User:Egm6341.s10.team3.heejun/Meeting 28 transcript

 '''MTG 28: Thu. 4 Mar 10'''


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Note:  Typeset of Transparencies, not lecture transcript  Heejun Chung 20:59, 11 August 2010 (UTC)


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  [[media: Egm6341.s10.mtg28.djvu | Page 28-1]]  

[[media: Egm6341.s10.mtg21.djvu | Page 21-2]]:


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&= [P_{2}g^{(1)} + P_{4}g^{(3)} + ... + P_{2l}g^{(2l-1)}]^{+1}_{-1} - \int^{+1}_{-1} P_{2l}(t) g^{(2l)}(t) dt \\\end{align} $$
 * $$ \displaystyle \begin{align} & E = \int^{+1}_{-1} P_{1}(t) g^{(1)}(t)dt \\
 * $$ \displaystyle \begin{align} & E = \int^{+1}_{-1} P_{1}(t) g^{(1)}(t)dt \\
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 * $$ \displaystyle \color{blue}{A} = [P_{2}(t)g^{(1)}(t)]^{+1}_{-1} = P_{2}(+1)g^{(1)}(+1) - P_{2}(-1)g^{(1)}(-1) $$
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 * $$ \displaystyle P_{2}\,\, even\,\, function\,\, \Rightarrow\,\, P_{2}(-1) = P_{2}(+1) $$
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 * $$ \displaystyle P_{2i}\,\, even\,\, function $$
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 * $$ \displaystyle \color{blue}{A} = P_{2}(+1) [g^{(1)}(+1) - g^{(1)}(-1)] $$
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 * $$ \displaystyle E = \sum ^{l}_{r=1} P_{2r}(+1) [g^{(2r-1)}(+1) - g^{(2r-1)}(-1)] $$
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[[media: Egm6341.s10.mtg28.djvu | Page 28-2]] 


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 * $$ \displaystyle - \int^{+1}_{-1} P_{2l} (t) g^{(2l)}(t) dt $$
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HW: Derive $$ E_n^1 = \sum_{r=1}^{l} \bar{d}_{2r}\big[ f^{2r-1}(b) - f^{2r-1}(a) \big] - \frac{h^{2l}}{2l} \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} p_{2l}(t_k(x))f^{(2l)}(x) \,dx$$ Reference:    (1)   [[media:Egm6341.s10.mtg27.djvu|p.27-1]]
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HW: Bernoulli numbers $$\displaystyle x\,coth(x) = \sum_{r=0}^{\infty}\underbrace{\frac{B_{2r}}{(2r)!}}_{\color{blue}{-\bar{d_{2r}}}}x^{2r} $$ i) Verify $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ by comparing the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$. ii) Compute $$\displaystyle \bar{d_{8}}, \; \bar{d_{10}}$$ iii) Compare to $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(+1)}{2^{2r}} $$   Reference:     (3)   [[media:Egm6341.s10.mtg27.djvu|p.27-1]]
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HW: Redo steps in proof of Trapezoidal error by trying to cancel terms with odd order of derivative of $$ \displaystyle g $$. (i.e., instead of even order)
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