Numerit[WIN32][1700][1703]Q qffffff)@j@fffffvq@?@ffffff9@?@ffffff9@ffffff)@ffffff)@         Times New RomanArialSymbol Courier New G Ea@\(V@yxdatayyxxfit>@ >@>@>@>@>@>@ >@ >@ >@>@>@>@>@>@ >@  >==>>??0.010.010.010.01T MH@!""""@@c Data coeff.b Fit coeff.vvvpvvvpvvvpvvvpvvvpvvvpvvv  vvv !  q8ffffff)@j@fffffvq@?@ffffff9@?@ffffff9@ffffff)@ffffff)@         Times New RomanArialSymbol Courier New 9` This sample program demonstrates a general linear least:` squares fit by fitting a set of basis functions to a set` of data points.&` -> Run and see results in the Report` Define a set of points x = -10 to 10%c = 1,2.2,-0.5,-3,1.5 ` coefficientsFy = c[1]+c[2]*cos(x)+c[3]*sin(x)+c[4]*cos(2*x)+c[5]*sin(2*x)+rand(x/4).` (remove the random term to get an exact fit) ` Dimensions of the basis matrix m = length(x)*n = 5 ` number of desired basis functions bb[m,n] = 0H` Define the basis matrix (each column corresponds to a basis function).E` Note that the basis matrix is defined specifically for the given x. bb[*,1] = 1bb[*,2] = cos(x)bb[*,3] = sin(x)bb[*,4] = cos(2*x)bb[*,5] = sin(2*x);` Find the coefficients that fit the set of points in (x,y)3` (remember that the x-dependency is already in bb)b = fitbasis(bb,y)` Define the fitxx = -10..10 ` 101 pointsAyy = b[1]+b[2]*cos(xx)+b[3]*sin(xx)+b[4]*cos(2*xx)+b[5]*sin(2*xx)d:\num\num1.5\release\pro\  x c y mnbbb xx yy   G e   G G 5 4 4 B@ 4 B@ 4  BB@ 4  BB@ C@    4  4  4  4  4  B 4  B   G d 4 4 B@ 4 B@ 4  BB@ 4  BB@ 10$@1?2.2@0.5?3@1.5?2@4@5@0