Even out a numerical curve

Question

I have a set of values that increase / decrease exponentially thus:

1.0008230452674898
1.5012345679012344
2.251851851851852
3.3777777777777778
5.066666666666666
7.6
11.4
17.1
25.650000000000002
38.475
57.7125

(they are created dynamically) Is there a simple equation that can 'even it out' ie convert it to 1,2,3,4,5,6,7,8,9,10,11 for example? I am not a mathmetician so any help appreciated!

Answer 1

Let $t = 1, 2, \ldots, 11$, and let your numbers be $y$.

Since you mentioned the $y$ numbers are expected to be exponential, I assume that $y = Ae^{rt}$ and computed the $\ln y$ values, because

$$y=Ae^{rt} \implies \ln y = \ln A +rt$$

so hopefully $\ln y$ and $t$ are in a linear relation.

ln y
0.0008227067515
0.4062878149
0.811752923
1.217218031
1.622683139
2.028148247
2.433613355
2.839078464
3.244543572
3.65000868
4.055473788

Plotting $\ln y$ against $t$ in Google Sheets gives a pretty straight trend line:

$$\begin{align*} \ln y &= -0.4046424014 + 0.4054651081 t\\ y &= e^{-0.4046424014 + 0.4054651081 t}\\ &= 0.6672153635 e^{ 0.4054651081 t} \end{align*}$$

Answer 2

You can do this with Desmos. For a table with $x_1$ ($1, 2, 3, \cdots, 11)$ and $y_1$ (your values), typing in $y_1 \sim ab^{x_1}$ will plot the exponential curve of best fit.

From this Desmos graph, $a = 0.0667215$, $b = 1.5$, so the curve that best fits is $y = 0.0667215x^{1.5}$. Since $R^2=1$, and plotting the residuals gives $0$ for all the points, this curve is also an exact fit.