Primes as roots of same function

eljose
Messages
484
Reaction score
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:

f(x)=\prod_{p}(1-xp^{-s})=\sum_{n=0}^{\infty}\frac{\mu(n)}{n^{s}}x^{n}

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:

ln[f(s)]/s=x\int_{2}^{\infty}dt\frac{\pi(t)}{t(t^{s}-x)}

and although is not perhaps completely correct i think f(x)=x/Li_{s}(x)
 
Physics news on Phys.org
I think f(x) doesn't converge at all, except at x = 0. (I was interested in a similar expression once, but I wanted zeroes at all the positive integers)


So, I suspect you'll need some exponential terms to get something that converges. Maybe something like:

<br /> \prod_{i = 1}^{+\infty} (1 - x / p_i) e^{x / p_i}<br />

The general exponential term in an infinite product can be more complicated, but the simplest nontrivial term was good enough for my purposes.
 
well using the same trick by Euler..i managed to prove that:

\prod_{p}(1-xp^{-s})^{-1}=Li_{s}(x)

where necessarily |xp^{-s}|&lt;1and LI is the Polylogarithm..if we extend the radius of convergence of Polylogarithm to |x|>1 we could get that p^{s} for p prime is a root of the function 1/Li for a certain s.
 
eljose said:
well using the same trick by Euler..i managed to prove that:

\prod_{p}(1-xp^{-s})^{-1}=Li_{s}(x)

I don't think you did. Take x=0, the left hand side will be 1, the right hand side will be 0.

If you expand the terms of the left hand side using geometric series, you get a product over primes of sums 1+(x/p^s)+(x/p^s)^2+... If you expand this product (taking care when this is valid), the power of x that appears with 1/n^s is not in general going to be n like you would need to equal the polylogarithm. It will be the sum of the exponents appearing in the prime decomposition of n.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
Back
Top