11) Let a = 3.55 and plot f^3(x), f^4(x), f^5(x), f^6(x), f^7(x), f^8(x) along with the replacement line y = x. List how many periodic points f(x) has of every period less than or equal to 8. Do you think that f(x) has any higher periodic orbits (when a = 3.55?) Why or why not?

Input := 

ClearAll[a, f, g, h]

Input := 

a = 3.55;
f[x_] = a x ( 1 - x );
g[x_] = f[f[x]];
h[x_] = g[g[x]];
i[x_] = h[h[x]];

Input := 

Plot[{x, f[x], g[x], g[f[x]], h[x]}, {x, 0, 1},
 AspectRatio -> Automatic,
 PlotStyle -> {AbsoluteThickness[0], AbsoluteThickness[4],
               AbsoluteThickness[3], AbsoluteThickness[2],
               AbsoluteThickness[1]}]

Output =

-Graphics-

Input := 

Plot[{x, h[f[x]], h[g[x]], h[g[f[x]]], i[x]}, {x, 0, 1},
 AspectRatio -> Automatic,
 PlotStyle -> {AbsoluteThickness[0], AbsoluteThickness[4],
               AbsoluteThickness[3], AbsoluteThickness[2],
               AbsoluteThickness[1]}]

Output =

-Graphics-

Input := 

Plot[{x, i[x]},
   {x, 0.34, 0.38},
   PlotRange -> {0.34, 0.38},
 AspectRatio -> Automatic,
 PlotStyle -> {AbsoluteThickness[0],
               AbsoluteThickness[2]}]

Output =

-Graphics-

If there is a period-8 orbit, then the graph above should show three crossings -- the middle fixed point corresponding to a period-4 point and the other two corresponding to period-8 points. We further note that the slope of the period-8 map is less than one at the period-8 points, so we conjecture that there are no more doubled periods since the period-8 orbit is attracting. So, there are 8 period-8 points (one period-8 orbit) which is attracting, 4 period-4 points (one period-4 orbit) which is repelling, 2 period-2 points (one period-2 orbit) which is repelling, and 2 fixed points which are repelling as well. There are no periodic points of any other period.