I was consider the set of linear operators:
$$O_{a,k} = \frac{f(ax^k) - f(x)}{ax^k - x} $$'
Particularly I am looking for the closed forms of the eigenfunctions of this operator, that is the functions f (dependent on parameters, a,k,x) such that
$$ O_{a,x}(f) = f$$
Work So Far:
It's trivial to show that $O_{a,k}(c) = 0$ for any constant c, and thus we intuitively find that: $O_{a,k}(x) = 1$. This naturally motivates the intuition that
$$ f = 1 + x + q_1(x) + q_2(x) + ... $$
Such that
$$ O_{a,k}[q_i(x)] = q_{i-1}(x) \ \text{and} \ O_{a,k}[q_1(x)] = x $$
Furthermore from the linearity of O it follows that $O_{a,k}(tf) = tO_{a,k}(f)$ for constants t.
We thus consider:
$$ O_{a,k}^{-1}(x^\mu) $$
Which by definition is the solution to
$$O_{a,k} = \frac{g(ax^k) - g(x)}{ax^k - x} = x^{\mu} $$
From here it follows that
$$g(ax^k) - g(x) = ax^{\mu + k} - x^{\mu+1}$$
Which trivially leads to
$$ g = \left( a\left( \frac{x}{a}\right)^{\frac{\mu + k}{k}} - \left( \frac{x}{a}\right)^{\frac{\mu + 1}{k}} \right) + \left( a\left( \frac{ \left(\frac{x}{a} \right)^{\frac{1}{k}}}{a}\right)^{\frac{\mu + k}{k}} - \left( \frac{\left(\frac{x}{a}\right)^{\frac{1}{k}}}{a}\right)^{\frac{\mu + 1}{k}} \right) ... $$
That reduces to
$$ g = \sum_{i=1}^{\infty} \left[ \frac{1}{a}\left( \frac{x^{\frac{\mu}{k^{i+1}}}}{a^{\frac{\left( \frac{1}{k} - 1 \right)^{i+1}}{\frac{1}{k}-1}}} - \frac{x^{\frac{\mu}{k^{i}}}}{a^{\frac{\left( \frac{1}{k} - 1 \right)^{i}}{\frac{1}{k}-1}}} \right)x^{\frac{1}{k^i}} \right]$$
This has an extremely elegant series formation which is due to the "power series in the power" structure of the term coefficients.
Interestingly every item in this series is itself a constant times a power of x and therefore the same formula can be recursively re-applied to each individual term, infinitely many times , for the case $\mu = 1$ to recover the definition of $f$ that we were originally considering.
Other Observations:
The case of k = 1 recovers back the the Jackson Q derivative and can thus be held down by the theory of Q series. If k = 1, and a= 1 we recover the traditional derivative from Calculus I.
The eigenfunction f is respectively the q exponential and traditional exponential for each.
