Hadamard Product

  • Thread starter eljose
  • Start date
492
0

Main Question or Discussion Point

let be the product:

[tex] \prod_{\sigma}(1-s/\sigma)e^{s/\sigma}=g(s) [/tex] where the product is over the non-trivial zeroes of Riemann function

then we take Logarithms to both sides so we have the equality:

[tex]Lng(s)=\sum_{\sigma}Ln(1-s/\sigma)+s/\sigma [/tex]

then we define the function M(x) in the way that gives the number of non trivial zeroes of riemann function up to x so how could we obtain using the same method that is applied to the product


[tex] \prod_p(1-p^{-s}) [/tex] ot get the integral equation for M(x)?...

this can be useful as if Riemann hypothesis is true we will have that M(z) is only non zero when Re(z)=1/2
 

Answers and Replies

matt grime
Science Advisor
Homework Helper
9,394
3
several things spring to mind.
namely the product series expansion of the zeta function isn't valid when Re(s)=1/2 nad even if it were we know there are other zeroes to the zeta fucntion, the riemann hypothesis only being abotu those that lie in the critical strip (the so called non-trivial ones)
 
492
0
why is not valid?....
 
matt grime
Science Advisor
Homework Helper
9,394
3
you do know the series and product expansions converge for Re(s)>1 (i hope i'm right on this one, the summation certainly only is valid for Re(s)>1), thuogh it depends on which expansion you use.



[tex]\prod_p(1-p^{-s})[/tex]

exists iff

[tex]\sum_p -p^{-s}[/tex]

exists which is iff Re(s)>1
 

Related Threads for: Hadamard Product

Replies
2
Views
589
Replies
3
Views
4K
  • Last Post
Replies
2
Views
3K
Replies
22
Views
1K
  • Last Post
Replies
11
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
3K
Top