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Homework Help: Infinite series, power series problem

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    In a water purification process, one-nth of the impurity is removed in the first stage. In each succeeding stage, the amount of impurity removed is one-nth of that removed in the preceding stage. Show that if n=2, the water can be made a pure as you like, but if n=3, at least one-half of the impurity remains no matter how many stages are used.


    2. Relevant equations

    may be relevant,


    Sn = (ao(1-rn)/(1-r))


    Sn-rSn = (ao-aor^n)


    3. The attempt at a solution

    well i figured out the series expansion for both


    SUM (1/(n^2 -n)) (sorry i couldn't figure out the text thing)


    n=2 gives = 1/2 + 1/6 + 1/12 + 1/20 + ...

    n=3 gives = 1/6 + 1/12 + 1/20 + 1/36 + ...

    I know the sum of n=2 is pretty much 1 and n=3 is pretty much 1/2. My real problem is i can't figure out how to get it into:


    Sn = (ao(1-rn)/(1-r))


    so I can differentiate to get the sum. Any help would be appreciated and thanks for helping a noob
     
  2. jcsd
  3. Sep 14, 2010 #2

    jgens

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    Can you explain how you got these sums? They don't model the situation that the problem describes.
     
  4. Sep 15, 2010 #3

    i got those sums, by finding the equation


    1/(n-1) - (1/n) = 1/(n^2 - n)

    This is what i did for n=2 n=2 wolfram

    n=3 n=3 wolfram
     
    Last edited: Sep 15, 2010
  5. Sep 15, 2010 #4

    ehild

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    Calculate the removed impurity.
    At the first step, 1/n part of the impurity is removed.
    In the second and every subsequent step, 1/n times the previously removed quantity is removed.

    So the removed amount in k steps is

    1/n +1/n^2 +1/n^3 +...1/n^k. What do you get after infinite steps?

    ehild
     
  6. Sep 15, 2010 #5
    i'm sorry but i messed up my second post, i'm not sure if you were responding to that one, but i made a mistake in my equation.
     
  7. Sep 15, 2010 #6

    jgens

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    If you follow ehild's lead, you'll solve the problem pretty quickly. I'm not sure what reasoning your using to get your sums, but it's not correct.
     
  8. Sep 16, 2010 #7
    can someone please just give me a solution, because this is apparently the easiest part of my assignment, and there are 3 questions like this, meanwhile i have completed all the apparent harder ones. If i could see one example of this sort, i should be able to figure it out.
     
  9. Sep 16, 2010 #8

    jgens

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    It's against PF policy to give complete solutions. We can however, guide you towards a solution.
     
  10. Sep 16, 2010 #9
    okay that is something i did not know.

    Well can you tell me why i am wrong, because i'm really lost

    all i have to work with for equations are;

    Sn = (a(1-rn))/(1-r)

    S = lim (n[tex]\rightarrow[/tex][tex]\infty[/tex]) Sn

    S = a/(1-r)

    for this section, so this is what i'm assuming should help solve this problem.
    I'm not sure how to find r or a, and i guess that is my major problem.
     
  11. Sep 16, 2010 #10

    jgens

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    Use this ...

     
  12. Sep 16, 2010 #11
    1/n^2 + 1/n^3 + 1/n^4 + ... + 1/n^(k-1) = Sn

    1/n^3 + 1/n^4 + 1/n^5 + ... + 1/n^k-1 + 1/n^k = rSn

    Sn - rSn = 1/n^3 - 1/n^k = a(1 - r^k)

    a(1 - r^k) = (1/n^3)(1 - (1/n^k)) no k-3 since k is going to infinity?

    Sn(1 - r) = Sn(1- (1/n))


    Sn = ((1/n^3)(1 - (1/n^k))) / (1 - (1/n))


    S = lim (k approaches infinity) Sn

    S = (1/n^3)/(1 - (1/n))

    Is this it? I think i got it
     
  13. Sep 17, 2010 #12

    jgens

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    What happened to the term 1/n?
     
  14. Sep 17, 2010 #13
    whoops, but if i add that it should work?

    i need n=2 to infinity to ~= 1

    and n=3 to infinity to ~=1/2
     
  15. Sep 17, 2010 #14

    jgens

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    You, if you redo your work including that term, it should work out.
     
  16. Sep 18, 2010 #15
    okay screw it, i'll try it your way once more, and if i don't get it... then i'll just skip it

    Thanks for trying to help me, I can't help i'm border line retarded sometimes
     
  17. Sep 18, 2010 #16

    ehild

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    I think you mix the "n" in the problem with the "running index" of a sequence. Forget Sn. It is the sum of the first n term of a sequence. You need the sum of infinite terms which is S=a1/(1-r) for a geometric sequence with first term a1 and ratio r.

    Consider the removed amount of the impurity. Suppose its original amount is 1 in some unit. In the first step, you remove 1/n of it. In the next step, you remove the n-th part of the previously removed amount, so you remove (1/n)^2. In the third step, you remove (1/n)^3, in the k-th step, the removed amount is (1/n)^k. You continue the process endlessly, and you get the following sequence

    1/n, (1/n)^2, (1/n)^3....... (1/n)^k.....

    If n=2, it is 1/2, 1/4, 1/8, 1/16........

    For n=3: 1/3, 1/9, 1/27, 1/81........

    What kind of sequence is it and what is the sum of infinite terms?

    This sum is the theoretical amount of impurity you can remove in infinite steps.

    ehild.
     
  18. Sep 21, 2010 #17
    you know that is what i did exactly and right after my last post i realized that, and got the problem figured out. Thank you very much for the help and not getting mad because i was so stubborn
     
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