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I am recovering from an injury and I go for a run each morning. The distance that I run does not change but the time is getting shorter each time, i.e. I am improving. I am recording my times and would like to fit an exponential of the form $y(x)=A+Be^{Cx}$ to the data such that I can predict my future minimum time for that journey when I am at full fitness. i.e. I want to know the $A$ value. $y=Time(min)$ and $x=Days$.

Here is my data: $(Day,Time)={(0,78),(1,76),(2,76),(3,70),(4,68),(5,69),(6,66),(7,66),(8,65),(9,66),(10,64),(11,64),(12,63)}$

When I plot these data there does seem to be an exponential improvement forming. But I haven't done it long enough yet to reach my plateauing time.

My calculator has an exponential function but it assumes an $A=0$, which cannot be true. How do I estimate or calculate my $A$ value?

Kantura
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There are many places on this site where you can see the problem of the fit of exponential functions like $$y=a+b\, e^{cx}$$ Just have a look here.

Using that, you should find (using your data) $$y=61.7695 +17.0408 e^{-0.196822 x}$$ corresponding to $R^2=0.999742$.

The key problem is that you need "reasonable" estimates. You could have obtained them in a first step using $a\approx 60$ and then consider $$y-60=b \,e^{cx} \implies \log(y-60)=\log(b)+cx\implies z=d+ cx$$

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Claude Leibovici
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  • Thanks Claude Leibovici. But that result does not make any sense. A normal person could run that distance in say 30 to 40 minutes. Your solution suggests that the minimum time that I could obtain would be 62 minutes. I ran it in 63 minutes this morning and I have only started this training regime. I can't believe that I will only improve by one more minute. – Kantura Jan 14 '18 at 09:31
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    @Derek. I considered the general problem. Now, if you have constraints, the problem can be slightly different. – Claude Leibovici Jan 14 '18 at 09:34
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    @Derek. My "result does not make any sense" but you asked for the fit of such a model. What it could mean is that, for a few days, you would not improve much and, at a time, change your performances. By the way, a linear regression is almost as good. I wish you a good and prompt recovery. – Claude Leibovici Jan 14 '18 at 10:09
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    @Derek. Seriously, I would like you continue recording your performances for a while. Update the data. I would like to see how these could be analyzed. Cheers. – Claude Leibovici Jan 14 '18 at 10:11
  • Thanks again Claude Leibovici. I will update the data later. I agree that a linear regression works reasonably well at this early stage. But based on the nature of the data it can't continue as a linear function as it is impossible for me to run that distance in zero time. There has to be a horizontal asymptote somewhere. But I guess it's just too early in the analysis to estimate that. I'll supplement the data later. Cheers. – Kantura Jan 14 '18 at 13:58
  • @ClaudeLeibovici How can you guess a "reasonable" initial value for $a$ value if you don't have any so called "domain specific information" on the data? – V. Vancak Jan 14 '18 at 14:05
  • @ClaudeLeibovici To be sure - your suggestion is to use $(60, \exp{\widehat{\log b}}, \hat{c})$ as initial values for the non-linear regression, right? If so, are there any general results on what will happen if we start from some "ridiculous" point like $(-10000, \exp{\widehat{\log b}}\ \hat{c})$ for this specific model? – V. Vancak Jan 14 '18 at 14:18