Lecture Notes on Discrete Mathematics doc

Lecture Notes on Discrete Mathematics A K Lal September 26, 2012 Contents Preliminaries 1.1 Basic Set Theory 1.2 Properties of Integers 1.3 Relations and Partitions 16 1.4 Functions 21 Counting and Permutations 2.1 27 Distinguishable Balls 27 2.1.2 Indistinguishable Balls and Distinguishable Boxes 36 2.1.3 Indistinguishable Balls and Indistinguishable Boxes 39 2.1.4 Round Table Configurations 40 Lattice Paths 41 2.2.1 Catalan Numbers 43 Some Generalizations 46 2.3.1 2.3 27 2.1.1 2.2 Principles of Basic Counting 50 Miscellaneous Exercises Advanced Counting 53 3.1 Pigeonhole Principle 53 3.2 Principle of Inclusion and Exclusion 58 Polya Theory 63 4.1 Groups 63 4.2 Lagrange’s Theorem 73 4.3 Group Action 78 4.4 The Cycle Index Polynomial 82 4.4.1 84 Polya’s Inventory Polynomial 86 4.4.2 Applications CONTENTS Generating Functions and Its Applications 93 5.1 Formal Power Series 93 5.2 Applications to Recurrence Relation 100 5.3 Applications to Generating Functions 106 Chapter Preliminaries We will use the following notation throughout these notes The empty set, denoted ∅, is a set that has no element N := {0, 1, 2, }, the set of Natural numbers; Z := { , −2, −1, 0, 1, 2, }, the set of Integers; Q := { p : p, q ∈ Z, q = 0}, the set of Rational numbers; q R := the set of Real numbers; and C := the set of Complex numbers For the sake of convenience, we have assumed that the integer 0, is also a natural number This chapter will be devoted to understanding set theory, relations, functions and the principle of mathematical induction We start with basic set theory 1.1 Basic Set Theory We have already seen examples of sets, such as N, Z, Q, R and C at the beginning of this chapter For example, one can also look at the following sets Example 1.1.1 {1, 3, 5, 7, }, the set of odd natural numbers {0, 2, 4, 6, }, the set of even natural numbers { , −5, −3, −1, 1, 3, 5, }, the set of odd integers { , −6, −4, −2, 0, 2, 4, 6, }, the set of even integers {0, 1, 2, , 10} CHAPTER PRELIMINARIES {1, 2, , 10} Q+ = {x ∈ Q : x > 0}, the set of positive rational numbers R+ = {x ∈ R : x > 0}, the set of positive real numbers Q∗ = {x ∈ Q : x = 0}, the set of non-zero rational numbers 10 R∗ = {x ∈ R : x = 0}, the set of non-zero real numbers We observe that the sets that appear in Example 1.1.1 have been obtained by picking certain elements from the sets N, Z, Q, R or C These sets are example of what are called “subsets of a set”, which we define next We also define certain operations on sets Definition 1.1.2 (Subset, Complement, Union, Intersection) Let A be a set If B is a set such that each element of B is also an element of the set A, then B is said to be a subset of the set A, denoted B ⊆ A Two sets A and B are said to be equal if A ⊆ B and B ⊆ A, denoted A = B Let A be a subset of a set Ω Then the complement of A in Ω, denoted A′ , is a set that contains every element of Ω that is not an element of A Specifically, A′ = {x ∈ Ω : x ∈ A} Let A and B be two subsets of a set Ω Then their (a) union, denoted A ∪ B, is a set that exactly contains all the elements of A and all the elements of B To be more precise, A ∪ B = {x ∈ Ω : x ∈ A or x ∈ B} (b) intersection, denoted A ∩ B, is a set that exactly contains those elements of A that are also elements of B To be more precise, A ∩ B = {x ∈ Ω : x ∈ A and x ∈ B} Example 1.1.3 Let A be a set Then A ⊆ A The empty set is a subset of every set Observe that N ⊆ Z ⊆ Q ⊆ R ⊆ C As mentioned earlier, all examples that appear in Example 1.1.1 are subsets of one or more sets from N, Z, Q, R and C Let A be the set of odd integers and B be the set of even integers Then A ∩ B = ∅ and A ∪ B = Z Thus, it also follows that the complement of A, in Z, equals B and vice-versa Let A = {{b, c}, {{b}, {c}}} and B = {a, b, c} be subsets of a set Ω Then A ∩ B = ∅ and A ∪ B = {a, b, c, {b, c}, {{b}, {c}} } Definition 1.1.4 (Cardinality) A set A is said to have finite cardinality, denoted |A|, if the number of distinct elements in A is finite, else the set A is said to have infinite cardinality 1.1 BASIC SET THEORY Example 1.1.5 The cardinality of the empty set equals That is, |∅| = Fix a positive integer n and consider the set A = {1, 2, , n} Then |A| = n Let S = {2x ∈ Z : x ∈ Z} Then S is the set of even integers and it’s cardinality is infinite Let A = {a1 , a2 , , am } and B = {b1 , b2 , , bn } be two finite subsets of a set Ω, with |A| = m and B| = n Also, assume that A ∩ B = ∅ Then, by definition it follows that A ∪ B = {a1 , a2 , , am , b1 , b2 , , bn } and hence |A ∪ B| = |A| + |B| Let A = {a1 , a2 , , am } and B = {b1 , b2 , , bn } be two finite subsets of a set Ω Then |A ∪ B| = |A| + |B| − |A ∩ B| Observe that Example 1.1.5.4 is a particular case of this result, when A ∩ B = ∅ Let A = {{a1 }, {a2 }, , {am }} be a subset of a set Ω Now choose an element a ∈ Ω such that a = , for any i, ≤ i ≤ n Then verify that the set B = {S ∪ {a} : S ∈ A} equals {{a, a1 }, {a, a2 }, , {a, am }} Also, observe that A ∩ B = ∅ and |B| = |A| Exercise 1.1.6 Does there exist unique sets X and Y such that X − Y = {1, 3, 5, 7} and Y − X = {2, 4, 8}? In a class of 60 students, all the students play either football or cricket If 20 students play both football and cricket, determine the number of players for each game if the number of students who play football is (a) 14 more than the number of students who play cricket (b) exactly times more than the number of students who play only cricket (c) a multiple of and and leaves a remainder of when divided by (d) is a factor of 90 and the number of students who play cricket is a factor of 70 Definition 1.1.7 (Power Set) Let A be a subset of a set Ω Then a set that contains all subsets of A is called the power set of A and is denoted by P(A) or 2A Example 1.1.8 Let A = ∅ Then P(∅) = {∅, A} = {∅} Let A = {∅} Then P(A) = {∅, A} = {∅, {∅}} Let A = {a, b, c} Then P(A) = {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} Let A = {{b, c}, {{b}, {c}}} Then P(A) = {∅, {{b, c}}, {{{b}, {c}}}, {{b, c}, {{b}, {c}}} } CHAPTER PRELIMINARIES 1.2 Properties of Integers Axiom 1.2.1 (Well-Ordering Principle) Every non-empty subset of natural numbers contains its least element We will use Axiom 1.2.1 to prove the weak form of the principle of mathematical induction The proof is based on contradiction That is, suppose that we need to prove that “whenever the statement P holds true, the statement Q holds true as well” A proof by contradiction starts with the assumption that “the statement P holds true and the statement Q does not hold true” and tries to arrive at a contradiction to the validity of the statement P being true Theorem 1.2.2 (Principle of Mathematical Induction: Weak Form) Let P (n) be a statement about a positive integer n such that P (1) is true, and P (k + 1) is true whenever one assumes that P (k) is true Then P (n) is true for all positive integer n Proof On the contrary, assume that there exists n0 ∈ N such that P (n0 ) is not true Now, consider the set S = {m ∈ N : P (m) is false } As n0 ∈ S, S = ∅ So, by Well-Ordering Principle, S must have a least element, say N By assumption, N = as P (1) is true Thus, N ≥ and hence N − ∈ N Therefore, the assumption that N is the least element in S and S contains all those m ∈ N, for which P (m) is false, one deduces that P (N − 1) holds true as N − < N ≤ Thus, the implication “P (N − 1) is true” and Hypothesis imply that P (N ) is true This leads to a contradiction and hence our first assumption that there exists n0 ∈ N, such that P (n0 ) is not true is false n(n + 1) Solution: Verify that the result is true for n = Hence, let the result be true for n Let Example 1.2.3 Prove that + + · · · + n = us now prove it for n + That is, one needs to show that + + · · · + n + (n + 1) = (n + 1)(n + 2) Using Hypothesis 2, + + · · · + n + (n + 1) = n(n + 1) n+1 + (n + 1) = (n + 2) 2 Thus, by the principle of mathematical induction, the result follows 1.2 PROPERTIES OF INTEGERS n(n + 1)(2n + 1) Solution: The result is clearly true for n = Hence, let the result be true for n and one (n + 1)(n + 2) (2(n + 1) + 1) needs to show that 12 + 22 + · · · + n2 + (n + 1)2 = Using Hypothesis 2, Prove that 12 + 22 + · · · + n2 = n(n + 1)(2n + 1) + (n + 1)2 n+1 = (n(2n + 1) + 6(n + 1)) (n + 1)(n + 2) (2n + 3) n+1 = 2n2 + 7n + = 6 Thus, by the principle of mathematical induction, the result follows 12 + 22 + · · · + n2 + (n + 1)2 = Prove that for any positive integer n, + + · · · + (2n − 1) = n2 Solution: The result is clearly true for n = Let the result be true for n That is, + + · · · + (2n − 1) = n2 Now, we see that + + · · · + (2n − 1) + (2n + 1) = n2 + (2n + 1) = (n + 1)2 Thus, by the principle of mathematical induction, the result follows AM-GM Inequality: Let n ∈ N and suppose we are given real numbers a1 ≥ a2 ≥ · · · ≥ an ≥ Then Arithmetic Mean (AM) := √ a1 + a2 + · · · + an ≥ n a1 · a2 · · · · · an =: (GM) Geometric Mean n Solution: The result is clearly true for n = 1, So, we assume the result holds for any collection of n non-negative real numbers Need to prove AM ≥ GM, for any collections of non-negative integers a1 ≥ a2 ≥ · · · ≥ an ≥ an+1 ≥ a1 + a2 + · · · + an + an+1 So, let us assume that A = Then, it can be easily verified that n+1 a1 ≥ A ≥ an+1 and hence a1 − A, A − an+1 ≥ Thus, (a1 − A)(A − an+1 ) ≥ Or equivalently, A(a1 + an+1 − A) ≥ a1 an+1 (1.1) Now, let us assume that the AM-GM inequality holds for any collection of n non-negative numbers Hence, in particular, for the collection a2 , a3 , , an , a1 + an+1 − A That is, a2 + · · · + an + (a1 + an+1 − A) ≥ n a2 · · · an · (a1 + an+1 − A) = GM (1.2) n a2 + a3 + · · · + an + (a1 + an+1 − A) But = A Thus, by Equation (1.1) and Equation (1.2), n one has AM = An+1 ≥ (a2 · a3 · · · · · an · (a1 + an+1 − A)) · A ≥ (a2 · a3 · · · · · an ) a1 an+1 Therefore, we see that by the principle of mathematical induction, the result follows 10 CHAPTER PRELIMINARIES Fix a positive integer n and let A be a set with |A| = n Then prove that P(A) = 2n Solution: Using Example 1.1.8, it follows that the result is true for n = Let the result be true for all subset A, for which |A| = n We need to prove the result for a set A that contains n + distinct elements, say a1 , a2 , , an+1 Let B = {a1 , a2 , , an } Then B ⊆ A, |B| = n and by induction hypothesis, |P(B)| = 2n Also, P(B) = {S ⊆ {a1 , a2 , , an , an+1 } : an+1 ∈ S} Therefore, it can be easily verified that P(A) = P(B) ∪ {S ∪ {an+1 } : S ∈ P(B)} Also, note that P(B) ∩ {S ∪ {an+1 } : S ∈ P(B)} = ∅, as an+1 ∈ S, for all S ∈ P(B) Hence, using Examples 1.1.5.4 and 1.1.5.6, we see that |P(A)| = |P(B)| + |{S ∪ {an+1 } : S ∈ P(B)}| = |P(B)| + |P(B)| = 2n + 2n = 2n+1 Thus, the result holds for any set that consists of n + distinct elements and hence by the principle of mathematical induction, the result holds for every positive integer n We state a corollary of the Theorem 1.2.2 without proof The readers are advised to prove it for the sake of clarity Corollary 1.2.4 (Principle of Mathematical Induction) Let P (n) be a statement about a positive integer n such that for some fixed positive integer n0 , P (n0 ) is true, P (k + 1) is true whenever one assumes that P (n) is true Then P (n) is true for all positive integer n ≥ n0 We are now ready to prove the strong form of the principle of mathematical induction Theorem 1.2.5 (Principle of Mathematical Induction: Strong Form) Let P (n) be a statement about a positive integer n such that P (1) is true, and P (k + 1) is true whenever one assume that P (m) is true, for all m, ≤ m ≤ k Then, P (n) is true for all positive integer n Proof Let R(n) be the statement that “the statement P (m) holds, for all positive integers m with ≤ m ≤ n” We prove that R(n) holds, for all positive integers n, using the weak-form of mathematical induction This will give us the required result as the statement “R(n) holds true” clearly implies that “P (n) also holds true” 97 5.1 FORMAL POWER SERIES Determine the sum of the squares of the first N positive integers Solution: Using Example 5.1.10.2, observe that k = [xk−1 ] Therefore, using (1 − x)2 Example 5.1.10.4, one has N 1 · 1−x (1 − x) k = [xN −1 ] k=1 = [xN −1 ] N Determine a closed form expression for N −1+3−1 N −1 = (1 − x)3 k=1 x Therefore, using the (1 − x)2 kxk = k≥0 k≥0 k xk = x  k≥0 Thus, by Example 5.1.10.4 N k2 = [xN ] k=1 = =  x (1 − x)2 k2 xk−1  = xD x(1 + x) · 1−x (1 − x) N (N + 1) k2 Solution: Using Example 5.1.10.3, observe that differentiation operator, one obtains  = = [xN −1 ] (1 − x)4 = x(1 + x) (1 − x)3 + [xN −2 ] (5.2) (1 − x)4 N −1+4−1 N −2+4−1 + N −1 N −2 N (N + 1)(2N + 1) N Determine a closed form expression for k3 k=1 k xk = Solution: Using Equation (5.2), observe that k≥0 k≥0  k xk = x  Thus, by Example 5.1.10.4 N k3 = [xN ] k=1 k≥0  k3 xk−1  = xD x(1 + x) (1 − x)3 x(1 + x) So, (1 − x)3 = x(1 + 4x + x2 ) (1 − x)4 x(1 + 4x + x2 · (1 − x)4 ) 1−x + [xN −2 ] + [xN −3 ] 5 (1 − x) (1 − x) (1 − x)5 N −1+5−1 N −2+5−1 N −3+5−1 +4 + N −1 N −2 N −3 = [xN −1 ] = = N (N + 1) 2 Hence, we observe that we can inductively use this technique to get a closed form expression N for k=1 kr , for any positive integer r 98 CHAPTER GENERATING FUNCTIONS AND ITS APPLICATIONS n2 + n + n! n≥0 Solution: As we need to compute the infinite sum, Cauchy product cannot be used Also, Determine a closed form expression for one needs to find a convergent series, which when evaluated at some x0 , gives the required xn expression Therefore, recall that the series ex = , converges for all x ∈ R and n≥0 n! n evaluating ex at x = 1, gives = e Similarly, = [xn ] (xD(ex )) = [xn ] (xex ) and n! n! n≥0 n2 = [xn ] (xD(xDex )) = [xn ] (x + x2 )ex Thus, n! n≥0 n2 + n + = (x + x2 )ex + xex + 6ex n! = 9e x=1 Let n and r be two fixed positive integers Then determine the number of non-negative integer solutions to the system x1 + x2 + · · · + xn = r? Solution: Recall that this number was already computed in Lemma 2.1.2 and equals r+n−1 r In this chapter, we can think of the problem as follows: the above system can be interpreted as coming from the monomial xr , where r = x1 +x2 +· · · +xn That is, the problem reduces to finding the coefficients of y xk of a formal power series, for non-negative integers xk ’s Now, recall that the terms y xk appear with a coefficient in the expression = yi − y i≥0 Hence, the question reduces to computing [y r ] (1 − y)(1 − y) · · · (1 − y) = [y r ] = (1 − y)n r+n−1 r We now look at some examples that may require the use the package “MATHEMATICA” or “MAPLE” to obtain the exact answer So, in the examples that we give below, we are interested in getting a formal power series and then its coefficients give the answer to the questions raised Example 5.1.11 Let n and r be two fixed positive integers Then determine the number of non-negative integer solutions to the system x1 + 2x2 + · · · + nxn = r? Solution: Note that using the ideas in Example 5.1.10.9, one needs to consider the for1 mal power series xki that equals Hence, the question reduces to computing the − xk i≥0 coefficient of xr in (1 − x)(1 − x2 ) · · · (1 − xn ) Determine the number of solutions in non-negative integer to the system x1 +x2 +· · ·+x5 = n such that x1 ≥ 4, x4 ≤ 10 and for r = 1, 4, xr is a multiple of r Solution: Note that the condition x1 ≥ corresponds to looking at xk , for k ≥ 4, forcing us to look at the formal power series k≥4 xk Similarly, x4 ≤ 10 gives the formal power 99 5.1 FORMAL POWER SERIES 10 series k=0 series xk and the condition xr is a multiple of r, for r = 1, 4, gives the formal power xrk So, we are interested in computing the coefficient of xn in the product k≥0   k≥4   10 xk · k=0 xk · k≥0  x2k · k≥0  x3k · k≥0  x5k  = x4 (1 − x11 ) (1 − x)2 (1 − x2 )(1 − x3 )(1 − x5 ) Determine the number of ways in which 100 voters can cast their 100 votes for 10 candidates such that no candidate gets more than 20 votes Solution: Note that we are assuming that the voters are identical So, we need to solve the system in non-negative integers to the system x1 + x2 + · · · + x10 = 100, with ≤ xi ≤ 20, for ≤ i ≤ 10 So, we need to find the coefficient of x100 in 10 20 xk = k=1 (1 − x21 )10 = (1 − x)10 (−1)i = i=0 10 (−1)i i=0 10 21i x i 10 109 − 21i · i  · j≥0  10 + j − j  x j Exercise 5.1.12 Let m, n, and r be fixed positive integers Then prove that the following problems are equivalent? Determine the number of solutions in non-negative integer to the system x1 + x2 + · · · + xn = r, with m ≤ xi ≤ 2m? Determine the number of ways of putting r indistinguishable balls into n distinguishable boxes so that the number of balls in each box is a number between m and 2m (endpoints included)? What is the coefficient of xr in the formal power series xmn (1 − xm+1 )n ? (1 − x)n Before moving to the applications of generating functions/formal power series to the solution of recurrence relations, let us list a few well known power series The readers are requested to get a proof of their satisfaction 100 CHAPTER GENERATING FUNCTIONS AND ITS APPLICATIONS ex = (1 + x)a xk k≥0 k! = (1 − x)a x−r (1 − x)n a k r≥0 k≥0 = k≥−r = sinh(x) = n r+k xk , xk , |x| < |x| < (−1)r x2r+1 r≥0 (2r + 1)! x2r+1 r≥0 (2r + 1)! 5.2 1− 1 − 4x = k≥0 = k≥0 cosh(x) − 4x = 2x √ = k≥0 n − 4x 2x = k≥0 xk , |x| < k≥1 k k≥0 = √ − = cos(x) 1− √ = 1−x √ − 4x xn (1 − x)n+1 xk , |x| < a+k−1 k = sin(x) log(1 − x) xk , |x| < 2k k n k xk , |x| < xn , n ≥ 0, |x| < (−1)r x2r (2r)! r≥0 x2r r≥0 (2r)! 1 2k k x , |x| < k+1 k 2k + n k x , |x| < k Applications to Recurrence Relation This section contains the applications of generating functions to solving recurrence relations Let us try to understand it using the following examples Example 5.2.1 Determine a formula for the numbers a(n)’s, where a(n)’s satisfy the recurrence relation a(n) = a(n − 1) + 2n, for n ≥ with a(0) = a(n)xn Then using Example 5.1.10.3, one has Solution: Define A(x) = n≥0 a(n)xn = a0 + A(x) = n≥0 = 3x n≥1 So, A(x) = a(n)xn = + n≥1 a(n − 1)x n−1 n≥1 n +2 (3 a(n − 1) + 2n) xn nx + = 3xA(x) + n≥1 x + (1 − x)2 + x2 1 = − − Thus, (1 − 3x)(1 − x) 2(1 − 3x) 2(1 − x) (1 − x)2 a(n) = [xn ]A(x) = n · 3n − − − (n + 1) = − (n + 1) 2 2 Determine a generating function for the numbers f (n) that satisfy the recurrence relation f (n) = f (n − 1) + f (n − 2), for n ≥ with f (0) = and f (1) = Hence or otherwise find a formula for the numbers f (n) 101 5.2 APPLICATIONS TO RECURRENCE RELATION f (n)xn Then one has Solution: Define F (x) = n≥0 f (n)xn = + x F (x) = n≥0 n≥2 = 1+x+x n≥2 (f (n − 1) + f (n − 2)) xn f (n − 1)xn−1 + x2 n≥2 f (n − 2)xn−2 = + xF (x) + x2 F (x) √ √ 1+ 1− Let α = Therefore, F (x) = and β = Then it can be checked − x − x2 2 and that (1 − αx)(1 − βx) = − x − x   α β F (x) = √ − =√  αn+1 xn − β n+1 xn  − αx − βx 5 n≥0 n≥0 Therefore, f (n) = [xn ]F (x) = √ αn+1 − β n+1 n≥0 As β < and |β| < 1, we observe that f (n) ≈ √ √ 1+ n+1 Remark 5.2.2 The numbers f (n), for n ≥ are called Fibonacci numbers It is related with the following problem: Suppose a couple bought a pair of rabbits (each one year old) in the year 2001 If a pair of rabbits starts giving birth to a pair of rabbits as soon as they grow years old, determine the number of rabbits the couple will have in the year 2025 Determine a formula for the numbers a(n)’s, where a(n)’s satisfy the recurrence relation a(n) = a(n − 1) + 4a(n − 2), for n ≥ with a(0) = and a(1) = c, a constant a(n)xn Then Solution: Define A(x) = n≥0 a(n)xn = a0 + a1 x + A(x) = n≥0 a(n)xn = + cx + n≥2 = + cx + 3x n≥2 a(n − 1)x n−1 n≥2 + 4x an−2 x (3 a(n − 1) + a(n − 2)) xn n−2 n≥2 = + cx + 3x(A(x) − a0 ) + 4x2 A(x) So, A(x) = + (c − 3)x + (c − 3)x = 2) (1 − 3x − 4x (1 + x)(1 − 4x) and hence an = [xn ] A(x) = 4n − 4x 1+c 4−c (b) If c = then A(x) = · + · and hence − 4x 1+x n n (4 − c) (1 + c) (−1) an = [xn ] A(x) = + 5 (a) If c = then A(x) = 102 CHAPTER GENERATING FUNCTIONS AND ITS APPLICATIONS Determine a sequence, {a(n) ∈ R : n ≥ 0}, such that a0 = and for all n ≥ n k=0 a(k)a(n − k) = n+2 , a(n)xn Then, using the Cauchy product, one has Solution: Define A(x) = n≥0 n A(x)2 = n≥0 Hence, A(x) = (1−x)3/2 k=0 a(k)a(n − k) xn = −3/2 n and thus a(n) = = n≥0 n+2 n x = (1 − x)3 · · · · · (2n + 1) , for all n ≥ 2n n! Determine a generating function for the numbers f (n, m), n, m ∈ Z, n, m ≥ that satisfy f (n, m) = f (n − 1, m) + f (n − 1, m − 1), (n, m) = (0, 0) with (5.1) f (n, 0) = 1, for all n ≥ and f (0, m) = 0, for all m > Hence or otherwise, find a formula for the numbers f (n, m) Solution: Note that in the above recurrence relation, the value of m need not be ≤ n Method 1: Define Fn (x) = m≥0 f (n, m)xm Then, for n ≥ 1, Equation (5.1) gives f (n, m)xm = Fn (x) = m≥0 m≥0 (f (n − 1, m) + f (n − 1, m − 1)) xm m = m≥0 f (n − 1, m)x + m≥0 f (n − 1, m − 1)xm = Fn−1 (x) + xFn−1 (x) = (1 + x)Fn−1 (x) = · · · = (1 + x)n F0 (x) Now, using the initial conditions, F0 (x) = and hence Fn (x) = (1 + x)n Thus, f (n, m) = [xm ](1 + x)n = n m Method 2: Define Gm (y) = n≥0 if ≤ m ≤ n and f (n, m) = 0, for m > n f (n, m)y n Then, for m ≥ 1, Equation (5.1) gives f (n, m)y n = Gm (y) = n≥0 = n≥0 n≥0 f (n − 1, m)y n + (f (n − 1, m) + f (n − 1, m − 1)) y n n≥0 f (n − 1, m − 1)y n = yGm (y) + yGm−1 (y) y Gm−1 (y) Now, using initial conditions, G0 (y) = and 1−y 1−y ym ym n hence Gm (y) = Thus, f (n, m) = [y n ] = [y n−m ] = m , (1 − y)m+1 (1 − y)m+1 (1 − y)m+1 whenever ≤ m ≤ n and f (n, m) = 0, for m > n Therefore, Gm (y) = 103 5.2 APPLICATIONS TO RECURRENCE RELATION Determine a generating function for the numbers S(n, m), n, m ∈ Z, n, m ≥ that satisfy S(n, m) = mS(n − 1, m) + S(n − 1, m − 1), (n, m) = (0, 0) with (5.2) S(0, 0) = 1, S(n, 0) = 0, for all n > and S(0, m) = 0, for all m > Hence or otherwise find a formula for the numbers S(n, m) Solution: Define Gm (y) = n≥0 S(n, m)y n Then, for m ≥ 1, Equation (5.2) gives S(n, m)y n = Gm (y) = n≥0 n≥0 = m n≥0 (mS(n − 1, m) + S(n − 1, m − 1)) y n S(n − 1, m)y n + n≥0 S(n − 1, m − 1)y n = myGm (y) + yGm−1 (y) Therefore, Gm (y) = ym Gm−1 (y) Using initial conditions, G0 (y) = and hence − my Gm (y) = where αk = ym = ym (1 − y)(1 − 2y) · · · (1 − my) m k=1 αk , − ky (5.3) (−1)m−k km , for ≤ k ≤ m Thus, k! (m − k)! m S(n, m) = [y n ] y m k=1 m m αk kn−m = = k=1 = Therefore, S(n, m) = m! k=1 m [y n−m ] = k=1 αk − ky (−1)m−k kn k! (m − k)! m−k n (−1) m αk − ky k k=1 m (−1)k (m − k)n m! k=1 m k m k = m! m k=1 (−1)k (m − k)n m and m! S(n, m) = m k (−1)k (m − k)n k=1 (5.4) m k The above expression was already obtained earlier (see Equation (2.1) and Exercise 6) This identity is generally known as the Stirling’s Identity Observation: S(n, m)xm is not considered But verify that (a) Hn (x) = m≥0 Hn (x) = (x + xD)n · as H0 (x) = Therefore, H1 (x) = x, H2 (x) = x + x2 , · · · Hence, it is difficult to obtain a general formula for its coefficients But it is helpful in showing that the numbers S(n, m), for fixed n, first increase and then decrease (commonly called unimodal) The same holds for the sequence of binomial coefficients n m , m = 0, 1, , n 104 CHAPTER GENERATING FUNCTIONS AND ITS APPLICATIONS (b) Since there is no restriction on the non-negative integers n and m, the expression Equation (5.4) is also valid for n < m But, in this case, we know that S(n, m) = m (−1)m−k kn−1 Hence, verify that = 0, whenever n < m k=1 (k − 1)! (m − k)! Bell Numbers: For a positive integer n, the nth Bell number, denoted b(n), is the number of partitions of the set {1, 2, , n} Therefore, by definition, b(n) = n S(n, m), for m=1 n ≥ and by convention (see Stirling Numbers), b(0) = Thus, for n ≥ 1, n b(n) = m S(n, m) = m=1 m≥1 kn = k≥1 k! m≥1 k=1 (−1)m−k (m − k)! m≥k (−1)m−k kn−1 (k − 1)! (m − k)! S(n, m) = = e k≥1 kn k! (5.5) kn , we compute its exponential generating function (see k! xn b(n) then Exercise 2.1.20.9) Thus, if B(x) = n! n≥0 As, b(n) has terms of the form B(x) = + n≥1 = 1+ = 1+ e e xn b(n) =1+ n! k≥1 k≥1 k! kn n≥1 xn n! n≥1  1 e =1+ k≥1 e 1 ekx − = + k! e  k n  xn k! n! k≥1 k! k≥1 n≥1 (kx)n n! (ex )k − k! k! ex x = 1+ e − − (e − 1) = ee −1 e Recall that ee x −1 (5.6) is a valid formal power series (see Remark 5.1.5) Now, let us derive the recurrence relation for b(n)’s Taking the natural logarithm on both the sides of Equaxn tion (5.6), one has Ln b(n) = ex − Now, differentiation with respect to x n! n≥0 xn−1 gives · b(n) = ex Therefore, after cross multiplication and an xn n≥0 (n − 1)! b(n) n! n≥0 multiplication with x, implies     n n m n b(n)x x x   x = xex b(n) = x · b(n)  (n − 1)! n! m! n! n≥1 n≥0 m≥0 n≥0 Thus, b(n) = [xn ] (n − 1)! b(n)xn n≥1 (n − 1)! n  = [x ] x xm m≥0 m!  · b(n) n≥0 xn n!  = n−1 b(m) · (n − − m)! m! m=0 105 5.2 APPLICATIONS TO RECURRENCE RELATION n−1 Hence, it follows that b(n) = m=0 n−1 m b(m), for n ≥ with b(0) = Determine the number of ways of arranging n pairs of parentheses (left and right) such that at any stage the number of right parentheses is always less than or equal to the number of left parentheses Solution: Recall that this number equals Cn , the nth Catalan number (see Page 44) Let us obtain a recurrence relation for these numbers and use it to get a formula for Cn ’s Let Pn denote the arrangements of those n pairs of parentheses that satisfy “at any stage, the number of right parentheses is always less than or equal to the number of left parentheses” Also, let Qn denote those elements of Pn for which, “at the 2k-th stage, for k < n, the number of left parentheses is strictly greater than the number of right parentheses” We now claim that Qn = and for n ≥ 2, Qn = Pn−1 Clearly Q1 = Note that, for n ≥ 2, any element of Qn , necessarily starts with two left parentheses and ends with two right parentheses So, if we remove the first left parenthesis and the last right parenthesis then one obtains as element of Pn−1 In a similar way, if we add one left parenthesis at the beginning and a right parenthesis at the end of an element of Pn−1 , one obtains an element of Qn Hence, Qn = Pn−1 Let n ≥ and consider an element of Pn Then, for some k, ≤ k ≤ n, the first k pairs of parentheses will form an element of Qk and the remaining (n − k) pairs of parenthesis will form an element of Pn−k Hence, if we take P0 = Q1 = 1, one has n Pn = n Qk Pn−k = k=1 k=1 Pk−1 Qn−k , for n ≥ Now, define P (x) = Pn xn Then n≥0 n n P (x) = Pn x = + n≥0 = + x k≥1 Pk−1 Qn−k n≥1 Pk−1 xk−1 = + x (P (x))2 Thus, Pn x = + n≥1  xP (x)2 n n≥k  k=1  Pn−k xn−k  = + x P (x) − P (x) + = and P (x) = 1± √ xn k≥1  Pk−1 xk−1  − 4x Therefore, using P0 = 1, one 2x 106 CHAPTER GENERATING FUNCTIONS AND ITS APPLICATIONS obtains P (x) = Pn = = = = 1− √ − 4x Hence, 2x √ [xn ]P (x) = · [xn+1 ] − − 4x 1 1 −1 − ··· −n 2 2 − · (−4)n+1 (n + 1)! · (−1) · (−3) · (−5) · · · (1 − 2n) · · · · · (2n − 1) 2(−4)n · = 2n n+1 (n + 1)! (n + 1)! 2n , the nth Catalan Number n+1 n We end this section with a set of exercises Exercise 5.2.3 In each of the following, obtain a recurrence relation for the numbers an ’s Hence or otherwise, determine a formula for the an ’s (a) Let an denote the number of ways of climbing a staircase in which “at each step, the climber climbs either one stair or two stairs (b) Let an denote the number of sequences of length n that consist of the digits 0, 1, 2, , that not have two consecutive appearances of 9’s (c) Let an denote the number of ways in which a person can spend n rupees to buy either a toffee worth rupee or a chocolate worth rupees or an ice-cream worth rupees Let f (n, k) denote the number of k-element subsets that can be selected from the set {1, 2, , n} that not contain two consecutive integers Determine a recurrence relation for f (n, k)’s and hence determine the value of f (n, k) Suppose the numbers {1, 2, , n} are arranged in a round table Let g(n, k) denote the number of k-element subsets that can be selected from this round table with the condition that no two consecutive integers appear Find a recurrence relation for g(n, k)’s and hence determine the value of g(n, k) [Hint: g(n, k) = f (n − 1, k) + f (n − 3, k − 1).] Let a(0) = a(1) = and let a(n) = a(n − 1) + (n − 1)a(n − 2), for n ≥ Compute the xn exponential generating function A(x) = a(n) Hence or otherwise, compute a(n)’s n! n≥0 n Let a(0) = and let k=0 5.3 a(k)a(n − k) = 1, for all n ≥ Determine the sequence a(n) Applications to Generating Functions The ideas learnt in the previous sections will be used to get closed form expressions for sums arising out of binomial coefficients To so, recall the list of formal power series that appear on Page 100 107 5.3 APPLICATIONS TO GENERATING FUNCTIONS Example 5.3.4 Find a closed form expression for the numbers a(n) = k≥0 a(n)xn Then Solution: Define A(x) = k n−k n≥0 a(n)xn = A(x) = n≥0 = k≥0 n≥0  xk  n≥k   k≥0    k k  xn =  xn  n−k n−k k≥0 n≥0  k xn−k  = n−k xk (1 + x)k = k≥0 x(1 + x) = k≥0 Therefore, Example 5.2.1.2 implies a(n) = [xn ]A(x) = [xn ] Fibonacci number Find a closed form expression for the polynomials a(n, x) = − x(1 + x) = Fn , the n-th − x(1 + x) ⌊n⌋ k=0 a(n, x)y n Then Solution: Define A(x, y) = k n−k k (−1)k xn−2k n≥0 a(n, x)y n = A(x, y) = n≥0 n≥0    ⌊n⌋ k=0  n−k (−1)k xn−2k  y n k  n−k (xy)n−2k  k k≥0 n≥2k   t = (−1)k y 2k (xy)−k  (xy)t  k (−1)k y 2k  = k≥0 t≥k (−y )k (xy)−k = k≥0 = where α = x+ (xy)k = · k+1 (1 − xy) − xy k≥0 −y − xy k 1 1 · = = , − xy − −y2 − xy + y (1 − αy)(1 − βy) 1−xy √ √ x2 − x − x2 − and β = Thus, 2 a(n, x) = [y n ]A(x, y) = [y n ] = = 1 = [y n ] − xy + y α−β αn+1 − β n+1 α−β  √  x+ x −4 √ x2 − n+1 − x− √ x2 α β − − αy − βy −4 n+1   Since α and β are the roots of y − xy + = 0, α2 = αx − and β = βx − Therefore, verify that the a(n, x)’s satisfy the recurrence relation a(n, x) = xa(n − 1, x) − a(n − 2, x), for n ≥ 2, with initial conditions a(0, x) = and a(1, x) = x 108 CHAPTER GENERATING FUNCTIONS AND ITS APPLICATIONS Let A = (aij ) be an n×n matrix, with aij = 1, whenever |i−j| = and 0, otherwise Then A is an adjacency matrix of a tree T on n vertices, say 1, 2, , n with the vertex i being adjacent to i + 1, for ≤ i ≤ n − It can be verified that if a(n, x) = det(xIn − A), the characteristic polynomial of A, then a(n, x)’s satisfy the above recurrence relation The polynomials a(n, 2x)’s are also known as Chebyshev’s polynomial of second kind Let us now substitute different values for x to obtain different expressions and then use them to get binomial identities √ 1 z 2n+2 − (a) Let x = z + Then x2 − = z − and we obtain a(n, z + ) = Hence, z z z (z − 1)z n n ⌊2⌋ z 2n+2 − n−2k n−k (−1)k z + we have = Or equivalently, k z (z − 1)z n k=0 ⌊n⌋ n−k (−1)k z + k k=0 n−2k 2k z = z 2n+2 − z2 − (b) Writing x in place of z , we obtain the following identity ⌊n⌋ k=0 n−k xn+1 − (−1)k (x + 1)n−2k xk = = k x−1 n xk k=0 (c) Hence, equating the coefficient of xm in (5.7) gives the identity ⌊n⌋ n−k k (−1)k k=0 (d) Substituting x = in (5.7) gives n − 2k m−k ⌊n⌋ (−1)k k=0 n−k k if ≤ m ≤ n; 1, = 0, otherwise 2n−2k = n + Determine the generating function for the numbers a(n, y) = k≥0 a(n, y)xn Then Solution: Define A(x, y) = n+k 2k yk n≥0 A(x, y) = n≥0 = k≥0 =   k≥0 y x k  n + k k n y x = 2k k≥0 x2k = 2k+1 (1 − x) 1−x y x k≥0 k n≥k n + k n+k x 2k yx (1 − x)2 k 1−x (1 − x)2 − xy (a) Verify that if we substituting y = −2 then k≥0 n+k 1−x (−2)k = [xn ] A(x, −2) = [xn ] = (−1)⌈n/2⌉ 2k (1 + x2 ) (5.7) 109 5.3 APPLICATIONS TO GENERATING FUNCTIONS (b) Verify that if we substituting y = −4 then n+k 1−x (−4)k = [xn ] A(x, −4) = [xn ] = (−1)n (2n + 1) 2k (1 + x)2 k≥0 (c) Let f (n) = k≥0 n+k 2k 2n−k and let F (z) = f (n)z n Then verify that n≥0 1 − 2z 1 F (z) = A(2z, ) = = · + · (1 − z)(1 − 4z) − 4z − z Hence, f (n) = [z n ]F (z) = · 4n 22n+1 + + = 3 We end this chapter with the following set of exercises Exercise 5.3.5 n k Let n be a non-negative integer and define a(n, y) = k≥0 A(x, y) = n≥0 a(n, y)xn Then prove that A(x, y) = √ spectively replace y with (a) k≥0 (b) 2k k y k and 1 ·√ Hence, re1 − x − 4xy 1−x −1 −1 and , to obtain the following identities: n k 2k k (−4)n−k = n k 2k k 2n n (−2)n−k = n n/2 , if n is even; 0, otherwise This identity is popularly known as the (Reed Dawson’s Identity) k≥0 Let m, n ∈ N Then prove that k≥0 n+k m+2k 2k k (−1)k = k+1 Fix positive integers m, n ∈ N and define a(n, y) = m k and c(n, y) = k≥0 and C(x, y) = m+n−k k m k k≥0 y k Compute A(x, y) c(n, y)xn to show the following: n−1 m−1 n+k k y k , b(n, y) = k≥0 a(n, y)xn , B(x, y) = n≥0 m k n k yk b(n, y)xn n≥0 n≥0 (a) k≥0 (b) k≥0 (c) m k n+k m m k n k m+n n (d) k≥0 (e) k≥0 = k≥0 m k (−1)k = k≥0 n k m k 2k = k≥0 · m+n−k m m k · m+n−k m (−2)k = [xn ] (1 + x)m+n = [xn ] (1 + x)m (x + 1)n = k≥0 m k · m+n−k m m k · n+k m Prove that k≥0 (−1)k = (−1)k = (−1)m m+1 k m+n−k m (−1)k = 1, if n = 0, 0, if n > m k n k = [xn ] (1 − x)m+1 110 CHAPTER GENERATING FUNCTIONS AND ITS APPLICATIONS (−1)k Prove that k≥1 n k k−1 m = (−1)m+1 Determine whether or not the following statement is correct n k=1 (−1)k+1 n k k n = k=1 k Determine the exponential generating function for the numbers a(n)’s that appeared in Exercise 14 Notes: Most of the ideas for this chapter have come from book [8] Bibliography [1] J Cofman, “Catalan Numbers for the Classroom?”, Elem Math., 52 (1997), 108 - 117 [2] D I A Cohen, Basic Techniques of Combinatorial Theory, John Wiley and Sons, New York, 1978 [3] William Dunham, Euler, The Master of Us All, Published and Distributed by The Mathematical Association of America, 1999 [4] G E Martin, Counting: The Art of Enumerative COmbinatorics, Undergraduate Texts in Mathematics, Springer, 2001 [5] R Merris, Combinatorics, 2th edition, Wiley-Interscience, 2003 [6] J Riordan, Introduction to Combinatorial Analysis, John Wiley and Sons, New York, 1958 [7] R P Stanley, Enumerative Combinatorics, vol 2, Cambridge University Press, 1999 [8] H S Wilf, Generatingfunctionology, Academic Press, 1990 111 ... respectively Then prove that the functions f and g ◦ f are one-one but g is not one-one Solution: By definiton, it is clear that f is indeed one-one and g is not one-one But g ◦ f (x) = g(f (x)) = g(2x)... collection of n + balls that contains at least one green ball From this collection, pick a collection of n balls that contains at least one green ball Then by the induction hypothesis, this collection... alphabet (a) with ONLY consonants? (b) with ONLY vowels? (c) with a consonant as the first letter and a vowel as the second letter? (d) if the vowels appear only at odd positions? Determine the