Odd results when using lsqnonlin. Please offer advice.

Hello,

I'm new to Matlab and the concepts behind non-linear optimization, so I would like to ensure that have a good understanding of what my code is doing. I have two sets of data x and y and am to use lsqnonlin to fit the following function:

 if x(k) < a 
    y = c*(1.5*(x/a)-0.5*(x(/a)^3);

 else 
    y(k) = c;

To do so, I use the main script:

x = vrgdata(:,1);
v = vrgdata(:,1);
lsqnonlin(@(B) eqn2(B,x,v),[0.2 0.2],x,v)

which calls the function:

function y = eqn2(B,x,v)
a2 = B(1);
c2 = B(2);
for k = 1:length(x)
if x(k) < a2 
    y(k) = c2*(1.5*(x(k)/a2) - 0.5*(x(k)/a2)^3);

else 
    y(k) = c2;
end 
end

When I run the code I do receive an output; however, there are some issues I have with my results. Firstly, I had expected that my function would return to me a vector B containing the coefficients a2 and c2. Secondly, the results that I received were accompanied by warnings that the lengths of the lower and upper bounds were greater than the length of x. I'd really appreciate it if someone could explain to me what's causing these issues and also let me know if I am using the lsqnonlin function improperly.

Answer 1

[Walter Roberson]

lsqnonlin(@(B) eqn2(B,x,v),[0.2 0.2],x,v) is using x and v twice, once passed in by the anonymous function, and once as explicit arguments in the lower-bound and upper-bound parameter positions.

[Vicky]

Once I remove the extraneous x and v located in the lower and upper bound positions, the warnings are gone; however, the results that I receive are still not what I expect. For example the estimates I have above are [0.2 0.2], but when I run my function I simply get [0.2 0]. If replace the estimate with [5 5], I simply get [5 0] in return. Why is this?

[Marc]

It is hard to tell what lsqnonlin is trying to do since you have an if/else within your function. Did you try plotting your data with the regressed results? It is very easy to get "unexpected" results with functions with logic due to discontinuity. Anyway you can show the data you are trying to fit

[Vicky]

This is the plot of my x versus v.

[Torsten]

You forgot to subtract your experimental v-data from the function values. Thus at the moment, your v-values do not influence the computation at all.

Best wishes

Torsten.

[Walter Roberson]

Good catch, Torsten.