Using least square method , determining unknown values ?

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osman
osman 2011-12-30
Hello, I have a question which i couldn't solve. Please help me.
The table shows the experimental results of the measured Coulomb Force, F, between two charges, (q1 and q2) corresponding to distance r. General form of the Coulomb Force is:
F = (1/4*pi*ε0)*q1*q2/(r^n)
Write a program to determine values of ε0 and n using least-square method.
r(cm) | F (N)
-------+------------
40 | 2.5 +- 0.1
30 | 4.5 +- 0.1
20 | 10.1 +- 0.2
10 | 40.5 +- 0.4
5 |162.0 +- 0.9
q1 = 5 μC q2 = 9 μC
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Wayne King
Wayne King 2011-12-30
What have you done to solve your homework problem?
Show the MATLAB code you have written and any error messages you are getting.
osman
osman 2011-12-30
I used curve fitting tool to determine n and epsilon and i did but i couldn't write a program determining it. So i can't show any program . Any clue about writing this program would be appreciated.

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Friedrich
Friedrich 2011-12-30
Hi,
is this even doable with normal least square? This is a nonlinear problem. Normally you would try to write the problem as
Ax = b
and than do
(A'A)^-1 A'b
But in your case you can't write the problem in that form since you have it like this
(1/4*pi*q1*q2) * (ε0/r^n) = F
Where r changed from equation to equation. You would need to use the logarithm to get the n but than the ε0 gets into a logarithm statement.
You would need to get it in a form which has a form like this
A * [ε0;n] = F
(sorry for the bad formatting, but writing formulas is pretty bad on answers). Where A is a matrix and F a vector.
I think one can't split ε0 and n up in such a way. So this seem that some algorithm for a non linear problem must be used here.
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Friedrich
Friedrich 2011-12-30
I never stated that n and ε0 change. But with a small but not very good program (used optimization toolbox) I get ε0 = 114.6834 n = 2.0005 as coeffs. The result strongly depends on the start values.
I created a function which depends on ε0 and n in which I calculate ||f(ε0,n) -F||.
function err = my_func( x )
%x(1) = ε0
%x(2) = n;
fix = 1/4*pi*5*9*x(1); %1/4*pi*q1*q2*ε0
%calc f - F
f_F = [ fix/40^x(2) - 2.5
fix/30^x(2) - 4.5
fix/20^x(2) - 10.1
fix/10^x(2) - 40.5
fix/5^x(2) - 162];
err = norm(f_F);
end
Than I used fminunc to calculate the minimum:
%guess 150 for ε0 and 1 for n
[coeefs,fval] = fminunc(@my_func,[150,1])
%test if the values are okay
disp([num2str(1/4*pi*5*9*coeefs(1)/40^coeefs(2)) ' = 2.5'])
disp([num2str(1/4*pi*5*9*coeefs(1)/30^coeefs(2)) ' = 4.5'])
disp([num2str(1/4*pi*5*9*coeefs(1)/20^coeefs(2)) ' = 10.1'])
disp([num2str(1/4*pi*5*9*coeefs(1)/10^coeefs(2)) ' = 40.5'])
disp([num2str(1/4*pi*5*9*coeefs(1)/5^coeefs(2)) ' = 162'])
Which leads to a warning of the solver and these results:
coeefs =
114.6834 2.0005
fval =
0.0368
2.5287 = 2.5
4.4961 = 4.5
10.1182 = 10.1
40.4867 = 40.5
162.0019 = 162
Friedrich
Friedrich 2011-12-30
if you have the global optimization toolbox you can use the follow code where the startpoint doesn't matter:
rs = RandomStartPointSet('NumStartPoints',1000); %generate 1000 startpoints
options = optimset('TolFun',1e-16,'TolX',1e-16,'MaxFunEvals',1000,'MaxIter',1000);
problem = createOptimProblem('fminunc','objective',@my_func,'x0',[1,100],'options',options) %choose x0 which is far from the actual solution
[xmin,fmin,flag,outpt,allmins] = run(MultiStart,problem,rs)

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