4
$\begingroup$

Define a sigmoid as any bounded, odd, increasing function from $\mathbb{R} \rightarrow \mathbb{R}$, and a pretty sigmoid as a sigmoid which is convex over $\mathbb{R^-}$ and concave over $\mathbb{R^+}.$

Let $P$ be the smallest set of functions from $\mathbb{R} \rightarrow \mathbb{R}$ containing polynomials and closed under exponentiation and product.

If there a function $f$ in $P$ such that $\lim_{x\rightarrow -\infty} f(x) = -\infty$, $\lim_{x\rightarrow\infty} f(x) = 0$ and $e^f - \frac{1}{2}$ is a (pretty) sigmoid.

Improve this question
$\endgroup$
11
  • 1
    $\begingroup$ What do you mean, $S$ is "the smallest set of functions"? There are no "seed" elements. For instance, $S$ being the empty set, or the set of constant functions, both satisfy your constraint. Does e.g. $S$ necessarily contain $f(x) = x$? $\endgroup$
    – Alex Meiburg
    Commented Feb 7, 2017 at 3:56
  • $\begingroup$ Sorry I meant to add that x is in S $\endgroup$
    – Arthur B
    Commented Feb 7, 2017 at 4:56
  • $\begingroup$ Is there still a mistake in this question? Surely for $x>0$ one wants $f(x)>f(0)$ and $f''(x)<0$, contradicting the last hypothesis in the question -- or have I misunderstood what a sigmoid is? $\endgroup$
    – Kevin Buzzard
    Commented Feb 7, 2017 at 9:48
  • 5
    $\begingroup$ Isn't the error function a pretty sigmoid function that extends to an entire function? $\endgroup$
    – Gro-Tsen
    Commented Feb 9, 2017 at 19:30
  • 1
    $\begingroup$ This version of the question doesn't seem equivalent to the one you asked originally (at least not obviously). Is there a reason for the change? $\endgroup$
    – Christian Remling
    Commented Feb 9, 2017 at 22:16

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .