19

Given a rational function $R(x)=P(x)/Q(x)$ with real coefficients, is it true that there exists an $M>0$ such that, for every $k\geq 0$, the restrictions $R^{(k)}|_{(-\infty,-M]}$ and $R^{(k)}|_{[M,\infty)}$ of the $k$-th derivatives $R^{(k)}(x)$ are all monotonic? Or, in other words, is all the interesting stuff happening inside $[-M,M]$ to all orders?

If we take the $k$-th derivative of $P/Q$, it can be proven by induction that it will be of the form $H/Q^{2^k}$ with $\deg H\leq p+(2^k-1)q-k$ where $p=\deg P$ and $q=\deg Q$. Idk if this is of any help.

(Edit: notice that we're asking if $\exists M\forall k \ldots$ and not just if $\forall k \exists M_k\ldots$; the latter is easy because a polynomial changes sign only finitely many times)

MathematicsStudent1122
  • 1
Qfwfq
  • 992
  • 1
    Hi! To avoid down-votes and close-votes, please provide us some context for this question, such as: (a) Is this homework? (b) If so, what course are you taking? (c) What specific topic are you covering at the moment? (d) What do you know that you think might be connected? (e) If you're stuck, what are you stuck on? For example, do you know what to apply, but don't know how to apply it, or do you not know what to apply? Please [edit] these facts into your original post, not as responses to this comment, as comments may be deleted without warning. – Brian Tung Jun 02 '24 at 01:05
  • Hi Brian. No, it's not homework, but I've no idea how "obious" (or not) my question is. I'm not adding this to the body of the question. – Qfwfq Jun 02 '24 at 01:10
  • There may be something useful, if only a good search term, at https://mathoverflow.net/questions/86738/which-functions-have-all-derivatives-everywhere-positive – Gerry Myerson Jun 02 '24 at 01:12
  • 1
    @Qfwfq: Oh, I'm not trying to bust your chops or anything. I've just noticed that in the past few years, Math.SE users have been more prone to down-voting and close-voting questions that are just problem statements. Yours has the advantage of being interesting, of course. How did you come to think of it? – Brian Tung Jun 02 '24 at 01:25
  • @Brian Tung: no problem! The backstory is: fragment of research problem of a friend who is a physicist; they asked me a question about a computation with some integrals; lo and behold I was led to ask myself a couple of questions about Schwartz class functions (such as some $R(e^x)$), whose answer might (or might not) be well known. But I think this one is a natural question to ask regardless from the motivation about Schwartz functions. – Qfwfq Jun 02 '24 at 01:40
  • @Gerry Myerson: thanks for the pointer: that may be related. Though, the condition in my OP might be weaker than "complete monotonicity", as I only ask for derivatives to not change sign (after $M$ and before $-M$), not necessarily to all have the same sign for all $k$. – Qfwfq Jun 02 '24 at 01:47
  • That's the sort of context that could go into the OP. For next time! :-) – Brian Tung Jun 02 '24 at 02:21

1 Answers1

28

This is false. It is easy to check that the derivatives of $f(x)=\frac{x-1}{x^2}$ have the form $$f^{(n)}(x)=C_n \frac{x-n-1}{x^{n+2}}$$

In particular, $f^{(n)}(x)$ changes sign about $x=n+1$. Thus, $f^{(n-1)}$ fails to be monotone in any open set containing $n+1$. So there's no absolute constant $M>0$ such that, for all $n$, $f^{(n)}$ is monotone in $[M, \infty)$.

MathematicsStudent1122
  • 1