# How to derive formula from a exponential scattered plot

I am doing some experiments and i got some results which i plot into a graph. The graph at small x values (x<15) gave random scatter plot but in large values more than 1000 it lookes like an negative exponential plot (similar to figure). How i can find equation for this scatter graph? can this method given in this tutorial can be used for finding equation in scatter graph?

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Can you show a plot of the data? That looks like an exponential curve.

Anyway, there are a number of ways to fit $a e^{-bt}$ to data to get values for $a$ and $b$. An easy way to start is to take the log of the data and do a linear fit of whatever type you like (least squares, $L_1$, ...). This can give you an initial estimate for $a$ and $b$ which can be put into a non-linear fitting routine to fit $a e^{-bt}$.

If the initial data points are "bad", you can remove them when you do the fitting.

Something I discovered many years ago (in the context of Newton's law of cooling) is that if we have three equally spaced (in time) data points, we can fit an exponential to them. Suppose the points are $(t-d, p), (t, q),$ and $(t+d, r)$, and we want to fit $a e^{bt}+c$ to them. We have $a e^{bt}+c = q$, $a e^{b(t-d)}+c = p$, and $a e^{b(t+d)}+c = r$.

Subtracting the first two, $q-p =a (e^{bt}-e^{b(t-d)}) =a e^{bt}(1-e^{-bd})$. Subtracting the last two, $r-p =a (e^{b(t+d)}-e^{bt}) =a e^{bt}(e^{bd}-1)$.

Dividing these two, $\frac{r-p}{q-p} = \frac{e^{bd}-1}{1-e^{-bd}}$. Since we know $p, q, r,$ and $d$, we can solve for $b$. We can then get $a$ and $c$. (This is modulo any errors on my part.)

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