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Math Help - Finding A Polynomial Model

  1. #1
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    Finding A Polynomial Model

    Hi everybody!

    I'm provided a table of values:

    x: 0 1 2 3 4 5 6
    y: 0 5 14 33 68 125 210

    I used the Polynomial difference theorum to find that by subtracting consecutive y values to find the difference 3 times, they equalled out to 6's. So, I knew that it was degree 6, or ax^3 + bx+2 + cx + d.

    So far so good.

    Next, I'm instructed to use Matrices to find the coefficients. This is what's giving me trouble.

    I set up the system:

    T(4): 33 = 64a + 16b + 4c + d
    T(3): 14 = 27a + 9b + 3c + d
    T(2): 5 = 8a + 4b + 2c + d
    T(1): 0 = 0a + 0b + 0c + 0d


    Puting that into matrices, I have:

    A: (4 by 4)

    64, 16, 4, 1
    27, 9, 3, 1
    8, 4, 2, 1
    0, 0, 0, 0

    ---

    B: (4 by 1):

    33,
    14,
    5,
    0


    When I multiply the inverse of matrix A times matrix B, I get an error message: "ERROR 03 SINGULAR MAT"


    I know I can find the coefficients by using STAT, typing in the x values and the y values and using cubic regression, but I'm supposed to solve this using matrices. Any idea why this isn't working?

    Thanks,
    Ethan


    Edit:
    Sorry about the formatting, I had the data values spaced out nicely, but they clumped back together after I posted!
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  2. #2
    Behold, the power of SARDINES!
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    T(1): 0 = 0a + 0b + 0c + 0d

    This is incorrect it should be

    T(1): 0 = 0a + 0b + 0c + d

    or d =0
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  3. #3
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    Re

    Thanks,

    That was my bad.

    If I replace that 0 with a 1 in the matrix, and I multiply the inverse of A times B, I get:

    a = .70833
    b = -1.375
    c = 2.416
    d = 0

    So, that's telling me my cubic regression should be:
    y = .70833x^3 - 1.375x^2 + 2.416x

    However, when I put the x values and the y values into Stat under L1 and L2, and I choose cubic regression, I get much NICER values:

    a= 1
    b=-1
    c=5
    d=0

    So, according to STAT, my regression equation should be:

    y = 1x^3 - 1x^2 + 5x

    If they are both degree 3, why am I getting different results? Again, I think I am doing something wrong when I solve by Matrices...

    Thanks for the help,
    Ethan
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  4. #4
    Behold, the power of SARDINES!
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    T(4): 33 = 64a + 16b + 4c + d
    T(3): 14 = 27a + 9b + 3c + d
    T(2): 5 = 8a + 4b + 2c + d
    T(1): 0 = 0a + 0b + 0c + 0d


    The Equations should be ...

    T(3): 33 = 27a + 9b + 3c + d
    T(2): 14 = 8a + 4b + 2c + d
    T(1): 5 = a + b + c + d
    T(0): 0 = 0a + 0b + 0c + 1
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  5. #5
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    Re

    Thank you very much!
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