Interpolation of Symmetric Data For symmetric data 
$(x_i,y_i), i=-n,-n+1,..., n-1, n$ 
such that 
$x_{-i}=-x_i$ and $y_{-i}=-y_i, i=0,1,...n$ 
what is the required degree for an interpolating polynomial $p$? Since there are $2n+1$ data points (one for each $\pm n$ plus one for when $i=0$) would the  degree be at most $2n$? Or would only the points when $i\ge0$, be considered for a degree at most $n$?
Also, to show that the polynomial is odd (which is information given) so $-p(x)=p(-x)$, is it sufficient to show that since $p(x_{-i})=p(-x_i)$, then $p(-x_i)=-p(x_i)$?
 A: Well, that is frequently a confusing issue. Let me try and organize it in this way.
In general, $2n+1$ points will be interpolated by a polynomial of degree at most $2n$.  
When, like in this case,  the $2n+1$ points
$$
\left( {x_{\,i} ,y_{\,i} } \right)\quad \left| {\;\left\{ \matrix{
  i =  - n, \ldots ,0, \ldots ,n \hfill \cr 
  x_{\,i}  =  - x_{\,i}  \hfill \cr 
  y_{\,i}  =  - y_{\,i}  \hfill \cr}  \right.} \right.\quad  \to \quad \left( {x_{\,0} ,y_{\,0} } \right) = \left( {0,0} \right)
$$
constitute an anti-symmentric pattern, then they will be interpolated by a odd polynomial
$$
y(x)\quad \left| {\;y( - x) =  - y(x)} \right.\quad  \to \quad y(x) = \sum\limits_{1\, \le \,k\, \le \;n} {c_{\;2k - 1} x^{\,2k - 1} } 
$$
of degree (at most) $2n-1$ , thus having only $n$ undetermined coefficients, which you will set by imposing to pass through $n$ points.
These can be obviously the points with $i = 1, \ldots ,n$ (but not necessarily, they could also be chosen as $
i =  \pm k\quad \left| {\;k = 1, \ldots ,n} \right.$ ).
The $n+1$ even coefficient $c_{\;0} ,c_{\;2} ,\, \ldots ,c_{\;2n} $ are null due to the anti-simmetry.
If the points where instead symmetric then the polynomial would be of degree (<=) $2n$, with $n+1$ undetermined coefficients. The additional indetermination ($c_{\;0} $) being due to that in this case ${y_{\,0} }$ is not fixed.
In conclusion, you can always apply the general method of interpolation.
But when there is a symmetry, you can profitably reduce the interpolation and consider only the "independent" points.
A: Thanks to G Cab for their help!
Here's an alternative solution that I pieced together with help from my professor:

Let $p(x)$ be the interpolating polynomial of degree at most $2n$ for these data. Define a new polynomial
            \begin{equation}
      q(x) := p(x) + p(-x).
     \end{equation}
            Then, $q(x)$ is also a polynomial of degree at most $2n$. Observe that $p(x)$ is odd on all of our abscissas; i.e., for all $x_i$ with $i=-n,-n+1,...,n-1,n$, we have
            \begin{equation}
      p(x_i) = y_i = -y_{-i} = -p(x_{-i}) = -p(-x_i).
     \end{equation}
            That is, for all $x_i$, we have
            \begin{align}
      p(x_i) + p(-x_i) = 0.
     \end{align}
            Therefore, all of the abscissas are roots of $q(x)$. This means that $q(x)$ has $2n+1$ roots. But, $q(x)$ is a polynomial of degree at most $2n$. The only polynomial that has a greater number of roots than its exact degree is the zero polynomial. Therefore, it must be that $q(x) \equiv 0$. This implies that $p(x)$ is odd for all real values of $x$, since
            \begin{align}
      q(x) = 0 & \implies \\
      p(x) + p(-x) = 0 & \implies \\
      p(x) = -p(-x).
     \end{align}
            Finally, we note that since $p(x)$ is odd, it cannot be of exact degree $2n$, because $2n$ is an even number. Therefore, $p(x)$ must be of degree at most $2n-1$.
