Candidate Functions f(x) for Domain R, Image (-1,1)

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The discussion focuses on finding candidate functions f(x) that meet specific criteria: defined for all real numbers, with an image of (-1,1), smooth and continuous, having an undefined first derivative at x=0, and approaching 1 as x approaches infinity and -1 as x approaches negative infinity. Suggestions include exploring rational functions, sigmoid curves, and transformations of arctan, although arctan is ultimately deemed unsuitable due to its behavior at zero. A proposed solution involves using the cube root of arctan, which aligns with the desired properties. The conversation highlights the challenge of balancing the function's behavior at zero with the required asymptotic limits. The thread concludes with a successful suggestion for a suitable function.
longrob
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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
 
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Well the horizontal asymptotes are at 1 and - 1, so a suitable rational function should do the trick.
 
Thanks, but I've not been able to find one. Any suggestions ?
 
check sigmoid curves
 
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.
 
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}}
 
snipez90 said:
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}
 

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g_edgar said:
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.
 

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