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Fourier series are Taylor series in complex z, so I'm wondering if there is any kind of series that represents analytic almost everywhere functions that is foundationally distinct from Taylor series?

Or is every series that matters for analytic functions a Taylor series?

futurebird
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  • This has a Taylor Series but it's meaningless https://en.wikipedia.org/wiki/Non-analytic_smooth_function – user29418 May 05 '18 at 02:01
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    Not sure what you mean by "0 replies 0 retweets 0 likes" but I think that question would be equivalent to asking "Are there any basis of the space of complex analytical functions that are not polynomial in nature?" – mol3574710n0fN074710n May 05 '18 at 02:03
  • I asked this on math twitter at first and didn't get a response. removed the "retweets" stuff. – futurebird May 05 '18 at 02:04
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    Try Dirichlet series! They show up all the time when studying zeta-functions and $L$-functions, and they are analytic in a (right) half-plane, but these series are not power series. – KCd May 05 '18 at 02:05
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    Lambert series are interesting and important also. – Somos May 05 '18 at 02:07
  • And there's the famous series for cotangent, which there are many questions about on this site, for example here: https://math.stackexchange.com/questions/1308948/series-identity-for-cotangent – Hans Lundmark May 05 '18 at 09:07
  • "Fourier series are Taylor series in complex z," I'm not sure what that means. "analytic almost everywhere functions " Not sure about that either. – zhw. Jun 13 '18 at 16:15
  • holomorphic except at a countable set of point. "AAEF" – futurebird Jun 13 '18 at 21:19
  • http://dev.ipol.im/~coco/website/taylorfourier.html – futurebird Jun 13 '18 at 21:19

3 Answers3

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How about the Riemann zeta function? $$ \zeta(s):=\sum_{n\in\mathbb N}n^{-s}\qquad \operatorname{re} s>1 $$

This is manifestly not a Taylor series, yet it is analytic on its domain of definition (and can be continued to all $\mathbb C\setminus\{0\}$).

See also Laurent series, Puiseux series, etc.

AccidentalFourierTransform
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  • OK, but that is not an example of what the OP is calling an "almost everywhere analytic function" – zhw. Jun 14 '18 at 18:20
  • @zhw. You'll have to elaborate. I do think it is a perfectly good example of what OP is looking for. – AccidentalFourierTransform Jun 14 '18 at 18:33
  • The OP wanted an "almost everywhere analytic function", which the OP in a comment defined as a function analytic on $\mathbb C \setminus E,$ where $E$ is a discrete subset of $\mathbb C.$ – zhw. Jun 14 '18 at 18:45
  • @zhw. I'm not sure where that comment is, but anyway. I don't think that is relevant to my example: the Riemann zeta function is analytic on $\mathbb C\setminus{0}$, even if the particular representation above is only valid for $\mathrm{re}, s>1$. There are other series representations that converge everywhere away from the singularity. – AccidentalFourierTransform Jun 14 '18 at 18:57
  • It's the second to last comment of the OP. – zhw. Jun 14 '18 at 19:25
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There are also the series $$\sum_{(a,b) \in \mathbb{Z}^2 \setminus\{(0,0)\}} \frac{1}{(a\tau + b)^{2k}},$$ where $k \geq 2$. Though these converge on the open set Im$(\tau) > 0$. They are important in the theories of modular forms and elliptic functions. They satisfy infinitely many symmetries !

Dzoooks
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Let $f(z) = 1/z$ in $\mathbb C\setminus \{0\}.$ Note $f(z)$ equals the infinite series $f(z) + 0+0+0+\cdots$ everywhere in its domain. But it can't equal a power series in this domain, because such a power series would have to converge in all of $\mathbb C.$ This would imply $f$ has a removable singularity at $0,$ contradiction.

zhw.
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