User:Egm6322.s12.team2.Xia/RP10.8

=R 10.8 -Lagrange Interpolation =

Given
The following Lagrange interpolation: $$\displaystyle p(x)\thickapprox p^{h}(x):=\sum_{j=1}^{n}p(x_{j})\ell_{j}(x)\in\mathcal{P}_{n-1} $$ (10.8.1)

where Lanrage basis function is $$\displaystyle \ell_{j}(x):=\frac{\displaystyle \prod_{k=1,k\neq j}^{n}(x-x_{k})}{\displaystyle \prod_{k=1,k\neq j}^{n}(x_{j}-x_{k})}\in\mathcal{P}_{n-1}$$. (10.8.2) Define $$\displaystyle d(x):=p(x)-p^{h}(x)\in\mathcal{P}_{n-1}$$. $$\displaystyle $$

Find
1.Verify $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$. 2.Show that: $$\displaystyle d(x_{i})=0$$, for $$\displaystyle i=1,\ldots,n$$ and thus $$\displaystyle d(x)=0,\forall x\Rightarrow p(x)\equiv p^{h}(x)$$

Solution
1, From the equation 10.8.2, when $$\displaystyle i\neq j$$, we have $$\displaystyle {\displaystyle \ell_{j}(x_{i}):=\frac}=\displaystyle \prod_{k\neq j}\frac{x_{i}-x_{k}}{x_{j}-x_{k}}=\frac{(x_{i}-x_{0})}{(x_{j}-x_{0})}\cdots\frac{\overbrace{(x_{i}-x_{i})}^{=0}}{(x_{j}-x_{i})}\cdots\frac{(x_{i}-x_{k})}{(x_{j}-x_{k})}=0$$ an when $$\displaystyle i=j$$ $$\displaystyle \ell_{i}(x_{i})=\ell_{j}(x_{j}):=\prod_{k\neq j}\frac{x_{j}-x_{k}}{x_{j}-x_{k}}=1$$ Therefore $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$ 2, We know that $$\displaystyle d(x):=p(x)-p^{h}(x)$$ Therefore $$\displaystyle d(x_{i}):=p(x_{i})-p^{h}(x_{i})=p(x_{i})-\sum_{j=1}^{n}p(x_{j})\ell_{j}(x_{i})$$ From above solution, we know that $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$,thus $$\displaystyle d(x_{i})=p(x_{i})-\sum_{j=1}^{n}\underbrace{p(x_{j})\delta_{ji}}_{=0,\, if\, i\neq j}=p(x_{i})-p(x_{i})=0$$ and thus $$\displaystyle d(x)=0,\forall x\Rightarrow p(x)\equiv p^{h}(x)$$ $$\displaystyle $$ $$\displaystyle $$