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Evaluate the sum

  1. Dec 15, 2012 #1

    sharks

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    Gold Member

    1. The problem statement, all variables and given/known data
    $$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]$$

    2. Relevant equations
    Multiple-sigma notation.


    3. The attempt at a solution
    I agree this seems like basic summation stuff, but i do not agree with the given answer. So, here is what i've done.

    $$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]
    \\=2\sum_{i=1}^{m} (x_i-x_{i-1})-3\sum_{j=1}^{n}(y_j-y_{j-1})
    \\=2(x_m-x_0)-3(y_n-y_0)$$

    However, the given answer for this problem is:
    $$2n(x_m-x_0)-3m(y_n-y_0)$$
     
  2. jcsd
  3. Dec 15, 2012 #2

    Zondrina

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    Homework Helper

    Do you know what the partition of the interval is? Was there an interval given in the question?
     
  4. Dec 15, 2012 #3

    Dick

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    Homework Helper

    The terms that are indexed with i are independent of j. So when you apply the j sum they will get added to themselves n times. So they will pick up a factor of n. Similarly for the terms indexed with j.
     
  5. Dec 15, 2012 #4

    Mark44

    Staff: Mentor

    You're skipping some steps above.

    After the first line you should have this:
    $$ \sum_{i=1}^{m} \left(\sum_{j=1}^{n}[2(x_i-x_{i-1}] - \sum_{j=1}^{n}[3(y_j-y_{j-1})]\right)$$
    The first sum isn't affected by j, so will simplify to n * the expression being summed. Something similar happens for the outer summation.

     
  6. Dec 15, 2012 #5

    sharks

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    Gold Member

    Yes, there are two intervals given, but in this case, the answer still wouldn't be correct, so i omitted that part from the stated problem in the 1st post. Anyway, here they are:

    Let ##P_1 = \{ x_0, x_1, x_2,..., x_m \} ## be a partition of ##[a_1, a_2]##
    Let ##P_2 = \{ y_0, y_1, y_2,..., y_n \} ## be a partition of ##[b_1, b_2]##

    That just cleared up all my confusion! Clear and concise explanation. Thanks, Dick!

    I had intentionally skipped those steps, as all the terms in i are independent of j, and all the terms in j are independent of i. So, in the expansion, i had wrongly assumed that expanding all the i terms with respect to j and all the j terms with respect to i, would give 1. Thanks for the help.
     
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