Introductory case: Assume that we are given an integer $k\geq 1$, a real number $a$, and a polynomial $P\in\mathbb R[X]$ which satisfies $$P(a)=a,\quad P^{(i)}(a)=0 \text{ for every } 1\leq i<k.$$ The goal is to find another polynomial $Q\in\mathbb R[X]$ which satisfies the following:
- A stronger property than the one satisfied by $P$, namely $$Q(a)=a,\quad Q^{(i)}(a)=0 \text{ for every } 1\leq i<\ell,$$ where $\ell$ is some integer greater than $k$.
- For every $t\in\mathbb R$, the value of $Q(t)$ should be easily computable* given the value of $P(t)$.
This is for instance achievable by taking $Q:=(P-a)^2+a$ and $\ell:=2k$.
Question: Does this generalise to two values $a$ and $b$, for instance? More precisely, assume we are given an integer $k\geq 1$, two real numbers $a\neq b$, and a polynomial $P\in\mathbb R[X]$ which satisfies $$P(a)=a,\quad P(b)=b,\quad P^{(i)}(a)=0 \text{ for every } 1\leq i<k,\quad P^{(i)}(b)=0 \text{ for every } 1\leq i<k$$ The goal is to find another polynomial $Q\in\mathbb R[X]$ which satisfies the following:
- A stronger property than the one satisfied by $P$, namely $$Q(a)=a,\quad Q(b)=b,\quad Q^{(i)}(a)=0 \text{ for every } 1\leq i<\ell,\quad Q^{(i)}(b)=0 \text{ for every } 1\leq i<\ell,$$ where $\ell$ is some integer greater than $k$.
- For every $t\in\mathbb R$, the value of $Q(t)$ should be easily computable* given the value of $P(t)$.
It would then be natural to see if this also works for three values, or more generally, for $n$ values. But I believe that if we are able to understand the case with two values, we might be able to generalise.
*Easily computable means trying to minimise the number of required multiplications.