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Andras Gilyen - 8 years ago 2017-09-07 19:58:34
gilyen@clayoquot.swat.cwi.nl
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    Let $\rho'\in\mathbb{R}[p]^{2^n}$ be a vector of polynomials, and let $\text{rank}(\rho')$ be defined in the following way: 
 
    $$\text{rank}(\rho'):=\min_{b\in\{0,1\}^n}\left( |b|+ \text{maximal } k\in\mathbb{N} \text{ such that } p^k \text{ divides } \rho'_b\right).$$
 
	Clearly for any $\rho'$ we have that $\text{rank}(\rho' M_{(n)})\geq \text{rank}(\rho') + 1$. Another observation is, that all elements of $\rho'$ are divisible by $p^{\text{rank}(\rho')-n}$.
 
    We observe that for the initial $\rho$ we have that $\text{rank}(\rho)=n$, therefore $\text{rank}(\rho*(M_{(n)}^k))\geq n+k$, and so $\rho*(M_{(n)}^k)*\mathbbm{1}$ is obviously divisible by $p^{k}$. This implies that $a_k^{(n)}$ can be calculated by only looking at $\rho*(M_{(n)}^1)*\mathbbm{1}, \ldots, \rho*(M_{(n)}^k)*\mathbbm{1}$.
 
    
 
\newpage
 
\section{Proving the strong cancellation claim}
 
\section{Proving that $a_k^{(k+1)}=a_k^{(n)}$ for all $n>k$}
 
It is useful to introduce some new notation. We will consider variations of the Markov Chains:
 
\begin{itemize}
 
    \item $\P^{(n)}$ refers to the original process on the length-$n$ cycle.
 
    \item $\P^{[a,b]}$ or $\P^{[n]}$ refers to a similar Markov Chain but on a finite chain ($[a,b]$ or $[1,n]$).
 
\end{itemize}
 
The process on the finite chain has the following modification at the boundary: if a boundary site is resampled, it can only resample its single neighbour so it draws only two new bits. 
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