User:Egm6322.s12.team2.Xia/RP10.4

=R*10.4 -Weight Function =

Given
The following equation: $$\displaystyle w_{j}=\frac{-2}{(n+1)P_{n}'(x_{j})P_{n+1}(x_{j})}$$ (10.4.1)

where the Legendre polynomial is $$\displaystyle P_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)!$$. (10.4.2)

Find
Verify that $$\displaystyle w_{2}=1$$

Solution
According to the 10.4.2, we have $$\displaystyle {\displaystyle P_{2}(x)=\frac{1}{2}(3x^{2}-1)}$$ (10.4.3) $$\displaystyle P_{3}(x)=\frac{1}{2}(5x^{3}-3x)$$ thus $$\displaystyle P_{2}^{'}(x)=3x$$ and the roots of 10.4.3 are $$\displaystyle x_{1}=\frac{1}{\sqrt{3}},\quad x_{2}=-\frac{1}{\sqrt{3}} $$ Interms of 10.4.1 $$\displaystyle w_{2}=\frac{-2}{(2+1)P_{2}'(x_{2})P_{2+1}(x_{2})}$$ then $$\displaystyle w_{2}=\frac{-2}{(2+1)\cdot3\cdot\frac{-1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\left(5\cdot(\frac{-1}{\sqrt{3}})^{3}-3\cdot\frac{-1}{\sqrt{3}}\right)}$$ $$\displaystyle =\frac{-2}{(2+1)\cdot3\cdot\frac{-1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\left(\frac{4\sqrt{3}}{9}\right)}$$ $$\displaystyle =1$$