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