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

Mtg 30: Mon, 14 Mar 11


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page30-1

$${\color{blue} \underline {Appl \ of \ HOTR \  p.29-2:}} \ {\color{red} \underline {{\color{blue}Corrected \ Trap. \ rule}s}}$$

$${\color{blue}CT_(n)}={\color{blue}\underset{CT_(n)}{\underbrace}}+ {a}^_h^+Q(h^)$$

$$I=CT_(n)+Q(h^)$$

$${\color{blue}CT_(n)}={\color{blue}CT_(n)}+{a}^_h^+Q(h^)$$

$$I=CT_(n)+Q(h^)$$

HW*5.6:See HW*2.4 p.7-3, HW*5.4, p.29-6</STRONG></P> <P><STRONG>Compare I<SUB>n</SUB> using CT<SUB>k</SUB><SUB> </SUB>(n),, for n=2,4,</STRONG></P> <P><STRONG>8,16,... until error I-I<SUB>n</SUB>=Q(10<SUP>-6</SUP>).</STRONG></P> <P><STRONG><U>HW</U><SUP>*</SUP><U>5.</U><U>7</U><U>:</U>Discuss ppros and cons of intergration methods</STRONG></P> <P><STRONG>1)Taylor series3)Compare Simpson5)CT<SUB>k</SUB>(n)</STRONG></P> <P><STRONG>2)Compare Trap.4)Romberg(Richardson)</STRONG></P> <P><STRONG></STRONG> </P>

page30-2

<U>Pf of HOTRE p.29-2:</U>

$${E}^{T}_{n}=I-I_{n}= \int_{a}^{b}f(x)dx-T_(n)$$

$$=\sum_{k=0}^{n-1}[\int_{x_{k}}^{x_{k+1}}f(x)dx-\frac{h}{2}{\color{blue} \left\{{\color{black}f(x_{k})+f(x_{kH})} \right\}}] \ {\color{red}(1)}$$

$$x_{k}:=a+kh \, \ h:=(b-a)/n \ {\color{red}(2)}$$

$$Transfer \int_{x_{k}}^{x_{kH}}-{\color{blue}dx} \ to \ \int_{-1}^{+1}-{\color{blue}dt} $$

$$x(t):=\frac{x_{k}+x_{k+1}}{2}+t\frac{h}{2}, \ t\in [-1,+1] \ {\color{red}(3)}$$

$$ {\color{blue}\begin{cases} & \ {\color{black}x(-1)=x_{k}} \\ & \ {\color{black}x(0)=(x_{k}+x_{k+1})/2} \ {\color{red}(4)} \\ & \ {\color{black}x(+1)=x_{k+1}} \end{cases}}$$

$${E}^{T}_{n} = \frac{h}{2} \sum_{k=0}^{n-1}[\int_{-1}^{+1}g_{k}(t)dt-{\color{blue} \left\{{\color{black}{g}^{(-1)}_{k}+ {g}^{(+1)}_{k}} \right\}}] \ {\color{red}(5)}$$

$${g}^{(t)}_{k}:=f(x(t)) \ st \ x\in[x_{k},x_{k+1}] \ {\color{red}(6)}$$

page30-3

$$ \int_{-1}^{+1} {\color{blue}\underset{P_{1}(t)}{ \underbrace}}g^(t)dt \overset{=}{\color{blue} \underset{:=E}{ \underbrace}} \ {\color{red}(1)}$$

<U>Step2a:</U> Some motivation

HW<SUP>*</SUP>5.8:

$${\color{blue} \underset{\frac{d^{i}}{dt^{i}}{g}^{(t)}_{k}}{ \underbrace}}=(\frac{h}{2})^{i}f^{\color{blue}(i)}(x(t)) \ {\color{red}(2)}$$

$$x\in[x_{k},x_{k+1}]$$

$$To \ obtain \ {\color{blue} \underline {\color{black}higher \ powers \ of \ h}} \ {\color{blue}\Rightarrow} \ Take \ higher$$

$$derivation \ of \ g_{k}(t) \ {\color{blue}\Rightarrow} \ {\color{blue} \underline {\color{black}successive \ int. \ by \ parts}}$$

$$(recall \ pf \ of \ Taylor \ series \ {\color{blue}Mtg5 \ and \ Mtg6})$$

$${\color{red}(1)\Rightarrow}E=\int_{-1}^{+1} {\color{blue} \underset{P_{1}(t)}{ \underbrace}}{\color{black}g^}dt$$

$$P_{2}(t) \overset{:=}c_{1}\frac{t^{2}}{2!}+c_{3} \, \ c_{2}\overset{=}0$$

page30-4

<U>Step2b:</U>Intergration by parts

$$E=[P_(t)g^{(1)}(t)]^{+1}_{-1}-{\color{blue}\overset{A \overset{2b}{=}}{ \overbrace}}+\int_{-1}^{+1}P_g^(t)dt \ {\color{red}(3)}$$

$$P_\overset{=} \int P_(t)dt=-\frac{t^{3}}{6}+ \alpha t +\beta {\color{blue}(new \ intergration \ constant)}=c_{1}\frac{t^{3}}{3!}+C_{3}t+C_{4} \ (C_{2}\overset{=}0)$$

Pb:Find(α, β)=(C<SUB>3</SUB>, C<SUB>4</SUB>) to<U> cancel terms with</U>g<SUP>(2)</SUP><SUP> </SUP>(<U>even</U>order derivation of g, i.e., oddpower of h ; here h<SUP>3</SUP>)

$$i.e., \ make $$

$$\Rightarrow$$

(recall P<SUB>1</SUB>)

$$Consider \ P_ \ st \ P_=0 \, \ P_(+1 \ and \ -1)=0 \ {\color{red}(6)}$$

page30-5


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$$P_{3}(0)=0\Rightarrow C_{4}=0=\beta \ {\color{red}(1)}$$

$$P_{3}(-1)=0=-\frac{(-1)^{3}}{6}+\alpha(-1)\Rightarrow C_{3}=\alpha=\frac{1}{6} \ {\color{red}(2)}$$

$$P_{3}(+1) \overset{=} -P_{3}(-1)=0 \ {\color{red}(3)}$$

<U>Summary, Steps2ab:</U>

<U>step3a:</U> (3)&amp;(5) p.30-4 ; intergration by parts

$$E=[P_(t)g^{(1)}(t)]^{+1}_{-1}+\int_{-1}^{+1}P_(t)g^(t)dt$$

$$=[P_(t)g^{(1)}(t)]^{+1}_{-1}+[P_(t)g^{(3)}(t)]^{+1}_{-1}-\int_{-1}^{+1}P_(t)g^(t)dt \ {\color{red}(5)}$$

page30-6

$$P_(t)=\int P_(t)dt=-\frac{t^{4}}{4!}+\frac{t^{2}}{12}+\alpha$$

$$=C_{1}\frac{t^{4}}{4!}+C_{3}\frac{t^{2}}{2!}+ \underset{C_{5}} \ {\color{red}(1)}$$

$$C_{1} \ and \ C_{3} \ in \ {\color{red}(4)}{\color{blue}p.30-5}$$

<U>step3b:</U>Intergration by parts

$${\color{red}(5)}{\color{blue}p.30-5}\Rightarrow \ E=[P_g^+P_g^]^{+1}_{-1}$$

$$[P_g^]^{+1}_{-1}{\color{red}+}\int_{-1}^{+1}P_g^dt \ {\color{red}(2)}$$

$$P_(t)=\int P_(t)dt=\frac{t^{5}}{120}+\frac{t^{3}}{36}+\alpha{\color{black}t+}{\color{blue} \underset{C_{6}}{ \underbrace}} \ {\color{red}(3)}$$

$$Want \ eliminate \ g^ \ (thus \ h^);$$

$$So \ make $$

$$\Rightarrow$$

(recall P<SUB>1</SUB>, P<SUB>3 </SUB>oddfuncs)

page30-7

Similar to (6)p.30-4:

$$Consider \ P_(t) \ st \ P_(0)=0, \ P_(+1 \ and \ -1)=0 \ {\color{red}(1)}$$

$$\Rightarrow \ C_{6}=0 \, \ C_{5}=-\frac{7}{360} \ {\color{blue}HW^{*}5.{\color{red}8}} \ {\color{red}(2)}$$

<U>Summar, Steps3ab:</U>

$${\color{red}(3)}$$

$${\color{red}(4)}$$

Compare to first part in(5)p.30-5

$$E=[P_(t)g^(t)]^{+1}_{-1} \ {\color{red}+} \ P_(t)g^(t)$$
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