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June 14th, 2018, 07:03 PM   #1
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Function for saturating logarithmic growth

Hi everyone,

After searching for a bit, I'm wondering: is there a class of functions which exhibit logarithmic growth and then saturate? An integrable form would be ideal. It seems like such a simple scenario, but I'm coming up short.

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June 15th, 2018, 01:07 PM   #2
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arctan(x) for x > 0 might work. At x=0 it is linear, but then "saturates" to $\frac{\pi}{2}$.
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June 15th, 2018, 07:34 PM   #3
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The function can't be strictly logarithmic growth if it is to saturate.

Can you separate the domain into an area where you want true logarithmic growth and then one that can tolerate the saturation?
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June 18th, 2018, 08:36 AM   #4
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Thanks for the responses!

mathman--This is an interesting suggestion. arctan(x) does exhibit the sort of saturation that I'm interested in describing, but below saturation the growth is linear on a log-log plot, whereas I'm looking for a function which is linear on a linear y / log x plot (i.e., logarithmic growth).

romsek--Yes, there are two domains, though they should grade into each other (not a break in slope, but a gradual transition).

I may have cobbled together a workable solution: y = ln(1 - exp[-x])
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June 21st, 2018, 03:27 AM   #5
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Sounds to me like you are talking about the "logistic function": Logistic Functions
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June 22nd, 2018, 09:41 AM   #6
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I don't think so, only because logistic functions begin to grow in an exponential manner (increasing growth rate with time) before they begin to saturate. What I'm describing is function which is always decreasing in growth rate but which then saturates completely (unlike traditional logarithimic growth, which increases forever, though ever more slowly).

That said, you could be right--there could be a subcategory of logistic fxn which behaves like you describe that I'm unfamiliar with.
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June 22nd, 2018, 10:06 AM   #7
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I'd pick some point $x_b$ you're happy with being the boundary of the domains. Then

$f(x) = \begin{cases}

c_1 \log(a x) &x <= x_b \\ \\

c_2 \left(1-e^{b x}\right) &x_b < x

\end{cases}$

where $c_1 \log(a x_b) = c_2 \left(1-e^{b x_b}\right)$

You'll need to tweak $a,~b,~c_1$ to get exactly the behavior you want.

You may also want to provide a shift for $x$ i.e. make $f(x) \to f(x-x_0)$ where you choose $x_0$ to keep things away from $x<1$

I think you'll find this does the trick.
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June 22nd, 2018, 01:34 PM   #8
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Thanks very much, romsek. This is a really clean solution!
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