write a matrix in the form AX=RBX
Show older comments
% Write matrix of form AX=R*B*X with X contains A(k) and B(k) are eigenvectors
% conditions k = 1 to 10; n = 1 to k and n not equal to k
-(2+(pi*h/H)^2)*A(1)+A(2)-(h*R/2)*B(2) = 0
-(2+(pi*h/H)^2)*B(1)+B(2)+(h/2)*A(2)+(2*H*h/pi^2*xi)*symsum(n*(A(n+1)-A(n-1)),n,1,k and n not equal to k) = 0
for k =2:8
A(k-1)-(2+(k*h*pi/H)^2)*A(k)+A(k+1)-(h*R/2)*(B(k+1)-B(k-1)) = 0
B(k-1)-(2+(k*h*pi/H)^2)*B(k)+B(k+1)+(h/2)*(A(k+1)-A(k-1))+(2*H*h/pi^2*xi)*symsum(n*k*(A(n+1)-A(n-1)),n,1,k and n not equal to k) = 0
end
A(8)-(2+(9*h*pi/H)^2)*A(9)+(h*R/2)*B(8) = 0
B(8)-(2+(9*h*pi/H)^2)*B(9)-(h/2)*A(8)+(2*H*h/pi^2*xi)*symsum(9*n*(A(n+1)-A(n-1)),n,1,k and n not equal to k) = 0
4 Comments
Steven Lord
on 16 Oct 2023
It's not clear to me what you're trying to compute and what you already know. Your expressions include variables h, H, A, R, B, and xi (I think I got all of them.) Which of these are known quantities and which of them are you trying to compute? What are the known or desired sizes for these variables?
MADHVI
on 16 Oct 2023
Yash
on 25 Oct 2023
Hi Madhvi,
There is still some confusion regarding the problem that you are trying to solve. Based on the information provided, here is what I have understood:
- There is a matrix equation of the form: A*X = R*B*X.
- There are two variables to iterate, "n" and "k". "k" varies from 1 to 10 and "n" varies from 1 to k-1.
- Variables "h", "H", "A", "R", "B", and "xi" are known, and the goal is to compute the value of "R".
- There is a summation expression that you are attempting to code.
However, the code you provided is not clear or understandable. To assist you further, it would be helpful if you could provide a properly documented problem statement. Since the expression is not readable, kindly provide a documented (handwritten or typed) expression that you are trying to code.
MADHVI
on 25 Oct 2023
Answers (0)
Categories
Find more on Creating and Concatenating Matrices in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!