# Understanding the coefficient matrix of Hermite Interpolation

I was not sure whether this is a computer science question or a math question, so I posted it here, hope that it is alright.

I am trying to learn the technique of Hermite interpolation. I do understand that it is a generalization of Newton polynomial interpolation, which also considers derivatives to ensure continuity of the interpolated curve at the provided points.

In the book Computer Animation — Algorithms and Techniques by Rick Parent, there is a definition of Hermite interpolation that looks like this:

Hermite interpolation generates a cubic polynomial from one point to another. In addition to specifying the beginning and ending points $$(P_i,P_{i+1})$$, the user needs to supply beginning and ending tangent vectors $$(P'_i,P'_{i+1})$$ as well.

Here, just the first derivatives are considered, and the aim is to find a polynomial of degree 3 by finding appropriate coefficients $$a_3,a_2,a_1,a_0$$ such that we have $$a_3x^3 + a_2x^2 + a_1x + a_0$$. These are found by solving a system of linear equations. So far, so good.

Now, in the book, the matrices look like this:

The general matrix form for a curve is $$P(u) = U^TMB$$. In the case of Hermite interpolation, $$U^T = \begin{bmatrix} u^3 & u^2 & u & 1 \end{bmatrix} \\ M = \begin{bmatrix} 2 & -2 & 1 & 1 \\ -3 & 3 & -2 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \\ B = \begin{bmatrix} P_i \\ P_{i+1} \\ P'_i \\ P'_{i+1} \end{bmatrix}$$

I do not understand the matrix $$M$$, I do not see where the values are coming from. Are these fixed, or can I somehow compute them?

• @Yuval Filmus : thank you for taking the time to edit my post and replacing the pictures I took with this nice looking format. It is better this way and I will do so myself for my future posts Jan 30 at 10:53

We are looking for a curve $$P(u) = a_3 u^3 + a_2 u^2 + a_1 u + a_0$$ which satisfies the following properties:

1. $$P(0) = P_i$$.
2. $$P(1) = P_{i+1}$$.
3. $$P'(0) = P'_i$$.
4. $$P'(1) = P'_{i+1}$$.

In terms of the coefficients $$a_0,a_1,a_2,a_3$$, these are:

1. $$a_0 = P_i$$.
2. $$a_3 + a_2 + a_1 + a_0 = P_{i+1}$$.
3. $$a_1 = P'_i$$.
4. $$3a_3 + 2a_2 + a_1 = P'_{i+1}$$.

In matrix form, this reads: $$\begin{pmatrix} P_i \\ P_{i+1} \\ P'_i \\ P'_{i+1} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 3 & 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} a_3 \\ a_2 \\ a_1 \\ a_0 \end{pmatrix}$$ Inverting the matrix, we get $$\begin{pmatrix} a_3 \\ a_2 \\ a_1 \\ a_0 \end{pmatrix} = \begin{pmatrix} 2 & -2 & 1 & 1 \\ -3 & 3 & -2 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} P_i \\ P_{i+1} \\ P'_i \\ P'_{i+1} \end{pmatrix}$$ Therefore the equation of the curve is $$P(u) = \begin{pmatrix} u^3 & u^2 & u & 1 \end{pmatrix} \begin{pmatrix} a_3 \\ a_2 \\ a_1 \\ a_0 \end{pmatrix} = \begin{pmatrix} u^3 & u^2 & u & 1 \end{pmatrix} \begin{pmatrix} 2 & -2 & 1 & 1 \\ -3 & 3 & -2 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} P_i \\ P_{i+1} \\ P'_i \\ P'_{i+1} \end{pmatrix}$$

• thank you very much. Your answer is very helpful. I see it now. Have a nice day! :) Jan 30 at 10:58