coefficient of operator for $B_{n,k}^{x^2}(x)$

Question

We start with the following: $$ (x+z)^2 - x^2 = \sum_{n \geq 1} \frac{z^n}{n!} \frac{d^n}{dx^n}[x^2] $$

$$ (x+z)^2 - x^2 = z(2x+z) $$

$$ z^k(2x+z)^k = \sum_{n \geq k} Y^{\Delta}(n,k,x)z^n $$

Where $$ Y^{\Delta}(x) = \frac{k!}{n!}B_{n,k}^{x^2}(x) $$ Therefore: $$ (2x)^k [z^{n-k}]\left(1+\frac{z}{2x}\right)^k $$

$$ = (2x)^k [z^{n-k}] \sum_{j=0}^k {k \choose j} \left(\frac{z}{2x}\right)^k $$

$$ = (2x)^{2k-n} {n-1 \choose n-k} $$ Is this correct? I feel like this is very wrong.

Answer 1

Only the last calculation is somewhat wrong (see Hints below). Everything else is ok.

Let's start from \begin{align*} \left[(x+z)^2-x^2\right]^k=z^k(2x+z)^k = \sum_{n \geq k} Y^{\Delta}(n,k,x)z^n\tag{1} \end{align*} Then we know from this paper by V. Kruchinin, that \begin{align*} Y^{\Delta}(n,k,x) = \frac{k!}{n!}B_{n,k}^{x^2}(x) \end{align*}

(Hint: $Y^{\Delta}$ has three arguments)

We obtain \begin{align*} \frac{k!}{n!}B_{n,k}^{x^2}(x)&=[z^n]\sum_{n \geq k} Y^{\Delta}(n,k,x)z^n\\ &=[z^n]z^k(2x+z)^k\\ &=(2x)^k[z^{n-k}]\left(1+\frac{z}{2x}\right)^k\\ &=(2x)^k[z^{n-k}]\sum_{j=0}^k\binom{k}{j}\left(\frac{z}{2x}\right)^j\tag{2}\\ &=(2x)^k\binom{k}{n-k}\left(\frac{1}{2x}\right)^{n-k}\\ &=(2x)^{2k-n}\binom{k}{n-k}\tag{3} \end{align*}

(Hint: The exponent of $\frac{z}{2x}$ in (2) is $j$, not $k$)

We conclude from (3)

The Bellpolynomials $B_{n,k}^{x^2}(x)$ are \begin{align*} B_{n,k}^{x^2}(x)=\frac{n!}{k!}\binom{k}{n-k}(2x)^{2k-n}\tag{4} \end{align*}

We can generalise the problem from $x^2$ to $x^N$ with $N>0$ a positive integer. A nice by-product is, when specializing the result to $N=2$ we will derive a binomial identity.

Generalisation:

We consider the $k$-th power of $Y(x+z)=(x+z)^N-x^N$ and obtain \begin{align*} [Y(x,z)]^k&=[(x+z)^N-x^N]^k\\ &=x^{Nk}\left[\left(1+\frac{z}{x}\right)^N-1\right]^k\\ &=x^{Nk}\sum_{j=0}^k\binom{k}{j}\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}\\ \end{align*}

Similarly to (1) we observe \begin{align*} [(x+z)^N-x^N]^k=x^{Nk}\sum_{j=0}^k\binom{k}{j}\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}=\sum_{n \geq k} Y^{\Delta}(n,k,x)z^n \end{align*}

Again, we know from this paper by V. Kruchinin, that \begin{align*} Y^{\Delta}(n,k,x) = \frac{k!}{n!}B_{n,k}^{x^N}(x) \end{align*}

It follows \begin{align*} \frac{k!}{n!}B_{n,k}^{x^N}(x)&=[z^n]\sum_{n \geq k} Y^{\Delta}(n,k,x)z^n\\ &=[z^n]x^{Nk}\sum_{j=0}^k\binom{k}{j}\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}\\ &=x^{Nk}\sum_{j=0}^k\binom{k}{j}[z^n]\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}\\ &=x^{Nk}\sum_{j=0}^k\binom{k}{j}\binom{Nj}{n}x^{-n}(-1)^{k-j}\\ &=x^{Nk-n}\sum_{j=0}^k\binom{k}{j}\binom{Nj}{n}(-1)^{k-j} \end{align*}

We conclude, the Bellpolynomials $B_{n,k}^{x^N}(x)$ are $$B_{n,k}^{x^N}(x)=x^{Nk-n}\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\binom{Nj}{n}(-1)^{k-j}$$

The special case $N=2$ results in $$B_{n,k}^{x^2}(x)=x^{2k-n}\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\binom{2j}{n}(-1)^{k-j}$$

Comparing this result with (4) above we observe

the following binomial identity is valid for $n\geq 1$

\begin{align*} \sum_{j=\lfloor\frac{n}{2}\rfloor}^k\binom{k}{j}\binom{2j}{n}(-1)^{k-j}=2^{2k-n}\binom{k}{2k-n}\qquad\quad 1\leq k \leq n \end{align*}