Taylor Series centered at $\alpha$. Is $\alpha$ arbitrary? 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?
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 $t,\beta\in\Bbb R$.
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.