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I saw a bit different definition of derivative fiven by $f(x+\epsilon) = f(x)+f'(x)\epsilon + {\cal O}(\epsilon^2)$, where the ${\cal O}(\epsilon^2)$ stands for any function $g$ which satisfies that for all $x$ sufficiently large exists $c>0$ in which we have $g(x)\leq c\epsilon^2$. The defition I know is given by $r(\epsilon)$ instead of ${\cal O}(\epsilon^2)$, and satisfies $\lim_{\epsilon \to 0} r(\epsilon)/\epsilon = 0$. Could someone please clarify the other defition please.

piero
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  • Vaguely in both cases it states that the error of the output is vanishingly small compared to the error of the input. Landau notation avoids using limit notation so it's a little easier to typeset. https://en.wikipedia.org/wiki/Big_O_notation – CyclotomicField Jan 13 '24 at 00:05
  • @CyclotomicField but why $O(\epsilon^2)$? It shouldn't be $O(\epsilon)$? – piero Jan 13 '24 at 00:17
  • The problem with $O(\epsilon)$ is that sometimes you end up with $\lim_{\epsilon \to 0} r(\epsilon)/\epsilon= 1$. – CyclotomicField Jan 13 '24 at 00:23
  • @CyclotomicField but you also have functions that are both $O(\epsilon^2)$ and $O(\epsilon)$ – piero Jan 13 '24 at 00:31
  • $\epsilon$ is $O(\epsilon)$ and I think you can see why $\epsilon / \epsilon \neq 0$. Think of $r(\epsilon)/\epsilon$ in the case when it's an indeterminant form like $0/0$ and by requiring it to be $O(\epsilon^2)$ we avoid this case. – CyclotomicField Jan 13 '24 at 00:49
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    @piero You mean $o(\epsilon)$. – Ted Shifrin Jan 13 '24 at 02:28

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The general definition is with lower case $o(\epsilon)$ as remainder term, which is defined as the class of functions behaving like $r$. $O(\epsilon^2)$ works almost everywhere for most functions that one practically encounters, but there are examples that fall in-between.

Lutz Lehmann
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