What is $a$ in the Tayor series? So I'm trying to learn some calculus II on my own ahead of my first (online) college semester. I'm studying Taylor and power series right now through Paul's math notes.
Although I understand how to get the coefficients of the power series as the author nicely outline, I'm confused about what the variable '$a$' represents:

Is '$a$' just some value close to '$x$'? If so why?
The specifically claim that the formula for $f(x)$ above is the Taylor series for $f(x)$ about $x = a$. Can someone explain what this means?
Knowing this would be useful because later on they try to introduce Maclaurin series as the Taylor series about a = 0 and x = 0...
Any guidance is thoroughly appreciated!
 A: The idea of Taylor Series expanding a function is to take some point $a$ and take information about the function at that point. That information includes the value of the function at that point, and how the function is "changing" at that specific point (which is why derivatives are involved). We then use this information to "replicate" the function through a (potentially infinite) polynomial.
Ultimately, you must choose what the value of $a$ is you want to take. If ${a=0}$ - we are taking information about the function at the point ${x=0}$.
Note that "$a$ being some point close to $x$" in this context wouldn't make sense, because $a$ will be a constant, static variable. What I mean to say is that you pick the value of $a$ first, and then you get the polynomial you are interested in - and $x$ is just some free variable. For example, if I wanted to expand some function ${f(x)}$ about ${x=5}$, I would write
$${f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(5)(x-5)^n}{n!}}$$
If you want to calculate ${f(20)}$, you get
$${f(20)=\sum_{n=0}^{\infty}\frac{f^{(n)}(5)(20-5)^n}{n!}}$$
I could also expand around, say, ${a=3}$ and get:
$${f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(3)(x-3)^n}{n!}}$$
Hence also
$${f(20)=\sum_{n=0}^{\infty}\frac{f^{(n)}(3)(20-3)^n}{n!}}$$
(provided some conditions, anyway).
