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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:
$$\displaystyle l_{0}:v\mapsto v(0)$$

$$\displaystyle \phi_{0} = - 5 x^{3} + 10 x^{2} - 6 x + 1$$

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

$$\displaystyle \phi_{1} = x \left(5 x^{2} - 5 x + 1\right)$$

This DOF is associated with vertex 1 of the reference element.
$$\displaystyle l_{2}:v\mapsto v(\tfrac{1}{2} - \tfrac{\sqrt{5}}{10})$$

$$\displaystyle \phi_{2} = \frac{5 x \left(2 \sqrt{5} x^{2} - 3 \sqrt{5} x - x + 1 + \sqrt{5}\right)}{2}$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{3}:v\mapsto v(\tfrac{\sqrt{5}}{10} + \tfrac{1}{2})$$

$$\displaystyle \phi_{3} = \frac{5 x \left(- 2 \sqrt{5} x^{2} - x + 3 \sqrt{5} x - \sqrt{5} + 1\right)}{2}$$

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