I have xyz data from which i want to formulate an equation

hello everyone, so previously i had used this xyz data to make a 3D plot. However i now want to formulate an equation out of this data?

i want to solve for z inserting x and y

any idea how i could do that

this is the code i have used for the 3D plot

T1 = readtable('solarcol3.csv');
VarNames = T1.Properties.VariableNames; 
x = T1{:,1};
y = T1{:,2};
z = T1{:,3};
[Ux,iax,ixx] = unique(x);
[Uy,iay,ixy] = unique(y);
N = 25;
xv = linspace(min(x), max(x), N);
yv = linspace(min(y), max(y), N);
[Xm,Ym] = ndgrid(xv, yv);
Zm = griddata(x, y, z, Xm, Ym, 'cubic'); % specify cubic interpolation
Zm = smoothn(Zm,1000); % Fex : https://fr.mathworks.com/matlabcentral/fileexchange/25634-smoothn?s_tid=ta_fx_results

Unrecognized function or variable 'smoothn'.

Answer 1

[Walter Roberson]

No, it can be shown mathematically (and I have posted proofs before) that given any finite set of finitely-represented points, that there are an uncountable infinity of functions that mathematically pass exactly through all of those points.

Because there is an infinite number of potential equations to choose from, the probability that any one equation is the "right" equation is 1/infinity ... which is zero.

The best you can do is pre-determine a finite list of models that have no more parameters than the number of points you have, and do fitting to determine the best model parameters for each possibility, and then check to see which of the models had the "best" fit.

However, it is common to discover that if you do tests by generating points according to a known model and then add even a tiny amount of noise, that some other completely "wrong" model will fit the points better than the known right model form does. For example you might discover that where your actual model was polynomial degree 6 plus a small bit of noise, that a sum-of-exponentials fit better than polyfit(x,y,6)... because by accident there was a small bit of long-term correlation in the noise that was added.

[Walter Roberson]

Suppose you sort your x values in increasing order. Then find the difference between adjacent values. Under the assumption that the x values are finitely represented, you can find a greatest common divisor the divides all of those.

If you were able to find an equation z = f(x,y) that fit the points, and you were to construct f2(x) = f(x,y) + sin(pi*F*(x-min(x))) where F is 1/greatest_common_divisor, then it would follow that the sine() would contribute 0 at each x location. Therefore if f(x,y) perfectly describes the available data, the f2(x) perfectly describes it too.

Therefore if you have finitely-represented coordinates there cannot be a unique equation that models the values: there are an infinite number of equations with higher frequency sine or cosine imposed on them that you cannot tell apart from the original given only the original data .

[Walter Roberson]

Suppose you can show that over the range of your x values, -1 to +1, that the polynomial

x^17/355687428096000 + x^16/20922789888000 + x^15/1307674368000 + x^14/87178291200 + x^13/6227020800 + x^12/479001600 + x^11/39916800 + x^10/3628800 + x^9/362880 + x^8/40320 + x^7/5040 + x^6/720 + x^5/120 + x^4/24 + x^3/6 + x^2/2 + x + 1

fits you data perfectly to to within the (single) precision of your inputs.

So that would be it, right? The polynomial would be "the" equation, right?

Ah, but that polynomial is taylor(exp(x), x, 0, 'order', 17) -- so to within the error of your data exp(x) might be an equally valid solution. How to decide? Is the equation "really" the degree-17 polynomial, or is the equation "really" exp(x) ?

No system operating on finitely represented values can determine the difference.

[Alex Sha]

Walter gives a very good explanntion. One of functions looks like bellow:

z = (p1+p2*x+p3*y+p4*x*y)/(1+p5*x+p6*y+p7*x*y)+p8;

Sum Squared Error (SSE): 0.00885938009296813
Root of Mean Square Error (RMSE): 0.0156873800065957
Correlation Coef. (R): 0.992629641433234
R-Square: 0.985313605051871

Parameter	Best Estimate       
---------	-------------       
p1       	-0.428781064157635  
p2       	0.0594834002179455  
p3       	0.000475515732748892
p4       	-4.72419933735348E-5
p5       	-0.00466058481058043
p6       	0.00305034598701402 
p7       	-0.0004153284867253 
p8       	0.43159235394028