17
$\begingroup$

Let $f$ be the entire function on $\mathbb C$ defined by $$ f(z)=\frac{z-\sin z}{z}. \tag{1}\label{1}$$ It is easy to see that $f$ is positive on $\mathbb R^*$ and has a zero of order 2 at 0. Does there exist an entire function $g$ such that $f=g^2$?

$\endgroup$
4
  • 1
    $\begingroup$ In case the notation $g^2$ refers to $x\mapsto g(x)^2$ and not $x\mapsto g(g(x))$: Then I would write down the Taylor series for $f$, write $g(x)=\sum a_i x^2$, try to solve for the $a_i$. I would bet that these $a_i$'s still go to zero fast enough, so that $g$ is defined on the whole of $\mathbb{C}$. $\endgroup$ Jul 11, 2023 at 10:22
  • 10
    $\begingroup$ That would be the most incredible coincidence in mathematics then :) but alas, most probably it is not true and here's how you can check it: find (numerically) some zero of $f$ off the real line, and then take a small circle around it and compute the logarithmic integral $\frac{f'(z)}{f(z)}dz$ along this circle. If the roots were doubled then we should get an even mutliple of $2\pi i$, but you most likely get something around $2\pi i$ so the root is simple and so there's no such $g$. $\endgroup$ Jul 11, 2023 at 10:26
  • 10
    $\begingroup$ I'm slightly puzzled by the poor reception of this question (currently 2 downvotes, 1 close vote). While it may be true, as explained by Aleksei, that no seems much more likely as the answer than yes, I don't see why this would make the question invalid. It still seems perfectly reasonable to ask for a proof of this fact. $\endgroup$ Jul 11, 2023 at 17:32
  • 4
    $\begingroup$ The function $f(z)$ has zeros very near to $$ z = \pm 7.497676277776385498272325 \pm 2.768678282987321532495314i $$ which are necessarily simple because (as already noted) $f'(z) = 1 - \cos z$ has only real zeros. Therefore $f$ does not have an analytic square root. $\endgroup$ Jul 12, 2023 at 21:19

2 Answers 2

15
$\begingroup$

$f$ is an entire function of exponential type (that is, $|f(z)|\lesssim e^{C|z|}$), and this implies that there are other zeros, in addition to $z=0$: if not, then its Hadmard factorization would be of the form $f(z)=Az^2e^{az}$, but clearly $f$ does not have such a representation.

However, as already explained by Mikhail in his answer (when the computational error is corrected), these additional zeros will be simple: if $f(z_0)=f'(z_0)=0$, then $\sin z_0=z_0$, $\cos z_0=1$, and taking squares and then the sum shows that $1=1+z_0^2$, so $z_0=0$.

Thus there is no entire $g$ with $g^2=f$.

$\endgroup$
2
  • 1
    $\begingroup$ For an explicit simple zero, experiments suggest that there is one close to 7.5+2.8i. More precisely, based on both the plot and computations, I believe that $f(z)$ should look like $\pm iw$ for $z=2\pi n+\pi/2\pm i\log(4\pi n)+w$ and $w$ small. I would assume that it can be made precise in a sense strong enough that we would be able to apply the Cauchy formula, and there would be a zero corresponding to every point of the form $2\pi n+\pi/2\pm i\log(4\pi n)$ for $n$ large enough. $\endgroup$
    – Pierre PC
    Jul 12, 2023 at 13:22
  • 1
    $\begingroup$ Even more explicitly, for $z_n=2\pi n+\pi/2+i\log(4\pi n)$, the sequence of functions $f_n:w\mapsto\sqrt nf(z_n-iw/\sqrt n)$ converges uniformly to the identity over compact sets, so for $n$ large enough it must have a zero of modulus at most $1$; this means that $f$ must have a zero within distance $1/\sqrt n$ of $z_n$. Since $\overline{f(z)}=f(\overline z)$, we also get zeros close to all the $\overline {z_n}$ for large $n$. $\endgroup$
    – Pierre PC
    Jul 12, 2023 at 13:54
3
$\begingroup$

If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Then $\sin^2z = 1-\cos^2 z =0$ and therefore $z=\sin z=0$, so that the only double zero is at the origin. If $f$ had no simple zeros, then the only zero would be the double zero at $z=0$, so that we can write $f(z)=z^2e^{h(z)}$. Since $\sin z$ grows at most exponentially, the growth of $Re\; h(z)$ cannot be quadratic or higher in $|z|$. Hence $h$ must be a linear polynomial. But the function $f$ is clearly not of that form. [I am grateful to Emil for providing this argument.] See Borel-Caratheodory.

Hence the square root cannot be an entire function. The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.

$\endgroup$
9
  • 2
    $\begingroup$ Kats Thanks for your interest in my question. If $1=\cos z$ and $w=e^{iz}$, you get $w+\frac{1}{w}=2$ (not $2i$). $\endgroup$
    – Bazin
    Jul 11, 2023 at 14:23
  • 3
    $\begingroup$ Can you elaborate on since $f$ is a transcendental function, it must hit every value infinitely many times? This is not true in general: e.g., the transcendental function $ze^z$ has exactly one zero. What follows from Great Picard is that a meromorphic function that is not a rational function attains all but possibly one values infinitely many times. $\endgroup$ Jul 12, 2023 at 10:54
  • 1
    $\begingroup$ No, that’s the point. I just gave you a counterexample: $ze^z$. Or more generally, $p(z)e^z$ for a nonconstant polynomial $p(z)$. This attains all values, but it attains zero only finitely many times. $\endgroup$ Jul 12, 2023 at 12:41
  • 3
    $\begingroup$ I don’t see how you’d do that, but then I’m no expert in complex analysis. However, Christian Remling’s argument is not really complicated. You don’t need any results on the existence of Hadamard factorization: the assumption that $f(z)$ has no zero besides the double zero at $z=0$ already gives that $f(z)=z^2e^{g(z)}$ for some entire $g$, and then you just need to argue that since $\operatorname{Re}g(z)=O(|z|)$, $g$ must be a linear polynomial. $\endgroup$ Jul 12, 2023 at 13:08
  • 1
    $\begingroup$ I should have mentioned that the last thing in my previous comment is an application of the Borel–Carathéodory theorem. $\endgroup$ Jul 13, 2023 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.