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Fit a polynomial function

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Katerina Rippi
Katerina Rippi 2015-6-21
评论： dpb 2015-6-21
Does someone know how it is possible to fit a polynomial function whent the x value is a vector? In other words, if we want to fit a polynomial function with output data y and input parameters x where x=[x1,x2,x3,....,xn]. Because until now the only thing that I have found is only if x is a single parameter.

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the cyclist
the cyclist 2015-6-21
编辑：the cyclist 2015-6-21
The easiest way to do this, if you have the Statistics Toolbox, is to use the fitlm function. Here is some example code:
load carsmall
tbl = table(Weight,Acceleration,MPG,'VariableNames',{'Weight','Acceleration','MPG'});
% % The formula from the example
% formula = 'MPG ~ Weight + Acceleration';
% A formula with 2nd- and 3rd-order polynomials
formula = 'MPG ~ Weight^3 + Acceleration^2';
lm = fitlm(tbl,formula)
A few things to notice:
This code is based on the example in that documentation.
I have included -- but commented out -- the formula in which Weight and Acceleration are included to first order.
The model I fit includes Acceleration to 2rd order and Weight to 3rd order. Notice that the way I specified the model, MATLAB automatically included the lower-order terms (including the intercept).
You could also include cross terms like Acceleration*Weight, but I did not.
Remember that a "linear" model is one that is linear in the coefficients. Even though this model includes terms like Acceleration^2, it is still linear because the coefficient of that term will be linear.
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Katerina Rippi
Katerina Rippi 2015-6-21
I thought that by giving the order of the polynomial and of course the data, Matlab would be capable of fitting this polynomial. But, anyway, I wrote an approximate formula like "cyclist" sail that I thought would best fit and it worked. So, thanks everyone. Problem solved, I think. If any other implications I may ask again
dpb
dpb 2015-6-21
But, again, ML has to have some way to "know" what you intend and with multiple variables the possibilities are endless so there's no attempt to try to generalize beyond some lower-order special cases that are fairly frequent but you kept dancing around the question of whether you did or did not, think interaction terms would be important for starters...

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Image Analyst
Image Analyst 2015-6-21
See my polyfit demo, attached below this image it creates
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Katerina Rippi
Katerina Rippi 2015-6-21
Thanks but I have already seen these options. My question was if y=f(x1, x2,x3,...,xn) then how we make the regression of the polynomial? Do we have this possibility or the option is only if y=f(x)?
Image Analyst
Image Analyst 2015-6-21
Attached is an example. Consider x1 to be the horizontal direction, and x2 to be the orthogonal (vertical) direction. It fits the data (models it) to a 4th order polynomial in both directions. For each (x1, x2) pair, I have a value f(x1,x2) which is the intensity of the image. Then I fit a 2D 4th order polynomial surface to those values.

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dpb
dpb 2015-6-21
There are several regression and curve fitting routines if you have the Statistics and/or Curve Fitting toolboxes; if you don't you can use the "backslash" operator that will do a least squares solution to an overdetermined system. In this case you write the explicit model as the design matrix (not forgetting to include the column of Ones for the intercept term, of course) and a vector of the observation values as
X=[ones(length(y),1) x1 x2 ... xN];
and solve as
c=X\y;
where each xi is the (column) vector of the values of the associated independent variable and y is your vector of observations.
For example, if you were to try to fit a model of the form z=f(x,y) with the cross term, you could write
X=[ones(size(z)) x x.*y y];
c=X\z;
Again, the above assumes column vectors x,y,z are the two independent variables and z is the observation. c will be the coefficients of the the model
z=c(1) + c(2)*x + c(3)*x.*y + c(4)*y;

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