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Mar 29, 2021 at 15:47 comment added Anixx @Davey Bernoulli numbers as function of the order are essentially, zeta function.
Mar 29, 2021 at 15:45 comment added Anixx Then you should prove that Hurwitz Zeta is non-elementary.
Jun 12, 2020 at 10:38 history edited CommunityBot
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May 14, 2020 at 4:03 comment added Davey This does not quite make sense to me: each individual Bernoulli polynomial $B_n(x)$ is of course elementary, but why is $B_n = B_n(0)$ elementary as a function of $n$? If it is, then what's a closed form expression for it? We can't use a series that involves binomial coefficients since I'm pretty sure those aren't elementary (they seem strictly more complicated than the factorial function, right?).I don't think we can even say that IF the factorial is elementary, then $B_n$ is elementary.
Sep 24, 2016 at 11:18 history edited Vincenzo Oliva CC BY-SA 3.0
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Aug 13, 2015 at 19:37 comment added Vincenzo Oliva @Dave: Nice, thank you. And +1, by the way.
Aug 13, 2015 at 19:32 comment added Dave L. Renfro In fact, the gamma function doesn't even satisfy an algebraic differential equation -- see Expanded concept of elementary function?.
Aug 13, 2015 at 19:19 history edited Vincenzo Oliva CC BY-SA 3.0
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Aug 13, 2015 at 19:13 comment added Vincenzo Oliva In fact there's no need of seeking for a contradiction, to prove Claim 2.
Aug 13, 2015 at 19:13 history edited Vincenzo Oliva CC BY-SA 3.0
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Aug 12, 2015 at 9:10 vote accept Akiva Weinberger
Aug 12, 2015 at 8:55 history edited Vincenzo Oliva CC BY-SA 3.0
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Aug 12, 2015 at 8:28 history edited Vincenzo Oliva CC BY-SA 3.0
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Aug 12, 2015 at 8:02 history edited Vincenzo Oliva CC BY-SA 3.0
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Aug 12, 2015 at 7:54 history edited Vincenzo Oliva CC BY-SA 3.0
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Aug 12, 2015 at 7:45 history answered Vincenzo Oliva CC BY-SA 3.0