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Unformatted text preview: b = [pi/9, sqrt(3)/2 sqrt(3)/2 pi/9]; N = length(d); x = zeros(N,1); for k = 2:N d(k) = d(k)  u(k1)*l(k)/d(k1); b(k) = b(k)  b(k1)*l(k)/d(k1); end x(N) = b(N)/d(N); for k = N1:1:1 x(k) = (b(k)  u(k)*x(k+1))/d(k); end display(x) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Section 2.7 3 The answer from matlab is below x = 0.018336695058824 0.035006775977942 0.048317587771715 0.056523986874421 0.057801073272040 0.050215676439535 0.031696149551397 The code is here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5 %% section 2.7 3 clear clc close all M = 8; N = M1; h = 1/M; l = 1*ones(N,1); l(1) = 0; d = ones(N,1)*(2+h^2); u = 1*ones(N,1); u(end) = 0; b = (h:h:N*h)*h^2; x = zeros(N,1); for k = 2:N d(k) = d(k)  u(k1)*l(k)/d(k1); b(k) = b(k)  b(k1)*l(k)/d(k1); end x(N) = b(N)/d(N); for k = N1:1:1 x(k) = (b(k)  u(k)*x(k+1))/d(k); end display(x) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Created by Jianjian Gao 2010/10/12...
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 Spring '08
 Rottman

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