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Infinite sum of product of vector elements

  1. Jan 16, 2010 #1

    diazona

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    I've been trying to simplify a formula that looks something like this:
    [tex]X = \sum_{k=0}^{\infty} v_{x[k]} v_{y[k]}[/tex]
    where [itex]v_{x[k]}[/itex] and [itex]v_{y[k]}[/itex] are the horizontal and vertical components of a velocity vector. Both components are positive for all k, and
    [tex]v_{x[k]} < v_{x[k-1]}[/tex]
    [tex]v_{y[k]} < v_{y[k-1]}[/tex]
    for all [itex]k \ge 1[/itex]. I also know (or, I'm assuming) that there is a linear recurrence relation that expresses [itex]v_{x[k]}[/itex] and [itex]v_{y[k]}[/itex] in terms of [itex]v_{x[k-1]}[/itex] and [itex]v_{y[k-1]}[/itex]:
    [tex]\begin{pmatrix}v_{x[k]} \\ v_{y[k]}\end{pmatrix} = \begin{pmatrix}C_{xx} & C_{xy} \\ C_{yx} & C_{yy}\end{pmatrix}\begin{pmatrix}v_{x[k-1]} \\ v_{y[k-1]}\end{pmatrix}[/tex]
    Each of the elements of the matrix C satisfies [itex]0 \le C_{ij} \le 1[/itex]. (More specifically: [itex]0 < C_{xx} \le 1[/itex], [itex]0 < C_{yy} < 1[/itex], I have reason to suspect that [itex]C_{xx},C_{yy} > C_{xy},C_{yx}[/itex], and it's quite possible that [itex]C_{yx} = 0[/itex].)

    I managed to figure out without any trouble that if the matrix C is diagonal, the series sums to
    [tex]X = \frac{v_{x[0]}v_{y[0]}}{1 - C_{xx}C_{yy}}[/tex]
    But when C has off-diagonal terms, I haven't been able to do anything meaningful with the sum. So far I've been playing with a structure like
    [tex]X = \sum_{k=0}^{\infty} \vec{v}_{x[0]}^T(C^T)^k \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}C^k \vec{v}_{y[0]}[/tex]
    What I'd like, ideally, is to find some way to express the sum X in closed form when the matrix is not diagonal. Even if that's not possible (as I'm suspecting), anything I can say about the sum could potentially be useful. I wanted to see if someone who knows linear algebra better than I do could suggest anything.

    (this is not homework)
     
  2. jcsd
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