Exp Fit Error: Inf computed by model function, fitting cannot continue. Try using or tightening upper and lower bounds on coefficients.
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I have successfully fit multiple similar sets with no problem but the attached data gives the error in the title. I tried fitting both first and second order exponentials but they both fail. If anyone can tell me what to do to resolve this I would greatly appreciate it.
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John D'Errico
2016-8-2
编辑：John D'Errico
2016-8-2
You need to look at your data. THINK about what is happening. Don't just throw a modeling tool at it and expect magic to always work.
[min(x),max(x)]
ans =
1.3776 1.3811
So your data is over a TINY range. But exponentials tend to have problems with huge numbers. So you also need to be VERY careful as to the multiplier of x in there.
min(y)
ans =
-6.4096
If you plot log(y - 6.45), as a function of x, except for some crap down at the bottom, the fit is pretty good. So a simple exponential model will do pretty well, IF you are careful.
polyfit(x,log(y + 6.45),1)
ans =
-1514.5 2091.1
This tells us that a model something like this is a good idea.
y = a + b*exp(-c*(x-1.38))
where a good starting value for c is approximately 1500.
I've subtracted 1.38 off, since exp(-1500*1.38) will fail in double precision. This is surely why you are having problems. Well, it does not fail. But it underflows. You get a pure zero as a result.
exp(-1500*1.38)
ans =
0
So you very much need to shift the problem. If you do that, and then use 1500 as a starting value for the multiplier, you can get a model of
y = -4.208 + 2.0602*exp(-1670.7*(x-1.38))
This fits your data fairly well.
plot(x,y,'o',x,-4.208 + 2.0602*exp(-1670.7*(x-1.38)),'b-')
So, hey, you don't need to do these things. But if you want to find a valid model that fits your data, it is a good idea. Otherwise, garbage will result.
2 个评论
Craig Siebe
2016-8-2
halo
2021-2-20
Can you explain the logic behind how you are deriving your inferences like 1500 is a good starting point for c?
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