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eljose

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- Thread starter eljose
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eljose

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- #2

Hurkyl

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There is no "smallest positive number", and the smallest possible error term is O(0).

- #3

eljose

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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...

-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...

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- #4

Hurkyl

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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)

- #5

eljose

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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...

- #6

shmoe

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eljose said:..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...

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...

eljose said:( i know this exist i have seen proofs with results a+e being e an infinitesimal number.

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 dependence explicit, but not usually).

eljose said: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.

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...

- #7

HallsofIvy

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eljose said:

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...

? The time you take to do calculate something has nothing to do with "error" in calculating. Do you mean "how

- #8

eljose

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[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.

- #9

shmoe

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eljose said:-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.

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.

- #10

eljose

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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...

[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...

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- #11

Zurtex

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eljose said: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}(Ln\zeta(bnq))/n[/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...

: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.

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- #12

eljose

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- #13

shmoe

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eljose said: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:

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?

- #14

eljose

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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¡¡ (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).

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¡¡ (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).

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- #15

shmoe

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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?

- #16

eljose

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[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

- #17

shmoe

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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.

- #18

Gokul43201

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Why, oh why, aren't these threads in the IR section ?

- #19

eljose

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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.

[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.

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- #20

CRGreathouse

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eljose said:I have the "exact" formula for Pi(x^a)[...] in fact this method is new as it calculates Pi(x^a) instead of Pi(x).

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!

- #21

shmoe

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- #22

eljose

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[tex]\int_0^{\infty}dx\pi(x^a)e^{-st}=bR(sb)/s[/tex] it yields to a triple integral for the Pi(x^a) depending on b the bigger b is the bigger you get the error if we set O(x^d) then d=f(b) wiht f a increasing function so for bigger a (small b) the error is smaller...

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CRGreathouse

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eljose

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Hurkyl

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[tex]\int_0^{\infty}dx\pi(x^a)e^{-st}=bR(sb)/s[/tex]

It should be clear that this is not a convergent integral...

- #26

Zurtex

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You don't knoweljose said:

a) How to write a program to do the sort of things you claim you can do

b) How to calculate your own integral numerically

And you expect people to take you seriously?

- #27

matt grime

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I go away for a week and look at the fun I missed.

- #28

shmoe

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eljose said:but i have given an expression for [tex]\pi(x^a)[/tex] in the form:

You didn't seem to get it last time, so I'll repeat- all you've done is a change of variables which does not reduce the error or the time it takes to calculate these integrals (to some accuracy). At least that's the nearest I can figure that you've done- as always your presentation leaves a lot of guesswork.

I've suggested before that you go take some number theory classes at your nearest university, and I'll suggest it again now. You've been trying for I don't know how long to understand this stuff, and you're still making utterly ridiculous claims involving these error terms. For some reason you're incapable of grasping the basics and we've apparently failed at teaching you, so I suggest you go somewhere you can get some face-to-face help from an expert. You're obviously not making progress attempting to learn on your own and then ranting on an internet forum about how unfair the mathematical community is, so I humbly suggest a change in tactics.

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