Let us consider the Riemann zeta function $\zeta(s)$, where $s$ can take on values on the domain $\mathbb{R}_{>1}$:

$$\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}} .$$

I wonder what is known about the functional square root(s) of the Riemann zeta function defined on the aforementioned domain (*). In other words, I'm curious about the properties of the function(s) $f$ such that $$f(f(s)) = \zeta(s). \qquad \qquad (1)$$


  1. Has a closed-form solution been found for $f$ in equation $(1)$ ?
  2. If not (which I expect), have partial results been found for such a function? Properties like existence, (non)uniqueness, continuity, or results about the functional square root of the partial sums? $$f(f(s)) = \sum_{n=1}^{k} \frac{1}{n^{s}} \qquad \qquad k \in \mathbb{Z}_{>0}$$
  3. If so, I would be grateful if you have some pointers to relevant articles or other sources.

(Cross-post from MSE.)

(*) Edit as per Gerald Edgar's answer, this condition should be changed. We must define $f$ on $(0, \infty) \cup X$ for some subset $X \subset \mathbb{R} \setminus (0,\infty) $. Then $f$ must map $(1,\infty)$ bijectively onto $X$, and $X$ iself onto $(0,\infty)$. Under these conditions, there is still a possibility that $f$ is both continuous and real-valued. I am interested in the properties of such an $f$.

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    $\begingroup$ Are you sure this is the question you want to ask? The zeta function on the domain you consider takes values outside this domain, so it cannot have a functional square root. In fact, for z with real part greater than 1, $\zeta(z)$ may have negative real part. See e.g. arxiv.org/pdf/1112.4910.pdf. $\endgroup$ May 1, 2020 at 23:37
  • $\begingroup$ Well, it turns out that one needs to deal with fixpoints. I no longer have all the references on a website, but take a look at mathoverflow.net/questions/45608/… and my self-answers. I can see that there is a fixpoint in the reals between 1 and, say, 10. $\endgroup$
    – Will Jagy
    May 2, 2020 at 3:35
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    $\begingroup$ @DmitryVaintrob I should also include that I'm only considering real $s>1$, so the values $\zeta(s)$ take are positive and real too $\endgroup$
    – Max Muller
    May 2, 2020 at 8:17
  • $\begingroup$ For real $s > 1$, $\zeta(s)$ maps $(1,\infty)$ bijectively onto itself. It is continuous and decreasing. A functional square root can therefore not be real-valued and continuous. $\endgroup$ May 2, 2020 at 11:27
  • $\begingroup$ @GeraldEdgar Could you please elaborate on that? I don't see why such a root can't be real-valued and continuous at once $\endgroup$
    – Max Muller
    May 2, 2020 at 22:00

1 Answer 1


Note that $\zeta$ maps $(1,\infty)$ bijectively onto itself, $\zeta$ is continuous, $\zeta$ is decreasing on $(1,\infty)$.

Suppose $\zeta = f \circ f$ where $f$ also maps $(1,\infty)$ onto itself. A continuous injective function $f$ from an interval to an interval is either increasing everywhere or decreasing everywhere. (This is from the intermediate value theorem.) But in either case, $f\circ f$ is increasing, so $f \circ f \ne \zeta$.

Now of course we can define $f$ on some larger set, say $(1,\infty) \cup X$, where $X$ has the power of the continuum. Let $f$ map $(1,\infty)$ bijectively onto $X$ and $X$ bijectively onto $(1,\infty)$. Easy.

  • $\begingroup$ Ahh, I see. Thank you! I'll modify the question according to the answer you've given, because I'm still curious about the existence of such an $f$ defined on a larger set. $\endgroup$
    – Max Muller
    May 3, 2020 at 16:46

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