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(EDIT: The variable $z$ is changed to $d$ so as not to be confused with generating function notation)

I have derived this formula involving the Stirling numbers that I now feel confident is correct (at least for non-negative integers $m$, $n$ and $d$).

$$\frac{\Delta^d m^n}{d!} = \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$$

(where the difference is taken with respect to $m$), giving the specific case for the plain exponent $$m^n = \sum_{k} \left[ m \atop k \right] { {k+n} \brace m}(-1)^{m+k}$$ I am hoping that someone can provide a nice concise proof, as my proof involves a relatively lengthy path using a two dimensional induction argument just to show the formula for $m^n$. From there a straight application of the difference $$\Delta^d m^n = \Delta^{d-1}(m+1)^n - \Delta^{d-1}m^n$$ along with the identities $$\left[ m+1 \atop k \right] = m\left[ m \atop k\right] + \left[ m \atop k-1\right]$$ and $${ k+n \brace m+d} = { k+n+1 \brace m+d+1} - ( m+d+1) { k+n \brace m+d+1} $$ leads to the general $d$-th difference formula.

This is in my opinion the nicest formula that I have seen involving both types of Stirling numbers, since it completely eradicates the powers. It is not a transformation from regular powers to falling powers/factorials, so maybe the formula could be used in situations where removing the exponent at the cost of adding the Stirling numbers is convenient.

Does anyone care to make an attempt at giving a nice proof (or a reference)?
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  • $\begingroup$ Notice that this is the inversion formula when $n=0$ and $z=0$, which gives the correct restult of $1$. $\endgroup$
    – adam W
    Mar 27, 2014 at 4:26

3 Answers 3

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What follows is a proof that is assembled from various pieces which we'll link to in order not to have to repeat well-established material.

We seek to evaluate $$f_n = \sum_{k=1}^m \left[m\atop k\right] {k+n\brace m} (-1)^{m+k}.$$ Following Wilf, we introduce the generating function $$f(z) = \sum_{n\ge 0} f_n z^n.$$ We will have succeeded if we can show that $$f(z) = \frac{1}{1-mz}$$ since $$[z^n] \frac{1}{1-mz} = m^n.$$

Observe that $$f(z) = \sum_{n\ge 0} z^n \sum_{k=1}^m \left[m\atop k\right] {k+n\brace m} (-1)^{m+k} = \sum_{k=1}^m \left[m\atop k\right] (-1)^{m+k} \sum_{n\ge 0} {k+n\brace m} z^n.$$

We will do the inner sum first. Recall that the bivariate generating function of the Stirling numbers of the second kind is given by $$G(z, u) = \exp\left(u (\exp(z)-1)\right).$$ This implies for the sum that $$\sum_{n\ge 0} {k+n\brace m} z^n = \sum_{n\ge 0} z^n (k+n)! [z^{n+k}] \frac{1}{m!} (\exp(z)-1)^m.$$

Note that the part after the coefficient extraction operator is the generating function $$\sum_{q\ge m} {q\brace m}\frac{z^q}{q!}.$$

Now this is the moment where we need a subtle observation concerning generating functions. If we extract the coefficient $[z^{n+k}]$ from an exponential generating function and multiply by $(k+n)!$ we get the value generated by the EGF at $k+n$. If we then multiply by $z^n$ and sum over all $n$ we get the ordinary generating function of these values divided by $z^k.$ (Note that the exp term starts at $z$ and $m\ge k$ so it is safe to divide by $z^k.$ Furthermore we start extracting coefficients at $k$ and $k\le m$ so we can be sure that we get all of them.) This is a bit of a cognitive leap but by no means voodoo of any sort. We have simply observed how a coefficient extraction operator combined with a summation can turn an EGF into an OGF.

This OGF is well known however and is calculated e.g. at this MSE link. It is given by $$z^m \prod_{q=1}^m \frac{1}{1-qz}.$$

Returning to $f(z)$ we thus obtain $$f(z) = \sum_{k=1}^m \left[m\atop k\right] (-1)^{m+k} \frac{z^m}{z^k} \prod_{q=1}^m \frac{1}{1-qz}.$$ Moving parts that do not depend on $k$ to the front we get $$(-1)^m z^m \times \prod_{q=1}^m \frac{1}{1-qz} \times \sum_{k=1}^m \left[m\atop k\right] \frac{(-1)^k}{z^k}.$$

Now note that the remaining sum is the ordinary generating function of $\left[m\atop k\right]$ with respect to $k$ evaluated at $-1/z.$ This generating function is given by (consult e.g. Wikipedia) $$\prod_{q=0}^{m-1} (x+q).$$

This gives for our sum $$(-1)^m z^m \times \prod_{q=1}^m \frac{1}{1-qz} \times \prod_{q=0}^{m-1} \left(-\frac{1}{z}+q\right) = (-1)^m \times \prod_{q=1}^m \frac{1}{1-qz} \times \prod_{q=0}^{m-1} \left(-1+qz\right) \\ = \prod_{q=1}^m \frac{1}{1-qz} \times \prod_{q=0}^{m-1} \left(1-qz\right) \\ = \frac{1}{1-mz}$$ and we are done. Very nice identity indeed.

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  • $\begingroup$ Quite cool and interesting (+1). It may take me some time to feel like I fully understand your answer though (after studying more with generating functions of Stirling numbers). I will add later the path which led me to the formula, it was a fun one. $\endgroup$
    – adam W
    Mar 28, 2014 at 17:35
  • $\begingroup$ I have posted more details on the MO Question $\endgroup$
    – adam W
    Mar 29, 2014 at 21:58
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I have found a simple induction argument. For convenience let us introduce notation for the sum when $d=0$ by definition as $$ \Theta_m^n = \sum_k \left[ m \atop k \right]{k+n \brace m} (-1)^{m+k}$$

The following will validate the recurrence $$ m \Theta_m^n = \Theta_m^{n+1} + \sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k}$$ so that what will remain in order to prove $\Theta_m^n = m^n$ is to show the second sum is zero. (Note that $\Theta_m^0 = 1$ is true by the inversion formula which will complete the induction.)

Use the identity for the Stirling numbers of the first kind $$ m\left[ m \atop k\right] = \left[ {m+1} \atop m\right] - \left[ m \atop {k-1}\right]$$ and we see that $$ \begin{align} m\Theta_m^n &= \sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} - \sum_k \left[ m \atop k-1 \right]{ k+n \brace m}(-1)^{m+k} \\ &= \sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} - \sum_k \left[ m \atop k \right]{ k+n+1 \brace m}(-1)^{m+k+1} \tag{1}\\ &= \sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} + \sum_k \left[ m \atop k \right]{ k+n+1 \brace m}(-1)^{m+k}\\ &= \sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} + \Theta_m^{n+1} \\ \end{align}$$

Can the sum be shown to be zero? I thought it was easy but not quite. See my other question $$\sum_k \left[ {m+1} \atop k \right]{ k \brace m}(-1)^{m+k} = 0$$ The case $n=0$ is easy since the only non-zero terms for this are when $k=m$ and $k=m+1$ $$\left[ m+1 \atop m\right] { m \brace m} - \left[ m+1 \atop m+1\right] { m+1 \brace m} = \left[ m+1 \atop m\right] \cdot 1 - 1\cdot { m+1 \brace m}$$ This is zero since it is well known that $$\left[ m+1 \atop m\right] = { m+1 \brace m}$$

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Consult the book Concrete Mathematics, Ronald L. Graham (Author), Donald E. Knuth (Author), Oren Patashnik (Author)

I'm almost certain that this or a very similar identity was in that book. They will not only prove but show how it is related to 15-thousand other identities you didn't think could possibly be proven! (it's a difficult and good read by the way... everyone should take some time to learn from that book IMAO)

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    $\begingroup$ I should have mentioned I am reading the book right now, and have not seen the identity. It has a fairly similar one $\sum_{k} \left[ n \atop k\right] { k \brace m } = {n \choose m}(n-1)^\underline{n-m}$ on page 265. I would be interested in an actual proof as opposed to "probably there is a proof/could be done". Good book though, yes. $\endgroup$
    – adam W
    Mar 27, 2014 at 4:24

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