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Assume that growth in a bacterial population has the following properties:

  1. At the beginning of every hour, two new bacteria are formed for each bacteria that lived in the previous hour.
  2. During the hour, all bacteria that have lived for two hours die.
  3. At the beginning of the first hour, the population consists of 100 bacteria.

Derive a recurrence relation for the number of bacteria.

I know that if the bacteria wouldn't die $A_{n} = 2A_{n - 1}$, but now I have no idea what to do.

Felix Marin
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arina
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2 Answers2

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We know $A_1 = 100$, and in the second hour there will be $200$ newborn and $100$ that will die the third hour. So, $A_2 = 300$. Those $300$ alive in the second hour mean that in the third hour there will be $600$ newborn and $200$ still alive; $A_3 = 800$.

Following this logic, we can define: $$A_n = 2 \, A_{n-1} + 2 \, A_{n-2}.$$

Leucippus
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Medicine
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  • When I calculate this by hand I get $A_1=100, A_2=300, 800, 2100, 5500, 14400,...$. The recurrence relation above with: $A_1=100, A_2=300$, gives: $100,300,800,2200,6000,16400,...$. But the relation $A_n=3A_{n-1}-A_{n-2}$ matches values calculated by hand. And it translates to English description better: all bacteria alive at a particular hour generate two progeny, which are added to the still living parents. The number of still living parents is the (# alive at the previous hour) - (# alive two hours ago). Hence $3\cdot$ (number # at the previous hour) - (# alive 2 hrs ago). – Χpẘ Jun 16 '17 at 17:34
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At time $n$:

  • The bacteria that have been there at the previous time (time $n-1$) give rise to 2 bacteria each: $A_{n-1} \to 2A_{n-1}$

  • The bacteria that were already there at time $n-2$ die: $A_{n-2} \to 0$

Put together, this means you get $$ A_n = 2 A_{n-1} - A_{n-2}. $$

Clement C.
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  • That isn't completely true, because at time An-2 there are some bacteria of age 1 year, that already died in An-1 – arina May 20 '16 at 19:13
  • Sorry, I don't understand what you mean. At time $n-1$, you have $A_{n-1}$ bacteria, that you can write as $A_{n-1} = A_{n-2} + \Delta_{n-1}$ (the "at least one step old", and the new.) Then, at time $n$ you get $A_n = 2A_{n-1} - A_{n-2}$ (all $A_{n-1}$ bacteria give rise to two bacteria, i.e. each stays and also gives a new one; and then, well, the $A_{n-2}$ old ones die). – Clement C. May 20 '16 at 19:16
  • I think I understand know your answer, thanks – arina May 20 '16 at 19:19