Taylor Series centered at $\alpha$. Is $\alpha$ arbitrary?

Question

Let's say I have a function $f \in C^{\infty}$ represented by a Taylor Series

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(\alpha)}{n!}(x-\alpha)^n$$

What relevance does $\alpha$ have to $f$? Can we just choose any $\alpha$ to be a value where we know how to evaluate $f^{(n)}(\alpha)$?

i.e. Do we just choose a $\alpha$ such that we know the value of the function $f$ and all it's derivatives at that point $\alpha$? If so is $\alpha$ just purely arbitrary?

Answer 1

First we need to note we have to be careful about the hypotheses, because $f\in C^\infty(\Bbb R)$ does not imply that $f$ is represented by its Taylor series. Yes, $\alpha$ is arbitrary, in the sense that if there exists an $\alpha$ that works for every $t$ then any other $\alpha$ also works:

Theorem. Suppose that $f(t)=\sum\frac{f^{(k)}(\alpha)}{k!}(t-\alpha)^k$ for every $t\in\Bbb R$. Then $f(t)=\sum\frac{f^{(k)}(\beta)}{k!}(t-\beta)^k$ for every $t,\beta\in\Bbb R$.

Answer 2

No, It's not arbitray. Because if you choose an alpha and in that alpha, f(alpha) or f'(alpha) or ... Become infinite it won't have meaning. For better understanding try expand f(x)=sqrt(x) and choose alpha=0, you will understand why we can't use any alpha.

Answer 3

It depends. First, you must choose an $\alpha$ such the radius of convergence of the power series is greater than 0. Second, if you are using the power series to compute values of the function for specific values of $x$, then you would want to choose a value of $\alpha$ close to that of $x$ to minimize the number of terms in the series needed to achieve a given accuracy.