University of Florida/Egm6321/f09.team1.gzc/Mtg22

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page22-1 Template:Font $h=\frac{(b-a)}{n}\leftarrow (3)p.7-4$ File:Mtg22.pg1.fig1.svg $E_n^{T(Trap.)}\underset{(1)p.7-4}{:=}I-I_n^T=\underset{a}{\overset{b}{\int }}f(x)dx-h[\frac{1}{2}f(x_0)+f(x_1)+\cdots +f(x_{n-1})\frac{1}{2}f(x_n)](1)$ $=\sum _{i=1}^n\left\{\underset{x_{i-1}}{\overset{x_i}{\int }}f(x)dx-\frac{h}{2}[f(x_{i-1})+f(x_i)]\right\}$ $\left|E_n^T\right|⩽\frac{h^3}{12}\underset{\underset{\stackrel{(3)}{=:\overline{M_2}}}{\underbrace{\xi \in ]x_{i-1},x_i[}}}{\sum _{i=1}^n(max\left|f^{(2)}(\xi )\right|)}(2)$ $\overline{M_2}⩽M_2(4)$ $\left|E_n^T\right|⩽\frac{(b-a{)}^3}{12n^3}n{M}_2=\frac{(b-a{)}^3}{12n^2}{M}_2=\frac{(b-a)h^2}{12}{M}_2$ $A.P.253$ $(5)$

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$\left|E_n^S\right|⩽\frac{(b-a{)}^5}{2880n^4}\underset{(4)p.17-1}{\underbrace{M_4}}=\frac{(b-a)h^{(4)}}{2880}{M}_4(1)$

Q(10^-6), and compare to number results.

(lin. regression).

 

File:Nm1.s11.Mtg22.pg2.fig1.svg$E=a{h}^k+Q(h^{k+1}),$ $logE=loga+klogh$

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$1)G(t):=e(t)-t^{4}e(1)(1)$

$2)G(t):=e(t)-t^{6}e(1)(2)$

Print out where pf breaks down.

see what happen.

 

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$(3)I=\underset{0}{\overset{\pi }{\int }}{e}^{x}sinxdx$(use Template:Fontto find I)