TITLE: trying to understand the incremental form of the Taylor series expansion
QUESTION [5 upvotes]: original form of Taylor series
$f(x)= f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2+ \cdots\\\\$
By making a substitution, we can find something that resembles the incremental form of the Taylor series:
$\text{let } x-a = h, \text{ so } x=h+a\\\\$
$f(x) = f(a+h)= f(a) + \frac{f'(a)}{1!}(h) + \frac{f''(a)}{2!}(h)^2+ \cdots \\\\$
If we now replace $a$ with $x$, we arrive at the incremental form of the Taylor series:
$ f(x+h)= f(x) + \frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2+ \cdots \\\\$
It looks as if we're expanding at some unspecified point $x$ instead of $a$ and using $h$ as an error term that will better approximate the value at $f(x)$ as we use higher order terms. Could someone confirm whether this thinking is correct? This explanation makes sense to me intuitively, but I'm not sure.

REPLY [3 votes]: Yes, your thinking is correct, with a minor correction as follows: "...using ${h^n \over n!}$ as an error term that will better approximate the value at $f(x+h)$ as we use higher order terms."  Once you made the substitution "...replace $a$ with $x$,...", you suddenly have a different $x$ value from your starting point.