Searching for an asymptotic to exp[0.5]

[JmsNxn]

Well not from Hadamard alone , but from the asymptotes of the zero's yes.

Maybe I'm misinterpreting your response but a huge lemma that hadamard uses in his proof is that:

if \( |f(z)| < C e^{\rho|z|^{\sigma}} \)

then NECESSARILY, (by doing some magic with jensen's formula and some other neat complex analysis), if \( a_j \) are the zeroes of f

\( \sum_{j=1}^\infty |a_j|^{-(\sigma + \epsilon)} < \infty \)

Very intresting.

Well I was simply saying that it did not follow from what your wrote alone.
Hadamard's proof is very intresting.

By the way the constant C is simply f(0).

Let t_n be the absolute value of the zero's.
So if we show that prod (1 + z/t_n) grows slower than exp(x^a) for any real a > 0 then we have

fake exp^[0.5] = f(0) prod (1 + z/t_n).

And I believe that to be the case.

The logic behind that is simply this :

f(0) exp(Az) prod [ (1 + z/t_n) exp(-z/t_n) ]

Now the exp terms in the prod are just there to make the prod convergant IFF that is not yet the case.
But here it already is so we rewrite :

f(0) exp(Az) prod [ (1 + z/t_n) exp(z/t_n) ] =
f(0) exp((A+B)z) prod [ (1 + z/t_n) ]

where B = 1/t_1 + 1/t_2 + ...

prod (1 + z/t_n) grows slower than exp(x^a) for any real a > 0 then we have

A+B = 0

=> f(0) prod [ (1 + z/t_n) ]

And the sky is blue

regards

tommy1729

That's exactly what I was hoping for! I wasn't sure if you could say it was order zero though, that makes a lot of sense.