$$\Large \text{Question}$$
Question
Let's say that $y^{\left( \alpha \right)}\left( x \right)$ is the $\alpha$.th derivative of $y\left( x \right)$ with respect to $x$ ($x \in \mathbb{R} \vee x \in \mathbb{C}$ and $\alpha \in \mathbb{N}_{0}$). How to solve \begin{align*} y\left( y^{\left( 1 \right)}\left( y^{\left( 2 \right)}\left( y^{\left( 3 \right)}\left( \cdots \left( y^{\left( n \right)}\left( x \right) \right) \right) \right) \right) \right) &= f\left( x \right)\\ \end{align*} for $y\left( x \right)$?
We could also write this equation as $\left( y \circ y^{\left( 1 \right)} \circ y^{\left( 2 \right)} \circ y^{\left( 3 \right)} \circ \cdots \circ y^{\left( n \right)}\right)\left( x \right) = f\left( x \right)$ or $y_{m + 1}\left( x \right) = \left( y_{m} \circ y^{\left( m + 1 \right)}\left( x \right) \right)\left( x \right) \wedge y_{0}\left( x \right) = y\left( x \right) \wedge y_{n}\left( x \right) = f\left( x \right)$ if $m \geq 0$.
I don't necessarily want a closed form solution. I would also be interested in solutions in non-closed form, numerical solutions, series solutions, ..., or even just possible solution methods. Special cases would also be fine.
Why this question
The question came to my mind in the form of special cases. They seem interesting to me, but in most cases I am unable to find a solution (which made it even more interesting for me). I didn't find anything about it online (the closest thing was questions about functional roots) and online calculators like Mathematica didn't help me either. Hence the question, how do you solve this equation?
$$\Large \text{My approach}$$
Special Cases
If $f$ is constant
Let's say $f\left( x \right) = \text{c}$ where $\text{c}$ is some constant then $y\left( x \right) = \text{c}$ is a solution. Let's say $f\left( x \right) = 0$ then $y\left( x \right) = \sum_{k = 0}^{n - 1}\left[ y_{k} \cdot x^{k} \right]$ (where $y_{k}$ are constant terms) would also be a solution. Or more general: If $y^{n}\left( x \right) = \text{c}$ then $f\left( x \right)$ is also a constant given in terms of $\text{c}$s.
But all this is trivial.
If $f\left( x \right) = c \cdot x^{d}$
Let's say $f\left( x \right) = x$ (easy example). Let us assume that a solution of the form $y\left( x \right) = a \cdot x^{b}$ exists (guess), then (using $y^{\left( \alpha \right)}\left( x \right) = a \cdot \frac{\left( b + 1 \right)}{\Gamma\left( b - \alpha + 1 \right)} \cdot x^{b - \alpha}$):
| $n$ | $a$ | $b$ |
|---|---|---|
| $0$ | $1$ | $1$ |
| $1$ | $b^{-\frac{b}{1 + b}}$ | $\frac{1 \pm \sqrt{5}}{2}$ |
| $2$ | $\left( \left( b - 1 \right)^{\left( b - 1 \right) \cdot b} \cdot b^{b + \left( b - 1 \right) \cdot b} \right)^{-\frac{1}{1 + b + \left( b - 1 \right) \cdot b}}$ | $2.3247... \vee 0.33764... \pm 0.56228... \cdot i$ |
| $3$ | $\left( \left( b - 2 \right)^{\left( b - 2 \right) \cdot \left( b - 1 \right) \cdot b} \cdot \left( b - 1 \right)^{\left( b - 1 \right) \cdot b + \left( b - 2 \right) \cdot \left( b - 1 \right) \cdot b} \cdot b^{b + \left( b - 1 \right) \cdot b + \left( b - 2 \right) \cdot \left( b - 1 \right) \cdot b} \right)^{-\frac{1}{1 + b + \left( b - 1 \right) \cdot b + \left( b - 2 \right) \cdot \left( b - 1 \right) \cdot b}}$ | $1.5 \pm 0.4052... \cdot i \vee -0.13224 \vee 3.1322$ |
| $\vdots$ | $\vdots$ | $\vdots$ |
| $n$ | $\left( \prod\limits_{k = 0}^{n - 1}\left[ \left( b - k \right)^{\sum\limits_{g = k + 1}^{n}\left[ \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - g + 1 \right)} \right]} \right] \right)^{-\frac{1}{1 + \sum\limits_{g = 1}^{n}\left[ \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - g + 1 \right)} \right]}}$ | roots of $\prod\limits_{k = 0}^{n}\left[ b - k \right] = 1$ |
One could probably generalize this for $f\left( x \right) = c \cdot x^{d}$ by saying that $b$ is the solution to $\prod\limits_{k = 0}^{n}\left[ b - k \right] = d$ and $a = \left( c \cdot \prod\limits_{k = 0}^{n - 1}\left[ \left( b - k \right)^{\sum\limits_{g = k + 1}^{n}\left[ \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - g + 1 \right)} \right]} \right] \right)^{-\frac{1}{1 + \sum\limits_{g = 1}^{n}\left[ \frac{\Gamma\left( b + 1 \right)}{\Gamma\left( b - g + 1 \right)} \right]}}$ (if the term for a (above) is correct).
My thinking behind is that $f\left( x \right)$ is a monomial of degree $d$ and both sides of the equation must have a monomial of same degree. The monomial on the LHS has the degree $\prod\limits_{k = 0}^{n}\left[ b - k \right]$ thus $\prod\limits_{k = 0}^{n}\left[ b - k \right] = d$. So we just have to find the factor in front of the $x$ (on the LHS) and get $c$ out. So we could set the factor in front of it equal to $c$ (but finding a solution in closed form would almost never be possible), or we set $a = \sqrt[\text{exponent}]{c \cdot \frac{1 }{b\text{-term}}}$ where $b\text{-term}$ is the term involving $b$ in front of the $x$ and $\text{exponent}$ is the exponent $a$. The root cancels the exponent and $\frac{1}{b\text{-term}}$ cancels the factor in before $x$ (only leaving $c$). What remains is $c \cdot x^{d}$.
But this only worked, because of a lucky guess.
Reduction
Assuming $y^{\left( \alpha \right)}\left( x \right) = y^{\left( \alpha + k \cdot m \right)}\left( x \right)$ where $\left\{ \alpha,\, m \right\} \in \mathbb{N} \wedge k \in \mathbb{N}_{0}$
If $y^{\left( \alpha \right)}\left( x \right) = y^{\left( \alpha + k \cdot m \right)}\left( x \right)$ where $\left\{ \alpha,\, m \right\} \in \mathbb{N} \wedge k \in \mathbb{N}_{0}$ and $g\left( x \right) := \left( y \circ y^{\left( 1 \right)} \circ \cdots \circ y^{\left( m - 1 \right)} \right)\left( x \right)$ we could rewrite the equation to: \begin{align*} \left( g \circ \cdots \circ g \circ y \circ y^{\left( 1 \right)} \circ \cdots \circ y^{\left( n \right)}\right) &= f\left( x \right)\\ \left( g^{\left[ k \cdot m \right]} \circ y^{\left( k \cdot m \right)} \circ y^{\left( k \cdot m + 1 \right)} \circ \cdots \circ y^{\left( n \right)} \right) &= f\left( x \right)\\ \left( g^{\left[ k \cdot m \right]} \circ y \circ y^{\left( 1 \right)} \circ \cdots \circ y^{\left( n - k \cdot m \right)} \right) &= f\left( x \right)\\ \end{align*}
or
\begin{align*} g^{\left[ k \cdot m \right]}\left( y\left( y^{\left( 1 \right)}\left( \cdots\left( y^{\left( n - k \cdot m \right)} \right) \right) \right) \right) &= f\left( x \right)\\ \end{align*} where $g^{\left[ m \right]}$ is the $m$.th iteration of $g$. Taking the functional $k \cdot m$.th-root: \begin{align*} g\left( y\left( y^{\left( 1 \right)}\left( \cdots\left( y^{\left( n - k \cdot m \right)}\left( x \right) \right) \right) \right) \right) &= f^{\left[ \frac{1}{k \cdot m} \right]}\left( x \right)\\ y\left( y^{\left( 1 \right)}\left( \cdots\left( y^{\left( m \right)}\left( y\left( y^{\left( 1 \right)}\left( \cdots\left( y^{\left( n - k \cdot m \right)}\left( x \right) \right) \right) \right) \right) \right) \right) \right) &= f^{\left[ \frac{1}{k \cdot m} \right]}\left( x \right)\\ \end{align*}
Wich may not look simpler but it reduces the order from $n$ to $n - k \cdot m$ or $\text{order} \leq m$. Of course, $f^{\left[ \frac{1}{k \cdot m} \right]}\left( x \right)$ sometimes has no solution, one solution, multiple solutions, no closed-form solution, closed-form Solutions, ... E.G. if $\frac{n}{k \cdot m} \in \mathbb{N}$ then $y$ would be a Mittag-Leffler function (in some real interval) or / and $f$ would be an iteration of the same Mittag-Leffler function.
But that does not solve the problem...