Say that you have $n+1$ points on the interval $[a,b]$, let's call them $\{x_0,\dots,x_n\}$. Take any two different $y_1, y_2$, points on $[a,b]$. My goal is to show that there exists a polynomial $p$ of any degree such that:
- $p(y_1) \neq p(y_2)$
- $p'(x_j) = 0 $ for all $j=0,\dots,n$.
This question, although fairly intuitive is not so simple to prove. My plan is to define $p$ by using its derivative $p'(x) = [\Pi_{j=0}^n (x-x_j)] \cdot [\Pi_{j=1}^m (x-z_j)]$ for some $z_j$ that I still need to think how to choose so that I alter the behavior of the polynomial. However, there is no simple step from these calculation to actually showing that $p(y_1) \neq p(y_2)$.
Is there a simpler way to show this?
We first define the polynomials $$ q_i(x)=\prod_{j\ne i}(x-x_j)^2, \quad i=1,\ldots,n. $$ Then the polynomial $$ p(x)=\sum_{i=1}^n\frac{q_i(x)}{q_i(x_i)} \left(w_i-\frac{q_i'(x_i)}{q_i(x_i)}w_i(x-x_i)\right), $$ satisfies: $$ p(x_i)=w_i\quad\text{and}\quad p'(x_i)=0, $$ for all $i=1,\ldots,n$.