reemplazando los valores \( \xi_{1} = -1 \), \( \xi_{2} = 0 \) y \( \xi_{3} = 1 \) $$ \begin{align*} \alpha_{0} + \alpha_{1} (-1) + \alpha_{2} (-1)^{2} + \alpha_{3} (-1)^{3} + \alpha_{4} (-1)^{4} + \alpha_{5} (-1)^{5} &= v_{1} \\ \alpha_{1} + 2 \alpha_{2} (-1) + 3 \alpha_{3} (-1)^{2} + 4 \alpha_{4} (-1)^{3} + 5 \alpha_{5} (-1)^{4} &= \theta_{1} \\ \alpha_{0} + \alpha_{1} (0) + \alpha_{2} (0)^{2} + \alpha_{3} (0)^{3} + \alpha_{4} (0)^{4} + \alpha_{5} (0)^{5} &= v_{2} \\ \alpha_{1} + 2 \alpha_{2} (0) + 3 \alpha_{3} (0)^{2} + 4 \alpha_{4} (0)^{3} + 5 \alpha_{5} (0)^{4} &= \theta_{2} \\ \alpha_{0} + \alpha_{1} (1) + \alpha_{2} (1)^{2} + \alpha_{3} (1)^{3} + \alpha_{4} (1)^{4} + \alpha_{5} (1)^{5} &= v_{3} \\ \alpha_{1} + 2 \alpha_{2} (1) + 3 \alpha_{3} (1)^{2} + 4 \alpha_{4} (1)^{3} + 5 \alpha_{5} (1)^{4} &= \theta_{3} \end{align*} $$
simplificando $$ \begin{align*} \alpha_{0} - \alpha_{1} + \alpha_{2} - \alpha_{3} + \alpha_{4} - \alpha_{5} &= v_{1} \\ \alpha_{1} - 2 \alpha_{2} + 3 \alpha_{3} - 4 \alpha_{4} + 5 \alpha_{5} &= \theta_{1} \\ \alpha_{0} &= v_{2} \\ \alpha_{1} &= \theta_{2} \\ \alpha_{0} + \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} + \alpha_{5} &= v_{3} \\ \alpha_{1} + 2 \alpha_{2} + 3 \alpha_{3} + 4 \alpha_{4} + 5 \alpha_{5} &= \theta_{3} \end{align*} $$
en forma matricial $$ \begin{equation*} \begin{bmatrix} 1 & -1 & 1 & -1 & 1 & -1 \\ 0 & 1 & -2 & 3 & -4 & 5 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 & 5 \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \alpha_{4} \\ \alpha_{5} \end{bmatrix} = \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \end{equation*} $$
resolviendo $$ \begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \alpha_{4} \\ \alpha_{5} \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & \frac{1}{4} & -2 & 0 & 1 & -\frac{1}{4} \\ -\frac{5}{4} & -\frac{1}{4} & 0 & -2 & \frac{5}{4} & -\frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{4} & 1 & 0 & -\frac{1}{2} & \frac{1}{4} \\ \frac{3}{4} & \frac{1}{4} & 0 & 1 & -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \end{equation*} $$
reemplazando $$ \begin{align*} v &= \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} & \xi^{4} & \xi^{5} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \alpha_{4} \\ \alpha_{5} \end{bmatrix} \\ &= \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} & \xi^{4} & \xi^{5} \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & \frac{1}{4} & -2 & 0 & 1 & -\frac{1}{4} \\ -\frac{5}{4} & -\frac{1}{4} & 0 & -2 & \frac{5}{4} & -\frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{4} & 1 & 0 & -\frac{1}{2} & \frac{1}{4} \\ \frac{3}{4} & \frac{1}{4} & 0 & 1 & -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{4} \xi^{2} (3 \xi + 4) (\xi - 1)^{2} \\ \frac{1}{4} \xi^{2} (\xi + 1) (\xi - 1)^{2} \\ (\xi - 1)^{2} (\xi + 1)^{2} \\ \xi (\xi - 1)^{2} (\xi + 1)^{2} \\ -\frac{1}{4} \xi^{2} (3 \xi - 4) (\xi + 1)^{2} \\ \frac{1}{4} \xi^{2} (\xi - 1) (\xi + 1)^{2} \end{bmatrix}^{\mathrm{T}} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \\ &= \begin{bmatrix} N_{1} \\ N_{2} \\ N_{3} \\ N_{4} \\ N_{5} \\ N_{6} \end{bmatrix}^{\mathrm{T}} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \end{align*} $$