Left and right acting derivatives

Saw an interesting notation in this Math Stackexchange answer. I quote -

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Eq. $(1)$ may be rewritten as $𝐷 (𝑓 \cdot 𝑔) = 𝑓 (\overset{⃖}{D} + \vec{D}) 𝑔$ , where $\overset{⃖}{D}$ acts on the left and $\vec{D}$ on the right, hence

𝐷^{𝑛} (𝑓 \cdot 𝑔) = {𝑓 (\overset{⃖}{D} + \vec{D})}^{𝑛} 𝑔 = \sum_{𝑘 = 0}^{𝑛} (\binom{𝑛}{𝑘}) 𝐷^{𝑛 - 𝑘} 𝑓 \cdot 𝐷^{𝑘} 𝑔

by the binomial theorem $-$ which can be applied straightforwardly, since $\overset{⃖}{D}$ and $\vec{D}$ commute. Now, this result can be extended to non-integer $𝑛$ by using [generalized binomial coefficients][1], leading to the following von Neumann series :

D^{𝛼} \equiv \sum_{𝑘 = 0}^{\infty} (\binom{𝛼}{𝑘}) {\overset{⃖}{D}}^{𝛼 - 𝑘} {\vec{D}}^{𝑘}

where the equivalent symbol means that the this relation has to be considered when applied to products of functions. Eq. $(3)$ corresponds with the case $𝛼 = 1$ .

As a final note, let’s remark that the geometric expansion is recovered due to the fact that the exponential function is an eigenvector of the derivative operator, whence

𝐷^{- 1} (𝑒^{𝑎 𝑥} 𝑞) = 𝑒^{𝑎 𝑥} {(\overset{⃖}{D} + \vec{D})}^{- 1} 𝑞 = 𝑒^{𝑎 𝑥} {(𝑎 + 𝐷)}^{- 1} 𝑞

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