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

Mtg 20: Wed, 16 Feb 11

 Pf of SSET p.19-3 cont'd ${\displaystyle {\color {red}(1)}{\color {blue}p.19-3\ Rolle's\ thm}\ \Rightarrow \ \exists \ \varsigma _{1}\in \ ]\ 0,1\ [\ st\ G^{\color {blue}(1)(\varsigma _{1})}{\overset {\color {red}(1)}{=}}0}$ ${\displaystyle Now\ G^{(1)}(0)\ {\overset {\color {red}(2)}{=}}0\ {\color {red}why?}}$ ${\displaystyle {\color {red}(1)}\ {\color {blue}p.19-1:}\ G^{(1)}(t){\overset {\color {red}(3)}{=}}e^{(1)}(t)-5t^{4}e(1)}$ ${\displaystyle {\color {red}(5)}{\color {blue}p.18-3:}\ e(t)=A(t)-{A}_{2}^{L}(t)\ {\color {red}(4)}}$ ${\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e^{(1)}(t)=A^{(1)}-{A}_{2}^{L{\color {red}(1)}}(t)\ {\color {red}(5)}}$ ${\displaystyle A(t)=\int _{+t}^{-t}-=\int _{k}^{-t}-+\int _{t}^{k}-}$ ${\displaystyle A^{(1)}{\overset {\color {red}(1)}{=}}F(-t)+F(t)}$ ${\displaystyle k\in {\color {red}]}-t,t{\color {red}[}\ ({\color {red}kisconstant})}$ (4)p.20-1& (5)p.18-3: ${\displaystyle A_{2}^{L{\color {red}(1)}}(t){\overset {\color {red}(2)}{=}}{\frac {1}{3}}[F(-t)+4F(0)+F(t)]+{\frac {t}{3}}[F^{(1)}(-t)+F^{(1)}(t)}$ (5) p.20-1: ${\displaystyle e^{(1)}(0)=A^{(1)}(0)-A_{2}^{L(1)}(0)=2F(0)-[2F(0)+0]=0\ {\color {red}(3)}}$ ${\displaystyle {\color {red}(3)\ and\ (3)}{\color {blue}p.20-1}\Rightarrow {\color {red}(2)}{\color {blue}p.20-1}}$ Recall: ${\displaystyle G^{(1)}(\xi _{1})=0\ {\color {red}(1)}{\color {blue}p.20-1}}$ ${\displaystyle G^{(1)}(0)=0\ {\color {red}(2)}{\color {blue}p.20-1}}$ ${\displaystyle Rolle'stheorem\Rightarrow \ \forall \xi _{2}\in ]0,\xi _{1}[st\ \ G^{(2)}(\xi _{2})=0\ {\color {red}(1)}}$ ${\displaystyle Again,\ G^{(2)}(0){\overset {\color {red}(2)}{=}}0\ {\color {blue}HW^{*}4.3}}$ ${\displaystyle {\color {red}(1)\ and\ (2)}\ Rolle'stheorem\ \Rightarrow \ \forall \xi _{3}\in ]0,\xi _{2}[\ G^{(2)}(\xi _{2})=0\ st\ \ G^{(3)}(\xi _{3}){\overset {\color {red}(2)}{=}}0}$ ${\displaystyle {\color {red}(1)}{\color {blue}p.19-1:}\ G^{(3)}(t){\overset {\color {red}(4)}{e}}^{(3)}(t)-{\color {red}{\underset {(\xi )(4)(3)}{\underbrace {\color {black}t^{2}} }}}e(1)}$ ${\displaystyle e^{(3)}(t){\underset {\color {blue}HW^{*}4.4}{=}}-{\frac {t}{3}}[F^{(3)}(t)-F^{(3)}(-t)]\ {\color {red}(5)}}$ ${\displaystyle G^{(3)}(\xi _{3})=-{\frac {\xi _{3}}{3}}{\color {green}{\overset {Apply\ DMVT}{\overbrace {\color {black}[F^{(3)}(\xi _{3})-F^{(3)}(-\xi _{3})]} }}}-60(\xi _{3})^{2}e(1){\underset {{\color {blue}from}{\color {red}(3)}}{\overset {\color {red}(6)}{=}}}0}$ ${\displaystyle {\underset {\color {red}(f)}{\overset {\color {blue}DMVT}{=}}}-{\frac {\xi _{3}}{3}}[{\color {blue}{\underset {\xi _{3}-(-\xi _{3})}{\underbrace {\color {black}2\xi _{3}} }}}F^{(4)}(\xi _{4})]-60(\xi _{3})^{2}e(1)\ \xi _{4}\in \ ]-\xi _{3},\xi _{3}[}$