Logarithmic trendline equation of data

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Cem Eren Aslan
Cem Eren Aslan 2024 年 7 月 31 日
編集済み: David Goodmanson 2024 年 8 月 1 日
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 件のコメント
Sam Chak
Sam Chak 2024 年 7 月 31 日
Yes, single input, single output, you can use the 'fit()' command from Curve Fitting Toolbox to achieve this.
The desired equation (fit model) can be selected from the library using the 'fittype()' command.
But is an exponential function, thus it does not exhibit a logarithmic trend.
Stephen23
Stephen23 2024 年 7 月 31 日

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David Goodmanson
David Goodmanson 2024 年 8 月 1 日
編集済み: David Goodmanson 2024 年 8 月 1 日
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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