This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 1 : Introduction We will start with a simple combinatorial problem. Consider { 1 , 1 } 1000 . How many elements x { 1 , 1 } 1000 satisfy fl fl fl fl 1000 X i =1 x i fl fl fl fl 50? More generally, for any n N and > 0 how many elements x { 1 , 1 } n satisfy fl fl fl fl n X i =1 x i fl fl fl fl n ? The answer is given by the binomial distribution. We are only seeking approximations. This is a question that we will spend a fair deal of time on this quarter. Today we will be satisfied with a crude upper bound. Fact: For any r R ( r + 1) 2 + ( r 1) 2 = ( r 2 + 2 r + 1) + ( r 2 2 r + 1) = 2( r 2 + 1) . For x { 1 , 1 } n we write S n ( x ) n i =1 x i and for m < n we write x  m for the restriction of x to the first m terms. Lemma 1.0.1 X x { 1 , 1 } n ( S n ( x )) 2 = n 2 n . Proof: By induction. It is easy to check that it is true for n = 1. Assume it is true for n ....
View
Full
Document
 Spring '09
 Reed

Click to edit the document details