How can I non linear fit this equation?

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Pengju 2016-4-19

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关闭： MATLAB Answer Bot 2021-8-20

fit.txt

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Dear all,

I know the equation which is used for fitting already,something like:

y=E-109736.623./[x1.*(x-p1).^2+x2.*(x-p2).^2+x3.*(x-p3).^2]

I have x value, y value, x1, x2, x3 value, respectively, they are corresponding to fit.txt file column1, column2, column3, column4, column5, respectively.

How can I get E, p1, p2, p3 by using this equation?

Thanks

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Walter Roberson 2016-4-19

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The easiest way is the Curvefitting Toolbox

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Walter Roberson 2016-4-20

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There is an excellent fit near

 [33690.9417099533603 .. 33690.9417107163827, 3.14040272505204721 .. 3.14040351360411218, 2.63427756239212973 .. 2.63427798866314777, 1.34093810668077085 .. 1.34093907346367525]

In the time I allowed my searching to run, the best location was

[33690.941710391533, 3.14040320712546128, 2.63427777512871719, 1.34093862330272717]

Code:

data = xlsread('error_fit1.xlsx');
x=data(:,1);
y=data(:,2);
x1=data(:,3);
x2=data(:,4);
x3=data(:,5);
c=109736.623;
NP = size(x,1);
Normk = @(Z,k) sqrt(sum(Z.^2,k));
repparm = @(var) repmat(permute(var, [ndims(var)+1, 1:ndims(var)]), [NP, 1]);
repinp = @(var1,var2) repmat(var1, [1, size(var2)]);
obj = @(E,p1,p2,p3) permute(Normk(repparm(E) - c./(repinp(x1,E).*(repinp(x,E) - repparm(p1)).^2 + repinp(x2,E) .* (repinp(x,E)-repparm(p2)).^2 + repinp(x3,E) .* (repinp(x,E) - repparm(p3)).^2) - repinp(y,E),1), [2:ndims(E)+1,1]);

obj is now a fully-vectorized objective function that can be applied to 4 dimensional grids of E, p1, p2, p3 values, giving one error residue value per grid location. You could wrap this in another function

obj1 = @(Ep1p2p3) obj(Ep1p2p3(1), Ep1p2p3(2), Ep1p2p3(3), Ep1p2p3(4))

and now obj1 can be used in a minimizer such as fminunc

Pengju 2016-4-22

Dear Walter, I am sorry, I still can not figure out the minima.

Best

Walter Roberson 2016-4-22

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minidx = min(residues(:));
min_is_near = [gE(minidx), gp1(minidx), gp2(minidx), gp3(minidx)];

The larger the matrix you use originally, the more likely it is that you will find a good minima.

The code that I posted does make copies of these arrays, so you end up with about 6 arrays that are 84 by (19 x 19 x 19 x 19). The copies themselves add up, so when you do not have a lot of memory, you might want to break the grid up into sections to search and find the best of the results. For example, if you used 20 x 20 x 20 x 20 then you could search first on (1:5, :, :, :, :) then on (6:10, :, :, :, :) and so on.

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