# 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 ....

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Although I believe I understand the problem and will try to write an answer, I ultimately find myself unable to figure out a reasonable way to edit the original question. In the future please consider using grammar, spelling, capitalization, and more readable math so that more than just a handful of people can understand you. – anon Aug 4 '11 at 15:02
some paragraphs are gone... – RenHoek Aug 4 '11 at 15:23

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.

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why are calculation here $p(x) = \begin{pmatrix}x^2 & x & 1\end{pmatrix}\begin{pmatrix}f\\g\\h\end{pmatrix}=\begin{pmatrix}x^2 & x & 1\end{pmatrix}\begin{pmatrix} 1 & 1 & 1\\ 4 & 2 & 1\\ 16 & 4 & 1\end{pmatrix}^{-1}\begin{pmatrix}4\\ 6\\ 1\end{pmatrix}$ the inverse matrix ??? – RenHoek Aug 4 '11 at 15:33
@RenHoek: If you have $Af=b$, in order to find $f$, you need to use the inverse with $f=A^{-1}b$. This holds true with matrices as it does with scalars. – anon Aug 4 '11 at 15:38
okay,.. it's and matrix "division" like $x^{-1} = \frac{1}{x}$ – RenHoek Aug 4 '11 at 15:40
@RenHoek: Yes. Do you know what a matrix inverse is? I suggest you familiarize yourself with basic linear algebra more if your class is going to use it this much consistently. – anon Aug 4 '11 at 15:43
i calculated a lot to test your result way... it complete the right way :) thanks a lot ! you made my day – RenHoek Aug 4 '11 at 16:19