Other Things & Some Applications

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1 Chapter 4 Other Things & Some Applications 4. Unimodality, Log-concavity & Real-rootedness (4.) Denition Let A = (a 0, a,..., a n ) be a sequence of positive real numbers. The sequence A is said to be: unimodal if a 0 a a d a n for some d N 0 ; log-concave if a 2 i+ a ia i+2 for all 0 < i < n; strictly log-concave if a 2 i+ > a ia i+ for all 0 < i < n. (4.2) Remark a 2 i+ a i a i+2 log a i+ log a i + log a i+2 2 (4.3) Example A = (, 7, 0, 3, 2) is unimodal but not log-concave ( ). A = (2, 5, 2, 5, 0, 4, ) is unimodal & log-concave & strictly log-concave. (4.4) Proposition If A (R + ) n+ is log-concave, then it is unimodal. Proof. If the sequence is not unimodal, then there exists 0 < r < n such that a r > a r < a r+ a 2 r < a r a r+ A is not log-concave. The contrapositive of this statement is the result. (4.5) Lemma If the zeros of a polynomial p(x) (of degree n > ) are all real and negative, then the zeros of p (x) are all real and negative. Proof. Suppose p(x) = c 0 + c x c n x n = a(x + r ) m (x + r k ) m k, r i, c i, a R, m i N, r i > 0 Suppose also that 0 < r i < < r k. The polynomial p (x) has degree n. Since (x + r i ) m i is a factor of p(x), we know (x + r i ) m i is a factor of p (x). From this r,..., r k are zeros of p (x) of multiplicities m,..., m k, respectively. Now, the 8

2 CHAPTER 4. function p(x) is continuously dierentiable of R. If i < k then p( r i ) = p( r i+ ) = 0. By Rolle's theorem, there exists ri ] r i+, r i [ such that p (ri ) = 0. Thus, there are k distinct numbers r,..., r k, none equal to { r,..., r k }, that are zeros of p (x). We have accounted for (m ) + + (m k ) + (k ) = m + + m k k + k = n zeros of p (x). Since deg(p (x)) = n, all roots are real and negative. (4.6) Lemma Let f(x, y) = c 0 x n + c x n y + + c n y n be a polynomial whose zeros ( x y ) are real. Let g(x, y) = i j x i f(x, y) yj If g(x, y) 0, then all zeros of g(x, y) are real. Proof. Suppose f(x, y) has k distinct roots λ,..., λ k. Then f(x, y) = a(x + λ y) m (x + λ k y) m k for a R, m i N. By repeated use of the previous lemma, all non-zero partial derivatives of f(x, y) will have real roots. (4.7) Theorem Let p(x) = c 0 + c x c n x n be a polynomial whose zeros are all real and negative. Then the sequence (c 0, c,..., c n ) is strictly log-concave. Proof. Since the zeros of p(x) are real and negative, the sequence of coecients is strictly positive (expand a(x + r ) m (x + r k ) m k). If p(x) = c 0 + c x c n x n, let Partially dierentiate f(x, y): We have Now do f(x, y) = c 0 x n + c x n y c n y n m n m 2 x m f(x, y) yn m 2 m x m f(x, y) = (n) mc 0 x n m + (n ) m c x n m y + + (m) m c n m 2 y n m 2 n m 2 y n m 2 ( m ) f(x, y) xm = i c i (n i) m (i) m x n m i y 2 (n m i) g(x, y) In the sum, i {n m 2, n m, n m}. Thus g(x, y) = [ 2 (m+)!(n m )! cn m m + (n m)y2 + c n m 2xy + c ] n m 2 (m + 2)x2 n m 82

3 CHAPTER 4. Let c j = ( n j) pj for all j, then g(x, y) = 2 n! [ p n m y 2 + 2p n m xy + p n m 2 x 2] According to the previous lemma, this polynomial has real roots, since it is nonzero. Thus the discriminant is 0, i.e. (2p n m ) 2 4(p n m p n m 2 ) 0 p 2 n m p n m p n m 2 c 2 n m m+2 n m m+ n m c n mc n m 2 > c n m c n m 2 Thus the sequence is strictly log-concave. (4.8) Remark {constants of neg. real rooted polynom.} {strictly log-concave} {log-concave} {unimodal} (4.9) Example Is { ( n i) } n i=0 strictly log-concave? Well ( ) ( ) ( ) n n n p(x) = + x + + x n = ( + x) n 0 n which has the single negative root x =. By the last theorem, the answer is yes. (4.0) Proposition The sequence of Stirling numbers of the 2 nd kind (S(n, ),..., S(n, n)) is log-concave. Proof. The generating function of these numbers is We know By dierentiating, we have i.e. B n (x) = n S(n, k)x k k= S(n, k) = S(n, k ) + ks(n, k) B n (x) = xb n (x) + xb n (x), n > 0, B 0 (x) = (4.) B (x) = x, B 2 (x) = x + x 2, B 3 (x) = x + 3x 2 + x 3, Multiply each side of (4.) by e x e x B n (x) = xe x B n (x) + xe x B n (x) = x d dx (ex B n (x)) 83

4 CHAPTER 4. Let f n (x) = e x B n (x). Then f n (x) = xf n (x) We claim that all zeros of f n (x), of which there are n, are real, distinct and negative (except for x = 0). This is true for n =, n = 2. Assume true for f (x),..., f n (x). since B n (x) has n distinct zeros, then (e x B n (x)) has n 2 distinct zeros that lie between the zeros of B n (x). Multiplying by x gives another zero at x = 0. Finally, since e x B n (x) 0 as x 0, there is another zero of (e x B n (x)) in ], r[ where r is the smallest root of B n (x). Thus f n (x) has n distinct real and negative (except at x = 0) roots/zeros. Hence, the sequence of coecients is log-concave. 4.2 Transfer Matrix Method-Theory (4.) Denition A directed graph (digraph) D consists of vertices V, edges E and an orientation map φ, D = (V, E, φ). v 2 v v 3 v 4 v 7 v 6 v 5 V = {v,..., v 7 } E = {{v, v 2 }, {v, v 3 },, {v 7, v 7 }} Given {a, b} E, e.g. φ(v, v 2 ) = (v 2, v ) φ(a, b) = (int(a, b), n(a, b)), For every edge e = {v, v } we dene int(e) = v and n(e) = v. A walk Γ in D of length n is a sequence of edges (e,..., e n ) such that n(e i ) = int(e i+ ) for all i n. If n(e n ) = int(e ) then the walk is called closed. (4.2) Denition Let w : E R be a weight function on the edges of the digraph D = (V, E, φ). For any walk Γ = (e,..., e n ) we dene w(γ) = w(e )w(e 2 ) w(e n ) Dene A ij (n) = Γ w(γ) where the sum is over all walks Γ of lengths n in D with int(γ) = v i and n(γ) = v j. Dene A ij (0) = δ ij The adjencency matrix of D is the matrix A = [A ij ], A ij = e E int(e) = v i n(e) = v j w(e) 84

5 CHAPTER 4. (4.3) Example v w(loop) = a w(edge) = b v 2 v 3 A = w(loop) = a; A 2 = b; A 3 = b A 2 = b; A 22 = a; A 23 = 0; A 3 = 0; A 32 = b; A 33 = a a b b A = b a 0 0 b a (4.4) Theorem (4.2..) For n N, the (i, j) entry of A n is A ij (n). Proof. The (i, j) entry of A n is (A n ) ij = A ii A i i 2 A in j where the sum is over all possible sequences (i,..., i n ). The sum will equal to zero if there is no walk of length n from v i v j. If an entry in the sum is non-zero, then the product A ii A i i 2 A in j equals the weight of the corresponding walk. (4.5) Denition From this, Dene the generating function (A n ) ij = A ij (n) F ij (D, λ) = A ij (n)λ n (All information about weighted walk from v i to v j.) For a matrix T, let (T : j, i) be the matrix obtained by removing row j and column i of T. (4.6) Theorem (4.2.2.) If then In other words F ij (D, λ) = A ij (n)λ n F ij (D, λ) = ( )i+j det(i λa : j, i) det(i λa) F ij (D, λ) = ((I λa) ) ij 85

6 CHAPTER 4. (4.7) Remark deg F ij (D, λ) < n 0 = multiplicity of the eigenvalue 0 of A Proof. Obvious using simple linear algebra. F ij (D, λ) is the (i, j) entry of the matrix λ n A n = (I λa) and then use expression for adj(i λa) If c A (λ) = det(a λi) = ( ) p (α p n0 λ n α λ p + λ p ) is the characteristic polynomial of A, then det(i λa) = α p n0 λ p n α λ + where n 0 is the multiplicity of eigenvalue 0 of A. So, deg(det(i λa)) = p n 0, deg(det(i λa : j, i)) p This gives deg(det(i λa : j, i)/ det(i λa)) p (p n 0 ) = n 0 < n 0 (4.8) Example Let Then f(n) = #{(a,..., a n ) : a i {, 2, 3}, (a i, a i+ ) / {(, ), (2, 3)}} f() = {(), (2), (3)} = 3 f(2) = {(, 2), (, 3), (2, ), (2, 2), (3, ), (3, 2), (3, 3)} = 7 f(3) =... = 6 Let D be the directed graph on {, 2, 3} with (i,j) as an edge if j is allowed to follow i in the sequence. 2 3 For any edge e, let w(e) =. Adjacency matrix 0 A = 0 86

7 CHAPTER 4. The sequence 7 {}}{ (, 2, 2,, 3,, 3) is represented by the walk The quantity we want is thus Now (I λa) = (, 2)(2, 2)(2, )(, 3)(3, )(, 3) }{{} 6 edges f(n) = A (n ) + A 2 (n ) + A 3 (n ) = A ij (n ) = (A n ) ij i,j= 2λ λ 2 + λ 3 and, e.g. the,-entry is equal to More precisely i,j= ( λ) 2 λ λ( λ) λ( λ) λ λ 2 λ 2 λ λ( + λ) λ λ 2 ( λ) 2 F (D, λ) = 2λ λ 2 + λ 3 = A (n)λ n = f (n + )λ n n 0 n λ λ 2 + F ij (D, λ) = 2λ λ 2 + λ 3 = f(n + )λ n i,j= and we have obtained the generating function of f(n + ). The o.g.f. of f(n) is thus f(n)λ n = + f(n + )λ n+ = + λ( + λ λ2 ) 2λ λ 2 + λ 3 = λ 2λ λ 2 + λ 3 If our attention is restricted to closed walks in D of length n, then let C D (n) = A (n) + + A pp (n) = Tr(A n ) (4.9) Theorem (4.2.3.) Let Q(λ) = det(i λa). Then n= C D (n)λ n = λq (λ) Q(λ) Proof. Let w,..., w q be the non-zero eigenvalues of A. Then So n= C D (n)λ n = C D (n) = Tr(A n ) = w n + + w n q n= (w n + + wq n )λ n = w λ w λ + + w qλ w q λ Since Q(λ) = ( w λ) ( w q λ), it is easy to see that n= C D (n)λ n = w λ w λ + + w qλ w q λ = λq (λ) Q(λ) 87

8 CHAPTER Transfer Matrix Method Applications (4.20) Example Let f(n) bet the number of sequences (x,..., x n ) of length n that have elements in {a, b} and such that () (x i, x i+ ) (a, a) i < n (2) x = a, x n = b. Let g(n) be the total number of sequences that satisfy (). Let h(n) be the total number of sequences that satisfy () and (x n, x ) (a, a). Invoke the transfer matrix method: D digraph: a b w(edge) =, (v, v 2 ) = (a, b) Adjacency matrix: Note [ ] [ ] A A 2 0 A = = A 2 A 22 [ ] λ I λa = λ λ We have that Using theorem (4.6) (i.e ) f(n) = A 2 (n ) F 2 (D, λ) = A 2 (n)λ n = ( )3 ( λ) λ λ 2 = λ λ λ 2 i.e. Thus ( f(n)λ n = λ λ λ λ 2 f(n + )λ n = λ λ λ 2 ) + = λ2 + λ λ 2 λ λ 2 = λ λ λ 2 For g(n) two ways to do this. First way: g(n) = A (n ) + A 2 (n ) + A 2 (n ) + A 22 (n ) Since deg I λa = 2 (quadratic in λ), we know 88 (A (n) + A 2 (n) + A 2 (n) + A 22 (n))λ n = P (λ) λ λ 2

9 CHAPTER 4. where deg(p ) < 2. Let P (λ) = a + bλ. So P (λ) λ λ 2 = (a + bλ)( + (λ + λ2 ) + (λ + λ 2 ) 2 + ) = a + λ(a + b) + O ( λ 2) = 2 + 3λ + O ( λ 2) (get 2 and 3 by counting paths, giving a = 2 and b =. Hence see original sum) g(n + )λ n = 2 + λ λ λ 2 i.e. Second way: We have (I λa) = g(n)λ n = λ λ 2 + λ λ λ 2 [ ] λ λ λ So (A (n) + A 2 (n) + A 2 (n) + A 22 (n))λ n = (( λ) + (λ) + (λ) + ()) λ λ2 = 2 + λ λ λ 2 as before, giving g(n)λ n = + λ λ λ 2 For h(n): h(n) = A (n) + A 22 (n) (since sequences of length n satisfying () and (x n, x ) (a, a) are isomorphic to sequences of length n+ satisfying () and x = x n check the function that sets x n+ = x [inverse: removes x n+ ], it's a bijection because of ()). Using theorem (4.9) (i.e ): C D (n) = A (n) + A 22 (n) = h(n) thus n= C D (n)λ n = λ( λ λ2 ) λ λ 2 = λ + 2λ2 λ λ 2 89

10 CHAPTER 4. (4.2) Example Dene g(n) = #{(a,..., a n ) : a i {, 2, 3}, (a i, a i+ ) / {(, ), (2, 3)}, i n} where a n+ = a. Transfer (adjacency) matrix 0 A = 0 We have g(n) = A (n) + A 22 (n) + A 33 (n) = C D (n) and g(n)λ n = λq (λ Q(λ) = 2λ + 2λ2 3λ 3 2λ λ 2 + λ 3 =... since Q(λ) = det(i λa) = 2λ λ 2 + λ 3 (4.22) Example Let f(n) = #{(a,..., a n ) : a i {, 2, 3}, (a i, a i+ ) (, 2), i < n (a i, a i+, a i+2 ) / {(2,, 3), (2, 2, 2), (2, 3, ), (3,, 3)}, i < n }} To apply TMM, look at the digraph on V = {(, ), (, 2),..., (3, 3)}. Insert a directed edge from (a, b) (b, c) if we are allowed the sequence (a, b, c) (,) (,2) (,3) (2,2) (2,) (2,3) (3,) (3,2) (3,3) 90

11 CHAPTER 4. Using the order {, 2, 3, 2, 22, 23, 3, 32, 33}, the transfer matrix becomes A = Now and Then where deg P < 8 and deg Q = 8. Q(λ) = det(i λa) f(n) = 9 A ij (n 2) i,j= f(n)λ n = P (λ) Q(λ) 4.4 Transfer Matrix Method Application to Ising Model Consider the -dimensional Ising model (with periodic boundary conditions). [ Fig. Circle with spins..., s n, s, s 2, s 3,..., s i {, +} (like magnets) ] There are 2 n possible spin congurations, s = (s,..., s n ) For any such conguration we associate an energy, given by the Hamiltonian H(s) = J(s s 2 + s 2 s s n s ) }{{} internal energy + h(s + + s n ) }{{} due to external eld J and h are parameters of this `system', showing how much weight is given to a unit of each type of energy. In statistical mechanics we are interested in systems that represent `something', solving them and comparing experimental and theoretical results. The central object of study is 9

12 CHAPTER 4. the partition function, Z n (J, h, x) = x H(s) all possible s=(s,...,s n) (In physics: x = e βc, β = T.) From this we calculate the free energy f(j, h, x) = lim n + n log Z n(j, h, x) Every spin conguration can be represented as a walk of length n on the directed graph D(V, E, φ) where V = {v, v 2 } = {+, }, + (s i, s i+ ) {(+, +), (+, ), (, +), (, )}, 4 edges. To the edge (s i, s i+ ) we associate the weight w(s i, s i+ ) = x J(s is i+ )+ h 2 (s i+s i+ ) Then Z n (J, h, x) = all s x H(s) = s x J(s s 2 +s 2 s 3 + +s ns )+h(s + +s n) = s = s x J(s s 2 )+ h 2 (s +s 2 ) x J(s 2s 3 )+ h 2 (s 2+s 3) x J(sns )+ h 2 (sn+s ) w(s, s 2 )w(s 2, s 3 ) w(s n, s ) = Tr(A n ) where A is the transfer matrix of D: [ ] [ ] A A 2 x J+h x J A = = A 2 A 22 x J x J h By theorem (4.9) (i.e ) we have where Thus n= Z n (J, h, x)λ n = λq (λ) Q(λ) Q(λ) = det(i λa) = λ(x J+h + x J h ) + λ 2 x 2J x 2J Z n (J, h, x) = [λ n λ(x J+h + x J h ) 2λ 2 x 2J ] λ(x J+h + x J h ) + λ 2 x 2J x 2J In order to calculate the fre energy, we must look at the eigenvalues of A x J+h µ x J 0 = det(a µi) = x J x J h µ = µ 2 2µx J cosh(h log x) + 2 sinh(2j log x) 92

13 CHAPTER 4. giving ] µ = x [cosh(h J log x) ± sinh 2 (h log x) + x 4J Let µ be the larger of the eigenvalues and µ be the smaller of the eigenvalues. By diagonalizing A, we have Tr(A n ) = (µ ) n + (µ ) n Thus so f(j, h, x) = lim n + = lim n + n log Z n(j, h, x) = n log(µ ) n = log(µ ) lim n + n log((µ ) n + (µ ) n ) [ ]) f(j, h, x) = log (x J cosh(h log x) + sinh 2 (h log x) + x 4J is the free energy of the -d Ising model. 93