Suppose . Then we can find and s.t. is exact for . Pick to be roots of where is the th Orthogonal polynomials for this weight function, and . First, by picking we find because has degree , so all are positive.

All roots of are distinct and in .

Proof

Suppose has roots . Define . If then is orthogonal to , so . But is always positive because they change signs at exactly the same points so this is impossible.

Now for any polynomial we can write it as where and have degrees at most . Hence

but is exact by choice of .