From ecd7e8aa6daf3a5cb6c87cae4d217f54245ba415 2017-06-09 17:02:58
From: Tom Bannink <tom.bannink@cwi.nl>
Date: 2017-06-09 17:02:58
Subject: [PATCH] Add comment on a^(3)_k = (k+2)(k+1)/6

---

diff --git a/main.tex b/main.tex
index 943d2bd865a982b579dd94ffab5d08ebc1450288..0e2ad6e4fb928aaf6062a2d09a86d687f2bbae9d 100644
--- a/main.tex
+++ b/main.tex
@@ -183,6 +183,9 @@
         R^{(5)}(p) &= \frac{p(90-300p+435p^2-325p^3+136p^4-36p^5+6p^6)}{15(1-p)^5(6-2p+p^2)}\\
                    &= \frac{(1-q)(6+5q+6q^2+21q^3+46q^4+6q^6)}{15q^5(5+q^2)}
     \end{align*}
+    For $n=3$ the system becomes very simple because regardless of the current state, the probability of going to $111$ is always equal to $(1-p)^3$. Therefore the expected number of resamplings is simply the expectation of a geometric distribution. This gives the formula for $R^{(3)}(p)$ as shown above. Note that the $k$-th coefficient of the powerseries of a function $f(p)$ is given by $\frac{1}{k!}\left.\frac{d^k f}{dp^k}\right|_{p=0}$, i.e. the $k$-th derivative to $p$ evaluated at $0$ divided by $k!$. For the function $R^{(3)}(p) = (1-p)^{-3} - 1$ this yields $a^{(3)}_k = (k+2)(k+1)/6$ for $k\geq 1$ and $a^{(3)}_0=0$.
+
+    ~
 
 	If statements \ref{it:pos}-\ref{it:lim} are true, then we can define the function 
 	$$R^{(\infty)}(p):=\sum_{k=0}^{\infty}a^{(k+1)}_k p^k,$$