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Fire Ant mounds are rising in my yard: if a mound can have only one queen, and she lays $50$ viable eggs per day, and if $2\%$ of the population dies per day, what is the limit of population growth? Obviously, when the death rate equals $50$ per day, no further growth can occur. My solution attempt:

  • population on day 1: $0.98*50$
  • population on day 2: $0.98*50 + 0.98^2*50$ $\quad$ (new hatchlings + survivors from day 1)
  • population on day 3: $0.98*50 + 0.98^3*50 + 0.98^2*50$
  • Generalizing: $P = \sum_{t=1}^{?} 0.98^t*50$
Mike Pierce
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BigTex
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  • There's a problem with your bullet points. Let $r=0.98$. If on day one there are $N$ ants, on day two there are $rN+50$. On day three are $r(rN+50)+50$. On day four $r(r(rN+50)+50)+50$. See the pattern. – Gregory Grant Apr 02 '15 at 21:51
  • I assume hatchlings are subject to the mortality rate of the general population, even on their first day of life. Thus the first day, not including the queen, N = 50r = 49; second day, including new hatch, N = 50r + 50r^2 – BigTex Apr 02 '15 at 23:00
  • I don't think you are supposed to assume that. The eggs aren't laid right after midnight. – Gregory Grant Apr 02 '15 at 23:09
  • Thank you Gregory; I hadn't considered the eggs are probably laid throughout the 24 hours. – BigTex Apr 03 '15 at 12:23
  • Thank you, Gregory. I hadn't considered that the eggs are likely laid round the clock. – BigTex Apr 03 '15 at 12:31

1 Answers1

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A recursion in this form $$ x_0 = k\\ x_{n+1} = p + \alpha x_n $$ where $\alpha\neq0$, is solved by the explicit formula

$x_n = \frac{1-\alpha^n}{1-\alpha}p+\alpha^nk$

If, moreover, $-1<\alpha<1$, there exists the asymptotic limit

$\lim_{n\rightarrow+\infty} x_n= \lim_{n\rightarrow+\infty}(\frac{1-\alpha^n}{1-\alpha}p+\alpha^nk)= \frac{p}{1-\alpha}$

In your problem, $x_n$ represents the number of non-queen ants on the n-th day.

The mathematisation you ask for has $\alpha=0.98,\ k=0,\ p=0.98*50=49$, and the formula tells you that your colony asymptotically attends $2450$ individuals, excluding the queen.

P.S.: If, on the other hand, you wanted to assume Gregory Grant's point of view (which makes more sense to me, but whatever), you should put $\alpha=0.98,\ k=0,\ p=50$ and run the same calculation, obtaining $2500$ instead of $2450$.