Fit that curve!
True or false :
We can fit a cubic through any nine points in R 2 .
Clarification : A cubic is a curve in R 2 given by an equation of the form c 1 + c 2 x + c 3 y + c 4 x 2 + c 5 x y + c 6 y 2 + c 7 x 3 + c 8 x 2 y + c 9 x y 2 + c 1 0 y 3 = 0 , where at least one of the coefficients c k is nonzero.
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1 solution
Yes, exactly! (+1) Basic linear algebra does the trick.
Just to make sure that everybody fully understands this solution: Why can't we always fit a cubic through ten points?
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This is from linear algebra, where a "system of linear equations" can be described in matrix form
A x = b
where if A is a non-singular n × n matrix, x is a column vector with n unknowns and b is a non-zero column vector with n values, then there is an unique solution. So, the coefficient c 1 would make up the b column vector, making the matrix A a 9 × 9 one. Having an extra row would generally not allow a solution.
To illustrate with a simple example, a line will go through any two distinct arbitrary points, but not three.
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I would put it this way: If we plug the 9 points into the given formula of a cubic, we end up with 9 homogeneous linear equations in 10 variables, c 1 , . . . , c 1 0 . Since we have more variables than equations, there will be non-trivial solutions (with at least one of the c k being nonzero), thus giving us a cubic through the given 9 points.
However, if we would attempt the same thing with 10 points, we would end up with 10 equations with 10 unknowns. This system might only have the trivial solution, with all the c k being 0, in which case there would be no cubic through the given 10 points.
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@Otto Bretscher – n linear equations in the form
A x = b
where A is a n × n non-singular matrix, etc., can only have one non-trivial solution is easily shown as follows
A − 1 ⋅ A x = 1 ⋅ x = x = A − 1 b
so that given the column vector b , there can only be one x that the inverse transformation matrix A − 1 will map back to. Hence, the solution is unique, if non-trivial. This is just a general finding that goes beyond curve-fitting. "Non-singular matrix" means that there exists an inverse.
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@Michael Mendrin – Your remarks about the case of a non-singular square matrix don't help us much with our curve fitting problem, though. For one thing, when fitting a cubic to nine points, the matrix isn't square but 9 × 1 0 . When trying to fit a cubic to 10 points, the ensuing 1 0 × 1 0 matrix may well be singular.
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@Otto Bretscher – What happens when, given non-zero
x 1 , y 1 , x 2 , y 2 , x 3 , y 3
we try to solve for c 1 , c 2 , c 3 in this system of equations?
c
1
x
1
+
c
2
y
1
+
c
3
=
0
c
1
x
2
+
c
2
y
2
+
c
3
=
0
c
1
x
3
+
c
2
y
3
+
c
3
=
0
We end up with c 1 , c 2 , c 3 = 0 as one trivial solution, and therefore the only one.
This question is actually somewhat poorly defined, which is part of the problem. It takes up space to fully explore this issue. But, as I said, intuitively speaking, we can ask ourselves, "Given three arbitrary points on the plane, can we always find a straight line passing through all of them?" It's the same thing with cubics and ten arbitrary points.
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@Michael Mendrin – You are missing one key condition, "where at least one of the coefficients c k is nonzero." Thus we must reject the solution c 1 = c 2 = c 3 = 0 ; we are looking for non-zero solutions only.
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@Otto Bretscher – Okay, I did miss that one key condition.
Exactly the straight line passes through these points.
Given any 9 points, we can solve for 9 simultaneous linear equations, which will always have a solution or solutions. Non-uniqueness and degeneracy can frequently be expected, but nonetheless, even just some straight lines can be considered to be a cubic. For instance, a single straight line passes through points ( 0 , 0 ) , ( 1 , 0 ) , ( 2 , 0 ) , etc.