Mtg 12: Wed, 26 Jan 11
page12-1
Appl. of LIET p.11-2
L e t f n L ( t ) b e L a g r a n g e i n t e r p o l a t i o n o f o r d e r n t o f ( t ) . ⇒ f n L ( . ) ∈ P n ⏟ s e t o f p o l y o f d e g r e e ⩽ n {\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}} }}}
n = 1 f 1 L ( . ) i n t e r p o l a t i o n e x a c t l y {\displaystyle {\color {blue}n=1}\ {f}_{1}^{L}(.)\ interpolation\ exactly}
P 1 ( x ) → s t r a i g h t l i n e s ( l i n e a r f u n c t i o n s ) i . e . , e 1 ( P 1 , t ) ≡ 0 ∀ t {\displaystyle {\color {blue}P_{1}(x)\rightarrow }straight\ lines\ (linear\ functions)\ i.e.,\ e_{1}(P_{1},t)\ \equiv \ 0\ \forall \ t}
n = 2 f 2 L ( . ) i n t e r p o l a t i o n e x a c t l y p a r a b o l a s ( p o l y n o m i a l o f o r d e r ⏟ d e g r e e ) 2 {\displaystyle {\color {blue}n=2}\ {f}_{2}^{L}(.)\ interpolation\ exactly\ parabolas\ (polynomial\ of\ {\color {blue}{\underset {degree}{\underbrace {\color {black}order} }}})\ 2}
n = 3 c u b i c p o l y n o m i a l {\displaystyle {\color {blue}n=3}\ cubic\ polynomial}
⋮ {\displaystyle \color {blue}\vdots }
n f n L ( . ) i n t e r p o l a t i o n e x a t c l y p o l y n o m i a l o f d e g r e e n ⏟ a c t u a l l y d e g r e e ⩽ n ⏞ ∈ P n {\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} }}} }}}
i . e . , e n ( p n ; t ) = 0 ∀ t {\displaystyle i.e.,\ e_{\color {blue}n}(p_{\color {blue}n};t)\ =0\ \forall t}
page12-2
LIET: (2) p.11-3
I F f = p ∈ P n ⇒ f ( n + 1 ) ≡ 0 {\displaystyle IF\ f\ =\ p\in \ P_{n}\ \Rightarrow \ f^{(n+1)}\ \equiv \ 0}
( 2 ) p .11 − 3 ⇒ e n ( p , t ) = 0 ∀ t {\displaystyle {\color {red}(2)}\ {\color {blue}\ p.11-3}\ \Rightarrow \ e_{\color {blue}n}(p,t)=0\ \forall t}
i . e . , f n L ( . ) i n t e r p o l a t i o n e x a c t l y p ∈ P n {\displaystyle i.e.,\ {f}_{n}^{L}(.)\ interpolation\ {\color {red}exactly}\ p\in P_{n}}
M e t h o d t o c o m p . { w i , n ; i = 0 , . . , n } g i v e n { x i ; i = 0 , . . , n } n o t _ n e c e s s a r i l y e q u i d i s t a n t . {\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.}
j = 0 _ c o n s i d e r f = p j = p 0 ∈ P 0 c o n s t a n t {\displaystyle {\color {blue}{\underline {j=0}}}\ consider\ f=p_{\color {blue}j}=p_{0}\in P_{0}\ {\color {red}constant}}
e n ( p 0 ⏟ ; t = 1 ) = f ( t ) ⏟ 1 − f n ( t ) L = 1 − f n L ( t ) = 0 {\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}
⇒ f n L ( t ) ⏟ ∑ i = 0 n l i , n ( t ) f ( x i ) ⏟ 1 = 1 ⇒ ∫ a b ( ∑ i = 0 n l i , n ( t ) . 1 ⏟ f n L ( t ) ) d t = ∫ a b 1 d t ⏟ b − a {\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
⇒ ( ∫ a b l i , n ( t ) d t ) ⏟ w i , n . 1 = b − a {\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 }
j = 1 _ l e t f = p j = p 1 ∈ P 1 {\displaystyle {\color {blue}{\underline {j=1}}}\ let\ f=p_{\color {blue}j}=p_{1}\in P_{1}}
c h o o s e f ( x ) = p 1 ( x ) = x ∈ P 1 {\displaystyle choose\ f(x)=p_{1}(x)=\ {\color {red}x}\in P_{1}}
e n ( p 1 ⏟ = x ; t ) = f ( t ) ⏟ t − f n L ( t ) = t − f n ( t ) = 0 {\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}
⇒ f n L ( t ) ⏟ ∑ i = 0 n l i , n ( t ) f ( x i ) ⏟ x i = t ⇒ ∫ a b ( ∑ i = 0 n l i , n ( t ) . x i ) ⏟ d t f n L ( t ) = ∫ a b t d t ⏟ ( b 2 − a 2 ) / 2 {\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}} }}}}
