iconEuler Examples

Positive Base for Quadratic Polynomials

by R. Grothmann

There is the problem in polynomial theory, if there is a positive
base for a given space.

We discuss the following case on the interval [-1,1]: Is it possible
to find finitely many non-negative polynomials of degree 2 such
that each non-negative polynomial can be expressed by a non-negative
linear combination of these polynomials?

We need only be able express poynomials with integral equal to
1 over [-1,1]. We try to plot this set. So we define a function
to compute the minimum on [-1,1] for polynomials with integral
1.
>:: p := a*x^2+b*x+c;
>:: solve(integrate(p,x,-1,1)=1,c), pab := at(p,%)
                                       2 a - 3
                                [c = - -------]
                                          6


                                2         2 a - 3
                             a x  + b x - -------
                                             6

To get the integral equal to 1, c depends on a and b. The minimum
of course, can easily be computed.
>:: solve(diff(pab,x)=0,x)
                                          b
                                  [x = - ---]
                                         2 a

Now we can write the Euler function computing the minimum of p
in [-1,1], if p has integral 1. We have to compare the boundary
values at -1 and 1 with the value at the minimum.
>function map pmin (a,b) ...
  c:=(3-2a)/6;
  s:=min(a+b+c,a-b+c);
  if a~=0 then return s; endif;
  x:=-b/(2a);
  if -1<=x and x<=1 then
    return min(s,a*x^2+b*x+c);
  else
    return s;
  endif;
endfunction
Here is an example. The polynomial x^2+x+1/6 has integral 1 and
minimum -0.83...
>pmin(1,1), plot2d("x^2+x+1/6",-1,1);
  -0.08333333333333 
>gauss("x^2+x+1/6",-1,1)
                  1 
Now, we are able to plot the region of a and b, where the polynomial
ax^2+bx+c with integral 1 is positive. This set represents the
quadratic positive polynomials with integral 1.
>plot2d("pmin",r=2,niveau=0,hue=1); insimg;

Positive Basis for Quadratic Polynomials

Looking at that region we see that it is not the convex hull of
finitely many points. But, if there would exist a positive base
of polynomials, this would be the case.

Imagine the set of all non-negative linear combinations of our
base. This would be a finitely generated cone. Intersected with
the polynomials of degree 2, it is still a finitely generated
cone. The intersection of this cone with the polynomials with
integral 1 would be the convex hull of finitely many points.

We plot the 3D graph of pmin(a,b) for comparison.
>plot3d("pmin",r=2,user=1,scale=1.3,xlabel="a",ylabel="b"); insimg;

Positive Basis for Quadratic Polynomials

Here is a cut with b=0.
>plot2d("pmin";0,r=2);
There is a well known non-negative base for the quadratic polynimals,
the Bernstein polynomials.

We like to see, which quadratic polynomials can be expressed as
a non-negative linear combination of the Bernstein polynomials.


To answer this, we can simply normalize the Bernstein polynomials
to integral 1, and insert the coefficients into our figure.
>:: function c(p) := block ( q:expand(p/integrate(p,x,-1,1)), ...
>:: [coeff(q,x,2),coeff(q,x,1)] );
>p1:=mxmeval("c((x+1)^2)")
              0.375                0.75 
>p2:=mxmeval("c(x^2-1)")
              -0.75                   0 
>p3:=mxmeval("c((x-1)^2)")
              0.375               -0.75 
>plot2d("pmin",r=2,niveau=0,hue=1); ...
>plot2d([p1[1],p2[1],p3[1]],[p1[2],p2[2],p3[2]],add=1,points=1);
The answer to our problem is the convex hull of these three points.
>plot2d([p1[1],p2[1],p3[1],p1[1]],[p1[2],p2[2],p3[2],p1[2]],add=1); insimg;

Positive Basis for Quadratic Polynomials

The function on the right edge (maximal a) is the x^2*3/2, and
can obviously not be expressed by the Bernstein polynomials in
any non-negative way, since it vanishes in 0.

The right part of the above plot looks like an ellipse. Let us
study this. A little bit of contemplation shows, that this are
belongs to polynomials of the type a*(x-d)^2, where -1<=d<=1,
and a is such that the integral is 1.

We use the coordinate function above, mapping polynomials to their
highest and second highest coefficients [a,b].
>:: s := c((x-d)^2), define(vx(d),s[1]); define(vy(d),s[2]);
                                1          2 d
                            [--------, - --------]
                                2   2       2   2
                             2 d  + -    2 d  + -
                                    3           3

Indeed!
>d:=-1:0.01:1; x:=mxmeval("vx(d)"); y:=mxmeval("vy(d)"); ...
>  plot2d(x,y,r=2); insimg;

Positive Basis for Quadratic Polynomials

Let us check, if this is an ellipse. First, we compute the center
as the midpoint between v(0)=[3/2,0] and v(inf)=[0,0].
>:: 1/2*(vx(0)+limit(vy(d),d,inf))
                                       3
                                       -
                                       4

So the small semiaxis is 3/4.

Then we can compute the large semiaxis.
>:: solve(vx(d)=3/4,d); at(vy(d),%[1]) | ratsimp
                                       3
                                   ---------
                                   2 sqrt(3)

Now we can check the equation, and find that the curve is indeed
an ellipse with the general equation

(x-x0)^2/a^2+(y-y0)^2/b^2=1.
>:: (vx(d)-3/4)^2/(3/4)^2+vy(d)^2/(sqrt(3)/2)^2 | ratsimp
                                       1

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