User:Egm6341.s10.team3.sa/Mtg19

Mtg 19: Thu, 04 Feb 10

 Typeset of transparencies, not lecture transcript. Subramanian Annamalai 13:53, 12 August 2010 (UTC)

[[media: Egm6341.s10.mtg19.djvu | Page 19-1]] 

Continuing with Richardson extrapolation on [[media: Egm6341.s10.mtg18.djvu | Page 18-3]]

$$\displaystyle 2^{2 \cdot 2} * (Eqn\, 3) - (Eqn\, 2)$$ $$\displaystyle = (2^4 - 1)I - 2^4 T_1(2n) +T_1(n) $$ $$\displaystyle = (2^4 - 1)a_2h^4 + (2^4 - 1)a_3h^6 + \theta(h^8) $$  $$\displaystyle \therefore I = \underbrace{\frac{2^{2.2}T_1(2n) - T_1(n)}{2^{2 \cdot 2} - 1}}_{T_2(n)} + a_3h^6 + \theta(h^8)$$ 

Recall from (Eqn 1) of [[media: Egm6341.s10.mtg18.djvu | Page 18-3]] that,

$$\displaystyle I = \underbrace{\frac{2^{2.1}T_0(2n) - T_0(n)}{2^{2 \cdot 1} - 1}}_{T_1(n)} + a_3h^4 + \theta(h^6)$$ 

Therefore in general, $$\displaystyle I = T_k(n) + a_{k+1} h^{2(k+1)} + \theta(h^{2(k+2)}) $$

[[media: Egm6341.s10.mtg19.djvu | Page 19-2]] 

$$\displaystyle T_k(n) = \frac{2^{2.k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2 \cdot k} - 1} $$ 


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HW: This problem is continued from the HW on [[media:Egm6341.s10.mtg6.pdf|p.6-5]]
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i) Modify matlab code to make the computation of $$\displaystyle \color{blue} T_{0}(2^{j}) $$ efficient, i.e. $$\displaystyle \color{blue} T_{0}(2^{j}) = T_{0}(2^{(j-1)}) + \cdot\cdot\cdot $$. 

ii) Construct Romberg table and compare to previous results, i.e. compare Romberg table to Taylor, Trap. and Simpson's rules. 


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[[media: Egm6341.s10.mtg19.djvu | Page 19-3]] 

THEOREM: Trapezoidal error continued from [[media:Egm6341.s10.mtg18.djvu|Pg.18-3]]  $$\displaystyle a_i = d_i \bigg[ f_{(2i-1)}(b) - f{(2i -1)}(a)\bigg]$$ $$\displaystyle d_i = -\frac{B_{2i}}{(2i)!} $$ Note here that $$\displaystyle B_2i $$ are called the Bernoulli numbers. $$\displaystyle d_i = -\frac{1}{12} \;\; d_2 = \frac{1}{720} \;\; d_3 = =\frac{1}{30240}$$ 

CORRECTED TRAPEZOIDAL RULES: $$\displaystyle \underbrace{CT_1(n)}_{First\;correction} = \underbrace{T_0(n)}_{CT_0(n)\;No\;correction} + a_1h^2 + \theta(h^4) $$