Question

A model for the number of lobsters caught per year is based on the assumption that the number of lobsters caught in a year is the average of the number caught in two previous years. (Show your work, circle or put in bold each intermediate and final answer.)
a. Find a recurrence relation for Ln -- the number of lobsters caught in year n.
b. Solve this recurrence relation (find Ln), if 100,000 lobsters were caught in year 1 and 300,000 lobsters were caught in year 2.

Answer 1

Answer: a. Ln = $\frac{1}{2} (L_{n-1} + L_{n-2})$

b. $L_{n} = 233333.4(1)^{n}+266666.7(-\frac{1}{2} )^{n}$

Step-by-step explanation: Recurrence Relation is a polynomial that relates a term in a sequence with its previous terms.

a. The number of lobster is the average of two previous years. Average is the sum of the elements of a set divided by the total number of the set, so for Ln:

Ln = $\frac{L_{n-1}+L_{n-2}}{2}$ = $\frac{1}{2}(L_{n-1}+L_{n-2})$

where $L_{n-1} and L_{n-2}$ are the two previous years.

b) Ln = $\frac{1}{2}(L_{n-1}+L_{n-2})$

$\frac{1}{2}(L_{n-1})+\frac{1}{2} (L_{n-2}) - L_{n} = 0$

To solve this recurrence, find the characteristic polynomial:

$r^{2} - \frac{1}{2}r - \frac{1}{2} = 0$

Solve for r:

The polynomial can be rewritten as:

$(r-1)(r+\frac{1}{2} )=0$

so, r = 1 and r = -1/2

The expression for the recurrence relation will be:

$L_{n} = \alpha_{1} .(1)^{n} + \alpha_{2}.(-\frac{1}{2} )^{n}$

In year 1, there were 100,000 lobsters, so, when n=1:

$\alpha_{1} .(1)^{1} + \alpha_{2}.(-\frac{1}{2} )^{1} = 100,000$

In year 2, when n=2, Ln = 300,000:

$\alpha_{1} .(1)^{2} + \alpha_{2}.(-\frac{1}{2} )^{2} = 300,000$

Solving the system of equations:

$\alpha_{1} .(1)^{1} + \alpha_{2}.(-\frac{1}{2} )^{1} = 100,000$

$\alpha_{1} = 100,000 + \frac{1}{2}.\alpha_{2}$

$\alpha_{1} .(1)^{2} + \alpha_{2}.(-\frac{1}{2} )^{2} = 300,000$

$\alpha_{1} = 300,000 - \frac{1}{4}.\alpha_{2}$

Finding $\alpha _{2}$:

$100,000+\frac{1}{2}.\alpha_{2} = 300,000 - \frac{1}{4}.\alpha_{2}$

$\frac{3}{4}.\alpha_{2} = 200,000$

$\alpha_{2} =$ 266666.7

With $\alpha _{2}$, plug in an equation to find $\alpha_{1}$:

$\alpha_{1} = 100,000 + \frac{1}{2}.\alpha_{2}$

$\alpha_{1} = 100,000 + \frac{266666.7}{2}$

$\alpha_{1} =$ 233333.4

The solved equation for this recurrence relation is:

$L_{n} = 233333.4(1)^{n} + 266666.7(-\frac{1}{2})^{n}$