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Binomial Multiplicative Measure 的Matlab数值实现: sanshiphy 2014-9-17 23:13; 2014 国庆第一次修改 BMM 是展示多重分形的一个简单数学模型，具体介绍参考 Jens Feder, Fractals, 1988. 本文只是简单介绍如何用 Matlab 数值实现该模型。方法一： % construct of BinomialMultiplicative Process clear;clc; n=11; p0=0.25; p1=1-p0; N=2^n; b=0:N-1; p=zeros(1,N); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 下面一段是代码的核心，其思想是在第 n 代二重分裂后，共有 2^n 个子线段，将该子线段从左至右依次编号为 0, 1, 2, …, 2^n-1 ，然后将每个编号转换成二进制数，注意有些二进制数前面需要补 0 ，保证有 n 个位数。数每个 2 进制数的 0 的个数（包括补上的 0 的个数），记做 k ，则每一线段对应的概率是 (p^k)*((1-p)^(n-k)). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% for i=1:length(b) a=dec2bin(b(i)); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% dec2bin 的作用是将十进制数转换为二进制数，二进制数是以字数窜形式存储 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% temp=zeros(1,n); temp=num2str(temp,'%d'); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %d 的作用是保证数值转换为字符串时，字符之间无空格 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% temp(end-length(a)+1:end)=a; % 补 0 k=sum(temp=='0'); % 数 0 的个数 p(i)=(p0^k)*(p1^(n-k)); clear k tempa; end 方法二：以上绿色部分代码可以用下面更简单的代码代替： a=fi(b',0,n,0); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 生成一个对象，对象中每个元素前无符号（第一个 0 ），有 n 位数字（中间的参数 n ），这 n 位数字中，没有小数位（最后一个 0 ）。 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% temp=bin(a); % 将对象用二进制表示出来 temp=temp'; k=sum(temp(1:n,:)== '0' ); % 数 0 的个数 p=(p0.^k).*(p1.^(n-k)); 结果： 1) Binominal Multiplicative Measure （p0=0.25，n=11） 2）魔鬼阶梯 3）质量指数谱（数据点是数值模拟结果，实线是理论值） 4) 多重分形谱（实线是理论结果，点是根据模拟数据计算的结果，计算方法参见Chhabra and Jensen, PRL, 62: 1327-1330, 1989, 模拟数据的p0=0.7, n=14）; 个人分类: 学习笔记|2769 次阅读|0 个评论

A proof of inclusion-exclusion principle: Yaleking 2014-8-2 22:48; In this post,I prove inclusion-exclusion principle inductively.This principle can be found at Exercise 33 in Terence Tao's post 245A, Notes 3: Integration on abstract measure spaces, and the convergence theorems. (Inclusion-exclusion principle) Let $ {(X, {\mathcal B}, \mu)}$ be a measure space, and let $ {A_1, \ldots, A_n}$ be $ {{\mathcal B}}$-measurable sets of finite measure. Show that $ {\displaystyle \mu\left( \bigcup_{i=1}^n A_i \right) = \sum_{J \subset \{1,\ldots,n\}: J \neq \emptyset} (-1)^{|J|-1} \mu\left( \bigcap_{i \in J} A_i \right).}$ Proof: Prove by induction.When $ n=1$,the identity clearly holds.Suppose when $ n=k$,the identity also holds.Then when $ n=k+1$,let's see sets $ A_1,\cdots,A_n,A_{n+1}$,we know that $ {\displaystyle \mu\left( \bigcup_{i=1}^n A_i \right) = \sum_{J \subset \{1,\ldots,n\}: J \neq \emptyset} (-1)^{|J|-1} \mu\left( \bigcap_{i \in J} A_i \right). }$ And clearly, $ {\displaystyle \mu\left(\bigcup_{i=1}^{n+1}A_i\right)=\mu\left(\bigcup_{i=1}^nA_i\right)+\mu\left(A_{n+1}\right)-\mu\left(A_{n+1}\bigcap\left(\bigcup_{i=1}^nA_i\right)\right). }$ Now we calculate $ \mu\left(A_{n+1}\bigcap(\bigcup_{i=1}^nA_i)\right)$.According to the Morgan law,we have $ {\displaystyle A_{n+1}\bigcap \left(\bigcup_{i=1}^nA_i\right)=\bigcup_{i=1}^n\left(A_i\bigcap A_{n+1}\right). }$ And $ {\displaystyle \mu\left(\bigcup_{i=1}^n\left(A_i\bigcap A_{n+1}\right)\right)=\sum_{J\subset\{1,\cdots,n\};J\neq\emptyset}(-1)^{|J|-1}\mu\left(\bigcap_{i\in J}\left(A_i\bigcap A_{n+1}\right)\right). }$ So $ {\displaystyle \mu\left(A_{n+1}\bigcap\left(\bigcup_{i=1}^nA_i\right)\right)=\sum_{J\subset\{1,\cdots,n\};J\neq\emptyset}(-1)^{|J|-1}\mu\left(\bigcap_{i\in J}\left(A_i\bigcap A_{n+1}\right)\right). }$ So $ {\displaystyle \mu\left(\bigcup_{i=1}^{n+1}A_i\right)=\mu\left(\bigcup_{i=1}^nA_i\right)+\mu(A_{n+1})-\sum_{J\subset\{1,\cdots,n\};J\neq\emptyset}(-1)^{|J|-1}\mu\left(\bigcap_{i\in J}\left(A_i\bigcap A_{n+1}\right)\right), }$ i.e, $ {\displaystyle \mu\left(\bigcup_{i=1}^{n+1}A_i\right)=\sum_{J \subset \{1,\ldots,n\}: J \neq \emptyset} (-1)^{|J|-1} \mu\left( \bigcap_{i \in J} A_i \right)+\mu(A_{n+1})-\sum_{J\subset\{1,\cdots,n\};J\neq\emptyset}(-1)^{|J|-1}\mu\left(\bigcap_{i\in J}\left(A_i\bigcap A_{n+1}\right)\right)=\sum_{J \subset \{1,\ldots,n,n+1\}: J \neq \emptyset} (-1)^{|J|-1} \mu\left( \bigcap_{i \in J} A_i \right). }$; 个人分类: 测度论|2943 次阅读|0 个评论

Geometric measure theory: ChinaAbel 2010-1-18 11:16; What is it? Geometric measure theory is concerned with investigating the structure of surfaces from a measure-theoretic viewpoint. Since the notion of a surface (in an appropriately general sense) appears in many different settings in mathematics, it is unsurprising that GMT has applications in many areas of modern mathematics including: PDEs, Harmonic analysis and variational problems. GMT is traditionally considered to be a hard subject. This is primarily because, although many of the ideas involved are simple, in order to work at the level of generality that we do, a lot of technicalities need to be considered and understood in order to prove useful results. To gain a brief overview of the subject, you may like to look at an article that the paper online from Springer.; 个人分类: 非线性科学论文集|4157 次阅读|0 个评论

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