Let $n \in \mathbb{N}$ be fix, $a>0$ and $f:[-a,a] \rightarrow \mathbb{R}$ be an even funciton. Consider the polynomial $p(x) = \sum\limits_{i =0}^n c_ix^i \in \mathbb{P}_n$ that interpolates $f$ at the interpolation nodes $x_j = -a +2a \frac{j}{n}$ for $j = 0,...,n$. Prove that $c_i = 0$ for all odd $i$.
My attempt:
We assume without loss of generality that $n$ is odd. Since $f(x_0 = -a) = f(x_n = a)$ it follows $p(x_0) = p_(x_n) \Leftrightarrow \sum\limits_{i=0}^n c_ix_0^i = \sum\limits_{i=0}^n c_ix_n^i$. Since $(x)^a = (-x)^a$ for all even $a$ we can simplify it to
$\sum\limits_{i=0}^{\frac{n-1}{2}} c_{2i+1}(x_0)^{2i+1} = \sum\limits_{i=0}^{\frac{n-1}{2}} c_{2i+1}(x_n)^{2i+1} \Leftrightarrow \sum\limits_{i=0}^{\frac{n-1}{2}} c_{2i+1}(x_0)^{2i}(x_0) = \sum\limits_{i=0}^{\frac{n-1}{2}} c_{2i+1}(x_n)^{2i}(x_n) \Leftrightarrow \sum\limits_{i = 0}^{\frac{n-1}{2}}c_{2i+1}(x_0)^{2i}(x_0-x_n) = 0$
Since $x_0 - x_n ≠ 0$ we get
$\sum\limits_{i = 0}^{\frac{n-1}{2}}c_{2i+1}(x_0)^{2i}=0.$
And now I'm stuck, any help would be much appreciated.
$p$ is the unique polynomial of degree less than $n$ and such that $p(x_j)=f(x_j)$ for all $j$. Proving that $c_i=0$ for all odd $i$ is equivalent to proving that $p$ is even. To this end, let $q(x)=p(-x)$, it is a polynomial of degree less than $n$ and $q(x_j)=p(-x_j)=p(x_{n-j})=f(x_{n-j})=f(-x_j)=f(x_j)$ because $f$ is even. Therefore $q=p$, i.e. $p(-x)=p(x)$ i.e. $p$ is even.