How can I choose powers of each variable separately and get it's value?
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Hi every one I solved a equation leading to function like this f=x^a*y^b*z^c; but I don't know a,b and c.I want matlab to show me these power constants but have trouble making matlab select them. How can I choose powers of each variable separately and get it's value?
3 Commenti
Risposte (2)
Matt J
il 23 Mag 2014
If there is no error in the equations, you can take logs and turn it into a linear equation in a,b, and c
logf=a*logx+b*logy+c*logz
1 Commento
Matt J
il 23 Mag 2014
Modificato: Matt J
il 23 Mag 2014
You can also use the linearization to develop an initial guess for the fminsearch approach (See Star Strider's answer).
Star Strider
il 23 Mag 2014
You can do a nonlinear fit with fminsearch:
f = @(p,x,y,z) x.^p(1) .* y.^p(2) .* z.^p(3); % Function
x = 3; % Define variables
y = 5;
z = 7;
s = 13; % Define f(x,y,z,) = s
objfcn = @(p) f(p,x,y,z) - s; % Objective function = 0 when f(x,y,z,) = s
[b, fval] = fminsearch(objfcn, [1 1 1])
produces:
b =
11.3000 -3.0500 -25.5000
fval =
-13.0000
4 Commenti
Matt J
il 23 Mag 2014
Modificato: Matt J
il 23 Mag 2014
If you don't include the norm(), your error function is signed. fminsearch will then look for the 'p' that makes the error function as negative as possible.
As an example, consider the alternative data x=13,y=1,z=1, s=13. With
objfcn = @(p) f(p,x,y,z) - s;
I obtain the false solution,
b =
-18.8889 6.2222 5.4278
fval =
-13
whereas when the norm() is included, fminsearch correctly detects that the initial point [1 1 1] is a solution,
b =
1 1 1
fval =
0
Star Strider
il 23 Mag 2014
Noted. I’ve only done minimisation or least-squares curve fitting with fminsearch, so never encountered that.
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