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Need help in explaining the calculation of n-th power of a summation

  1. Apr 29, 2010 #1

    I got to the following equation while going through a book. I can't figure out how the second line comes from the first. Can anyone please help me understand?

    (1/2*\sum_{q=-Q}^Q V_s,q .H(w_q) .exp(iw_q t))^n is written as,

    1/2^n * \sum_{q1=-Q}^Q \sum_{q2=-Q}^Q ....... \sum_{qn=-Q}^Q V_s,q1 . V_s,q2 .... V_s,qn .H(w_q1).H(w_q2).....H(w_qn).exp(i(w_q1 +w_q2 +....w_qn).t)

    Please forgive my poor latex style writing.
  2. jcsd
  3. May 2, 2010 #2
    Well, I can forgive it, but unfortunately I can't read it.
  4. May 2, 2010 #3
    [itex](1/2*\sum_{q=-Q}^Q V_{s,q} .H(w_q) .exp(iw_q t))^n[/itex]

    is written as,

    [itex]1/2^n * \sum_{q_1=-Q}^Q \sum_{q_2=-Q}^Q ....... \sum_{q_n=-Q}^Q V_{s,q_1} . V_{s,q_2} .... V_{s,q_n} .H(w_{q_1}).H(w_{q_2}).....H(w_{q_n}).exp(i(w_{q_1} +w_{q_2} +....w_{q_n}).t)[/itex]

    The above is hopefully closer. I don't know if its just me, but I see quite a few posts with unprocessed Latex.
    Last edited: May 2, 2010
  5. May 2, 2010 #4
    And if you have

    [itex](A\sum_{i=r}^s a_i)^n=A^n(a_r+a_{r+1}\dots a_s)(a_r+a_{r+1}\dots a_s)\dots (a_r+a_{r+1}\dots a_s)\text{ (to }n\text{ factors})[/itex]

    you get

    [itex]A^n(\Sigma a_{i_1}a_{i_2}\dots a_{i_n}})[/itex]

    where the sum on the second line runs through all different ways of selecting each [itex]a_{i_k}[/itex] from the corresponding bracket on the rhs of the first line.

    This can be written as

    [itex]A^n\sum_{i_1=r}^s\sum_{i_2=r}^s\dots \sum_{i_n=r}^s a_{i_1}a_{i_2}\dots a_{i_n}}[/itex]

    and as the indices run from r to s in each of the multiple sums, all ways of selecting [itex]a_{i_1}a_{i_2}\dots a_{i_n}[/itex] will occur exactly once.

    With appropriate values inserted from your example this gives you the transition between the first and second lines.
  6. May 3, 2010 #5
    Makes Perfect sense. Thank you so much for the explanation :D and special thanks for forgiving my latex :P
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