Conference rusure::math

I was always taught that 2 points determines a line, 3 points determines a
parabola, 4 points determines a a cubic etc.

So, I don't get what you're saying either.

For example, suppose we want to determine the equation

	Ax^3 + Bx^2 + Cx + D = y

that passes through the points

	(1,1) and (2,7) and (3,3) and (4,0)  [I picked these arbitrarily]

We've got 4 unknowns A,B,C,D and we can make 4 equations from the 4 points,
and hence we have a regular simultaneous system of equations.

My understanding has always been that if we add, say, another point, now we'll
have 5 equations, so no longer is there guarantee of a solution.

/Eric

        If the x coordinates of the points are all distinct, then n+1
        points will determine a unique y = a(n) x^n + a(n-1) x^(n-1) +
        ... + a(1) x + a(0).
        
        Whether or not the x coordinates are not distinct, you can try
        to fit a curve f(x,y) = 0 through all of the points.  For example,
        a circle x^2 + y^2 - 2 = 0 can be fit through the four points
        (+/- 1, +/- 1).
        
        Perhaps it was talking about the degree of a polynomial in x
        and y together.
        
        Dan
        

>>        Perhaps it was talking about the degree of a polynomial in x
>>        and y together.

That seems likely, since the number of terms in such a polynomial of degree
n is 1+n(n+3)/2.

	n=0	a					 1 coeff
	n=1	a + bx + cy				 3 coeffs
	n=2	a + bx + cy + dx^2 + exy + fy^2		 6 coeffs
	n=3	above + gx^3 + hx^2y + ixy^2 + j y^3	10 coeffs

But if the equation is of the form F(x,y)=0, then you can arbitrarily
set one nonzero coefficient to 1, so there are really only about n(n+3)/2
unknowns.

>    
>    "a curve of the nth order is not always determined by 1/2  n(n+3)
>    points. so that 9 points may not uniquely determine a cubic while
>    10 would be too many."
>    

Here is one interpretation (I have not heard of this paradox before).

Assuming we a dealing with nth order curves in 2 dimensions that are
represented by the general form f(x,y)=0 where

f(x,y) = SIGMA [C(i)* x^j * y^k] = 0   (j+k not greater than n)
         
Taking the first three orders, for...

n=1,    Ax + By + C = 0   (P=2)

n=2,    Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0   (P=5)  

n=3,    Ax^3 + Bx^2y + Cxy^2 + Dy^3 + Ex^2 + Fy^2 + Gxy + Hx + Iy + J = 0
                                                                      (P=9)

The number of 0-order terms in each polynomial is 1
              1-order terms .................  is 2
              2-order terms ........           is 3
              3-order terms ........           is 4
              n-order terms in each polynomial is n+1,

Total number of terms is (n+1)(n+2)/2  = n(n+3)/2 + 1
              
One of these parameters is redundant because f(x,y) could be divided by a
non-zero co-efficient.  Thus P=n(n+3)/2 points would generate P simultaneous
linear equations in P unknowns to solve for all the co-efficients for a unqiue
curve.  I think this part is obvious to all.

To understand the paradox, I looked at the quadratic as a good example since
I can't count much over order 2.

5 points in the plane will uniquely determine a quadratic curve (taking into
consideration transformations like rotated parabolas, etc.) from the set of
all quadratic curves. So the paradox to analyze is

    "A 2-order curve is not always determined by 5 
    points. So that 5 points may not uniquely determine a quadratic while
    6 would be too many."

Perhaps the paradox occurs in degenerate cases.  Take f(x,y)=x^2-y^2=0
which plots as two intersecting lines y=x and y=-x.

The *4* points (1,1), (-1,1), (-1,-1), (1,-1) clearly do not uniquely determine
the "curve". (could be a circle) but adding point (0,0) the *5* points _do_
uniquely define the curve.  If the fifth point were (sqrt(2),0), a circle is
forced. A sixth point is easy to find so that no quadratic curve is possible.

The challenge to illustrate this paradox is to find 5 points that yield an
ambiguous quadratic.  !!  ???  Is it possible.  Or to find 5 points that
will not produce any quadratic curve.  Is that possible??


Allen

>
>5 points in the plane will uniquely determine a quadratic curve (taking into
>consideration transformations like rotated parabolas, etc.) from the set of
>all quadratic curves. So the paradox to analyze is
>
>    "A 2-order curve is not always determined by 5 
>    points. So that 5 points may not uniquely determine a quadratic while
>    6 would be too many."
>
...

>The challenge to illustrate this paradox is to find 5 points that yield an
>ambiguous quadratic.  !!  ???  Is it possible.  Or to find 5 points that
>will not produce any quadratic curve.  Is that possible??

Perhaps there is one more interpretation...that 5 points may not determine
a quadratic.

So take any 5 points on a straight line... NO quadratic (i.e. its order-1)
Take any other 6th point not on the line....TOO MANY, no quadratic.

Allen

    Well, I poked around some reference books over lunch looking for
    something which might be the Cramer Paradox, and didn't find anything.
    However, I did find that Cram�r, best known for his work on
    mathematical statistics, has his name associated with some results
    in algebras on lattices.  So maybe the problem is that we have been
    assuming that this is supposed to apply to Real (or Complex)
    polynomials.  Perhaps this is over some class of integral equations.

                                   Topher