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