Prime products

[eljose]

Is this equaltiy exact?:

$$\Pi(a_p+b_p)= \Pi(a_p)+\Pi(b_p)$$

where both products a_p,b_p and (a_p+b_p) converge

another qeustion $$\Pi 1=1$$ ?

all products are made respect to all primes..

 

[matt grime]

your first conjecture is vacuous i think, though i haven't checked properly: there is no situation where all three products can exist for the usual definitions of convergent products, namely, prod(1+x_n) converges iff sum(x_n) converges. and besides, does it even seem reasonable? if the index set were finite it is false.

and yes for the second one.

 

[Hurkyl]

The equality is not exact -- it's not in the ballpark. It's impossible for all three of those infinite products to converge.

Yes, the infinite product of the constant 1 is, in fact, 1, as can be easily seen by taking the limit of the partial products.

 

[eljose]

how about $$a_p=1 , b_p=\frac{1}{1-p^{-s}}$$they both converge, if the equality is exact the sum:

$$a_p-b_p=\frac{p^{-s}}{1-p^{-s}}$$ the product would converge to

$$1-\zeta(s)$$ being R(s) riemann,s zeta function..is that true?..

 

[matt grime]

did you not read the posts above?

 

[eljose]

yes i have read the post, i have given two converging products and i would like to know if the equality $$1-\zeta(s)=\Pi\frac{p^{-s}}{1-p^{-s}}$$ is exact, because you can see that all the products

a(p)=1 b(p)=1/1-p^{-s} and a(p)+b(p) converge for certain values of s and how about?...

$$1+1/\zeta(s)=\Pi(2+p^{-s}$$

 

[matt grime]

so, you'vwe been told that if prod(1+x_n) is convergent, and so is prod(1+y_n) then the prod(1+x_n+1+y_n) cannot be convergent. so why are you still asking if it is possible?

let p+n be the n'th prime.

let 1+y_n = a_{p_n}-b_{p_n} and 1+x_n= b_{p_n}. you are claiming these both converge, therefore, for whatever value of s this correpsonds to, to a_{p_n} cannot converge. but a_p_n=1 so actually (at least one of) the other two products cannot exist.

 

[eljose]

could you help me with these product and say if they converge and what would be their value?..

$$\Pi\frac{2+p^{-s}}{1+p^{-s}} \Pi2-p^{-s}$$

 

[matt grime]

it converges iff the sum of 1/(1+p^{-s}) work out when this can happen, eg by the sandwich principle say. i think you'll find it's approximately when s<-1 (assuming s real) which as we know is out of the range of the region of convergence for the zeta function.


edit: now you've changed it. and $\prod 2$ doesn't make sense, really. what's the insdex? why is there a stray p^-s involved? why am i bothering to help you since you never listen to me anyway?

 

[Zurtex]

it converges iff the sum of 1/(1+p^{-s}) work out when this can happen, eg by the sandwich principle say. i think you'll find it's approximately when s<-1 (assuming s real) which as we know is out of the range of the region of convergence for the zeta function.


edit: now you've changed it. and $\prod 2$ doesn't make sense, really. what's the insdex? why is there a stray p^-s involved? why am i bothering to help you since you never listen to me anyway?

Reading his code, not that it really matters anyway, it's meant to be displayed as:

$$\prod \left(\frac{2+p^{-s}}{1+p^{-s}}\right) \prod \left(2 - p^{-s}\right)$$

Although to me it still seems quite ill-defined, but never mind.

 

[matt grime]

well, that never exists since the second product never exists; the terms either converge to zero or diverge to (minus) infinity.

 

[eljose]

can you help me i need to find a product so:

$$1+1/\zeta(s)=\prod f(p)$$

 

[matt grime]

almost certainly not since there isn't even a product for zeta(s) that works for all s.