07/01/2014, 10:10 PM
Lets assume t_n - the absolute value of the zero's a_n - eventually(!) grows faster than n^A for any positive real A.
then if exp(B z) ~ (1+z/t_1)(1+z/t_2)...
We have either B is positive , negative or 0.
Now positive and negative have the same order , so lets assume B is positive.
Then we consider the order by taking Re(z) >> 1.
So we are allowed to consider exp(B z) ~ (1+z/t_1)(1+z/t_2)...
and then just take the log of it
( since B > 0 , Re(z) >> 1 => (1 - z/a_n) is far from 0 )
B z + C ~ ln(1 + z/t_1) + ln(1 + z/t_2) + ...
Now replace t_n with n^A to get an upper bound
B z + C ~ ln(1 + z/1^A) + ln(1 + z/2^A) + ...
Now take the derivative with respect to z :
B + o(D) ~ 1 / (1^A + z) + 1 / (2^A + z) + ...
However now take the limit Re(z) -> oo on the right side and replace A with 2 + A^2.
Then we get B + o(D) = 0.
So B goes to 0.
And thus if t_n > n^A then we have order 0.
That settles the case if that assumption t_n > n^A is true.
However Im not sure if sheldon or anyone else has a proof of that assumption WITHOUT assumptions about the Hadamard product for the fake exp^[1/2] ... to avoid circular reasoning.
It seems the attention is now 100% at the zero's as far as product expansions are concerned.
regards
tommy1729
then if exp(B z) ~ (1+z/t_1)(1+z/t_2)...
We have either B is positive , negative or 0.
Now positive and negative have the same order , so lets assume B is positive.
Then we consider the order by taking Re(z) >> 1.
So we are allowed to consider exp(B z) ~ (1+z/t_1)(1+z/t_2)...
and then just take the log of it
( since B > 0 , Re(z) >> 1 => (1 - z/a_n) is far from 0 )
B z + C ~ ln(1 + z/t_1) + ln(1 + z/t_2) + ...
Now replace t_n with n^A to get an upper bound
B z + C ~ ln(1 + z/1^A) + ln(1 + z/2^A) + ...
Now take the derivative with respect to z :
B + o(D) ~ 1 / (1^A + z) + 1 / (2^A + z) + ...
However now take the limit Re(z) -> oo on the right side and replace A with 2 + A^2.
Then we get B + o(D) = 0.
So B goes to 0.
And thus if t_n > n^A then we have order 0.
That settles the case if that assumption t_n > n^A is true.
However Im not sure if sheldon or anyone else has a proof of that assumption WITHOUT assumptions about the Hadamard product for the fake exp^[1/2] ... to avoid circular reasoning.
It seems the attention is now 100% at the zero's as far as product expansions are concerned.
regards
tommy1729

