How to find operator with Fibonacci eigenvalues?

Question

How can I find the operator that satisfies this equation?

$$F_nx^n=Dx^n$$

Summing over $n$ we can rewrite this as $$\frac1{1-x-x^2}=D\frac1{1-x}$$ I am unsure whether this can be solved.

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I am trying to solve it in a similar way to something like this: $$nx^n=Dx^n$$ has solution $$D=x\frac{\mathrm d}{\mathrm dx}$$ I am not sure really how else to describe it. Other than the fact that $D$ can't really depend on $n$ which might make this equation unsolvable.

Answer 1

Using Binet's formula: $$F_n=\frac{1}{\sqrt{5}}\left(\varphi^n-\psi^n\right)$$

where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$.

Then $$(Df)(x) = \frac{1}{\sqrt{5}}\left(f(\varphi x)-f(\psi x)\right)$$

If you write the scaling operators $(S_\alpha f)(x)=f(\alpha x)$, then you can write the above as:

$$D=\frac{1}{\sqrt{5}}\left(S_\phi-S_\psi\right)$$

I suspect there is no closed form in terms of just $\frac{d}{dx}$ and $x$.

Note that the $S_\alpha$ have the following properties:

$$\begin{align}\frac{d}{dx}S_\alpha &= \alpha S_\alpha \frac{d}{dx}\\ S_\alpha S_\beta &= S_{\alpha\beta}\\ xS_\alpha &= \frac{1}\alpha(S_\alpha x)\end{align}$$

Answer 2

Note: There are (at least) two ways of indexing the Fibonacci series. Below I work with the $F_0 = F_1 = 1$ convention rather than the $F_0 = 0, F_1 = F_2 = 1$ convention.

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I assume you want $D$ to be a differential operator, given the tags. Then we have

$$D = \sum_{k = 0}^\infty f_k(x) \frac{d^k}{dx^k}$$

for some functions $f_k(x)$, and we want

$$F_n x^n = D x^n = \sum_{k = 0}^\infty f_k(x) \frac{d^k}{dx^k} x^n = \sum_{k = 0}^n (n \; P \; k) \; f_k(x) \; x^{n-k}$$

Let's suppose $f_k(x) = c_k x^k$ for some constant $c_k$, making this say

$$F_n x^n = \sum_{k = 0}^n (n \; P \; k) c_k x^n$$

or just

$$F_n = \sum_{k = 0}^n (n \; P \; k) c_k = \sum_{k=0}^n \frac{n!}{(n-k)!} c_k$$

This gives a recursive definition of the $c_k$. Now if we find the first few $c_k$ we see that they're fractions with denominators that look like k!, so let's try defining $a_k := k! c_k$ so that our expression becomes

$$F_n = \sum_{k=0}^n \frac{n!}{(n-k)! k!} k! c_k = \sum_{k=0}^n \binom{n}{k} a_k$$

Calculating the first several $a_k$ we see that they are $$1, 0, 1, -1, 2, -3, 5, -8, 13, -21, \ldots$$ To me, that looks like $a_0 = 1$ and $a_k = (-1)^k F_{k-2}$ for $k > 0$. I imagine you can check this actually works using Pascal's identity or something, though I didn't actually check it. This would then give the formula $$D = 1 + \sum_{k=1}^\infty \frac{(-1)^k F_{k-2}}{k!} \; x^k \; \frac{d^k}{dx^k},$$ at least assuming I got all the indices correct.