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I'm stuck how to evaluate this $$\sum_{j=0}^n \sum_{k=0}^n p_j p_k (-1)^k i^{j+k} $$ where $i = \sqrt{-1}$ and $p_k$ are just variables. Can this be expressed as squares of $p_k$'s like $(p_0 - p_2 + p_4 -p_6 \pm \cdots)^2 + (p_1 - p_3 + p_5 \mp \cdots)^2$?

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$(-1)^ki^{j+k}=i^{j+3k}=i^{j-k}$, so you can write it $\sum_{j=0}^n\sum_{k=0}^np_jp_ki^{j-k}$, but I doubt that this will do any good. –  Brian M. Scott Sep 19 '12 at 7:58
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up vote 3 down vote accepted

The double sum decomposes into a product of single sums:

$$ \begin{align} \sum_{j=0}^n \sum_{k=0}^n p_j p_k (-1)^k\mathrm i^{j+k} &= \left(\sum_{j=0}^np_j\mathrm i^j\right)\left(\sum_{k=0}^n p_k (-\mathrm i)^k\right)\;. \end{align} $$

If the $p_j$ are real, then this is the squared absolute value of $\sum_jp_j\mathrm i^j$, which is a component of the discrete Fourier transform of the $p_j$ if $4\mid n+1$. For general $p_j$, it's the product of two different Fourier components.

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