Aha! Better proof of the first formula

I have my Power point presentation all assembled, and in the compilation process I have organized a better proof. Below you will find the slides from my Power point.

The last substitution involved this same grouping [A(n-4) + A(n-5)]. Instead of working this out again, we can simply substitute in A(n-2).

The LHS = RHS, and this confirms the hypothesis of the first formula.

I will be presenting this Power point (which includes more slides) to the Mathematics department here at Northern Michigan University.

Proof of the first formula

Let Ak represent the number of newborn rabbits in month k.
Let Bk represent the number of one-month old rabbits in month k.
Let Ck represent the number of two-month old rabbits in month k.

Then Ak = B(k-1) + C(k-1) , because only one-month and two-month old rabbits produce offspring the following month.

Also, the rabbits will age. So we have Bk = A(k-1) , and Ck = B(k-1).

Therefore, the total number of rabbits in month k is the sum of the number of newborns, one-month, and two-month old rabbits.

Tk = Ak + Bk + Ck

Then, Tk = Ak + A(k-1) + B(k-1).

Furthermore, Tk = Ak + A(k-1) + A(k-2).


How many newborn rabbits (An) are in the nth month?
Well, let's start at the beginning. We are given one pair of newborn rabbits in the first month, so A1 = 1. (Bk = 0 and Ck = 0)

We are also told that this pair will not produce any offspring yet, but they will age. So, in the second month A2 = 0. (Bk = 1 and Ck = 0)

Now, that original pair can produce offspring, so in the third month we have A3 = 1.
(Bk = 0 and Ck = 1)

In the fourth month, the original pair of rabbits will produce offspring and then die, as specified in the conditions. Also, the newborns from the third month have just reached maturity in this fourth month, so they are not yet able to produce any offspring and we now have A4 = 1. (Bk = 1 and Ck = 0)

In the fifth month, we have only one pair of rabbits that can produce offspring and one pair that has just reached maturity. So A5 = 1. (Bk = 1 and Ck = 1)

In the sixth month, we have two pairs of rabbits that can produce offspring, one of which will die, and one pair that has reached maturity. So A6 = 2. (Bk = 1 and Ck = 1)

In the seventh month, we have two pairs of rabbits that can produce offspring, one of which will die, and two pairs that have reached maturity. So A7 = 2. (Bk = 2 and Ck = 1)

In the eighth month, we have three pairs of rabbits that can produce offspring, one of which will die, and two pairs that have reached maturity. So A8 = 3. (Bk = 2 and Ck = 2)

This pattern continues as time goes on. To see this data in a table, please refer to Table 6.

With this, the proof by mathematical induction (the previous post) is possible.

Proof by induction on the first formula

Current Project
Modified Fibonacci sequence
3-month life-span

My original intention for proving the first formula to be true turned out to only loop on itself; the proof depended on the conjecture being true in the first place. Therefore, I am going to try a different approach. I want to use mathematical induction, a method which can be described much like falling dominoes. That is, the first domino is knocked over by me, knocking over the second domino, which knocks over the third domino, and so on. In theory, I could also begin by knocking over the 10th, which will knock over the 11th, then the 12th, and so on.

Applying this same idea to my sequence, the 3MLS (three-month life-span)sequence, I should be able to begin in any month and generate the sequence forward. Knowing how to do this will allow me to change my starting point (like in the domino arrangement), which will then make mathematical induction possible.


Q: What do I need to know about my rabbits in any given month in order to determine the population for the following month?

A: I need to know how many newborn, one-month old, and two-month old rabbits are in the current population. The table below shows how many of each age group are present in the first twelve months.

As you can see, the 3MLS sequence is present in each column. For the sake of comparison, let's view a similar table pertaining to the Fibonacci sequence.

Once again, the sequence (this time the Fibonacci sequence) is present in each column.

Therefore, each term in both sequences can now be represented as the sum of three terms rather than two. We have Tn = Tn-3 + Tn-4 + Tn-5. Now, let's use this to prove the first formula: Tn = Tn-2 + Tn-3.


Given: Tn = Tn-3 + Tn-4 + Tn-5
Prove: Tn = Tn-2 + Tn-3

Step 1: Show true for the smallest case
T4 = T2 + T1
2 = 1 + 1
True

Step 2: Assume true for all cases from the smallest to k, that is 4<=n<=k. Tk = Tk-2 + Tk-3 assumed true

Step 3: Show true for case k+1
T(k+1) = T(k+1)-2 + T(k+1)-3
Simplify
T(k+1) = Tk-1 +Tk-2

Use what is given, so
T(k+1) = T(k+1)-3 +T(k+1)-4 +T(k+1)-5
Simplify
* T(k+1) = Tk-2 + Tk-3 + Tk-4

Because we assumed true for all cases 4 <= n <= k this includes k-1. So, we know T(k-1) = T(k-1)-2 + T(k-1)-3 T(k-1) = Tk-3 + Tk-4 Substitute this into the * equation and we have
T(k+1) = Tk-2 + Tk-1

The proof is now concluded, and Tn = Tn-2 + Tn-3 has been found to be true.