# University of Florida/Egm6321/f09.team1.gzc/Mtg12

Mtg 12: Wed, 26 Jan 11

Appl. of LIET p.11-2

${\displaystyle Let\ {f}_{n}^{L}(t)\ be\ Lagrange\ interpolation\ of\ order\ n\ to\ f(t).\ \Rightarrow \ {f}_{n}^{L}(.)\ \in \ \color {blue}{\underset {set\ of\ poly\ of\ degree\ \leqslant \ n}{\underbrace {\color {black}P_{n}} }}}$

${\displaystyle {\color {blue}n=1}\ {f}_{1}^{L}(.)\ interpolation\ exactly}$

${\displaystyle {\color {blue}P_{1}(x)\rightarrow }straight\ lines\ (linear\ functions)\ i.e.,\ e_{1}(P_{1},t)\ \equiv \ 0\ \forall \ t}$

${\displaystyle {\color {blue}n=2}\ {f}_{2}^{L}(.)\ interpolation\ exactly\ parabolas\ (polynomial\ of\ {\color {blue}{\underset {degree}{\underbrace {\color {black}order} }}})\ 2}$

${\displaystyle {\color {blue}n=3}\ cubic\ polynomial}$

${\displaystyle \color {blue}\vdots }$

${\displaystyle {\color {blue}n}\ {f}_{n}^{L}(.)\ interpolation\ exatcly\ \color {blue}{\overset {\in P_{n}}{\overbrace {\color {black}polynomial\ of\ \color {blue}{\underset {actually\ degree\ \leqslant \ n}{\underbrace {\color {black}degree\ n} }}} }}}$

${\displaystyle i.e.,\ e_{\color {blue}n}(p_{\color {blue}n};t)\ =0\ \forall t}$

LIET: (2)  p.11-3

${\displaystyle IF\ f\ =\ p\in \ P_{n}\ \Rightarrow \ f^{(n+1)}\ \equiv \ 0}$

${\displaystyle {\color {red}(2)}\ {\color {blue}\ p.11-3}\ \Rightarrow \ e_{\color {blue}n}(p,t)=0\ \forall t}$

${\displaystyle i.e.,\ {f}_{n}^{L}(.)\ interpolation\ {\color {red}exactly}\ p\in P_{n}}$

${\displaystyle \color {blue}Method\ to\ comp.\ \left\{w_{i,n};\ i=0,..,n\right\}\ given\ \left\{x_{i};\ i=0,..,n\right\}\ {\underline {not}}\ necessarily\ equidistant.}$

${\displaystyle {\color {blue}{\underline {j=0}}}\ consider\ f=p_{\color {blue}j}=p_{0}\in P_{0}\ {\color {red}constant}}$

${\displaystyle e_{\color {blue}n}\left({\underset {\color {blue}=1}{{\color {blue}\underbrace {\color {black}p_{\color {red}0}} };t}}\right)\ ={\color {blue}{\underset {1}{\underbrace {\color {black}f(t)} }}}-{f}_{{\color {blue}n}(t)}^{L}={\color {blue}1}-{f}_{n}^{L}(t)=0}$

${\displaystyle \Rightarrow {\color {blue}{\underset {\sum _{i=0}^{n}l_{i,n}(t){\color {red}{\underset {1}{\underbrace {\color {blue}f(x_{i})} }}}}{\underbrace {\color {black}{f}_{n}^{L}(t)} }}}=1\ \Rightarrow \ {\color {red}\int _{a}^{b}}{\color {blue}\left({\underset {{f}_{n}^{L}(t)}{\underbrace {\color {black}\sum _{i=0}^{n}l_{i,n}(t).{\color {red}1}} }}\right)}dt\ =\ {\color {blue}{\underset {b-a}{\underbrace {{\color {red}\int _{a}^{b}}{\color {black}1dt}} }}}}$

${\displaystyle \Rightarrow \ {\color {blue}{\underset {w_{i,n}}{\underbrace {\left({\color {red}\int _{a}^{b}}l_{i,n}(t)dt\right)} }}.{\color {red}1}}=b-a}$

${\displaystyle \Rightarrow }$

 ${\displaystyle \sum _{i=0}^{n}w_{i,n}=b-a}$

${\displaystyle {\color {blue}{\underline {j=1}}}\ let\ f=p_{\color {blue}j}=p_{1}\in P_{1}}$

${\displaystyle choose\ f(x)=p_{1}(x)=\ {\color {red}x}\in P_{1}}$

${\displaystyle e_{\color {blue}n}{\color {blue}\left({\underset {{\color {blue}=}{\color {red}x}}{\underbrace {\color {black}p_{\color {red}1}} }}{\color {black};t}\right)}\ =\ {\color {blue}{\underset {\color {red}t}{\underbrace {\color {black}f(t)} }}}-{f}_{\color {blue}n}^{L}(t)={\color {blue}t}-f_{n}(t)=0}$

${\displaystyle \Rightarrow {\underset {\color {blue}\sum _{i=0}^{n}l_{i,n}(t){\underset {\color {red}x_{i}}{\underbrace {\color {blue}f(x_{i})} }}}{\color {blue}\underbrace {\color {black}{f}_{n}^{L}(t)} }}={\color {red}t}\ \Rightarrow \ {\color {red}\int _{a}^{b}}{\color {blue}{\underset {{f}_{n}^{L}(t)}{\underbrace {\left({\color {black}\sum _{i=0}^{n}l_{i,n}(t).}{\color {red}x_{i}}\right)} dt}}}={\color {blue}{\underset {(b^{2}-a^{2})/2}{\underbrace {{\color {red}\int _{a}^{b}t}{\color {black}dt}} }}}}$

${\displaystyle \Rightarrow }$

 ${\displaystyle \sum _{i=0}^{n}w_{i,n}\ x_{i}\ =(b^{2}-a^{2})/2}$