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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 11, 2010 at 21:37 comment added Kevin O'Bryant Okay, now I see that the question includes $0<a<1$.
May 11, 2010 at 21:37 comment added Kevin O'Bryant Well, it motivates the nonexistence of an algebraic formula valid over the complexes. There are other "closed expressions" that may qualify, particularly if Henry isn't thinking of this as a complex power series. For example, if the coefficients are in the binary field $F_2$, then the series is algebraic.
May 11, 2010 at 3:25 vote accept Henry Yuen
May 11, 2010 at 3:17 vote accept Henry Yuen
May 11, 2010 at 3:25
May 11, 2010 at 3:10 comment added Wadim Zudilin Agreed. My point is that Mahler's series are already special functions (as far as I know $f(z)$ does not satisfy an algebraic differential equation but only the functional equation relating $f(z^d)$ to $f(z)$).
May 11, 2010 at 3:01 comment added Qiaochu Yuan I think a simpler motivation is that the function has a natural boundary. Anything one might naively call an elementary function won't.
May 11, 2010 at 2:55 history answered Wadim Zudilin CC BY-SA 2.5