> #OK to post Homework
> #Timothy Nasralla, 9/6/21, Homework 1
> #For question 2, I wrote the equation found in question 1 in a recursive format that calculates the number of females birthed per year based on probabilities, the initial amount of females who can give birth, and the time passed. 
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> F := proc(c0, c1, c2, p0, p1, p2, n) option remember; if n = 0 then c0; elif n = 1 then c1; elif n = 2 then c2; else expand(p0*F(c0, c1, c2, p0, p1, p2, n - 1) + p1*F(c0, c1, c2, p0, p1, p2, n - 2) + p2*F(c0, c1, c2, p0, p1, p2, n - 3)); end if; end proc;
> #For question 3, since c0 = c1 = c2 = 1, meaning there are 3 females, I tested the probabilities where p1=p2=p3 and started with 1/3 probability. This led to a stable population.
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> F(1, 1, 1, 1/3, 1/3, 1/3, 100);
                               1


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> F(1, 1, 1, 1/3, 1/3, 1/3, 90);
                               1


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> F(1, 1, 1, 1/3, 1/3, 1/3, 110);
                               1


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> #When p > 1/3, the population increased dramatically and eventually exploded.
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> F(1, 1, 1, 0.35, 0.35, 0.35, 100);
                          11.21045116


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> F(1, 1, 1, 0.37, 0.37, 0.37, 100);
                          180.1832529


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> F(1, 1, 1, 0.40, 0.40, 0.40, 100);
                          9277.369156


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> F(1, 1, 1, 0.40, 0.40, 0.40, 90);
                          3675.512923


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> F(1, 1, 1, 0.40, 0.40, 0.40, 110);
                          23417.02514


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> #When p < 1/3, the population declined until virtual extinction.
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> F(1, 1, 1, 0.31, 0.31, 0.31, 100);
                         0.02847413326


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> F(1, 1, 1, 0.29, 0.29, 0.29, 100);
                         0.001122348542


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> F(1, 1, 1, 0.27, 0.27, 0.27, 100);
                        0.00003646262147


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> #So, 3i is < 1/3, 3ii is = 1/3, and 3iii is > 1/3
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