Exponential curve fit matlab

Question

I have the following equation:

$y = \exp(Au^2 + Ba^3)$

I want to do a exponential curve fitting using MATLAB for the above equation, where y = f(u,a). y is my output while (u,a) are my inputs. I want to find the coefficients A,B for a set of provided data.

I know how to do this for simple polynomials by defining states. As an example, if states= (ones(size(u)), u u.^2), this will give me L+Mu+Nu^2, with L, M and N being regression coefficients.

However, this is not the case for the above equation. How could I do this in MATLAB?

You could do a polynomial regression by fitting to the log(y) data instead of the y data. — eigenchris
– eigenchris, Commented Jan 23, 2015 at 15:06

Community · Accepted Answer · 2017-02-09 15:09:45Z

3

Building on what @eigenchris said, simply take the natural logarithm (log in MATLAB) of both sides of the equation. If we do this, we would in fact be linearizing the equation in log space. In other words, given your original equation:

$y = \exp(Au^2 + Ba^3)$

We get:

$\ln y = Au^2 + Ba^3$

However, this isn't exactly polynomial regression. This is more of a least squares fitting of your points. Specifically, what you would do is given a set of y and set pair of (u,a) points, you would build a system of equations and solve for this system via least squares. In other words, given the set y = (y_0, y_1, y_2,...y_N), and (u,a) = ((u_0, a_0), (u_1, a_1), ..., (u_N, a_N)), where N is the number of points that you have, you would build your system of equations like so:

$\begin{align*} \ln y_0 = Au_0^2 + Ba_0^3\\\ln y_1 = Au_1^2 + Ba_1^3\\\ldots\\\ldots\\\ln y_N = Au_N^2 + Ba_N^3\end{align*}$

This can be written in matrix form:

$\begin{bmatrix} \ln y_0 \\ \ln y_1 \\\ldots \\ \ln y_N \end{bmatrix} = \begin{bmatrix} \ u_0^2 & a_0^3 \\ u_1^2 & a_1^3 \\ \ldots \\ u_N^2 & a_N^3 \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix}$

To solve for A and B, you simply need to find the least-squares solution. You can see that it's in the form of:

Y = AX

To solve for X, we use what is called the pseudoinverse. As such:

X = A^{*} * Y

A^{*} is the pseudoinverse. This can eloquently be done in MATLAB using the \ or mldivide operator. All you have to do is build a vector of y values with the log taken, as well as building the matrix of u and a values. Therefore, if your points (u,a) are stored in U and A respectively, as well as the values of y stored in Y, you would simply do this:

x = [u.^2 a.^3] \ log(y);

x(1) will contain the coefficient for A, while x(2) will contain the coefficient for B. As A. Donda has noted in his answer (which I embarrassingly forgot about), the values of A and B are obtained assuming that the errors with respect to the exact curve you are trying to fit to are normally (Gaussian) distributed with a constant variance. The errors also need to be additive. If this is not the case, then your parameters achieved may not represent the best fit possible.

See this Wikipedia page for more details on what assumptions least-squares fitting takes:

http://en.wikipedia.org/wiki/Least_squares#Least_squares.2C_regression_analysis_and_statistics

edited Feb 9, 2017 at 15:09

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answered Jan 23, 2015 at 15:28

rayryeng

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15 Comments

rayryeng Over a year ago

@A.Donda - ok. I'll remove them then. That's very true. The second and third lines were unnecessary.

rayryeng Over a year ago

@A.Donda - Nope! I deal with linear algebra and numerical methods almost every day. It just so happens that I have found someone who knows more about it than I do :). I certainly welcome your edits. Please suggest more changes if there are more!

Ben Over a year ago

Thanks to both of you; your discussion was very informative for me haha +1 each :)

rayryeng Over a year ago

@A.Donda - I embarrassingly forgot to include the assumption that the errors come from a normal distribution as you have stated in your post. I've amended my post some more.

A. Donda Over a year ago

The OP will be well educated by our combined efforts. :-)

|

A. Donda · Accepted Answer · 2015-01-23 15:24:22Z

3

One approach is to use a linear regression of log(y) with respect to u² and a³:

Assuming that u, a, and y are column vectors of the same length:

AB = [u .^ 2, a .^ 3] \ log(y)

After this, AB(1) is the fit value for A and AB(2) is the fit value for B. The computation uses Matlab's mldivide operator; an alternative would be to use the pseudo-inverse.

The fit values found this way are Maximum Likelihood estimates of the parameters under the assumption that deviations from the exact equation are constant-variance normally distributed errors additive to A u² + B a³. If the actual source of deviations differs from this, these estimates may not be optimal.

edited Jan 23, 2015 at 15:24

answered Jan 23, 2015 at 15:18

A. Donda

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3 Comments

rayryeng Over a year ago

ah shoot. I didn't see this answer here. I spent all this time typing up the equations in LaTeX lol. Nice job. I'll leave my answer in so someone can see how the above code is justified.

A. Donda Over a year ago

@rayryeng, yours has just been accepted by the OP. :-) But have a look at my comment, there are some minor problems in your answer.

rayryeng Over a year ago

Oops :) Sorry to do that lol.... but I do thank you for your comments. I'll amend my post.

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