How can I do this L1 integral minimization?

Greetings,

I have the following integral

where kdx = [pi/16, pi/2], and

I want to minimize the integral above and solving corresponding a_j. But I have no clue how do to it. Can anyone give me some hint?

Thanks.

Answer 1

[Bruno Luong]

The problem of linear L1 fit (your case); meaning

argmin_x | M*x - y |_l1
argmin sum abs(M*x - y)

can be reformulated and solved by linear programming (opt toolbox required) using slack variables trick as following

n = length(y);
Aeq = [M speye(n) -speye(n)];
Aeqpr=nonzeros(Aeq);
beq = y(:);
c = [zeros(1,size(M,2)) ones(1,2*n)];
LB = [-inf(1,size(M,2)) zeros(1,2*n)];
UB = [];
c = c(:);
LB = LB(:);
UB = UB(:);
x0 = zeros(size(c)); % guess vector
[sol, f, exitflag] = linprog(c,[],[], Aeq, beq, LB, UB, x0);
x = sol(1:size(M,2));

You just need to build M with sin(k*j*dx) and log(dx).

[Sijie Huang]

Thanks for the reply. I am not sure about the one you said: build M with sin(k*j*dx) and log(dx).

I am thinking maybe do change of variable first, sth like y=ln(kdx), plug this into sin(jkdx), and then build the matrix?

[Bruno Luong]

I don't know the detail on how you compute the integral, but I imagine it comes down to be a sum of discrete points values times a step.

Write it down then you'll see how M(i,j) would be.

[Sijie Huang]

does this

linprog

contains the integral inherently?

[Bruno Luong]

No.

[Sijie Huang]

So you assume if the error inside the integral is minimum, the integral will be minimum?

[Bruno Luong]

???? Sorry I can't get your question.

I don't assume anything like that.

What I assume is this:

The integral is ~ | M*x - y |_l1.

It is up to you to find a matrix M and y to approximate the integral, the detail how to do that depends on (k*Deltax) which you know , not me.

[Sijie Huang]

Oh, I see. Sorry I didn't understand your subscript _l1.

[Bruno Luong]

I don't know how to type a curly "l" (lowercase L), which is the right notation.

Answer 2

[Matt J]

Use MINL1LIN from the File Exchange.