lscurve fit problem issue in Money Rivlin model

Two independent variables may require one or two dependent variables, however with only one dependent variable, it will need to be duplicated to two columns (assuming all data are column-oriented) —

fun1=@(a,x1) a(1).*(2*x1-2./x1.^2)+a(2).*(2-2./x1.^3)+a(3).*(6* x1.^2-6*x1-6./x1.^4+6./x1.^3 +6./x1.^2-6);
fun2=@(a,x2) a(1).*(2*x2)+a(2).*(2*x2)+a(3).*(4*x2.^3);
fun12 = @(a,x1x2) [fun1(a,x1x2(:,1)) fun2(a,x1x2(:,2))];
x1x2 = rand(12,2);
y1y2 = rand(12,2);
B0 = rand(3,1);
B = lsqcurvefit(fun12, B0, x1x2, y1y2)                          % Two Dependent Variables
Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
B = 3×1
   -0.0744
    0.0176
   -0.0004
y1y1 = [1 1].*y1y2(:,1);                                        % Duplicate First Column
B = lsqcurvefit(fun12, B0, x1x2, y1y1)                          % One Dependent Variable, Duplicated
Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
B = 3×1
   -0.0865
    0.0210
   -0.0004

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Star Strider 2023-2-20

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I am not certain that you can sum them (amd I am not certain what that means) if they have different numbers of elements. The approach that I use here estimates the parameters for the two functions with the same data and same parameters.

One possibility —

fun1=@(a,x1) a(1).*(2*x1-2./x1.^2)+a(2).*(2-2./x1.^3)+a(3).*(6* x1.^2-6*x1-6./x1.^4+6./x1.^3 +6./x1.^2-6);
fun2=@(a,x2) a(1).*(2*x2)+a(2).*(2*x2)+a(3).*(4*x2.^3);
fun12 = @(a,x1x2) sum([fun1(a,x1x2(:,1)) fun2(a,x1x2(:,2))].^2,2);      % Sum Squares Of Functions
x1x2 = rand(12,2);
y1y1 = rand(12,1);
B0 = rand(3,1);
B = lsqcurvefit(fun12, B0, x1x2, y1y1)                                  % Two Dependent Variables
Local minimum possible.
lsqcurvefit stopped because the size of the current step is less than
the value of the step size tolerance.
B = 3×1
    0.0157
   -0.0014
    0.0000

This computes ‘fun12’ as ‘fun1^2 + fun2^2’. There may be other ways to square them and sum them as well, depending on what you want to do. The order of summing and squaring (summing then squaring or squaring then summing) is important, and produce different results. The summed-and-squared functions, regardless of the order, produce a one-column result, so that requires a single dependent variable to fit.

Star Strider 2023-2-20

在 MATLAB Online 中打开

The functions have to return the same size vectors if you are to work with them as I described earlier.

One option would be to vertically concatenate them instead of horizontally concatenating them —

fun1=@(a,x1) a(1).*(2*x1-2./x1.^2)+a(2).*(2-2./x1.^3)+a(3).*(6* x1.^2-6*x1-6./x1.^4+6./x1.^3 +6./x1.^2-6);
fun2=@(a,x2) a(1).*(2*x2)+a(2).*(2*x2)+a(3).*(4*x2.^3);
% fun12 = @(a,x1x2) sum([fun1(a,x1x2(:,1)); fun2(a,x1x2(:,2))].^2,2);      % Sum Squares Of Functions
fun12 = @(a,x) [fun1(a,x(1:6)); fun2(a,x(7:15))];
x1 = rand(6,1);
x2 = rand(9,1);
y1 = rand(6,1);
y2 = rand(9,1);
x1x2 = [x1; x2];
y1y2 = [y1; y2];
B0 = rand(3,1);
B = lsqcurvefit(fun12, B0, x1x2, y1y2)                                  % Two Dependent Variables
Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
B = 3×1
    1.3650
   -1.0124
    0.0700

The problem with this is that summing and squaring ceases to make sense in this context (since summing and squaring a vector returns a scalar, and that will not work with regressing vector independent and dependent variables). In order to do that, they should have the same sizes, and use the earlier description of ‘fun12’.

The only other option is to truncate the longer independent and dependent vectors to the length of the shorter vectors, and then use the earlier implementation that uses equal-length vectors. (How you create the shortened vectors is entirely up to you.) This is likely the best approach if adding and squaring continue to be required.

The alternative of adding (concatenating) elements of the shorter vectors to equal the lengths of the longer vectors has the undesirable effect of ‘weighting’ the duplicated elements disproportionately, resulting in estimated parameters that do not reflect the actual data.

SATYA PAL 2023-2-23

Thanks a lot Sir

Star Strider 2023-2-23

As always, my pleasure!

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