Function $f: \mathbb{Z} \to \mathbb{Z}^n$ related to $\sum_{k=1}^{x} k^n$.

Question

The sequence $\{a_0,a_1,...a_x\}$ has closed form $a_n=\sum_{i=0}^{\infty} \Delta^i(0) {n \choose i}$ where $\Delta a_n$ denotes the operation mapping $a_n$ to $a_{n+1}-a_n$ and $\Delta^i(0)$ is shorthand for this operation iterated $i$ times on $a_n$ and then evaluated at $n=0$. Furthermore $\sum_{n=0}^{x} {n \choose k}={x+1 \choose k+1}$ and a polynomial of degree $n$ can be determined uniquely from $n+1$ points. Combining these results we may get:

$$\sum_{k=1}^{x} k^1=1{x+1 \choose 2}$$

$$f(1)=1$$

$$\sum_{k=1}^{x} k^2=1{x+1 \choose 2}+2{x+1 \choose 3}$$

$$f(2)=(1,2)$$

$$\sum_{k=1}^{x} k^3=1{x+1 \choose 2}+6{x+1 \choose 3}+6{x+1 \choose 4}$$

$$f(3)=(1,6,6)$$

Question

Is there expression for $f$ without the use of $\Delta^i(0)$?

Answer 1

See https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind .

$\Delta^i x^n=\sum\limits_{j=0}^i (-1)^{i-j}\binom{i}{j}(x+j)^n$

Set $x:=0$.

E.g.: $f(3)=(\sum\limits_{j=0}^1 (-1)^{1-j}\binom{1}{j}j^3,\sum\limits_{j=0}^2 (-1)^{2-j}\binom{2}{j}j^3,\sum\limits_{j=0}^3 (-1)^{3-j}\binom{3}{j}j^3)$

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It’s a method of the “Discrete Mathematics”. Be $f(x)$ a function without poles.

Identity operator $I$: $If(x):=f(x)$

Shift operator $E$: $Ef(x):=f(x+1)$

Important: $I$ and $E$ are linear operators.

Difference Operator $\Delta$: $\Delta:=E-I$

=> $\Delta f(x)=(E-I)f(x)=Ef(x)-If(x)=f(x+1)-f(x)$

Therefore $\Delta^i f(x)=(E-I)^i f(x)=\sum\limits_{k=0}^i (-1)^{i-k}\binom{i}{k}E^k f(x)=\sum\limits_{k=0}^i (-1)^{i-k}\binom{i}{k}f(x+k)$.

Set $f(x):=x^n$ (here: $x\in\mathbb{R}_0^+$), then you get the formula for $\Delta^i x^n$.