# Notation: Is $(\Delta x)^2 = \Delta x^2$?

I read this in a book and was wondering whether it's valid or not: I thought $\Delta x^2$ would mean 'change in $x^2$', which would be quantitatively different to $(\Delta x)^2$; no?

• I guess change in $x^2$ would be rather $\Delta(x^2)$ – Adrian Jan 10 '16 at 11:16
• @addy2012 But this is just Classical Geometry, not calculus. Usually $\delta x$ represents an infinitesimal change - as far as I'm aware. – Luke Collins Jan 10 '16 at 11:17
• @addy2012 That was to your previous comment - sorry. Yeah true, I guess so. Perhaps $\Delta x^2$ just seems ambiguous and the book decides to stick to that meaning of it? – Luke Collins Jan 10 '16 at 11:19
• I would guess that $\Delta(x^2)$ would mean $(x+\Delta x)^2-x^2=2x\Delta x+(\Delta x)^2$, which is quite different from $(\Delta x)^2$ (which seems to be what is meant in your textbook). – Akiva Weinberger Jan 10 '16 at 12:11
• Just sloppy notation in the book, correct is $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ – vonbrand Jan 10 '16 at 15:10

This is just notation. It is a typical convention that $\Delta x^2 = (\Delta x)^2$.
You are right that it seems ambiguous, but it is consistent in the calculus literature that I have seen that whenever they write $\Delta x^2$, they mean $(\Delta x)^2$.
• Right. And it makes sense insofar as $\Delta x$ is basically a single symbol, not, like $\Delta$ multiplied by $x$. It's rather that $\Delta$ is an operator applied to $x$, and then I would parse $\Delta x^2$ as $(\Delta(x))^2$. Of course, by this logic, $\sin x^2$ would also have to be parsed as $(\sin x)^2$, which IMO would actually make sense too (certainly better than $\sin^2 x$). But I'm afraid most people would read it as $\sin(x^2)$ instead. – leftaroundabout Jan 10 '16 at 12:39
Yes, it is different from $(\Delta x)^2$. $(\Delta x)^2$ means square of change in $x$. Whereas $\Delta(x^2)$ means change in square of $x$.