Let $Df$ be an operator such that $Df = f(x+1) - f(x)$
Observe that
$$ D(e^{2\pi ix} ) = e^{2\pi i(x+1)} - e^{2 \pi i x} = 0$$
So if we attempt to place it in the basis $(1,x, \frac{1}{2}x(x-1), \frac{1}{3!}x(x-1)(x-2) ... )$
We naively assume that
$$ e^{2\pi i x} = e^{2\pi i (0)} + D(e^{2\pi i x})[0] x + \frac{1}{2}D^2 (e^{2\pi i x} )[0]x(x-1) + ...$$
But this tells us
$$ e^{2\pi i x} = 1 + 0 + 0 + 0 ... $$
Which has me very confused. How do I find cofficients $c_0, c_1 ... $ such that
$$ \sum_{k=0}^{\infty} c_k \begin{pmatrix} x \\ k \end{pmatrix} = e^{2\pi i x} $$
If the operator falls through, does this mean its EVEN possible?