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Conjecture. An infinitely differentiable function $f:\Bbb R \to \Bbb R$ at some $x\in \Bbb R$ has all its derivatives nonzero with identical or alternative sign if and only for any polynomial $p$ the function $f(x)-p(x)$ has at most $n+1$ roots, where $n$ is the degree of $p$.

It was show in [1] that the function $f(x)=e^x$ has the property that

Property. Every polynomial $p$ of degree $n$ intersects $f$ at no more than $n+1$ points.

What is more, every infinitely differentiable function $f:\Bbb R \to \Bbb R$ satisfying

Condition. All the derivatives of $f$ are nonzero with identical or alternative sign.

has the Property, as can be seen by any applying the result in [2] to the function $g(x)\equiv f(x)-p(x)$, $g(x)\equiv -f(x)-p(x)$, $g(x)\equiv f(-x)-p(x)$, or $g(x)\equiv -f(-x)-p(x)$.

Is the Condition necessary for the Property?

Pavel Kocourek
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  • Isn't $f(x) = e^{-x}$ a simple counterexample? It has the property, but does not satisfy the condition. – Martin R Jan 16 '23 at 19:30
  • Btw, your “condition” is related to “complete monotonicity,” compare https://mathworld.wolfram.com/CompletelyMonotonicFunction.html and https://en.wikipedia.org/wiki/Bernstein%27s_theorem_on_monotone_functions. – Martin R Jan 16 '23 at 19:46
  • $f(x) = -e^x$ is an even more trivial counterexample. A more proper condition might be that $f$ and all its derivatives are non-zero and have identical or alternating signs. – Martin R Jan 16 '23 at 19:58
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    @MartinR Thanks for the observations! I edited the question accordingly, I hope that now it makes sense. – Pavel Kocourek Jan 16 '23 at 21:10

1 Answers1

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I can give you a handwaivy argument why this should be a necessary condition at least for analytic functions $f$. Also w.l.o.g. $f(0) = 0$.

Lets first look at a simple case: assume $f'(x_0) = 0$ (again w.l.o.g. $x_0 = 0$) and $f'(x) > 0 ~\forall x\neq 0$ (otherwise it's even simpler). You now have to note, that $f^{(n)}(0) = 0$ too for all $n>1$ and thus at $x=0$ we have a saddlepoint meaning that locally $f$ looks like a $3$rd-degree polynomial. Indeed it looks like $ax^3$ locally, and you can thus easily construct a linear function $mx$ that interscts $f$ $3$ times. (Note: if we had not assumed $f'(x) > 0 ~\forall x\neq 0$ then we could also have an extrem point, but, as long as the condition is only broken locally, extrem points have to come in pairs, so that makes it even simpler.)

This generalizes the following way. If $n$ is the smallest natural number s.t. $f^{(n)}(x_0) = 0$ (w.l.o.g. $x_0 = 0$) and $f^{n}(x) > 0 ~\forall x\neq 0$ then around $x= 0$ $f$ looks like $p$ with \begin{align*} p(x) = ax^{n+2}+\sum_{k=0}^n a_kx^k \end{align*} You should now be able to find a polynomial of degree $n$ that intesects $f$ $(n+2)$ times.

StiftungWarentest
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  • Let me clarify what a necessary condition means when discussing implications in mathematics: The Condition being necessary means that every function satisfying the Property has to also satisfy the Condition. What you argue is that if we don't assume the Condition, then we can find a function violating the Property. – Pavel Kocourek Jan 16 '23 at 16:53
  • you are totally right, my bad, I've edited my answer to something more useful – StiftungWarentest Jan 17 '23 at 11:43
  • Thanks for the edit! I struggle to follow the proof, but perhaps I just need to take a nap. Btw, saddle point refers to a point of $f(x,y)$ at which $f$ is convex around some point in one direction and convex in another. What you describe is an inflection point. – Pavel Kocourek Jan 17 '23 at 12:13
  • I think you can call inflection points saddle points, too. At least that's how we do it in german. Anyway if you'd tell me more specifically which questions you have, I could explain my reasoning – StiftungWarentest Jan 17 '23 at 12:17