# Stirling Number Generating Function Relationship

In "Generatingfunctionology" by Herbert Wilf, there is a section where he derives explicit formulas for Stirling numbers. (Please see images below).

I'm wondering how he arrives at the relationship in eq. 1.6.8 (I do not think I understand what how the derivative w.r.t 'y' is operating here).

Appreciate any input - thanks! Equations / Relations

## 2 Answers

This is a use of a standard identity that appears throughout the book: $$ky^k = y(ky^{k-1})=y\frac d{dy}(y^k).$$ So, $$\sum_k k\begin{Bmatrix}n-1\\k\end{Bmatrix} y^k = \sum_k \begin{Bmatrix}n-1\\k\end{Bmatrix} y\frac d{dy}(y^k) = y\frac d{dy}\sum_k \begin{Bmatrix}n-1\\k\end{Bmatrix} y^k = y\frac d{dy}A_{n-1}(y).$$

We have $$\begin{eqnarray*} A_n(y)= y \sum_{k} { n-1 \brace k-1} y^{k-1} +y \sum_{k} { n-1 \brace k} \color{red}{ k y^{k-1}} . \end{eqnarray*}$$ Now use $$\begin{eqnarray*} \frac{d}{dy} y^k = k y^{k-1} \end{eqnarray*}$$ and interchange the order of the sum & derivative.