# Analytic Combinatorics for Computing Seeding Probabilities
## Appendix A  
Here, we show that in the error model of Section 4.2, the _moduli_ of the singularities of \(S_{\gamma}(z)\) expressed in (10) get closer to each other as \(p\) decreases. More specifically, \(|z_{j}|\sim|z_{m}|\) (\(p\downarrow 0\)) for any two singularities \(z_{j}\) and \(z_{m}\).  
Recall that the singularities of \(S_{\gamma}(z)\) are the roots of \(Q(z)=1-pz\big{(}1+qz+\ldots+(qz)^{\gamma-1}\big{)}\), where \(q=1-p\). Let \(z_{j}\) be a root of \(Q\) and rearrange the terms of the equation \(Q(z_{j})=0\) to obtain \(z_{j}\big{(}1+qz_{j}+\ldots+(qz_{j})^{\gamma-1}\big{)}=1/p\). As \(p\downarrow 0\), the right-hand side tends to \(+\infty\) so the left-hand side must also tend to \(+\infty\), imposing \(\lim_{p\downarrow 0}|z_{j}|=+\infty\).  
Multiply \(Q(z)=0\) by \((1-qz)\) and use \(1+qz+\ldots+(qz)^{\gamma-1}=\big{(}1-(qz)^{\gamma}\big{)}/(1-qz)\), where \(z\
eq 1/q\) to see that every singularity \(z_{j}\) solves the equation \((1-qz)Q(z)=1-z+pq^{\gamma}z^{\gamma+1}=0\) or equivalently  
\[1-1/z_{j}=pq^{\gamma}z_{j}^{\gamma},\quad z_{j}\
eq\frac{1}{q}.\]  
Since \(\lim_{p\downarrow 0}|z_{j}|=+\infty\), taking the limit of the equation above yields  
\[\lim_{p\downarrow 0}\big{(}qp^{1/\gamma}z_{j}\big{)}^{\gamma}=1,\quad\text{i.e.},\quad|z_{j}|\sim\frac{1}{qp^{1/\gamma}}\quad(p\downarrow 0). \tag{10}\]  
This is sufficient to prove that \(|z_{j}|\sim|z_{m}|\) (\(p\downarrow 0\)) for any two singularities \(z_{j}\) and \(z_{m}\), but we can further show that  
\[z_{j}\sim\frac{e^{2i(j-1)\pi/\gamma}}{qp^{1/\gamma}}\quad(p\downarrow 0),\quad j =1,2,\ldots,\gamma. \tag{11}\]  
From (10), we see that the terms \(qp^{1/\gamma}z_{j}\) tend to \(\gamma\)-th roots of unity, which have \(\gamma\) possible values. If we prove that \(Q\) has \(\gamma\) distinct roots, then each of them must correspond to a different \(\gamma\)-th root of unity and (11) will follow. Since \(Q\) is a polynomial of degree \(\gamma\), we must prove that all its roots have multiplicity \(1\), i.e., that they do not solve \(Q^{\prime}(z)=0\).  
Let \(V(z)=(1-qz)Q(z)=1-z+pq^{\gamma}z^{\gamma+1}\), so \(V^{\prime}(z)=-1+(\gamma+1)pq^{\gamma}z^{\gamma}\), and compute the greatest common divisor of \(V(z)\) and \(V^{\prime}(z)\):  
\[\gcd\big{(}V(z),V^{\prime}(z)\big{)}=\gcd\big{(}V(z)-\frac{z}{\gamma+1}V^{ \prime}(z),V^{\prime}(z)\big{)}=\gcd\big{(}1-\frac{\gamma z}{\gamma+1},V^{ \prime}(z)\big{)}.\]  
Up to a constant factor independent of \(z\), the greatest common divisor is either \(1\) or \(1-\gamma z/(\gamma+1)\). If \(z=1+1/\gamma\) is not a root of \(V^{\prime}(z)\), then \(V(z)\) and \(V^{\prime}(z)\) are relatively prime, so \(V(z)\) does not have any double roots and neither does \(Q(z)\).  
If \(z=1+1/\gamma\) is a root of \(V^{\prime}(z)\), then the greatest common divisor is \(1-\gamma z/(\gamma+1)\) so \(z=1+1/\gamma\) is a root of \(V(z)\) with multiplicity \(2\). This case arises when \(p=1/(\gamma+1)\), which implies that \(z=1+1/\gamma=1/q\). In \(V(z)=(1-qz)Q(z)\), the factor \(1-qz\) contributes one occurrence of the root, so \(Q(z)\) contributes the other occurrence and \(z=1+1/\gamma\) is thus a single root of \(Q(z)\).