Is every series that matters a Taylor series?

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?

• This has a Taylor Series but it's meaningless en.wikipedia.org/wiki/Non-analytic_smooth_function Commented May 5, 2018 at 2:01
• 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?" Commented May 5, 2018 at 2:03
• I asked this on math twitter at first and didn't get a response. removed the "retweets" stuff. Commented May 5, 2018 at 2:04
• 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
Commented May 5, 2018 at 2:05
• Lambert series are interesting and important also. Commented May 5, 2018 at 2:07

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.

• OK, but that is not an example of what the OP is calling an "almost everywhere analytic function"
– zhw.
Commented Jun 14, 2018 at 18:20
• @zhw. You'll have to elaborate. I do think it is a perfectly good example of what OP is looking for. Commented Jun 14, 2018 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.
Commented Jun 14, 2018 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. Commented Jun 14, 2018 at 18:57
• It's the second to last comment of the OP.
– zhw.
Commented Jun 14, 2018 at 19:25

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 !

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.