How to make a good fittings

x =

   1.0e-08 *

         0
    0.0430
    0.0770
    0.1061
    0.1309
    0.1522
    0.1705
    0.1918
    0.2123
    0.2302
    0.2471
    0.2611
    0.2743
    0.2852
    0.2962
    0.3064
    0.3170
    0.3271
    0.3386
    0.3484
    0.3609
    0.3715
    0.3801
    0.3789
y =

   1.0e-05 *

         0
    0.0008
    0.0031
    0.0063
    0.0107
    0.0159
    0.0214
    0.0263
    0.0315
    0.0372
    0.0432
    0.0498
    0.0564
    0.0631
    0.0698
    0.0769
    0.0839
    0.0910
    0.0979
    0.1054
    0.1124
    0.1199
    0.1275
    0.1347
    
g=fittype(@(a,x) 4/3.*a.*(10.^(-9)*10).^0.5.*x.^(3/2),'coefficient',{'a'},'independent',{'x'},'dependent',{'y'});
myfit=fit(x,y,g);
plot(myfit,x,y);

These are my codes for fitting my data with the function of a*x^2/3 finding the coefficient 'a'.

However,

the results are gone wrong. My fitting function (Orange) looks really deviated from the data points (Blue).

'a' coefficient should be about 30*10^10 value to be overlapped with each other like the below image.

Could you give me the advice to improve my fitting codes?

Thanks.

Best regards,

qwertypo.

Answer 1

[Mathieu NOE]

hello

sorry I don't have the curve fitting toolbox but such a simple fit can be easily done with fminsearch

the constant you are looking for is

a_sol = 3.5829e+10

clear;
x = 1.0e-08 *[0
    0.0430
    0.0770
    0.1061
    0.1309
    0.1522
    0.1705
    0.1918
    0.2123
    0.2302
    0.2471
    0.2611
    0.2743
    0.2852
    0.2962
    0.3064
    0.3170
    0.3271
    0.3386
    0.3484
    0.3609
    0.3715
    0.3801
    0.3789];

y =   1.0e-05 *[0
    0.0008
    0.0031
    0.0063
    0.0107
    0.0159
    0.0214
    0.0263
    0.0315
    0.0372
    0.0432
    0.0498
    0.0564
    0.0631
    0.0698
    0.0769
    0.0839
    0.0910
    0.0979
    0.1054
    0.1124
    0.1199
    0.1275
    0.1347];

const = 4/3.*(10.^(-9)*10).^0.5;

% equation model y = a*const*x^(3/2)
f = @(a,x) (a*const*x.^(3/2));
obj_fun = @(params) norm(f(params(1), x)-y);

% IC guessed 
a_ic = (y(end)/(const*x(end).^(3/2)))

a_ic = 4.3315e+10

sol = fminsearch(obj_fun, a_ic);
a_sol = sol(1)

a_sol = 3.5829e+10

y_fit = f(a_sol, x);

Rsquared = my_Rsquared_coeff(y,y_fit); % correlation coefficient

figure(1)
plot(x,y,'rd',x,y_fit,'b-');
title(['Power Fit / R² = ' num2str(Rsquared) ], 'FontSize', 15)
ylabel('y', 'FontSize', 14)
xlabel('x', 'FontSize', 14)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Rsquared = my_Rsquared_coeff(data,data_fit)
    % R² correlation coefficient computation
    
    % The total sum of squares
    sum_of_squares = sum((data-mean(data)).^2);
    
    % The sum of squares of residuals, also called the residual sum of squares:
    sum_of_squares_of_residuals = sum((data-data_fit).^2);
    
    % definition of the coefficient of correlation is
    Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;
    
    
end

[Mathieu NOE]

seems to me the data matches a x^(5/2) law better than a x^(3/2) law

clear;
x = 1.0e-08 *[0
    0.0430
    0.0770
    0.1061
    0.1309
    0.1522
    0.1705
    0.1918
    0.2123
    0.2302
    0.2471
    0.2611
    0.2743
    0.2852
    0.2962
    0.3064
    0.3170
    0.3271
    0.3386
    0.3484
    0.3609
    0.3715
    0.3801
    0.3789];

y =   1.0e-05 *[0
    0.0008
    0.0031
    0.0063
    0.0107
    0.0159
    0.0214
    0.0263
    0.0315
    0.0372
    0.0432
    0.0498
    0.0564
    0.0631
    0.0698
    0.0769
    0.0839
    0.0910
    0.0979
    0.1054
    0.1124
    0.1199
    0.1275
    0.1347];

const = 4/3.*(10.^(-9)*10).^0.5;

% equation model y = a*const*x^(3/2)
% f = @(a,x) (a*const*x.^(3/2));
f = @(a,x) (a*const*x.^(5/2));
obj_fun = @(params) norm(f(params(1), x)-y);

% IC guessed 
% a_ic = (y(end)/(const*x(end).^(3/2)))
a_ic = (y(end)/(const*x(end).^(5/2)))

a_ic = 1.1432e+19

sol = fminsearch(obj_fun, a_ic);
a_sol = sol(1)

a_sol = 1.0981e+19

y_fit = f(a_sol, x);

Rsquared = my_Rsquared_coeff(y,y_fit); % correlation coefficient

figure(1)
plot(x,y,'rd',x,y_fit,'b-');
title(['Power Fit / R² = ' num2str(Rsquared) ], 'FontSize', 15)
ylabel('y', 'FontSize', 14)
xlabel('x', 'FontSize', 14)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Rsquared = my_Rsquared_coeff(data,data_fit)
    % R² correlation coefficient computation
    
    % The total sum of squares
    sum_of_squares = sum((data-mean(data)).^2);
    
    % The sum of squares of residuals, also called the residual sum of squares:
    sum_of_squares_of_residuals = sum((data-data_fit).^2);
    
    % definition of the coefficient of correlation is
    Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;
    
    
end

[Youngki Yeo]

I really appreciated your help. Thanks! However, I still wonder that what is wrong with my codes. Could you give me some explanation why my codes gone wrong?

[Mathieu NOE]

hello

thanks for accepting my answer !

As I said before, unfortunately I don't have (and practice) the Curve Fitting Toolbox

maybe someone else in this forum will be keen to answer this specific topic.

[Youngki Yeo]

Hi, Mathieu.

Okay. I see. Thanks!