# Thread Subject: Right hand values of linear systems

 Subject: Right hand values of linear systems From: Ryan Date: 29 Nov, 2010 03:55:04 Message: 1 of 1 for a tridiagonal matrix, where the matrix is nxn, the right hand side is: b_i = 6/h^2*(y_i+1-2*y_i+y_i-1), i = 1...n where y_i = cos(i*h), i = 0...n+1 and h = pi/(n+1). Instead I just wrote the equation as b_i+1 = 6/h^2*(cos((i+2)*h) - 2*cos((i+1)*h) + cos(i*h), where h = pi/(n+1) since it seemed easier to organize. I want to solve the tridiagonal system and analyze what happens when n increases. But I'm not sure how to do this. Here's what I have so far: array=zeros(1,4) for i=1:4 h=pi/(i+1) array(i)=h end array2=zeros(1,4) for ii=1:4 d=6./array.^2.*(cos((i+2).*array)-2.*cos((i+1).*array)+cos(i.*array)) array2(ii)=d end Please help. Thanks so much!

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Tag Applied By Date/Time
iteration Ryan 28 Nov, 2010 22:59:05
loops Ryan 28 Nov, 2010 22:59:05
matrix Ryan 28 Nov, 2010 22:59:05
tridiagonal Ryan 28 Nov, 2010 22:59:05
numerical analysis Ryan 28 Nov, 2010 22:59:05
array indexing Ryan 28 Nov, 2010 22:59:05