# I'd like to force my exponential fit to start at [0 0]

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Stefan on 24 Apr 2013
So, I've got some noisy data. When averaged, the data fit nicely along a double exponential fit, starting off with with a positive slope that decays to near 0 (horizontal line).
However, when I want to fit the same curve to the individual traces, things aren't so neat. Sometimes, the slope is initially negative and shallows out, other times, it's flat and then increases exponentially at the end.
Some of them look great, like this: http://i35.tinypic.com/23si5bs.jpg
And some of them fit poorly, like this: http://i34.tinypic.com/2rfdaxe.jpg
How can I force the exponential fit that I want? It seems like 'Startpoint' is a good starting point (heh).
Here's my code:
[u,gof] = fit(time,y,'exp2');
And this doesn't seem to work:
[u,gof] = fit(time,y,'exp2','Startpoint',[0 0]);
??? Index exceeds matrix dimensions.
Any thoughts would be appreciated.
Thanks, Stefan
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### Answers (1)

Joe S on 24 Apr 2013
Not sure if it will fix it, but remember an exponential function can never equal zero unless your A1 & A2 are both 0 (which you don't want). Try a small positive number, say 0.001 for your y-start..
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Walter Roberson on 24 Apr 2013
For any exponential function, f(x+1)/f(x) is always a constant (dependent on the exponential being used.) If any f(x+1) was exactly 0 for finite x, then f(x) would have to have been 0 as well, and by induction f(x) would have to be identical to 0. Therefore no exponential function can ever be exactly 0 at any finite location.
You can force an exponential to fit exactly through any finite non-zero location, but not through 0.
Stefan on 24 Apr 2013
Ok. I haven't been able to implement a start point at all for using 'fit,' and there aren't examples of its use outside of the cftool.
Do you know of any good examples of using code to restrain the startpoints?

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