Last time | Next time |
We can think of it as a function $a(x)$, but with integers for arguments, and real numbers for values: \[ a: \mathbb{N} \rightarrow \mathbb{R} \]
It might be better simply to write it as $a(n)$, but mathematicians have adopted the alternative notation $a_n$, perhaps to make clear that this function has a domain restricted to integers.
And this is one of the reasons that we are not too surprised when we find ourselves comparing series and improper integrals.
Better to look at a specific example: consider the cubic Taylor polynomial.
\[ T_3(x) = \frac{f^{(0)}(a)}{0!} (x-a)^0 + \frac{f^{(1)}(a)}{1!} (x-a)^1 + \frac{f^{(2)}(a)}{2!} (x-a)^2 + \frac{f^{(3)}(a)}{3!} (x-a)^3 \] or \[ T_3(x) = f(a) + f'(a) (x-a) + \frac{f''(a)}{2!} (x-a)^2 + \frac{f'''(a)}{3!} (x-a)^3 \] Then the derivative of $T_3(x)$ is \[ T_3'(x) = \sum_{k=1}^{3} \frac{f^{(k)}(a)}{(k-1)!} (x-a)^{k-1} \] or \[ T_3'(x) = f'(a) + \frac{f''(a)}{1!} (x-a)^1 + \frac{f'''(a)}{2!} (x-a)^2 \] Notice how we lost the $k=0$ term -- it was a constant, and when we differentiated we lost it. Continuing on, \[ T_3''(x) = f''(a) + \frac{f'''(a)}{1!} (x-a)^1 \] and \[ T_3'''(x) = f'''(a) \]
Remember that it's all about the tail. In terms of convergence, we don't care about any handful of terms at the head of a series: \[ \sum_{k=1}^\infty a_k = \sum_{k=1}^m a_k + \sum_{k=m+1}^\infty a_k \]