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I'm just playing around with the Pochhammer notation where $$n^{\underline{k}} \equiv n(n-1)\ldots(n-k+1).$$ I've established the formulas $$\begin{split} n &= n^{\underline{1}}\\ n^2 &= n^{\underline{2}} + n^{\underline{1}}\\ n^3 &= n^{\underline{3}} + 3n^{\underline{2}} + n^{\underline{1}}\\ n^4 &= n^{\underline{4}} + 6n^{\underline{3}} + 7n^{\underline{2}} + n^{\underline{1}}\\ n^5 &= n^{\underline{5}} + 10n^{\underline{4}} + 25n^{\underline{3}} + 15n^{\underline{2}} + n^{\underline{1}}. \end{split} $$ Does it exist a general formula of the form $$ n^k = \sum_{i=k}^1 c(i,k)n^{\underline{i}}, $$ for such expansions? What are in that case the closed form of the coefficients $c(i,k)$? Does this expansion have a name?

9cco
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    I asked a somewhat similar question a few years ago but involving forward differences instead of the falling factorial: 1591890. I wonder if the method shown there, or some flavor of it, might help here. – user170231 Dec 16 '20 at 16:43
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    Actually Pochhammer symbol gives $(n)_5=n (n+1) (n+2) (n+3) (n+4)$ https://en.wikipedia.org/wiki/Pochhammer_k-symbol You are talking of falling factorials https://en.wikipedia.org/wiki/Falling_and_rising_factorials – Raffaele Dec 16 '20 at 20:25
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    $$n^{\underline{k}} \equiv n(n-1)\ldots(n-k+1)=n!/(n - k)!$$ I'm not convinced that $n^3=n^{\underline{3}}+2n^{\underline{2}}+n^{\underline{1}}$. I get $(n-2) (n-1) n+2 (n-1) n+n=n^3-n^2+n$ – Raffaele Dec 16 '20 at 20:36
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    @Raffaele: Yes, that should be $$n^3=n^{\underline3}+3n^{\underline2}+n^{\underline1},.$$ (By the way, the Pochhammer symbol is also sometimes used for the falling factorial.) – Brian M. Scott Dec 16 '20 at 21:32
  • Yes I did a mistake, but computing it by using the formula $$nn^{\underline{k}} = (n-k)n^{\underline{k}} + kn^{\underline{k}} = n^{\underline{k+1}} + kn^{\underline{k}},$$ the above answer is correct. Incidentally then $n^4$ and $n^5$ are wrong too. $$n^4 = n^{\underline{4}} + 6n^{\underline{3}} + 7n^{\underline{2}} + n^{\underline{1}}$$ – 9cco Dec 16 '20 at 21:44
  • Indeed it is very related to the $\Delta f(n) = f(n+1) - f(n)$ operation since my motivation for looking at this was because you have $\Delta n^{\underline{k}} = kn^{\underline{k-1}}$. With the fundamental theorem of calculus on sum form $$\sum_{n=a}^{b-1}f(n) = f(b) - f(a),$$ I can now prove any summation formula of the form $\sum_{n=1}^Nn^k,$ by using $$\sum_{n=a}^{b-1}n^{\underline{k}} = \frac{1}{k+1}\Big[b^{\underline{k+1}} - a^{\underline{k+1}}\Big]$$ – 9cco Dec 16 '20 at 21:46
  • As to whether it has a name, it's sometimes called a Newton series: https://en.wikipedia.org/wiki/Finite_difference#Newton's_series – Qiaochu Yuan Dec 17 '20 at 00:55
  • As to your motivation, yes, this does let you sum $k^{th}$ powers, and you'll end up getting a formula equivalent (although not obviously so) to Faulhaber's formula: https://en.wikipedia.org/wiki/Faulhaber%27s_formula – Qiaochu Yuan Dec 17 '20 at 00:56

2 Answers2

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The general expression for $n\ge 0$ is

$$x^n=\sum_{k=0}^n{n\brace k}x^{\underline k}\,,$$

where $n\brace k$ is the Stirling number of the second kind that counts the partitions of $[n]$ into $k$ non-empty subsets. This is easily proved by induction on $n$ once you have the identity

$$x\cdot x^{\underline k}=x^{\underline{k+1}}+kx^{\underline k}$$

and the recurrence

$${n\brace k}=k{{n-1}\brace k}+{{n-1}\brace{k-1}}\,.$$

The induction step is:

$$\begin{align*} x^n&=x\sum_k{{n-1}\brace k}x^{\underline k}\\ &=\sum_k{{n-1}\brace k}x^{\underline{k+1}}+\sum_k{{n-1}\brace k}kx^{\underline k}\\ &=\sum_k{{n-1}\brace{k-1}}x^{\underline k}+\sum_k{{n-1}\brace k}kx^{\underline k}\\ &=\sum_k\left(k{{n-1}\brace k}+{{n-1}\brace{k-1}}\right)x^{\underline k}\\ &=\sum_k{n\brace k}x^{\underline k} \end{align*}$$

Brian M. Scott
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    There is also a direct bijective proof. The LHS is the number of functions from a set of size $n$ to a set of size $x$, and the RHS counts functions sorted by the size of the image: a function whose image has size $k$ determines and is determined by a surjection from the domain to a set of size $k$, of which there are ${n\brace k}$, together with an injection from the set of size $k$ to the codomain, of which there are $(x)_k$ (I prefer the subscript notation here). – Qiaochu Yuan Dec 17 '20 at 00:47
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    @QiaochuYuan: To finish off the argument, both sides are polynomials, and they agree on $\Bbb N$, so they must be identical. (I go the other way: I dislike the subscript notation and find the bar notation far more intuitive.) – Brian M. Scott Dec 17 '20 at 00:52
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We have that

$$n^k = \sum_{q=0}^k {k\brace q} n^\underline{q}.$$

This is because we have for the RHS

$$\sum_{q=0}^k {k\brace q} q! {n\choose q} = k! [z^k] \sum_{q=0}^k (\exp(z)-1)^q {n\choose q}.$$

Now $\exp(z)-1= z+\cdots$ so that $(\exp(z)-1)^q = z^q+\cdots$ and the coefficient extractor enforces the upper limit of the sum and we find

$$k! [z^k] \sum_{q\ge 0} (\exp(z)-1)^q {n\choose q} = k! [z^k] \exp(nz) = n^k.$$

We can also prove this combinatorially. The left side counts all $k$-tuples with component values ranging from $1$ to $n.$ The right is a classification by the number $q$ of different values that appear. We choose these in ${n\choose q}$ ways, then we partition the slots of the tuple into $q$ non-empty sets in ${k\brace q}$ ways. To conclude note that we can assign the $q$ chosen values to the sets of slots in $q!$ ways. This is essentially the comment by @QuiaochuYuan.

Marko Riedel
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