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Aug 10, 2020 at 1:36 comment added Alexandre Eremenko You are right: that answer with Taylor series makes no sense. The correct one is the first answer.
Aug 9, 2020 at 22:58 comment added Anna Thank you very much for the reference! Anver Kiro suggested such an example in the explicit form of the Taylor series here: mathoverflow.net/q/191663. However, I am confused with the fact that this series is divergent for |z|>1 if the notation log^n (...) is assumed to be (log(...))^n. But an entire function must be represented by the Tailor series which are convergent everywhere. If I misunderstand the notation I would be grateful to learn what does this mean.
Aug 9, 2020 at 12:27 comment added Alexandre Eremenko This question was answered here: mathoverflow.net/questions/190837/… If this is not enough, state your question more precisely.
Aug 9, 2020 at 10:57 history edited R.P. CC BY-SA 4.0
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Aug 9, 2020 at 9:56 review First posts
Aug 9, 2020 at 10:58
Aug 9, 2020 at 9:52 history asked Anna CC BY-SA 4.0