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when toggle format what by license comment
S Feb 28, 2021 at 14:43 history suggested J. W. Tanner CC BY-SA 4.0
corrected spelling
Feb 28, 2021 at 14:30 review Suggested edits
S Feb 28, 2021 at 14:43
Feb 28, 2021 at 13:22 comment added Christoff_ferland I got your idea. Thanks a lot.
Feb 28, 2021 at 13:00 comment added Fedor Petrov well, $\mu(n)=\int y^{-n} d\mu(y)$, $\bar{\mu}(n)=\int x^{n}d\mu(x)$, multiply to get $|\mu(n)|^2=\int \int (x/y)^{n} d\mu(x)d\mu(y)$
Feb 28, 2021 at 12:44 comment added Christoff_ferland Why is the first equality right? I'm not quite sure of this, please let me know.
Feb 28, 2021 at 12:01 history answered Fedor Petrov CC BY-SA 4.0