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2. The first is an example of exponential growth, whereas the second is an example of exponential decay. In the first, the steepness of the graph increases with time; in the second, the graph flattens out over time. 3) will always lead to the first behavior when α > 1 and to the second when 0 < α < 1. 6) n = 0, 1, 2, . , is called a first-order linear difference equation. 3) is of this form with β = 0. 3) will enable us to solve this equation as well. Namely, xn = αxn−1 + β = α(αxn−2 + β) + β = α2 xn−2 + β(α + 1) = α2 (αxn−3 + β) + β(α + 1) = α3 xn−3 + β(α2 + α + 1) ..

D) Plot wn versus n for n = 0, 1, 2, . . , 150. From the plot, guess lim wn . 4. 2. 5% per year, but the lake can support no more than 10,000 pike. 5 number of pike n years from now and suppose p0 = 1000. (a) Use the inhibited growth model to write a difference equation which describes how the population changes from year to year. (b) Using the difference equation from part (a), compute pn for n = 1, 2, 3, . . 50. (c) How many years will it take for the population to double? To triple? (d) Plot pn versus n for n = 0, 1, 2, .

In this case, by assuming that a given animal of this species has a constant 10% chance of dying in any given year, we have implicitly assumed that there is no fixed upper bound to its life-span. 748 × 10−46 ) and, hence, is not actually ever going to happen. Since we have assumed that these animals cannot reproduce in their first year of life, we have P1 = 0. To compute P2 , we note that 90% of all such females will live through their first year and that 20% of these will then have offspring successfully.