4

I just finished proving that this statement holds for all positive integers $r$ (through a combinatorial argument) $$r^n=\sum_{k=1}^{n}{n\brace k}(r)(r-1)\ldots (r-k+1)$$ (where the curly braces indicate the Stirling set numbers) However, I'm being asked to show that this holds for all numbers $x$. That is $$x^n=\sum_{k=1}^{n}{n\brace k}(x)(x-1)\ldots (x-k+1)$$ The hint we received was that we should note that the second equation has at most $n$ solutions. However, I'm not sure how that will help me here. Any help would be greatly appreciated.

user3803721
  • 43
  • Let $p_l(x) = x^n$ and $p_r$ be the right hand side above. Both are polynomials of degree $n$ and $p_l(k) = p_r(k)$ for $k=0,...,n$, hence they are equal. – copper.hat Apr 03 '16 at 20:36
  • Okay, I see that's a good approach too. For me, I think Brian explained it very clearly as well down below. – user3803721 Apr 03 '16 at 22:01

2 Answers2

3

I’ll expand on the hint. Write the second equation in the form

$$f(x)=x^n-\sum_{k=1}^n{n\brace k}\prod_{i=0}^{k-1}(x-i)\;;$$

$f(x)$ is a polynomial, and you’ve shown that $f(r)=0$ for each $r\in\Bbb Z^+$. What polynomial has infinitely many zeroes?

Brian M. Scott
  • 616,228
  • Ah, I see the value of proving the first statement now. From the first statement, we know that $$f(x)=x^n-\sum_{k=1}^n{n\brace k}\prod_{i=0}^{k-1}(x-i)$$ must have infinitely many zeros, and therefore $f(x)$ must be the zero polynomial. From there, we're looking at $$x^n-\sum_{k=1}^n{n\brace k}\prod_{i=0}^{k-1}(x-i)=0$$ and we can just bring the summation to the other side. Thank you :) – user3803721 Apr 03 '16 at 21:37
  • 1
    @user3803721: You’re welcome. This is a cute trick that you’ll probably see again if you study much combinatorics. – Brian M. Scott Apr 04 '16 at 00:09
0

Suppose we seek to verify that $$x^n = \sum_{k=1}^n {n\brace k} x^{\underline{k}}$$ for all $x$ including $x$ complex.

We start with the basic

$$x^{\underline{k}} = \sum_{q=1}^k \left[k\atop q\right] (-1)^{k+q} x^q.$$

This follows from the defining recurrence of the signed Stirling numbers of the first kind and certainly holds for any $x.$

Substitute this into the target sum to get

$$x^n = \sum_{k=1}^n {n\brace k} \sum_{q=1}^k \left[k\atop q\right] (-1)^{k+q} x^q \\ = \sum_{q=1}^n x^q \sum_{k=q}^n {n\brace k} \left[k\atop q\right] (-1)^{k+q}.$$

We thus have to show that $$\sum_{k=q}^n {n\brace k} \left[k\atop q\right] (-1)^{k+q} = \begin{cases} 1 \quad\text{if}\quad q=n \\ 0 \quad\text{if}\quad 1\le q\lt n.\end{cases}.$$

The first of these follows by inspection. For the second recall the species for set partitions which is $$\mathfrak{P}(\mathcal{U} \mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ which gives the generating function $$G(z, u) = \exp(u(\exp(z)-1)).$$

and the species of cycle decompositions $$\mathfrak{P}(\mathcal{U} \mathfrak{C}_{\ge 1}(\mathcal{W}))$$

which gives the generating function $$H(w, u) = \exp\left(u\left(\log\frac{1}{1-w}\right)\right).$$

Substitute these into the sum to get

$$n! [z^n] \sum_{k=q}^n \frac{(\exp(z)-1)^k}{k!} k! [w^k] (-1)^{k+q} \frac{1}{q!} \left(\log\frac{1}{1-w}\right)^q.$$

Now the exponential term makes a zero contribution to the coefficient extractor at the front when $k\gt n$ so we may extend $k$ to infinity. The logarithmic term does not contribute to the second coefficient extractor when $k\lt q$ so we may start $k$ at zero. This yields

$$n! (-1)^q [z^n] \sum_{k\ge 0} (\exp(z)-1)^k (-1)^{k} [w^k] \frac{1}{q!} \left(\log\frac{1}{1-w}\right)^q.$$

Now what we have here in $w$ is an annihilated coefficient extractor (ACE) that simplifies to

$$n! (-1)^q [z^n] \frac{1}{q!} \left(\log\frac{1}{1-(1-\exp(z))}\right)^q = n! (-1)^q [z^n] \frac{1}{q!} (-z)^q.$$

This is zero unless $q=n$ when it is one as claimed and we are done.

Remark. If there are any issues with the last simplification here it can be verified using formal power series only and the proof is at this MSE link.

Remark, II. There are several more examples of the technique of annihilated coefficient extractors at this MSE link I and at this MSE link II and also here at this MSE link III.

Community
  • 1
Marko Riedel
  • 61,317