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