Prove that the $n$th difference of $x^n$ is $\sum _{i=0}^n(-1)^i\binom{n}{i}(x-i)^n$

Question

Prove that the $n$th difference of $x^n$ is $\displaystyle \sum \limits _{i=0}^n(-1)^i\binom{n}{i}(x-i)^n$ (here the first difference is $\Delta (x^n):=(x)^n-(x-1)^n$ and the $n$th difference is defined recursively as $\Delta ^n(x^n):=\Delta ^{n-1}((x)^n)-\Delta ^{n-1}((x-1)^n)$).

I tried using induction, but just proving $\Delta ^n(x^n)$ is given by the indicated formula for all $n$ doesn't seem to work. Instead, I may need a more general formula involving $\Delta ^n(x^m)$, where $m$ and $n$ are positive integers. However, I'm not sure how to come up with this more general formula.

The pattern is not hard to establish, do $\Delta (x^n), \Delta^2(x^n),...$. — copper.hat, Jul 30 '22 at 21:43
@Gord452 FYI, using an Approach0 search, I found the AoPS thread nth Difference Sequence of nth Powers (in particular, message #$4$), plus quite a few closely related questions on this site, especially $k$th difference of $1,2^k,3^k,...$. There are many other associated questions, e.g., proving the expression is $n!$, eg., such as at the AoPS thread ... — John Omielan, Jul 30 '22 at 21:49
I imagine it should be $(x+i)^n$ above? And $(-1)^i$ should be $(-1)^{n-i}$, I believe. — copper.hat, Jul 30 '22 at 21:53
@Gord452 (cont.) An interesting identity, and on this site there's Binomial Sum: Values, Why does this process generate the factorial of the exponent?, Repeatedly taking differences on a polynomial yields the factorial of its degree?, Expressing a factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$?, Combinatorial proof of $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!$, using inclusion-exclusion, etc. — John Omielan, Jul 30 '22 at 21:55
it remands me https://math.stackexchange.com/questions/4298639/binomial-rule-and-repeatedly-differentiating this question I think differentiating repeatedly (x-i)^n will help — emil agazade, Jul 30 '22 at 22:02
It is much more natural to define $\Delta^n$ as $\Delta\circ \Delta^{n-1}$. $(\Delta p)(x)=p(x+1)-p(x), (\Delta^2 p)(x)=p(x+2)-2p(x+1)+p(x), (\Delta^3 p)(x) = p(x+3)-3 p(x+2)+3 p(x+1)-p(x)$ and in general $$(\Delta^n p)(x) = \sum_{k=0}^{n}\binom{n}{k}(-1)^k p(x+n-k)$$ — Jack D'Aurizio, Jul 30 '22 at 23:26
@JackD'Aurizio for reference this was question B5 of the 1976 putnam exam. An answer I got says the result is $n!$ because it's the nth difference of $x^n$. Is the logic behind that wrong then? — Gord452, Jul 31 '22 at 03:32
I am not quite getting the point of the formula, the $n$th difference is $n!$, so why are you focused on the above formula? — copper.hat, Jul 31 '22 at 18:23

score 2 · Answer 1 · answered Jul 31 '22 at 13:55

This problem refers to the backward difference operator \begin{align*} \Delta\left(x^n\right):= x^n-(x-1)^{n} \end{align*} We can prove the formula \begin{align*} \color{blue}{\Delta^n\left(x^n\right)=\sum _{i=0}^n(-1)^i\binom{n}{i}(x-i)^n}\tag{1} \end{align*} using operator methods. We introduce the shift operator $E$ and identity operator $I$ with \begin{align*} Ef(x)&=f(x+1)\\ If(x)&=f(x) \end{align*} We can write the delta operator $\Delta$ as \begin{align*} \color{blue}{\Delta}\left(x^n\right)&=x^n-\left(x-1\right)^n\\ &=I\left(x^n\right)-E^{-1}\left(x^n\right)\\ &=\color{blue}{\left(I-E^{-1}\right)}\left(x^n\right)\tag{2} \end{align*} Since shift operator $E$ and identity operator $I$ are linear operators which commute we obtain according to (2) using the binomial theorem

\begin{align*} \color{blue}{\Delta^n\left(x^n\right)}&=\left(I-E^{-1}\right)^n\left(x^n\right)\\ &=\sum_{i=0}^n\binom{n}{i}(-1)^iE^{-i}I^{n-i}\left(x^n\right)\\ &=\sum_{i=0}^n\binom{n}{i}(-1)^iE^{-i}\left(x^n\right)\\ &\,\,\color{blue}{=\sum_{i=0}^n\binom{n}{i}(-1)^i\left(x-i\right)^n} \end{align*} and the claim (1) follows.

score 0 · Answer 2 · answered Jul 31 '22 at 07:52

Define $(Rf)(x) = f(x)-f(x-1)$. Note that $R$ is linear. It is straightforward using induction to show that $(R^nf )(x) = \sum_{k=0}^n \binom{n}{k} (-1)^k f(x-k)$.

Abusing notation slightly, we see that for any constant $c$ we have $Rc = 0$ and that $Rx^p$ is a polynomial of degree $p-1$ with coefficient $p = \binom{p}{1}$. In particular, $R^n x^n = n!$ and so $ \sum_{k=0}^n \binom{n}{k} (-1)^k (x-k)^n = n!$.

Prove that the $n$th difference of $x^n$ is $\sum _{i=0}^n(-1)^i\binom{n}{i}(x-i)^n$

2 Answers2

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