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Claim: Let $n\in \mathbb N$, and let $f:\mathbb R \to \mathbb R$ be such that its $n$-th derivative $f^{(n)}(x)>0, \ \forall x\in \mathbb R$, then $f$ has at most $n$ roots.

Context: The fundamental theorem of algebra states (when only the real numbers are concerned) that an $n$-th degree polynomial has at most $n$ real roots. Since an $n$-th degree polynomial can be characterized as a function $f:\mathbb R \to \mathbb R$ with constant $n$-th derivative, the fundamental theorem of algebra can be equivalently stated as:

Let $n\in \mathbb N$, and let $f:\mathbb R \to \mathbb R$ has constant $n$-th derivative, then $f$ has at most $n$ roots.

Analogy: Thus the claim I'm questioning can be interpreted as a version of the fundamental theorem of algebra with the condition that

the function's $n$-th derivative is constant

replaced by the condition that

the function's $n$-th derivative is strictly positive.

Special cases:

  • for $n=1$ the claim follows from the fact that strictly increasing functions have at most one root;
  • likewise for $n=2$ the claim follows from the fact that strictly increasing functions have at most two roots (strictly convex function is strictly negative between any two roots, thus there can not be a third root).

Does the claim hold for $n\geq 3$ as well?

Pavel Kocourek
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    Hint: if $f$ has $\ge k$ roots then (by Rolle) $f'$ has $\ge k-1$ roots. – Anne Bauval Jan 12 '23 at 17:48
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    What you called the fundamental theorem of algebra, namely that a polynomial of degree has at most roots, is true for arbitrary fields, not just the real numbers. What is usually called the fundamental theorem of algebra, in contrast, is an existence theorem for roots, specific for the field of complex numbers. (See the Wikipedia page linked in your question.) – Andreas Blass Jan 13 '23 at 00:44
  • I just wanted to note that the claim could be stated in a more general way as follows: If $f:\mathbb R \to \mathbb R$ has nonzero $n$-th derivative, then it has at most $n$ roots. This way it is indeed a generalization in comparison to assuming that $f^{(n)}$ is a nonzero constant.

    Proof: From Darboux theorem it follows that nonzero $n$-th derivative implies that either $f^{(n)}>0$ or $f^{(n)}<0$. In the second case we can apply the original claim to $-f$ instead of $f$.

     – Pavel Kocourek Jan 13 '23 at 14:32

1 Answers1

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As long as we're talking about distinct roots (not roots with multiplicity), then the answer is yes. We can deduce it from the following fact:

If $f(x)$ is differentiable and has at least $k$ distinct roots, then $f'(x)$ has at least $k-1$ distinct roots. (Proof: let $x_1<x_2<\cdots<x_k$ be roots of $f(x)$; then by Rolle's theorem, $f'(x)$ has a root in each of the disjoint intervals $(x_1,x_2)$, $(x_2,x_3)$, $\dots$, $(x_{k-1},x_k)$.)

The contrapositive of that fact is: If $f'(x)$ has at most $\ell$ distinct roots, then $f(x)$ has at most $\ell+1$ distinct roots. The OP's claim now follows by induction.

(Note that in the context of general functions, rather than only polynomials, the multiplicity of a root isn't a nice quantity; in particular it doesn't have to be an integer even if it is well defined.)

Greg Martin
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  • It also works if you count roots with multiplicity--see my answer here. – Eric Wofsey Jan 12 '23 at 17:50
  • @GregMartin Thank you very much for your solution. I was actually contemplating about this type of argument, but initially I intended to assume only that the $n$-th derivative is weakly positive (which however would allow for functions identical to zero on an interval); if the $n$-th derivative was allowed to be zero, then finding a point $x_0$ such that $f^{(n)}(x_0)=0$ (by using Rolle's theorem iteratively) would not yet lead to a contradiction. I just reformulated the claim to weaken its assumptions here: https://math.stackexchange.com/q/4617227/1134951 – Pavel Kocourek Jan 12 '23 at 21:27
  • @EricWofsey You're right; I'll move the comment. – Andreas Blass Jan 13 '23 at 00:42