##
Lab #4

** Goal:**
Using the orbit diagram, we have seen in the previous lab
that the quadratic function **
Q**_{c}(x) = x^{2} + c and
the logistic function **F**_{c}(x) = cx(1-x) both
undergo a sequence of period-doubling bifurcations as the parameter
tends to the chaotic regime. We have also seen that magnifications of
the orbit diagram tend to look "the same." In this experiment, we
will see that there really is some truth to this.

** Procedure:** In this experiment you will work with
both of the above families of functions.
We first need a definition:

** Definition.** Suppose **x**_{0} is a critical point
for a function **F**,
that is, **F'(x**_{0}) = 0. If **x**_{0} is
also a periodic point of **F**
with period **n**,
then the orbit of **x**_{0} is called * superstable *. Note
that (**F**^{n}
)' (x_{0}) = 0.

Recall that the critical point for **Q**_{c} is 0. Similarly, the
critical point of **F**_{c} is 1/2.

In this experiment you will first
determine the c-values at which **Q**_{c} and ** F**_{c}.
have superstable cycles of periods 1, 2, 4, 8, 16, 32, and
64. The easiest way to do this is to use the Orbit Diagram applet
to determine
experimentally the
c-values at which the function has a superstable point of the given
period. This can be done by zooming in on the appropriate branch of
the curve representing cycles with the right period. By zooming in
over and over and then using the ``Current Location'' feature of the lab,
you should be able to determine the parameter value for which there is
a superstable cycle. You should try to determine this value to five
or six digits of accuracy. Recall that the
parameter values are plotted horizontally on the screen and the
x-coordinates along the orbit are plotted vertically.
Once you have found the appropriate parameter value, you may check
your results using the list feature of the Function Iterator or
Lab 1. Simply use the critical
point as the initial seed for the iteration and enter the desired
number of iterations. If your parameter value is correct, you should
see that the critical point is very close to being periodic with the
right prime period. You can try to get better accuracy by changing
the parameter slightly to try to make the critical point exactly
periodic (up to the accuracy of the computer). Note that the software
stores
more digits for the parameter value than it displays.
After finding the seven c-values for your function, record these
numbers in tabular form:

**c**_{0} = ____________________ = c-value for period 2^{0}

**c**_{1} = ____________________ = c-value for period 2^{1}

**c**_{2} = ____________________ = c-value for period 2^{2}

**c**_{3} = ____________________ = c-value for period 2^{3}

**c**_{4} = ____________________ = c-value for period 2^{4}

**c**_{5} = ____________________ = c-value for period 2^{5}

**c**_{6} = ____________________ = c-value for period 2^{6}

Now use a calculator to compute the following ratios:

** f**_{0} = (c_{0}
- c_{1})/(c_{1} - c_{2}) = ______________________

** f**_{1} = (c_{1}
- c_{2})/(c_{2} - c_{3}) = ______________________

** f**_{2} = (c_{2}
- c_{3})/(c_{3} - c_{4}) = ______________________

** f**_{3} = (c_{3}
- c_{4})/(c_{4} - c_{5}) = ______________________

** f**_{4} = (c_{4}
- c_{5})/(c_{5}
- c_{6}
) = ______________________

List these numbers in tabular form, too. Do you notice any
convergence? You should, at least if you have carried out the above
search to enough decimal places.