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# Degree 4 Lagrange on a interval

◀ Back to Lagrange definition page 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: $$\displaystyle l_{0}:v\mapsto v(0)$$

$$\displaystyle \phi_{0} = 14 x^{4} - 35 x^{3} + 30 x^{2} - 10 x + 1$$

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

$$\displaystyle \phi_{1} = x \left(14 x^{3} - 21 x^{2} + 9 x - 1\right)$$

This DOF is associated with vertex 1 of the reference element. $$\displaystyle l_{2}:v\mapsto v(\tfrac{1}{2} - \tfrac{\sqrt{21}}{14})$$

$$\displaystyle \phi_{2} = \frac{7 x \left(- 28 x^{3} + 2 \sqrt{21} x^{2} + 56 x^{2} - 35 x - 3 \sqrt{21} x + \sqrt{21} + 7\right)}{6}$$

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

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

This DOF is associated with edge 0 of the reference element. $$\displaystyle l_{4}:v\mapsto v(\tfrac{\sqrt{21}}{14} + \tfrac{1}{2})$$

$$\displaystyle \phi_{4} = \frac{7 x \left(- 28 x^{3} - 2 \sqrt{21} x^{2} + 56 x^{2} - 35 x + 3 \sqrt{21} x - \sqrt{21} + 7\right)}{6}$$

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