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(Don’t let this notation mislead you into inverting the full matrix W(x) + λS. You only need to solve for some y the linear system (W(x) + λS) · y = R, and then substitute y into both the numerators and denominators of 18.6.12 or 18.6.13.) Equations (18.6.12) | 818 Chapter 18. Integral Equations and Inverse Theory Don t let this notation mislead you into inverting the full matrix W x AS. You only need to solve for some y the linear system W x AS y R and then substitute y into both the numerators and denominators of 18.6.12 or 18.6.13. Equations 18.6.12 and 18.6.13 have a completely different character from the linearly regularized solutions to 18.5.7 and 18.5.8 . The vectors and matrices in 18.6.12 all have size N the number of measurements. There is no discretization of the underlying variable x so M does not come into play at all. One solves a different N x N set of linear equations for each desired value of x. By contrast in 18.5.8 one solves an M x M linear set but only once. In general the computational burden of repeatedly solving linear systems makes the Backus-Gilbert method unsuitable for other than one-dimensional problems. How does one choose A within the Backus-Gilbert scheme As already mentioned you can in some cases should make the choice before you see any actual data. For a given trial value of A and for a sequence of x s use equation 18.6.12 to calculate q x then use equation 18.6.6 to plot the resolution functions b x x as a function of x . These plots will exhibit the amplitude with which different underlying values x contribute to the point u x of your estimate. For the same value of A also plot the function Varp x using equation 18.6.8 . You need an estimate of your measurement covariance matrix for this. As you change A you will see very explicitly the trade-off between resolution and stability. Pick the value that meets your needs. You can even choose A to be a function of x A A x in equations 18.6.12 and 18.6.13 should you desire to do so. This is one benefit of solving a separate set of equations for each x. For the chosen value or values of A you now have a quantitative understanding of your inverse solution procedure. This can prove invaluable if once you are processing real data you need to .
