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

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June 8

Function meeting a requirement

What possible expressions could I use to define the function $f(x)$ such that $f(0)=1$, $\underset{x\to \pm \mathrm{\infty }}{\lim }f(x)=0$, $f(x)>f(x+\mathrm{\Delta }x)$ for all $x$ over $[0,\mathrm{\infty })$, and $f(x)<f(x+\mathrm{\Delta }x)$ for all $x$ over $(-\mathrm{\infty },0]$? 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 ☎ 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 $\mathrm{\Delta }x$ 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:

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

$f(x)=\frac{2}{e^{x^2}+\cos (\log (1+x^2))}$

Icek (talk) 11:42, 9 June 2014 (UTC)

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