Wikipedia:Reference desk/Archives/Mathematics/2014 June 8

= June 8 =

Function meeting a requirement
What possible expressions could I use to define the function $$f(x)$$ such that $$f(0)=1$$, $$\lim_{x \to \pm \infty}f(x)=0$$, $$f(x)>f(x+\Delta x)$$ for all $$x$$ over $$[0,\infty)$$, and $$f(x)<f(x+\Delta x)$$ for all $$x$$ over $$(-\infty,0]$$? I know that $$f(x)=e^{-x^2}$$ satisfies those conditions, but I feel like I've seen others that do as well. Possible one involving cosine. — Melab±1 &#9742; 03:49, 8 June 2014 (UTC)


 * This one
 * $$f(x)=\frac 1{1+x^2}$$.
 * Bo Jacoby (talk) 05:31, 8 June 2014 (UTC).


 * Your requirement is somewhat unclear - what if $$\Delta x$$ is negative, or is greater than |x| ? So I'm going to assume your requirement is satisfied if $$f(x)$$ has strictly positive gradient when x is negative and has strictly negative gradient when x is positive. An example involving cosine could be


 * $$f(x) = \cos \left ( \frac {\pi x^2}{2x^2 + 1} \right )$$


 * Gandalf61 (talk) 09:43, 8 June 2014 (UTC)


 * $$\frac{1}{\cos(ix)}$$? —Tamfang (talk) 09:14, 9 June 2014 (UTC)


 * Other random suggestions:
 * $$f(x) = \cos\left(\frac{\pi}{2} \left(1 - e^{-x^2}\right)\right)$$
 * $$f(x) = \frac{2}{e^{x^2} + \cos(\log(1+x^2))}$$
 * Icek (talk) 11:42, 9 June 2014 (UTC)

Edited cos and log to \cos and \log. :) --CiaPan (talk) 17:15, 9 June 2014 (UTC)