Accepting multiple values for a function. I want my function to accept multiple values for beta

function x = backsub(U,b) 
%FORWARDSUB Solve a lower triangular linear system
%Input: 
% U = Upper triangular matrix (n by n)
% b = right-hand side vector (n by 1)
% Output:
%Solution of Ux=b (n by 1 vector)
n =  length(U);
x= zeros(n,1);
for i = 1:3
    x(i) =( b(i) - U(i, 1:i-1)*x(1:i-1) ) / U(i,i);
end
end
alpha= 0.1; composition of specific matrix
beta= 1e1 (here lies the problem, i want the code to accept values of [10, 100, 1000 to 10^12])
U = eye(5)+ diag([-1 -1 -1 -1],1);
U(1,[4 5]) = [ alpha-beta, beta ];
x_exact = ones(5, 1);
b = [alpha;0;0;0;1];
x=backsub(U,b)

[Bruno Luong]

Is it recursive function (you call backsub in backsub)? How beta change between two recursion? When the recusion stops?

[Walter Roberson]

I am pretty sure it is not intended to be recursive -- I think they posted the function and then the script to drive the function.

Answer 1

[Walter Roberson]

U(1,[4 5]) = [ alpha-beta, beta ];

when beta is a vector, then you have a problem: you need different 5 x 5 U matrices for each different value of beta. You are not operating on "the same U matrix but different beta values" each time: you are operating on different U matrices each time.

So you will need to either switch to 3D calculations, with U being 5 x 5 x length(beta), and appropriate adjustment for the backsub() function -- or else you will need to loop your code.

[Muhideen Ogunlowo]

How do I loop the code?

[Walter Roberson]

Looping...

But I didn't fix any bugs in your code.

format short g
alpha= 0.1; %composition of specific matrix
betas = 10.^(1:12);
baseU = eye(5)+ diag([-1 -1 -1 -1],1);

for K = 1 : length(betas)
    U = baseU;
    beta = betas(K)
    U(1,[4 5]) = [ alpha-beta, beta ]
    x_exact = ones(5, 1);
    b = [alpha;0;0;0;1];
    x=backsub(U,b)
end

beta = 
    10

U = 5×5

1 -1 0 -9.9 10 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
   100

U = 5×5

1 -1 0 -99.9 100 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
        1000

U = 5×5

1.0e+00 * 1 -1 0 -999.9 1000 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
       10000

U = 5×5

1.0e+00 * 1 -1 0 -9999.9 10000 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
      100000

U = 5×5

1.0e+00 * 1 -1 0 -1e+05 1e+05 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
     1000000

U = 5×5

1.0e+00 * 1 -1 0 -1e+06 1e+06 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
    10000000

U = 5×5

1.0e+00 * 1 -1 0 -1e+07 1e+07 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
   100000000

U = 5×5

1.0e+00 * 1 -1 0 -1e+08 1e+08 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
        1e+09

U = 5×5

1.0e+00 * 1 -1 0 -1e+09 1e+09 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
        1e+10

U = 5×5

1.0e+00 * 1 -1 0 -1e+10 1e+10 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
        1e+11

U = 5×5

1.0e+00 * 1 -1 0 -1e+11 1e+11 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

beta = 
        1e+12

U = 5×5

1.0e+00 * 1 -1 0 -1e+12 1e+12 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1 -1 0 0 0 0 1

x = 5×1

0.1 0 0 0 0

function x = backsub(U,b) 
%FORWARDSUB Solve a lower triangular linear system
%Input: 
% U = Upper triangular matrix (n by n)
% b = right-hand side vector (n by 1)
% Output:
%Solution of Ux=b (n by 1 vector)
    n =  length(U);
    x= zeros(n,1);
    for i = 1:3
        x(i) =( b(i) - U(i, 1:i-1)*x(1:i-1) ) / U(i,i);
    end
end

[Torsten]

for i = 1:n
    x(i) =( b(i) - U(i, 1:i-1)*x(1:i-1) ) / U(i,i);
end

Why do you use a code for forward substitution if you need backward substitution ?