University of Florida/Egm6321/f09.team1.gzc/Mtg20

Mtg 20: Wed, 16 Feb 11


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Pf of SSET p.19-3 cont'd

$${\color{red}(1)}{\color{blue}p.19-3 \ Rolle's \ thm} \ \Rightarrow \ \exists \  \varsigma_{1} \in \  ] \ 0,1 \ [ \ st \ G^{\color{blue}(1)( \varsigma_{1})}  \overset{=}0$$

$$Now \ G^{(1)}(0) \ \overset{=}0 \ {\color{red}why ?}$$

$${\color{red}(1)} \ {\color{blue}p.19-1:} \ G^{(1)}(t)\overset{=}e^{(1)}(t)-5t^{4}e(1)$$

$${\color{red}(5)} {\color{blue}p.18-3:} \ e(t)=A(t)-{A}^{L}_{2}(t) \ {\color{red}(4)}$$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e^{(1)}(t)=A^{(1)}-{A}^{L{\color{red}(1)}}_{2}(t) \ {\color{red}(5)}$$

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$$A(t)= \int_{+t}^{-t} -=\int_{k}^{-t} -+\int_{t}^{k}-$$

$$A^{(1)} \overset{=}F(-t)+F(t)$$

$$k\in{\color{red}]}-t,t{\color{red}[} \ ({\color{red}k is constant})$$

(4)p.20-1&amp; (5)p.18-3:

$$A^{L{\color{red}(1)}}_{2}(t)\overset{=} \frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[F^{(1)}(-t)+F^{(1)}(t)$$

(5) p.20-1:

$$e^{(1)}(0)=A^{(1)}(0)-A^{L(1)}_{2}(0)=2F(0)-[2F(0)+0]=0 \ {\color{red}(3)}$$

$${\color{red}(3) \ and \ (3)}{\color{blue}p.20-1} \Rightarrow{\color{red}(2)}{\color{blue}p.20-1}$$

Recall:

$$G^{(1)}( \xi_{1})=0 \ {\color{red}(1)}{\color{blue}p.20-1}$$

$$G^{(1)}( 0)=0 \ {\color{red}(2)}{\color{blue}p.20-1}$$
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$$Rolle's theorem \Rightarrow \ \forall\xi_{2}\in]0,\xi_{1}[ st \ \ G^{(2)}(\xi_{2})=0 \ {\color{red}(1)}$$

$$Again, \ G^{(2)}(0)\overset{=}0 \ {\color{blue}HW^{*}4.3}$$

$${\color{red}(1) \ and \ (2)} \ Rolle's theorem \ \Rightarrow \ \forall\xi_{3}\in]0,\xi_{2}[ \ G^{(2)}(\xi_{2})=0 \ st \ \ G^{(3)}(\xi_{3})\overset{=}0$$

$${\color{red}(1)}{\color{blue}p.19-1:} \ G^{(3)}(t)\oversete^{(3)}(t)-{\color{red} \underset{(\xi)(4)(3)}{ \underbrace}}e(1)$$

$$e^{(3)}(t) \underset{=}-\frac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)] \ {\color{red}(5)}$$

$$G^{(3)}(\xi_{3})=-\frac{\xi_{3}}{3}{\color{green} \overset{Apply \ DMVT}{ \overbrace}}-60(\xi_{3})^{2}e(1) \underset{ \overset{=}}0 $$

$$ \underset{ \overset{=}}-\frac{\xi_{3}}{3}[{\color{blue} \underset{\xi_{3}-(-\xi_{3})}{ \underbrace{\color{black}2\xi_{3}}}}F^{(4)}(\xi_{4})]-60(\xi_{3})^{2}e(1) \ \xi_{4}\in \ ]-\xi_{3},\xi_{3}[$$