What kind of functions cannot be described by the Taylor series? Why is this? It's true that I'm not familiar with too many exotic functions, but I don't understand why there exist functions that cannot be described by a Taylor series?  What makes it okay to describe any particular functions with such a series? Is there any difference for different number sets? In the case of complex numbers maybe? 
Could somebody provide an example?
 A: I think the intuition you want is the fact that functions that are not complex-differentiable* (also known as holomorphic) are not described by a Taylor series.  
And to give another example that is perhaps even more unexpected than the one given by Andrew:
$$f(z) = \begin{cases} e^{-\frac{1}{z}} && \text{if } z > 0 \\ 0 && \text{otherwise}\end{cases}$$
This function is smooth and zero over an infinitely long interval, and yet nonzero, because it is not holomorphic.

*If you're not familiar with complex differentiation, it's like real differentiation, with $h$ complex:
$$f'(z) = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h}$$
For details, see here.
A: The existence of functions that cannot be described by Taylor series is actually completely intuitive; take the indicator function of the rational numbers viewed as a subset of the reals, for example. Try to keep in mind that functions can be really... arbitrary.
Much more subtle is the existence of smooth functions that aren't analytic; Thomas Andrews gives the standard example of such a beast. Fwiw, my understanding of why this is possible is that okay, there's functions that change behaviour suddenly at a point, BUT the change in behaviour at that point is so gradual, so gentle, so smooth, that none of the function's derivatives can see the change happening; therefore, the Taylor series can't, either.
A: In addition to all the comments here, I would like to add the curious Weierstrass function, which is known for its quality of being nowhere differentiable despite the fact that it is continuous everywhere:
$$ W(x) = \sum_{n=0}^\infty a^n\cos(b^n\pi x)$$
Consequently, it does not have a Taylor series.
You can find a visualization of $W$ here.
A: We have the somewhat famous function:
$$f(x)=\begin{cases}e^{-1/x^2}&x\neq 0\\
0&x=0
\end{cases}$$
is infinitely differentiable at $0$ with $f^{(n)}(0)=0$ for all $n$, so, even though the function is infinitely differentiable, the Taylor series around $0$ does not converge to the value of the function for any $x>0$.

Technically, any function that is infinitely differentiable at $a$ has a Taylor series at $a$. Whether you find that Taylor series useful depends on what you want the series to do.
For example, if given a $g$ infinitely differentiable at $0$, the we know that there exists $C,\epsilon>0$ such that:
$$\left|g(x)-\sum_{k=0}^{n} \frac{g^{(k)}(0)}{k!}x^k\right|<Cx^{n+1}$$
for all $|x|<\epsilon$.
So the finite terms of the Taylor series are in some sense always the "best" polynomial for agreeing with function.
So what happens to our function $f$ above is that $f(x)$ converges to $0$ faster than any function $x^n$.
What we don't always get, for real functions, is a Taylor series that converges to the function in the interval. 

In complex numbers, things become intriguing. It turns out, if you define differentiation on complex functions in a relatively simple way, then any function which is differentiable at a point is infinitely differentiable at that point, and the Taylor series converges in some "ball" centered on that point. 
A: If the limit of the Lagrange Error term does not tend to zero (as $n \to \infty $), then the function will not be equal to its Taylor Series.
You can also read more on this in Appendix $1$ in Introduction to Calculus and Analysis $1$ by Courant and John. Hope it helps. 
