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Oct 14, 2021 at 21:17 vote accept Analyst_311419
Oct 14, 2021 at 21:14 comment added Iosif Pinelis @Analyst_311419 : The term is $k\delta^s$, as written, not $(k\delta)^s$. Since your posted question has been fully answered, please mark it accordingly, to keep things in good order. If you have further questions on the additional statement in the parentheses, you can post them separately, and then I will answer them.
Oct 14, 2021 at 20:32 comment added Analyst_311419 Is there any way to show this is the infimum?
Oct 14, 2021 at 20:25 comment added Analyst_311419 Do ypu mean $(k\delta)^s$ in the last equality?
Oct 14, 2021 at 20:01 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 14, 2021 at 19:51 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 14, 2021 at 19:45 history answered Iosif Pinelis CC BY-SA 4.0