University of Florida/Egm6321/f09.team1.gzc/Mtg30

Mtg 30: Mon, 14 Mar 11

page30-1 ${\displaystyle {\color {blue}{\underline {Appl\ of\ HOTR\ p.29-2:}}}\ {\color {red}{\underline {{\color {blue}Corrected\ Trap.\ rule}s}}}}$ ${\displaystyle {\color {blue}CT_{\color {red}1}(n)}={\color {blue}{\underset {CT_{\color {red}0}(n)}{\underbrace {\color {black}T_{\color {red}0}(n)} }}}+{a}_{\color {blue}1}^{\color {red}0}h^{\color {blue}2}+Q(h^{\color {red}4})}$ ${\displaystyle I=CT_{\color {red}1}(n)+Q(h^{\color {red}(4)})}$ ${\displaystyle {\color {blue}CT_{\color {red}2}(n)}={\color {blue}CT_{\color {red}1}(n)}+{a}_{\color {blue}2}^{\color {red}0}h^{\color {blue}4}+Q(h^{\color {red}6})}$ ${\displaystyle I=CT_{\color {red}2}(n)+Q(h^{\color {red}6})}$ ${\displaystyle {\color {blue}CT_{\color {red}k}(n)}={\color {blue}CT_{\color {red}k-1}(n)}+{a}_{\color {blue}k}^{\color {red}0}h^{\color {blue}2k}+Q(h^{\color {red}2(k+1)})}$ HW*5.6:See HW*2.4 p.7-3, HW*5.4, p.29-6 Compare In using CTk (n), 8,16,... until error I-In=Q(10-6). HW*5.7:Discuss ppros and cons of intergration methods 1)Taylor series3)Compare Simpson5)CTk(n) 2)Compare Trap.4)Romberg(Richardson)   page30-2 Pf of HOTRE p.29-2: ${\displaystyle {E}_{n}^{T}=I-I_{n}=\int _{a}^{b}f(x)dx-T_{\color {red}0}(n)}$ ${\displaystyle =\sum _{k=0}^{n-1}[\int _{x_{k}}^{x_{k+1}}f(x)dx-{\frac {h}{2}}{\color {blue}\left\{{\color {black}f(x_{k})+f(x_{kH})}\right\}}]\ {\color {red}(1)}}$ ${\displaystyle x_{k}:=a+kh\ ,\ h:=(b-a)/n\ {\color {red}(2)}}$ ${\displaystyle Transfer\int _{x_{k}}^{x_{kH}}-{\color {blue}dx}\ to\ \int _{-1}^{+1}-{\color {blue}dt}}$ ${\displaystyle x(t):={\frac {x_{k}+x_{k+1}}{2}}+t{\frac {h}{2}},\ t\in [-1,+1]\ {\color {red}(3)}}$ ${\displaystyle {\color {blue}{\begin{cases}&\ {\color {black}x(-1)=x_{k}}\\&\ {\color {black}x(0)=(x_{k}+x_{k+1})/2}\ {\color {red}(4)}\\&\ {\color {black}x(+1)=x_{k+1}}\end{cases}}}}$ ${\displaystyle {E}_{n}^{T}={\frac {h}{2}}\sum _{k=0}^{n-1}[\int _{-1}^{+1}g_{k}(t)dt-{\color {blue}\left\{{\color {black}{g}_{k}^{(-1)}+{g}_{k}^{(+1)}}\right\}}]\ {\color {red}(5)}}$ ${\displaystyle {g}_{k}^{(t)}:=f(x(t))\ st\ x\in [x_{k},x_{k+1}]\ {\color {red}(6)}}$ page30-3 ${\displaystyle \int _{-1}^{+1}{\color {blue}{\underset {P_{1}(t)}{\underbrace {\color {black}(-t)} }}}g^{\color {red}(1)}(t)dt{\overset {\color {blue}HW^{*}5.{\color {red}8}}{=}}{\color {blue}{\underset {:=E}{\underbrace {\color {black}\int _{-1}^{+1}g(t)dt-[g(-1)+g(+1)]} }}}\ {\color {red}(1)}}$ Step2a: Some motivation HW*5.8: ${\displaystyle {\color {blue}{\underset {{\frac {d^{i}}{dt^{i}}}{g}_{k}^{(t)}}{\underbrace {\color {black}{g}_{k}^{\color {blue}(i)}(t)} }}}=({\frac {h}{2}})^{i}f^{\color {blue}(i)}(x(t))\ {\color {red}(2)}}$ ${\displaystyle x\in [x_{k},x_{k+1}]}$ ${\displaystyle To\ obtain\ {\color {blue}{\underline {\color {black}higher\ powers\ of\ h}}}\ {\color {blue}\Rightarrow }\ Take\ higher}$ ${\displaystyle derivation\ of\ g_{k}(t)\ {\color {blue}\Rightarrow }\ {\color {blue}{\underline {\color {black}successive\ int.\ by\ parts}}}}$ ${\displaystyle (recall\ pf\ of\ Taylor\ series\ {\color {blue}Mtg5\ and\ Mtg6})}$ ${\displaystyle {\color {red}(1)\Rightarrow }E=\int _{-1}^{+1}{\color {blue}{\underset {P_{1}(t)}{\underbrace {\color {black}(-t)} }}}{\color {black}g^{\color {blue}(1)}}dt}$ ${\displaystyle P_{2}(t){\overset {\color {red}(1)}{:=}}c_{1}{\frac {t^{2}}{2!}}+c_{3}\ ,\ c_{2}{\overset {\color {red}(1)}{=}}0}$ page30-4 Step2b:Intergration by parts ${\displaystyle E=[P_{\color {blue}2}(t)g^{(1)}(t)]_{-1}^{+1}-{\color {blue}{\overset {A{\overset {2b}{=}}}{\overbrace {\color {black}[P_{\color {red}3}g^{\color {blue}(2)}(t)]_{-1}^{+1}} }}}+\int _{-1}^{+1}P_{\color {red}3}g^{\color {red}(3)}(t)dt\ {\color {red}(3)}}$ ${\displaystyle P_{\color {red}3}{\overset {\color {red}(4)}{=}}\int P_{\color {blue}2}(t)dt=-{\frac {t^{3}}{6}}+\alpha t+\beta {\color {blue}(new\ intergration\ constant)}=c_{1}{\frac {t^{3}}{3!}}+C_{3}t+C_{4}\ (C_{2}{\overset {\color {red}(2)}{=}}0)}$ Pb: cancel terms withg(2) (evenorder derivation of g, i.e., odd power of h ;  here  h3) ${\displaystyle i.e.,\ make}$ ${\displaystyle {\color {blue}A_{2b}{\overset {\color {red}(5)}{=}}0}}$ ${\displaystyle \Rightarrow }$ ${\displaystyle want\ P_{\color {red}3}(t)\ {\color {blue}odd}\ functions}$ (recall P1 ) ${\displaystyle Consider\ P_{\color {red}3}\ st\ P_{\color {red}3}=0\ ,\ P_{\color {red}3}(+1\ and\ -1)=0\ {\color {red}(6)}}$ page30-5 ${\displaystyle P_{3}(0)=0\Rightarrow C_{4}=0=\beta \ {\color {red}(1)}}$ ${\displaystyle P_{3}(-1)=0=-{\frac {(-1)^{3}}{6}}+\alpha (-1)\Rightarrow C_{3}=\alpha ={\frac {1}{6}}\ {\color {red}(2)}}$ ${\displaystyle P_{3}(+1){\overset {\color {blue}P_{3}\ odd}{=}}-P_{3}(-1)=0\ {\color {red}(3)}}$ Summary, Steps2ab: ${\displaystyle P_{\color {blue}2}=C_{1}{\frac {t^{2}}{2!}}+C_{3}}$ ${\displaystyle P_{\color {red}3}=C_{1}{\frac {t^{3}}{3!}}+C_{3}t}$ ${\displaystyle C_{1}=-1,\ C_{3}={\frac {1}{6}}}$ ${\displaystyle {\color {red}(4)}}$ ${\displaystyle {\color {blue}{\underline {steps3ab:}}}\ (P_{\color {blue}4},P_{\color {red}5})}$ step3a: (3)&(5) p.30-4 ; intergration by parts ${\displaystyle E=[P_{\color {blue}2}(t)g^{(1)}(t)]_{-1}^{+1}+\int _{-1}^{+1}P_{\color {blue}3}(t)g^{\color {blue}(3)}(t)dt}$ ${\displaystyle =[P_{\color {blue}2}(t)g^{(1)}(t)]_{-1}^{+1}+[P_{\color {red}4}(t)g^{(3)}(t)]_{-1}^{+1}-\int _{-1}^{+1}P_{\color {red}4}(t)g^{\color {red}(4)}(t)dt\ {\color {red}(5)}}$ page30-6 ${\displaystyle P_{\color {red}4}(t)=\int P_{\color {blue}3}(t)dt=-{\frac {t^{4}}{4!}}+{\frac {t^{2}}{12}}+\alpha }$ ${\displaystyle =C_{1}{\frac {t^{4}}{4!}}+C_{3}{\frac {t^{2}}{2!}}+{\underset {\color {blue}=\alpha }{C_{5}}}\ {\color {red}(1)}}$ ${\displaystyle C_{1}\ and\ C_{3}\ in\ {\color {red}(4)}{\color {blue}p.30-5}}$ step3b:Intergration by parts ${\displaystyle {\color {red}(5)}{\color {blue}p.30-5}\Rightarrow \ E=[P_{\color {blue}2}g^{\color {blue}(1)}+P_{\color {blue}4}g^{\color {blue}(3)}]_{-1}^{+1}}$ ${\displaystyle [P_{\color {red}5}g^{\color {blue}(4)}]_{-1}^{+1}{\color {red}+}\int _{-1}^{+1}P_{\color {red}5}g^{\color {red}(5)}dt\ {\color {red}(2)}}$ ${\displaystyle P_{\color {red}5}(t)=\int P_{\color {blue}4}(t)dt={\frac {t^{5}}{120}}+{\frac {t^{3}}{36}}+\alpha {\color {black}t+}{\color {blue}{\underset {C_{6}}{\underbrace {\color {black}\beta } }}}\ {\color {red}(3)}}$ ${\displaystyle Want\ eliminate\ g^{\color {blue}(4)}\ (thus\ h^{\color {red}5});}$ ${\displaystyle So\ make}$ ${\displaystyle {\color {blue}A_{3b}{\overset {\color {red}(4)}{=}}0}}$ ${\displaystyle \Rightarrow }$ ${\displaystyle want\ P_{\color {red}5}\ {\color {blue}odd}\ functions}$ (recall P1, P3 oddfuncs) page30-7 Similar to (6)p.30-4: ${\displaystyle Consider\ P_{\color {red}5}(t)\ st\ P_{\color {red}5}(0)=0,\ P_{\color {red}5}(+1\ and\ -1)=0\ {\color {red}(1)}}$ ${\displaystyle \Rightarrow \ C_{6}=0\ ,\ C_{5}=-{\frac {7}{360}}\ {\color {blue}HW^{*}5.{\color {red}8}}\ {\color {red}(2)}}$ Summar, Steps3ab: ${\displaystyle P_{\color {red}4}(t)=C_{1}{\frac {t^{4}}{4!}}+C_{3}{\frac {t^{2}}{2!}}+C_{5}}$ ${\displaystyle P_{\color {red}5}(t)=C_{1}{\frac {t^{5}}{5!}}+C_{3}{\frac {t^{3}}{3!}}+C_{5}{\color {red}t}}$ ${\displaystyle C_{1}=-1\ ,\ C_{3}={\frac {1}{6}}\ ,\ C_{5}={\frac {-7}{360}}}$ ${\displaystyle {\color {red}(3)}}$ ${\displaystyle E=[P_{\color {blue}g^{\color {blue}(1)}}+P_{\color {blue}4}g^{\color {blue}(3)}]_{-1}^{+1}{\color {red}(+)}\int _{-1}^{+1}P_{\color {red}5}g^{\color {red}(5)}dt}$ ${\displaystyle {\color {red}(4)}}$ Compare to first part in(5)p.30-5 ${\displaystyle E=[P_{\color {blue}2}(t)g^{\color {blue}(1)}(t)]_{-1}^{+1}\ {\color {red}+}\ P_{\color {blue}3}(t)g^{\color {blue}(3)}(t)}$