Long Term Behavior -- Cycles

You should use one of the computer algebra systems below with this module. Click on the appropriate icon for your preferred CAS and then arrange your screen so that you can easily move back-and-forth between this window and your CAS window. Click on the appropriate help button for help.

You may recall one of the logistic models we looked at.

p(1) = 50

p(n + 1) = 3.4 [1 - .001 p(n)] p(n)

        1   50.00        6  840.11       11  454.66       16  842.65
        2  161.50        7  456.69       12  843.01       17  450.81
        3  460.42        8  843.62       13  449.97       18  841.77
        4  844.67        9  448.54       14  841.49       19  452.85
        5  446.08       10  841.00       15  453.51       20  842.44

In this model the population seems to be settling in a pattern -- going back-and-forth between two different populations -- roughly 450 and 840. This kind of pattern in which the same numbers repeat over and over again is called a cycle. In this case because two numbers are repeating it is called a 2-cycle. When the repeating pattern has three numbers, it is called a 3-cycle and so forth. Sometimes we say a cycle of period 2 or a cycle of period 3 instead of a 2-cycle or a 3-cycle.

Cycles have some similarities with equilibrium points. We find equilibrium points for a dynamical system with the change equation

p(n + 1) = f(p(n))

by solving the equation

p = f(p)

Graphically, solving this equation finds the points at which the curve y = f(p) crosses the diagonal line y = p.

We find 2-cycles by solving the equation

p = f(f(p))

Notice that equilibrium points are also a solution of this equation.

f(f(p)) = p

where

f(p) = 3.4 [1 - .001 p] p.

Recall that an equilibrium point of a dynamical system may or may not be a limit of the sequence generated by that dynamical system. In a similar way a 2-cycle or, more generally an n-cycle may or may not attract a sequence. Consider, for example, another logistic model, the model

p(1) = 50

p(n + 1) = 4.0 [1 - .001 p(n)] p(n)

        1   50.00        6  645.70       11  999.65       16  316.37
        2  190.00        7  915.09       12    1.39       17  865.11
        3  615.60        8  310.82       13    5.57       18  466.77
        4  946.55        9  856.84       14   22.14       19  995.58
        5  202.38       10  490.67       15   86.59       20   17.59

f(f(p)) = p

where

f(p) = 4 [1 - .001 p] p

Check Your Understanding

• This is an open-ended assignment. Find as many examples as you can of cycles of different periods.

  answer

• This is an open-ended assignment. Find as many examples as you can of cycles of different periods that are or are not "attracting cycles."

  answer