Mtg 20: Wed, 16 Feb 11
page20-1
Pf of SSET p.19-3 cont'd
(1)p.19−3 Rolle′s thm ⇒ ∃ ς1∈ ] 0,1 [ st G(1)(ς1)=(1)0{\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}
Now G(1)(0) =(2)0 why?{\displaystyle Now\ G^{(1)}(0)\ {\overset {\color {red}(2)}{=}}0\ {\color {red}why?}}
(1) p.19−1: G(1)(t)=(3)e(1)(t)−5t4e(1){\displaystyle {\color {red}(1)}\ {\color {blue}p.19-1:}\ G^{(1)}(t){\overset {\color {red}(3)}{=}}e^{(1)}(t)-5t^{4}e(1)}
(5)p.18−3: e(t)=A(t)−A2L(t) (4){\displaystyle {\color {red}(5)}{\color {blue}p.18-3:}\ e(t)=A(t)-{A}_{2}^{L}(t)\ {\color {red}(4)}}
e(1)(t)=A(1)−A2L(1)(t) (5){\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e^{(1)}(t)=A^{(1)}-{A}_{2}^{L{\color {red}(1)}}(t)\ {\color {red}(5)}}
page20-2
A(t)=∫+t−t−=∫k−t−+∫tk−{\displaystyle A(t)=\int _{+t}^{-t}-=\int _{k}^{-t}-+\int _{t}^{k}-}
A(1)=(1)F(−t)+F(t){\displaystyle A^{(1)}{\overset {\color {red}(1)}{=}}F(-t)+F(t)}
k∈]−t,t[ (kisconstant){\displaystyle k\in {\color {red}]}-t,t{\color {red}[}\ ({\color {red}kisconstant})}
(4)p.20-1& (5)p.18-3:
A2L(1)(t)=(2)13[F(−t)+4F(0)+F(t)]+t3[F(1)(−t)+F(1)(t){\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:
e(1)(0)=A(1)(0)−A2L(1)(0)=2F(0)−[2F(0)+0]=0 (3){\displaystyle e^{(1)}(0)=A^{(1)}(0)-A_{2}^{L(1)}(0)=2F(0)-[2F(0)+0]=0\ {\color {red}(3)}}
(3) and (3)p.20−1⇒(2)p.20−1{\displaystyle {\color {red}(3)\ and\ (3)}{\color {blue}p.20-1}\Rightarrow {\color {red}(2)}{\color {blue}p.20-1}}
Recall:
G(1)(ξ1)=0 (1)p.20−1{\displaystyle G^{(1)}(\xi _{1})=0\ {\color {red}(1)}{\color {blue}p.20-1}}
G(1)(0)=0 (2)p.20−1{\displaystyle G^{(1)}(0)=0\ {\color {red}(2)}{\color {blue}p.20-1}}
page20-3
Rolle′stheorem⇒ ∀ξ2∈]0,ξ1[st G(2)(ξ2)=0 (1){\displaystyle Rolle'stheorem\Rightarrow \ \forall \xi _{2}\in ]0,\xi _{1}[st\ \ G^{(2)}(\xi _{2})=0\ {\color {red}(1)}}
Again, G(2)(0)=(2)0 HW∗4.3{\displaystyle Again,\ G^{(2)}(0){\overset {\color {red}(2)}{=}}0\ {\color {blue}HW^{*}4.3}}
(1) and (2) Rolle′stheorem ⇒ ∀ξ3∈]0,ξ2[ G(2)(ξ2)=0 st G(3)(ξ3)=(2)0{\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}
(1)p.19−1: G(3)(t)e(4)(3)(t)−t2⏟(ξ)(4)(3)e(1){\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)}
e(3)(t)=HW∗4.4−t3[F(3)(t)−F(3)(−t)] (5){\displaystyle e^{(3)}(t){\underset {\color {blue}HW^{*}4.4}{=}}-{\frac {t}{3}}[F^{(3)}(t)-F^{(3)}(-t)]\ {\color {red}(5)}}
G(3)(ξ3)=−ξ33[F(3)(ξ3)−F(3)(−ξ3)]⏞Apply DMVT−60(ξ3)2e(1)=(6)from(3)0{\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}
=DMVT(f)−ξ33[2ξ3⏟ξ3−(−ξ3)F(4)(ξ4)]−60(ξ3)2e(1) ξ4∈ ]−ξ3,ξ3[{\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}[}