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# Degree 3 Lagrange on a interval

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In this example:
• $$R$$ is the reference interval. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$1$$, $$x$$, $$x^{2}$$, $$x^{3}$$
• $$\mathcal{L}=\{l_0,...,l_{3}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:v\mapsto v(0)$$

$$\displaystyle \phi_{0} = - \frac{9 x^{3}}{2} + 9 x^{2} - \frac{11 x}{2} + 1$$

This DOF is associated with vertex 0 of the reference element.
$$\displaystyle l_{1}:v\mapsto v(1)$$

$$\displaystyle \phi_{1} = \frac{x \left(9 x^{2} - 9 x + 2\right)}{2}$$

This DOF is associated with vertex 1 of the reference element.
$$\displaystyle l_{2}:v\mapsto v(\tfrac{1}{3})$$

$$\displaystyle \phi_{2} = \frac{9 x \left(3 x^{2} - 5 x + 2\right)}{2}$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{3}:v\mapsto v(\tfrac{2}{3})$$

$$\displaystyle \phi_{3} = \frac{9 x \left(- 3 x^{2} + 4 x - 1\right)}{2}$$

This DOF is associated with edge 0 of the reference element.