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Math Help - Recurrence relations - Iterative Method

  1. #1
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    Recurrence relations - Iterative Method

    Could somebody please solve y_t+1=0.5y_t+3 for me by using the iterative method (t=0,1,2 & 3 - please write down every step)? I don't understand it!

    I thank you in advance.
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  2. #2
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    Hello, valal!

    If no one is responding, it's because we don't understand the problem.
    Could you state the original problem?


    Could somebody please solve y_t+1=0.5y_t+3 for me (?)

    by using the iterative method (*)

    t=0,1,2 & 3 (#)


    (?) .What do you mean by "solve"?
    . . . .Do they want a "closed form"?

    Also, a few spaces and parentheses would help us read what you wrote.
    . . I would guess that it is: . y_{t+1} \:=\:\frac{1}{2}y_t + 3


    (*) .We need to know what your teacher/course/textbook
    . . . .means by the "iterative method".


    (#) .This is strange.
    . . . .We're involved with only the first four terms?


    I would conjecture that they wants us to list the first four terms.
    . . . .But they are expected to give us the first term, t_o

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  3. #3
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    I want to solve the recurrence relation/equation - find the "common" solution. (contra the "exact" solution. I'm not sure if these terms are the right names in English)

    And yes, it is y_{t+1} \:=\:\frac{1}{2}y_t + 3 which I mean.

    Example of iterative (= repeated) method:

    y_{t+1} = y_t + 2

    t=0: y_1 = y_0 + 2

    t=1: y_2 = y_1 + 2 = (y_0 + 2) + 2 = y_0 + 2*(2)

    t=2: y_3 = y_2 + 2 = [y_0 + 2*(2)] + 2 = y_0 + 3*(2)

    t=3: y_4 = y_3 + 2 = [y_0 + 3*(2)] + 2 = y_0 + 4*(2)

    From this you can see that:

    y_t = y_0 + t*(2) = y_0 + 2t <--- (solution)



    If you use my system, what is then the solution for y_{t+1} = 0.5y_t + 3?

    In advance thank you.
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  4. #4
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    Nobody who is able to help? It's just the basics ...
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