Problem 258. linear least squares fitting

Solution 188736

Submitted on 9 Jan 2013 by Alfonso Nieto-Castanon

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Test	Status	Code Input and Output
1	Pass	%%% first test: fit to a constant x = [1,2,3,4]'; y = rand(4,1); f{1} = @(x) ones(size(x)); aref=mean(y); assert(norm(fit_coefficients(f,x,y)-aref)<1e-6)
2	Pass	%%% second test: fit to a straight line (linear regression) x = [1,2,3,4,5]' + randn(5,1); y = [1,2,3,4,5]' + randn(5,1); f{1} = @(x) ones(size(x)); f{2} = @(x) x; aref(2) = sum((x-mean(x)).(y-mean(y)))/sum((x-mean(x)).^2); aref(1) = mean(y)-aref(2)mean(x); assert(norm(fit_coefficients(f,x,y)-aref')<1e-6)
3	Pass	%%% third test: polynomial fit x = [1:10]' + randn(10,1); y = -10+0.2x-0.5x.^2+0.4x.^3+0.001log(x) + 0.2*randn(10,1); f{1} = @(x) ones(size(x)); f{2} = @(x) x; f{3} = @(x) x.^2; f{4} = @(x) x.^3; aref = fliplr(polyfit(x,y,3)); assert(norm(fit_coefficients(f,x,y)-aref')<1e-6)
4	Pass	%%% fourth test: non-polynomial fit (yes, we are that crazy) x = [0:0.1:2pi]'; y = 0.123 + 0.456sin(x).exp(0.1x); f{1} = @(x) ones(size(x)); f{2} = @(x) sin(x).exp(0.1x); aref=[0.123 0.456]'; assert(norm(fit_coefficients(f,x,y)-aref)<1e-6)

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