Vandermonde and curve interpolation I hesitate here because of an understanding with a calculation problems.
I want to calculate an interpolation using the Vandermonde matrix.
see: http://en.wikipedia.org/wiki/Vandermonde_matrix
My polynomial has degree 2. I have the following coordinates p1 (1,4), p2 (2,6), p3 (4,1).
I want a formula exhibition that interpolates these points. An interpolation point is difend as $p(u,g_i)$
The general matrix form is like this: T*M*vector_g
$q(u)=
\begin{matrix} 
(u ^ {2} & u & 1)

\end{matrix}
 
*
\begin{matrix} 
a_1 & a_2 &a_3 
\\b_2 & b_2 &b_2 
\\c_1 & c_1 &c_2 
\end{matrix}
*
\begin{matrix} 
g_0\\g_1\\g_2
\end{matrix} $
Explenation:
Monomvector T = ($u^{2}$  u  1 )
geometry vector vector_g = $\begin{matrix} 
g_0\\g_1\\g_2
\end{matrix} $
Vanemonde Matrix M = $\begin{matrix} 
a_1 & a_2 &a_3 
\\b_2 & b_2 &b_2 
\\c_1 & c_1 &c_2 
\end{matrix} $
Now my understanding problem:
When i use the given parameter p1..p2 i have a linear equation like
p1 (1,4) => a + b + c = 4
p2 (2,6) => 4a + 2b + c = 6
p3 (4,1) => 16a + 4b + c = 4
and the result are the coefficients for a,b,c are $\frac{-3}{2} u^2 $\frac{13}{2}u -1$
But my Professor wants the general matrix form like above ....
all i have is following :
q(u) = ($u^{2}$  u  1) * $(\frac{-3}{2} u^2 $\frac{13}{2}u -1)$ = g0
But when i calculate 
p1 (1,4) => a + b + c = $g_0$
p2 (2,6) => 4a + 2b + c = $g_1$
p3 (4,1) => 16a + 4b + c = $g_2$
i get a wrong matrix but in the right form ....
if you need more information just tell
 A: The purpose of polynomial interpolation is to find a polynomial $p(x)$ whose graph touches a given set of points and therefore gives a reasonable curve to approximate data. In this case, we want a polynomial with the following conditions:
$$p(1)=4 , \quad p(2)=6, \quad p(4)=1.$$
This comes from the fact that if $(x,y)$ is a point on the graph, $p(x)=y$ is a condition on the polynomial. The number of points is $3$, so the degree of our polynomial is at most $2$ (quadratic). If we let $p(x)=fx^2+gx+h$, then (plugging this into the three conditions) this gives the system
$$f+g+h=4$$
$$4f+2g+h=6$$
$$ 16f+4g+h=1.$$
Putting this in matrix form:
$$\begin{pmatrix} 1 & 1 & 1\\ 4 & 2 & 1\\ 16 & 4 & 1\end{pmatrix} \begin{pmatrix}f\\ g\\ h\end{pmatrix} = \begin{pmatrix}4\\ 6\\ 1\end{pmatrix}.$$
Hence
$$ p(x) = \begin{pmatrix}x^2 & x & 1\end{pmatrix}\begin{pmatrix}f\\g\\h\end{pmatrix}=\begin{pmatrix}x^2 & x & 1\end{pmatrix}\begin{pmatrix} 1 & 1 & 1\\ 4 & 2 & 1\\ 16 & 4 & 1\end{pmatrix}^{-1}\begin{pmatrix}4\\ 6\\ 1\end{pmatrix}$$
$$=\begin{pmatrix}x^2 & x & 1\end{pmatrix}\begin{pmatrix} 1/3 & -1/2 & 1/6\\ -2 & 5/2 & -1/2\\ 8/3 & -2 & 1/3\end{pmatrix}\begin{pmatrix}4\\ 6\\ 1\end{pmatrix}= \begin{pmatrix}x^2 & x & 1\end{pmatrix}\begin{pmatrix}-3/2\\13/2\\-1\end{pmatrix}$$
$$ = -\frac{3}{2}x^2+\frac{13}{2}x-1.$$
You can check that this is the correct polynomial. If you have any questions about the matrix manipulations, such as how I calculated the inverse, I'll see what I can do.
