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Math Help - Series/Sequence Problem

  1. #1
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    Series/Sequence Problem

    I'm having a problem with a question

    The question is: A chemical plant discharges D units of chemical molecules into a river each week. At the end of the week 90% of the chemical molecules have dissipated.


    a) Find the number of units of the molecule after n weeks: I got this
    an=D+.01(an-1) or an=a1+.01(an-1)

    b) Estimate the amount of the molecule in the river for a very long time.
    I know it involves the series of what I found in part a but how do I write that.


    c) If the toxic level of the molecule is T units, how large an amount of the molecule can the plant discharge each week?
    I don't know what do here.

    Any ideas will help, Thanks
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  2. #2
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    Quote Originally Posted by Imuell1 View Post
    I'm having a problem with a question

    The question is: A chemical plant discharges D units of chemical molecules into a river each week. At the end of the week 90% of the chemical molecules have dissipated.


    a) Find the number of units of the molecule after n weeks: I got this
    an=D+.01(an-1) or an=a1+.01(an-1)

    b) Estimate the amount of the molecule in the river for a very long time.
    I know it involves the series of what I found in part a but how do I write that.


    c) If the toxic level of the molecule is T units, how large an amount of the molecule can the plant discharge each week?
    I don't know what do here.

    Any ideas will help, Thanks
    A_1=D
    A_2=D+\frac{1}{10}\left(A_1 \right) = D+\frac{1}{10}\left(D\right)= D\left(1+\frac{1}{10}\right)
    A_3=D+\frac{1}{10}\left(A_2 \right) = D\left(1+\frac{1}{10}+\frac{1}{100}\right)
    If we continue the pattern we get
    A_n=D\sum_{i=1}^{n}\frac{1}{10^{n-1}}

    This should get you started good luck.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    A_1=D
    A_2=D+\frac{1}{10}\left(A_1 \right) = D+\frac{1}{10}\left(D\right)= D\left(1+\frac{1}{10}\right)
    A_3=D+\frac{1}{10}\left(A_2 \right) = D\left(1+\frac{1}{10}+\frac{1}{100}\right)
    If we continue the pattern we get
    A_n=D\sum_{i=1}^{n}\frac{1}{10^{n-1}}

    This should get you started good luck.
    Thanks.

    But for part c...

    Would I test for convergence with the p test or something?
    or do I somehow take the limit then just set that to less than T?
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    A_1=D
    A_2=D+\frac{1}{10}\left(A_1 \right) = D+\frac{1}{10}\left(D\right)= D\left(1+\frac{1}{10}\right)
    A_3=D+\frac{1}{10}\left(A_2 \right) = D\left(1+\frac{1}{10}+\frac{1}{100}\right)
    If we continue the pattern we get
    A_n=D\sum_{i=1}^{n}\frac{1}{10^{n-1}}

    This should get you started good luck.
    I believe it would actually be

    A_n = D\sum_{i = 1}^n\frac{1}{10^{i - 1}}.

    This is a geometric series, with a = 1, r = \frac{1}{10} and n = n.

    So A_n = \frac{Da(1 - r^n)}{1 - r}.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    I believe it would actually be

    A_n = D\sum_{i = 1}^n\frac{1}{10^{i - 1}}.

    This is a geometric series, with a = 1, r = \frac{1}{10} and n = n.

    So A_n = \frac{Da(1 - r^n)}{1 - r}.
    So I would take the limit as n approaches infinity of \frac{Da(1 - r^n)}{1 - r}? Which I got 1.11D.

    Then for part C would I just set that limit to less than T?

    I think I've got it, since the limit = 1.11D, you set that to T and solve for D? giving D=9T/10
    Last edited by Imuell1; December 10th 2009 at 07:23 AM.
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  6. #6
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    Note the geometric converges for common ratio's less than 1, its the one of its most known properties. your common ration here is 1/10
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