How do I fit an exponential curve fit function using fminsearch. Program runs, but result is incorrect.

Attached is my code

yfirst=[.8967e07 1.6294e07 2.6587e07 3.2537e07 3.6136e07 3.8419e07 3.9706e07]'; 
xfirst=[1,2,4,6,8,10,15];

[estimates, model] = myfun(xfirst,yfirst);
[sse, FittedCurve] = model(estimates);

semilogx(xfirst,FittedCurve,'-g*'); hold on;
semilogx(xfirst,yfirst,'-rd');hold off;
 xlim([ 0 50 ]);

%%%%FUNCTION FILE 
function [estimates, model] = myfun(xdata, ydata)
% Call fminsearch with guessed starting point.
% start_point =[2.23e7;.005];
model = @expfun;
estimates = fminsearch(model,start_point);

    function [sse, FittedCurve] = expfun(params)
        A=params(1)
        lambda=params(2)
        FittedCurve =(A .* exp(lambda * xdata));
        ErrorVector = FittedCurve - ydata;
        sse = sum(ErrorVector .^ 2);
    end
end

I get a large error and the curve does not match when plotted together. Thanks

Answer 1

[Star Strider]

There are several problems with your code.

This works:

yfirst=[.8967e07 1.6294e07 2.6587e07 3.2537e07 3.6136e07 3.8419e07 3.9706e07]'; 
xfirst=[1,2,4,6,8,10,15]';

expfun = @(b,xdata) b(1) -b(2) .* exp(b(3) .* xdata);             % Objective Funciton

SSECF = @(b) sum((yfirst - expfun(b,xfirst)).^2);           % Sum-Squared-Error Cost Function
start_point =[4E+7; 2.23e7; -.005];
[B, SSE] = fminsearch(SSECF, start_point);

figure(1)
plot(xfirst, yfirst, 'bp')
hold on
plot(xfirst, expfun(B,xfirst), '-r')
hold off
grid
text(5.2, 1.75E+7, sprintf('f(x) = %9.2E - %9.2E\\cdote^{%9.2E\\cdotx}', B))

[carenar]

Do necessarily need the third constant term in the front?

[Star Strider]

In this instance (that is with your data), you do. It is an asymptotically-increasing exponential, so you have to have parameters for the asymptote, amplitude, and exponential rate. The integrated differential equation for the process that created your data would require values for all three parameters. (With a simple decaying exponential, you would only need parameters for the initial value and exponential rate.)