# Writing up a Recurrence Relation Problem...

• Oct 11th 2010, 11:40 AM
aamiri
Writing up a Recurrence Relation Problem...
I know how to solve Recurrence Relation problems, I just need help writing up the equation:

Assume that the deer population of Rustic County is 200 at time
n = 0 and 220 at time n = 1 and that the increase from time n-1 to time n is twice the increase from time n-2 to time n-1. Write a recurrence relation and an initial condition that de ne the deer population at time n and then solve the recurrence relation.

• Oct 11th 2010, 12:34 PM
emakarov
Let the population be $x_n$ a time $n$, $x_{n-1}$ at time $n-1$ and $x_{n-2}$ at time $n-2$. Using this notation, you know how to write the increase in population from time $n-1$ to $n$, don't you?
• Oct 11th 2010, 03:39 PM
Plato
• Oct 11th 2010, 05:55 PM
Soroban
Hello, aamiri!

Quote:

Assume that the deer population of Rustic County is 200 at time 0,
and 220 at time 1.
And that the increase from time $\,n-1$ to time $\,n$ is twice the increase
from time $\,n-2$ to time $\,n-1.$

(a) Write a recurrence relation and an initial condition
that define the deer population at time $\,n.$

(b) Solve the recurrence relation.

Let $P(t)$ = deer population at time $\,t.$

We are given: . $P(0) = 200,\;P(1) = 220$

We are told that: . $P(n) \;=\;P(n-1) + 2\!\cdot\!\text{(previous difference)}$

. . . . . . . . . . . . . $P(n) \;=\;P(n-1) + 2\!\cdot\![P(n-1) - P(n-2)]$

. . . . . . . . . . . . . $P(n) \;=\;P(n-1) + 2\!\cdot\!P(n-1) - 2\!\cdot\!P(n-2)$

. . . . . . . . . . . . . $P(n) \;=\;3\!\cdot\!P(n-1) - 2\cdot\!P(n-2)$ .(a)

Hence: . $P(n) - 3\cdot P(n-1) + 2\cdotP(n-2) \;=\;0$

Let $X^n =P(n)\!:\;\;X^n - 3X^{n-1} + 2X^{n-2} \;=\;0$

Divide by $X^{n-2}\!:\;\;X^2 - 3X + 2 \:=\:0 \quad\Rightarrow\quad (X-1)(X-2) \:=\:0
$

. . Hence: . $X \:=\:1,\,2$

The function is of the form: . $P(n) \;=\;A\!\cdot1^n + B\!\cdot\!2^n$

We know the first two terms of the sequence:

$\begin{array}{cccccccccc}P(0) = 200\!: & A + B &=& 200 & [1] \\
P(1) = 220\!: & A + 2B &=& 220 & [2] \end{array}$

Subtract [2] - [1]: . $B \,=\,20$

Substitute into [1]: . $A + 20 \:=\:200 \quad\Rightarrow\quad A \:=\:180$

Therefore: . $P(n) \;=\;180 + 20\!\cdot\!2^n$ (b)