# What's this function ?

1. Jun 26, 2009

### longrob

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

2. Jun 27, 2009

### snipez90

Well the horizontal asymptotes are at 1 and - 1, so a suitable rational function should do the trick.

3. Jun 27, 2009

### longrob

Thanks, but I've not been able to find one. Any suggestions ?

4. Jun 27, 2009

### trambolin

check sigmoid curves

5. Jun 27, 2009

### longrob

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.

6. Jun 27, 2009

### arildno

Well, you might try with something like:

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

7. Jun 27, 2009

### g_edgar

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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8. Jun 28, 2009

### longrob

Perfect. Thank you.