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!
Best Answer
You're right, in general $f$ is not determined by its derivatives at one single point. Functions satisfying this condition are called analytic. But not all smooth functions are analytic, for example
$$x\mapsto\left\{\begin{array}{c}e^{-\frac{1}{x^2}}, x>0\\0, x\leq 0\end{array}\right.$$ is a smooth function and the derivatives at zero are all zero, hence the Taylor series developed at zero does not determine the function.
Furthermore the exact statement of Taylor's theorem is quite different from what you said. It is as follows:
If $f\in C^{k+1}(\mathbb{R})$, then $$f(x)=\sum_{n=0}^k f^{(n)}(a)(x-a)^n\frac{1}{n!} + f^{(k+1)}(\xi)\frac{1}{(k+1)!}(x-a)^{k+1}$$
If you now take $k\rightarrow\infty$ it is in general not clear, that this error term converges to zero.