Recently I came across a rather interesting equivalence to finding the Maclurin Expansion of $e^x$ .
The Solution goes as follows;
Let $\mathbb{D^{-1}_{m}}$ be the antiderivative operator , where $m\in\mathbb{R}$
Assume that the antiderivate operator is bounded in $[0,x]\forall{x}\in{m}$.
We apply the operator as follows:
$\mathbb{D^{-1}}[e^x] = e^x - 1$
This then yields via a transpose of the equation that,
$e^x = \frac{1}{1-\mathbb{D^{-1}}}$
Notice that this formulation is equivalent to that of a geometric series, such that it can be rewritten that,
$e^x = \sum\limits_{k=1}^{\infty}\mathbb{D^{-1}}^{k-1}$
From here, we look at the first few initial terms of the series, namely;
$e^x = 1 + \mathbb{D^{-1}} + \mathbb{D^{-1}}\mathbb{D^{-1}} + \mathbb{D^{-1}}\mathbb{D^{-1}}\mathbb{D^{-1}} + ... $
Which can be written in integral form as,
$e^x = 1 + \int_{0}^{x}dx + \int_{0}^{x}\int_{0}^{x}dxdx + \int_{0}^{x}\int_{0}^{x}\int_{0}^{x}dxdxdx + ...$
This eventually results in a series that is equivalent to the Maclurin Expansion of $e^x$ namely,
$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...$
Which eventually leads to,
$e^x = \sum\limits_{k=0}^{\infty}\frac{x^k}{k!}$
Curiously this entails that the geometric series of the operator is equivalent to the Maclurin series of $e^x$ , I am curious to know if that there is a reason outside of pure chance, that the expansion yields this equivalence or if it is just by chance that the geometric series of the integral operator happens to be equivalent to the Maclurin Series expansion of $e^x$.