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    I dunno if I'm being really stupid, but I can't see how to get a solution given.

    Consider the affine map given by:

    x_{n+1} = a + bx_{n}

    Show the solution is given by:

    x_n = b^n + \frac{a}{1-b}

    The solution says to just make sure it satisfies the affine map, with the first line as:

    x_{n+1} = (x_0 - \frac{a}{1-b})b^{n+1} + \frac{a}{1-b}

    I can't see where that came from. Thanks!
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    You could treat it as a second order difference equation and solve it I think,

    i.e solve x_{n+1}-bx_n + 0x_{n-1} = a for x_n


    Wait just sub the solution in and show it satisfies the equation

    if x_n = b^n + \frac{a}{1-b} then what does x_{n+1}=? (just replace the ns for n+1s)
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    (Original post by TheJ0ker)
    You could treat it as a second order difference equation and solve it I think,

    i.e solve x_{n+1}-bx_n + 0x_{n-1} = a for x_n

    Wait just sub the solution in and show it satisfies the equation
    I thought I was going down the wrong road with that as I just got

    x_{n+1} = b^{n+1} + \frac{a}{1-b}

    I wasn't sure where to go from there.
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    (Original post by Brit_Miller)
    I thought I was going down the wrong road with that as I just got

    x_{n+1} = b^{n+1} + \frac{a}{1-b}

    I wasn't sure where to go from there.
    Yeah now sub that into the original equation, you have an expression for x_(n+1) and x_n so see if they satisfy the affine map, if it does x_n must be a solution
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    (Original post by TheJ0ker)
    Yeah now sub that into the original equation, you have an expression for x_(n+1) and x_n so see if they satisfy the affine map, if it does x_n must be a solution
    Ohh, I see. Thanks!
 
 
 
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