I have a problem with MatLab in Octave
I need to make a program, which solves a Triangular system of linear equations, using back substitution method (https://i.stack.imgur.com/ETPxn.png)
I have a problem with my code, Octave displays this error:
error: =: nonconformant arguments (op1 is 1x1, op2 is 2x2)
error: called from
backsub at line 6 column 10
prog31 at line 3 column 1
run at line 93 column 5
Here is my code:
Backsub function (backsub.m)
function X = backsub(A,B)
n = length(B);
X = zeros(n, 1);
X(n) = B(n)./A(n,n);
for k = n-1:-1:1
X(k) = (B(k)-A(k,k+1:n)+X(k+1:n))./A(k,k);
end
Main function (prog31.m)
A = [3 -2 1 -1; 0 4 -1 2; 0 0 2 3; 0 0 0 5];
B = [8; -3; 11; 15];
X = backsub(A, B);
disp(X);
You mistyped * as + in the for-loop, hence the error. This is the fix:
function X = backsub(A,B)
n = length(B);
X = zeros(n, 1);
X(n) = B(n)/A(n,n); % remove . because A(n,n) is scalar
for k = flip(1:n-1)
X(k) = (B(k)-(A(k,k+1:n)*X(k+1:n)))/A(k,k); % remove . and replace + with *
end % end the for loop
end
As a side note: The product in the for-loop is a dot product, because A(k,k+1:n) is a row vector while X(k+1:n) is a column vector of appropriate size.
If you want to use provided formula verbatim (with the sigma sum), it would be
X(k) = (B(k)-sum(A(k,k+1:n)'.*X(k+1:n)))/A(k,k);
Related
below is my code to perform jacobi iterations to solve Ax=b.
I try this code on the matrix A =[4 -1 1; 4 -8 1; -2 1 5] and b=[7 -21 15].
and x is a first guess 1 x 3 vector. Are not these dimensions correct? It gives me the error in the code that calculates: r = b - x*A
and M\(x*N + b)
What am I missing?!? how do I fix this? please help!
function [x, error, iter, flag] = jacobi(A, x, b, maxiter, tol)
%implement jacobi iterations
%[x, error, iter, flag] = jacobi(A, x, b, maxiter, tol)
%jacobi.m solves the linear system Ax=b using the Jacobi iteration
%
%
%INPUT A the matrix of the system Ax=b
% x the first guess vector Ax=b
% b the vector in the system
% maxiter the maximum number of iteration to perform
% tol the tolerance
%
%OUTPUT x the solution vector
% error error norm
% niter the number of iterations it took
% flag indicates whether a solution was found. 0 means there was a
% solution and 1 means there was not a solution
iter = 0;
flag = 0;
bnrm2 = norm(b);
if (bnrm2 == 0)
bnrm2 = 1.0;
end
r = b - x*A;
error = norm(r) / bnrm2;
if (error<tol)
return;
end
[m,n] = size(A);
M = diag(diag(A));
N = diag(diag(A)) - A;
for iter = 1:maxiter,
oldx = x;
x = M\(x*N + b);
error = norm(x - oldx) / norm(x);
if (error <= tol)
break;
end
end
if (error > tol)
flag = 1;
end
Since, in the code, you're solving what I'll call (not sure if it's proper since I never do it) the left-multiply problem, the operator and order of matrices are, in some sense, reversed.
If you were solving the problem A*x = b with the residual r = b - A*x (i.e., x and b are column vectors), you would perform right-vector multiplies and left-matrix divides. Therefore, the update line in the loop would be
x = M \ (N*x + b);
Conversely, if you were solving the problem x*A = b with the residual r = b - x*A (i.e., x and b are row vectors), you would perform left-vector multiplies and right-matrix divides. Therefore, the update line in the loop would be
x = (x*N + b) / M;
Note that \ resolves to the function mldivide and / resolves to mrdivide. There is no function distinction for the multiply.
It appears your current updater mixes the two, which is bad news bears for dimension matching.
I can't seem to find a fix to my infinite loop. I have coded a Jacobi solver to solve a system of linear equations.
Here is my code:
function [x, i] = Jacobi(A, b, x0, TOL)
[m n] = size(A);
i = 0;
x = [0;0;0];
while (true)
i =1;
for r=1:m
sum = 0;
for c=1:n
if r~=c
sum = sum + A(r,c)*x(c);
else
x(r) = (-sum + b(r))/A(r,c);
end
x(r) = (-sum + b(r))/A(r,c);
xxx end xxx
end
if abs(norm(x) - norm(x0)) < TOL;
break
end
x0 = x;
i = i + 1;
end
When I terminate the code it ends at the line with xxx
The reason why your code isn't working is due to the logic of your if statements inside your for loops. Specifically, you need to accumulate all values for a particular row that don't belong to the diagonal of that row first. Once that's done, you then perform the division. You also need to make sure that you're dividing by the diagonal coefficient of A for that row you're concentrating on, which corresponds to the component of x you're trying to solve for. You also need to remove the i=1 statement at the beginning of your loop. You're resetting i each time.
In other words:
function [x, i] = Jacobi(A, b, x0, TOL)
[m n] = size(A);
i = 0;
x = [0;0;0];
while (true)
for r=1:m
sum = 0;
for c=1:n
if r==c %// NEW
continue;
end
sum = sum + A(r,c)*x(c); %// NEW
end
x(r) = (-sum + b(r))/A(r,r); %// CHANGE
end
if abs(norm(x) - norm(x0)) < TOL;
break
end
x0 = x;
i = i + 1;
end
Example use:
A = [6 1 1; 1 5 3; 0 2 4]
b = [1 2 3].';
[x,i] = Jacobi(A, b, [0;0;0], 1e-10)
x =
0.048780487792648
-0.085365853612062
0.792682926806031
i =
20
This means it took 20 iterations to achieve a solution with tolerance 1e-10. Compare this with MATLAB's built-in inverse:
x2 = A \ b
x2 =
0.048780487804878
-0.085365853658537
0.792682926829268
As you can see, I specified a tolerance of 1e-10, which means we are guaranteed to have 10 decimal places of accuracy. We can certainly see 10 decimal places of accuracy between what Jacobi gives us with what MATLAB gives us built-in.
In Matlab I want to create the partial derivative of a cost function called J(theta_0, theta_1) (in order to do the calculations necessary to do gradient descent).
The function J(theta_0, theta_1) is defined as:
Lets say h_theta(x) = theta_1 + theta_2*x. Also: alpha is fixed, the starting values of theta_1 and theta_2 are given. Let's say in this example: alpha = 0.1 theta_1 = 0, theta_2 = 1. Also I have all the values for x and y in two different vectors.
VectorOfX =
5
5
6
VectorOfX =
6
6
10
Steps I took to try to solve this in Matlab: I have no clue how to solve this problem in matlab. So I started off with trying to define a function in Matlab and tried this:
theta_1 = 0
theta_2 = 1
syms x;
h_theta(x) = theta_1 + t2*x;
This worked, but is not what I really wanted. I wanted to get x^(i), which is in a vector. The next thing I tried was:
theta_1 = 0
theta_2 = 1
syms x;
h_theta(x) = theta_1 + t2*vectorOfX(1);
This gives the following error:
Error using sym/subsindex (line 672)
Invalid indexing or function definition. When defining a
function, ensure that the body of the function is a SYM
object. When indexing, the input must be numeric, logical or
':'.
Error in prog1>gradientDescent (line 46)
h_theta(x) = theta_1 + theta_2*vectorOfX(x);
I looked up this error and don't know how to solve it for this particular example. I have the feeling that I make matlab work against me instead of using it in my favor.
When I have to perform symbolic computations I prefer to use Mathematica. In that environment this is the code to get the partial derivatives you are looking for.
J[th1_, th2_, m_] := Sum[(th1 + th2*Subscript[x, i] - Subscript[y, i])^2, {i, 1, m}]/(2*m)
D[J[th1, th2, m], th1]
D[J[th1, th2, m], th2]
and gives
Coming back to MATLAB we can solve this problem with the following code
%// Constants.
alpha = 0.1;
theta_1 = 0;
theta_2 = 1;
X = [5 ; 5 ; 6];
Y = [6 ; 6 ; 10];
%// Number of points.
m = length(X);
%// Partial derivatives.
Dtheta1 = #(theta_1, theta_2) sum(2*(theta_1+theta_2*X-Y))/2/m;
Dtheta2 = #(theta_1, theta_2) sum(2*X.*(theta_1+theta_2*X-Y))/2/m;
%// Loop initialization.
toll = 1e-5;
maxIter = 100;
it = 0;
err = 1;
theta_1_Last = theta_1;
theta_2_Last = theta_2;
%// Iterations.
while err>toll && it<maxIter
theta_1 = theta_1 - alpha*Dtheta1(theta_1, theta_2);
theta_2 = theta_2 - alpha*Dtheta2(theta_1, theta_2);
it = it + 1;
err = norm([theta_1-theta_1_Last ; theta_2-theta_2_Last]);
theta_1_Last = theta_1;
theta_2_Last = theta_2;
end
Unfortunately for this case the iterations does not converge.
MATLAB is not very flexible for symbolic computations, however a way to get those partial derivatives is the following
m = 10;
syms th1 th2
x = sym('x', [m 1]);
y = sym('y', [m 1]);
J = #(th1, th2) sum((th1+th2.*x-y).^2)/2/m;
diff(J, th1)
diff(J, th2)
Consider a difference equation with its initial conditions.
5y(n) + y(n-1) - 3y(n-2) = (1/5^n) u(n), n>=0
y(n-1) = 2, y(n-2) = 0
How can I determine y(n) in Matlab?
Use an approach similar to this (using filter), but specifying initial conditions as done here (using filtic).
I'm assuming your initial conditions are: y(-1)=2, y(-2)=0.
num = 1; %// numerator of transfer function (from difference equation)
den = [5 1 -3]; %// denominator of transfer function (from difference equation)
n = 0:100; %// choose as desired
x = (1/5).^n; %// n is >= 0, so u(n) is 1
y = filter(num, den, x, filtic(num, den, [2 0], [0 0]));
%// [2 0] reflects initial conditions on y, and [0 0] those on x.
Here's a plot of the result, obtained with stem(n,y).
The second line of your code does not give initial conditions, because it refers to the index variable n. Since Matlab only allows positive integer indices, I'll assume that you mean y(1) = 0 and y(2) = 2.
You can get an iteration rule out of your first equation by simple algebra:
y(n) = ( (1/5^n) u(n) - y(n-1) + 3y(n-2) ) / 5
Code to apply this rule in Matlab:
n_max = 100;
y = nan(n_max, 1);
y(1) = 0;
y(2) = 2;
for n = 3 : n_max
y(n) = ( (1/5^n) * u(n) - y(n-1) + 3 * y(n-2) ) / 5;
end
This code assumes that the array u is already defined. n_max specifies how many elements of y to compute.
f1 = 1 ;
N = 1024 ;
fs = 200 ;
ts = 1/fs ;
t = -(N/(2*fs)):ts:(N/(2*fs)) ;
theta=rand(0:2*pi);
X = sin(2*pi*f1*t+theta) ;
plot(t,x)
grid
Error using +
Matrix dimensions must agree.
And how can i calculate the autocorrelation of x function Rxx(n) ?
Replace the theta line by
theta = 2*pi*rand; %// generates a random number between 0 and 2*pi
and the plot line by
plot(t, X); %// capital "X", as you have defined previously
For the autocorrelation, you can use conv (correlation for real signals is equivalent to convolution with a time-reversal):
c = conv(X,fliplr(X));
plot(-(N/fs):ts:(N/fs), c)
To add to what Luis Mendo has answered, the reason for the error message is that:
>> size(t)
ans =
1 1025
>> size(theta)
ans =
0 1 2 3 4 5 6
So you are trying to add two things which are not the same dimension in X = sin(2*pi*f1*t+theta), hence the error message.
Use Luis's suggestions to fix your code.