Given enough terms, does a taylor series become equivalent to the function it is approximating? I've recently started learning about Taylor/Maclauren series and I'm finding it a bit hard to wrap my head around a few things.
So, if $f(x)$ is not infinitely differentiable and we construct a polynomial $p(x)$ such that $p(a) = f(a)$, $f'(a) = p'(a)$, $f''(a) = p''(a)$ etc, for all possible derivatives of $f(x)$, can we say that the functions $f(x)$ and $p(x)$ are equivalent?
Likewise, if $f(x)$ is infinitely differentiable, does the taylor series $p(x)$ become equivalent to $f(x)$ when given an infinite number of terms?
 A: For your first question, such an $f$ will never be equal to its Taylor series for the simple fact that by assumption $f$ is not infinitely differentiable but its Taylor series (just a polynomial) is infinitely differentiable.
Your second question is more interesting.  If $f(x)$ is infinitely differentiable then it may not be equal to its Taylor series anywhere!  A classic example of this is the function$$
f(x) = \begin{cases}
e^{-1/x^2}; &x \ne 0 \\
0; &x = 0.
\end{cases}
$$
You can check that all of its derivatives at zero are 0.  This means the Maclaurin series is just the 0 polynomial.  However, in no neighborhood of 0 is $f$ always zero, i.e. $f$ does not equal its Maclaurin series in any neighborhood.  Functions that equal their Taylor series are called analytic and for real functions I don't know of an easy to check criteria for a function being analytic.  
For complex functions the situation is much nicer: if a complex function is differentiable then at any point $a$ it is equal to its Taylor series centered at $a$ on the largest disk about $a$ that does not contain a singularity of the function.
