13
$\begingroup$

I asked this on Math.SE some days ago, but without any success. For some application I need a formal definition of bell-shaped function. So I had the following idea:

Definition. A $C^\infty$-function $f:\Bbb R\to\Bbb R$ should be called bell-shaped if for all $n\geq 1$ the $n$-th derivative has exactly $n$ zeros (counted with multiplicity).

E.g. this works for $\exp(-x^2)$. As this is my try to formalize an informal term, I have no way to check if this is correct. However, I tried to prove that such functions must be bounded (for me, intuitively bell-shaped functions are always bounded). But I had no success.

Question: Is a bell-shaped function (in the sense above) always bounded?

Improve this question
$\endgroup$
3
  • $\begingroup$ What about $e^{x^2}$? $\endgroup$
    – Ori Gurel-Gurevich
    Commented Sep 28, 2017 at 16:16
  • $\begingroup$ @OriGurel-Gurevich The second derivative has no zeros. See the comments under the Math.SE post. $\endgroup$
    – M. Winter
    Commented Sep 28, 2017 at 16:17
  • $\begingroup$ Right, oops.... $\endgroup$
    – Ori Gurel-Gurevich
    Commented Sep 28, 2017 at 16:19

2 Answers 2

Reset to default
17
$\begingroup$

As suggested by Mateusz Kwaśnicki, the function $f : x \mapsto (1+x^2)^{s}$ is bell-shaped and unbounded for any $s \in (0,\frac{1}{2})$.

It is easy to see that $f^{(n)}(x) = P_n(x) (1+x^2)^{s-n}$ where $P_n$ is a polynomial of degree $\leq n$. Actually $$ P_{n+1}(x) = (1+x^2) P_n'(x) - 2(n-s)xP_n(x). $$ Let $a_n$ be the coefficient of $x^n$ in $P_n$. Then $a_{n+1} = n a_n - 2(n-s)a_n = (2s-n) a_n$. In particular $(-1)^{n-1} a_n > 0$ for $n \geq 1$, so that $P_n$ has degree exactly $n$ (this is where we use $s < \frac{1}{2}$).

One first checks that $f'$ has a single zero. We then assume that $f^{(n)}$ has exactly $n$ distinct simple zeroes $x_n < x_{n-1} < \dots < x_1$ for some $n \geq 1$. Then $$ (-1)^{n-1} f^{(n+1)}(x_1) = (-1)^{n-1} P_n'(x_1) (1+x_1^2)^{s-n} > 0 $$ since $(-1)^{n-1} P_n(x) > 0$ for $x > x_1$. But $(-1)^{n-1} f^{(n+1)}(x) < 0$ for large $x$ so $f^{(n+1)}$ has a zero in $(x_1,+\infty)$. Similarly $f^{(n+1)}$ has a zero in $(-\infty ,x_n)$. Together with the zeroes of $f^{(n+1)}$ between the $x_i$'s, we get $\geq n+1$ distinct zeroes (and therefore exactly $n+1$ zeroes).

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ For what it's worth, I think basically the same proof works for $log(1+x^2)$. I wonder if one can characterize a class of functions that look like these inverted bell shapes with slow (i.e., less than linear) growth to infinity. $\endgroup$
    – Aaron Bergman
    Commented Sep 29, 2017 at 0:16
  • $\begingroup$ @AaronBergman I think we can divide the "bell-shaped" functions from my question into the two classes bounded and unbounded and the latter ones are exactly the ones you want to characterize. When the function has a minimum at $x=0$ then $f''(0)>0$ and because there are zeros of $f''$ left and right of $x=0$ the second derivative is negative far from the origin. This means $f$ is finally concave, hence sub-linear. I am missing something? $\endgroup$
    – M. Winter
    Commented Sep 29, 2017 at 0:45
9
$\begingroup$

If I remember correctly, $\sqrt{1 + x^2}$ is bell-shaped, and it is not bounded. (The $n$-th derivative is a polynomial of degree $n$ times $(1 + x^2)^{1/2-n}$, and it has at least $n$ zeroes by the intermediate value theorem).

By the way, is there any progress on the characterisation of bell-shaped functions?

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Thank you. But a quick check showed no zeros of the second derivative. For your question, do you mean progress since my question on Math.SE? $\endgroup$
    – M. Winter
    Commented Sep 28, 2017 at 14:57
  • 3
    $\begingroup$ The $n$'th derivative is actually a polynomial of degree $n-2$ (not $n$) times $(1+x^2)^{1/2-n}$, provided that $n\geq 2$. This means that $y''$ is potentially bell-shaped, but $y''=(1+x^2)^{-3/2}$, which is bounded. $\endgroup$
    – Neil Strickland
    Commented Sep 28, 2017 at 15:05
  • 2
    $\begingroup$ Oh, indeed, then I remembered incorrectly. How about $(1+x^2)^s$ for $s \in (0, \tfrac{1}{2}$)? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 28, 2017 at 15:24
  • 1
    $\begingroup$ If we take $f(x)=(1+x^2)^{1/4}$ then numerical calculation suggests that $f^{(n)}(x)$ has precisely $n$ roots, all simple, for $n\leq 20$. $\endgroup$
    – Neil Strickland
    Commented Sep 28, 2017 at 15:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .