Great Theoretical Ideas In Computer Science Steven Rudich CS 15 251 Spring 2003 Lecture 11 Feb 14 2004 Carnegie Mellon University 1 X1 X2 X 3 X n 1 Xn Xn 1 1 X 1 Counting III Pascal s Triangle Polynomials and Vector Programs The Geometric Series X1 X2 X3 1 X1 X2 X 3 Xn 1 1 X The Infinite Geometric Series 1 X1 X2 X 3 Xn 1 x 1 x 1 1 x x x x 2 x2 1 1 X 1 X X 1 1 X1 X2 X 3 X n X1 X2 X 3 X n X n 1 1 X1 X2 X 1 X1 X2 X 3 Xn 1 3 X n 1 X n X n 1 1 1 aX1 a 2X2 a 3X 3 a n Xn 1 1 aX Geometric Series Linear Form 1 1 aX1 a2X2 an Xn 1 bX1 b 2X2 bn Xn 1 1 aX1 a2X2 an Xn 1 bX1 b 2X2 bn Xn 1 c1 X1 ck Xk 1 aX 1 bX Geometric Series Quadratic Form Suppose we multiply this out to get a single infinite polynomial What is an expression for Cn 1 aX1 a2X2 an Xn 1 bX1 b 2X2 bn Xn 1 c1 X1 ck Xk 1 aX1 a2X2 an Xn 1 bX1 b 2X2 bn Xn 1 c1 X1 ck Xk If a b then cn cn n 1 an a 0bn a 1 bn 1 a ibn i a n 1 b1 a n b0 a 0bn a 1 bn 1 a ibn i a n 1 b1 a n b0 a 0bn a 1 bn 1 a ibn i a n 1 b1 a n b0 a b a 0 bn bn 1 a b 1 aX1 a2X2 an Xn 1 bX1 b 2X2 bn Xn 1 c1 X1 ck Xk a 1 bn 1 a ibn i a n 1 b1 a n b0 a 1 bn ai 1bn i a 0 bn 1 a n 1 bn 1 a 1 bn a nb1 a n 1b0 a i 1bn i a n 1 b2 a nb1 a n 1 a n 1 bn 1 if a b then cn a n 1 bn 1 a b a 0bn a 1 bn 1 a ibn i a n 1 b1 a n b0 2 Previously we saw that 1 aX1 a2X2 an Xn 1 bX1 b 2X2 bn Xn n 0 1 or n 0 1 1 Polynomials Count 1 aX 1 bX a n 1 bn 1 a b n 1 an Xn Xn when a b Geometric Series Quadratic Form Choice tree for terms of 1 X 3 What is the coefficient of BA3 N2 in the expansion of B A N 6 1 1 1 The number of ways to rearrange the letters in the word BANANA The Binomial Formula 1 X n FG nIJ FG nIJ X FG nIJ X FG nIJ X FG nIJ X H 0K H 1K H 2K H kK H nK 2 k 1 X X X X 1 1 X X 1 X2 X X X 1 X X2 X2 X3 Combine like terms to get 1 3X 3X2 X3 The Binomial Formula n n n 1 x n x k k 0 k Binomial Coefficients binomial expression 3 One polynomial two representations n 1 x x k k 0 k n n Product form or Generating form Additive form or Expanded form Power Series Representation n n 1 x n x k k 0 k n k x Closed form or k 0 k Generating form n Since 0 if k n k Power series Taylor series expansion By playing these two representations against each other we obtain a new representation of a previous insight n n 1 x n x k k 0 k By varying x we can discover new identities n n 1 x n xk k 0 k Let x 1 n n 0 1 k k 0 k Let x 1 n n 2n k 0 k 123 Equivalently n The number of even sized subsets of an n element set is the same as the number of odd sized subsets n n 1 x n xk k 0 k Let x 1 n n 0 1 k k 0 k Equivalently n n n k k 2 k even k odd n k even The number of subsets of an n element set n n n k k 2 n 1 n 1 k odd n 1 x xk k 0 k n n We could discover new identities by substituting in different numbers for X One cool idea is to try complex roots of unity however the lecture is going in another direction 4 n n 1 x n xk k 0 k Proofs that work by manipulating algebraic forms are called algebraic arguments Proofs that build a 1 1 onto correspondence are called combinatorial arguments n n n n k k 2 k even n 1 k odd Let On be the set of binary strings of length n with an odd number of ones Let En be the set of binary strings of length n with an even number of ones We gave an algebraic proof that On En A Combinatorial Proof Let On be the set of binary strings of length n with an odd number of ones Let En be the set of binary strings of length n with an even number of ones An attempt at a correspondence Let f n be the function that takes an n bit string and flips all its bits fn is clearly a one to one and onto function but do even n work In f6 we have for odd n E g in f7 we have 110011 001100 0010011 1101100 101010 010101 1001101 0110010 Uh oh Complementing maps evens to evens A combinatorial proof must construct a one toone correspondence between On and En A correspondence that works for all n n 1 x xk k 0 k n n Let f n be the function that takes an n bit string and flips only the first bit For example 0010011 1010011 1001101 0001101 110011 010011 101010 001010 The binomial coefficients have so many representations that many fundamental mathematical identities emerge 5 The Binomial Formula Pascal s Triangle kth row are the coefficients of 1 X k 1 X 0 1 1 X 0 1 1 X 1 1 1X 1 X 1 1 1X 1 X 2 1 2X 1X2 1 X 2 1 2X 1X2 1 X 3 1 3X 3X2 1X3 1 X 3 1 3X 3X2 1X3 1 X 4 1 4X 6X2 4X3 1X4 1 X 4 1 4X 6X2 4X3 1X4 kth Row Of Pascal s Triangle n n n n n 0 1 2 k n Inductive definition of kth entry of nth row Pascal n 0 Pacal n n 1 Pascal n k Pascal n 1 k 1 Pascal n k 1 X 0 1 1 X 0 1 1 X 1 1 1X 1 X 1 1 1X 1 X 2 …
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