Is $\lg^{-1}n = (\lg n)^{-1}$?

Question

Today my professor said that $\lg^{-1}(n)=\lg(n)^{-1}$. As I'm used to know, $\lg^{-1}(n)$ denotes the inverse function ($2^n$) and $\lg(n)^{-1}$ the reciprocal (inverse number) $(\frac{1}{\lg(n)})$. Is there a situation where this notation may fail?

Answer 1

Unfortunately, there are two different concepts that are both commonly written as $f^n(x)$:

1. Iteration of the function $f$, in which for example $f^3(x)$ means $f(f(f(x))$.
2. Pointwise power, in which $f^3(x)$ means $f(x)\cdot f(x)\cdot f(x)$.

Each of these make independent sense -- the first of these concepts is exponentiation in the monoid of function and function composition, whereas the second is exponentiation in the ring of functions with pointwise addition and multiplication, and both of these structures are important and common.

In the particular case $n=-1$, the iteration concept leads to $f^{-1}(x)$ meaning the inverse function of $f$ -- that is, $f^{-1}(x)$ is the $y$ such that $f(y)=x$; whereas the pointwise power leads to $f^{-1}(x)$ meaning $\frac{1}{f(x)}$.

When one encounters a $f^n(x)$, one simply needs either to guess from the context which of the two meanings appear to make sense, or to be told explicitly by the writer which of them he intends to use. The latter of these was what your professor was doing when he defined that he's going to use $\lg^{-1}(x)$ in the sense of $\frac{1}{\lg x}$ rather than as the inverse function. This can be a useful choice to make if you're going to meet a lot of reciprocals of logarithms, especially since the inverse function already has the nice short notation $10^x$ anyway.