User:EGM6341.s11.team1.Chiu/Mtg16

Mtg 16: Mon, 7 Feb 11

[[media: Nm1.s11.mtg16.djvu| page16-1]]

Derivative MVT: (A. p.4 ; Apostol, Calculus, 1961, p.354)


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$$ f: \mathbb{R}\rightarrow \mathbb{R} \ $$ continuous differentiable $$ \Rightarrow \exists \ $$ at least one point $$ \xi \in [a,b] \ $$ at


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$$  \displaystyle \frac{f \left( b \right)-f \left( a \right)}{b-a}=f^{\left( 1 \right)} \left( \xi \right) \Rightarrow f \left( b \right)-f \left( a \right)=f^{\left( 1 \right)} \left( \xi \right) \left( b-a \right) $$
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End Derivative MVT

Rolle's Theorem:

$$ f \left( a \right)=f \left( b \right)= \ $$ constant (e.g., zero) $$ \Rightarrow \exists \ $$ at least $$ \xi \in [a,b] \ $$ at $$ f^{\left( 1 \right)} \left( \xi \right)=0 \ $$

End Rolle's Theorem

Example:


 * nm1.s11.Mtg16.pg1.fig2.svg$$ f^{\left( 1 \right)} \left( . \right) \ $$ has 3 zeros $$ \ \color{red} \left \{ \xi^{0}_{1}, \xi^{0}_{2}, \xi^{0}_{3} \right \} \ $$

[[media: Nm1.s11.mtg16.djvu| page16-2]]


 * nm1.s11.Mtg16.pg2.fig1.svg$$ f^{\left( 2 \right)} \left( . \right) \ $$ has 2 zeros $$ \ \color{magenta} \left \{ \xi^{1}_{1}, \xi^{1}_{2} \right \} \ $$


 * nm1.s11.Mtg16.pg2.fig2.svg$$ f^{\left( 3 \right)} \left( . \right) \ $$ has 1 zero $$ \ \color{red} \left \{ \xi^{2}_{1} \right \} \ $$

End Example

Proof of LIET continued [[media: Nm1.s11.mtg15.djvu| page15-4]] Apply Rolle's theorem to $$ G \left( . \right): \ $$ $$ G^{ \left( 0 \right)}: \ $$ has at least $$ \left( n+2 \right) \ $$ zeros $$ G^{ \left( {\color{red}1} \right)}: \ $$ has at least $$ \left( n+{\color{red}1} \right) \ $$ zeros

[[media: Nm1.s11.mtg16.djvu| page16-3]] $$ \color{blue} \vdots \ $$ $$ \ \ $$ $$ \color{blue} \vdots \ $$


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$$  \displaystyle G^{ \left( {\color{red}n+1} \right)} \ $$ has at least 1 zero (1)
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$$  \displaystyle \Rightarrow \exists \xi  \in {I}_{t} \ at \ G^{\left( n+1 \right)} \left( \xi \right)=0 $$     (2)
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[[media: Nm1.s11.mtg14.djvu| (1) p.14-3]] $$ G^{\left( n+1 \right)} \left( x \right) \underset{ \color{red} \underset{\left( 3 \right)}{\uparrow}}= e^{\left( n+1 \right)} \left( x \right)- \frac{q_{n+1}^{\left( n+1 \right)} \left( x \right)}{q_{n+1} \left( t \right)}e \left( t \right) \ $$


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[[media: Nm1.s11.mtg11.djvu| (2) p.11-3]], [[media: Nm1.s11.mtg9.djvu| (2) p.9-2]], [[media: Nm1.s11.mtg8.djvu| (2) p.8-3]] $$ {\color{blue} \Rightarrow} e^{\left( n+1 \right)} \left( x \right) \underset{ \color{blue}\underset{HW*3.5}{\uparrow}}= f^{\left( n+1 \right)} \left( x \right)- \underbrace{0}_{ \color{blue}HW*3.6 \ f^{L \left( n+1 \right)}_{n} \left( x \right)} $$     (4)
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$$  \displaystyle q_{n+1}^{\left( n+1 \right)} \left( x \right)= \left( n+1 \right)! $$     (5)
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(3)-(5):
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$$  \displaystyle G^{ \left( n+1 \right)} \left( x \right)=f^{\left( n+1 \right)} \left( x \right)- \frac{\left( n+1 \right)!}{q_{n+1} \left( t \right)}e \left( t \right) $$
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Let $$ x= \xi \ $$ in (2) :


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$$  \displaystyle G^{ \left( n+1 \right)} \left( \xi \right)=0=f^{\left( n+1 \right)} \left( \xi \right)- \frac{\left( n+1 \right)!}{q_{n+1} \left( t \right)}e \left( t \right) $$
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$$  \displaystyle \Rightarrow e \left( t \right)= \frac{q_{n+1} \left( t \right)}{\left( n+1 \right)!} f^{\left( n+1 \right)} \left( \xi \right) $$
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End Proof of LIET