Question about the Taylor theorem and different ways of expressing it In my lecture notes, it says that the Taylor theorem in one dimension is given by 
$$f(x+ \delta x) \simeq f(x)+\delta x f_{x} + \frac{1}{2!} (\delta x)^2 f_{xx}+ \dots$$
Conversely, when I was back at school, I was taught that the Taylor theorem, evaluated about some point $x=a$, in one dimension is given by 
$$f(x) \simeq f(a) + f'(a)(x-a) + f''(a) \frac{(x-a)^{2}}{2!} + \dots $$
Are these definitions equivalent? It would appear that they are if $\delta x = x-a$, but the second definition says that it is evaluated "about some point" whilst the second definition mentions nothing about where the function is evaluated.
Secondly, what does it mean that the function is evaluated about a certain point $x=a$? The Taylor expansion is still a function in terms of $x$, so why should it make a difference where you evaluate it? For example, the series expansion of $e^{x}$, evaluated at the point $x=0$ is given by
$$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$$
However, the series expansion of $e^{x}$, evaluated at the point $x=a$ is given by
$$e^{x}=e^{a}+e^{a}(x-a)+\frac{e^{a}(x-a)^{2}}{2!}+\frac{e^{a}(x-a)^{3}}{3!}+...$$
What is the practical difference between these expansions of $e^{x}$?
 A: 
$$f(x+ \delta x) \simeq f(x)+\delta x f_{x} + \frac{1}{2!} (\delta x)^2 f_{xx}+ \dots$$

This kind of notation is often used by physicists and engineers, who are usually not interested in mathematical niceties and use $x$ to mean a couple of different things at the same time.
As written the series is expanded around $x$, which is also the name of the variable used for the domain.

$$f(x) \simeq f(a) + f'(a)(x-a) + f''(a) \frac{(x-a)^{2}}{2!} + \dots $$

This is closer to a notation typically used by mathematicians as it more carefully delineates the point around which the series is being expanded.
This series is expanded around $x = a$. I would say the series is being evaluated at $x$.

Are these definitions equivalent? 

Yes, as per the translation your propose and the untangling of the use of $x$ in the first series.

What is the practical difference between these expansions of $e^{x}$?

If you want to find $e^{2.1}$ and you knew $e^2$ say, you would prefer to expand around $a = 2$ than around $a = 0$ as it will converge more quickly to $e^{2.1}$ or equivalently the size of the error for $n$ terms is materially smaller for the series expanded around $a = 2$ vs. $a = 0$.
A: Both expansions are correct. One practical difference is that usually when you approximate $f(x)$ around a particular point $a$, the farther away $a$ is from $x$, the more terms of the expansion you will need to compute to get an approximation with good precision.
E.g. you can compute $e^\pi$ expanding around $0$ with your first formula, or expanding around $a=3$, and the second approximation will converge much faster.
