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Suppose one has an entire complex analytic function $f(z)$ having no zeros on the real axis. Is it possible to find an analytic function $g(z)$ such that the coefficients of the power series expansion of $f(z)g(z)$ in $z$ are always real and positive (so that, e.g. "Descartes' rule of signs" can be used to "explain" the absence of zeros)? How can one find $g(z)$?

This is attempting be a kind-of generalization of the question Property of a polynomial with no positive real roots to complex analytic functions. The goal is to "explain" why $f(z)$ has no real roots, by studying the properties of $f(z)g(z)$ (or obtaining insights into $f(z)$ by studying the properties of $g(z)$). Are there (non-trivial) theorems that apply to complex analytic functions with no real zeros? Some Fredholm-alternative-like thingy?

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    $\begingroup$ What are "the coefficients" of an analytic function? Are you talking about the power series at some specific point? Also, on what domains are these analytic functions defined? If $f$ is analytic at $x_0$ and $f(z_0)\ne0$, then on a small enough disc around $z_0$, one can choose $g$ such that $f(z)g(z)$ is any analytic function you want. $\endgroup$
    – Greg Martin
    Dec 26, 2018 at 7:27
  • $\begingroup$ It's entire; the power series at $z=0$, as "usual". I'm not trying to be tricky, here. I don't want "any" analytic function; I want to understand why there are no zeros on the real axis. $\endgroup$
    – Linas
    Dec 26, 2018 at 7:58

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Let $g(z)=\overline{f(\overline z)}$. Then, regardless of whether $f$ has any zeroes on the real axis, the function $fg$ is real on the real axis, so all its derivatives at the origin are real.

I'm mystified what it might mean to explain why a function has no zeroes on the real axis, or what sort of "explanation" you're hoping for. But the existence of $g$ so that all the coefficients of $fg$ are real has absolutely nothing to do with whether $f$ has any zeroes on the real axis.

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  • $\begingroup$ Yes, you are right. My question arises from some late-night insanity on my part; I guess the correct protocol is to accept your answer. What I actually have is a function that has no zeros on any ray that is a rational multiple of $2\pi$. I was imaging that I could rotate each ray so that it lies along the real axis ... and deduce something clever. $\endgroup$
    – Linas
    Dec 26, 2018 at 17:19
  • $\begingroup$ Oh, and since I decided that my question is a "bad question", should I delete it, or let it live on? $\endgroup$
    – Linas
    Dec 26, 2018 at 17:21
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    $\begingroup$ @Linas Whether to delete it is up to you. I probably would, since I really don't see the point to it... $\endgroup$
    – David C. Ullrich
    Dec 26, 2018 at 17:31
  • $\begingroup$ OK, thanks, my apologies for wasting your time. I'm embarassed. $\endgroup$
    – Linas
    Dec 26, 2018 at 17:50

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