$r\rightarrow1/r$ invariant (Not sure the tags are appropriate, but can't think of better ones. Please suggest better.) Suppose you have a function $f(x,y,z,...;g(r))$ with the requirement that $r\rightarrow1/r$ leaves $f$ invariant. That works if $g(r)=\frac r{r^2+1}$. Are there other $g$'s such that $g(r)=g(1/r)$?
More generally, for any $h:\mathbb R\rightarrow \mathbb R$ (where in the above case $h=1/r$), how do you find the function $K:(\mathbb R\rightarrow\mathbb R)\longrightarrow(\mathbb R\rightarrow\mathbb R)$ such that $g=K(h)$ leaves $g(r)=g(h(r))$ invariant under $r\rightarrow h(r)$? That is, for example, in the above case $K(\frac1r)=\frac r{r^2+1}$.
So, this $K$ is kind of like the paradoxical combinator which, when applied to any expression (any $\lambda$-expression), renders a fixed-point of that expression (is that about right?). But I can't find quite what I'm asking for in $\lambda$-calculus books. Is this question already a textbook topic?
 A: One simple way to construct $g:\mathbb{R} \to \mathbb{R}$ such that $g(r)=g(1/r)$ is to pick an arbitrary function $g':[-1,1] \to \mathbb{R}$ and then define
$$g(x) = \begin{cases}
g'(1/x) &\text{if $x>1$ or $x<-1$}\\
g'(x)   &\text{if $-1 \le x \le 1$.}
\end{cases}$$
By construction, this satisfies your requirements.  As a bonus, if $g'$ is continuous, then $g$ will be continuous too.
How did I pick the interval $[-1,1]$?  Well, $-1$ and $1$ are the only fixpoints of the map $r \mapsto 1/r$, and $r \mapsto 1/r$ is continuous on $(0,1]$ and $[-1,0)$.  Most importantly, we can partition the domain $\mathbb{R}$ into $\mathbb{R} = (-\infty,-1) \cup [-1,0) \cup \{0\} \cup (0,1] \cup (1,\infty)$ where if $r,r'$ are in the same part of the partition, then $1/r$ and $1/r'$ are in the same part of the partition.  The fixpoints of $r \mapsto 1/r$ help you find the "cutpoints" of the partition.  It's possible this might help you generalize to continuous $h$, by looking at the fixpoints of $h$ and the regions where $h$ is monotonic.
