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Math Help - Population Growth

  1. #1
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    Population Growth

    Here's the problem

    The growth of a certain population is modelled by the recursion formula

     a_{n+2}=\frac{3}{2}a_{n+1}-\frac{1}{2}a_n

    and a_1=6000, a_2=7000

    for n \geq 1, where a_k is the size of the population after period k.

    I have already done the first part which is to show by induction that it is strictly increasing. The next part is

    Make a conjecture for a formula for a_n, and prove it by induction. Hint: Consider the difference between a_n and some suitable upper limit.

    How do I go about that? Will the formula be something like a_n = 6000+something to the power of n? Any help appreciated thank you.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Rombie View Post
    Here's the problem

    The growth of a certain population is modelled by the recursion formula

     a_{n+2}=\frac{3}{2}a_{n+1}-\frac{1}{2}a_n

    and a_1=6000, a_2=7000

    for n \geq 1, where a_k is the size of the population after period k.

    I have already done the first part which is to show by induction that it is strictly increasing. The next part is

    Make a conjecture for a formula for a_n, and prove it by induction. Hint: Consider the difference between a_n and some suitable upper limit.

    How do I go about that? Will the formula be something like a_n = 6000+something to the power of n? Any help appreciated thank you.
    Try:

    <br />
a_n=7000-4000 ~(1/2)^n<br />

    RonL
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  3. #3
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    Thanks, 8000-4000(1/2)^n works. 8000 must be that suitable upper limit. Since it says conjecture do I need to give some reasoning for how I got there? Or is it enough to propose this formula and then prove it by induction.

    How would I show that the population size is limited as n tends to infinity?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Rombie View Post
    Thanks, 8000-4000(1/2)^n works. 8000 must be that suitable upper limit. Since it says conjecture do I need to give some reasoning for how I got there? Or is it enough to propose this formula and then prove it by induction.

    How would I show that the population size is limited as n tends to infinity?
    You can try some experiments to see where the sequence is going, which
    should allow you to guess what the upper limit might be.

    As n goes to infinity if a_n=8000-4000(1/2)^n, then a_n goes to 8000 as
    (1/2)^n goes to zero. Also it approches this from below as (1/2)^n is
    strictly decreasing.

    RonL
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  5. #5
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    Hello, Rombie!

    I got your result . . .


    The growth of a certain population is modelled by the recursion formula

    a_{n+2}\;=\;\frac{3}{2}\,a_{n+1}-\frac{1}{2}\,a_n . and a_1 = 6000,\;a_2=7000

    for n \geq 1, where a_k is the size of the population after period k.

    Let a_n \:=\:X^n

    We have: . X^{n+2} \;=\;\frac{3}{2}\,X^{n+1} - \frac{1}{2}\,X^n

    Multiply by \frac{2}{X^n}\!:\;\;2X^2 \;=\;3X - 1

    Then: . X^2-3X + 1\:=\:0\quad\Rightarrow\quad(X - 1)(2X - 1) \:=\:0\quad\Rightarrow\quad X \:=\:1,\:\frac{1}{2}


    Form a linear combination of these roots.
    The function has the form: . a_n \;=\;A(1^n) + B\left(\frac{1}{2}\right)^n\;=\;A + \frac{B}{2^n}

    We are told that: . \begin{array}{ccccc}a_1 \: = \:6000\!: & A + \frac{B}{2} & = & 6000 & {\color{blue}[1]} \\<br />
a_2 \:= \:7000\!: & A + \frac{B}{4} & = & 7000 & {\color{blue}[2]}\end{array}

    Subtract [2] from [1]: . \frac{B}{4} \:=\:-1000\quad\Rightarrow\quad B \:=\:-4000

    Substitute into [1]: . A -\frac{4000}{2} \:=\:6000\quad\Rightarrow\quad A \:=\:8000

    . . Therefore: . \boxed{a_n \;=\;8000 - \frac{4000}{2^n}}


    We see that: . \lim_{n\to\infty}a_n \;=\;\lim_{n\to\infty}\left(8000 - \frac{4000}{2^n}\right) \;=\;8000

    The population has an upper limit.

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  6. #6
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    Thanks a lot, both of you.
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