MATLAB: Curve fitting tool with custom equation, odd power polynomial

Curve Fitting Toolboxcustom equationodd power polynomial fitpolynomial fit

I am trying to fit an odd power polynomial of the form (f(x) = p1*x^3 + p3*x) to a data set of with 38 points, which clearly shows a smooth cubic polynomial. I am using the custom equation type in the curve fitting tool to do this. I understand the custom fit is sensitive to the starting parameters, so I first run the standard 3rd order polynomial fit to my data. The fit includes even power terms, but I extract the odd order coefficients and use those as a starting point in my custom fit. It does not work well at all.

As a test, I plugged in the full 3rd order polynomial equation into the custom equation (i.e. f(x) = p1*x^3 + p2*x^2 + p3*x + p4), and set my starting values to the same order of magnitude solved for in the standard polynomial fit, and the custom equation fit still did not fit well at all. How do I fix this?

Alternatively, how can I best fit an odd power polynomial?

Best Answer

It seems like overkill to use the curve fit toolbox just to fit a 1D polynomial. And it should not depend on any starting parameter guess. For a polynomial fit, the coefficients should be the unique solution to a set of linear equations. Below is what I use. For you, the usage would be,

>> polyfitc(x,y,[3 1])
function p=polyfitc(x,y,nvec)
%Simple 1D polynomial fitting with particular coefficients constrained to
%zero
%




%    p=polyfitc(x,y,nvec)
%
%in: 
%
%  nvec: A vector of integer exponents present in the polynomial.
%    x: x data
%    y: y data
%
%out:
%
%   p: vector of polynomial coefficients. Order of polynomial is max(nvec).
           A=bsxfun(@power,x(:),nvec(:).');
           [QQ,RR]=qr(A,0);
           coeffs = RR\( QQ'*y(:) );
            p=zeros(1,max(nvec)+1);
           p(nvec+1)=coeffs;
           p=p(end:-1:1);

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MATLAB: Cftool two term exponential curve fit doesn’t make sense

It is good to think about the parameters that come back. Does what you got make sense? Never just accept what a computer gives you without applying a mental filter to it.

So, what has happened here? It turns out that IF you look carefully at the parameters, to full precision, you will see a difference. The differences between those terms is tiny, and with those huge coefficients, it tells me that the solver managed to overfit the problem. Sadly, this fit is meaningless, and I would argue useless. You might as well use a spline, or some other tool.

If you really are wedded to an exponential model BECAUSE the parameters are necessary for other purposes, then these coefficients are useless anyway. They have no meaning, EXCEPT that they managed to squeeze a slightly lower residual error out of the model, FOR that set of data. Yes, sometimes people need a rate parameter from a model, and this is why they use an exponential model. But those model parameters offer no meaning at all in that context.

I'd prefer to play with the data (or at least look at it) before trying to suggest a better model. But essentially, if your goal is purely to get a good fit, then use a tool that can fit your data well instead of trying to force an exponential model to fit by just throwing extra parameters at the curve fit. My suggestion would probably be a spline model, specifically my SLM toolbox (found on the File Exchange), because you can control the basic shape of that curve.

MATLAB: Coefficients of the polynomial – the least squares method

Sort of correct. But I see at least two subtle problems.

First, you have 9 data points. So a polynomial with as many coefficients as you have data points will be an interpolating polynomial. This is where you need to be careful though, as some people use the words order and degree to be synonyms. In fact, this used to be the case, that they were synonyms. But today it is common that the order of a polynomial is the number of coefficients, whereas the degree is the highest power of the independent variable. That means we would have

degree + 1 = order

So by order 8, that would tend to imply a polynomial of degree 7 (thus the highest power of x would be 7.) As such, it would be a least squares fit, not an interpolating polynomial on 9 data points (thus one more data point than you would have coefficients to fit.) And that is what you get by use of polyfit as you have done. So your use of polyfit was correct, if you intended to fit a polynomial of DEGREE 7.

For example, polyfit(x,y,1) will produce a polynomial of degree 1, with two coefficients, thus order 2. The model would be y=a!*x+a2.

The important issue is how does your teacher use the word order? What do they intend on this question?

The second question is how you generated xfit. We see this:

x_fit = 0:1.0:5.0;
y_fit = polyval(fit, x_fit);

How many points does the vector xfit have? Only 6 of them. In fact, it has fewer points than your original data, and they are more sparse. So when you use plot to plot a curve, you want to use LOTS of points, to fill in the curve. Instead, you plotted only a few points, and plot will connect them with straight lines. That is a bad idea. As well, you went down to 0, when your data only went down to 1 in x. So you are using polyval to extrapolate the polynomial. And extrapolating high order polynomials is a REALLY bad idea.

Instead, you want to do something like this:

x_fit = linspace(1,5,100);
y_fit = polyval(fit, x_fit);
figure;
plot(x,y);
hold on;
plot(x_fit, y_fit);
grid on;
xlabel('x');
ylabel('y');
legend('Data', 'Polyfit', 'Location', 'best');
title(sprintf('x = 3.25 y is %.2f & x = 1.85 y is %.2f',polyval(fit,3.25),polyval(fit,1.85)))

Another possible problem, is if you will use the curve fitting toolbox ever, then it is a bad idea to define a variable named fit, as that will make it difficult to use the FUNCTION named fit from the curve fitting toolbox.

Best Answer

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MATLAB: Cftool two term exponential curve fit doesn’t make sense

MATLAB: Coefficients of the polynomial – the least squares method

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