Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinite sum of product of vector elements

  1. Jan 16, 2010 #1


    User Avatar
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?