Hi, i have a problem regarding curve fitting. I have a set of data which is linear, but i want to fit a cos(k*l)^2 to this data and wants to find for which value of (k*l), for which the initial linear part of cosine curve fits my data?

%% Here is my code.

[data]=[0 0 0.050000000000000 1.108646244630E-01 0.100000000000000 2.217423074817E-01 0.150000000000000 3.325947375398E-01 0.200000000000000 4.434863433851E-01 0.250000000000000 5.543595496420E-01 0.300000000000000 6.652338361973E-01 0.350000000000000 7.761094191116E-01 0.400000000000000 8.869865144820E-01 0.450000000000000 9.978653384221E-01 0.500000000000000 1.108746107036E+00]';

x=data(:,1);

y=data(:,2);

k=0.3 %fixed

l=0.01 %variable, can have any value to fit the linear part

f=cos(k*l)^2 % function to fit

I will be highly grateful if anyone can help me on this problem. Thank you very much.

Answer 1

[Rik]

You can use the code below to choose a value for l that results in the least difference between the actual data and the output defined by f. (Also, you should avoid the use of the lowercase L as a variable name, because it looks very similar to a 1 or I (one, uppercase i))

In future, you should use the {}Code button to make sure your code in ready to copy and paste to Matlab.

data=[0  0;
    0.05 1.108646244630E-01;
    0.10 2.217423074817E-01;
    0.15 3.325947375398E-01;
    0.20 4.434863433851E-01;
    0.25 5.543595496420E-01;
    0.30 6.652338361973E-01;
    0.35 7.761094191116E-01;
    0.40 8.869865144820E-01;
    0.45 9.978653384221E-01;
    0.50 1.108746107036E+00];

% Objective function
k=0.3;%fixed
f=@(l,x) cos(k*l)^2.*x;% function to fit
x=data(:,1);
yx=data(:,2);
intial_b_vals=0.01;

% Ordinary Least Squares cost function
OLS = @(b) sum((f(b,x) - yx).^2);
opts = optimset('MaxFunEvals',50000, 'MaxIter',10000);
% Use 'fminsearch' to minimise the 'OLS' function
fitted_b = fminsearch(OLS, intial_b_vals, opts);

For me, this results in l having an optimal value of 0.

[Sumera Yamin]

Dear Rik, Thank you very much for your answer and for the tip to copy paste the ready to use code. Can you please explain one question regarding your code. What i understand is, when you define the objective function, you have used x values of data in the argument of cosine function. Is this correct? In reality my cosine function is independent of the data. Or to phrase in other words, i want to calibrate my cosine function (to find correct value of "k*l" for which my data fits the initial almost linear part of cosine curve). Please explain if i understood the code correctly.

[Rik]

You understood my code correctly.

You can't fit f=cos(k*l)^2 to an x-y curve, because they are unrelated. How do f, k and l relate to x and y? I can't tell you what is the optimal number of butterflies in Australia if you want to buy a laptop in Austria. If they don't share the same coordinate system, you can't fit it.

[Sumera Yamin]

Sorry for the misunderstanding, you are right. They are indeed related by y=cos(k*l)^2.x So knowing x and y, i want to find out the value of "k*l" (with k being fixed and l variable) which fits the data.

[Rik]

See my edited code. The optimal value will be 0, because cos(0) is 1, which results in a perfect fit. You can edit the function to be fitted yourself, as long as your only inputs are the value to be fitted and the x-values, you can enter anything you like. That fit-value is allowed to be a vector or matrix, as long as the output is a scalar and it works with an array (because the cost function enters the entire x-vector at once).

[Sumera Yamin]

Thank you for the editing code. I have two questions

1) You define @(b) as the difference of original data values and f(l,x) and evaluate the minimum (which is zero ofcourse for cosine function). This is fine but 'k' and 'l' are real physical quantities and they cannot be zero.

Also please note that should the following line in your code have f(l,x) instead of f(b,x)?

"OLS = @(b) sum((f(b,x) - yx).^2);" as we have not define f (b,x) before but using f(b,x) gives error. Please explain this line.

2) Also, where can we set the limit on the max value of 'l'. Does 'MaxFunEvals' does this job or we have to set 'TolFun' for it?

[Rik]

You need to read the doc for anonymous functions. Those should cover most of your question here. Setting a limit on the max value of l will not help, because it will go to 0 (unless you set a negative value as the max). Setting a constraint is not possible with this function, but you can add a term that sets the result to infinity:

OLS_con = @(b) sum((f(b,x) - yx).^2)+...
    1/(b>0.1)-1;%set cost function to inf when (b>0.1) is false

[Sumera Yamin]

Can we somehow put the limit that the argument of cosine function should be greater than 0 and less than pi/2 (0<k*l<pi/2) so that only only that portion of the curve which is of interest is considered.

[Rik]

Same idea as above:

OLS_con = @(b) sum((f(b,x) - yx).^2)+...
    1/(b>0.1)-1+...
    1/(0<k*b & k*b<pi/2)