In Taylor series, what's the significance of choosing the point of expansion $x=a$?

Solution 1:

Sticking to your example of $e^x$, if you can expand it around $a=1$, then you already know the value of $e^x$ at $x=1$. In other words, you would not use the Taylor expansion to approximate a function about a point you already can compute the value at.

Choosing the point for the expansion is largely a question of computational ease and what's available. It's a lot easier to compute the Taylor expansion of, say, $e^x$, $\sin(x)$, or $\cos (x)$ about the point $x=0$ then it would about the point $x=0.12345563$ or $x=\pi + 6.7$ for the simple reason that it's so easy to compute the value the derivatives attain at $x=0$, but less easy (and a lot more messy) at other points. Issues of suitably approximating the error are of importance here, as well as making a choice that will increase the speed of convergence could be relevant.

Also, when one tries to extrapolate a function from given empirical values you simply have to work with what you have. If you have more numerical information about a function and its derivatives at and about a point $a$ than you have at or about a point $b$, then use $x=a$ as the point for the Taylor expansion.

Solution 2:

Yes, it would matter, because the idea of the Taylor expansion is that you want to evaluate f(x), when x is very close to a. This means that the closer x is to a, you'll need to sum up less terms to get to a certain accuracy.

For example, if I wanted to calculate $ e^2 $, then it would be smart to expand around a=1, because 2 is closer to 1 than to 0, but if I wanted to calculate $e^{0.234325}$, then it would be smarter to expand around a=0.