Suppose we are given an odd number of data points $x_i$ and the corresponding values $f_i=f(x_i),i=1,...,n+1$($n$ is even), which are symmetric about the origin, i.e for each $x_i$ there is a $j$ such that $x_j=-x_i$ and for that pair $i,j,y_i=y_j$. For example, with $n=4$
x -2 -1 0 1 2
f(x) 10 8 3 8 10
Let $p_n(x)\in \Pi_n$ be the interpolating polynomial through these $n+1$ points. Define the polynomial $q_n\in\Pi_n$ as $q_n(x)=p_n(-x)$.
(a) Can you conclude that $q_n(x)=p_n(x)$, Why or why not?
(b) In the example above if we write $p_4(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$, can you conclude that $a_1=a_3=0$? Why or why not?
I can show that $q_n$ is also an interpolation polynomial through these $n+1$ points. Because $q_n(x_i)=p_n(-x_i)=f(-x_i)=f(x_i)$, so $q_n$ also passes through these $n+1$ points and $q_n$ is also an interpolation polynomial.
Thus, I intuitively think that $p_n(x), q_n(x)$ should be the same polynomial, since they are both interpolation polynomials and the nodes are symmetric about the y-axis.
However, I have trouble in rigorously prove or disprove it. Could someone kindly help? Thanks so much!