# Wilson theorem and formulas for pi(x)

1. Apr 20, 2006

### eljose

Hello..my question is from wilson theorem:

$$(p-1)!=-1mod(p)$$

how could you derive the formulae for $$pi(x)$$?..this should be impossible as you can,t solve the congruence exactly a proposed method would be to find the roots of:

$$f(x)=[\gamma(x)+1/x]-(\gamma(x)+1)/x$$

that precisely x=p for every prime and intege...

2. Apr 20, 2006

### matt grime

Do you not feel the slightest desire to define gamma?

3. Apr 20, 2006

Nor to use proper English sentences?

I mean what the hell is a "root of [equation] such that precisely x=p for every prime and intege..."

Last edited: Apr 20, 2006
4. Apr 21, 2006

### eljose

-Umm..sometimes i don,t know if you,re speaking seriously or if this some kind of British-American sarcastic humour:

$$\Gamma(x)=\int_{0}^{\infty}dxt^{x-1}e^{-t}$$

a "root" is just a root in the sense that is a number that satisfies for some function that f(c)=0 then c is a root of the function. for the primes and using Wilson,s theorem we would have that

$$Sin(\pi[\Gamma(x)+1]/x)=0$$ iff x is p for p prime.

and using this "Gamma" function Wilson,s theorem becomes

$$\Gamma(p)+1=0mod(p)$$ but how can you obtain the Prime number counting function from wilson,s theore?..thanks.

Last edited: Apr 21, 2006
5. Apr 21, 2006

### matt grime

it is weariness at seeing you repeatedly post articles that contain contradictory notation (using n for two different things in one equation for instance), symbols that you simply do not explain and careless abuse of mathematics: notice how your gamma is now an uppercase gamma when it started off as a lower case one.

notice how this produces a real number, for real x...

Does it? In what sense does it become this, since mod p is not something that makes sense for the real numbers? What is the point of introducing the gamma function if you can only use it for certain values in the domain where we already have a better description of its value (at these points)? So, what good does introducing the gamma function do? You can't use it, can you? No arguments, that (p-1)! is indeed Gamma(p), and that Wilson's theorem is that p-1!=-1 mod p, but why use Gamma instead of p-1!? The proof of Wilson's theorem is not one that uses real analysis, so why do this?

Last edited: Apr 21, 2006
6. Apr 21, 2006

### shmoe

$$\gamma$$ and $$\Gamma$$ look different to me, we aren't mind readers.

see http://mathworld.wolfram.com/PrimeFormulas.html (3)-(5) show how to get pi(x) using Wilson's theorem. Wilson's theorem is not at all practical for computing pi(x) like this, nor for locating primes by looking at zeros of that sin function. n! grows too fast. Besides, the version you have with real numbers for x in $$\sin (\pi(\Gamma(x)+1)/x)$$ has zeros at many real numbers that aren't prime integers.

Last edited: Apr 21, 2006
7. Apr 22, 2006

### eljose

the main question is not computing a prime this was not my intention..but perhaps to throw some of light into the problem of time primes in fact for m and m+2 twin primes:

$$4((m-1)!+1)+m=0mod(m(m+2)$$ so we could compute the density of twin primes in the form:

$$\pi_{2}(m)=\sum_{2}^{m}cos^{2}[\pi([4((m-1)!+1)+m/(m+2)m]-4((m-1)!+1)+m/(m+2)m)$$

if somehow we could invert this twin prime counting function t obtai the k-th twin prime in the form:

$$p_{n}=C+\sum_{k=2}^{f(n)}F(k,\pi_{2}(k))$$ we could prove that p-n is not bounded for p a twin prime so it must be an oo number of them.

8. Apr 22, 2006

### shmoe

If you mean something essentially like (3)-(5) from that link, then sure. What you have here has a few errors in the parenthesis so it's hard to tell what you mean. You would also want the floor of the cos terms at some point, otherwise this is most likely not even an integer.

You can do essentially the same as (10)-(11) from the mathworld link. As in the case of primes, this isn't a very enlightening expression, it's mostly a novelty.

9. Apr 22, 2006

### eljose

-I think that the "cos" term is enclosed by a floor function [cos] o this only can be 1 or 0.

-If wemanaged in the same case that for ordinary primes to prove that the k-th twin prime is:

$$p_{k}=\sum_{n=2}^{f(k)}G(n,\pi_{2}(n)+C$$

we would have proved "twin prime conjecture" as there would be the (k+1)-th prime so:

$$p_{k+1}-p_{k}=\sum_{f(k)}^{f(k+1)}G(n,\pi_{2}(n))+C$$

this formula somehow would use the [] function as the difference would be an integer so if this sum is $$p_{k+1}-p_{k}>0$$ then for every k exist a higher prime (k+1) wich is the next twin prime in the series.

10. Apr 22, 2006

### shmoe

Which we can do. If you have the counting function of a sequence, (10)-(11) can be used to give you the nth term of the sequence. However:

This isn't true. Using that expression to show there are in fact infinitely many twin primes will rely on already knowing that $$\pi_2(x)$$ goes to infinity as x does, i.e. you already need to know there are infinitely many twin primes.

Try using the formula from Hardy&Wright ((10)-(11) in the link) to prove there are infinitely many primes without already knowing that pi(x)->infinity as x->infinity (edit-more specifically, try proving that the p_n defined by this formula are in fact prime without already knowing there are infinitely many primes)

Last edited: Apr 22, 2006
11. Apr 23, 2006

### eljose

-umm...i see in this form (perhaps wrong) if we have the formula:

$$p_{k+1}-p_{k}=\sum_{f(k)}^{f(k+1)}G(n,\pi_{2}(n))+C$$

this means that for every k-th twin prime there is another consecutive (k+1)-th prime no matter what k is of course if the number of twin primes is finite you have that the maximum value of $$\pi_{2}(n)=a$$ your only chance for having a finite number of primes is that somehow the function inside the sum including the floor function would be <1 and form tis we could perhaps calculate the last twin prime so either there are infinite number of twin primes or the last number can be calculated.