Hello,
if i have polynomials
\(\displaystyle p(x) = c_0 + c_1x +c_2x^2 + c_3x^3\)
\(\displaystyle p'(x) = 0 + c_1 + 2c_2x + 3c_3x^2\)
\(\displaystyle P(x) = c_0x + 1/2c_1x^2 + 1/3c_2x^3 + 1/4c3x^4\)
p(x) is polynomial, p'(x) is derivation of p(x) and P(x) is integral of p(x) (c0 ,c1,c2,c3 are coefficients)
Now I have to find matrices \(\displaystyle A_{3} \in \mathbb{R} ^{3x4}, B_{2} \in \mathbb{R} ^{4x3}\) so that
\(\displaystyle c' = A_{3} \cdot c \) where c' is vector of coefficients of p'(x) and c is vector of coefficients of p(x) (in this case c is: (c0,c1,c2,c3)^T)
\(\displaystyle c^{*} = B_{2} \cdot c \) where c* is vector of coefficients of P(x) and c is vector of coefficients of p(x) (in this case c is: (c0,c1,c2,c3)^T)
I have to find matrices \(\displaystyle A_3 (derivation), B_2(integral)\)
My solution:
\(\displaystyle A_3 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}\) because
\(\displaystyle \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{pmatrix} \)= \(\displaystyle (c_1, 2 \cdot c_2, 3 \cdot c_3)^T\)
Now how to find \(\displaystyle B_2\)?
I have
\(\displaystyle B_2 = \begin{pmatrix} 0 & 0 & 0 \\ 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{pmatrix} \cdot \begin{pmatrix} c_{1} \\ c_{2} \\ c_{3} \end{pmatrix} = \)\(\displaystyle (0, 1/2c_1, 1/3c_2, 1/4c_3)^T\)
The product of A*B should be identity (from here https://demonstrations.wolfram.com/TheDerivativeAndTheIntegralAsInfiniteMatrices/#more) matrix but in my case it is not. So I assume B is not correct.
Can somebody help please, it should be easy but I dont get it..
if i have polynomials
\(\displaystyle p(x) = c_0 + c_1x +c_2x^2 + c_3x^3\)
\(\displaystyle p'(x) = 0 + c_1 + 2c_2x + 3c_3x^2\)
\(\displaystyle P(x) = c_0x + 1/2c_1x^2 + 1/3c_2x^3 + 1/4c3x^4\)
p(x) is polynomial, p'(x) is derivation of p(x) and P(x) is integral of p(x) (c0 ,c1,c2,c3 are coefficients)
Now I have to find matrices \(\displaystyle A_{3} \in \mathbb{R} ^{3x4}, B_{2} \in \mathbb{R} ^{4x3}\) so that
\(\displaystyle c' = A_{3} \cdot c \) where c' is vector of coefficients of p'(x) and c is vector of coefficients of p(x) (in this case c is: (c0,c1,c2,c3)^T)
\(\displaystyle c^{*} = B_{2} \cdot c \) where c* is vector of coefficients of P(x) and c is vector of coefficients of p(x) (in this case c is: (c0,c1,c2,c3)^T)
I have to find matrices \(\displaystyle A_3 (derivation), B_2(integral)\)
My solution:
\(\displaystyle A_3 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}\) because
\(\displaystyle \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{pmatrix} \)= \(\displaystyle (c_1, 2 \cdot c_2, 3 \cdot c_3)^T\)
Now how to find \(\displaystyle B_2\)?
I have
\(\displaystyle B_2 = \begin{pmatrix} 0 & 0 & 0 \\ 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{pmatrix} \cdot \begin{pmatrix} c_{1} \\ c_{2} \\ c_{3} \end{pmatrix} = \)\(\displaystyle (0, 1/2c_1, 1/3c_2, 1/4c_3)^T\)
The product of A*B should be identity (from here https://demonstrations.wolfram.com/TheDerivativeAndTheIntegralAsInfiniteMatrices/#more) matrix but in my case it is not. So I assume B is not correct.
Can somebody help please, it should be easy but I dont get it..