Counting Bug PopulationsDate: 12/03/98 at 09:04:16 From: Naveed Ahmed Subject: Probability, sequences, formulas We are given the following information: A happy bug splits into a sad bug and a blank bug. A sad bug splits into 2 happy bugs. A blank bug splits into a sad bug and a happy bug. The generation starts off with 1 happy bug which splits and forms the 2nd generation. The bugs die when they split. The bugs live the same amount of time. I have worked the first five generations to find how many of each kind of bugs there are, but I can't find the formula for the nth generation. Could also you tell me the formulas if the first generation starts with a sad bug and a blank bug? Thanks. Date: 12/03/98 at 11:39:23 From: Doctor Anthony Subject: Re: Probability, sequences, formulas The first step is to make a table, using the following notation: n = generation, H = happy bug, S = sad bug, and B = blank bug. n 1 2 3 4 5 6 7 8 9 10 11 12 ---------------------------------------------------------------- H 1 0 2 2 6 10 22 42 86 170 342 682 B 0 1 1 3 5 11 21 43 85 171 341 683 S 0 1 1 3 5 11 21 43 85 171 341 683 ----------------------------------------------------------------- Total 1 2 2^2 2^3 2^4 2^5 2^6 2^7 2^8 2^9 2^10 2^11 The pattern shows H is one more and one less alternatively than B and S as we move up the generations. The total for the nth generation is always given by 2^(n-1). Also each of H, B and S is close to 1/3 of the total for any generation. To take account of the pattern we can give two formulae depending on whether n is even or odd. A little juggling shows the following is the pattern: n EVEN n ODD --------- ----------- 2^(n-1)- 2 2^(n-1)+ 2 H = ----------- H = ------------ 3 3 2^(n-1)+ 1 2^(n-1)- 1 B and S = ---------- B and S = ------------ 3 3 If you start with a first generation of S + B, you get the same table as above except that n-1 is replaced by n. So the new formulae can be produced by using n instead of n-1 and swapping over the n = even and n = odd labels: n ODD n EVEN ----------- --------------- 2^n - 2 2^n + 2 H = --------- H = ---------- 3 3 2^n + 1 2^n - 1 B and S = -------- B and S = --------- 3 3 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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