Are there closed form solutions for functions

$f(z),g(z)$

Such that $A)$

$$ \frac{f(exp(z))}{f(z)} = g(z) $$

$B)$ $g(z),f(z)$ are both analytic and nonconstant.

$C)$ $g(z)$ is analytic at $z$ Whenever $f(z) = 0$.

$D)$ the RHS is shorter (simplified).

——

To clarify here is an analogue where $exp$ is replaced by $z^2$.

$$\frac{A(z^2)}{A(z)} = B(z)$$

$$ A(z) = z-1, B(z) = z + 1$$

Notice $A,B$ are analytic and nonconstant. And When $z = 1$ Then $A(z) = 0 $ and $B(z)$ is analytic.

Also $D)$ has been satified ; the RHS ( $B(z)$ ) is shorter / simplified version of the LHS.

—-

Also I do not accept the Abel function of $exp$ as a closed form.

I Will add tetration as a tag though, Because it relates.

  • Ummm...can't you just pick any particular $f$ with a closed form, and then you've given a closed form for $g$? – Eric Wofsey Feb 27 at 21:47
  • See the edit I made. – mick Feb 27 at 22:15
  • 1
    I am not sure what kind of clean formula do you expect, but if $\alpha$ solves $\alpha = e^{\alpha}$ then you may let $f(z) = z - \alpha$ so that $$g(z)=\frac{e^z - e^{\alpha}}{z-\alpha}$$ has removable singularity at $z=\alpha$. – Sangchul Lee Feb 27 at 22:44

EDIT: My answer works for a domain, but it appears the OP wants $f$ and $g$ to be entire functions. In this case, see Bumblebee's answer.

First, recognize that the composition of two analytic functions is analytic. This means that $f(\exp(z))$ is analytic as long as $f$ is.

Next, recognize that the quotient of two analytic functions is analytic whenever the denominator is non-zero. A proof can be found on pages 10-11 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch10.pdf

Finally, this implies that for a domain $\Omega$, as long as $f(\exp(z))$ and $f(z)$ are defined everywhere on $\Omega$ and $f$ is nowhere zero on $\Omega$, we can take: $$g(z) = \frac{f(\exp(z))}{f(z)}$$ This $g$ will be analytic, and non-constant so long as $f$ is non-constant. EDIT: this won't necessarily be non-constant, but it probably will be.

For example, take $\Omega = \{z: |z| < 1\}$ and $f(z) = z^2 + 1$. We get: $$g(z) = \frac{e^{2z} + 1}{z^2 + 1}$$ This pair satisfies.

  • In my analogue example we devide by $z-1$ yet our resulting function $ z + 1$ is analytic even When $ z - 1 = 0 $. – mick Feb 27 at 21:50
  • It's not obvious that $g$ is nonconstant so long as $f$ is nonconstant. It certainly will be nonconstant in most cases though. – Eric Wofsey Feb 27 at 21:51
  • Oh if you want a total function that's a little harder then. hang on let me think. – Joe Feb 27 at 21:52
  • 3
    @mick: If you want to require specific things about the domains of $f$ and $g$, you definitely need to mention that in the question itself... – Eric Wofsey Feb 27 at 21:53
  • 1
    @mick I really don't think "shorter/simplified" is well defined in this way. I think what you want is that the resulting $g$ is an entire function, in which case Bumblebee's answer is satisfactory. – Joe Feb 27 at 22:18

Here I assume all functions are entire:

To make $g(z)$ analytic, we need both $f(z)$ and $f(\exp(z))$ have the same zeros or both have no zeros.
In the case where $f(z)$ has no zeros, it is of the form $\exp(F(z))$ for some entire function $F$ and then simply choose $g(z)=\exp(F(\exp(z))-F(z)).$

  • I edited. Sorry for the confusion. – mick Feb 27 at 22:16

Let $f$ and $g$ be non-constant entire functions such that

$$ \frac{f(e^z)}{f(z)} = g(z) \quad \text{on} \ \mathbb{C}.$$

If $Z = \{z \in \mathbb{Z} : f(z) = 0\}$ denotes the zero set of $f$, then the above identity tells that $\exp(Z) \subseteq Z$. Now there are three possibilities:

  1. $Z = \varnothing$. This case reduces to Bumblebee's answer.

  2. If $Z$ is bounded, then $Z$ must be finite by the identity theorem. By partitioning $Z$ according to the orbits of $\exp$, we can find $(z_k)_{k=1}^{n} \subset \mathbb{C}$ and $(m_k)_{k=1}^{n} \subset \mathbb{Z}_{\geq 1}$ such that $z_k = \exp^{\circ m_k}(z_k)$ for each $k = 1,\cdots, n$ and

    $$Z = \bigcup_{k=1}^{n} \{ z_k, \exp(z_k), \cdots, \exp^{\circ(m_k-1)}(z_k) \} $$

    Of course, we may assume that all the elements appearing in the above formula distinct. Then there exists an entire function $F$ such that

    $$ f(z) = e^{F(z)} \prod_{k=1}^{n} \prod_{j=0}^{m_k-1} (z - \exp^{\circ j}(z_k)). \tag{*}$$

    Conversely, any $f$ of the form above induces an entire $g$ satisfying the functional equation.

  3. If $Z$ if unbounded, then I suspect that $f$ cannot be written in terms of elementary functions, based on the guess that $\exp$-orbits in $Z$ will blow up so fast and such growth speed cannot be replicated by elementary functions. On the other hand, the behavior of orbits of $\exp$ seems quite complicated as demonstrated in this article, so I am not sure about this case.

To be honest, I have absolutely no idea about Case 3 except that there is indeed an entire function $f$ falling into Case 3 (which can be constructed by Weierstrass factorization theorem). So my bet is that $\text{(*)}$ will cover all the 'closed-form solutions of $f$ (by allowing the empty product so as to include Case 1 as a special case).

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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