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Extra resources for Analysis, Calculus. Difference-eq To Differential eq
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.