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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?

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    $\begingroup$ @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 $\endgroup$ – sschimper Jan 30 at 10:53
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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 $$ P(u) = \begin{pmatrix} u^3 & u^2 & u & 1 \end{pmatrix} \begin{pmatrix} a_3 \\ a_2 \\ a_1 \\ a_0 \end{pmatrix} = \begin{pmatrix} u^3 & u^2 & u & 1 \end{pmatrix} \begin{pmatrix} 2 & -2 & 1 & 1 \\ -3 & 3 & -2 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} P_i \\ P_{i+1} \\ P'_i \\ P'_{i+1} \end{pmatrix} $$

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

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