Has the Pi(x) Function Challenge Been Conquered?

[eljose79]

Pi(x) function solved?...

I recently sent a paper to arxiv.org in the link

http://arxiv.org/list/math.GM/recent

You can find a title "An approach to Pi(x) function" hope it can be interesting i think i hve found the solution to the problem of getting PI(x)

 

[eljose79]

By the way the link is:

http://arxiv.org/ftp/math/papers/0402/0402259.pdf

and i would like to thank arxiv.org webpage for giving this oportunity, also thanks to the snobish journal that rejected this works and the teachers of my university whom i sent this work and i did not get any answer at all...thanks for being so snob and arrogant.

No the die is cast time will say is my formula is correct.

 

[matt grime]

A quick glance at your paper reveals one key issue. you do not produce a formula for pi(x), merely an estimate without any attempt to find the error term. As you yourseld say, your ideal formula is self referential - it would require you to know pi(x) to compute anything.

And that doesn't even require one to decide if any of the mathematics is correct, which may, or may not be true, and is quite hard to decide becuase you don't state clearly what's going on.

Do you want a list of the misprints, typos etc too?

 

[suyver]

I really like your abstract:

"In this paper i pretend to give an approach to the Pi(x) function ..."

 

[eljose79]

To Matt Grime:

First of all thanks for your reply in fct my paper is an approach to the Pi(x) function , yes you know to know some values of it Pi(1),PI(2) ...pi(n) and so on to the formula would be self referential but i have applied a series transform to make the series converge faster let,s see the n-th term goes like ^D**n(a0)exp(-sn)/1+exp(-s) as n tends to infinite Pi(n) diverges like n/ln(n) whereas exp(-sn) tends to 0 so nexp(-sn) tends to 0 with s>0 so nexp(-ns) goes to 0 for big terms.

In fact as i have just said before..this is ONLY an approach to Pi(x) function to know the Pi(x) exactly yhou would need to know all the primes but perhaps with my solution with knowing the values for x<1000 of Pi(x), which is a simple task you would have the Pi(x) function and a(n) function a(n)=1 iif n is prime or 0 if composite so you would only need to introduce a number into the expresion of a(n) and see if is 1 or 0 to know the primes..

 

[matt grime]

So you do not produce any solution to pi(x), merely some approximation. Perhaps, then, saying 'pi(x) solved?' should be rephrased as 'another approx to pi(x)'

As such it doesn't meet the standards required for a journal as you do not demonstrate that you've done anything new. Perhaps proving it better that li(x) in some sense would be a start. However, I see nothing new or novel in this approach even then, and the presentational standard is way below that required. You shouldn't take journal rejection so seriously, it happens a lot to lots of very good mathematicians.

In short, in order to compute pi(x) you need to know pi(x), can you really not see why that doesn't help?

 

[eljose79]

As an example i will tell this, let be the integral equation

f(x)-g(x)=Int(a,b)K(x,y)f(y)dy now let,s take instead of integral a sum

f(x)-g(x)=sum(yj)K(x,yj)f(yj) or f(x)=g(x)+sum(yj)K(x,y)f(yj)

to solve all f(yj) just form the system of equation

f(xj)=g(xj)+sum(yj)K(xj,yj)f(yj) now you have solving the f(yj) solving this system...do you need to know the f(x) to get an approach?..no the same thing happens with my formula.

Solutions for Pi(x) are needed however they need a lot of computations whereas my forumla gives an integral representation for Pi(x), only for big n the Pi(x) is known x/ln(x).

Another question do you know lots of formulae for a(n)=Pi(n)-Pi(n-1) which is 1 if n is prime and o elsewhere?...yes but they are more complicate than mine.

 

[matt grime]

Until such time as you have worked out a way of calculating the sum you want without knowing every prime then you have not SOLVED anything. You've just rearranged things and not provided a closed form for the solution that is in anyway easier to calculate. Being able to provide some solution for small x when pi(x) is known doesn't constitute a new idea. That is perhaps why you got the rejections. Tone down your claims and perhaps you will get someone to take an interest.

Deciphering your post, you seem to confuse explicit formulae with approximations. pi(x) is not ln(x)/x for x large, but the ratio between them tends to 1, which isn't the same thing. Just attempt to figure out the error terms in your calculations for a start.

 

[eljose79]

the question is...if you would like to know the sum over all primes of:

exp(-sp)...would you need to know all the primes to get an accurate aproximation?or perhaps you should only need the 1000 first primes to get and error of 0.01?..

That is the spirit of my work the exact formula is very complicte but perhaps a good approach could be made to Pi(x) function by knowing only a few primes..do not know if you will have a computer but you could try to compute my expresions for x large or small...i think that for the two cases my formula is valid..

 

[HallsofIvy]

You seem to be missing the point: You titled this thread "Pi(x) function solved?" and you titled the paper "An Approach to Pi(x)" (which is not at all the same as "solving" it but might be interpreted as meaning some things that might eventually lead to a formula for Pi(x)). In fact, you are saying that you have done neither. What you are giving is a way of approximating Pi(x). There are, in fact, many such approximations. They don't help in determining exactly what Pi(x) is for large x- which is what most people would think "Pi(x) solved" means. In any case, until you can give some sort of error estimate for your approximation, you haven't shown that it is a good estimate and no one is going to bother with it.

 

[eljose79]

in fact i do not know if it is valid for big n but i think that after applied the Euler,s transform to the series it will converge with only a few terms for any n hte error term would be

diff(n)a0eps(-sn-s).(1+exp(-s))**-(n+1) htat for large n it tends to 0
is that enough for you ?..inb fct although my formula was correct these snobish teachers and journals wouldn,t like it because i am an unknown author.

By hteh way someone has tried computing it for high primes?..thanks.

 

[matt grime]

No, that is not remotely good enough to be considered an approximation of the error, indeed it isn't clear what the thing you've written down is.

What you need is to truncate the sum after R terms, and estimate the error in some way between pi(x) and what ever number it returns. All I believe you've stated is something about the radius of convergence of the infinite sum. (its terms go to zero - you've not even demonstrated the sum converges)

The reason why 'snobbish' teachers and journals have turned you away is because you've not proven anything. You've just applied some transform to a function. You cannot and do not calculate the function or its transform. You do not estimate the error in this sum after terminating at some stage. Nor have you in my mind justified you are allowed to take the inverse transform for discontinuous functions (you may be; this isn't my area; cite a proof) - two functions can can have the same Fourier coeffs, yet differ by a non-trivial function of measure zero, for instance.

Moreover, it is not clear that the method you adopt is even new, or does anything useful. As you cannot prove anything, and even state that you do not know if the truncation is valid for large input, then how on Earth can you accuse these people of being snobbish in refusing to accept the result? Using transforms in number theory is not new. Try 'Fourier analysis on finite abelian groups' by Audrey Terras for some examples.

The reviewers and teachers don't like it for those reasons I would suggest, and for numerous stylistic and typographical errors in it (iif instead of iff, for example),and perhaps for your insistence you've done something amazing when you've just moved the furniture around. It is not up to them to see the merit in that, it is up to you to demonstrate it has some. Evidently their experience tells us they believe that nothing will come from this approach, or that it is already known.

 

[eljose79]

ok ik i have understood...:(:( :(:( :( now i feel sad to see what i have tried is useless.

at least i hpe the second part of it, the approach for the series sum of f(p) for all pirmes can be approached i the sense i give...have someone taken a look to it?... is my last chance to see if the Euler,s formula can be applied to the sum over primes of functions..that is the second part of the paper.

sum(0,infinite)(-1)**n(Pi(n)-Pi(n-1)+1)f(n)=Sum(0.infinite)(-1)**nf(n)+2f(2)-(sum over all pirmes of)f(p)

so we can apply euler,s transform to calculate it..is that correct?.

 

[matt grime]

Sums and products indexed by primes are very well studied. Look up Legendre's method and its related improvements for calculating pi(x), see also the Riemann Zeta function.

You were at least attempting to apply one technique from another area in number theory. That is how a lot of research gets done. As is the case with most first ideas it doesn't lead anywhere. Feynman is supposed to have given a talk saying 'hey, thanks for the prize for this really great idea I had, here are the 20 that came before it that were all stupid' or words to that effect.

 

[eljose79]

ok ik i have understood...:(:( :(:( :( now i feel sad to see what i have tried is useless.

at least i hpe the second part of it, the approach for the series sum of f(p) for all pirmes can be approached i the sense i give...have someone taken a look to it?... is my last chance to see if the Euler,s formula can be applied to the sum over primes of functions..that is the second part of the paper.

sum(0,infinite)(-1)**n(Pi(n)-Pi(n-1)+1)f(n)=Sum(0.infinite)(-1)**nf(n)+2f(2)-(sum over all pirmes of)f(p)

so we can apply euler,s transform to calculate it..is that correct?.