How to obtain regression polynomial equation with more than 2 independent variables with degree 5.

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Beh Poi Keat
Beh Poi Keat 2020-3-31
回答： David Wilson 2020-4-4
How to obtain regression polynomial equation with more than 2 independent variables with degree 5, because curve fitting tool in MATLAB only support 2 independent variables.
I want the equation with 3 independent variable with coefficient provided, also plot the graph.
Anyone can help me ? Thank you in advance.
  2 个评论
David Wilson
David Wilson 2020-3-31
Write out the Vandermonde matrix columns by hand, (you'll have quite a lot of them if using quintics), and then use backslash to get the polynomial coefficients.
Beh Poi Keat
Beh Poi Keat 2020-3-31
Hi, can you show one example ? I had no idea how to do it as if my equation need to have 3 independent variable. How can i know the equation ?
Thank you in advance.

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采纳的回答

David Wilson
David Wilson 2020-4-4
OK, the fitting is easy, the plotting less so. Since you didn't give us too many hints, not did you give us any data to work with, I'll make up my own.
Let's start with a fake 3D equation, where we know the true coefficients, and then try to recover them. I'll use something simple like,
Here you can see that my coefficients are, in the order given above, p = [1, -3, 1, -1]. If we get these back again, we know we are on the right track.
fun3D = @(x,y,z) x.^2 - 3*x.*y.^3 +x.^2.*z - z.^3
Now we will sample this function with 200 values, somewhat randomly.
N = 200; % # of points. I hope you have a lot !
x = rand(N,1); y = randn(N,1); z = rand(N,1)*2-1;
F = fun3D(x,y,z);
stem3(x,y,z) % only shows layout
For the moment, all I have done is given a plot of the independent variables, I haven't yet plotted the 4D function. (That's going to be always tricky, but since you asked for it, I hope you have the relevant 4D goggles!)
Now we are ready to recover the parameters. If I choose (somehow) a perfect model structure, then I should recover the parameters exactly. You may need to revise the maths behind the next two steps.
>> V = [x.^2, x.*y.^3, x.^2.*z, z.^3, ones(size(x))] % Vandemonde matrix
>> p = V\F % do LS fit
p =
1.0000
-3.0000
1.0000
-1.0000
0.0000
We got exactly the parameters what we were looking for. Let's do a plot. For more info, check out the help for isosurface etc.
col = colororder;
xi = linspace(0,1)'; yi = linspace(-2,2)'; zi = linspace(-1,1)';
[X,Y,Z] = meshgrid(xi, yi, zi);
Fi = p(1)*X.^2 - p(2)*X.*Y.^3 +p(3)*X.^2.*Z - p(4)*Z.^3;
fval = [-3:3];
for i=1:7
pa = patch(isosurface(X,Y,Z, Fi, fval(i)));
isonormals(X,Y,Z,Fi, pa)
pa.FaceColor = col(i,:);
pa.EdgeColor = 'none';
end
daspect([1 1 1])
view(3)
camlight; lighting phong
Now I hope that is meaningful for you. Most people wouldn't bother.
In reality, we wouldn't know the exact structure of the 3D function, so we might have to guess something like:
V = [x.^2, x, y.^2 y, z.^2, z, x.*y x.*z, y.*z ones(size(x))]
p = V\F

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