Mtg 20: Wed, 16 Feb 11
page20-1
Pf of SSET p.19-3 cont'd
( 1 ) p .19 − 3 R o l l e ′ s t h m ⇒ ∃ ς 1 ∈ ] 0 , 1 [ s t G ( 1 ) ( ς 1 ) = ( 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}
N o w G ( 1 ) ( 0 ) = ( 2 ) 0 w h y ? {\displaystyle Now\ G^{(1)}(0)\ {\overset {\color {red}(2)}{=}}0\ {\color {red}why?}}
( 1 ) p .19 − 1 : G ( 1 ) ( t ) = ( 3 ) e ( 1 ) ( t ) − 5 t 4 e ( 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 ) − A 2 L ( 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 ) − A 2 L ( 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 − + ∫ t k − {\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 [ ( k i s c o n s t a n t ) {\displaystyle k\in {\color {red}]}-t,t{\color {red}[}\ ({\color {red}kisconstant})}
(4)p.20-1& (5)p.18-3:
A 2 L ( 1 ) ( t ) = ( 2 ) 1 3 [ F ( − t ) + 4 F ( 0 ) + F ( t ) ] + t 3 [ 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 ) − A 2 L ( 1 ) ( 0 ) = 2 F ( 0 ) − [ 2 F ( 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 ) a n d ( 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 ) ( ξ 1 ) = 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
R o l l e ′ s t h e o r e m ⇒ ∀ ξ 2 ∈ ] 0 , ξ 1 [ s t G ( 2 ) ( ξ 2 ) = 0 ( 1 ) {\displaystyle Rolle'stheorem\Rightarrow \ \forall \xi _{2}\in ]0,\xi _{1}[st\ \ G^{(2)}(\xi _{2})=0\ {\color {red}(1)}}
A g a i n , G ( 2 ) ( 0 ) = ( 2 ) 0 H W ∗ 4.3 {\displaystyle Again,\ G^{(2)}(0){\overset {\color {red}(2)}{=}}0\ {\color {blue}HW^{*}4.3}}
( 1 ) a n d ( 2 ) R o l l e ′ s t h e o r e m ⇒ ∀ ξ 3 ∈ ] 0 , ξ 2 [ G ( 2 ) ( ξ 2 ) = 0 s t G ( 3 ) ( ξ 3 ) = ( 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 ) − t 2 ⏟ ( 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 ) = H W ∗ 4.4 − t 3 [ 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 ) ( ξ 3 ) = − ξ 3 3 [ F ( 3 ) ( ξ 3 ) − F ( 3 ) ( − ξ 3 ) ] ⏞ A p p l y D M V T − 60 ( ξ 3 ) 2 e ( 1 ) = ( 6 ) f r o m ( 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}
= D M V T ( f ) − ξ 3 3 [ 2 ξ 3 ⏟ ξ 3 − ( − ξ 3 ) F ( 4 ) ( ξ 4 ) ] − 60 ( ξ 3 ) 2 e ( 1 ) ξ 4 ∈ ] − ξ 3 , ξ 3 [ {\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}[}