- #1
eljose
- 492
- 0
It,s proven that there can not be any Polynomial that gives all the primes..but could exist a function to its roots are precisely the primes 8or related to them) if we write:
[tex] f(x)=\prod_{p}(1-xp^{-s})=\sum_{n=0}^{\infty}\frac{\mu(n)}{n^{s}}x^{n} [/tex]
wher for x=1 you get the classical relationship between the Riemann zeta function and the product of primes as you can see the equation above for x=p^{s} for p prime is 0 the problem is that due to we have to know the function \mu(x) this function is not computable using Abel,s sum formula:
[tex] ln[f(s)]/s=x\int_{2}^{\infty}dt\frac{\pi(t)}{t(t^{s}-x)} [/tex]
and although is not perhaps completely correct i think [tex] f(x)=x/Li_{s}(x) [/tex]
[tex] f(x)=\prod_{p}(1-xp^{-s})=\sum_{n=0}^{\infty}\frac{\mu(n)}{n^{s}}x^{n} [/tex]
wher for x=1 you get the classical relationship between the Riemann zeta function and the product of primes as you can see the equation above for x=p^{s} for p prime is 0 the problem is that due to we have to know the function \mu(x) this function is not computable using Abel,s sum formula:
[tex] ln[f(s)]/s=x\int_{2}^{\infty}dt\frac{\pi(t)}{t(t^{s}-x)} [/tex]
and although is not perhaps completely correct i think [tex] f(x)=x/Li_{s}(x) [/tex]