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 Comments
behzad
behzad on 23 May 2014
Edited: behzad on 23 May 2014
No. It was just an example.I want to do a numerical integration process using guass quadrature using equation ∫_A▒〖B_i ξ_1^(a_i ) 〗 ξ_2^(b_i ) ξ_3^(c_i )=B_i (a_i !b_i !c_i !)/((a+b+c+2)!) but the and I have a long polynomial including lots of these integrals.I want matlab to divide each term seperately term by term and identify the power constants ai,bi,ci in each term so I will be able to exert them into the right side of equation.

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Answers (2)

Matt J
Matt J on 23 May 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 Comment
Matt J
Matt J on 23 May 2014
Edited: Matt J on 23 May 2014
You can also use the linearization to develop an initial guess for the fminsearch approach (See Star Strider's answer).

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Star Strider
Star Strider on 23 May 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 Comments
Matt J
Matt J on 23 May 2014
Edited: Matt J on 23 May 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
Star Strider on 23 May 2014
Noted. I’ve only done minimisation or least-squares curve fitting with fminsearch, so never encountered that.

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