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Math Help - Maybe I'll Get It This TIME

  1. #1
    Member Ranger SVO's Avatar
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    Maybe I'll Get It This TIME

    One more time and maybe it will sink in.

    D(0) = 5

    D(1) = 2.25 = ((1/4)(D(0))+1

    D(2) = 1.5625 = ((1/4)(D(1))+1

    D(3) = 1.390625 = ((1/4)(D(2))+1

    D(n) = ????

    This is the final question on an assesment due Thursday. An explination is just as important as the answer. Every step is important so that I might explain the prosses correctly on the assesment

    Thank you for your time
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  2. #2
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    Quote Originally Posted by Ranger SVO
    One more time and maybe it will sink in.

    D(0) = 5

    D(1) = 2.25 = ((1/4)(D(0))+1

    D(2) = 1.5625 = ((1/4)(D(1))+1

    D(3) = 1.390625 = ((1/4)(D(2))+1

    D(n) = ????

    This is the final question on an assesment due Thursday. An explination is just as important as the answer. Every step is important so that I might explain the prosses correctly on the assesment

    Thank you for your time
    Notice ya got,
    D_0=5
    D_1=\frac{1}{4}D_0+1
    D_2=\frac{1}{4}D_1+1=\frac{1}{4}\left( \frac{1}{4}D_0+1 \right) +1= \frac{1}{4^2}D_0+\frac{1}{4}+1
    D_3=\frac{1}{4}D_2+1=\frac{1}{4}\left( \frac{1}{4^2}D_0+\frac{1}{4}+1 \right)= \frac{1}{4^3}D_0+\frac{1}{4^2}+\frac{1}{4}+1
    Notice da pattern?
    D_n=\frac{1}{4^n}D_0+\frac{1}{4^{n-1}}+...+\frac{1}{4}+1
    But, da series,
    1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{n-1}}
    is geometric.
    Thus, its sum iz,
    \frac{1-(1/4)^n}{1-1/4}=\frac{4}{3}-\frac{4}{3}\cdot \left( \frac{1}{4} \right)^n=\frac{4}{3}-\frac{1}{3\cdot 4^{n-1}}
    Thus, you are left with,
    \frac{1}{4^n}D_0+\frac{4}{3}-\frac{1}{3\cdot 4^{n-1}}
    But, D_0=5
    Thus,
    \frac{5}{4^n}+\frac{4}{3}-\frac{1}{3\cdot 4^{n-1}}
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  3. #3
    Member Ranger SVO's Avatar
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    It took a little rearranging but I see what I was sopposed to.

    It is a decaying (decreasing) exponential function with a limit of 4/3.
    Its rate of decay is 1/4^x with a base of 44/3.

    On this assesment the answer was not as important as what I needed to see. Below is a small part of that assesment

    "Both are graphs of decreasing exponential functions. Both functions have been transformed up. The horizontal asymptote represents the limit. It also represents how much the function has been shifted up.
    Y1 = 2 + 6 * (1/2)x
    Y2 = (4/3) + (44/3) * (1/4)x
    So L = d + c * ( 1- r )t"

    Thank you very much !!
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