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S May 28 at 3:07 history suggested cottontail CC BY-SA 4.0
mathjax improvements; show hidden graph
May 28 at 2:20 review Suggested edits
S May 28 at 3:07
Nov 12, 2022 at 16:48 comment added hosford42 Numerically stable formulation: f(x) = x^b/(x^b + (1-x)^b)
Jun 5, 2020 at 9:57 review Suggested edits
Jun 5, 2020 at 17:48
May 9, 2019 at 17:27 comment added wmsmith You can add an extra level of adjustment so that you can tune the position at which the function is equal to .5 by using y=1/(1+(x^r/(1-x^r))^-b). Then, to attain y=.5 at x0 set r=-log(2)/log(x0). Or, if you want to ensure that y=k for some k between 0 and 1 at x=x0 then set r=-log((1/k - 1)^(1/b)+1)/log(x0)
Jan 10, 2018 at 10:05 comment added Ismam Huda Hi Darkmoor, I got this equaion by playing around with the "inverse logit function". You can see that it resembles y=inverse logit (x) = 1 / (1 + e^-x) Because logit maps to (0,1) imgur.com/a/H0kGF
Jan 8, 2018 at 10:39 comment added Darkmoor Is this a standard function like the $tanh$ of $sin$? I am interested to identify it in the wed but I could not. Could you please give a reference?
Jul 8, 2017 at 8:57 review Late answers
Jul 8, 2017 at 9:18
Jul 8, 2017 at 8:47 review First posts
Jul 8, 2017 at 14:12
Jul 8, 2017 at 8:40 history answered Ismam Huda CC BY-SA 3.0