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I noticed the similarity between Taylor series and Newton series: $$\sum\limits_{n \geq 0} \frac{D^nf(0)}{n!} x^n $$

$$\sum\limits_{n \geq 0} \frac{\Delta^nf(0)}{n!} x^{\underline{n}} $$ where $Df = df/dx$, $\Delta f(x) = f(x+1)-f(x)$, $x^\underline{n} = x(x-1)\ldots(x-n+1)$.

The natural question that comes to mind is that we can generalize that for a more general linear operator $A$ and a family of functions $\{ p_n \}_{n\geq 0}$ which satisfies properties

$$ p_0 (x) = 1 $$ $$ p_n(0)= 0, \ n \geq 1$$ $$ A p_n = n p_{n-1} $$ $$ A1 = 0$$

So we will have the series

$$\sum\limits_{n \geq 0} \frac{A^nf(0)}{n!} p_n (x) $$

Has it been studied already? Which conditions should satisfy $A$, $\{ p_n \}_{n\geq 0}$ so the series converges to $f$? What are the other examples of such series?

ids
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  • Series of Binomial type can be written as series by means of a delta operator. –  Jul 26 '22 at 18:47
  • @DecarbonatedOdes Okay, thanks. I see that $p_n (x+y)$ is expanded to $\sum \binom{n}{k} p_k(x) p_{n-k}(y)$. But why does sequence $p_n$ have to be polynomial? – ids Jul 27 '22 at 08:08
  • And why $A$ have to be shift equivariant? What will break if it's not? – ids Jul 27 '22 at 08:21

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