# Cubic Ploynomial forms

1. Aug 24, 2007

### hotvette

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: $$y = a(x-x_0)^3 + b(x-x_0)^2 + c(x-x_0) + y_0$$

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

3 pt: $$y = \ ?$$

4 pt: $$y = \ ?$$

I don't think there are.

Last edited: Aug 24, 2007
2. Aug 24, 2007

### HallsofIvy

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 $(x_0,y_0)$. In your second example, different choices for a and b give different cubics through $(x_0,y_0)$ and $(x_1,y_1)$. (I think you don't really need the cubes in the last two fractions.)

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

for 4 points, the unique cubic is given by the LaGrange polynomial
$$y= \frac{y_0(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}+\frac{y_1(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}+\frac{y_2(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}+\frac{y_3(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}$$

3. Aug 24, 2007

### hotvette

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
4. Aug 24, 2007

### HallsofIvy

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.

5. Aug 24, 2007

### hotvette

Looking at it all now, it makes perfect logical sense. Thanks.

6. Aug 24, 2007

### hotvette

Actually, the same 4-pt form can be used for all situations (0 fixed points - 4 fixed points). See attachment.
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• ###### Least Squares2.pdf
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Last edited: Aug 25, 2007