Attractive parameters

quiz-prep

#1

Show that f_c(x) = x^2+c has an attractive orbit of period 2 when -5/4<c<-3/4.


#2

Someone tell me if I’m doing something wrong or misunderstanding something.

Pick f_{-1}(x)=x^2-1.

Note that -\frac{5}{4} < -1<-\frac{3}{4}.

Suppose f_{-1}(x) = x^2-1=x.

Then x^2-x-1=0.

This implies that x=\frac{1+\sqrt{5}}{2} or x = \frac{1-\sqrt{5}}{2}. These are the fixed points of f_{-1}(x).

Note that f_{-1}'(x) = 2x.

So then |f_{-1}'(\frac{1+\sqrt{5}}{2})| = |2(\frac{1+\sqrt{5}}{2})| = 1+\sqrt{5} > 1.

Also, |f_{-1}'(\frac{1-\sqrt{5}}{2})| = |2(\frac{1-\sqrt{5}}{2})| = |1-\sqrt{5}| >1.

So then f_{-1} doesn’t have an attractive fixed point.


#3

@Ed_Boy Yes, you are correct. I meant to type an attractive orbit of period 2. A simpler version of the problem would be "Show that f_c has an attractive fixed point when -3/4<c<1/4".


#4

(Answering the revised question)

Let f_c(x) = x^2 + c, where -5/4<c<-3/4.

So then x^2+c = x implies x^2-x+c = 0.

In order to show that f_c has an attractive orbit of period 2, we must investigate f_c^2 = f_c \circ f_c.

f_c^2 = f_c \circ f_c = (x^2+c)^2 +c = x^4 + 2cx^2 + (c^2 +c).

We can see that any point fixed by f_c will be fixed by f_c^2. So it follows that x^2-x+c is a factor of f_c^2 (see Parameterized Families of Functions in the textbook).

So through polynomial long division, we find that f_c^2 = (x^2-x+c)(x^2+x+(c+1)). Note that f_c^2=0 whenever either (x^2-x+c)=0 or (x^2+x+(c+1)) = 0. Solutions for x of the former term are our fixed points in f_c. So solutions for x of the latter term will give use our points who are in an orbit of period 2.

Through the quadratic formula, we find that x = \dfrac{-1\pm \sqrt{-(4c+3)}}{2}.

Now note that f'_c = 2x.

We will proceed by using Lemma 2.5.2 of the textbook to find the multiplier.

f'(\dfrac{-1+\sqrt{-(4c+3)}}{2})\cdot f'(\dfrac{-1-\sqrt{-(4c+3)}}{2})

= (-1+\sqrt{-(4c+3)}) \cdot (-1-\sqrt{-(4c+3)})

= 4c+4.

Now recall that -5/4 < c < -3/4

So then -5<4c<-3.

So -1<4c+4<1.

That is, |4c+4|<1 .

Therefore, f_c(x) = x^2+c has an attractive orbit of period 2 when -5/4 < c < -3/4.