The idea of the proof is captured by considering a 2-cycle |
First, note |
f2(x1) = f(f(x1)) = f(x2) = x1 |
and |
f2(x2) = f(f(x2)) = f(x1) = x2 |
Now to test the stability of x1 as a fixed point of f2(x),
we compute the derivative
|
By the chain rule |
(f2)'(x1) = f'(f(x1))⋅f'(x1) = f'(x2)⋅f'(x1) |
Similarly, |
(f2)'(x2) = f'(f(x2))⋅f'(x2) = f'(x1)⋅f'(x2) |
That is, |
Return to stability of cycles.