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!

  • Please explain your downvote. I am not a mathmatician so I do not know what is wrong with the question. – gavin Oct 13 at 23:58
  • @SaucyO'Path haha. I have no idea what you are talking about. I think it is maybe a conceptual question - is it possible? Can exponential values be 'de-exponentialised'? Thanks – gavin Oct 14 at 0:14
  • 1
    Forget it, I thought you wanted to round the numbers, but that was silly. – Saucy O'Path Oct 14 at 0:18
  • 1
    You can certainly sort the values and return the position of each in the list. That will get exactly the output you desire. I suspect you are looking for something else, but do not have any idea what it is. How should the output depend on the input values, so if the last value were $5000$ what answer would you like? – Ross Millikan Oct 14 at 0:27

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:

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

  • Thanks, I will look into this although I don't understand it right away! – gavin Oct 14 at 1:32

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.

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