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8 Notice that instead of approaching a single limiting value, the population is settling down to an oscillation between 344 and 591. 4. 48 It is possible to obtain limiting cycles of longer periods by increasing β. 48. Note that {xn } appears to be approaching a limiting cycle of period 4. With appropriate choices for β and x0 , it is in fact possible for the inhibited growth model to exhibit limiting cycles of any given period. This is related to the fact that it is possible for this model to behave chaotically.

9 million dollars. Thus the initial governmental expenditure of three million dollars results in approximately 30 million dollars, 10 times the initial amount, in new spending in the economy. This partially explains why deficit spending by the government in depressed times can be far more beneficial to the economy than the actual amount spent, and why such spending during other times can be highly inflationary. Example This example involves slightly more complicated probabilistic reasoning, as well as some additional algebraic simplification, before the problem is reduced to the summation of a geometric series.

Hence the sequence {xn } is monotone increasing and bounded, and so must have a limit. In Problem 8 you will be asked to verify that this limit is in fact M , as appeared to be the case in the previous example. If β > 1, it may be the case that there are values of n for which xn > M , in which case β xn (M − xn ) < 0 M and, as a consequence, xn+1 < xn . 5. 003xn (500 − xn ) 500 for n = 0, 1, 2, . .. 5 Note how the values increase rapidly (as we should expect with such a large value for β) to above the carrying capacity of 500, but then oscillate about 500, with the oscillations diminishing in size.

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