L to the Kernel of integral we would get the formula:

[tex]L[LnH(s)]=\pi(e^{e^{s}) [/tex] to obtain the operator L we would use the Green function theory
 
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  • #2
arildno
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Now, without gauging the validity of this, what would be the gain in doing so? :confused:
 
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is very simple we could write a closed formula for Pi(x) in the form:

[tex]\pi(e^{e^s})=a_{0}(s)lnH(s)+a_{1}(s)DlnH(s)+a_{2}(s)D^{2}LnH(s) [/tex]

we only need to know what a0,a1 and a2 are and we could construct the best exact formula for pi(x) (the function H(s) is defined in my first post)
 
  • #4
matt grime
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oh my god, please explain why this is 'best' in any meaningful sense...
 
  • #5
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is easy to respond to this matt:

-Pi(x) is expressed in a closed form, there is not a summation over x or a complex contour integration (you wouldn,t need to make any contour integration at all with no error term the worst it can happen is that any function a0,a1 or a2 is defined by an integral on x)..
-Pi(x) is an exact formula....
-In case there must be an integral to calculate pi(10^1000) you only would need to calculate the integral for 7.741787724 that is s=ln(ln(x)) in case there is an error term on x in the sense the error goes as O(x^a) we would have an error going like this O(ln(ln(s)) that is a better error rather than this that goes like O(x^a) for some a...
-i don,t know how to convince you this is the best exact formula that would be for pi(x)
 
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matt grime
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so a real integral is easier to work out (numerically) than one over a complex contour. right. ok, that doesn't seem a remotely reasonable assertion to make: it's still a numerical calculation for pity's sake of an integral and with an error term as ever in any numerical calculation.
 
  • #7
shmoe
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eljose said:
-In case there must be an integral to calculate pi(10^1000) you only would need to calculate the integral for 7.741787724 that is s=ln(ln(x)) in case there is an error term on x in the sense the error goes as O(x^a) we would have an error going like this O(ln(ln(s)) that is a better error rather than this that goes like O(x^a) for some a...[/tex]
You've made a claim like this before and it doesn't make any sense. The usual integrals expressing pi(x) are on an infinite contour, you aren't calculating them "up to x", the contour is the same irregardles of x. Your change of variables will not make the relevant integrals easier to calculate.

By the way, why does an "exact formula" have an error term? Do you mean the time it takes to compute pi(x) with your formula that you don't seem to even have yet you're already making some great claims about?

eljose said:
-i don,t know how to convince you this is the best exact formula that would be for pi(x)
It's simple- produce the formula you are declaring the "best" then prove it's correct and it's run time is asymptotically better than the current ones.
 
  • #8
Hurkyl
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And if that isn't feasible, an empirical demonstration would at least lend some credibility to the claim.

For example, you could take a selection of values of many different sizes, and then time how long Mathematica takes to compute π with its built-in function, vs how long it takes Mathematica to compute it with your implementation.
 
  • #9
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first of all i have not said that i get it ( i only spoke about the fact that i would like to do my phD in this subject)....

Yes my formula would be exact (if we could calculate the a0,a1 and a2) but unfortunately this includes calculating [tex]ln\zeta(e^{-s}) [/tex] and this only can be made by numerical methods that include an error term (is the only numerical calculation apart than for n=0,1 or 2 could have that:

[tex]a_{n}=\int_{a}^{x}dtf_{n}(t) [/tex] than can happen for 0,1 or 2
 
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  • #10
shmoe
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eljose said:
first of all i have not said that i get it ( i only spoke about the fact that i would like to do my phD in this subject)....
So you thought you'd make same claims about how great a formula is without even having it? This is a step of speculation beyond your usual.

eljose said:
Yes my formula would be exact (if we could calculate the a0,a1 and a2) but unfortunately this includes calculating [tex]ln\zeta(e^{-s}) [/tex] and this only can be made by numerical methods that include an error term (is the only numerical calculation apart than for n=0,1 or 2 could have that:
Sure, but what you've said about an error term still makes no sense. If you can compute pi(x) with an error term that's O(log(log(x))) then this still goes to infinity as x does and you'll be hard pressed to convince anyone that this is an "exact formula".

If you hope to calculate pi(x) exaclty, then your error term had better be +/- 0.5. For such an exact algorithm, what is relevant is the time it takes to achieve this error. This is what you would have to show is asymptotically superior in order to show you've got the "best". Alternatively show it's at least better than current methods up to a range that's still interesting and would be used in practice (the best asymptotic algorithms are sometimes inferior in the ranges used in practice, so a better "working" algorithm is still very interesting). Or do a practical comparison as Hurkyl suggests.
 
  • #11
Hurkyl
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Incidentally, I tried some of your older integral formulas today... Mathematica's numerical integration tool couldn't integrate them, even for small arguments. (At least, when used straight out of the box)
 

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