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Math Help - Linear recurrence sequence, closed form.

  1. #1
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    Linear recurrence sequence, closed form.

    Hi I need help with the following –

    Consider the linear recurrence sequence:

    x(subscript)1 = 5, x(subscript)n+1=1.2x(subscript)n-14 (n=1,2,3,…)

    Find a closed form for the sequence.
    Do I need to use the geometric progression to work out the closed form?

    Thanks.
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  2. #2
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    Quote Originally Posted by mrszone View Post
    Hi I need help with the following



    Do I need to use the geometric progression to work out the closed form?

    Thanks.
    The 14 messed that up. However, if we let

    x_n = y_n + k then y_{n+1} + k = 1.2(y_n + k )-14 and if we choose k = 70 then we have y_{n+1} = 1.2 y_n which is now geometric.
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    The general term of the sequence is given by
     <br />
T_n= 5(1.2)^{n-1} - \sum _{r=0} ^{n-2} {(1.2 )^r\times 14}<br />
........for n>=2

    =5(1.2)^{n-1} - \frac{14((1.2)^{n-2} -1)}{0.2}
    Last edited by ADARSH; February 20th 2009 at 08:08 AM.
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  4. #4
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    Quote Originally Posted by ADARSH View Post
    The general term of the sequence is given by
     <br />
T_n= 5(1.2)^{n-1} - \sum _{r=0} ^{n-2} {(1.2 )^r\times 14}<br />
........for n>=2

    T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-2} -1)}{0.2}
    I think you meant
    T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-1} -1)}{0.2}

    since what you have now T_1 \ne 5

    However, I think it's easier to write the answer as

    T_n=-65 (1.2)^{n-1} +70.
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by danny arrigo View Post
    I think you meant
    T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-1} -1)}{0.2}

    since what you have now T_1 \ne 5

    However, I think it's easier to write the answer as

    T_n=-65 (1.2)^{n-1} +70.
    Yeah it can further be simplified
    but the second step was

    T_n=5(1.2)^{n-1} - \frac{14((1.2)^{n-1} -1)}{0.2}
    (as you said)
    It was a typo, Sorry for that
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  6. #6
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    Hello, mrszone!

    Find the closed form: . x_1 = 5,\;x_{n+1} \:=\:1.2x_n - 14
    It is not a geometric series . . . . x_1=5,\;x_2 = \text{-}8,\;x_3 =\text{-}23.6 \;\hdots

    . . \begin{array}{cccccc}\text{We have:} & x_{n+1} &=& 1.2x_n - 14 & {\color{blue}[1]}\\<br />
\text{and:} & x_{n+2} &=& 1.2x_{n+1} - 14 & {\color{blue}[2]} \end{array}

    Subtract [2] - [1]: . x_{n+2} - x_{n-1} \;=\;1.2x_{n+1} - 1.2x_n \quad\Rightarrow\quad x_{n+2} - 2.2x_{n+1} + 1.2x_n \:=\:0

    Suppose the function is exponential: . x_n \:=\:Y^n

    . . The equation becomes: . Y^{n+2} - 2.2Y^{n+1} + 1.2Y^n \:=\:0

    . . which factors: . Y^n\left(Y^2 - 2.2Y + 1.2\right) \:=\:0 \quad\Rightarrow\quad Y^n(Y - 1)(Y - 1.2) \;=\;0

    . . and has nonzero roots: . Y \:=\:1,\:1.2


    \text{The function, }f(n)\text{, is either: }\:1^n\,\text{ or }\,1.2^n

    Form a linear combination of the two functions: . f(n) \;=\;A + B(1.2)^n


    From the first two values of the sequence:

    . . \begin{array}{cccccc}f(1)=5\!: & A + 1.2B &=& 5 & {\color{blue}[3]} \\<br />
f(2)=\text{-}8\!: & A + 1.44B &=&\text{-}8 & {\color{blue}[4]} \end{array}

    Subtract [4] - [3]: . 0.24B \:=\:\text{-}13 \quad \Rightarrow\quad\boxed{ B \:=\:-\frac{325}{6}}

    Substitute into [3]: . A + (1.2)\left(-\tfrac{325}{6}\right) \:=\: 5 \quad\Rightarrow\quad\boxed{ A \:=\:70}


    Therefore: . f(n) \;=\;70 -\frac{325}{6}(1.2)^n


    Edit: a silly typo! . . . fixed it.
    .
    Last edited by Soroban; February 21st 2009 at 09:56 AM.
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  7. #7
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, mrszone!

    It is not a geometric series . . . . x_1=5,\;x_2 = \text{-}8,\;x_3 =\text{-}23.6 \;\hdots

    . . \begin{array}{cccccc}\text{We have:} & x_{n+1} &=& 1.2x_n - 14 & {\color{blue}[1]}\\ \text{and:} & x_{n+2} &=& 1.2x_{n+1} - 14 & {\color{blue}[2]} \end{array}" alt="
    \text{and:} & x_{n+2} &=& 1.2x_{n+1} - 14 & {\color{blue}[2]} \end{array}" />

    Subtract [2] - [1]: . x_{n+2} - x_{n-1} \;=\;1.2x_{n+1} - 1.2x_n \quad\Rightarrow\quad x_{n+2} - 2.2x_{n+1} + 1.2x_n \:=\:0

    Suppose the function is exponential: . x_n \:=\:Y^n

    . . The equation becomes: . Y^{n+2} - 2.2Y^{n+1} + 1.2Y^n \:=\:0

    . . which factors: . Y^n\left(Y^2 - 2.2Y + 1.2\right) \:=\:0 \quad\Rightarrow\quad Y^n(Y - 1)(Y - 1.2) \;=\;0

    . . and has nonzero roots: . Y \:=\:1,\:1.2


    \text{The function, }f(n)\text{, is either: }\:1^n\,\text{ or }\,1.2^n

    Form a linear combination of the two functions: . f(n) \;=\;A + B(1.2)^n


    From the first two values of the sequence:

    . . \begin{array}{cccccc}f(1)=5\!: & A + 1.2B &=& 5 & {\color{blue}[3]} \\ f(2)=\text{-}8\!: & A + 1.44B &=&\text{-}8 & {\color{blue}[4]} \end{array}" alt="
    f(2)=\text{-}8\!: & A + 1.44B &=&\text{-}8 & {\color{blue}[4]} \end{array}" />

    Subtract [4] - [3]: . 0.24B \:=\:\text{-}13 \quad \Rightarrow\quad\boxed{ B \:=\:-\frac{325}{6}}

    Substitute into [3]: . A + (1.2)\left(-\tfrac{325}{6}\right) \:=\: 5 \quad\Rightarrow\quad\boxed{ A \:=\:70}


    Therefore: . f(n) \;=\;70 -\frac{325}{6}(1.2)^m

    Just a (very) small thing
    m = n
    f(n) \;=\;70 -\frac{325}{6}(1.2)^{n}

    And I think we got it same
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