In sigma notation: $$P_N(x) = \sum_{n=0}^{N} \frac{f^{(n)}(a)}{n!}(x-a)^n$$
Remember that (with respect to this formula):
The polynomial $P_N$ is called the Nth-order Taylor polynomial of $f(x)$ at $x=a$.
Also, notice that it's called the Nth-"order" Taylor polynomial, not the Nth-"degree" Taylor polynomial.
That is becuase the degree of $P_N$ could be less than $N$; for instance, if the term $f^{(n)}(a)$ = 0, then the final term in the sum $P_N$ would disappear (i.e. be equal to zero), and the degree of $P_N$ would be at most $N-1$.
The most important property of the Taylor polynomial is that: $P_N^{(n)}(a) = f^{(n)}(a)$ for all $n = 0, 1, ..., N$.
Meaning, the values of all the derivatives of $P_N$ and $f$ match when $x=a$, up to and including the Nth derivative.
But, all higher derivatives of $P_N$ must be zero everywhere.
The function $P_N$ is the distillation of all the information about $f$ which comes from its derivatives up to order $N$ at $x=a$.
These are my notes & observations from "[The Calculus Lifesaver](http://press.princeton.edu/video/banner/)," by Adrian Banner.
Try this out with $f(x) = e^x$:
By the Taylor Approximation Theorem, with $N=3$ and $a=0$:
$$P_3(x) = f(0) + f'(0)(x)+\frac{f{(2)}(0)}{2!}(x^2) + \frac{f^{(3)}}{3!}(x^3)$$Since all the derivatives of $f(x) = e^x$ (wrt $x$) are just $e^x$, $f(0), f'(0), f''(0), \text{ and } f^{(3)}(0)$ are all $e^0$. Also, $2! = 2$ and $3! = 6$, so you're left with:
$$P_3(x) = 1 + x + \left( \frac{1}{2} \right)x^2 + \left( \frac{1}{6} \right) x^3$$
In [8]:
import sympy as sp
from sympy.mpmath import exp
%matplotlib inline
# Customize figure size
plt.rcParams['figure.figsize'] = 25, 15
#plt.rcParams['lines.linewidth'] = 1
#plt.rcParams['lines.color'] = 'g'
#plt.rcParams['font.family'] = 'monospace'
plt.rcParams['font.size'] = '16.0'
plt.rcParams['font.monospace'] = 'Anonymous Pro, serif'
plt.rcParams['text.hinting'] = 'either'
f = lambda x: exp(x)
p = lambda x: 1 + x + (1/2)*(x**2)+ (1/6)*(x**3)
sp.mpmath.plot([f, p], xlim=[-5,5], ylim=[-5,15], points=500)