University of Florida/Egm6321/f09.team1.gzc/Mtg12
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University of Florida
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Egm6321
< f09.team1.gzc
Mtg 12: Wed, 26 Jan 11
page12-1
Appl. of LIET p.11-2
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{\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}} }}}
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{\displaystyle {\color {blue}n=1}\ {f}_{1}^{L}(.)\ interpolation\ exactly}
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{\displaystyle {\color {blue}P_{1}(x)\rightarrow }straight\ lines\ (linear\ functions)\ i.e.,\ e_{1}(P_{1},t)\ \equiv \ 0\ \forall \ t}
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{\displaystyle {\color {blue}n=2}\ {f}_{2}^{L}(.)\ interpolation\ exactly\ parabolas\ (polynomial\ of\ {\color {blue}{\underset {degree}{\underbrace {\color {black}order} }}})\ 2}
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{\displaystyle {\color {blue}n=3}\ cubic\ polynomial}
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{\displaystyle \color {blue}\vdots }
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{\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} }}} }}}
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{\displaystyle i.e.,\ e_{\color {blue}n}(p_{\color {blue}n};t)\ =0\ \forall t}
page12-2
LIET
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(2)
p.11-3
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{\displaystyle IF\ f\ =\ p\in \ P_{n}\ \Rightarrow \ f^{(n+1)}\ \equiv \ 0}
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{\displaystyle {\color {red}(2)}\ {\color {blue}\ p.11-3}\ \Rightarrow \ e_{\color {blue}n}(p,t)=0\ \forall t}
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{\displaystyle i.e.,\ {f}_{n}^{L}(.)\ interpolation\ {\color {red}exactly}\ p\in P_{n}}
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{\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.}
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{\displaystyle {\color {blue}{\underline {j=0}}}\ consider\ f=p_{\color {blue}j}=p_{0}\in P_{0}\ {\color {red}constant}}
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{\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}
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{\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}} }}}}
page12-3
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{\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}
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{\displaystyle \Rightarrow }
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{\displaystyle \sum _{i=0}^{n}w_{i,n}=b-a}
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{\displaystyle {\color {blue}{\underline {j=1}}}\ let\ f=p_{\color {blue}j}=p_{1}\in P_{1}}
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{\displaystyle choose\ f(x)=p_{1}(x)=\ {\color {red}x}\in P_{1}}
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{\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}
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{\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}} }}}}
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{\displaystyle \Rightarrow }
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{\displaystyle \sum _{i=0}^{n}w_{i,n}\ x_{i}\ =(b^{2}-a^{2})/2}
page12-2
LIET: (2) p.11-3
page12-3
