Logarithmic trendline equation of data
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Cem Eren Aslan
le 31 Juil 2024
Modifié(e) : David Goodmanson
le 1 Août 2024
Hello eveyone,
I try to get a trendline equation of data. This equation must be like "e^x". How can i do that. are there any tools?
Thanks
Cem
2 commentaires
Sam Chak
le 31 Juil 2024
Yes, single input, single output, you can use the 'fit()' command from Curve Fitting Toolbox to achieve this.
But
is an exponential function, thus it does not exhibit a logarithmic trend.
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David Goodmanson
le 1 Août 2024
Modifié(e) : David Goodmanson
le 1 Août 2024
Hello Cem,
As has been mentioned you can use the curve fitting toolbox for this job, but if you don't have it you can try a simple linear fit. If
y = b*e^(ax), then log(y) = log(b) + ax
and the linear fit is of log(y) vs. x. Here is a quick example. Too be somewhat realistic, to the exponential y = e^(ax) is added a nonzero baseline (which is pretty common), a function that is proportonal to 20x near the origin, and some noise. The curve fitting toolbox will do better, but this fit is better than I thought it would be.
Whether the fit is great or not, I think a linear fit to log(y) is always worth doing, just to see if the idea of fitting an exponential to the given data is in the ballpark in the first place.
x = 0:.01:12;
a = .3;
b = 6;
r = .3*randn(size(x));
r(r<-b) = 0;
% exponential with some added stuff
y = b/3 + 20*x./(1+x) + b*exp(a*x) + b*r;
logy = log(y);
p = polyfit(x,logy,1)
ylogfit = x*p(1)+p(2);
figure(1)
grid on
plot(x,logy,x,ylogfit)
figure(2)
grid on
plot(x,y,x,exp(ylogfit))
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