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Find Cubic Equation from Four Points?

  1. May 18, 2014 #1
    I am working on finding the area of a solid object. I have 4 points that I need to calculate a cubic equation from. I have tried relentlessly but to no avail I always get the wrong answer.

    The four points are;(0,2.7) (0.5, 2.9) (1,3.2) (1.9, 3.4)

    Using excel, the formula should be; -0.24728x^3 + 0.57093x^2 + 0.17636x + 2.7

    If anyone can provide working out on how you got the equation it would be much obliged! No matrices please just substitution.
     
  2. jcsd
  3. May 18, 2014 #2

    maajdl

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    Gold Member

    Excel does that by the least squares method:

    http://en.wikipedia.org/wiki/Least_squares

    It reduces to the solution of a linear system.

    Another way to proceed for your specific problem is based on the Lagrange polynomials:

    http://en.wikipedia.org/wiki/Lagrange_polynomial

    This is a very simple method. It is applicable only because you have 4 data and 4 unknowns.
    In your case, this gives the following polynomial:

    P(x) =
    2.7 ((x-0.5)(x-1)(x-1.9)) / ((0-0.5)(0-1)(0-1.9)) +
    2.9 ((x-0)(x-1)(x-1.9)) / ((0.5-0)(0.5-1)(0.5-1.9)) +
    3.2 ((x-0)(x-0.5)(x-1.9)) / ((1-0)(1-0.5)(1-1.9)) +
    3.4 ((x-0)(x-0.5)(x-1)) / ((1.9-0)(1.9-0.5)(1.9-1))

    =
    (32319 + 2111 x + 6834 x^2 - 2960 x^3)/11970
     
  4. May 18, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Here's another way to do that: a cubic function can be written in the form [itex]y= ax^3+ bx^2+ cx+ d[/itex] and determining the function means determining the four values of a, b, c, and d. Setting x and y equal to their values in those four points gives four equations:
    [itex]d= 2.7[/itex]
    [itex]0.125a+ .25b+ .5c+ d= 2.9[/itex]
    [itex]a+ b+ c+ d= 32[/itex]
    [itex]6.859a+ 3.61b+ 1.9c+ d= 3.4[/itex]

    Solve those equations for a, b, c, and d.

    maajdl's "Lagrange polynomial" is simpler. But "least squares" is a method for fitting a curve closest to a larger number of points that you cannot get a single curve pass through. It would be appropriate if we had more than four points we wanted to fit a cubic to.
     
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