$k$th difference of $1,2^k,3^k,...$

Question

I read, in an exposition of Euler's proof of Fermat's theorem on sums of squares, that the $k$-th order finite forward difference of the function $f(x_i)=x_i^k$, relatively to the nodes $x_i=i$ where $i=1,2,3,...$, defined by the recurrence $\Delta^n f(x_i):=\Delta^{n-1} f(x_{i+1})-\Delta^{n-1} f(x_{i}),\quad \Delta^1 f(x_i)=\Delta f(x_i):=f(x_{i+1})-f(x_i),\quad n=1,2,...$

is always $k!$ for any natural number $i=1,2,3,...$.

For $x_{i+1}=x_i +h$ Wikipedia gives the formula (easily proved by induction) $\Delta^{k}f(x_i)=\sum_{j=0}^k (-1)^j \binom{k}{j}f(x_i+(k-j)h)$, which, in our case where $x_i=i$, $h=1$ and $f(x)=x^k$, gives $\Delta^k f(i)=\sum_{j=0}^k (-1)^j\binom{k}{j} (i+k-j)^k$. But I am not able to use that to prove (I've tried by induction, in vain) that for all $i\in\mathbb{N}_{\geq 1}$, for $f(x)=x^k$, we have $\Delta^k f(i)=k!$. Neither am I able to prove it by recursion because it isn't a linear one...

Has anybody any idea on how to prove it? Thank you so much in advance!

score 4 · Accepted Answer · answered Mar 03 '14 at 12:01

4

First convince yourself that if $P(x)$ is a polynomial of degree $m$ with leading coefficient $a$ then $P(x+1)-P(x)$ is a polynomial of degree $m-1$ with leading coefficient $ma$. Now argue by induction on the degree.

answered Mar 03 '14 at 12:01

Gerry Myerson

179,216

Thank you so much, Gerry!!! – Self-teaching worker Mar 06 '14 at 14:02
1

There's a check mark next to the answer. You can click in it to indicate that you accept the answer, if you want to. – Gerry Myerson Mar 06 '14 at 21:40
Done! ;) Never noticed before...! – Self-teaching worker Mar 08 '14 at 10:09

$k$th difference of $1,2^k,3^k,...$

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