Are there closed form solutions for functions


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
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    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

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
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    @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
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    @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).

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