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# Degree 4 Gauss–Lobatto–Legendre 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: $$1$$, $$x$$, $$x^{2}$$, $$x^{3}$$, $$x^{4}$$
• $$\mathcal{L}=\{l_0,...,l_{4}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:v\mapsto v(0)$$

$$\displaystyle \phi_{0} = \frac{32 x^{4}}{3} - \frac{80 x^{3}}{3} + \frac{70 x^{2}}{3} - \frac{25 x}{3} + 1$$

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

$$\displaystyle \phi_{1} = \frac{x \left(32 x^{3} - 48 x^{2} + 22 x - 3\right)}{3}$$

This DOF is associated with vertex 1 of the reference element.
$$\displaystyle l_{2}:v\mapsto v(\tfrac{1}{4})$$

$$\displaystyle \phi_{2} = \frac{16 x \left(- 8 x^{3} + 18 x^{2} - 13 x + 3\right)}{3}$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{3}:v\mapsto v(\tfrac{1}{2})$$

$$\displaystyle \phi_{3} = 4 x \left(16 x^{3} - 32 x^{2} + 19 x - 3\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{4}:v\mapsto v(\tfrac{3}{4})$$

$$\displaystyle \phi_{4} = \frac{16 x \left(- 8 x^{3} + 14 x^{2} - 7 x + 1\right)}{3}$$

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