Help to find the best lower bound function for a given set of data, based in the natural logarithm function

Question

I am trying to find a lower bound function for a set of data I have, and I am struggling with it. In the following graph the blue color is the set of data and the red color is my lower bound function.

The data is bounded between $0$ and $1$, and looks a little bit like the inverse of the natural logarithm ($LN$), but with some initial noise, so my approach (by trial/error in Excel) was as follows (the red-colored line).

$f(n) = 0\quad,\quad n \in [4,9000]$

$f(n) = 0.98-\frac{1}{LN(n-9000)}\quad,\quad n\ge9000$

So i was able to approach to the data, but the initial segment of data $[4,9000]$ was avoided, because the initial growing of the data is smoother (and noisy) than the initial growing of the function $f(n)$ I prepared, so the lower bound is "broken" in that segment for some $n$.

For instance, in the worst case I could do exactly the same approach and create two or three more $f_i(s_i)$ functions to split the initial pending segment of data $n=[4..9000]$ in two, three or more segments $s_i$ with a similar lower bound function $f_i(s_i)$ based on the inverse logarithm but adapted to the specific segment $s_i$ of data.

Please I would appreciate very much if somebody could give me an idea about how could I make using only one function a closer lower bound, like the one I have drawn in black color.

Initially I would like to be able to get a closer function by using the natural logarithm or a similar approach, not using a "pure" polynomial interpolation.

In other words, what I am looking for is a "parametrized" natural logarithm-based (or "similar to natural logarithm-based") lower bound function more than a "pure" parametrized polynomial interpolation, any solution is welcomed but I would appreciate very much a natural logarithm interpolation if possible. Thank you!

Answer 1

To my eye, the noise is about $\pm (10-20)\%$ of $(1-$ the curve). I would be tempted to smooth the data, fit a curve, then reduce it by a "three sigma" amount to produce a curve that is below "almost all" the points. It seems to me that would be a better representation of the data. Is that acceptable? The scare quotes are meant to indicate that a few points would fall below the curve you quote as a lower bound, but that may or may not be acceptable for your application. If you need a curve that is truly a lower bound, you will be farther from the "by eye" fit.

Answer 2

Well, finally I was able to find a better mix as follows:

The initial segments are just simple polynomial approximations, and finally I found a better option for the last segment, including a combination of the inverse of the natural logarithm and $\frac{1}{x}$. This could give ideas to more people about how to attack this kind of problems, so I am adding it to the solutions.

$f(n) = 0,\quad n\lt1918$

$f(n) = (\frac{n^2-n}{(3300^2-3300)}*0.63)\quad,\quad n\lt3300$

$f(n) = 0.63\quad,\quad n\lt6000$

$f(n) = 0.71\quad,\quad n\lt9000$

$f(n) = \frac{4.95-\frac{1}{n}+\frac{LN(n)}{2.7}}{10}\quad,\quad n\ge9000$