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Math Help - 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: . 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: . f(n) \:=\:\tfrac{4}{5}\!\cdot\!f(n\text{-}1) + 1

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


    \begin{array}{cccccc}\text{So we have:} & X^n &=& \tfrac{4}{5}X^{n-1} + 1 & [1] \\ \\[-3mm]<br /> <br />
\text{Write the next term:} & X^{n+1} &=& \tfrac{4}{5}X^n + 1 & [2] \end{array}


    Subtract [2] - [1]: . 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<br />

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

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


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


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


    Then we have:

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


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

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


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


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

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

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