Given f(x) = 3.2x(1-x), show:
1. x = 0 and x = 11/16 (which = .6875) are fixed points, and
2. x = [21 + sqrt (21)] / 32 and x = [21 - sqrt (21)]
/ 32 are period-2 points.
1. We must show that f(x) = x, or f(x) = -x for the given points.
f(0) = 3.2(0)(1-0) = 0.
f(.6875) = 3.2(.6875)(1-.6875)
= .6875
Thus, both points are fixed points for the given function.
2. We must show that f(f(x)) = x, or f(f(x)) = -x for the given points.
First, we note that [21 + sqrt (21)] / 32 ~ 0.800, and that [21 - sqrt (21)] / 32 ~ 0.513
Now, f(f([21 + sqrt (21)] / 32)) = f{3.2([21 + sqrt (21)] / 32)(1-[21 + sqrt (21)] / 32)} = f(0.513044509) = 3.2(.513044509)(1-.513044509) = 0.800072685 ~0.800. Thus, [21 + sqrt (21)] / 32 is a fixed point of period 2.
It then follows that f(f([21 - sqrt (21)] / 32)) ~ f(f(0.513)) ~ f(.800)
~ 0.513, so that [21 - sqrt (21)] / 32 is also a fixed point of period
2.
TO GO TO OUR MAIN GROUP WEB PAGE FOR MATH 5337, CLICK here
.