Finding Critical Points of the Model $x=\exp(-rx)$

Question

Given the model $x_{t+1} = e^{-rx_t}$, how do I find the fixed/critical points? I know that critical points are $x = f(x)$ but how would you go about that here? Either as originally put or $\frac{\ln(x)}x = -r$ but not sure how to interpret that.

Answer 1

Here is a geometric argument, that gives the counts and approximate positions of fixed points:

Put $f_r(x)=f(r,x)=e^{-rx}$. Then the fixed points of $f_r$ correspond to the intersection points of the graph of $f_r$ and the main diagonal $y=x$. First if $r=0$, then $f_r(x)=1$, so it has a unique fixed point (namely $x=1$). If $r>0$, then $f_r$ is strictly decreasing, and consequently again it has a unique fixed point (which is in $]0,1[$).

Note that $f'_r(x)=-re^{-rx}$, so if $r<0$, then both $f_r$ and $f_r'$ are strictly increasing. Thus there is a unique point $x_r$ such that $f'_r(x_r)=1$; namely

$$x_r = -\dfrac{1}{r}\ln\left(-\dfrac{1}{r}\right)=\dfrac{\ln(-r)}{r}.$$

Depending on the position of $x_r$, $f_r(x)$ will have none, exactly one, xor exactly two fixed points, as one could think of the graph of $f_r$ "bouncing off" of the line through $(x_r,f_r(x_r))$ with slope $1$, thus it is useful to understand the function $x_\bullet$ better. Taking its derivative (note that applying the Implicit Function Theorem to the real analytic $(r,x)\mapsto e^{-rx}$ gives that $r\mapsto x_r$ is real analytic, hence we may safely take derivatives) we have

$$\dfrac{dx_r}{dr}=\dfrac{1-\ln(-r)}{r^2},$$

consequently it has exactly one critical point (namely at $r=-e$). Computing $x_{-e}$ $=-\dfrac{1}{e}$ $<0=x_{-1}$, we have that $r=-e$ is a global minimum for $x_\bullet$; for $r\in ]-e,0[$, $x_r$ is strictly increasing; and for $r<-e$ we have that $x_r$ is strictly decreasing. An application of the L'Hôpital's rule gives that $\lim_{r\to-\infty} x_r=0$, consequently for $r<-e$, $x_r\in [-1/e,0[$. Also note that $\lim_{r\to\infty}x_r=\infty$, so in conclusion there is a unique number $r^\ast<0$ such that

• If $r<r^\ast$, then $f_r$ has no fixed points,
• $f_{r^\ast}$ has exactly one fixed point (which is $x_{r^\ast}$),
• If $r\in ]r^\ast,0[$, then $f_r$ has exactly two fixed points,
• If $r\geq 0$, then $f_r$ has exactly one fixed point.

We can also calculate $r^\ast$, as by definition it is the unique $r$ such that $f_r(x_r)=x_r$; expanding this equation gives $r^\ast=-\dfrac{1}{e}$.

Here are some accompanying graphs (the interactive graph is available at https://www.desmos.com/calculator/iuupxth2mn):