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Homework Help: Weighted Moving Average of Cubic

  1. Feb 2, 2013 #1
    1. Show that applying a second-order weighted moving average to a cubic polynomial will not change anything.
    [tex]X_t = a_0 + a_1t + a_2t^2 + a_3t^3[/tex] is our polynomial
    Second-order weighted moving average: [tex]\sum_{i=-L}^{i=L} B_iX_{t+i}[/tex]
    where [tex]B_i=(1-i^2I_2/I_4)/(2L+1-I_2^{2}/I_4)[/tex]
    where [tex]I_2=\sum_{i=-L}^{i=L} i^2[/tex]

    2. I did a similar problem where applying a linear moving average to a linear equation returns the same thing, but I'm not sure how to proceed here.
  2. jcsd
  3. Feb 2, 2013 #2

    Ray Vickson

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    What is L supposed to be in your case?
  4. Feb 2, 2013 #3
    L is any arbitrary number.
    For any L, this should be true.
  5. Feb 2, 2013 #4


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    What is I4 ?
  6. Feb 2, 2013 #5
    Sorry, it just follows the same pattern as I2:

    [tex]I_4=\sum_{i=-L}^{i=L} i^{4}[/tex]
  7. Feb 2, 2013 #6

    Ray Vickson

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    So you have weighting factors of the form
    [tex] B_i = A - B i^2, i=-L, \ldots, L.[/tex] What happens when you compute
    [tex] \sum_{i=-L}^{L} (A - B i^2) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3]?[/tex]
  8. Feb 2, 2013 #7
    I'm unsure as to how you rearranged the weights to get [tex]B_i = A - B i^2[/tex], would you mind clarifying what A and Bi2 are?
    Last edited: Feb 2, 2013
  9. Feb 2, 2013 #8
    I mean, one way to write what we have is
    [tex]\frac{1}{2L+1-I_2^{2}/I_4} \sum_{i=-L}^{i=L} (1-i^2\frac{I_2}{I_4}) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3][/tex]
  10. Feb 2, 2013 #9
    I THINK I may have got it. I basically looked at what we had, and what we need.

    In order for us to get back a0, for example, we need:

    [tex]\frac{1}{2L+1-\frac{I_2^{2}}{I_4}}\sum_{i=-L}^{i=L}a_0(1-i^2\frac{I_2}{I_4}) = a_0[/tex]
    Well, let's multiply through:

    [tex]\frac{a_0}{2L+1-\frac{I_2^{2}}{I_4}}\sum_{i=-L}^{i=L} 1 - a_0\frac{\sum_{i=-L}^{i=L} i^2\frac{I_2}{I_4}}{2L+1-\frac{I_2^{2}}{I_4}}[/tex]
    [tex]\frac{a_0(2L+1) - a_0\frac{I_2^{2}}{I_4}}{2L+1-\frac{I_2^{2}}{I_4}} = a_0[/tex]

    Then, do the same for the rest.
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