Let $f\colon \mathbb{R} \to \mathbb{R}$ be a differentiable function. Let us define the "discrete derivative"
$$ \Delta f( x) = f(x + 1) - f( x ) $$
and recursively $\Delta ^n f\left( x \right) = \Delta \left( {\Delta ^{n - 1} f\left( x \right)} \right)$ for all $n\geq 1$.
Is there some simple relation between $\Delta ^n f$ and the standard derivative $f^{(n)}$? Can we write a Taylor series expansion (say about 0) of $f$ in terms of discrete derivatives?
Yes, using the Stirling numbers one has $$ \frac{1}{k!} \frac{d^k}{dx^k} f(x) = \sum_{n=k}^{\infty} \frac{s(n,k)}{n!} \Delta^{n} f(x) $$ and $$ \frac{1}{k!} \Delta^{k} f(x) = \sum_{n=k}^{\infty} \frac{S(n,k)}{n!} \frac{d^n}{dx^n} f(x)$$
Edit: The Newton Series are defined as $$ f(x) = \sum_{k=0}^{\infty} \binom{x-a}{k} \: \Delta^k f(a) $$ some examples include: $$ \sum_{k=0}^{\infty} (-1)^k \binom{n}{2k} = 2^{\frac{n}{2}} \: \cos(\frac{\pi n}{4}) \\ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \binom{n}{k} = H_n $$ where $H_n$ are the Harmonic numbers.
