Breeding Bunnies: A Model for Population Growth

Suppose that rabbits breed in the following way:

·  Only mature rabbits can have offspring.

·  Rabbits reach maturity at the end of their 2nd year of life; in other words, they cannot reproduce during their first two years of life.

·  Each pair of mature rabbits will always have exactly two offspring each year.

·  (Incidental fact regarding graphics: baby rabbits have white fur that turns brown at the end of their first year of life.)

Using these seemingly simple guidelines for rabbit breeding, determine a function that will provide the population of rabbits (counted in pairs) by year, with f(0) = 1.

Time
(years)
t / Population
(pairs of rabbits)
f(t)
0 / 1
1
2
3
4
5
6
7
8
9
10

Extension: Let’s introduce mortality into our population of rabbits. Let’s say that each rabbit (pair) will live only 5 years, giving 3 years of breeding for each pair.

What will be the population of rabbits for each of the first five years?

Is there a pattern for population growth when mortality is included?

Discussion:

·  A strategy is essential when working out the solution to a question such as this. What bookkeeping technique did you use?
Did you determine a pattern in the population growth?
How much convincing does it take that the pattern is valid?

·  What are some variations on this question that might make it more interesting (i.e. difficult)?
More realistic?

·  How could you use technology to amplify this? For instance, how would you use Excel to determine the population for large numbers of years, or for the more “interesting” variations?
Could you check to see whether or not exponential growth is a good fit to the situation as originally posed?

·  Use the model you have discovered to predict how many buds might form on a tree that has branched 15 times.

There is a way to determine large terms in the Fibonacci sequence, using another series, called the Lucas sequence. Both series are intimately related to the Golden Mean, a number that shows up repeatedly in nature. The Golden Mean is a solution to the following equation:

If a and b are the two solutions to this quadratic equation, then a formula for the nth Fibonacci number may be written:

Similarly, the nth Lucas number may be written:

It’s not obvious from either formula, but both will always produce integers, for any positive integer n. Wow!

Here are the first few Fibonacci and Lucas numbers, to get you started.

n / 0 / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9
F(n) / 0 / 1 / 1 / 2 / 3 / 5 / 8 / 13 / 21 / 34
L(n) / 2 / 1 / 3 / 4 / 7 / 11 / 18 / 29 / 47 / 76

Notice that for both sets of numbers, you can find the next one in the sequence by adding the two preceding it. For instance, the 4th Fibonacci (or Lucas) number can be found by adding the second to the third:

Generally speaking,

However, this approach to finding Fibonacci and Lucas numbers is not so useful for finding large order numbers.

The following equations show how the Lucas and Fibonacci numbers interrelate:

While it is not easy to calculate Fibonacci or Lucas numbers of high order (i.e. high values of n), these formulas offer something like a shortcut. Try them out! These last four formulas are all you need to find any Lucas or Fibonacci numbers you desire.

EXAMPLE: Let’s say we want to calculate the 15th Fibonacci number, using only the equations provided and the tabulated values above.

From the table, F(7) = 13 and L(7) = 29

According to our formulas,

Substituting:

Finally, using the F(n+1) formula:

We have now found the 15th Fibonacci number: 610.

One last interesting fact:

As , the ratios and both approach the Golden Mean!

The Golden Mean can be found throughout nature. Mathematics is very mysterious...

Summarized from Dr. Rob’s response to a question on Fibonacci numbers at The Math Forum,

http://mathforum.org/library/drmath/view/52677.html