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Thread: Pn(x) Polynomial fitting

  1. #1
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    Pn(x) Polynomial fitting

    Hello, out there! I am Samuel.

    Please, I need your assistance to fit the following set of data points with a polynomial (could be Newtons divided difference polynomial):

    i 0 1 2 3 4 5 6 7 8 9 01 11
    xi 0.005 0.01 0.02 0.03 0.05 0.1 0.2 0.3 0.4 0.6 0.8 1.0
    f(xi) 0.376 0.41 0.428 0.486 0.537 0.624 0.556 0.512 0.450 0.378 0.274 0.106

    I need the polynomial to pass through all the data points.

    Looking forward to your kind response.

    Thanks.

    Last edited by saayo3; Nov 24th 2012 at 10:06 PM. Reason: Additional information
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Pn(x) Polynomial fitting

    Here. I'll create a polynomial that passes through for your first three points

    .005 f(.005) = .376
    .01 f(.01) .41
    .02 f(.02) = .428

    Since we have 3 points, assume that there exists a polynomial $\displaystyle p(x) = a_2 x^2 + a_1 x + a_0 $ such that $\displaystyle p(x_i) = f(x_i) $. Then
    p(.005) = $\displaystyle a_2 (.005)^2 + a_1 (.005) + a_0 = .376 $
    p(.01) = $\displaystyle a_2 (.01)^2 + a_1 (.01) + a_0 = .41 $
    p(.02) = $\displaystyle a_2 (.02)^2 + a_1 (.02) + a_0 = .428 $

    What we get is a system of linear equations with 3 unknowns and 3 equations.
    $\displaystyle \begin{bmatrix} (.005)^2 & (.005) & 1\\ (.01)^2 & (.01) & 1 \\ (.02)^2 & (.02) & 1 \end{bmatrix} \begin{bmatrix} a_2 \\ a_1 \\ a_0 \end{bmatrix} = \begin{bmatrix} .376 \\ .41 \\ .428 \end{bmatrix} $

    So solving, i get $\displaystyle a_2 = -333.33 $ $\displaystyle a_1 = 11.8 $ and $\displaystyle a_0 = .325 $ .

    So our polynomial is $\displaystyle p(x) = -333.33x^2 + 11.8x + .325 $

    So verification
    $\displaystyle p(.005) = -333.33(.005)^2 + 11.8(.005) + .325 = .376 $
    $\displaystyle p(.01) = -333.33(.01)^2 + 11.8(.01) + .325 = .41 $
    $\displaystyle p(.02) = -333.33(.02)^2 + 11.8(.02) + .325 = .428 $

    So There ya go. Just do this for all your data points and use this linear solver linked below to get the coefficients.
    Linear solver


    The linear solver seems to be able to only solve 9 linear equations in 9 variables. So i got for the first 9 points
    $\displaystyle -400000000.0x^8 + 438999999.99999994x^7 + -180000000.0x^6 + 34700000.0x^5 + -3319243.63x^4 + 156177.88x^3 + -3498.67x^2 + 37.53x^1 + 0.26x^0 + $

    Here are the test results
    Got Value $\displaystyle p(x_i): 0.38$ Expected Value $\displaystyle f(x_i): 0.376$
    Got Value $\displaystyle p(x_i): 0.41$ Expected Value $\displaystyle f(x_i): 0.41$
    Got Value $\displaystyle p(x_i): 0.43$ Expected Value $\displaystyle f(x_i): 0.428$
    Got Value $\displaystyle p(x_i): 0.49$ Expected Value $\displaystyle f(x_i): 0.486$
    Got Value $\displaystyle p(x_i): 0.53$ Expected Value $\displaystyle f(x_i): 0.537$
    Got Value $\displaystyle p(x_i): 0.18$ Expected Value $\displaystyle f(x_i): 0.624$
    Got Value $\displaystyle p(x_i): -14.35$ Expected Value $\displaystyle f(x_i): 0.556$
    Got Value $\displaystyle p(x_i): -106.13$ Expected Value $\displaystyle f(x_i): 0.512$
    Got Value $\displaystyle p(x_i): -360.17$ Expected Value $\displaystyle f(x_i): 0.450$

    As you can see every data point added makes the polynomial error grow at a very large rate, i guess that is to be expected
    Last edited by jakncoke; Nov 25th 2012 at 03:28 AM.
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