My Math Forum > Math Function for saturating logarithmic growth

 Math General Math Forum - For general math related discussion and news

 June 14th, 2018, 08:03 PM #1 Newbie   Joined: Dec 2016 From: California Posts: 9 Thanks: 0 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. Thanks!
 June 15th, 2018, 02:07 PM #2 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 658 arctan(x) for x > 0 might work. At x=0 it is linear, but then "saturates" to $\frac{\pi}{2}$. Thanks from hydronate
 June 15th, 2018, 08:34 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224 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? Thanks from hydronate
 June 18th, 2018, 09:36 AM #4 Newbie   Joined: Dec 2016 From: California Posts: 9 Thanks: 0 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])
 June 21st, 2018, 04:27 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 Sounds to me like you are talking about the "logistic function": Logistic Functions Thanks from hydronate
 June 22nd, 2018, 10:41 AM #6 Newbie   Joined: Dec 2016 From: California Posts: 9 Thanks: 0 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.
 June 22nd, 2018, 11:06 AM #7 Senior Member     Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224 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. Thanks from hydronate
 June 22nd, 2018, 02:34 PM #8 Newbie   Joined: Dec 2016 From: California Posts: 9 Thanks: 0 Thanks very much, romsek. This is a really clean solution!

 Tags function, growth, logarithmic, saturating

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Sonprelis Calculus 1 August 6th, 2014 09:50 PM bilano99 Algebra 2 October 4th, 2012 08:12 AM danield3 Calculus 10 May 1st, 2010 02:55 PM Shehab Calculus 2 April 30th, 2010 08:01 AM bilano99 Calculus 2 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top