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:

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


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


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



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


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

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 $\Delta^i x^n$.

  • $\begingroup$ Can you show how you derived $\Delta^i x^n$ please. $\endgroup$ – Ahmed S. Attaalla Jul 4 '16 at 12:59
  • $\begingroup$ Also so $f(n)=(1!s(n,1),2!s(n,2),3!s(x,3),...,n!s(n,n))$ correct? $\endgroup$ – Ahmed S. Attaalla Jul 4 '16 at 13:08
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    $\begingroup$ That's right. And I have added the explanation above. $\endgroup$ – user90369 Jul 4 '16 at 13:35
  • $\begingroup$ Made a typo $f(n)=(1!s(n,1),2!s(n,2),3!s(n,3),...,n!s(n,n))$ $\endgroup$ – Ahmed S. Attaalla Jul 4 '16 at 13:46
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    $\begingroup$ @Ahmed S. Attaalla : I hope that you mean with $s(n,k)$ the stirling numbers of the second kind - but they are written e.g. $S(n,k)$. Please don't mix it up with the stirling numbers of the first kind. $\endgroup$ – user90369 Jul 5 '16 at 10:30

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