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S Sep 12, 2020 at 13:00 history bounty ended CommunityBot
S Sep 12, 2020 at 13:00 history notice removed CommunityBot
Sep 6, 2020 at 8:04 comment added Chris Very interesting, I did not think about multiplication by such function to obtain simple zeroes. So indeed, it is not always possible. I just need to make this assumption then. I wonder if it says something interesting about the covariances functions that cannot be root squared that way.
Sep 6, 2020 at 4:55 comment added fedja And how do you expect to get an entire square root when the Fourier transform has simple complex zeroes (which isn't hard to arrange: take any positive and fast decaying on the line entire function of exponential type and multiply it by $z^2+\mu^2$ with some real $\mu$ such that $i\mu$ is not a zero of the original function)?
S Sep 4, 2020 at 11:30 history bounty started Chris
S Sep 4, 2020 at 11:30 history notice added Chris Draw attention
Aug 31, 2020 at 12:15 history edited Chris CC BY-SA 4.0
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Aug 31, 2020 at 6:42 history asked Chris CC BY-SA 4.0