My Math Forum Diffusion limited aggregation

 May 5th, 2017, 09:17 AM #1 Newbie   Joined: Jun 2016 From: France Posts: 4 Thanks: 0 Diffusion limited aggregation Hello! Firstly, sorry for my bad English. We consider a sequence $A_n$ of subsets of $\mathbb{Z}$, which increase according to the following property: at the step n, a particle is thrown at the origin and is moving according to a random walk - the probabilities to go at the left or at the right are 1/2. The particle is moving until it comes out of $A_n$ in a point $X_{n+1}$. We define : $A_{n+1} = A_n \cup X_n$. The model is initialized with $A_1=\{0\}$. $A_n$ is called aggregate at step n. I can understand that: If $A_1 = \{0\}$, then $A_2 = \{0,1\}$ with probability $1/2$, or $A_2 = \{-1,0\}$ with probability $1/2$. Indeed, the particle comes out of $A_1$ at the point $1$ (by the right) or $-1$ (by the left). If $A_2 = \{0,1\}$, $A_3 = \{0,1,2\}$ or $A_3 = \{-1,0,1\}$. If $A_2 = \{-1,0\}$, $A_3 = \{-1,0,2\}$ or $A_3 = \{-2,-1,0\}$ (always with probability $1/2$). I've done this, until the step 5: I have two questions: Could you explain why the "set" of occupied sites is an interval of integers and what its length? I really can't answer to this question... And we note $A_n = \{G_n, ..., D_n\}$ with $G_n$ and $D_n$ are respectively the leftmost and the rightmost points of the aggregate. We note $Z_n = D_n + G_n$. How we find 'again' $G_n$ and $D_n$ from $Z_n$ ? I've done this, until the step 5, but I don't see how we found $G_n$ and $D_n$ from $Z_n$: If someone could help me, just for one question, I will be very grateful to him! Have a nice day! Last edited by skipjack; August 7th, 2017 at 10:30 PM.
 August 7th, 2017, 10:32 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,372 Thanks: 2009 The English is okay.

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