- #1

Cathr

- 67

- 2

##d/dX (P) = P'## and ##d^2/dX^2 (P) = P''##.

Calculating the matrix in class, the teacher found the following matrix, call it ##A##:

\begin{bmatrix}

0 & 1 & 0 & 0 \\

0 & 0 & 2 & 0 \\

0 & 0 & 0 & 3 \\

0 & 0 & 0 & 0

\end{bmatrix}

This is the matrix for the first derivative, and it has the right property that when it's multiplied with itself, it gives the matrix of the second derivative:

\begin{bmatrix}

0 & 0 & 2 & 0 \\

0 & 0 & 0 & 6 \\

0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0

\end{bmatrix}

However, when I try to calculate the derivative of a polynomial, say ##P=X^3+ X^2 + X + 1##, I don't find the right answer.

I found that the transpose ##C## of the matrix ##A## works, so that if we multiply ##P## with the following matrix:

\begin{bmatrix}

0 & 0 & 0 & 0 \\

1 & 0 & 0 & 0 \\

0 & 2 & 0 & 0 \\

0 & 0 & 3 & 0

\end{bmatrix} *

I find ##P= 3X^2 + 2X +1##, which is the derivative.

However, this doesn't work for all vectors and, multiplied by itself, it doesn't give the matrix of the second derivative, so it's still wrong.

May you please give me a hint on finding the solution?