Consider a linear operator $$L: \lbrace f: \Bbb{R}\rightarrow \Bbb{R} \rbrace \rightarrow \lbrace f: \Bbb{R}\rightarrow \Bbb{R} \rbrace $$
For example
$$ L(f) = f(x+1) - f(x)$$
Define the $M-\text{kernel}$ as the set $\lbrace x: L(x) = M \rbrace$ One can seek the 0-kernel of $L$, call it $M_1$ and then ask for the $M_1-\text{kernel}$, call that result $M_2$ etc... If we consider one representative from each set we form a basis for our underlying space, An example basis for the set of functions is given below:
$$ 0, 1, x, \frac{1}{2}x(x-1), \frac{1}{6}(x)(x-1)(x-2) ... $$
A more familiar example if $L(x) = \frac{df}{dx}$ is the basis
$$ 0, 1, x, \frac{1}{2}x^2 , \frac{1}{6}x^3 ... $$
So here's the question,
If I'm given an order basis
$$ w_1(x), w_2(x), w_3(x) ... $$
How do I find a linear operator L such that $L(w_i(x)) = w_{i-1}(x)$ and $L(w_1(x)) = 0 $
For example if given: $$ 0, 1, x, \frac{1}{2}x^2 , \frac{1}{6}x^3 ... $$
One possible answer would be differentiation, but suppose Instead I give
$$ 0 ,1, x, \frac{1}{2}x(x-1), \frac{1}{6}x(x-1)(x-4), \frac{1}{24}x(x-1)(x-4)(x-9) ... $$
How do I find the linear operator that gives rise to this?