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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.

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    $\begingroup$ The formulas are all the same, just minor changes of notation, depending on what feature we want to emphasize. $\endgroup$ – André Nicolas Aug 12 '13 at 2:33
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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.

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