reemplazando los valores \( \xi = -1 \), \( \xi = 0 \) y \( \xi = 1 \) $$ \begin{align*} \alpha_{0} + \alpha_{1}(-1) + \alpha_{2}(-1)^{2} &= u_{1} \\ \alpha_{0} + \alpha_{1}(0) + \alpha_{2}(0)^{2} &= u_{2} \\ \alpha_{0} + \alpha_{1}(1) + \alpha_{2}(1)^{2} &= u_{3} \end{align*} $$
simplificando $$ \begin{align*} \alpha_{0} - \alpha_{1} + \alpha_{1} &= u_{1} \\ \alpha_{0} &= u_{2}\\ \alpha_{0} + \alpha_{1} + \alpha_{1}&= u_{3} \end{align*} $$
en forma matricial $$ \begin{equation*} \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{equation*} $$
resolviendo $$ \begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{equation*} $$
reemplazando $$ \begin{equation*} u = \begin{bmatrix} 1 & \xi & \xi^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} 1 & \xi & \xi^{2} \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{equation*} $$
reescribiendo \( u \) $$ \begin{equation*} u = \bigg( -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) u_{1} + (1 - \xi^{2}) u_{2} + \bigg( \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) u_{3} = N_{1} u_{1} + N_{2} u_{2} + N_{3} u_{3} \end{equation*} $$