1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member


    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Weighted Moving Average of Cubic
  1. Cubic Equation (Replies: 2)

  2. Cubic Splines (Replies: 1)

  3. Simplifying a cubic (Replies: 5)

  4. Moving Average Process (Replies: 1)

Loading...