I have data of 10000 points of nonlinear fit. I want to have an equation that gives me the trend of that data.

Equation that gives the trend of the data.

[the cyclist]

I suggest you read this tutorial and edit your question, to get a more useful response.

Answer 1

[the cyclist]

With the complete lack of detail you offer us, I can only give a generic response, which is to perhaps try the function nlinfit from the Statistics Toolbox.

[Sruthi Mantri]

Thank you

Answer 2

[Chad Greene]

p = polyfit(x,y,n) will give you that equation. If n=1, p will have a slope and a y-intercept value. If n=2, the first value in p will be coefficient for the squared term, the second value in p is the slope, and the third value in p is the y-intercept. Choose a value of n that gives you the best balance of fit versus simplicity.

[Sruthi Mantri]

Thank you very much for the quick reply. I am not familiar with the MATLAB software at all. I just installed it now. Could you explain me in a more detailed way if possible?

[Chad Greene]

The polyfit documentation explains it pretty well. Here's an example using 100 data points instead of your 10000, but it works the same.

xdata = 1:100; 
ydata = 0.1*xdata.^2 + 4*xdata - 4000 + 200*rand(size(xdata));

plot(xdata,ydata,'ko')
hold on

linearFit = polyfit(xdata,ydata,1); 
xfit = 0:5:100; 
yfitLinear = linearFit(1)*xfit + linearFit(2); 
plot(xfit,yfitLinear,'b')

secondOrderFit = polyfit(xdata,ydata,2); 
yfitSecondOrder = secondOrderFit(1)*xfit.^2 + secondOrderFit(2)*xfit + secondOrderFit(3); 
plot(xfit,yfitSecondOrder,'r')

legend('original data','linear fit','second order fit','location','northwest')
legend boxoff
box off

[the cyclist]

Chad, this is very nice, and helpful.

Sruthi, you have still given only a tiny amount of information about your problem. You should add a lot more detail in your question.

I encourage you to be very careful about what you mean by a "nonlinear fit". Usually, the word "linear" in "linear fit" refers to the parameters, NOT the variables. polyfit() actually does a linear fit (in the parameters).

So,

y = A + B1 * x + B2 * x.^2

which is what polyfit() does, is a linear fit.

But if you mean something like

y = A ./ (1 + exp(-B * x))

then that is a nonlinear fit (because the parameters do not come in linearly).

If that is more like what you have, then you will need nlinfit.