Curve fitting lsqnonlin Error? Optimization

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Jay 2019-4-9

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https://ww2.mathworks.cn/matlabcentral/answers/455276-curve-fitting-lsqnonlin-error-optimization

评论： Jay 2019-4-26

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Hello

I have done the curve fitting for the customized equation using the following method:

function res = FUN(x,xdata,ydata,zdata)
n  = 8
% A  = ydata
% B  = -xdata
% C  = xdata. - ydata
% H  = zdata
a1 = x(1)*ydata - x(3)*(xdata - ydata)
b1 = x(2)*(-xdata) - x(1)*(ydata)
c1 = x(3)*(xdata - ydata) - x(2)*(-xdata)
h1 = x(4)*(zdata)
I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6))
I2 = ((h1.^2)/(3))+((a1.^2 + b1.^2 + c1.^2)/(54))
th = acos(I3/(I2.^(3/2)))
v1 = ((2*th)+pi)/6
an1 = (abs(2*cos(v1)))^n
an2 = (abs(2*cos((2*th+3*pi)/6)))^n
an3 = (abs(2*cos((2*th+5*pi)/6)))^n
res = ((3*I2).^(n/2)) * (an1 + an2 + an3) - (2*(189.32)^8);
end
x_100 = [204.31,0,31.59,181.02,219.97,191.86,0,-31.811,-179.81,-202.56,-218.67,-191.13];
y_100 = [0,-203.68,-181.29,-33.38,193.82,222.11,205.42,182.49,33.18,0,-193.06,-220.79];
zdata = [0,0,0,0,0,0,0,0,0,0,0,0];
x0 = [1 1 1 0];      % initial values for a, b and c
[x] = lsqnonlin(@(x)FUN(x,x_100,y_100,zdata),x0)

I got the result as 0.8535 0.9247 0.9893

Then I plotted the same equation using the obtained coefficients using fimplicit and it looks like:

If I am using lsqnonlin, the implicit equation should pass through the points, but there is a gap between points and implicit equation.

Could anyone answer this, why the implicit equation is not passing through the points though I calculated the coefficients of the function using curve fitting with these points?

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Matt J 2019-4-26

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The code you've posted fails with

Matrix dimensions must agree.
Error in test>FUN (line 18)
I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6))
Error in test>@(x)FUN(x,x_100,y_100,zdata)
Error in lsqnonlin (line 206)
            initVals.F = feval(funfcn{3},xCurrent,varargin{:});
Error in test (line 6)
[x,r1,r2,exitflag] = lsqnonlin(@(x)FUN(x,x_100,y_100,zdata),x0)
Caused by:
    Failure in initial objective function evaluation. LSQNONLIN cannot continue.