I have been reading a book on elementary mathematics and have come to a problem where I don't understand where the solution comes from. The problem goes :
"Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be an odd function, such that $g(x) > 0$ whenever $x > 0$. Show that there exists a function $f : \mathbb{R} \rightarrow \mathbb{R}$ for which $g = f \circ f$."
In the answer key they just say :
"Take $f(x) = -g(x)$ for $x \leq 0$ and $f(x) = -x$ for $x \geq 0$."
I am wondering what the reasoning is behind this answer. The truth is that I just want to see a more detailed solution to the problem. Can anyone help with this ?
Interesting question, as an example consider the identity function for $g$, then $f$ is either trivially the identity function or more interestingly the negative the identity function (or more generally Babbage's functional equation, but that takes us off topic). What about $f(x)=x^3$? Well it's just as easy to solve the general case.
Now consider $f\circ f(x)=g(x)$ the naive attempt would be to let $f(x)=g(x)$ but then things wouldn't workout, so to fix the issue we need to define $f(x)=x$ sometimes, and $f(x)=g(x)$ other times. If we let $f(x)=x$ if $x>0$ then we get stuck because $f\circ f(x)=x\neq g(x)$ if $x>0$, so to avoid getting stuck we let $f(x)=-x$ if $x>0$, and once we do that we know that $f(x)$ must equal $g(-x)$ if $x<0$. so that $f\circ f(x)=g(-(-x))=g(x)$ if $x>0$, or $f\circ f(x)=-g(-x)=g(x)$ if $x<0$.(Since $g$ is odd)
Notice the solution is not unique, since we could have defined $f(x)=-x$ if $x<0$ then we would define $f(x)=-g(x)$ if $x<0$. Are there more?