1. ## sequences

Hi could someone please check my work

A pop size is modeled by the recurrance system in 2008:

A0=2400, An+1=1.107An-260 (n=0,1,2…)

The population size predicted in 2010 is

(2009) A1 =1.107 x 2400 - 260 =2396.8
(2010) A2= 1.107 x 2400 -260=2393.3

A closed form for the above

An=2400 (1.107) -260 (1.07^n-1) (n=0,1,2,....)

Population predicted 2020 =2290

2. Originally Posted by stewpot
Hi could someone please check my work

A pop size is modeled by the recurrance system in 2008:

A0=2400, An+1=1.107An-260 (n=0,1,2…)

The population size predicted in 2010 is

(2009) A1 =1.107 x 2400 - 260 =2396.8
(2010) A2= 1.107 x 2400 -260=2393.3
I presume you meant [tex]A_2= 1.107(2396.8)- 260= 2393.3.

A closed form for the above

An=2400 (1.107) -260 (1.107^n-1) (n=0,1,2,....)[/quote]
$\displaystyle a_1= 1.107(2400)- 260$
[tex]a_2= 1.107(1.107(2400)- 260)= 2400(1.107^2)- 260(1.07)- 260
$\displaystyle a_3= 1.107(2400(1.107^2)- 260(1.07)- 260)- 260$
$\displaystyle = 2400(1.107)^3- 260(1+ 1.107+ 1.107^2$
It looks to me like $\displaystyle a_n$ is 2400(1.107)^n[tex] minus 260 times the sum of a geometric series. Do you know the formula for that?
[tex]= 2400(1.107)^3

Population predicted 2020 =2290

I'm not totally happy about trying to work out the closed form formula, any help greatly appreciated.

4. Hello, stewpot!

If you let $\displaystyle n = 0$, you should get: .$\displaystyle A_0 = 2400$

A population size is modeled by the recurrance system in 2008:
. . $\displaystyle A_0 = 2400,\; A_{n=1} \:=\:1.107A_n -260$
We have: .$\displaystyle A_0 \:=\:2400$
and: .$\displaystyle A_1 \:=\:1.107(2400) - 260 \:=\:2396.8$

This recurrence has a constant term.
This is how I was taught to handle it . . .

$\displaystyle \begin{array}{ccccc}\text{We are given:} & A_{n+1} &=& 1.107A_n - 260 & {\color{blue}[1]} \\ \text{The "next term" is:} & A_{n+2} &=& 1.107A_{n+1} - 260 & {\color{blue}[2]} \end{array}$

Subtract [2] - [1]: .$\displaystyle A_{n+2} - A_{n+1} \;=\;1.107A_{n+1} - 1.107A_n$

. . . .$\displaystyle A_{n+2} - 2.107A_{n+1} + 1.107A_n \;=\;0$

Assume that $\displaystyle A^n \:=\:X^n$ . . . that the recurrence is an exponential function.

. . Then we have: .$\displaystyle X^{n+2} - 2.107X^{n+1} + 1.107X^n \;=\;0$

Divide by $\displaystyle X^n\!:\quad X^2 - 2.107X + 1.107 \;=\;0$ . . . a quadratic

. . which factors: .$\displaystyle (X - 1)(X - 1.107) \;=\;0$

. . and has roots: .$\displaystyle X \;=\;1,\:1.107$

Form a linear combination of these roots: .$\displaystyle A_n \;=\;a(1)^n + b(1.107)^n$
. . and we have: .$\displaystyle A_n \;=\;a + (1.107)^nb$

We know the first two terms of the sequence, $\displaystyle A_0 = 2400 \text{ and }A_1=2396.8$

. . $\displaystyle \begin{array}{ccccccc}n=0: & a + b &=& 2400 & {\color{blue}[3]} \\ n=1: & a + 1.107b &=& 2396.8 & {\color{blue}[4]} \end{array}$

Subtract [4] - [3]: .$\displaystyle 0.107b \:=\:-3.2 \quad\Rightarrow\quad\boxed{ b \:=\:-\frac{3200}{107}}$

Substitute into [3]: .$\displaystyle a - \frac{3200}{107} \:=\:2400 \quad\Rightarrow\quad\boxed{ a \;=\;\frac{260,\!000}{107}}$

Therefore: .$\displaystyle \boxed{{\color{blue}A_n \;=\;\frac{260,\!000}{107} - \frac{3200}{107}(1.107)^n }}$