I am told to find the stable period $4$ orbit of $f(x) = 13xe^{-x}$ for my discrete dynamical system through direct numerical iteration. However, I am a bit confused on what is meant by direct numerical iteration.
Usually I can just find $x$ such that $f^{(2)}(x)=x$, but for this function that is quite complicated to do. So how can I find the period $4$ equilibrium points through numerical iteration? I can plot $f$ and $y=x$ together and then vary $x$ and look at the cobweb diagram, but I don't think that is what I need.
Hint: If $x$ is a stable equilibrium point for $f^4$, what happens if you choose $y$ (hopefully reasonably close to $x$) and compute $$y, \, f^4(y), \, f^8(y), \, f^{12}(y), \, f^{16}(y), ... $$ ???
Followup: Let me expand my hint.
This is experimental mathematics: you pick values; you make guesses; you compute; you look for patterns; you try to prove your patterns.
Pick some starting value of $y$. Any value. Let's pick $y=1$. To remember that it is a starting value, let's put a subscript on it: $y_0=1$.
Now let's start computing. Mathematica might be useful. But, not having it at my fingertips, I searched for a web calculator, and settled on http://web2.0calc.com $$f(y_0) = f(1) = 13 \times 1 \times e^{-1} \approx 4.7824 $$ $$f^2(y_0) \approx f(4.7824) = 13 \times 4.7824 \times e^{-4.7824} \approx .52072457 $$ $$f^3(y_0) \approx f(.5207245) = 13 \times .52072457 \times e^{-.52072457} \approx 4.0216418 $$ $$f^4(y_0) \approx f(4.0216418) = 13 \times 4.0216418 \times e^{-4.0216418} \approx .937065 $$ That's your first iteration, let's call it $y_1$: $$y_1 \approx .937065 $$ Now on to your second iteration: $$f(y_1) \approx 4.772554 $$ $$f^2(y_1) \approx f(4.772554) \approx .524807 $$ $$f^3(y_1) \approx f(.524807) \approx 4.036660 $$ $$f^4(y_1) \approx f(4.036660) \approx 0.926544 $$ And that's your second iteration, let's call it $y_2$: $$y_2 \approx 0.926544 $$ So far, we have computed the first three terms of a sequence $y_0,y_1,y_2,y_3,y_4,...$ the totality of which is defined by recursion: $y_0=1$, $y_{n+1} = f^4(y_n)$ for all $n \ge 0$.
So far there's not too much of a pattern, other than $y_0 > y_1 > y_2$. But that's not bad actually, because a decreasing sequence of positive numbers must converge. Of course, you don't know for sure yet whether the entire infinite sequence is decreasing.
Shall we try to prove rigorously, perhaps by an induction argument that the sequence must decrease? Shall we compute a few more terms to get a better feel for the pattern? Shall we decide that our first pick $y_0=1$ was not very helpful and try another pick? Only time will tell...