User:EGM6341.S11.team5.cavalcanti/Mtg13

Mtg 13: Fri, 28 Jan 11

[[media: Nm1.s11.mtg13.djvu | Page 13-1]]

Computation $$\color{blue}{\left\lbrace {w}_{i,n};i=\mathrm{0,1,.}\mathrm{..},n\right\rbrace }$$cont'd p.12-3

$$\color{blue}{j=1}$$$$\mathrm{\colon }f={p}_{\color{blue}{j}}\in {P}_{\color{blue}{j}}$$

choose

$$\begin{array}{c}f\left(x\right)={p}_{\color{blue}{j}}\left(x\right)={x}^{\color{blue}{j}}\\ {e}_{\color{blue}{n}}\left({p}_{\color{blue}{j}};t\right)=\underbrace{f\left(t\right)}_{\color{blue}{t^j}}-{f}_{\color{blue}{n}}^{L}\left(t\right)=\color{blue}{{t}^{j}} - \color{black}{{f}_{n}^{L}\left(t\right)}\\ \Rightarrow {f}_{n}^{L}\left(t\right)={t}^{j}\Rightarrow \color{red}{{\int }_{a}^{b}} \color{black}{\left(\sum _{i=0}^{n}{l}_{i,n}\left(t\right){ {\color{red}{\left({x}_{i}\right)}}^{\color{blue}{j}}}\mathit{dt}\right)}\\ \Rightarrow \sum _{i=0}^{n}{W}_{i,n}{\left({x}_{i}\right)}^{j}= \\ \frac{\left({b}^{j+1}-{a}^{j+1}\right)}{\left(j+1\right)}\\ j=\mathrm{0,1,.}\mathrm{..},n\end{array}$$

[[media:Nm1.s11.mtg13.djvu | Page 13-2]] $$\begin{bmatrix} 1 & 1 & ... & 1 & ... & 1\\ x_{0} & x_{1} & ... & x_{j} & ... & x_{n}\\ \vdots & \vdots & & \vdots &  & \vdots\\ x^{j}_{0} & (x_{1})^{j} & ... & (x_{j})^{j} & ... & (x_{n})^{j}\\ \vdots & \vdots & & \vdots &  & \vdots\\ x_{0}^{n} & x_{1}^{n} & ... & (x_{j})^{n} & ... & (x_{n})^{n} \end{bmatrix} \begin{Bmatrix} W_{0,n}\\ W_{1,n}\\ \vdots\\ W_{j,n}\\ \vdots\\ W_{n,n} \end{Bmatrix} = \begin{Bmatrix} b-a\\ (b^{2}-a^2)/2\\ \vdots\\ (b^{j+1}-a^{j+1})/(j+1)\\ \vdots\\ (b^{n+1}-a^{n+1})/(n+1) \end{Bmatrix}$$ $$x_{0}<x_{1}<x_{2}<...<x_{n}$$ ____________________________________________________________________________________

Note: Comp. error by Taylor Series and Lagrange interpolation (5) p.3-3: $$e_{n}^{T}(f;x)=f(x)-\underbrace{f_{n}^{T}(x;\widehat{x})}_{\color{blue}{TS \ of \ order \ n \ about \ \widehat{x}}}=\frac{(x-\widehat{x})^{\color{blue}{n+1}}}{(n+1)!}f^{(n+1)}(\xi)$$ (2) p.11-3 $$ \Rightarrow$$$$e_{n}^{L}(f;x)=\frac{q_{n+1}(x)}{(n+1)!}f^{(n+1)}(\xi)$$ Comp. $$(x- \widehat{x})^{n+1}$$ to $$ q_{n+1}(x)$$ Lagrange interpolation better than Taylor Series over [a,b]. "End Note" $$\left | \left ( x- \widehat{x} \right )^{n+1} \right | \to \infty \, as \left |  \left ( x- \widehat{x} \right ) \right | \to \infty$$ $$q_{n+1}(x_{i})=0, i=0,1,...,n$$