Talk:Product Form of Sum on Completely Multiplicative Function

It occurs to me that this is only valid if the sum inside the product is absolutely convergent. Needs more work to make it rigorous. --Matt Westwood 05:31, 17 April 2009 (UTC)

... there, I think I've covered it. --Matt Westwood 07:14, 17 April 2009 (UTC)


 * Still some convergence issues I think...


 * Take it away, then, maestro ... --prime mover 13:49, 2 March 2011 (CST)


 * My mistake; forgot that $f$ is multiplicative, so $|f(p)| < 1 \implies$ abs. convergence --Linus44 19:33, 2 March 2011 (CST)


 * This proof seems fairly vague to me. I am not sure that anybody would no what you meant unless they already new how the proof went.
 * "Here we may freely rearrange the terms, so expanding the product we see that each term has the form:"
 * I know what you mean, but am still confused.
 * Here are the steps to formalize it.
 * Product over a sum is the sum over the cartesian products of the products.
 * $ \prod_{a \in A} \sum_{b \in B_a} t_{a,b} = \sum_{c \in \prod_{a \in A} B_a} \prod_{a \in A}t_{a,c_a} $

Product over a sum is the sum over the cartesian products of the products.


 * where the product of sets $\prod_{a \in A} B_a$ is taken to be a cartesian product.

Now consider the finite product of absolutely convergent series:
 * $\displaystyle P \left({x}\right) = \prod_{p \in \mathbb{P}}^{p \le x} \left({ \sum_{k=0}^{\infty} f \left({p}\right)^{v_p}} \right) = \sum_{v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\}}  \left({ \prod_{p \in P} f \left({p}\right)^{v_p}} \right)$


 * $ \prod_{p \in \mathbb{P}} \sum_{k=0}^{\infty} p^{-ks} = \sum_{v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\}} \prod_{p \in P}p^{-v_ps} = \sum_{v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\}}  (\prod_{p \in P}p^{-v_p})^{-s}$


 * Change the summing variable using,
 * $ \sum_{v \in V } g(f(v)) = \sum_{w \in \{f(v) : v \in V \} }  g(w)$
 * which is valid only if f is one to one. This is true by Fundamental theorem of arithmetic, as every number has a unique factorization. This gives,


 * $ \prod_{k=0}^{\infty} \sum_{p \in \mathbb{P}} p^{-ks} =  \sum_{n \in \{\prod_{p \in P} p^{-v_p} : v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\} \} } n^{-s} $


 * But the positive natural numbers are given by:
 * $ \mathbb{N+} = \{\prod_{p \in P} p^{-v_p} : v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\} \}$


 * $ \prod_{k=0}^{\infty} \sum_{p \in \mathbb{P}} p^{-ks} =  \sum_{n \in \mathbb{N+}} n^{-s} =  \sum_{n = 1}^{\infty} n^{-s} $


 * as the series converges absolutesly, and therefor the convergence to the same limit, no matter what the order. I prefer to use this method, because it is clearer. The alternative approach is set up upper limits for p and k to consider the difference between the zeta function and the Euler Product. That is perhaps more rigorous, but it has about 6 extra steps, and is less readable.
 * Peter Driscoll (talk) 17:35, 10 September 2017 (EDT)


 * I like a lot the idea to formalize this by exhibiting the bijection given by FTA. I say go ahead (as long as you don't replace valid steps by alternative arguments, in which case you can set up multiple proofs instead). Watch out for convergence issues, but I suppose you're aware of that. --barto (talk) 13:05, 10 September 2017 (EDT)


 * As a sidenote, remember to sign your posts so we know who wrote what. --barto (talk) 13:07, 10 September 2017 (EDT)