# The $n$-th derivative has $n$ zeros. Can such a function be unbounded?

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.

• what about $x^{2n}$ ?
– zwim
Sep 24, 2017 at 21:02
• @zwim See my edit. I meant that the definition holds for all $n\in\Bbb N_0$ simultaneously. Sep 24, 2017 at 21:03
• doesn't $\exp(x^2)$ satify the same property for the zeros ?
– zwim
Sep 24, 2017 at 21:17
• @zwim I checked. The second derivative of $\exp(x^2)$ has no zero. Sep 24, 2017 at 21:19
• 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 $s\in(0,1/2)$. Sep 29, 2017 at 13:49

## 1 Answer

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.