# Curve fitting f(x,y) result

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Hi,

I used Curve fitting tool to get the equation of the regression line of my area. So I have some X and Y which give some Z.

The curve fitting tool provide me a linear model like this :

Linear model Poly54:

f(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + p21*x^2*y+ p12*x*y^2 + p03*y^3 + p40*x^4 + p31*x^3*y + p22*x^2*y^2+ p13*x*y^3 + p04*y^4 + p50*x^5 + p41*x^4*y + p32*x^3*y^2+ p23*x^2*y^3 + p14*x*y^4

But when I try to obtain a Z value from this equation it's not good at all. For example with X = 60 and Y = 60 , Z = 300 but with the formula I have a result like xxxE+14

Actually, I'm looking for an equation which can provide me a Z value whatever X and Y such as f(x,y) = z The area looks like this : http://www.hostingpics.net/viewer.php?id=329914example.jpg

Thank you if you can help

##### 0 Comments

### Answers (7)

Richard Willey
on 20 Apr 2012

I'm attaching code that shows a couple different ways to solve your problem.

I prefer the second option. The R^2 is slightly better and the model is simplier. (Note that option 2 as coded uses Statistics Toolbox rather than Curve Fitting Toolbox)

%%Input some data

x = [0 10 20 30 35 40 45 50 55 60];

y = [60 120 180 240 300];

z = [299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;

299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;

299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;

299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;

299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1];

% Observe that the data is constant in one dimension.

% The variable Y has no predictive power

% This means we can treat this as a curve fitting problem rather

% than a surface fitting problem

x = x';

z = mean(z, 1)';

% Visualize our data

scatter(x,z)

%%Generate a fit

% Option one

[foo, GoF] = fit(x,z, 'poly2')

hold on

plot(foo)

%%Option two: Fit a piecewise linear model

% Looking at the data, it appears as if we might generate a better fit by

% assuming a piecewise linear model with a "kink" at the 7th data point

% Start by centering the data such that the 7th data point is equal to zero

New_x = x - 45;

% Create a dummy variable that flags whether we are before or after the 7th

% data point

dummy = New_x > 0

% Create an interaction term

interaction = New_x .* dummy

% Generate our regression model

X = [ones(size(x)), New_x, interaction];

b = regress(z, X)

plot(x, X*b)

##### 2 Comments

Ibrahim Maassarani
on 28 Feb 2019

Is b the function?

I got:

b =

1.0e+03 *

7.6023

0.0043

-0.0144

How should I read this?!

Walter Roberson
on 28 Feb 2019

You should give the command

format long g

and then you should display b again.

Richard Willey
on 20 Apr 2012

Hi there

The "Poly54" fit type is specifying a 5th order polynomial in one direction and a 4th order polynomial in the other.

From the looks of things, you have roughly 40 data points. Your model has 20 terms (many of which are high order). I suspect that you're overfitting like crazy, which means that your model can perform very poorly if used to generate a prediction for out of sample data.

Based on the chart that you provided, you can probably cobble together a simple custom equation with a linear model in one direction and a cubic in the other. Something like

Y = b0 + b1*X1 + b2*X2^3

##### 0 Comments

Richard Willey
on 20 Apr 2012

Hi ZazOufUMl

A "Poly31" model is different than the one that I suggested. (Poly31 will contain a number of cross terms).

Any chance that you can post your raw X, Y, and Z data?

##### 1 Comment

Walter Roberson
on 20 Apr 2012

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