# I have discovered a formula for PI(x)

1. Feb 9, 2005

### eljose

Let be PI(x) the prime number counting function ,i have discovered an integral representation for it the integral is on CxCxR space with bounds

(c-i8,c+i8)x(0,8)x(c-i8,c+i8) where 8=infinite

I have submitted to several webpages but i always have been rejected and i don,t know why if my formula is correct....

File size:
45.5 KB
Views:
199
2. Feb 9, 2005

### matt grime

Because, as with most things you "discover", it is probably neither practically useful, nor theoretically interesting, being either very old or a facile observation about the existence of some constants of integration? Just a suggestion. I have a fromula too:

let I(x) be one if x is prime, zer other wise, then pi(x) satisfies the recurrence relation

p(x) = pi(x-1) + I(x)

but that is completely pointless, obviously, only I seem to realize the difference.

3. Feb 9, 2005

### eljose

but my integral includes some known formulae as Riemann,s Zeta function...you only have to do the integral (presented at the end of integral.doc) numerically or by residues and obtain a result.....

your formula need to knowteh value of I(x) for example what would be the value of I(x) if x=12345678901234567890123456789001928374656362237783484858584787373737377347373474747757766798007060505040400404304301121341234132412354352135323213532512532151?

4. Feb 9, 2005

### arildno

eljose, I'm no mathematician, but I have a question:
Can your integral representation of $$\pi(x)$$ simplify some of the standard proofs in which $$\pi(x)$$ is utilized by using your representation rather than the standard one?
That is, can your representation be considered a TOOL in proofs, rather than something which cannot be used (or even, easily computed)?.

5. Feb 9, 2005

### eljose

Yes,my representation can help to obtain a closed formula for series of the kind:
sum (over all primes)f(p) equating it to an integral (a fourth dimensional integral) as
sum(over all primes)=-Int(0,8)Pi(x)f(x)dx where the means derivation and 8=infinite.

My formula can give Pi(x) for any real number only by evaluating a triple integral (there are lots and lots of methods to do that) .

6. Feb 9, 2005

### Zurtex

0, that was easy, do I get a cookie?

7. Feb 10, 2005

### matt grime

For the love of God, eljose, I'm not saying what you've got is incorrect I'm saying that it doesn't appear to be useful, and so far none of the people who understand number theory think it is new. Proof: you've yet to actually evaluate pi(x) for any x using it, and had it rejected by lots of number theorists: if it were something revolutionary then you can't believe that they'd all be turning down a tool that could help them win a million dollars do you?

Pi(x) satisfies a whole range of relations, some complicated, some not so (did you not see the bit where I said the one I wrote down is useless? You didn't have to explain to me why it was useless). We have exact formulae for pi(x), we have Meissel's Algorithm, we convert using Mellin transforms all the time, we have L-functions, zeta-functions, modular forms, Euler sums.

You say that all you need is to know some poles - do you even know where the poles are? They occur at the zeroes of the zeta function. Do you know where they all lie?

You've just written down some passably interesting manipulations that anyone who knows 3 basic facts could do. You've not proven any of the sums converge, the integrals exist (they all do, I believe, but you've not said why) plus you've misspelled Riemann. And you've not explained how this is NEW, or implementable, or how you intend to calculate numerically those integrals (assuming you don't evaluate the residue at the poles), and show that the numerical triple integral over infinite ranges will be correct to the integral part of pi(x)/x^4.

8. Feb 11, 2005

### eljose

i guesss that if my formula were discovered by Riemann or Gauss you wouldn,t disagree so much with it looking for its defects.

my formula can evaluate Pi(x) only by using a triple integral...please tell me an easier way to find PI(x) with an exact result for this function

The poles are the zeroes of the Riemann function,yes you are right i don,t know where they lie exactly but using a computer more than a billion of the zeroes have been calculated (is a billion enough accurate to it?) using a computer my tripel integral could be evaluated,or else try the change of variables s=c+iu q=d+iv and the integrands are all real numbers

9. Feb 11, 2005

### Zurtex

eljose show us how it's useful by actually computing values with ease then.

10. Feb 11, 2005

### Hurkyl

Staff Emeritus
I don't know how hard it is to evaluate that triple integral, so I couldn't tell you if it's easier or harder than any other method.

Mathworld gives the time and space complexity of several methods of (exactly) evaluating Pi(x). How does yours compare?

11. Feb 11, 2005

### shmoe

This isn't a conspiracy. I guarantee that Gauss and Riemann discovered many useless things. However, they were capable of judging the worth of their discoveries and only published the gems.

I don't think you really understand the difficulties of dealing with all the zeros of the zeta function at once and the difficulty you'll have evaluating that integral via residues. Riemann gave a very elegant exact explicit formula for the prime counting function involving a sum over the zeros of zeta. Trouble is, it's not a computationally feasible thing to work with due to those pesky zeros. You really need to understand the long term behavior of these zeros, which we don't (a few billion does not equal "long term"). You might want to have a look at the truncated form of this explicit formula (due to Landau) and see just how many zeros you'd need to calculate pi(x) accurately enough to detect primes when x gets large. I'd wager you'll run into similar problems if you hoped to estimate your integral via residues accurately enough to detect primes.

If you want some info on methods used to calculate pi(x), Andrew Odlyzko's web page is a great place to start. Check under the "analytic number theory" section and you'll find some nice papers about evaluating zeta and pi(x). The algorithms on evaluating zeta efficiently are of course directly relevant if you have any hopes of approximating you integral numerically.

12. Feb 11, 2005

### eljose

-A triple integral is only a triple integral and nothing more difficult,my formula is the easiest one to calculate Pi(x) because the number of step to calculate it does not depend on the value of x (other exact formulae are useful,but fail to calculate pi(x) when x is big

13. Feb 11, 2005

### shmoe

Then do it. If your method is so computationally efficient and independant on the size of x, you should have no problems finding primes with a staggeringly large number of digits. Do this and stun the mathematics community. I believe the largest known prime is currently the 41st (known) Mersenne prime, which is in the region of 7 million digits, you aught to be able to top that with ease. Heck, if you can start pumping out primes in the million digit range you'll be famous.

However I highly reccomend that you test out the answers you're getting for pi(x) against methods that are known to work. If yours is truly independant on the size of x, I'm willing to bet that it's wrong, but I tend to have a pessimisstic view of anything that claims a primality test with O(1) running time so don't take my word for it.

Last edited: Feb 11, 2005
14. Feb 11, 2005

### Hurkyl

Staff Emeritus
Yep, a tautology. What reason do you have to think this is an easy integral to approximate?

Not even close to true. Even without analyzing the complexity of approximating the integral, notice that your equation looks like:

Pi(t) = t^4 * triple integral

Which means that every time you double t, the error of approximation gets magnified 16 times. This means that you need to do extra work approximating the integral to decrease the error by a factor of 16.

In the case of a single integral and we were using Simpson's rule, this would mean that you need to double the number of subdivisions every time you double t, so that your algorithm is &Omega;(x), already worse than existing methods.

However, you're not using a single integral, you're using a triple integral, which takes much more effort to approximate well.

This is also for proper integrals: yours are improper, which are much harder to approximate well than ordinary integrals. One would probably truncate the domain of integration, which generates error. To get the greater accuracy needed with higher t, you will have to consider ever larger portions of the domain of integration.

Furthermore, the bigger t is, the bigger the integrand. In particular, the integrand grows much faster than t. The bigger the integrand, the more work you need to do to get a good approximation.

But wait, there's more. As the numbers get larger, the number of operations increases, and the greater precision required, you need to store more and more significant figures, which means more time per basic operation.

15. Feb 11, 2005

### Zurtex

Indeed, you'd be able solve the problem of RSA encryption slowly being broken by just generating stupidly large primes. You could become a multi-millionare.

Last edited: Feb 11, 2005
16. Feb 11, 2005

### mathwonk

please do not compare your situation with that of riemann and gauss; as contrary to what you may think, it does not increase your credibility.

17. Feb 12, 2005

### eljose

Hurkyl this also happens with the approach x/ln(x) for example if you evaluate Pi(x) for x=10^20 you get an error term of 10000 knowing ln(x) with a precision of error term 10^-16 and this is only an approximation.

pi(t)=t^4*Triple integral, then introduce the term t^4 inside the integral now we would have the integral in q with x^4-q now we make the cahnge of variables 4-q=ro so we would have the formula:

Pi(t)=Triple integral then now matter how hard is to evaluate only take the integer part of it (we could calculate simply with an error of 0.01 as we know that Pi(t) is alwayas an integer)

18. Feb 12, 2005

### matt grime

If that is true then it is even more unrealistic than I first thought: you're saying that ti costs as much to evaluate pi(2) with your method as pi(10^10)? And it takes no more effort to calculate any of the other ones.....

19. Feb 12, 2005

### Hurkyl

Staff Emeritus
Now, don't you think there's something fishy? You're essentially telling me that $\int K f(x) \, dx$ is easier to integrate to the desired precision than $K \int f(x) \, dx$, despite the fact they're the same thing...

Yep, he's saying that π(x) takes no longer to compute than π(2)... even if x is so large that it takes longer to just read the digits of x from memory than it does to compute π(2).

Last edited: Feb 12, 2005
20. Feb 12, 2005

### Hurkyl

Staff Emeritus
eljose: I presume you took calc 2 (integrals and series), and still have your book. Review the chapter on numerical approximation of definite integrals.

In particular, your book should have formulas for the bound on the error for approximations via various basic methods: left hand rule, right hand rule, trapezoidal rule, and Simpson's rule.

It probably has examples and problems where you compute the number of subdivisions needed to approximate the integral to the desired precision.

(For instance, you'll see that $K \int f(x) \, dx$ requires exactly the same number of subdivisions as $\int K f(x) \, dx$ to get the desired accuracy)

Now, this doesn't convey the entire source of complexity for numerically approximating an integral: (it doesn't deal with roundoff errors, complexity of basic operations, more sophisticated methods, or the fact that triple integrals are harder than single integrals), but it should at least help to clear up some of your major misconceptions.