Curve fitter limitation for equations in the form f(x,y)=a*x^n/(y+b)^m

Question

Boyan 2024-9-23

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2154950-curve-fitter-limitation-for-equations-in-the-form-f-x-y-a-x-n-y-b-m

评论： Sam Chak 2024-9-28

Dear Community Members,

When I use curve fitter for equation in the form f(x,y)=a*x^n/(y+b)^m, the results are very bad (R-square is very, very low).

After removing "b" or changing it with a random number (for example 8, then f(x,y)=a*x^n/(y+8)^m), R-square is very good.

The question is: Is this a limitation of the software or there is a way to overcome this problem and to use "b", which value wiil be calculated by Curve fitter.

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Alex Sha 2024-9-24

@Boyan show your data please.

Boyan 2024-9-25

编辑：Boyan 2024-9-25

Dear all,

The surface looks like fig. 2. In Figure 1 are given just the points. These are three exponential curves, laying in vertical planes.

fig. 1

fig. 2

The data is given below in figure 3. I'm sorry for giving it in picture, not in table, but I don't know how to insert table.

fig. 3

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Answer 1

Sam Chak 2024-9-25

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https://ww2.mathworks.cn/matlabcentral/answers/2154950-curve-fitter-limitation-for-equations-in-the-form-f-x-y-a-x-n-y-b-m#answer_1522750

在 MATLAB Online 中打开

Hi @Boyan

Given that there are only

data points, employing a bivariate polynomial of degree two in x and degree four in y will ensure a perfect fit, resulting in an R-squared value of 1.

% Data

x = [ 1 4 5 1 4 5 1 4 5 1 4 5 1 4 5];

y = [ 5 5 5 10 10 10 15 15 15 20 20 20 25 25 25];

z = [257 410 455 185 306 322 151 246 261 125 191 205 109 159 169];

% p1*x^2 + p2*x + p3

[p21, gof] = fit([1 4 5]', [257 410 455]', 'poly2');

[p22, gof] = fit([1 4 5]', [185 306 322]', 'poly2');

[p23, gof] = fit([1 4 5]', [151 246 261]', 'poly2');

[p24, gof] = fit([1 4 5]', [125 191 205]', 'poly2');

[p25, gof] = fit([1 4 5]', [109 159 169]', 'poly2');

% p1*y^4 + p2*y^3 + p3*y^2 + p4*y + p5

[q14, gof] = fit([5 10 15 20 25]', [p21.p1 p22.p1 p23.p1 p24.p1 p25.p1]', 'poly4');

[q24, gof] = fit([5 10 15 20 25]', [p21.p2 p22.p2 p23.p2 p24.p2 p25.p2]', 'poly4');

[q34, gof] = fit([5 10 15 20 25]', [p21.p3 p22.p3 p23.p3 p24.p3 p25.p3]', 'poly4');

plot3(x, y, z, 'rp'), grid on, hold on

xlabel('x'), ylabel('y'), zlabel('z')

% Surface

XX = linspace(x(1), x(end), 101);

YY = linspace(y(1), y(end), 101);

[X, Y] = meshgrid(XX, YY);

% z(x, y) = f(y)*x^2 + g(y)*x + h(y)

f = q14.p1*Y.^4 + q14.p2*Y.^3 + q14.p3*Y.^2 + q14.p4*Y + q14.p5;

g = q24.p1*Y.^4 + q24.p2*Y.^3 + q24.p3*Y.^2 + q24.p4*Y + q24.p5;

h = q34.p1*Y.^4 + q34.p2*Y.^3 + q34.p3*Y.^2 + q34.p4*Y + q34.p5;

Z = f.*X.^2 + g.*X + h;

s = surf(X, Y, Z, 'FaceAlpha', 0.25); s.EdgeColor = 'none';

title('Surface fitting with bivariate polynomial, poly24(x,y)')

view(145, 30)

hold off

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Sam Chak 2024-9-28

在 MATLAB Online 中打开

Hi @Boyan

For relatively few data points, the bivariate polynomial will guarantee a perfect fit, which is generally effective for prediction within the enclosed boundaries.

If a custom fit model is preferred, you should provide initial guesses for the parameters {a, b, m, n}. If the starting points are relatively close to the solution, they should yield the highest R-squared value.

If you find the codes helpful, please consider clicking 'Accept' ✔ on the answer and voting 👍 for it. Your support is greatly appreciated!

% your data

X = [ 1 4 5 1 4 5 1 4 5 1 4 5 1 4 5];

Y = [ 5 5 5 10 10 10 15 15 15 20 20 20 25 25 25];

Z = [257 410 455 185 306 322 151 246 261 125 191 205 109 159 169];

% flattening the matrices for fitting

xData = X(:);

yData = Y(:);

zData = Z(:);

% proposed fit model with parameters a, b, m, n

ft = fittype('(a*x^n)/(y + b)^m', 'independent', {'x', 'y'}, 'dependent', 'z');

% Initial guesses for a, b, m, n

% a b m n

% ↓ ↓ ↓ ↓

initialGuess = [12000, 14, 1, 0.5];

% Enter the fit command and view the Goodness of Fit

[fitResult, gof] = fit([xData, yData], zData, ft, 'StartPoint', initialGuess)

fitResult =

General model: fitResult(x,y) = (a*x^n)/(y + b)^m Coefficients (with 95% confidence bounds): a = 1.163e+04 (-1.567e+04, 3.893e+04) b = 13.72 (3.139, 24.3) m = 1.298 (0.7447, 1.852) n = 0.3395 (0.3155, 0.3634)

gof = struct with fields:

sse: 388.7688 rsquare: 0.9973 dfe: 11 adjrsquare: 0.9965 rmse: 5.9450

% Plot the original data and the fit

figure;

plot(fitResult, [xData, yData], zData);

xlabel('X-axis');

ylabel('Y-axis');

zlabel('Z-axis');

title('Surface Fitting');

view(145, 30)

Boyan 2024-9-28

Hi @Alex Sha,

I'm sorry but can't see your result in our discussion above. I see only your request to show my data. Maybe it's somehow hidden in the comments above. But anyway the problem is solved by the code of @Sam Chak.

Anyway, because you are mention it, could I ask what is this package "1stOpt"?

Sam Chak 2024-9-28

在 MATLAB Online 中打开

Hi @Boyan

I am not affiliated with the product "1stOpt," but it is a very powerful third-party tool, as demonstrated by @Alex Sha in multiple curve-fitting problems. The initial guesses were made after he published his 1stOpt solution. If you are interested, please take a look at the product page:

http://www.7d-soft.com/en/products.htm

I typically guess the starting point parameter values to be close to 0 (to avoid singularity), 1, or the midpoint between 0 and 1, and then gradually increase the magnitudes. In your case, since the z-axis data range is in the hundreds, an educated guess would be to set

. See example below:

% your data

X = [ 1 4 5 1 4 5 1 4 5 1 4 5 1 4 5];

Y = [ 5 5 5 10 10 10 15 15 15 20 20 20 25 25 25];

Z = [257 410 455 185 306 322 151 246 261 125 191 205 109 159 169];

% flattening the matrices for fitting

xData = X(:);

yData = Y(:);

zData = Z(:);

% proposed fit model with parameters a, b, m, n

ft = fittype('(a*x^n)/(y + b)^m', 'independent', {'x', 'y'}, 'dependent', 'z');

% Initial guesses for a, b, m, n

% a b m n

% ↓ ↓ ↓ ↓

initialGuess = [100, 1, 1, 1];

% Enter the fit command and view the Goodness of Fit

[fitResult, gof] = fit([xData, yData], zData, ft, 'StartPoint', initialGuess)

fitResult =

General model: fitResult(x,y) = (a*x^n)/(y + b)^m Coefficients (with 95% confidence bounds): a = 1.163e+04 (-1.567e+04, 3.893e+04) b = 13.72 (3.139, 24.3) m = 1.298 (0.7447, 1.852) n = 0.3395 (0.3155, 0.3634)

gof = struct with fields:

sse: 388.7688 rsquare: 0.9973 dfe: 11 adjrsquare: 0.9965 rmse: 5.9450

% Plot the original data and the fit

figure;

plot(fitResult, [xData, yData], zData);

xlabel('X-axis');

ylabel('Y-axis');

zlabel('Z-axis');

title('Surface Fitting');

view(145, 30)

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Curve fitter limitation for equations in the form f(x,y)=a*x^n/(y+b)^m

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Curve fitter limitation for equations in the form f(x,y)=a*x^n/(y+b)^m

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