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Lagrange Identity Sum Notation

  1. Oct 1, 2006 #1
    Hi, how do I interpret the last sum:
    http://planetmath.org/encyclopedia/LagrangesIdentity.html

    Sum (...)
    1<=k < j <= n

    Is it the double sum:

    Sum( Sum( (a_k*b_j - a_j*b_k)^2 from k = 1 to n) from j = 2 to n ) ?
     
  2. jcsd
  3. Oct 1, 2006 #2

    shmoe

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    In your notation it would be:

    Sum( Sum( (a_k*b_j - a_j*b_k)^2 from k = 1 to j-1) from j = 2 to n )

    as it's a sum of pairs (k,j) from {1,...,n}x{1,...,n} where k is strictly less than j.

    If you arrange the pairs of {1,...,n}x{1,...,n} in an nxn grid with k indexing the rows and j the columns like so:

    (1,1), (1,2), (1,3),...
    (2,1), (2,2), (2,3),...
    ...

    your sum is over the terms above the main diagonal, e.g. for n=3 it's the bold terms:

    (1,1), (1,2), (1,3)
    (2,1), (2,2), (2,3)
    (3,1), (3,2), (3,3)
     
  4. Oct 2, 2006 #3

    quasar987

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    I like to think of it as the sum over all vectors (k,i) where k and i can take any value btw 1 and n, BUT we consider only case vectors for which k<i.
     
  5. Oct 2, 2006 #4
    Thanks, I see what the notation means, and your way of visualizing the index pairs of the terms being summed was very helpful too, as I never thought to think about it like that.
     
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