(06/29/2014, 01:40 PM)JmsNxn Wrote: \( f(z) = C e^{Az + B} \prod_{j=1}^\infty (1-\frac{z}{a_j})e^{-\frac{z}{a_j}} \)Your equation is overdetermined I think.
for some \( A,B,C \in \mathbb{C} \) where \( a_j \) is all the zeroes counted with multiplicity.
\( f(z) = C e^{Az} \prod_{j=1}^\infty (1-\frac{z}{a_j})e^{-\frac{z}{a_j}} \)
Removing \( B \) was trivial and the other variable \( A \) is correct because of the genus theory / theorem in combination with below :
Quote:This also implies, quite adequately that at least \( \sum_{j=1}^\infty |a_j|^{-(1+\epsilon)} < \infty \) for \( \epsilon > 0 \).
Well not from Hadamard alone , but from the asymptotes of the zero's yes.
And in combination with the genus theory we get the result above.
I did assume an upper bound on multiplicity though.
Thanks for the post James.
regards
tommy1729

