User:Egm6341.s10.team3.sa/Mtg27

Mtg 27: Tue, 2 Mar 10

 Typeset of transparencies, not lecture transcript. Subramanian Annamalai 13:55, 12 August 2010 (UTC)

[[media: Egm6341.s10.mtg27.djvu | Page 27-1]] 


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HW:
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$$ \displaystyle \color{blue} i)\, Do\; steps\; 4a\; and\; 4b\; to\; determine\; p_{6}(t)\; and\; p_{7}(t)\; $$


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TRAPEZOIDAL ERROR

$$\displaystyle E_n^1 = I - T_0(n) = \sum_{r=1}^{l} h^{2r} \bar{d}_{2r} \bigg[ f^{(2r-1)(b) - f^{(2r-1)}(a)}\bigg] - \bigg( \frac{h}{2}\bigg )^{2l} \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} p_{2l}(t_k(x))f^{(2l)}(x) \,dx $$ $$\displaystyle (Eq. 1) $$

From (Eq.2) of [[media: Egm6341.s10.mtg21.djvu | Page 21-1]], $$\displaystyle x(t_k) = \frac{1}{2} \bigg( x_k + x_{k+1}\bigg) + t_k \frac{h}{2}$$ $$\displaystyle (Eq. 2) $$


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HW:
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Find $$\displaystyle \color{blue} {t}_{k}(x) $$ by the reversion of $$ x ({t}_{k}) $$

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$$ $$
 * $$\displaystyle \color{blue} x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle \color{blue} x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle\color{blue}(Eq. 1)
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$$
 * $$\displaystyle \color{blue} h=x_{k+1}-x_k
 * $$\displaystyle \color{blue} h=x_{k+1}-x_k
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Recalling Bernoulli numbers from [[media: Egm6341.s10.mtg19.djvu | Page 19-3]], $$\displaystyle \bar{d}_{2r} = d_i = \frac{p_{2r}(1)}{2^{2r}} = - \frac{B_{2r}}{(2r)!} $$ $$\displaystyle (Eq. 3) $$ Note: $$\displaystyle \bar{B}_{2r} $$ represents the Bernoulli numbers.

[[media: Egm6341.s10.mtg27.djvu | Page 27-2]] 

$$\displaystyle E_n^1 = \alpha - \beta $$ $$\displaystyle \alpha= \sum_{i=1}^{l} h^{2i} \bar{d}_{2i} \underbrace{\bigg[ f^{(2i-1)(b) - f^{(2i-1)}(a)} \bigg]}_{a_i}$$

$$\displaystyle a_i= d_i {\bigg[ f^{(2i-1)(b) - f^{(2i-1)}(a)} \bigg]} $$ $$\displaystyle \bar{d}_{2i} = d_i = -\frac{B_{2i}}{(2i)!}$$ $$\displaystyle \bar{d}_{2i} = \frac{p_{2i}}{2^{2i}} $$ which is (Eq.3) on [[media: Egm6341.s10.mtg27.djvu | Page 27-1]]


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HW:
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Find coefficients $$\displaystyle \color{blue} \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ using the following relations.


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d_{r} = \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}} $$
 * $$\displaystyle\color{blue}
 * $$\displaystyle\color{blue}
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