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Math Help - double sum designation question..

  1. #1
    MHF Contributor
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    double sum designation question..

    the telegan law basically states that the total sum of power is zero.
    my prof proved lik this:

    we choose a node (a point where more then one currents come together)

    and decide that the voltage on that node to be zero.
    we designate the voltages on the nodes to be e_k
    J_k is the current.

    v_kJ_k=(e_a-e_b)J_{ab}
    v_kJ_k=\frac{1}{2}[(e_b-e_a)J_{ab}+(e_a-e_b)J_{ab}]
    n_t is the number of nodes.[/tex]
    B is the number of branches.[/tex]
    \sum_{k}^{B}v_kJ_k=\frac{1}{2}\sum_{a=1}^{n_t}\sum  _{b=1}^{n_t}(e_a-e_b)J_{ab}
    each J that does not exist in the graph will be zero.
    \sum_{k}^{B}v_kJ_k=\frac{1}{2}\sum_{a=1}^{n_t}e_a\  sum_{b=1}^{n_t}J_{ab}-\frac{1}{2}\sum_{a=1}^{n_t}e_b\sum_{b=1}^{n_t}J_{a  b}=0

    because by kcl
    \sum_{b=1}^{n_t}J_{ab}=0

    my problem iswhen he sums for all nodes
    he uses
    \sum\sum sign which by me represents multiplication
    of the sums

    why not \sum+\sum,thus we can know that ist the sum of many similar equations.

    but how he did it doesnt represent a sum
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  2. #2
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    The double sum in this case represents a nested sum (if you're familiar with programming, it's like a nested for loop).
    In your problem:
    \frac{1}{2}\sum_{a=1}^{n_t}\sum_{b=1}^{n_t}(e_a-e_b)J_{ab}
    means for every iteration of a, we go through all n_t iterations of b. I like to think of it as sort of a series exponentiation.

    Formally if we define: F = \sum_{i=0}^{n} a_{i} and G = \sum_{j=0}^{n} b_{j}

    F * G = \sum_{i=0}^{n} a_{i}b_{n-i}

    F  G = \sum_{i=0}^{n}\sum_{j=0}^{n} a_{i}b_{j}<br />
    Will yield different answers. I'm not sure about the actual concepts, I'm not very well versed in physics, but your prof is using the nested case.
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