Is the following described $p(x), q(x)$ the same interpolation polynomial? 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?
My attempt:
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!
 A: Your intuition is correct. In general the interpolating polynomial of $p$ is of at most degree $n$ is unique. 
Suppose that $p_1$ and $p_2$ are polynomials of degree at most $n$ which interpolate a function $f$ on the $n+1$ same $n+1$ distinct nodes. Then $d = p_1 - p_2$. This is a polynomial of degree at most $n$ which has at least $n+1$ zeros. By the fundamental theorem of algebra $d \equiv 0$ and $p_1 = p_2$. 
Now in your special case you have established that $p_n$ and $q_n$ of degree at most $n$ both interpolate $f$ at $n+1$ nodes. It follows that they are both identical and that $p_n$ is an even function, i.e. 
\begin{equation}
p_n(x) = p_n(-x).
\end{equation}
Even functions are said to have even parity, while odd functions are said to have odd parity. Differentiation switches the parity of a function. Ex. if $g$ is differentiable and odd, i.e.
\begin{equation}
g(-x) = - g(x)
\end{equation}
then the derivative is even. This is seen by considering the appropriate difference quotients. We have
\begin{equation}
\frac{g(x+h) - g(x)}{h} \rightarrow g'(x), \quad h \rightarrow 0, \quad h \not = 0,
\end{equation}
because $g$ is differentiable at $x$. Simultanously, we have
\begin{multline}
\frac{g(x+h) - g(x)}{h} = \frac{-g(-x-h) + g(-x)}{h} \\ = \frac{g(-x+(-h)) - g(-x)}{(-h)} \rightarrow g'(-x), \quad (-h) \rightarrow 0, \quad (-h) \not = 0,
\end{multline}
because $g$ is odd and differentiable at $-x$. It follows that 
\begin{equation}
g'(x) = g'(-x)
\end{equation}
or equivalently that $g'$ is even.
Now suppose the polynomial
\begin{equation}
p(x) = \sum_{k=0}^n a_k x^k
\end{equation}
is an even function. We claim that all the odd numbered coefficients, i.e. $a_{2j+1}$ are zero. Since $p$ is even, $p'$ is odd, hence $p'(0) = 0$. However, $p'$ is simply
\begin{equation}
p'(x) = \sum_{k=1}^n k a_k x^{k-1} = a_1 + 2 a_2 x + 3 a_3 x^2 + \dotsc nx^{n-1}.
\end{equation}
In particular, $p'(0) = a_1 = 0$. Since $p'$ is odd, $p''$ is again even and we are all set to continue an inductive process.
