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Math Help - linear recurrence sequences

  1. #1
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    linear recurrence sequences

    i am having trouble with this sequence:

    x1=3 xn+1=0.8 xn-5 (n=1,2,3....)

    how do i find a closed form, also in the tenth term correct to 4SF

    Can someone explain how can i describe a long term behaviour of the sequence

    many thanks
    rich
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  2. #2
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    One way is to work it out. Write down a few terms and see what emerges in your mind.

    3
    -2.6
    -7.08
    ...

    Okay, that is totally not what I meant.

    n=1) 3

    n=2) 3*(4/5) - 5

    n=3) (4/5)[3*(4/5) - 5] - 5 = 3*(4/5)^2 - 5(4/5) - 5

    I'm beginning to see it. Are you? Let's just guess what the next one will be.

    n=4) 3*(4/5)^3 - 5*(4/5)^2 - 5(4/5) - 5

    And the nth one? (Feel free to prove this by induction.)

    n) 3*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5

    Make a small adjustment and add them up.

    n) 8*(4/5)^(n-1) - 5*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5

    Some algebra

    n) 8*(4/5)^(n-1) - 5*[(4/5)^(n-1) + (4/5)^(n-2) + ... + (4/5)^2 + (4/5) + 5]

    A little more algebra

    n) 8*(4/5)^(n-1) - 5*[(1 - (4/5)^n)/(1-(4/5))

    Simplify

    n) x_{n}\;=\;35\left[\frac{4}{5}\right]^{n} - 25

    I think I'm done. Of course, you're not. You still have to prove the formula.

    Note: There is a better way. It starts with simply assuming the form of the final result and solving for a few parameters. In this case, there is a hint. Rewrite the formula a little:

    x_{n+1}\;=\;\frac{4x_{n}-25}{5}

    Compare this to the final result. From now on, with a first order recursion and a constant

    x_{n+1}\;=\;\frac{a}{b}x_{n} + c\;=\;\frac{ax_{n}+bc}{b}

    You say to yourself, it must look like this.

    x_{n}\;=\;D\cdot \left[\frac{a}{b}\right]^{n} + bc

    Use your initial values to solve for D and you are rather done.
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  3. #3
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    trying to figure it out meself
    i presumed that u1=5*(0.4)n-1

    so find the 12tth term would be u12= 5(0.4)12

    i guess i am totally off the mark with this one
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  4. #4
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    help

    can anyone help please?
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  5. #5
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    <br /> <br />
x_{n+1}=0.8x_{n}-5\rightarrow \vec{y}_{n+1}=<br />
\begin{bmatrix} x_{n+1} \\ 1 \end{bmatrix}=<br />
\begin{bmatrix} 0.8x_{n}-5 \\ 1 \end{bmatrix}=<br />
\begin{bmatrix} 0.8 &-5 \\  0 &1 \end{bmatrix}<br />
\begin{bmatrix} x_{n} \\ 1 \end{bmatrix}=<br />
A\vec{y}_{n}<br /> <br />

    <br /> <br />
f_{A}(\lambda)=\det\begin{bmatrix} A-\lambda I_{2} \end{bmatrix}=0<br />
\rightarrow \lambda_{1,2}=\{~0.8~,~1~\}<br /> <br />

    <br />
\vec{v}_1\in \ker(A-\lambda_{1}I_{2})=span(\vec{e}_1)~,~<br />
 \vec{v}_2\in \ker(A-\lambda_{2}I_{2})=span(\begin{bmatrix} -25 \\ 1 \end{bmatrix})<br />

    <br /> <br />
\vec{y}_{1}=\begin{bmatrix} 3 \\ 1 \end{bmatrix}=<br />
\begin{bmatrix} \vec{v}_{1} &\vec{v}_{2}  \end{bmatrix}<br />
(\begin{bmatrix} \vec{v}_{1} &\vec{v}_{2}  \end{bmatrix}^{-1}\vec{y}_{1})=28\vec{v}_{1} + \vec{v}_{2}<br /> <br />

    <br /> <br />
\vec{y}_{n}=A^{n-1}\vec{y}_{1}=<br />
A^{n-1}(28\vec{v}_{1} + \vec{v}_{2})=<br />
28\lambda_{1}^{n-1}\vec{v}_{1} + \lambda_{1}^{n-1}\vec{v}_{2}=<br />
28(0.8)^{n-1}\vec{e}_{1}+1^{n-1}\begin{bmatrix} -25 \\ 1 \end{bmatrix}<br /> <br />

    <br /> <br />
=\begin{bmatrix} 28(0.8)^{n-1}-25 \\ 1 \end{bmatrix}~,~<br />
x_{n}=28(0.8)^{n-1}-25<br /> <br />
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  6. #6
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    As this is part of question 2 on the first Assignment with the MST121 course at the open uni I am guessing you are on this course, it would be worth contacting your tutor for support. That is what they are there for..
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