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. --Prime.mover 05:31, 17 April 2009 (UTC)

... there, I think I've covered it. --Prime.mover 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.
 * $\ds \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 $\ds \prod_{a \in A} B_a$ is taken to be a cartesian product.

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


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


 * Change the summing variable using,
 * $\ds \sum_{v \in V } g(f(v)) = \sum_{w \in \{f(v) : v \in V \} }  g(w)$
 * 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,


 * $\ds \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:
 * $\ds \mathbb{N+} = \{\prod_{p \in P} p^{-v_p} : v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\} \}$


 * $\ds \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)


 * I have updated the proof. This version relies on equalitity of infinite sets. If you are uncomfortable with that, the proof may be completed using the usual limit of the difference of sums method. The use of multiplicative functions instead of $n^{-s}$ was unformaliar to me. For the Riemann Zeta function, s is complex. For that case the imaginary part of the power has magnitude 1. I am unclear as to whether or not f is a complex function. If f is not complex the proof is straight forward.
 * Please roll back if I have stepped on any toes. I am not in the toe stepping business.
 * I have endeavoured to emulate house style.
 * Peter Driscoll (talk) 02:14, 11 September 2017 (EDT)


 * Minor tweaks here and there.
 * We have a strict one-sentence-per-line policy which is inflexible, and we deliberately insert an extra line break between sections.
 * OK,
 * We are strict about use of commas where colons are appropriate, for example, and all mathematical symbols (including simple letters of the alphabet) are to be enclosed in $\LaTeX$ delimiters (we allow no exceptions).
 * So if I use a letter, such as $p$ in the text, it must be in $\LaTeX$ delimiters. I thought I had done this. I will check.
 * Other stylistic details can be found in the Help pages, which are fairly comprehensive.
 * Grammatical and linguistic accuracy are also considered important on.
 * I will study the style and try to emulate it.
 * In particular I wonder whether there might be a way to improve the presentation of the complex summation limits. The current way of doing it (putting the product notation in the subscript) makes it difficult to read.
 * I have changed this to improve it. I use $Q(a, K)$ as the summation set, right from the start.
 * I also note the unusual style of arranging conditions on summations and products where one of the conditions goes to the bottom and one to the top. The usual technique is to place all such multiple conditions at the bottom, and use the bottom-to-top to specify a range.
 * They are both conditions $p \in \mathbb(P) \wedge p \le A$ but to put them both at the bottom makes that wider. Also $p \le A$ has a simmilar role to the top index, in $\sum_{k=1}^{n}$. So logically you may be correct, but for readability I think it helps. I think I stole the idea from somewhere.
 * And as I have said elsewhere: use of \mathop in the conditions helps present those conditions more readably, due to an annoying quirk in MathJax.
 * I thought I had done all these. I will check. --prime mover (talk) 02:47, 11 September 2017 (EDT)


 * OK. I will go through these.
 * Peter Driscoll (talk) 11:06, 12 September 2017 (EDT)