= Solution
Use coordinates in which the current spacing is $h$. On the first child subgrid, the polynomial candidate is
$$
Q_+(x)=\tfrac34P(x-h/4)+\tfrac14P(x+3h/4);
$$
on the second it is
$$
Q_-(x)=\tfrac14P(x-3h/4)+\tfrac34P(x+h/4).
$$
For arbitrary sufficiently long uniformly spaced data, all refined vertices can lie on one degree-$n$ polynomial only if these two candidate polynomials agree. They agree with $P$ when $n=0,1$. For $P(x)=ax^2+bx+c$, expansion gives
$$
Q_+(x)=Q_-(x)=P(x)+\frac{3ah^2}{16}.
$$
Thus quadratics preserve their degree, although not the same values.
For a polynomial of degree $n\geq3$, subtract the two finite Taylor expansions. The first-derivative term cancels, while the leading nonzero differential term is
$$
Q_+(x)-Q_-(x)=\frac{h^3}{32}P^{(3)}(x)+\text{terms involving higher odd derivatives}.
$$
If the leading coefficient of $P$ is $a_n\ne0$, this difference has nonzero coefficient $a_n n(n-1)(n-2)h^3/32$ at degree $n-3$; the higher derivatives have lower degree and cannot cancel it. Therefore the <polynomial degree preservation under Chaikin subdivision> holds exactly for
$$
\boxed{n=0,1,2.}
$$
This is a general refinement property, independent of accidental interpolation through a small finite number of vertices.
Back to article page
