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Thread: First order difference equation

  1. #1
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    First order difference equation

    I need to solve this using iteration method



    y <t> = 0.8 y <t-1> +1
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  2. #2
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    Hello, molimoli!

    Is there any more information . . . like the first term?


    I need to solve this using iteration method: .$\displaystyle y_{t} \:=\: 0.8y_{t-1} +1$
    I'm not familiar with the term "interation method", but I can solve it.
    . . I'll change the notation to a more convenient form (for me).

    We have: .$\displaystyle f(n) \:=\:\tfrac{4}{5}\!\cdot\!f(n\text{-}1) + 1$

    We conjecture that $\displaystyle f(n)$ is an exponential function: .$\displaystyle f(n) \:=\:X^n$


    $\displaystyle \begin{array}{cccccc}\text{So we have:} & X^n &=& \tfrac{4}{5}X^{n-1} + 1 & [1] \\ \\[-3mm]

    \text{Write the next term:} & X^{n+1} &=& \tfrac{4}{5}X^n + 1 & [2] \end{array}$


    Subtract [2] - [1]: .$\displaystyle X^{n+1} - X^n \:=\:\tfrac{4}{5}X^n - \tfrac{4}{5}X^{n-1} \quad\Rightarrow\quad X^{n+1} - \tfrac{9}{5}X^n + \tfrac{4}{5}X^{n-1} \:=\:0
    $

    Multiply by 5, divide by $\displaystyle X^{n-1}\quad 5X^2 - 9X + 4 \:=\:0$

    Factor and solve: .$\displaystyle (X - 1)(5X - 4) \:=\:0 \quad\Rightarrow\quad X \:=\:1,\:\tfrac{4}{5}$


    Hence, the function is of the form: .$\displaystyle f(n) \;=\;A + B\left(\tfrac{4}{5}\right)^n $


    Suppose the initial term is: .$\displaystyle f(0) = a$
    . . Then: .$\displaystyle f(1) \:=\:\tfrac{4}{5}a + 1$


    Then we have:

    $\displaystyle \begin{array}{ccccccccccc}f(0)\:=\:a\!: & &A + B(\frac{4}{5})^0 &=&a & \Longrightarrow & A + B &=& a & [3] \\ \\[-3mm]
    f(1) \:=\:\frac{4}{5}a+1\!: & &A + B(\frac{4}{5})^1 &=& \frac{4}{5}a+1 & \Longrightarrow & A + \frac{4}{5}B &=& \frac{4}{5}a + 1 & [4]
    \end{array}$


    Subtact [3] - [4]: .$\displaystyle \tfrac{1}{5}B \:=\:\tfrac{1}{5}a - 1 \quad\Rightarrow\quad B \:=\:a-5$

    Substitute into [3]: .$\displaystyle A + (a-5) \:=\:a \quad\Rightarrow\quad A \:=\:5$


    Therefore: .$\displaystyle \boxed{f(n) \;=\;5 + (a-5)\left(\tfrac{4}{5}\right)^n} $


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    In the original notation: .$\displaystyle y_t \;=\;5 + (y_0-5)\left(\tfrac{4}{5}\right)^t $

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