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In analogy with Taylor series, we have

$$ f(x + \alpha) = \sum_{n=0}^{\infty} \frac{\alpha^{\underline{n}}}{n!} \Delta^{n}f(x) = \sum_{n=0}^{\infty} \binom{\alpha}{n} \Delta^{n}f(x). $$

We may take $\alpha$ to be any real number, with suitable definitions for $\alpha^{\underline{n}}$ and $\binom{\alpha}{n}$. Given a sample of points $f(k)$ at $k=0, 1, 2, \ldots$ we may use this to interpolate and get $f(x)$ to for any real $x$. Surprisingly, this works for a variety of analytic non-polynomial functions.

However, we know this does not recover all possible functions, even if they are analytic. For example, $f(x) = \sin(2\pi x)$ has $f(0) = 0, f(1) = 0, f(2) = 0, f(3) = 0, \ldots$, but Newton's interpolation formula returns $f = 0$. However, if I plug $f(x) = \sin(x)$, the formula seems to provide accurate results for $f(x+\alpha) = \sin(x+\alpha)$.

Is there some sort of a non-trivial characterization of functions for which inputting $f$ into the formula returns the same function $f$? I am able to derive a remainder formula for the partial sums, but it's inelegant to write down and it doesn't offer much insight to me at the moment.

MaximusIdeal
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  • So $\alpha^{\underline{n}} = \alpha(\alpha - 1)\cdots(\alpha - n + 1)$? – Paul Sinclair Aug 09 '24 at 17:51
  • There isn't a nice way of characterizing which $C^\infty$ functions have a Taylor series about a point $a$ which converges to the function. So I wouldn't hold out much hope for its discrete equivalent being any better. – Paul Sinclair Aug 09 '24 at 17:59
  • @PaulSinclair "So $\alpha^{\underline{n}} = \alpha(\alpha - 1)\cdots(\alpha - n + 1)$?" Yes, it's the falling factorial. – MaximusIdeal Aug 09 '24 at 21:11

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