By Kenneth I. Joy
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Let x(t) be the population t days after we begin the experiment, so that JC(O) is the initial population, x(1) is the population after one day, and so on. 1. The change in the population during the first day is the population at the end of one day minus the population at the start, or JC(1) - x ( 0 ) . Thus we find that the change during the first day is *(1) - x(0) = 1 010 000 - 1 000 000 = 10 000. Similarly, the change during the second day is x(2) -x(l) = 1 020 100 - 1010000 = 10 100. 1. During each day any particular bacterium has a certain chance of reproducing and a certain chance of dying.
As an example, let us look at 7(5), and set 7(5) = x. First let us estimate x between consecutive integers. In our previous work, we found that 2 < x < 3. Then we set where xx > 1. We obtain 2 = xi , x We try to estimate xx between consecutive integers. Let us compare 7(5) - 7(4) with 7(2) = 1. We see that 2 [7(5) - 7(4)] = 27(5) - 27(4) =7(5 2 ) - 7(4 2 ), which is less than 7(2) if 7(5 2 )< [7(4 2 )+7(2)] =7(4 2 X 2). Is this so? 04. If we want to refine this estimate, we put JC! = 3 + 1 x2 where x2 > 1, and obtain the equation - /(5 3 )] = 7(5) - /(2 2 ).
What is the worst that can happen? What is the maximum risk in asking question (0)? Would it be better to ask question (b) first? Now what is the worst that can happen? What is the maximum risk in asking question (b)l As a first question, which is better, (a) or (&)? Can you think of a better question than either? Can you think of a question which minimizes the maximum risk? Can you think of another? Can such a question be improved? Now can you see why our strategy worked? Return to exercise 28 and try it again.