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Math Help - sequences

  1. #1
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    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)

    My answers to check

    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
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  2. #2
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    Quote Originally Posted by stewpot View Post
    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)

    My answers to check

    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]
    a_1= 1.107(2400)- 260
    [tex]a_2= 1.107(1.107(2400)- 260)= 2400(1.107^2)- 260(1.07)- 260
    a_3= 1.107(2400(1.107^2)- 260(1.07)- 260)- 260
    = 2400(1.107)^3- 260(1+ 1.107+ 1.107^2
    It looks to me like 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
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  3. #3
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    Sequence reply

    I'm not totally happy about trying to work out the closed form formula, any help greatly appreciated.
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  4. #4
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    Hello, stewpot!

    Your closed form doesn't work.

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


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

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


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


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

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


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

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

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

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

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


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


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

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

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

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



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

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  5. #5
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    Sequence Reply

    Many thanks for breaking down the chain of events but just one more question, how do you end up with 260000? I can't see where it comes from,

    Once again thanks
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