User:Egm6341.s10.team3.heejun/Meeting 12 transcript

 MTG 12: Thu. 21 Jan 10 [[media: Egm6341.s10.mtg12.djvu | Page 12-1]] 

$$ \displaystyle G(t)=0\,\,\,\,\,\, \Rightarrow G(\cdot)\,\, has\,\, (n+2)\,\, zeros\,\, \underbrace{(t, x_{0},..., x_{n})}_{\color{blue}{(n+2)\,\, points}} $$

$$ \displaystyle G(x_{i})=0\,\,\,\, where\,\, i=0,...,n $$

Rolle's Theorem: $$ \displaystyle G^{(1)}(\cdot)\,\, has\,\, \color{red}{at\,\, least\,\, (n+1)}\,\, \color{black}{zeros.} $$


 * Rolle's Thm. is  a particular case of Deriv. MVT:


 * Deriv. MVT means $$ \displaystyle f:\mathbb{R} \rightarrow \mathbb{R}\,\, cont.\,\, diff.\,\, \exists \,\, at\,\, least\,\, one\,\, point\,\, \xi \in [a,b] \,\, such\,\, that\,\, \frac{f(b)-f(a)}{b-a} = f^{1}(\xi)\,\, or\,\, f(b)-f(a) = f^{1}(\xi)(b-a) $$



Rolle's Theorem: $$ \displaystyle f(a)=f(b)= constant\,\,\,\,\,\, \Rightarrow \exists \,\, at\,\, least\,\, \xi \in [a,b]\,\, such\,\, that\,\, f^{1}(\xi)=0 $$



$$ f_{(1)}(\cdot) \,\, has \,\, 3\,\, zeros\,\,\,\, \color{red} $$

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$$ f_{(2)}(\cdot) \,\, has \,\, 2\,\, zeros\,\,\,\, \color{red} $$



$$ f_{(3)}(\cdot) \,\, has \,\, 1\,\, zero\,\,\,\, \color{red} $$


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 * * $$ \displaystyle Apply\,\, Rolle's\,\, Thm.\,\, to\,\, G(\cdot): $$
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$$ \displaystyle G^{(1)}(\cdot) \,\, has\,\, \color{red}{at\,\,least}\,\, \color{black}{(n+1)}\,\, zeros $$ $$ \cdot \cdot \cdot $$ $$ \displaystyle G^{(n+1)}(\cdot) \,\, has\,\, \color{red}{at\,\,least}\,\, \color{black}{1}\,\, zero\,\, \Rightarrow \exists \xi \in I_{t}\,\, such\,\, that\,\, G^{(n+1)}(\xi)=0 $$
 * $$ \displaystyle G^{(0)}(\cdot) \,\, has\,\, \color{red}{at\,\,least}\,\, \color{black}{(n+2)}\,\, zeros $$
 * $$ \displaystyle G^{(0)}(\cdot) \,\, has\,\, \color{red}{at\,\,least}\,\, \color{black}{(n+2)}\,\, zeros $$
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Note: $$ E^{(n+1)}(x)=f^{(n+1)}(x) - \underbrace{0}_{\color{red}{f_{n}^{(n+1)}(x)}} $$


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HW: Lagrange Interpolation Error - $$ \displaystyle (n+1)^{th} $$
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Prove that $$ \displaystyle E^{n+1}(x) = f^{n+1}(x) $$
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Recall, $$ f_{n} \in \mathbb{P}_{n} $$

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 * $$ \displaystyle q_{n+1}^{(n+1)}(x)=(n+1)! $$
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HW: Proof: $$ \displaystyle (n+1)^{th} $$ derivative of $$ \displaystyle q_{n+1}(x) = (n+1)! $$ Show that $$ \displaystyle q_{n+1}^{(n+1)}(x) = (n+1)! $$
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 * $$ \displaystyle G^{(n+1)}(x) = f^{(n+1)}(x) - \frac {(n+1)!}{q_{n+1}(t)} \underbrace{E(t)}_{\color{red}{constant}}\,\,\,\,\,\, \Rightarrow t\,\, is\,\, not\,\, a\,\, variable,\,\, only\,\, x\,\, is\,\, a\,\, variable $$
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 * $$ \displaystyle Let\,\, x= \xi $$
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 * $$ \displaystyle G^{(n+1)}(\xi) = 0 = f^{(n+1)}(\xi) - \frac{(n+1)!}{q_{n+1}(t)} E(t) $$
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 * $$ \displaystyle E(t)= \frac {q_{n+1}(t)}{(n+1)!}f^{(n+1)}(\xi) $$
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HW: Error analysis of function $$ f(x)=log(x) $$ [[media: Egm6341.s10.mtg11.djvu | P.11-2]] Consider the function $$\displaystyle f(x)=log(x) $$ and set $$\displaystyle t=2, x_{0}=3, x_{1}=4, x_{2}=5, \cdot \cdot \cdot, x_{6}=9$$
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(i) Plot $$\displaystyle f(x) $$ and $$\displaystyle f_{n}(x)$$

(ii) Plot $$\displaystyle l_{i,n} $$ when $$\displaystyle n = 3 $$

(iii) Plot $$\displaystyle q_{n+1}(x) $$

(iv) Obtain $$\displaystyle G(x) $$ for $$\displaystyle x = 5.5 $$
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