solve overdetermined linear system

Hello. I need to solve overdetermined linear system A*x=B, where x is [a1 a2 a3 a4 a5]. I got a solution with command

    inv(A'*A)*A'*B;

but the answer is not what I need, because i have a theoretical solution for this problem. Can I somehow set the initial approximation to get the solution I need?

Answer 1

[John D'Errico]

First of all, DON'T USE THAT SOLUTION!!!!!!!! That is a bad idea, because it is terrible when applied to moderately ill-posed problems.

Instead, use

coef = A\b;

This will be more accurate.

You say that the answer is not what you wanted. Too bad. You cannot set an initial start point for a problem that has a unique solution.

An alternative solution is to use lsqr.

coef = lsqr(A,b);

You can set the start point for lsqr. From the help for lsqr...

X = lsqr(A,B,TOL,MAXIT,M1,M2,X0) specifies the P-by-1 initial guess. If
  X0 is [] then lsqr uses the default, an all zero vector.

But as long as A is full rank, then you will get the same solution, no matter what the start point.

So, is A of less than full rank? If so, then the solution is non-unique, and there are infinitely many equally good solutions.

If you have additional information about the problem, then you need to explain what it is.

[Alexandr Fedorov]

Yes, rank(A)=2, but i got same solution. I tried this:

x01 = 30*ones(nr-1,1);
lsqr(A,B,[],30,[],[],x01)

I have theoretical solution. This is K1=a1*sqrt(2*pi), so a1=K1/sqrt(2*pi)=31,... Becouse theoretical solution is 31, i set x0 as vector of 30, but solution is 1.2639

[John D'Errico]

You have told me only that A has rank of 2. Not what the size of A is. IS A FULL RANK? How many times must I ask that?

Ok, so your call to lsqr tells me that A has nr-1 columns. Of course, you do not tell me what nr is.

Then you tell me that A has rank 2. Before, you told me the problem is over determined. Therefore there are more than nr-1 rows in A. But is nr 3? If it is, then A is of full rank.

IF A IS FULL RANK, THEN THERE IS ONLY ONE SOLUTION. IT IS UNIQUE. So it really does not matter what method you use on a full rank matrix A. Just wanting a different solution is irrelevant, at least unless you possess good wanding skills. How is your magic?

So if you want help, then best would be to provide the matrix A and vector B. Put them in a .mat file, and attach it to a comment using the paper clip. Then explain why you think that the solution you have posed is a solution. You have claimed theory here, but that does not mean I have any reason to believe or accept your claim. Perhaps your theory is even correct, but you created the matrices A or B incorrectly.

[Alexandr Fedorov]

Sorry, i understand you, i have bad English. Size of A is 30 or 40 by 2 or 3 or 4, depends on the number of points and on the number of terms in the sum. tochki2 - points from experimental picture

No, nr is number of members in sum, so nr is 3, 4 or 5

[John D'Errico]

I cannot run your code though, because I do not know what input to provide.

Your .mat file does not contain A and B, which is all that we need here for the moment.

But if A has rank 2 AND it has 2 columns, then this explains why you get the same answer from both backslash and lsqr. It will be of full rank. If A had 3 or 4 columns and 30-40 rows, and still was rank 2, then backslash and lsqr can return different results, because there will be infinitely many results. So I think you are trying the case where A has 2 columns.

[Alexandr Fedorov]

But how i can make the rank of A is 3 or 4, when i take like 3 members in sum? I take for example 40 point, 3 members and 223 MPa of stress. Ok. I finding K1 which is mode I stress-intensity factors by solve this system

The left side of expression is stress from ANSYS and F1n is (q=n/2):

. K1=a1*sqrt(2*pi). In theory K1=Sigma*sqrt(2*pi*l), where l is lenght of fracture, Sigma is the tensile load value. So theta and r i got from picture of stress intensity from ANSYS.

This theory is described in Interference-optical Methods of Solid Mechanics by Razumovsky, Igor A.