Question: Given a $C^\infty$-function $f:\Bbb R\to\Bbb R$ for which the $n$-th derivative has exactly $n$ zeros (counted with multiplicity) for all $n\in \Bbb N_0$. Can such a function be unbounded?

The motivation comes from another question of mine. I conjectured that functions with such zero-patterns look "bell-shaped". Examples might be $$\exp(-x^2)\quad\text{and}\quad \frac1{1+x^2}.$$

To construct an unbounded example, I had the following idea: take an intuitively bell-shaped function (like one of the above) which vanishes at infinity. Now, replace the converging tails with something that does diverge to $-\infty$ instead. The divergence must be sufficiently slow so that the zero pattern is preserved. I was not successful so far.

For another idea, take once more a function with the desired zero-pattern. Then, add an unbounded function but pay attention to not destroy the zero pattern. This too turned out to be very tricky.

  • $\begingroup$ what about $x^{2n}$ ? $\endgroup$
    – zwim
    Sep 24, 2017 at 21:02
  • $\begingroup$ @zwim See my edit. I meant that the definition holds for all $n\in\Bbb N_0$ simultaneously. $\endgroup$
    – M. Winter
    Sep 24, 2017 at 21:03
  • $\begingroup$ doesn't $\exp(x^2)$ satify the same property for the zeros ? $\endgroup$
    – zwim
    Sep 24, 2017 at 21:17
  • $\begingroup$ @zwim I checked. The second derivative of $\exp(x^2)$ has no zero. $\endgroup$
    – M. Winter
    Sep 24, 2017 at 21:19
  • 1
    $\begingroup$ An answer to this question was provided by a later post on MO. Unbounded examples are $\log(1+x^2)$ or $(1+x^2)^s$ for $s\in(0,1/2)$. $\endgroup$
    – M. Winter
    Sep 29, 2017 at 13:49

1 Answer 1


Let $$f(x) = 1 + \log(\sqrt{1 + x^2}).$$ This has no zeros.

Then $$f'(x) = \frac{x}{x^2 + 1},$$ which is defined for all real $x,$ has exactly one zero, and asymptotically approaches zero as $x\to\infty$ or $x\to-\infty.$

Observe that $f'(x)$ is the real part of $g(x)$ (notated $f'(x) = \Re[g(x)]$) where $$g(z) = \frac{z+i}{z^2+1} = \frac{1}{z-i}$$ for $z \in \mathbb C, z \neq i,$ and that for $n \geq 1,$ $$g^{(n)}(z) = \frac{d^n}{dx^n}\left(\frac{1}{z-i}\right) = \frac{(-1)^n \,n!}{(z - i)^{n+1}} = (-1)^n \,n!\frac{(z + i)^{n+1}}{(z^2 + 1)^{n+1}}.\tag1$$

If we define $g^{(0)} = g,$ that is, the "zeroth derivative" of a function is the function itself, then Equation $(1)$ holds for all $n \geq 0.$

Since $g(z)$ is analytic for $z \neq i,$ in particular, at any real $z,$ $$ f^{(n+1)}(x) = \Re[g^{(n)}(x)] = (-1)^n n! \frac{\binom n0 x^{n+1} - \binom n2 x^{n-1} + \binom n4 x^{n-3} - \binom n6 x^{n-5} \pm \cdots} {(x^2 + 1)^{n+1}}.\tag2 $$ for $x \in \mathbb R, n \geq 0.$ That is, $f^{(n+1)}(x)$ is a rational function in which the denominator is always positive and is a higher-degree polynomial than the numerator, so $f^{(n+1)}(x)$ is defined for all real $x$ and approaches zero as $x\to\infty$ or $x\to-\infty.$ Because every derivative of $f$ has these asymptotes, $f^{(n+1)}(x)$ has one more root than $f^{(n)}(x)$ does, not counting multiplicity of roots; it follows by induction that $f^{(n+1)}(x)$ has $n+1$ roots without counting multiplicity, and since it can only have $n+1$ roots with multiplicity, it has exactly that many roots with multiplicity. (Every root has multiplicity $1.$)

I think we can conclude that $f(x) = 1 + \log(\sqrt{1 + x^2})$ is a $C^\infty$ function $f:\mathbb R\to\mathbb R$ for which the $n$th derivative has exactly $n$ zeros (counting with multiplicity) for every non-negative integer $n.$ And it is easy to see that $1 + \log(\sqrt{1 + x^2})$ is unbounded.

The inspiration to look for a function of this form is that $1 + \log(\sqrt{1 + x^2})$ is bell-shaped, but "upside down," so that the tails are not bounded by the $x$-axis. If you want a "right-side up" bell, you can take a function such as $-(1 + \log(\sqrt{1 + x^2})),$ which is below the $x$-axis and therefore still can have unbounded tails. I believe a "right-side up" bell above the $x$-axis must be bounded, since it is only permitted to have one horizontal tangent and is not permitted to have any zeros.


You must log in to answer this question.

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