TITLE: Doubt about Taylor series: do successive derivatives on a point determine the whole function?
QUESTION [24 upvotes]: I'm currently relearning Taylor series and yersterday I thought about something that left me puzzled. As far as I understand, whenever you take the Taylor series of any function $f(x)$ around a point $x = a$, the function is exactly equal to its Taylor series, that is:
$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n $$
For example, if we take $f(x) = e^x$ and $x = 0$, we obtain: $ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $
My doubt is: the only variables in the Tayor series formula are $f(a), f'(a), f''(a),$ etc., that is, the successive derivatives of the function $f$ evaluated in one point $x = a$. But the Taylor series of $f(x)$ determine the whole function! How is it possible that the successive derivatives of the function evaluated in a single point determine the whole function? Does this mean that if we know the values of $f^{(n)}(a)$, then $f$ is uniquely determined? Is there an intuition as to why the succesive derivatives of $f$ on a single point encode the necessary information to determine $f$ uniquely?
Maybe I'm missing a key insight and all my reasoning is wrong, if so please tell where is my mistake.
Thanks!

REPLY [4 votes]: HumanStampedist has adequately answered the question. I'd like to mention that, just as there are continuous functions that are nowhere differentiable, such as the Weierstrass function, there are smooth functions (all $n$th derivatives exist at every point) that are nowhere analytic, i.e. at no point does the Taylor series converge to the original function. An example is the Fabius function.