How do I solve min ||y - A(x)||2 efficiently when A(x) is function (fourier ish) and x is large?

How do I solve for min ||y - A(x)||2 efficiently when A(x) is function? A(x) is a fourier transform plus a few other little things (ie ifftshifts, multiplication). The number of unknowns is very large (x is large)

Ideally, I could use x = A\y. But I can't figure out how to do this. Even fminlbfgs_Kroon is super slow!

Thanks so much for your help!

Christine

[Matt J]

Christine, please use Comment boxes instead of Answer boxes to have a dialog with people answering your post. As you can see, Star Strider and Sean are exhausting themselves moving your remarks to Comment boxes, where they belong.

[SwishSwisher]

This is my first time using this platform. Will do next time.

Answer 1

[Bjorn Gustavsson]

1, start rewriting your function A without calls to the fft-family functions - simply replace them with multiplications with the Fourier matrix and its inverse (transpose..., look at dftmtx), that should make it a fair bit clearer how linear your problem is, I'm a bit concerned about the .* in the first and 2nd last line of the A function. outsize of that it seems pretty linear, and those seems to be just different weighting/filter factors?

If that works out and you end up with a linear problem it might end up a "rather big one". But try that anyway...

[SwishSwisher]

That is correct. Those are just weighting factors.

Thanks so much for your help! I will try this.

Answer 2

[Matt J]

Do you really need the normalization by max(Us(:)) in

   y = Us./max(Us(:));

I assume it's because you normalize the measured 'y' data in a similar way. If you stop normalizing, it looks like your model becomes linear and you could solve it just using a series of filtering steps.

If you include the normalization, the model equation becomes non-differentiable, which creates issues for lsqcurvefit, and other solvers.

[SwishSwisher]

I don't need the normalization. I can just remove the normalization of the measured data y.

[SwishSwisher]

Thanks so much!!!! I will try this.