Lsqnonlin to determine coefficient

I am looking to fit data to a complex equation using lsqnonlin solver. I am not sure where I am going wrong. I keep getting a response "not enough input arguments"

% Approach
% 1. Creating the following function named fit_simp.m which uses the time and Ar data(Ar = At/Ainf). 
% 2. time and Ar are passed into lsqnonlin as input arguments. 
% 3. Use the time and n data to calculate values values (Ar) the diffusion equation, and subtract the original Ar data from this. 
% 4. The result will be the difference between the experimental data and the calculated values.
% 5. The lsqnonlin function will minimize the sum of the squares of the differences.
% 6. Condensation of the diffusion equation: 
            % a=0.0008;
            % n1 = 2.43;
            % n2= 1.4;
            % Lambda = 950;
            % theta = 45; 
            % gama = (2*n1*pi*sqrt ((sin(theta))^2-(n2/n1)^2))/(Lambda)
% Above entered in: 1-(8*gama/pi*(1-exp(-2*gama*a)))*((exp(-D*(2*n+1)^2*pi^2*t/4*a^2))*((-1)^n*2*gama +(((2*n+1)*pi)/(2*a))*exp(-2*gama*a))/((2*n+1)*((4*gama^2)+((2*n+1)*pi/2*a)^2)));

function diff = fit_simp(x,~,~) % This function is called by lsqnonlin.
                                    % x is a vector which contains the coefficient of the equation. 
                                    % time and Ar are the option data sets that are passed to lsqnonlin
        
 % Defining the data sets that you are trying to fit the function to.
Ar = [0.3 0.2 0.28 0.318 0.421 0.492 0.572 0.55 0.63 0.61 0.73 0.8 0.81 0.84 0.93 0.91]';
l = length(Ar);
t = [0:l-1]';
plot(t,Ar,'ro')
title('Data points')

% D is the coefficient we are looking to determine

%*******************************************************************************
a=0.0008;
n1 = 2.43;
n2=1.4;
Lambda = 950;
theta = 45;
d= x(1);
    gama = (2*n1*3.14*sqrt ((sin(theta))^2-(n2/n1)^2))/(Lambda);
     time=12;
     
    for t = 1:time
  
        for  n=0:time
r = 1-((8*gama/pi)*(1-exp(-2*gama*a)))*((exp(-d.*(2*n+1)^2*pi^2*t/4*a^2))*((-1)^n*2*gama +(((2*n+1)*pi)/(2*a))*exp(-2*gama*a))/((2*n+1)*((4*gama^2)+((2*n+1)*pi/2*a)^2)));
       
        end
         result(t) = r;
    end
%*******************************

diff = result - Ar;


  
% Initialize the coefficients of the function.
X0=[1];

% Calculate the new coefficients using LSQNONLIN.
x=lsqnonlin(@fit_simp,X0,t,Ar);


% Plot the original and experimental data.


    for t = 1:time
  
        for  n=0:time
Ar_new = 1-(8*gama/pi*(1-exp(-2*gama*a)))*((exp(-x(1).*(2*n+1)^2*pi^2*t/4*a^2))*((-1)^n*2*gama +(((2*n+1)*pi)/(2*a))*exp(-2*gama*a))/((2*n+1)*((4*gama^2)+((2*n+1)*pi/2*a)^2)))
        end
        Ar_newv(t)=Ar_new;
    end
plot(time,Ar,'+r',time,Ar_newv,'b')
 
 

[Star Strider]

What is the code supposed to do?

If you want to fit data, use lsqcurvefit. It takes care of all the details.

If you want to fit data while integrating and fitting parameters to a differential equation, that is also relatively straightforward.

.

[Anand Ra]

Thanks for the response.

The code is suppose to spit out the value for the coefficient 'd' as a result of best fit of the data to the given equation.

My problem: Fitting a set of data ( array) to an equation to derive the coefficient(d) in the equation.

Should I use lscurvefit for my problem?

[Star Strider]

‘Should I use lscurvefit for my problem?’

I would. It’s simply easier.

[Matt J]

@Anand Rathnam Note that your loop

        for  n=0:time
            r = 1-((8*gama/pi)*...
        end

is not doing anything except repeatedly over-writing r. It is not updating r in any way. You have a similar loop later in your posted code with the same problem.