I am having trouble understanding the way applying a Taylor expansion.
Say I want to approximate $e^2$.
According to the first formula in the Wikipedia page I have $e^2 = e^2 + e^2(x-2)+\frac{e^2}{2!}(x-2)^2...$
How come this is useful? We still need $e^2$ in the right hand side of the equation.
The idea of a taylor expansion is to use known values of a function and some number of its derivatives "nearby" to an unknown value to get a polynomial approximation for the function at the unknown point. The nearby is important as the error will ultimately be weighted by the distance between the unknown value and the known.
For your example (taylor expanded incorrectly fyi) let us leverage the values of $e^0=1$, and note that $(e^x)'=e^x$, then $$ e^2=e^0+e^0(2-0)+\frac{e^0}{2}2^2+\frac{e^0}{3!}2^3+\text{error}=19/3+\text{error} $$ Which is off by about 1. Continuing, we can get as close as we would like.
Edit: In response to question on the error, if you expand to degree $n$, then your error is given by $$ \frac{f^{(n+1)}(\xi)}{(n+1)!} $$ Where $\xi$ is some value in the interval between the value you know and are leveraging, and the unknown. Bad news: you don't know this value $\xi$, good news, if you have a bound on the $n+1$ derivative on the interval, you get a bound on this error.