How can I choose powers of each variable separately and get it's value?

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behzad
behzad 2014-5-23
编辑： Matt J 2014-5-23
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?
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behzad
behzad 2014-5-23
编辑：behzad 2014-5-23
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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回答（2 个）

Matt J
Matt J 2014-5-23
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
Star Strider
Star Strider 2014-5-23
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
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Matt J
Matt J 2014-5-23
编辑：Matt J 2014-5-23
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 2014-5-23
Noted. I’ve only done minimisation or least-squares curve fitting with fminsearch, so never encountered that.

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