I was attempting to help answer a question, but I am curious about the following.
Suppose we have an analytic function $f(z)\equiv\sum\limits_{k=0}^{\infty} a_k(z-z_0)^k $ on some suitable disk of convergence in $\mathbb{C}$, if $g(z)$ is some other function, when is the taylor series $\sum\limits_{k=0}^{\infty} a_k(g(z)-z_0)^k $ equal to $f(g(z))$?
The example I was attempting to address was the radius of convergence in the real taylor series in the following question:
Range of convergence for Taylor's series (about 0) for e^(sin x)
but another one I am curious is the following:
$e^z\equiv\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}$ for all $z \in \mathbb{C}$.
The exponential function is entire. But suppose we compose with another function e.g. going back to $\mathbb{R}$ with the $\mid x\mid$ function, we could sub in to the taylor series about and get the following series:
$\sum\limits_{n=0}^{\infty}\frac{\mid x\mid^n}{n!}$
but would this converge to $e^{\mid x\mid}$, $\forall x \in \mathbb{R}$.
I expect it would but it shouldn't be a taylor series for the function $e^{\mid x\mid}$, because it isn't analytic at $x=0$ right?
If $f(z) = g(z)$ for $z$ in some set of complex numbers, you can substitute any expression $w$ for $z$, and the equation $f(w) = g(w)$ will still be true as long as $w$ is in that same set. There's nothing special about series here.