A problem while using lsqnonlin

I want to find the parameter x using nonlinear solver lsqnonlin.

I think the two functions are the same, but ErrorFunc2 doesn't seem to work properly with the optimization process.

Can you tell me what the problem is?

d = linspace(0,3);
y = exp(-1.3*d) + 0.05*randn(size(d)); 

ErrorFunc = @(r)exp(-d*r)-y;


x0 = 4;
x = lsqnonlin(ErrorFunc,x0);
x2 = lsqnonlin(@ErrorFunc2,x0);

figure()
hold on
plot(d,y,'ko')              % DATA
plot(d,exp(-x*d),'b-')      % BEST FIT
plot(d,exp(-x2*d))      % BEST FIT
legend('Data','x1 fit', 'x2 fit')
xlabel('t')
ylabel('exp(-tx)')


function [error] = ErrorFunc2(x)

  d = linspace(0,3);
  y = exp(-1.3*d) + 0.05*randn(size(d));  

  error = exp(-d*x) -y;

end

Answer 1

[Stephan]

Use the same random numbers for the same results:

rng('default')
d = linspace(0,3);
y = exp(-1.3*d) + 0.05*randn(size(d)); 

ErrorFunc = @(r)exp(-d*r)-y;


x0 = 4;
x = lsqnonlin(ErrorFunc,x0);
x2 = lsqnonlin(@ErrorFunc2,x0);

figure()
hold on
plot(d,y,'ko')              % DATA
plot(d,exp(-x*d),'b-')      % BEST FIT
plot(d,exp(-x2*d),'r--')      % BEST FIT
legend('Data','x1 fit', 'x2 fit')
xlabel('t')
ylabel('exp(-tx)')


function [error] = ErrorFunc2(x)
  rng('default')
  d = linspace(0,3);
  y = exp(-1.3*d) + 0.05*randn(size(d));  

  error = exp(-d*x) -y;

end

Another approach would be to generate random numbers only one time and use them as input to func2:

d = linspace(0,3);
y = exp(-1.3*d) + 0.05*randn(size(d));
ErrorFunc = @(r)exp(-d*r)-y;

x0 = 4;
x = lsqnonlin(ErrorFunc,x0);
x2 = lsqnonlin(@(x)ErrorFunc2(x,y),x0);

figure()
hold on
plot(d,y,'ko')              % DATA
plot(d,exp(-x*d),'b-')      % BEST FIT
plot(d,exp(-x2*d),'r--')      % BEST FIT
legend('Data','x1 fit', 'x2 fit')
xlabel('t')
ylabel('exp(-tx)')

function [error] = ErrorFunc2(x,y)
d = linspace(0,3);
%   y = exp(-1.3*d) + 0.05*randn(size(d));
error = exp(-d*x) -y;
end

[Caritas]

I thought the original expression exp(-1.3*x) would be the final result even if the random numbers (noise) were different, but it was wrong. Thank you for your answer!