# Combinatorial generating function networks

I am working through the following paper (ref 1), and do not understand how to generate equation 6.

The generating function for the degree distribution of a network is

$$G_0(x) = \sum_{k=0}^{\infty} p_kx^k$$

The probability that $i$ of the $k$ edges of a node are occupied is given by the binomial distribution

$$\begin{pmatrix} k \\ i \end{pmatrix} x^i (1-x)^{k-i}$$ If we select a node at random then the probability of $i$ of it's edges being occupied is $$\sum^\infty_{k=i} p_k \begin{pmatrix} k \\ i \end{pmatrix} x^i (1 - x ) ^{k-i}$$ So far so good. Now, I do not understand how this becomes $$\frac{(-x)^i}{i!}\frac{d^i}{dx^i}G_0(1-x)$$ References

1 Li, Wang. Generating function technique in complex networks. Journal of Physics: Conference Series 604 (2015) 012013

• $d^{i}/dx^{i}(1 - x)^{k} = (-1)^{i}\frac{k!}{(k - i)!}(1 - x)^{k - i}$ for $k \geq i$ and $0$ otherwise. – Alex Zorn Nov 14 '17 at 22:44

We have \begin{eqnarray*} G_0(x) =\sum_{k=0}^{\infty} p_k x^k. \end{eqnarray*} differentiating $i$ times gives \begin{eqnarray*} \frac{d^i}{dx^i} G_0(x) =\sum_{k=0}^{\infty} p_k \frac{k!}{(k-i)!} x^{k-i} . \end{eqnarray*} Now substitute $x \rightarrow 1-x$ and we have \begin{eqnarray*} (-1)^i \frac{d^i}{dx^i} G_0(1-x) =\sum_{k=0}^{\infty} p_k \frac{k!}{(k-i)!} (1-x)^{k-i} . \end{eqnarray*}