I have been reading this blog concerning holomorphic functional calculus in Banach algebras. I am a bit stuck in Section 5: General Analytic functions of the blog.
5. General Analytic functions.
Define $\mathcal{O}(a)$ to be the algebra of all functions which are analytic in some neighbourhood of $\sigma(a).$ Our goal is to define $f(a)$ for $f \in \mathcal{O}(a).$ Working in this generality requires considerably more effort than the above cases of polynomials, rational functions with poles off $\sigma(a),$ and power series.
For example, suppose that $\sigma(a)$ is equal to the circle $\{z: > |z|=1\}.$ Let $U = \{z: 3/4<|z|<5/4\}.$ There are many functions analytic in $U$ which are neither rational nor represented by power series. Let $f$ be such a function. Since $f$ is not a rational function we do not have the natural algebraic formulas given in the rational case. We know that there is a power series $\sum c_n(z_0)(z-z_0)^n$ around every point in $z_0 \in U,$ that converges to $f$ in a small neighbourhood of $z_0.$ Should we plug a in one of these power series? Which one? It turns out that it is senseless to plug a in one of these power series (think about this). What we need is some formula into which we can plug a. What formulas do we have at our disposal?
It says that there are many functions analytic in the annulus U which are neither rational nor represented by power series. Can anyone give me an example of such functions?
Can anyone tell what the author means when he says: "Should we plug a in one of these power series? Which one? It turns out that it is senseless to plug a in one of these power series (think about this). What we need is some formula into which we can plug a. What formulas do we have at our disposal?"
He has tried to explain the same in the comments to the post but I am unable to comprehend the same.
For the first question, you could take for example $$F(z)=\frac{1}{1-e^z}$$ This is not rational but is holomorphic on $U$ since its poles are at integer multiples of $2\pi i$ (note that $|2\pi i|>5/4)$. In particular, $0$ is a pole hence no power series can be defined everywhere on $U$.
As for the second question, I think the point they are trying to make is that for a given holomorphic function on $U$, there is not necessarily one default power series that works for every point. A proto-typical example of a "nice" function is $e^z$, which is holomorphic on $U$. If I want to calculate $e^z$, I can always use the series $\sum\frac{z^n}{n!}$ regardless of which $z$ I happen to pick. However, not every holomorphic function is like this. For example, consider $F(z)$ or perhaps the simpler $f(z)=1/z$. Note that $f$ is clearly holomorphic on $U$, however the power series expansion for $f$ has very different coefficients depending on whether I expand around $z=1$ , $z=-1$ , $z=i+\frac{\pi}{163}$ , or what have you. If this doesn't seem obvious to you, try computing the coefficients \begin{align*}\frac{1}{z}=\sum a_n(z-1)^n&&\text{and}&&\frac{1}{z}=\sum b_n(z+1)^n\end{align*} The reason is that power series are always defined on an open ball centered at some point, and any open ball containing $U$ must contain the origin (why?). But $f$ is not holomorphic at the origin, so there cannot be one series that works for every point in $U$ simultaneously.