Is Taylor Series changing from an uncountable basis to a countable basis? Say we've got an analytic function $f(x)$ from $\mathbb R$ to $\mathbb R$. It has an uncountable number of components in this basis, since there is one value of $f(x)$ for each $x$ and $x$ varies continuously. 
When we do a Fourier transform and hence change the variable to $w$, there is still an uncountable number of components, because $w$ too varies continuously. 
But, in case of a Taylor series, the function can be represented by a countable number of components: $[f(0)\,\, f'(0)\,\, f''(0)\,\, f'''(0)\, \cdots]$. In this case, the number of components changes from uncountable infinite to countably infinite, when we change the basis. 
Am I interpreting this right? The number of components never changes when we change the basis of a finite-dimensional vector space. Why does the number of components change here?

It is true that the domain ${\mathbb R}$ of a function $f:\>{\mathbb R}\to{\mathbb R}$ has uncountably many points. If you want to produce the function values   by a random number generator independently for each point then you would have to call this routine an uncountable number of times. But already a continuous function, let alone an analytic function, is completely determined by its values on a dense subset of ${\mathbb R}$, e.g., the countable set of rationals: Given a counting ${\mathbb Q}=\bigl\{ r_n\bigm| n\geq1\bigr\}$ it is sufficient to store the list $(y_n)_{n\geq1}$ of function values $y_n:=f(r_n)$. In the case of an (entire) analytic function you can store the list of Taylor coefficients $a_n:={1\over n!} f^{(n)}(0)$ $(n\geq0)$ and have enough information to produce the exact value of $f(z)$ even for all $z\in{\mathbb C}$. Note that there are some very special functions where even a few dozen "tokens" suffice to define them at all points $x\in{\mathbb R}$, e.g., $\exp(x):=\sum_{k=0}^\infty{1\over k!} x^k$.
