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New to Computational Science, I hope I'm on the right exchange network for this question.

I have a time series data set that contains the sum of a source data set representing an exponential decay curve at discrete points in time. In other words, the time series data set is the integral over discrete time bins. An example, written in R:

expFunc <- data.table(year=seq(1,20,1),
                      value=2*exp((-0.1*seq(1,20,1))+4)-2,
                      variable="Desired Exponential Function")
fiveYear <- data.table(year=c(1,5,10,15,20),
                       value=c(expFunc[1,value],
                               sum(expFunc[2:5,value]),
                               sum(expFunc[6:10,value]),
                               sum(expFunc[11:15,value]),
                               sum(expFunc[16:20,value])),
                       variable="Data That I Have")

demonstrated in this plot plot of example code

The data that I have follows the time steps in my example, where the first value is always the value of the desired function, and the remaining values represent the sum of the desired function over sequential time bins. Essentially, the data that I have I think can be represented as

$$ \int_{t_{1}}^{t_{2}} a e^{-bt+c}dt $$

over different time intervals $t_{1} \rightarrow t_{2}$. I was imagining that I could solve for variables $a$, $b$, and $c$ using three equations, three unknowns but I'm either stuck in the algebra or am missing something.

Does anyone have an idea of how could obtain the underlying exponential function, either as

  • a discrete data set at a one year time resolution, or
  • an explicit function?

Thanks in advance for your ideas.

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    $\begingroup$ If you have F(t)= $\int^t f(\tau) d\tau$ then you take the derivative of F and that will give you f(t). Then you can fit it by an exponential by optimizing the parameters a,b,c. $\endgroup$ – Maxim Umansky Sep 16 '20 at 3:07
  • $\begingroup$ Does this thread answer your question? stackoverflow.com/a/31854405/11304107 $\endgroup$ – cdalitz Oct 2 '20 at 9:25

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