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Matrix Vector Transpose

  1. Mar 15, 2012 #1
    To find the least squares polynomial of degree 2 to
    approximate points (X,Y) given in the table

    X_____________Y


    1_____________36
    1.9_____________-49
    2.9_____________23
    4_____________-38
    5.1_____________-14


    the normal equations to be solved have the form
    A c = v, where A is the matrix (NOT NECESSARILY DISPLAYED TO THE
    PRECISION USED FOR THE CALCULATIONS)
    _ _
    | |
    | _____ 1______ _____ 2______ _____ 3______ |
    | _____ 4______ _____ 5______ _____ 6______ |
    | _____ 7______ _____ 8______ _____ 9______ |
    |_ _|

    and c is the coefficient vector for the polynomial;
    and v is the transpose of the vector:
    _ _
    | |
    | _____10______ _____11______ _____12______ |
    |_ _|


    The solution vector c (FOUND USING DOUBLE PRECISION
    CALCULATIONS with the ORIGINAL DATA above) is
    the transpose of vector:
    _ _
    | |
    | _____13______ _____14______ _____15______ |
    |_ _|


    2. Relevant equations

    Ac=v

    3. The attempt at a solution


    I believe I've got the first part; i.e. solving the matrix:


    5.1________14.9_________55.03
    14.9_______55.03________228.899
    55.03______228.899______1017.2803



    Now, I'm clueless as to how to proceed from here on.
     
  2. jcsd
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