Coefficient of operator and how to do it

Question

This question stems from this $$ \frac{1}{x+z}- \frac{1}{x} = \sum_{k=0}^\infty \frac{z^k}{k!}\frac{d^k}{dx^k}[\frac{1}{x}] $$ Now, i need to find the Bell Polynomial of $\frac{1}{x}$,

$$ [z^n]\left(\frac{(-z)^k}{x^k(x+z)^k}\right) = [z^n]\sum_{n \geq k} Y^{\Delta}(n,k,x) z^n = \frac{k!}{n!} B_{n,k}(\frac{d}{dx}[\frac{1}{x}], \cdots,\frac{d^{n-k+1}}{dx^{n-k+1}}[\frac{1}{x}]) $$

$$ \frac{(-1)^k}{x^k} [z^n] z^k (x+z)^k = \frac{(-1)^k}{x^{k}} [z^n] (\frac{x}{z}+1)^{-k} $$

$$ \frac{(-1)^k}{x^{k}} [z^n] (\frac{x}{z}+1)^{-k} = \frac{(-1)^k}{x^{k}} [z^n] \sum_{j=0}^\infty {-k \choose j} (\frac{x}{z})^j $$

How do i take the coefficient of this power series?

Answer 1

Here's the calculation of $[z^n]$:

\begin{align*} [z^n]\frac{(-z)^k}{x^k(x+z)^k}&=[z^n]\frac{(-z)^k}{x^{2k}\left(1+\frac{z}{x}\right)^k}\\ &=\frac{(-1)^k}{x^{2k}}[z^n]\frac{z^k}{\left(1+\frac{z}{x}\right)^k}\\ &=\frac{(-1)^k}{x^{2k}}[z^{n-k}]\frac{1}{\left(1+\frac{z}{k}\right)^k}\tag{1}\\ &=\frac{(-1)^k}{x^{2k}}[z^{n-k}]\sum_{j=0}^{\infty}\binom{-k}{j}\left(\frac{z}{x}\right)^j\tag{2}\\ &=\frac{(-1)^k}{x^{2k}}[z^{n-k}]\sum_{j=0}^{\infty}\binom{k+j-1}{k-1}(-1)^j\left(\frac{z}{x}\right)^j\tag{3}\\ &=\frac{(-1)^k}{x^{2k}}\binom{n-1}{k-1}(-1)^{n-k}\left(\frac{1}{x}\right)^{n-k}\\ &=\frac{(-1)^n}{x^{n+k}}\binom{n-1}{k-1} \end{align*}

Comment:

• In (1) we use the relationship $[z^n]z^k=[z^{n-k}]$ of the coefficient of operator

• In (2) we apply the binomial series expansion $(1+t)^\alpha=\sum_{j=0}^{\infty}\binom{\alpha}{j}t^j$ with $\alpha=-k$

• In (3) we apply the binomial coefficient identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$