General question on Taylor Series The Taylor Series comes from an assumption that a function has an expression as power series. Given such assumption we can then say that the $n$-th derivative and evaluate them at $x = a$, it can give us the coefficient on the $n$-th term when the function converges on $a$. How do we know that every function has a power series?
 A: Verily, the story begins at the Taylor polynomials. 
Let $a \in \mathbb{R}$; let $N(a)$ be a 1-neighborhood of $a$; let $f: N(a) \to \mathbb{R}$; and let $D^{n}f(a)$ exist. Then it can be shown that (try to prove it) the polynomial $p_{n}: x \mapsto f(a) + \sum_{1}^{n}D^{k}f(a)(x-a)^{k}/k!$ on $N(a)$ is the unique choice such that
$$
D^{k}p_{n}(a) = D^{k}f(a)
$$
for all $0 \leq k \leq n$. 
And $p_{n}$ is called the Taylor polynomial of degree $n$ generated by $f$ at $a$.
The problem is about the remainder, or the error term $E_{n}: x \mapsto f(x) - p_{n}(x)$ on $N(a)$. Thanks to a theorem neighboring on the mean-value theorem for integrals, under suitable conditions we can write $E_{n}$ in the so-called Lagrange form 
$$\frac{D^{n+1}f(c)(x-a)^{n-1}}{(n+1)}.$$
And it can shown that, under suitable conditions, if there is some $M \geq 0$ such that $\sup_{x \in N(a)}|D^{k}f(x)| \leq M^{k}$ for all $k \geq 1$, then the error term vanishes as $k$ grows, and hence we can express $f$ as its Taylor polynomial to any degree at $a$. 
A: We don't. For example, the Weierstrass function will have no power series that works everywhere on the real numbers. This is true for any non-differentiable function, e.g. $f(x) = |x|$ will need different power series for different domains, but no one single one will work...
