How can I fit an exponential curve?

Below is an example of finding a fit with only one term of exponential term but I dont know how to find the fit of the curve when it has 2 degree of exponential term, i.e.[y = a*e^(bx) + c*e^(dx)]

example for y = a*e^(bx)

phi =[ones(size(xx)),xx];
aa=phi\log(yy);
yfit = exp(phi*aa);
plot(xx, yy, ’ro’, xx, yfit, ’k-’) ;
 s=sprintf(’y=%8fexp(%8fx)’,exp(aa(1)),aa(2)); 

Answer 1

[Star Strider]

Try this:

filename1 = 'x2.mat';
D1 = load(filename1);
x = D1.x2;
filename2 = 'y2.mat';
D2 = load(filename2);
y = D2.y2;

fcn = @(b,x) b(1).*exp(b(2).*x) + b(3).*exp(b(4).*x);

[B,fval] = fminsearch(@(b) norm(y - fcn(b,x)), ones(4,1));

figure
plot(x, y, 'p')
hold on
plot(x, fcn(B,x), '-')
hold off
grid

The fitted parameters are:

B =
          1.13024777245481
         -2.75090020576997
         -2.09865110252493
         -5.48051415288241

EDIT — Added plot figure.

[fengsen huang]

Hi,

the code doesnt run, i copy and pasted exactly.

sorry i am new to matlab, please be patient with me :)

[Image Analyst]

Did you see any error messages in the command window? Or absolutely nothing at all happened (impossible I think)?

[Star Strider]

@fengsen huang — That it ‘doesn’t run’ does not give me any useful information. My code ran successfully for me, and yielded acceptable parameter estimates and a good fit to your data. (I am running R2018b, however this code should work for the past several releases, and perhaps for all releases.)

What was the problem?

Did the load not work?

Did my code throw an error? If so, please copy and paste all the red text from your Command Window so I can understand the error.

The fminsearch function is part of core MATLAB, so I know everyone has it.

[Image Analyst]

I get the plot, but when it hits fminsearch(), it says:

Exiting: Maximum number of function evaluations has been exceeded

- increase MaxFunEvals option.

Current function value: 0.854365

The fit does seem reasonable, though perhaps not as good for higher x as in my answer which used fitnlm().

[Star Strider]

I got the same message. I chose to ignore it.

Changing the fminsearch call to:

[B,fval] = fminsearch(@(b) norm(y - fcn(b,x)), [1; -1; -1; -1])

eliminates the message, and gives a slightly better (lower) residual norm, with:

B =
        0.0123997313717212
          6.04789376996529
         -1.14123029526501
         -12.6780841992101

[fengsen huang]

Hi,

it worked now but is it possible to use the method I listed to find a fit.

Also how can I sprintf the equation out on the grapgh so I can see all the coefficient with all the exp.

[Star Strider]

‘it worked now but is it possible to use the method I listed to find a fit.’

Unfortunately, no. This is the sum of two nonlinear exponential functions, so even a transformed linear function will not be an appropriate model.

‘Also how can I sprintf the equation out on the grapgh so I can see all the coefficient with all the exp.’

To print it:

eqn = sprintf('y = %.3f * exp(%.3f * x) %+.3f * exp(%.3f * x)', B)                              % Printed

to display it on the plot:

figure
plot(x, y, 'p')
hold on
plot(x, fcn(B,x), '-', 'LineWidth',1.5)
hold off
grid
text(0.22, -0.42, sprintf('y = %.3f\\cdote^{%.3f\\cdotx} %+.3f\\cdote^{%.3f\\cdotx}', B))       % Displayed On Plot

You can position it anywhere you want.

[fengsen huang]

fcn = @(b,x) b(1).*exp(b(2).*x) + b(3).*exp(b(4).*x);
[B,fval] = fminsearch(@(b) norm(y - fcn(b,x)), ones(4,1))

Hi, can you kindly explain what those 2 lines mean

also how do you know this fit is the most appropriate i.e R^2 value or something?

Thank you so much appreciate it

[Star Strider]

This line:

fcn = @(b,x) b(1).*exp(b(2).*x) + b(3).*exp(b(4).*x);

is the objective function, the expression that describes the function to fit to the data.

This line:

[B,fval] = fminsearch(@(b) norm(y - fcn(b,x)), ones(4,1))

calls the fminsearch function to fit the function to the data. The norm function compares the function output to the data and returns a single scalar value (the square root of the sum of squares of the difference between the function evaluation and the data here), that fminsearch uses. I refer you to the documentation on fminsearch (link) for details on how it works.

The value would be calculated as:

Rsq = 1 - sum((y - fcn(B,x)).^2) / sum((y - mean(y)).^2)

returning:

Rsq =
         0.980214434988184

We are not comparing models, so this is the only statistic available. There are several ways to compare models, a subject much more involved than I will go into here. See any good text on nonlinear parameter estimation for details.

@fengsen huang —

If my Answer helped you solve your problem, please Accept it!

Answer 2

[Image Analyst]

Here's another way using fitnlm(). I get

coefficients =
        0.0124001386786833
          6.04782212479857
         -1.14123018111715
         -12.6780685424329
formulaString =
    'Y = 0.012 * exp(6.048 * X) + -1.141 * exp(-12.678 * X)'
    

Quite a bit different than Star's numbers.

Answer 3

[Arturo Gonzalez]

Per this answer, you can do it with the following matlab code

https://math.stackexchange.com/a/3808325/605948

clear all;
clc;
% get data
dx = 0.001;
x = (dx:dx:1.5)';
y = -1 + 5*exp(0.5*x) + 4*exp(-3*x) + 2*exp(-2*x);
% calculate n integrals of y and n-1 powers of x
n = 3;
iy = zeros(length(x), n);
xp = zeros(length(x), n+1);
iy(:,1) = cumtrapz(x, y);
xp(:,1) = x;
for ii=2:1:n
iy(:, ii) = cumtrapz(x, iy(:, ii-1));
xp(:, ii) = xp(:, ii-1) .* x;
end
xp(:, n+1) = ones(size(x));
% get exponentials lambdas
Y = [iy, xp];
A = pinv(Y)*y;
Ahat = [A(1:n)'; [eye(n-1), zeros(n-1, 1)]];
lambdas = eig(Ahat);
lambdas
% get exponentials multipliers
X = [ones(size(x)), exp(lambdas'.*x)];
P = pinv(X)*y;
P
% show estimate
y_est = X*P;
figure();
plot(x, y); hold on;
plot(x, y_est, 'r--');