Help me to show that exist only one interpolation polynomial with such properties

Question

Let $f:[a,b]\to \Bbb R$, $f \in C^2[a,b]$ and $x_0,x_1,x_2 \in [a,b]$ where $ x_0 \neq x_2$. Show that exists only one polynomial $p \in P_3$ for which $$p(x_0)=f(x_0),\ p\prime(x_1)=f\prime(x_1),\ p\prime\prime(x_1)=f\prime\prime(x_1),\ p(x_2)=f(x_2)$$ Find form of polynomial and error.

Answer 1

There are many ways to skin a cat.

I followed the first comment, which is equivalent as "show the following matrix is invertible"

$$ \begin{bmatrix} x_0^3 & x_0^2 & x_0 & 1 \\ x_2^3 & x_2^2 & x_2 & 1 \\ 3x_1^2 & 2x_1 & 1 & 0 \\ 6x_1 & 2 & 0 & 0 \end{bmatrix} $$

which leads to two obvious solutions. You can get the determinant, possibly using a computational tool, and immediately find that the determinant can factorize into $2(x_1-x_0)^3-2(x_1-x_2)^3$ which in turn factors into $2(x_2-x_0)(x_0^2+x_0x_2+x_2^2)$ which equals $0$ iff $x_2=x_0$.

Or you can try to convert the matrix using elementary operations into an upper triangular matrix. It's more obvious than how it writes out to be. First move the existing $0$s into lower left. Next see that if $x_1$ is zero, you get a trivial case. Next, you will rely on $x_2-x_0\neq0$ and $x_1(x_2-x_0)\neq0$ to get the upper triangular matrix with all non-zero diagonal terms, which demonstrates invertibility.

EDIT: I missed the question about the the form. Since the form is required, you should definitely follow the hint of the second comment.