Could someone clarify whether this is correct?
If $$D(x) = \frac{5x + 8}{x - 15},$$
would $$D(x^2) = \frac{5x^2 + 8}{x^2 - 15},$$
and $$D(x)^2 = \frac{5x + 8}{x - 15}\times\frac{5x + 8}{x - 15}?$$
Clarification appreciated!
Could someone clarify whether this is correct?
If $$D(x) = \frac{5x + 8}{x - 15},$$
would $$D(x^2) = \frac{5x^2 + 8}{x^2 - 15},$$
and $$D(x)^2 = \frac{5x + 8}{x - 15}\times\frac{5x + 8}{x - 15}?$$
Clarification appreciated!
On
The parenthesis behind a function identifier isolate its argument. So
$$D(x^2)$$ perforce denotes the function $D$ evaluated at $x^2$.
On the other hand,
$$D(x)^2$$ cannot be interpreted as $$D((x)^2)$$ because the inner parenthesis would be superfluous and the outer ones missing. Hence you have the choice between
meaningless, or
$(D(x))^2$.
An alternative notation is $D^2(x)$, though in some rare cases (which you will be aware of by context), this designates $D(D(x))$.
Yes, you are completely right. However, it might be worth noting a few other things:
• Trigonometric functions do not follow this pattern. $\sin^2(x)$ means $(\sin x)^2$, not $\sin(\sin(x))$.
• Some notations are more readable than others. In my opinion, $(\log x)^2$ is preferable to $\log(x)^2$, though they both unambiguously mean the same thing. (Incidentally, some authors adopt a trigonometric-type convention for logarithms, which is awful!)