Iteratively defining and solving equations

Ok so basically I want to solve k4 = [a,b,c,d,....,] (<-M variables) for a system of M equations that go something like this:

k4(1) = k4(2)-2k4(1)+k4(M)

k4(2) = k4(3)-2k4(2)+k4(1)

.

.

.

k4(M-1) = k4(M)-2k4(M-1)+k4(M-2)

k4(M) = k4(1)-2k4(M-1)+k4(M-2)

I am not quite sure how to go about defining each of the variables and then soving the set of M equations so that I can get a numerical solution for each element of k4. I assume I'd need a for loop but I don't know how to go about it?

[James Tursa]

How big is M and what is the size of k4? The obvious solution to this set of linear equations is k4 = 0 (all elements). Is the system underdetermined and you want a description of all possible solutions? Or ...?

Side note: Indexes start at 1 in MATLAB, not 0.

[AA]

M is arbitrary but it will be a large number like 100. The size of k4 is also M. This is the simpler version of the equation I want to solve. Essentially I want a way to find solutions for a,b,c..... from this set of equations but I can't write them all out because M is large

Also I noticed a typo for k4(0) - I've changed it now to k4(1)

[James Tursa]

So it is going to boil down to whether your MxM system of equations is full rank or not. If it is full rank then the unique solution is k4 = 0 all elements. If not full rank then you will have other solutions. Just write everything on the left hand side with 0's on the right hand side, figure out the matrix of coefficients, and go from there. Is that last equation correct? It doesn't fit the pattern.

Answer 1

[John D'Errico]

It is a LINEAR system of equations. You don't want to be solving them iteratively. You need to learn how to define the matrix that describes that system. Anyway, I'm not positive about the coefficients of your system. But it apears there are no constant terms in the problem. But your question is not very clear, so there may be. ANd I'm not confident about the coefficients in your system. So it is difficult to be sure if the problem is rank deficient or not.

But if the right hand side constants are all zero, then the solution is as James says. You will solve it using null. NOT using an iterative solution.

So start by learning how to create a matrix of coefficients. If M is large, then you may need to use tools like sparse. And it LOOKs like your matrix will be a circulant matrix. In that case, you could use either the circulant matrix tool I posted on the File Exchange, or even easier, just use gallery.

M = 8;
R = [-3,1,zeros(1,M-3) , 1];
A = gallery('circul',R)

A = 8×8

-3 1 0 0 0 0 0 1 1 -3 1 0 0 0 0 0 0 1 -3 1 0 0 0 0 0 0 1 -3 1 0 0 0 0 0 0 1 -3 1 0 0 0 0 0 0 1 -3 1 0 0 0 0 0 0 1 -3 1 1 0 0 0 0 0 1 -3

It looks vageuely like this is the matrix you want. But that is just a quck glance. In that case, this matrix would have full rank. For example:

rank(A)

ans = 8

Anyway, since it is full rank, then it has NO solution, except the vector of all zeros. But again, all of this is just a guess, since I'm not sure I am reading your system of equations propeerly.