Taylor expansion of an stirling identity

Question

I have been searching many ways for a week just to solve this, to no avail.

I'm still confused about how the Taylor expansion is produced.

It is so advanced compared to the subjects that I took.

I am currently taking advance researches or work/journals from other mathematicians but I still cannot do this:

$$\frac{(e^{w}-1)^{k}}{k!} = \sum_{n=k}^{\infty }{ n \brace k} \frac{w^{n}}{n!}.$$

Let us recall that the Stirling numbers satisfy the identities: $$\begin{array}{rcl} \displaystyle{ n \brace k} &=& \displaystyle \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j}j^n = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{j}{k \choose j}(k-j)^{n} \\\displaystyle{ n+1 \brace k} &=& \displaystyle k{ n \brace k}+ { n \brace k-1} \end{array}$$

and appear in the Taylor expansion: $$\frac{(e^{w}-1)^{k}}{k!} = \sum_{n=k}^{\infty }{ n \brace k}\frac{w^{n}}{n!}.$$

Answer 1

In the following we use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} \binom{n}{k}=[z^k](1+z)^n\qquad\text{and}\qquad k^n=n![z^n]e^{kz} \end{align*}

We obtain \begin{align*} \displaystyle{ n \brace k}&=\frac{1}{k!}\sum_{j=0}^k(-1)^{j-k}\binom{k}{j}j^n\\ &=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}(-1)^{j-k}n![z^n]e^{jz}\tag{1}\\ &=\frac{n!}{k!}[z^n]\sum_{j=0}^k\binom{k}{j}\left(e^z\right)^j(-1)^{k-j}\tag{2}\\ &=\frac{n!}{k!}[z^n](e^z-1)^k\tag{3}\\ &=n![z^n]\frac{(e^z-1)^k}{k!} \end{align*} and the claim $$\frac{(e^{z}-1)^{k}}{k!} = \sum_{n=k}^{\infty }{ n \brace k} \frac{z^{n}}{n!}$$ follows.

Comment:

• In (1) we apply the coefficient of operator.

• In (2) we do some rearrangements and use the linearity of the coefficient of operator.

• In (3) we apply the binomial theorem.

Answer 2

Here is the proof of the EGF from the recurrence. We have

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

with ${0\brace k} = 0,$ ${n\brace 0} = 0$ and ${0\brace 0} = 1.$

Introduce the mixed generating function

$$G(z, u) = 1 + \sum_{n\ge 1} \sum_{k\ge 1} {n\brace k} u^k \frac{z^n}{n!}.$$

Multiply the recurrence by $u^k \frac{z^n}{n!}$ and sum over $n\ge 0, k\ge 1$ to get

$$\sum_{n\ge 0} \sum_{k\ge 1} u^k \frac{z^n}{n!} (n+1)! [z^{n+1}] [u^k] G(z, u) \\ = \sum_{n\ge 0} \sum_{k\ge 1} k u^k \frac{z^n}{n!} n! [z^n] [u^k] G(z, u) + \sum_{n\ge 0} \sum_{k\ge 1} u^k \frac{z^n}{n!} n! [z^n] [u^{k-1}] G(z, u).$$

This yields

$$ \sum_{n\ge 0} (n+1) z^n [z^{n+1}] \sum_{k\ge 1} u^k [u^k] G(z, u) \\ = \sum_{n\ge 0} z^n [z^n] \sum_{k\ge 1} k u^k [u^k] G(z, u) + \sum_{n\ge 0} z^n [z^n] \sum_{k\ge 1} u^k [u^{k-1}] G(z, u).$$

Continuing we have

$$ \sum_{n\ge 0} (n+1) z^n [z^{n+1}] (G(z, u) - 1) \\ = u \sum_{n\ge 0} z^n [z^n] \sum_{k\ge 1} k u^{k-1} [u^k] G(z, u) + u \sum_{n\ge 0} z^n [z^n] \sum_{k\ge 1} u^{k-1} [u^{k-1}] G(z, u)$$

which finally yields

$$\frac{\partial}{\partial z} G(z, u) = u \sum_{n\ge 0} z^n [z^n] \frac{\partial}{\partial u} G(z, u) + u \sum_{n\ge 0} z^n [z^n] G(z, u)$$

or

$$\bbox[5px,border:2px solid #00A000] {\frac{\partial}{\partial z} G(z, u) = u \frac{\partial}{\partial u} G(z, u) + u G(z, u).}$$

Now introduce $G(z, u) = \exp H_1(z, u)$ to get

$$\exp H_1(z, u) \frac{\partial}{\partial z} H_1(z, u) = u \exp H_1(z, u) \frac{\partial}{\partial u} H_1(z, u) + u \exp H_1(z, u)$$

or

$$\frac{\partial}{\partial z} H_1(z, u) = u \frac{\partial}{\partial u} H_1(z, u) + u.$$

Putting $H_1(z, u) = H_2(z, u) - u$ we obtain

$$\frac{\partial}{\partial z} H_2(z, u) = u \frac{\partial}{\partial u} H_2(z, u).$$

In the last step we put $H_2(z, u) = u \exp H_3(z)$ to get

$$u \exp H_3(z) \frac{d}{dz} H_3(z) = u \exp H_3(z)$$

or $$\frac{d}{dz} H_3(z) = 1$$

and $H_3(z) = z + C.$

This yields for $G(z, u)$ the form

$$G(z, u) = \exp(u\exp(z+C)-u).$$

Now we have ${n\brace 1} = 1$ for all $n\ge 1$ which says

$$1 = n! [z^n] [u^1] \exp(u(\exp(z+C)-1)) \\ = n! [z^n] (\exp(z+C)-1) = \exp(C) n! [z^n] \exp(z) = \exp(C).$$

Therefore $C=0$ and

$$\bbox[5px,border:2px solid #00A000] {G(z, u) = \exp(u(\exp(z)-1)).}$$

Addendum. Apparently the above manipulations to solve the differential equation are not quite rigorous and the most general solution (which is easily verified) is given by

$$\exp(-u) F(u \exp(z))$$

with $F$ an arbitrary function. We now show how to determine $F$. Let $$F(w) = \sum_{q\ge 0} f_q \frac{w^q}{q!}.$$

Then we have for all $n\ge 0$

$$1 = {n\brace n} = n! [z^n] [u^n] \exp(-u) F(u \exp(z)).$$

We obtain

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

We claim $f_q=1$ for all $q$ with the values determined by interpreting the above as a recurrence. The base case for $n=0$ says $$1 = 1 \times 1 \times (-1)^0 \times f_0 \times 0^0 = f_0$$ and it holds.

For the induction step we get from the induction hypothesis that

$$1 = \frac{1}{n!} {n\choose n} (-1)^{n-n} n^n f_n + \frac{1}{n!} \sum_{k=0}^{n-1} {n\choose k} (-1)^{n-k} k^n \\ = \frac{1}{n!} n^n (f_n - 1) + \frac{1}{n!} \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} k^n.$$

Note however that

$$\frac{1}{n!} \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} k^n = \frac{1}{n!} \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} n! [v^n] \exp(kv) \\ = [v^n] \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} \exp(kv) = [v^n] (\exp(v)-1)^n = 1$$

because $\exp(v)-1$ starts at $[v^1].$ This finally yields

$$1 = \frac{1}{n!} n^n (f_n - 1) + 1 \quad\text{or}\quad 0 = \frac{1}{n!} n^n (f_n - 1)$$ and hence $f_n=1$ as claimed.

With $F(w)$ shown to be $\exp(w)$ we obtain once more

$$\bbox[5px,border:2px solid #00A000] {G(z, u) = \exp(u(\exp(z)-1)).}$$

We have used $0^0={0\choose 0}=1$ throughout.