Nonlinear regression using lsqcurvefit

Here is my code, along with my error message.

T = V22050100Emat;
A = table2array(T);

f = A(389:402, 6);
y = A(389:402, 5);
% axis([0 1010 0 1.6E-7])
% hold on
plot(f,y,'ro')
title('Data points')
% hold off




F = @(x,f)(x(1)./2)*(sin(x(2).*pi/2)./(cosh(x(2).*log(2*pi*f*x(3).).)+cos(x(2).*pi/2)));
x0 = [1e-6 0.16 0.04];

size(f)
size(F)
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,f,y);

hold on
plot(f,F(x,f))
hold off

My error message

Error using lsqcurvefit
Function value and YDATA sizes are not equal.

Error in nonlinfit (line 20)
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,f,y);
 

I have also tried the 'fit' function with mixed results. Thank you.

Answer 1

[Torsten]

F = @(x,f) x(1)/2 * sin(x(2)*pi/2)./(cosh(x(2)*log(2*pi*f*x(3)))+cos(x(2)*pi/2));

instead of

F = @(x,f)(x(1)./2)*(sin(x(2).*pi/2)./(cosh(x(2).*log(2*pi*f*x(3).).)+cos(x(2).*pi/2)));

What do you get if you type

size(f)
size(y)

?

Does f and/or y have Inf or NaN values ?

[Jason Yee]

f and y are finite arrays of number values. With your suggestions i get

f = A(389:402, 6);
y = A(389:402, 5);
% axis([0 1010 0 1.6E-7])
% hold on
plot(f,y,'ro')
title('Data points')
% hold off

size(f)
size(y)


F = @(x,f)(x(1)./2)*(sin(x(2).*pi/2)./(cosh(x(2).*log(2*pi*f*x(3).).)+cos(x(2).*pi/2)));
x0 = [1e-6 0.16 0.04];


size(F)
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,f,y);

hold on
plot(f,F(x,f))
hold off

and an output with error message

ans =

    14     1


ans =

    14     1


ans =

     1     1

Error using lsqcurvefit
Function value and YDATA sizes are not equal.

Error in nonlinfit (line 21)
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,f,y);
 
>> nonlinfit
File: nonlinfit.m Line: 16 Column: 67
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check
for mismatched delimiters.
 

Thank you.

[Torsten]

Why do you still use the faulty F definition although I provided the correct one ?

f = [3;4];
F = @(x,f)(x(1)./2)*(sin(x(2).*pi/2)./(cosh(x(2).*log(2*pi*f*x(3).).)+cos(x(2).*pi/2)));

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

F([1 4 7],f)

[Jason Yee]

My bad, i missread you answer. I now have this

T = V22050100Emat;
A = table2array(T);

f = A(389:402, 6);
y = A(389:402, 5);
% axis([0 1010 0 1.6E-7])
% hold on
plot(f,y,'ro')
title('Data points')
% hold off

size(f)
size(y)


F = @(x,f) x(1)/2 * sin(x(2)*pi/2)./(cosh(x(2)*log(2*pi*f*x(3)))+cos(x(2)*pi/2));
x0 = [1e-6 0.63 9e-5];


size(F)
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,f,y);

hold on
plot(f,F(x,f))
hold off

My output looks like this

Initial point is a local minimum.

Optimization completed because the size of the gradient at the initial point 
is less than the value of the optimality tolerance.

<stopping criteria details>

Optimization completed: The final point is the initial point.
The first-order optimality measure, 2.210162e-07, is less than
options.OptimalityTolerance = 1.000000e-06.

My thinking is to decrease my optimality tolerance.

[Torsten]

Tighten the tolerances (FunctionTolerance and StepTolerance) in the options structure for lsqcurvefit.

And/or multiply F by a large number.

[Jason Yee]

Thank you. This has solved my problem.