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longrob
Jun27-09, 12:33 AM
Hi all

What are some candidate functions f(x) that satisfy these conditions:
1. domain of f is R
2. image of f is (-1,1)
2. Smooth and continuous everywhere
3. first derivative undefined at x=0
4. f(x)-->1 as x--> inf
5. f(x)-->-1 as x--> -inf

Thanks
LR

snipez90
Jun27-09, 01:49 AM
Well the horizontal asymptotes are at 1 and - 1, so a suitable rational function should do the trick.

longrob
Jun27-09, 04:23 AM
Thanks, but I've not been able to find one. Any suggestions ?

trambolin
Jun27-09, 04:35 AM
check sigmoid curves

longrob
Jun27-09, 05:04 AM
Thanks. I considered arctan already, but since this function goes momentarily vertical zero arctan doesn't work. Same with a Gompertz function and Richards curve (I think). Also, this function appears to be odd, so that would rule out a Gompterz function also. Are there Sigmoid curves that are odd ? I don't know much about them, except in population models, and in those models a disappearing first derivative isn't too desirable I guess.

arildno
Jun27-09, 05:42 AM
Well, you might try with something like:

x>0: f(x)=C\sqrt{arctan(|x|)}
x<0: f(x)=-C\sqrt{arctan(|x|)}, C=\sqrt{\frac{2}{\pi}}

g_edgar
Jun27-09, 07:08 AM
Well the horizontal asymptotes are at 1 and - 1, so a suitable rational function should do the trick.

No, a rational function has the same asymptote at both ends.

So: in rejecting arctan, you say that you WANT it to be vertical at zero? (arctan has slope 1 at zero). Then take the cube root: (arctan(x))^{1/3}

longrob
Jun28-09, 07:37 PM
So: in rejecting arctan, you say that you WANT it to be vertical at zero? (arctan has slope 1 at zero). Then take the cube root: (arctan(x))^{1/3}

Perfect. Thank you.