What is the formula to set up a generalized Vandermonde matrix? THere is not a lot of material online. I was wondering how to set up a generalized Vandermonde matrix to find an interpolating polynomial of degree 6 or less if we have:
$$p(x_0)=p_0$$
$$p(x_1)=p_1$$
$$p'(x_1)=p'_1$$
$$p(x_2)=p_2$$
$$p'(x_2)=p'_2$$
$$p''(x_2)=p''_2$$
$$p(x_3)=p_3$$
Do we have to include all points in the matrix? Or do we have to ommit some of them? What is the general formula for setting up the matrix when we know the derivatives of only some of the points?
 A: If the fitting polynomial is
$$
p(x) = a_{\,0}  + a_{\,1} x + a_{\,2} x^{\,2}  +  \cdots 
$$
where the unknown are the $a$'s coefficients, then
$$
p'(x) = a_{\,0} 0 + a_{\,1}  + 2a_{\,2} x +  \cdots 
$$
and you know how to proceed with the 2nd derivative
Then the matrix will be
$$
\left( {\matrix{
   1 & {x_{\,1} } & {x_{\,1} ^{\,2} } &  \cdots   \cr 
   0 & 1 & {2x_{\,1} } &  \cdots   \cr 
   0 & 0 & 2 &  \cdots   \cr 
    \vdots  &  \vdots  &  \vdots  &  \ddots   \cr 
 } } \right)\left( {\matrix{
   {a_{\,0} }  \cr 
   {a_{\,1} }  \cr 
   {a_{\,2} }  \cr 
    \vdots   \cr 
 } } \right) = \left( {\matrix{
   {p(x_{\,1} )}  \cr 
   {p'(x_{\,1} )}  \cr 
   {p''(x_{\,1} )}  \cr 
    \vdots   \cr 
 } } \right)
$$
and you put all the points & derivatives that you have.
Concerning the fundamental point about the invertibility of the matrix above, let's consider
that in the normal Vandermonde Matrix($(0..n)\times(0..n)$ the determinant is
$$
D = \prod\limits_{0\, \leqslant \,k\, < \,j\, \leqslant \,n} {\left( {x_{\,k}  - x_{\,j} } \right)} 
$$
so that it is non null if the $n+1$ points are distinct( $x_k \ne x_j$).
Suppose now to replace one the rows, e.g. that in $x_0$, with a combination 
of the row in $x_1$ and a point very near to it ($x_1+\Delta x_1$).
That is with
$$
\mathop {\lim }\limits_{\Delta x_{\,1}  \to 0} \frac{{\left( {x_{\,1}  + \Delta x_{\,1} } \right)^{\,m}  - x_{\,1} ^{\,m} }}
{{\Delta x_{\,1} }} = mx_{\,1} ^{\,m - 1} 
$$
The determinant of this new matrix will be
$$
D' = \left. {\frac{\partial }
{{\partial x_{\,0} }}D\,} \right|_{\,x_{\,0}  = x_{\,1} } 
$$
and we can easily see that it is non-null if the points are distinct.
Therefore the "derivative row" can also be taken together with "normal" Vandermonde rows
(including the one at the same point) to consitute a base.
And inductively we can proceed for the further derivatives.
