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Cubic Ploynomial forms

  1. Aug 24, 2007 #1

    hotvette

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    Here's an interesting question. I'm aware of closed forms of cubic polynomials that go through 1 or 2 specific (x,y) points. Are there closed form versions for 3 or 4 points?

    1 pt: [tex]y = a(x-x_0)^3 + b(x-x_0)^2 + c(x-x_0) + y_0[/tex]

    2 pt: [tex]y = a(x-x_0)^2(x-x_1)\ +\ b(x-x_0)(x-x_1)^2 \ +\ \frac{y_0(x-x_1)^3}{(x_0-x_1)^3} \ +\ \frac{y_1(x-x_0)^3}{(x_1-x_0)^3}[/tex]

    3 pt: [tex]y = \ ?[/tex]

    4 pt: [tex]y = \ ?[/tex]

    I don't think there are.
     
    Last edited: Aug 24, 2007
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  3. Aug 24, 2007 #2

    HallsofIvy

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    Given any 4 points in the plane, there exist a unique cubic polynomial whose graph goes through those 4 points. Given 1, 2, or 3 points, there exist an infinite number of different cubics passing through those points. In your first example, yes, different choices for a, b, c give different cubics through [itex](x_0,y_0)[/itex]. In your second example, different choices for a and b give different cubics through [itex](x_0,y_0)[/itex] and [itex](x_1,y_1)[/itex]. (I think you don't really need the cubes in the last two fractions.)

    For 3 points, what's wrong with
    [tex]y= a(x-x_0)(x-x_1)(x-x_2)+\frac{y_0(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}+ \frac{y_1(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}+ \frac{y_2(x-x_0)(x-x_1)}{(x_2-x_1)(x_2-x_0)}[/tex]

    for 4 points, the unique cubic is given by the LaGrange polynomial
    [tex]y= \frac{y_0(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}+\frac{y_1(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}+\frac{y_2(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}+\frac{y_3(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}[/tex]
     
  4. Aug 24, 2007 #3

    hotvette

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    Thanks! I should have known the 4-pt. verison. Lagrange. Of course. The 3-pt version I haven't seen before. And you are right, don't need cubes in the last two fractions for the 2-pt version. Thanks.
     
    Last edited: Aug 24, 2007
  5. Aug 24, 2007 #4

    HallsofIvy

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    For the 3 pt version, I just extended what you did with 1 and 2 pts! The fractions are, of course, the Lagrange formula for a quadratic through the three points and the first term is a cubic that is 0 at each given point.
     
  6. Aug 24, 2007 #5

    hotvette

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    Looking at it all now, it makes perfect logical sense. Thanks.
     
  7. Aug 24, 2007 #6

    hotvette

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    Actually, the same 4-pt form can be used for all situations (0 fixed points - 4 fixed points). See attachment.
    .
     

    Attached Files:

    Last edited: Aug 25, 2007
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