0

I have ran into it a few time in my books and it seems to be a function that tends to $0$ as its argument tends to $0$. They are all written in the same notation ($\varepsilon(x)$) and we do not use indices as all the epsilon functions behave the same way. Is it related to O-notation?

I could not find any useful sources.

A example can be seen in the comments

  • 3
    Like this? "For every $x$ there exists $\varepsilon=\varepsilon(x)$ such that ..." This is a (rather old-fashioned) way to say that $\varepsilon$ depends on $x$. If that is not it, then (as doobdood says) we need an actual example of this notation. – GEdgar Mar 23 '21 at 21:18
  • 1
    Could you add a more detailed description of these epsilon functions? There are many many many functions that go to 0 as $x$ goes to 0, and they may have different behaviours near 0. – doobdood Mar 23 '21 at 21:19
  • 2
    It would be best if you can exactly name books with appropriate pages. – zkutch Mar 23 '21 at 21:31
  • @GEdgar It is not used for continuouity and limits but rather to define things like differentiability i.e we say that $f$ is differentiable in $a$ with gradient $c$ if $\Delta f = c\Delta x + \epsilon (\Delta x) \left | \Delta x \right |$ where epsilon here is a epsilon function. –  Mar 23 '21 at 21:34
  • 1
    @zkutch It is a danish text book for my course but I have written an example here. –  Mar 23 '21 at 21:36
  • Even if it is in Danish, if you show pages, then together with your, or the help of Google, we will translate the places you are interested in and, I hope, we can help a little. – zkutch Mar 23 '21 at 21:42
  • I suspect this is indeed just a way to talk about $o$. Related: https://math.stackexchange.com/questions/4073417/symbol-for-very-small-variable/4073465#4073465 – Ethan Bolker Mar 23 '21 at 22:05

1 Answers1

1

Most likely it is little-$o$ notation. For example, in John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), page 69 you can find definition for differentiation for function of one real variable: $$\Delta y=\frac{dy}{dx}\Delta x + \varepsilon \Delta x$$ where $\varepsilon$ is infinitesimal, as wrote author, i.e. $\varepsilon(\Delta x)\to 0$, when $\Delta x \to 0$.

Farther, in same book, in page 267 for definition for differentiation for multiple variables is used formula: $$\Delta u = \frac{\partial u}{\partial x}\Delta x+\frac{\partial u}{\partial y}\Delta y +\varepsilon_1 \Delta x + \varepsilon_2 \Delta y$$ where again $\varepsilon_1, \varepsilon_2$ are infinitesimals i.e. tending toward zero as $\Delta x$ and $\Delta y$ tend toward zero.

At end let me bring one version of definition little-$o$: $$o(f),x\to x_0=\{g: \exists \varepsilon(x), \lim\limits_{x\to x_0} \varepsilon(x) = 0, \exists U_\delta(x_0), \forall x \in U_\delta(x_0), g(x)=f(x) \varepsilon(x) \}$$ where $U_\delta(x_0)$ is neighbourhood of $x_0$. This definition works as for single, so for multiple, real variables.

zkutch
  • 13,410
  • I remember getting explained (in my example) that both terms tend to 0 as $\Delta x$ tends to $0$. Would it simply be that a $\epsilon$-function tends to $0$ as its argument does? –  Mar 23 '21 at 22:36
  • Yes, it is in my answer - $\lim\limits_{x\to x_0} \varepsilon(x) = 0$. – zkutch Mar 23 '21 at 23:13