C M A ommunications in athematical nalysis Conference 03 (2011), pp. 23–41 Proceedings of “Analysis, Mathematical Physics and Applications” held in Ixtapa, Mexico, March 1-5, 2010 ISSN 1938-9787 www.math-res-pub.org/cma E IGENVALUES OF H ESSENBERG T OEPLITZ MATRICES GENERATED BY SYMBOLS WITH SEVERAL SINGULARITIES ∗ J. M. B OGOYA† Departamento de Matemáticas, CINVESTAV del I.P.N., Ciudad de México, Apartado Postal 14-740, 07000, México A. B ÖTTCHER‡ Fakultät für Mathematik, Technische Universität Chemnitz, Chemnitz, 09107, Germany S. M. G RUDSKY§ Departamento de Matemáticas, CINVESTAV del I.P.N., Ciudad de México, Apartado Postal 14-740, 07000, México E. A. M AKSIMENKO¶ Escuela de Fı́sica y Matemáticas, Instituto Politécnico Nacional, Ciudad de México, 07730, México (Communicated by Vladimir Rabinovich) Abstract. In a recent paper, we established asymptotic formulas for the eigenvalues of the n × n truncations of certain infinite Hessenberg Toeplitz matrices as n goes to infinity. The symbol of the Toeplitz matrices was of the form a(t) = t −1 (1−t)α f (t) (t ∈ T), where α is a positive real number but not an integer and f is a smooth function in H ∞ . Thus, a has a single power singularity at the point 1. In the present work we extend the results αk to symbols with a finite number of power singularities. To be more precise, we consider symbols of the form a(t) = t −1 f (t) ∏K k=1 (1 − t/tk ) (t ∈ T), where tk = eiθk , the arguments θk are all different, and the exponents αk are positive real numbers but not integers. Keywords. Toeplitz matrix, eigenvalue, Fourier integral, asymptotic expansion. MSC 2010. Primary 47B35. Secondary 15A15, 15A18, 47N50, 65F15. 1 Introduction and main results Given a function a ∈ L1 on the unit circle in the complex plane T, we denote by ak = ∗ We Z 2π ¡ ¢ a eiθ e−ikθ dθ/2π, k ∈ Z, 0 acknowledge support of this work by a grant of the DAAD, by CONACYT grants 219345, 80503, 102800, and by PROMEP (México) “Proyecto de Redes”. † E-mail address: mbogoya@math.cinvestav.mx ‡ E-mail address: aboettch@mathematik.tu-chemnitz.de § E-mail address: grudsky@math.cinvestav.mx ¶ E-mail address: maximenko@esfm.ipn.mx 24 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko the kth Fourier coefficient and by Tn (a) the n × n Toeplitz matrix (a j−k )nj,k=1 . We are interested in the behavior of the eigenvalues of Tn (a) as n goes to infinity. The function a is usually referred to as the symbol or the generating function of the sequence {Tn (a)}∞ n=1 . For real-valued functions a the matrices Tn (a) are all Hermitian and a number of results on the asymptotics of the eigenvalues of Tn (a) are available in this case: see, for example, [6], [12], [15], [17], [19], [20], [21], [22], [24], [25], [27], [28]. In this case the eigenvalues mimic in the one or other sense the distribution of the values of the function a at equispaced points on the unit circle. The picture is less complete for complex-valued symbols. Papers [10], [14], [18] are devoted to the limiting behavior of the eigenvalues of Tn (a) if a is a rational function, while papers [1] and [26] embark on the asymptotic eigenvalue distribution in the case of non-smooth symbols. In [23] and [26], it was observed that if a ∈ L∞ and the essential range R (a) does not separate the plane, then the eigenvalues of Tn (a) approximate R (a), which resembles the Hermitian case. Many of the results of the papers cited above can also be found in the books [5], [7], [8]. An extreme situation is the one where ak = 0 for k ≤ −1. Then, the matrices Tn (a) are lower triangular and hence the spectrum sp Tn (a) is just the singleton {a0 }. Note that a0 captures almost no information about the function a itself. The first interesting case beyond this trivial situation is the one where Tn (a) has an additional super-diagonal and thus is a Hessenberg Toeplitz matrix. Of course, this happens if and only if ak = 0 for k ≤ −2. Such symbols can be analytically continued into the punctured disk 0 < |z| < 1, which, as pointed out in [18] and [26], can result in an eigenvalues distribution along points and curves that are very different from the range R (a). On the other hand, the presence of singularities in the symbol causes the opposite tendency, that is, it somehow forces the eigenvalues to mimic the range [26]. In [4], we considered symbols with a singularity of the type (1 − t)α (t ∈ T) in order to illustrate certain instability phenomena in the eigenvalue distribution. The eigenvalues of the Hessenberg Toeplitz matrices generated by a(t) = t −1 (1 − t)α were studied in [2]. The recent papers [9] and [16] contain intriguing numerical experiments for individual eigenvalues of Toeplitz matrices whose symbols have a so-called Fisher-Hartwig singularity. These are special symbols that are smooth on T minus a single point but not smooth on the entire circle T; see [7], [8]. Papers [9] and [16] motivated us to take up the singularity (1 − t)α again, and in [3] we established fairly precise results on the eigenvalues of Tn (a) in the case where a(t) = t −1 (1 −t)α f (t) and f satisfies certain smoothness and analyticity requirements. In the present paper, we generalize these results to symbols with several singularities of the power type. Let H ∞ be the usual Hardy space of (boundary values of) bounded analytic functions in the unit disk D. Given a ∈ C(T), we denote by windλ (a) the winding number of a about a point λ ∈ C\ R (a) and by D (a) the set of all λ ∈ C\ R (a) for which windλ (a) 6= 0. In this paper we study the eigenvalues of Tn (a) for symbols a(t) = t −1 f (t) ∏Kk=1 (1 − t/tk )αk (t ∈ T), where f is a smooth function subject to additional conditions, the points tk = eiθk are all different, and the exponents αk are distinct positive real numbers but not integers. Thus, we require in particular that αk 6= α` for k 6= `. Our approach also works if two or more of the exponents αk coincide, although then a series of technical details emerges. To keep this paper within a reasonable volume, we decided not to embark on the case of coinciding exponents here. We enumerate the singularity points tk as follows: let t1 be such that α1 = min1≤k≤K {αk } and number the remaining tk counterclockwise. Let {Tk }Kk=1 be the connected components of T \ {t1 , . . . ,tK } and denote by clos Tk be the arc Tk together with its two endpoints. Let h(t):= a(t)t and h0 be its zeroth Fourier coefficient. We assume that a has the following properties. 1. h ∈ H ∞ and h0 6= 0. 2. f ∈ C∞ (T). 3. h can be analytically extended to an open neighborhood W of T \ {t1 , . . . ,tK } not containing the set {t1 , . . . ,tK }. 4. The derivative a0 (t) does not vanish for t ∈ T \ {t1 , . . . ,tK }, each a(clos Tk ) is a Jordan curve which surrounds the points in its interior clockwise, and for k 6= `, the interiors of the curves a(clos Tk ) and a(clos T` ) are disjoint. Figure 2 shows two concrete examples of such functions. Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 25 If f is identically 1, that is, if a(t) = t −1 ∏Kk=1 (1 − t/tk )αk , then properties 1 to 4 are satisfied if and only if σ:= < 2. To see this, let t revolve the unit circle once counterclockwise starting at t1 . We have ¶ K µ 1 − t/tk αk −1 σ a(t) = t (1 − t/t1 ) ∏ . k=2 1 − t/t1 ∑Kk=1 αk Taking into account that the argument of (1 − t/tk )/(1 − t/t1 ) is piecewise constant and that t −1 (1 − t/t1 )σ describes a loop that encircles the points in its interior exactly once clockwise if and only if σ < 2, it is not difficult to see that the range of a is a flower with K non-overlapping petals and that the petals surround their interiors exactly once clockwise if and only if σ < 2. Let Dn (a) denote the determinant of Tn (a). Thus, the eigenvalues λ of Tn (a) are the solutions of the equation Dn (a − λ) = 0. Our assumptions imply that Tn (a) is a Hessenberg matrix, that is, it arises from a lower triangular matrix by adding the super-diagonal. This circumstance together with the Baxter-Schmidt formula for Toeplitz determinants allows us to express Dn (a−λ) as a Fourier integral. The value of this integral mainly depends on λ and on the singularity of each (1 −¡t/tk )αk at the ¢ point tk . Let W0 be a small open neighborhood of zero in C. We show that for every point λ ∈ D (a) ∩ a(W ) \ W0 there is a unique point tλ ∈ / D such that a(tλ ) = λ. After exploring the contributions of λ and the singular points tk to the Fourier integral, we get the following asymptotic expansion for Dn (a − λ). Theorem 1.1. Let a(t) = t −1¡ h(t) be a symbol with properties 1 to 4. Then, for every small open neighborhood W0 of ¢ zero in C, every λ ∈ D (a) ∩ a(W ) \W0 , and every real positive µ, ! à A 1 k,`,s − ∑ + R1 (λ, n) , (1.1) Dn (a − λ) = (−h0 )n+1 n+2 0 tλ a (tλ ) (k,`,s)∈Lλs+1tkn nαk s+`+1 µ where Lµ is the collection of all the triples (k, `, s) such that k ∈ {1, . . . , K}, ` ∈ {0, 1, . . .}, s ∈ {1, 2, . . .}, and αk s + ` + 1 < µ; " #(`) sin(αk πs)Γ(αk s + ` + 1) f s (tk eiθ )gαk s (θ) ∏ j6=k (1 − eiθtk /t j )α j s Ak,`,s = , eiθ(s+1) i` πtks+1 `! θ=0 g(θ) = (eiθ − 1)/(iθ), and R1 (λ, n) = O (1/nµ ) as n → ∞, uniformly with respect to λ ∈ a(W ) \W0 . Of course, in Theorem 1.1 the superscript (`) means “take ` derivatives with respect to θ” and the subscript θ = 0 means “evaluate in θ = 0”. The order of the sum in (1.1) is 1/nα1 +1 . Thus, among the singularities of the symbol a, the factor (1 − t/t1 )α1 makes the biggest contribution to Dn (a − λ). Changing to the variable t/t1 in a, we can obtain a new symbol ã in which the first singularity point will be 1. Moreover, sp Tn (a) = sp Tn (ã); see [18] or [5, Section 11.1] for details. In order to simplify some of our forthcoming results, © we henceforth assume jwithoutªloss of generality that t1 = 1. Let ωn := exp(−2πi/n) and Jn := j ∈ {0, . . . , n − 1} : a(ωn ) ∈ / W0 , also let γ:= min1≤k≤K {αk : αk > α1 } and ζ:= min{1, α1 , γ − α1 }. As µ is any real positive number, we can develop (1.1) with an arbitrary error bound, but to make our calculations reasonable and readable, we use Theorem 1.1 with µ = 2ζ + α1 + 1 to obtain the following two results. Theorem 1.2. Let a(t) = t −1 h(t) be a symbol with properties 1 to 4. Then, for every small open neighborhood W0 of the origin in C and every j ∈ Jn , the equation Dn (a − λ) = 0 has a solution λ = λ j,n such that à ! à j [1+2ζ] 1 a2 (ωn ) j (α1 +1)/n m tλ j,n = ωn n 1 + ∑ log 2j 0 j m! nm A1,0,1 ωn a (ωn ) m=1 ! Ak,`,s 1 − + R2 ( j, n) , (1.2) ∑ n s−1 A1,0,1 (k,`,s)∈K tk a (ωnj )nαk s+`−α1 +1 26 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko where K is the collection of all triples (k, `, s) 6= (1, 0, 1) such that k ∈ {1, . . . , K}, ` ∈ {0, 1, . . .}, s ∈ {1, 2, . . .}, and αk s + ` < 2ζ + α1 . The remainder satisfies R2 ( j, n) = O (1/n2ζ+1 ) + O (log n/n2 ) as n → ∞, uniformly in j ∈ Jn . The terms logm (·)/(m! nm ) are large when ωn is close to one of the singularity points t j and are small when ωn is far from all the t j ’s. Thus, these terms correct the behavior of the eigenvalues close to each singularity point. j j Theorem 1.3. Let a(t) = t −1 h(t) be a symbol with properties 1 to 4. Then, for every small neighborhood W0 of zero in C and every j ∈ Jn , log n λ j,n = a(ωnj ) + (α1 + 1)ωnj a0 (ωnj ) n ! à [1+2ζ] 2 (ω j ) 1 a n + ωnj a0 (ωnj ) ∑ logm 2j 0 j A1,0,1 ωn a (ωn ) m! nm m=1 − j j Ak,`,s ωn a0 (ωn ) + R3 ( j, n), ∑ n s−1 A1,0,1 (k,`,s)∈K tk a (ωnj )nαk s+`−α1 +1 (1.3) where ζ and K are as in Theorem 1.2 and R3 ( j, n) = O (1/n2ζ+1 ) + O (log2n/n2 ) as n → ∞, uniformly in j ∈ Jn . Figures 1 and 2 illustrate Theorem 1.3. 0.875 0.87 R (a) 0.865 0.86 0.855 0.85 0.38 0.3805 0.381 0.3815 0.382 0.3825 Figure 1. The picture shows a piece of R (a) for the symbol a(t) = t −1 (1 − t)0.3 (1 − t/e2i )0.4 (1 − t/e4i )0.5 (solid blue line) located far from zero. The black dots are sp T4096 (a) calculated by Matlab. The red pluses, blue crosses, and green stars are the approximations obtained by using 2, 3, and 4 terms of (1.3), respectively. Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 27 a(t) = t −1 (1 − t)0.6 (1 + t)0.9 1.5 1 0.5 R (a) 0 −0.5 −1 −1.5 0 0.1 0.2 0.3 0.4 0.5 0.6 a(t) = t −1 (1 − t)0.4 (1 − t/e2i )0.6 (1 − t/e4i )0.7 1 0.5 R (a) 0 −0.5 −1 −1.5 −1.2 −0.9 −0.6 −0.3 0 0.3 0.6 Figure 2. The black dots and the green stars, are the spectrum of T128 (a) calculated with Matlab and formula (1.3) with 4 terms, respectively. 28 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko 2 Toeplitz determinant ¡ ¢ Consider the function b(λ) (t):= 1/ h(t) − λt where λ ∈ D (a) and t ∈ T. Lemma 2.1. Let a(t) = t −1 h(t) be a symbol with property 1. Then, for each λ ∈ D (a) and every n ∈ N, (λ) Dn (a − λ) = (−1)n hn+1 0 bn , (2.1) (λ) where bn stands for the nth Fourier coefficient of b(λ) and h0 for the zeroth Fourier coefficient of h. Proof. The Baxter-Schmidt formula, which can for example be found in [5, p. 37], says that if n, r ≥ 1 are integers and f is a function which is analytic and non-zero in some neighborhood of the origin, then −n f0−r Dn (t −r f ) = (−1)rn [1/ f ]−n 0 Dr (t / f ), where [ ]n denotes the nth Fourier coefficient. Because of property 1, the function f (t):= h(t)−λt satisfies the hypothesis of the Baxter-Schmidt formula. Finally, taking r = 1 we easily obtain the lemma. With the aid of expression (2.1) we will calculate the Toeplitz determinant Dn (a − λ) as a Fourier integral. As in the one singularity case [3], this is our starting point to find an asymptotic expansion for the eigenvalues of Tn (a). The major contributions to this integral comes from λ when λ is close to R (a) and from the singularity points tk . We analyze them in separate sections. 3 Contribution of λ to the asymptotic behavior of Dn Recall that (λ) bn 1 = 2π Z π −π ¡ ¢ b(λ) eiθ e−inθ dθ, is the nth Fourier coefficient of the function b(λ) . Lemma 3.1. Let a(t) = t −1 h(t) be a symbol satisfying properties 1, 3, and 4. Let W0 be a small open neighborhood of zero in C. Then, for each λ ∈ D (a) \ W0 sufficiently close to R (a), there is a unique point tλ in W \ D such that a(tλ ) = λ. Moreover, the point tλ is a simple pole for b(λ) . Proof. Enumerate the collection {Tk }Kk=1 in the following way: for 1 ≤ k < K let Tk be such that tk and tk+1 are its extreme points, and let TK be such that tK and t1 = 1 are its extreme points. The symbol a maps each arch Tk to a different petal Pk := a(Tk ) in R (a); see Figure 3. As h belongs to H ∞ and can be analytically extended to W , the map h can be thought of as a bounded and analytic function in D ∪ W . Since h0 = h(0) 6= 0, the function z−1 h(z) = a(z) is unbounded in D. Thus, the map a must send D \ {0} to the exterior of R (a), that is, the unbounded connected component of C \ R (a), and it must accordingly send W \ D to D (a) ∩ a(W ). By property 4, a0 (t) 6= 0 for every t ∈ Tk . Take S = {t ∈ Tk : a(t) ∈ / W0 }. As a0 is also analytic in W , for each t ∈ S (k) (k) there is an open neighborhood Vt ⊂ W of t such that a0 (t) 6= 0 for every t ∈ Vt . Then, there is an open neighborhood ¡ (k) ¢ (k) (k) (k) Ut ⊂ Vt of t such that a is a conformal map (and hence bijective) from Ut to a Ut . As each S is compact, we can © (k) ª S k (k) Uts . It follows that a is a conformal map (and hence bijective) take a finite sub-cover from Ut t∈S , say U (k) := Ns=1 ¡ ¢ ¡ ¢ from U (k) ⊃ S(k) to a U (k) ⊃ a S(k) . S Let U:= Kk=1 U (k) . The lemma holds for every λ ∈ a(U) ∩ (D (a) \W0 ). Finally, since a0 (tλ ) 6= 0, the point tλ must be a simple pole of b(λ) . Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 29 a(W ) W a(t) t2 W0 (2) Ut tλ 0 1 λ ¡ (2) ¢ a Ut t3 Figure 3. A typical range for the map a with 3 singularities over the unit circle. Lemma 3.1 allows us to write b(λ) (z) = 1 tλ a0 (tλ )(z − tλ ) + b̂(λ) (z), (3.1) where b̂(λ) is analytic with respect to z in W ¡and ¢uniformly bounded with respect to λ in a(W ) \ W0 . Taking Fourier coefficients and writing b̂(λ) (θ) instead of b̂(λ) eiθ , we easily obtain −1 (λ) bn = where 1 I := 2π tλn+2 a0 (tλ ) Z π −π +I, (3.2) b̂(λ) (θ)e−inθ dθ. The first term in (3.2) times (−1)n hn+1 is the contribution of tλ to the asymptotic expansion of Dn (a − λ); see (2.1). The 0 (λ) function b̂ has singularities at each θk , and we use this fact to expand I in the following Section. 4 Contribution of tk to the asymptotic behavior of Dn We start this Section by constructing a particular partition of the unity. Let δ be a small number satisfying 0 < δ < min j6=k {|θ j − θk |}/2 and take a function Φ0 ∈ C∞ [−π, π] which is supported in (−δ/2, δ/2) and is identically 1 in (−δ/4, δ/4). We may also suppose that R (Φ0 ) = [0, 1]. For each x ∈ [−π, π], let Φx (θ):= Φ0 (θ − x). The collection P := {Φθ1 , . . . , ΦθK , Φ∗ }, with Φ∗ (θ):= 1 − ∑Kk=1 Φθk (θ), is a partition of the unity for the interval [−π, π]. By pasting segments [−π, π] in both directions, we continue this partition P to the entire real line R. We will use the following well known asymptotic results, which are, for example, in [11, p. 47] and [13, p. 97], respectively. Theorem 4.1. If α < β, v ∈ CK [α, β], and v(s) (α) = v(s) (β) = 0 for 0 ≤ s ≤ K, then Z β α −inθ v(θ)e 1 dθ = (in)K Z β α v(K) (θ)e−inθ dθ = o(1/nK ) as n → ∞. 30 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko Theorem 4.2. Let β > 0, δ > 0, v ∈ C∞ [0, δ], and v(s) (δ) = 0 for all s ≥ 0. Then, for every K ∈ N, Z δ 0 θβ−1 v(θ)einθ dθ = K−1 (k) v (0)Γ(β + k)iβ+k ∑ k!nβ+k k=0 + RK,v (n), where |RK,v (n)| ≤ CK,v /nβ+K , the branch of the power β + k is the one corresponding to the argument in (−π, π], and Γ(z) is Euler’s Gamma function. If v depends on a parameter and the L∞ norms of the derivatives v(s) for 0 ≤ s ≤ K have bounds that do not depend on the parameter, then one can take a single constant CK,v for all parameters. Lemma 4.3. For every sufficiently small positive δ, we have I= 1 K ∑ 2π k=1 Z θk +δ θk −δ Φθk (θ)b(λ) (θ)e−inθ dθ + Q1 (λ, n), (4.1) where Q1 (λ, n) = o(1/n∞ ) as n → ∞, uniformly with respect to λ in a(W ) \W0 . Proof. Using the partition P , we may write I = I1 + · · · + IK + I ∗ where Ik := 1 2π Z θk +δ θk −δ for k = 1, . . . , K and I ∗ := 1 2π Z π −π Φθk (θ)b̂(λ) (θ)e−inθ dθ Φ∗ (θ)b̂(λ) (θ)e−inθ dθ. Taking v(θ):= Φ∗ (θ)b̂(λ) (θ), α:= θ1 , and β:= 2π + θ1 in Theorem 4.1 we easily get I ∗ = o(1/n∞ ) as n → ∞, uniformly with respect to λ ∈ a(W ) \W0 . Using (3.1), we arrive at Ik = Ik1 − Ik2 where Ik1 := 1 2π Z θk +δ θk −δ Φθk (θ)b(λ) (θ)e−inθ dθ (4.2) and Z 1 θk +δ Φθk (θ)e−inθ ¡ ¢ dθ. 2π θk −δ tλ a0 (tλ ) eiθ − tλ ¡ ¢ Finally, letting v(θ):= Φθk (θ)/ tλ a0 (tλ )(eiθ − tλ ) , α:= θk − δ, and β:= θk + δ in Theorem 4.1 we easily obtain Ik2 = o(1/n∞ ) as n → ∞, uniformly with respect to λ in a(W ) \W0 . Ik2 := Expression (4.1) says that the value of I basically depends on the integrand b(λ) (θ)e−inθ at the singularity arguments θk . As we can take δ as small as we desire, we may assume that in every integral of the sum of (4.1) the variable θ is arbitrarily close to θk . Keeping this idea in mind, we will develop an asymptotic expansion for b(λ) . For future reference, we rewrite (4.1) as I= K ∑ Ik1 + Q1 (λ, n), (4.3) k=1 where Q1 (λ, n) = o(1/n∞ ) as n → ∞, uniformly in λ ∈ a(W )\W0 . Writing h(θ) instead of h(eiθ ), we obtain the following lemma. Lemma 4.4. For every k ∈ {1, . . . , K} and every sufficiently small positive δ, Ik1 = −1 ∞ 1 ∑ λs 2πλ s=0 Z θk +δ Φθk (θ)hs (θ)e−inθ θk −δ eiθ(s+1) dθ. (4.4) Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 31 Proof. Note that b(λ) (θ) = 1 −1 1 = iθ · . iθ −1 h(θ) − λe λe 1 − λ e−iθ h(θ) Let k ∈ {1, . . . , K}. As |h(θ)| → 0 when |θ − θk | → 0, there is a small positive constant δk such that |λ−1 e−iθ h(θ)| < 1 for every |θ − θk | < δk . Let δ = min1≤k≤K {δk }. Thus, b(λ) (θ) = ∞ ¢s −1 ∞ ¡ −1 −iθ hs (θ) λ e h(θ) = − ∑ ∑ s+1 eiθ(s+1) λeiθ s=0 s=0 λ (4.5) for every k ∈ {1, . . . , K} and every |θ − θk | < δ. Finally, inserting (4.5) in (4.2) finishes the proof. We will use the notation Z θk +δ Φ (θ)hs (θ)e−inθ −1 θk dθ. (4.6) 2πλs+1 θk −δ eiθ(s+1) ¡ ¢ Once more, taking v(θ):= −Φθk (θ)/ 2πλeiθ , α:= θk − δ, and β:= θk + δ in Theorem 4.1 we easily obtain Ik1s |s=0 = o(1/n∞ ) as n → ∞, uniformly with respect to λ ∈ a(W ) \W0 . With the previous notation, we can rewrite (4.4) as Ik1s := ∞ Ik1 = ∑ Ik1s + Q2 (k, λ, n), s=1 where Q2 (k, λ, n) = o(1/n∞ ) as n → ∞, uniformly with respect to λ ∈ a(W ) \W0 . Now we use Theorem 4.2 to express Ik1s asymptotically. We recall that h(t) = f (t) ∏Kk=1 (1 − t/tk )αk , where tk = eiθk , the arguments θk are all different, and the exponents αk are positive reals but not integers, with α1 = min1≤k≤K {αk }. Lemma 4.5. Let f be a function with property 2 and µ be any positive real number. Then, for k ∈ {1, . . . , K}, Ik1 = Ak,`,s + Q7 (k, λ, n), s+1t n nαk s+`+1 λ k (`,s)∈L ∗ ∑ (4.7) µ where Lµ∗ is the collection of all pairs (`, s) such that ` ∈ {0, 1, . . .}, s ∈ {1, 2, . . .}, and αk s + ` + 1 < µ; sin(αk πs)Γ(αk s + ` + 1) Ak,`,s = i` πtks+1 `! " f s (tk eiθ )gαk s (θ) ∏ j6=k (1 − eiθtk /t j )α j s eiθ(s+1) #(`) , θ=0 g(θ) = (eiθ − 1)/(iθ), and Q7 (k, λ, n) = O (1/nµ ) as n → ∞, uniformly with respect to λ ∈ a(W ) \W0 . Proof. Changing θ to θ + θk in (4.6), we obtain −1 Ik1s = 2πλs+1 ¡ ¢α s ¡ ¢α s Z δ Φ0 (θ) f s (tk eiθ ) 1 − eiθ k ∏ j6=k 1 − eiθtk /t j j e−inθ −δ eiθ(s+1)tkn+s+1 dθ. It is easy to verify that 1 − eiθ = −iθg(θ), where g(θ):= 1 + iθ/2 + (iθ)2 /6 + O (θ3 ) as θ → 0. Thus, we can write Rδ α s Ik1s = −δ θ k v(θ)e−inθ dθ, where v(θ):= ¡ ¢α s −(−i)αk s Φ0 (θ) f s (tk eiθ )gαk s (θ) ∏ j6=k 1 − eiθtk /t j j 2πλs+1 eiθ(s+1)tkn+s+1 , 32 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko the branch of the power αk s being the one corresponding to the argument in (−π, π]. Note that for every sufficiently small positive δ we have g ∈ C∞ [−δ, δ] and g(0) = 1. Clearly, Ik1s = = Z 0 θαk s v(θ)e−inθ dθ + −δ Z δ Z δ 0 (−θ)αk s v(−θ)einθ dθ + θαk s v(θ)e−inθ dθ Z δ 0 0 where Ik1s1 := (−1)αk s ∈ C∞ [0, δ] Z δ 0 αk s θ θαk s v(θ)e−inθ dθ = Ik1s1 + Ik1s2 , inθ v(−θ)e dθ, Ik1s2 := Z δ 0 (4.8) θαk s v(θ)e−inθ dθ. Note that v(±θ) and = 0 for all s ≥ 0 because Φ0 ≡ 0 in a small neighborhood of ±δ. Applying Theorem 4.2 to Ik1s1 and Ik1s2 , we obtain v(s) (±δ) Ik1s1 = Ik1s2 = (−1)αk s+` v(`) (0)Γ(αk s + ` + 1)iαk s+`+1 + Q3 (s, k, L, λ, n), nαk s+`+1 `! `=0 L−1 ∑ L−1 (`) v (0)Γ(α −αk s−`−1 k s + ` + 1)i nαk s+`+1 `! ∑ `=0 + Q4 (s, k, L, λ, n), (4.9) for every L ∈ N, where Q3 and Q4 are O (1/nαk s+L+1 ) as n → ∞, uniformly in λ ∈ a(W ) \ W0 . Substitution of (4.9) in (4.8) yields L−1 (`) ¢ v (0)Γ(αk s + ` + 1) ¡ −αk s−`−1 i + (−1)αk s+` iαk s+`+1 α s+`+1 k n `! `=0 Ik1s = ∑ + Q5 (s, k, L, λ, n), for every L ∈ N, where Q5 (s, k, L, λ, n) = O (1/nαk s+L+1 ) as n → ∞, uniformly in λ ∈ a(W ) \W0 . At this point, one could be tempted to write à ! ∞ L−1 Ak,`,s Ik1 = ∑ ∑ s+1 n αk s+`+1 + Q5 (s, k, L, λ, n) + Q2 (k, λ, n) as n → ∞, (4.10) tk n s=1 `=0 λ where Ak,`,s equals " #(`) sin(αk πs)Γ(αk s + ` + 1) Φ0 (θ) f s (tk eiθ )gαk s (θ) ∏ j6=k (1 − eiθtk /t j )α j s . eiθ(s+1) i` πtks+1 `! θ=0 Note that we can drop the factor Φ0 (θ) above because Φ0 ≡ 1 in a neighborhood of θ = 0. However, representation (4.10) does not permit us to get an appropriate bound for the remainder of Ik1 . We therefore tackle the problem as follows. First notice that K h(θ + θk ) = f (θ + θk ) ∏ (1 − eiθtk /t j )α j j=1 = (1 − e ) f (θ + θk ) ∏(1 − eiθtk /t j )α j = O (θαk ) as θ → 0. iθ αk j6=k Thus, from (4.5) we obtain hs (θ + θk ) (λ) + fk,S (θ) i(θ+θ )(s+1) s+1 k e s=0 λ S−1 b(λ) (θ + θk ) = − ∑ (4.11) Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 33 (λ) for every S ∈ N and every k ∈ {1, . . . , K}. Here fk,S (θ) = O (θαk S ) as θ → 0, uniformly in λ ∈ a(W ) \ W0 . Inserting (4.11) in (4.2) and (4.3) we obtain S−1 1 Ik1 = ∑ Ik1s + 2π s=1 Z δ −δ (λ) Φ0 (θ) fk,S (θ)e−inθ dθ + Q2 (k, λ, n) S−1 L−1 S−1 Ak,`,s + ∑ s+1t n nαk s+`+1 ∑ Q5 (s, k, L, λ, n) k s=1 `=0 λ s=1 =∑ 1 + 2π Z δ −δ (λ) Φ0 (θ) fk,S (θ)e−inθ dθ + Q2 (k, λ, n) (4.12) (λ) for every L, S ∈ N. The function Φ0 (θ) fk,S (θ) belongs to C[αk S] [−δ, δ] and thus by Theorem 4.1, the integral on the right side of (4.12) is o(1/n[αk S] ) as n → ∞, uniformly in λ ∈ a(W ) \W0 . Fix S ∈ N such that [α1 S] > µ. Then, the integral on the right side of (4.12) is o(1/nµ ) as n → ∞, uniformly in λ ∈ a(W ) \W0 for every k ∈ {1, . . . , K}. Now fix L ∈ N such that α1 + L + 1 > µ. Thus, Q5 (s, k, L, λ, n) = O (1/nµ ) as n → ∞, uniformly in λ ∈ a(W ) \W0 for µ every k ∈ {1, . . . , K}. Therefore, the finite sum ∑S−1 s=1 Q5 (s, k, L, λ, n) is O (1/n ) as n → ∞, uniformly in λ ∈ a(W ) \W0 for every k ∈ {1, . . . , K}. In summary, S−1 L−1 A + Q6 (k, λ, n), Ik1 = ∑ ∑ s+1 n k,`,s tk nαk s+`+1 s=1 `=0 λ where Q6 (S, k, L, λ, n) = O (1/nµ ) as n → ∞, uniformly in λ ∈ a(W ) \W0 for every k ∈ {1, . . . , K}. Finally, avoiding the unnecessary terms of the sum we finish the proof. Proof of Theorem 1.1. Combine (2.1), (3.2), (4.3), and (4.7). 5 Individual eigenvalues In order to find the eigenvalues of the matrices Tn (a), we need to solve the equations Dn (a − λ) = 0. We start this Section by locating the zeros of Dn (a − λ). Let W0 be a small open neighborhood of zero in C and ωn := exp(−2πi/n). Let © ª (5.1) Jn := j ∈ {0, . . . , n − 1} : a(ωnj ) ∈/ W0 . Recall that λ = a(tλ ). Take an integer j ∈ Jn . Using the representations 1 tλ2 a0 (tλ ) = 1 2j j ωn a0 (ωn ) 1 + O (|tλ − ωnj |), a2 (tλ ) = 1 j a2 (ωn ) + O (|tλ − ωnj |), where tλ belongs to a small neighborhood of ωn , we see that the determinant Dn (a − λ) in (1.1) equals (−h0 )n+1 times ¯! ï ï ¯! ¯t − ω j ¯ ¯tλ − ωnj ¯ n¯ ¯λ T1 − T2 + O ¯ n ¯ + O + R1 (λ, n), (5.2) ¯ tλ ¯ nα1 +1 j where tλ belongs to a small neighborhood of ωn , j T1 := 1 tλn ωn a0 (ωn ) 2j j , T2 := ∑ (k,`,s)∈Lµ Ak,`,s as+1 (ωn )tkn nαk s+`+1 j = ¡ ¢ A1,0,1 1 + Q8 (λ, n) a2 (ωn )nα1 +1 j 34 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko with Q8 (λ, n) = O (1/nζ ) as n → ∞, uniformly with respect to λ ∈ a(W ) \W0 . Here Lµ , Ak,`,s , and ζ are as in Theorem j 1.1. Expression (5.2) makes sense only when tλ is sufficiently close to ωn and thus it is necessary to know whether there j iφ is a zero of Dn (a − λ) close to ωn . Let tλ := ρe . It is easy to verify that T1 − T2 = 0 if and only if à ρ= !1/n ¯ ¯ j |a(ωn )|2 ¯1 + Q9 (n)¯nα1 +1 (5.3) j |A1,0,1 a0 (ωn )| and à 1 φ = φs = arg n ¢! j ¡ a2 (ωn ) 1 + Q9 (n) 2j j A1,0,1 ωn a0 (ωn ) − 2πs n where s ∈ {0, . . . , n − 1} and Q9 (λ, n) = O (1/nζ ) as n → ∞, uniformly with respect to λ ∈ a(W ) \W0 . When n tends to infinity, (5.3) shows that ρ remains greater than 1 and ρ → 1. The function T1 − T2 has n zeros with respect to λ ∈ D (a) given by ¡ ¢ ¡ ¢ a ρeiφ0 , . . . , a ρeiφn−1 . As Lemma 3.1 establishes a 1-1 correspondence between λ and tλ , the function Dn (a − λ) is analytic with respect to λ ∈ a(W ) \ W0 , that is, analytic with respect to tλ ∈ W \ a−1 (W0 ). We can therefore suppose that T1 − T2 has n zeros with respect to tλ in the exterior of D given by z0 := ρeiφ0 , zn−1 := ρeiφn−1 . ..., We take the function “arg” in the interval (−π, π]. Thus, z j = eiφ j is the nearest zero to ωn . Consider the open neighborhood E j of z j sketched in Figure 4. The boundary of E j is Γ:= Γ1 ∪ Γ2 ∪ Γ3 ∪ Γ4 . We have chosen radial segments Γ2 and Γ4 so that their length is ε 1/n with ε ∈ (0, min{1, α1 , γ − α1 }) and γ = min{α j : α j > α1 } and all the points in Γ2 have the common argument (φ j+1 + φ j )/2, while all the points in Γ4 have the common argument (φ j−1 + φ j )/2. As we can see in Figure 4, these points run from the unit circle T to (1 + 1/nε )T. Note also that Γ1 ⊂ (1 + 1/nε )T and Γ3 ∈ T. Recall Jn from (5.1). We put diam(E j ):= sup{|z1 − z2 | : z1 , z2 ∈ E j }. j Γ1 Ej Γ2 zj ωn j+1 ωn j Γ4 Γ3 ωn j−1 T z j−1 z j+1 Figure 4. The neighborhood E j of z j in the complex plane. Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 35 Theorem 5.1. Suppose a(t) = t −1 h(t) is a symbol with properties 1 to 4. Let ε ∈ (0, min{1, α1 , γ − α1 }) be a constant. Then, there is a family of sets {E j } j∈Jn in C such that 1. {E j } j∈Jn is a family of pairwise disjoint open sets, 2. diam(E j ) ≤ 2/nε , 3. ωn ∈ ∂E j , ¡ ¢ 4. Dn a − a(tλ ) = Dn (a − λ) has exactly one zero in each E j . j Proof. Assertions 1, 2, and 3 can be deduced from the above construction. We prove assertion 4 by studying the behavior of |Dn (a − λ)| in dependence on tλ ∈ Γ. For tλ ∈ Γ1 we have, as n → ∞, |T1 |Γ1 µ ¶ µ ¶ 1 −n exp(−n1−ε ) exp(−n1−ε ) 1+ ε = +O , = j j n n2ε−1 |a0 (ωn )| |a0 (ωn )| 1 ¯ ¡ ¢ ¯¯ ¯A ¯ 1,0,1 1 + Q8 (n) ¯ |T2 |Γ1 = α +1 ¯ ¯, j ¯ n 1 ¯ a2 (ωn ) !¯ ¶ ¯¯ à ¯¯ µ ¶ µ j¯ tλ − ωn ¯ ¯¯ 1 exp(−n1−ε ) ¯ , ¯O , =O ¯ =O ¯ ¯ nε nα1 +1 nα1 +ε+1 1 ¯!¯ ¯ ï ¯ ¯t − ω j ¯ ¯ ¯ ¯λ n¯ ¯ ¯O ¯ n ¯ ¯ ¯ ¯ tλ ¯ ¯ Γ1 Γ1 ¯ ¯ and ¯R1 (n,tλ )¯Γ1 = O (1/nµ ). When n goes to infinity, the absolute value of T2 decreases at polynomial speed over Γ1 , while the absolute values of the remaining terms in (5.2) are smaller over Γ1 . Thus, ¯ ¯ ¯ ¯ ¶ µ ¯ A ¯ ¯ D (a − λ) ¯ 1 1 ¯ ¯ 1,0,1 ¯ ¯ n as n → ∞. ¯ = α1 +1 ¯ ¯+O ¯ ¯ ¯ a2 (ωnj ) ¯ ¯ hn+1 n nα1 +ε+1 0 Γ1 For tλ ∈ Γ3 , as n → ∞, we get |T1 |Γ3 = 1 j |a0 (ωn )| |T2 |Γ3 , ¯!¯ ¯ ï µ ¶ ¯t − ωj ¯ ¯ ¯ 1 n¯ ¯ ¯λ ¯ , ¯O ¯ n ¯ ¯ = O ¯ tλ ¯ ¯ ¯ n Γ3 ¯ ¡ ¢ ¯¯ ¯A ¯ 1,0,1 1 + Q8 (n) ¯ = α +1 ¯ ¯, j ¯ n 1 ¯ a2 (ωn ) 1 ¯ ï !¯ µ ¶ j ¯¯ ¯ ¯ ¯ − ω t 1 n ¯ ¯ λ , ¯O ¯ =O ¯ ¯ nα1 +1 nα1 +2 Γ3 ¯ ¯ and ¯R1 (n,tλ )¯Γ3 = O (1/nµ ). When n goes to infinity, the modulus of T1 remains constant over Γ3 , while the moduli of the remaining terms in (5.2) are smaller there. Consequently, ¯ ¯ µ ¶ ¯ D (a − λ) ¯ 1 1 ¯ ¯ n = + O as n → ∞. ¯ ¯ j 0 ¯ ¯ hn+1 n |a (ω )| n 0 Γ 3 As for the radial segments Γ2 and Γ4 , we start by showing that T1 and −T2 have the same argument there. Since z j is a zero of T1 − T2 , we deduce that à ! à ¡ ¢! A1,0,1 1 + Q8 (n) 1 arg = arg j 2j j a2 (ωn )nα1 +1 znj ωn a0 (ωn ) 36 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko as n → ∞ and thus à −nφ j + arg ! 1 ωn a0 (ωn ) 2j j à = arg ¡ ¢! A1,0,1 1 + Q8 (n) j a2 (ωn ) . (5.4) For tλ ∈ Γ2 we have à ! à n arg(T1 ) = arg = − (φ j−1 + φ j ) + arg 2j 0 j n 2 tλ ωn a (ωn ) à ¡ ¢! A1,0,1 1 + Q8 (n) n = (φ j − φ j−1 ) + arg j 2 a2 (ωn ) à ¡ ¢! A1,0,1 1 + Q8 (n) = arg(−T2 ). = π + arg j a2 (ωn ) ! 1 Here the third line is due to (5.4). In addition, as n → ∞, ¯!¯ ¯ ï µ ¶ ¯ ¯t − ωj ¯ ¯ 1 ¯ ¯λ n¯ ¯ = O , O ¯ ¯ n ¯ ¯ ¯ ¯ tλ ¯ ¯ nε |tλ |n Γ2 1 ωn a0 (ωn ) 2j j ¯ ï ¯ !¯ µ ¶ ¯ ¯tλ − ωnj ¯ ¯¯ 1 ¯ = O , O ¯ ¯ ¯ ¯ nα1 +1 nα1 +ε+1 Γ2 ¯ ¯ and ¯R1 (n,tλ )¯Γ2 = O (1/nµ ). Furthermore, ¯ ¯ ¯ ¯ µ ¶ ¶ µ ¯ D (a − λ) ¯ 1 1 ¯¯ A1,0,1 ¯¯ 1 1 ¯ ¯ n + O + = + O ¯ ¯ ¯ ¯ j ¯ ¯ hn+1 nε |tλ |n nα1 +1 ¯ a2 (ωnj ) ¯ nα1 +ε+1 |tλn a0 (ωn )| 0 Γ 2 over Γ2 as n → ∞. The situation is similar for the segment Γ4 . 1 j |a0 (ωn )| ≈ |A1,0,1 | nα1 +1 |a(ωn )|2 1 Γ Γ2 z T j Γ 4 n ωj Γ 3 ≈ Figure 5. The absolute value of Dn (a − λ)/hn+1 over E j . 0 j Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 37 Figure 5 resumes our analysis of |Dn (a − λ)/hn+1 0 |. From the previous study of |Dn (a − λ)| over Γ we infer that for every sufficiently large n we have ¯ ¯ 1 ¯¯ A1,0,1 ¯¯ |T1 − T2 |Γ ≥ α +1 ¯ ¯ 2n 1 ¯ a2 (ωnj ) ¯ and ¯! ¯ ¯ ï ï ¯! ¶ µ ¯ ¯ ¯t − ωj ¯ ¯tλ − ωnj ¯ 1 ¯ ¯ ¯λ n¯ . + O + R (n,t ) ≤ O O ¯ ¯ n ¯ 1 λ ¯ ¯ ¯ ¯ tλ ¯ nα1 +1 nα1 +ε+1 Γ Hence, by Rouché’s theorem, Dn (a − λ)/(−h0 )n+1 and T1 − T2 have the same number of zeros in E j , that is, a unique zero. As a consequence of Theorem 5.1, we can iterate the variable tλ in the equation Dn (a − λ) = 0, where Dn (a − λ) is j given by (1.1). In this fashion we find the unique eigenvalue of Tn (a) which is located close to a(ωn ). Proof of Theorem 1.2. The equation Dn (a − λ) = 0 with Dn (a − λ) given by (1.1) is equivalent to the equation   tλ−n =  A1,0,1tλ2 a0 (tλ )  Ak,`,s , 1 + 1 + Q (n,t ) 10 λ ∑  a2 (tλ )nα1 +1  A1,0,1 (k,`,s)∈L as−1 (tλ )tkn nαk s+`−α1 (5.5) µ (k,`,s)6=(1,0,1) where Q10 (n,tλ ) = O (1/nµ−α1 −1 ) as n → ∞, uniformly with respect to tλ ∈ W \ a−1 (W0 ). Recall from Theorem 1.1 that γ = min{α j : α j > α1 } and ζ = min{1, α1 , γ − α1 }. As µ is any real positive number, we can develop (5.5) with an arbitrary error bound, but to make our calculations reasonable and readable, we limit ourselves to µ = 2ζ + α1 + 1. Equation (5.5) is an implicit expression for tλ . We manipulate it to obtain a few asymptotic terms for tλ . Remember that λ belongs to D (a) \ W0 ; see Figure 3. We can choose W so thin that λ = a(tλ ), a0 (tλ ), and tλ are bounded and not too close to zero. After taking the nth root for the main branch specified by the argument in (−π, π] and expanding in (5.5),   tλ j,n =ωnj n(α1 +1)/n  1 + [1+2ζ] ∑ m=1 à logm a2 (tλ j,n ) A1,0,1tλ2 j,n a0 (tλ j,n ) !   1  + Q ( j, n) 11  m! nm    × 1 − Ak,`,s s−1 A1,0,1 (k,`,s)∈L a (tλ j,n )tkn nαk s+`−α1 +1 µ (k,`,s)6=(1,0,1) 1 ∑  + Q12 ( j, n) , (5.6) where Q11 and Q12 are O (1/n2ζ+1 ) as n → ∞, uniformly with respect to j ∈ Jn . After multiplying in (5.6) we obtain  tλ j,n à [1+2ζ]  a2 (tλ j,n ) m =ωnj n(α1 +1)/n  1 + log ∑  A1,0,1t 2 a0 (t m=1 λ j,n ! λ j,n ) 1 m! nm  − Ak,`,s s−1 A1,0,1 (k,`,s)∈L a (tλ j,n )tkn nαk s+`−α1 +1 µ (k,`,s)6=(1,0,1) 1 ∑  + Q13 ( j, n) , (5.7) 38 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko where Q13 (n,tλ ) = O (1/n2ζ+1 ) as n → ∞, uniformly with respect to tλ ∈ W \ a−1 (W0 ). Note that, as n → ∞, (α1 +1)/n n µ ¶ µ 2 ¶ log n log n log n = exp (α1 + 1) = 1 + (α1 + 1) +O . n n n2 (5.8) Thus, our first approximation for tλ j,n is tλ j,n = ωnj + Q14 ( j, n), where Q14 ( j, n) = O (log n/n) as n → ∞, uniformly with respect to j ∈ Jn . Replacing tλ j,n by this approximation in (5.7) we obtain  à ! [1+2ζ] 2 (ω j )  a 1 n m tλ j,n =ωnj n(α1 +1)/n  2j 0 j 1 + ∑ log A m! nm m=1 1,0,1 ωn a (ωn )  − Ak,`,s n A1,0,1 (k,`,s)∈L tk as−1 (ωnj )nαk s+`−α1 +1 µ (k,`,s)6=(1,0,1) 1 ∑  + R2 ( j, n) , where R2 ( j, n) = O (1/n2ζ+1 ) + O (log n/n2 ) as n → ∞, uniformly with respect to j ∈ Jn . Proof of Theorem 1.3. Inserting (5.8) in (1.2) we obtain  tλ j,n  log n =ωnj  1 + (α1 + 1) n à [1+2ζ] + ∑ m=1 log ! j a2 (ωn ) m 2j j A1,0,1 ωn a0 (ωn ) 1 m! nm  − 1 ∑ Ak,`,s A1,0,1 (k,`,s)∈L tkn as−1 (ωnj )nαk s+`−α1 +1 µ  + Q15 ( j, n) , (5.9) (k,`,s)6=(1,0,1) where Q15 ( j, n) = O (1/n2ζ+1 ) + O (log2n/n2 ) as n → ∞, uniformly with respect to j ∈ Jn . j j Since the symbol a is analytic in a small neighborhood of each tλ j,n , we have λ j,n = a(tλ j,n ) = a(ωn + z) = a(ωn ) + j a0 (ωn )z + O (|z|2 ). Thus, applying the symbol a to (5.9), we get log n λ j,n=a(ωnj ) + (α1 + 1)ωnj a0 (ωnj ) n ! à j [1+2ζ] 1 a2 (ωn ) j 0 j m + ωn a (ωn ) ∑ log 2j 0 j m! nm A1,0,1 ωn a (ωn ) m=1 j j Ak,`,stk−n ωn a0 (ωn ) − ∑ as−1 (ω j )nαk s+`−α1 +1 + ωnj a0 (ωnj )Q15 ( j, n) + Q16 ( j, n), A1,0,1 (k,`,s)∈ n Lµ (k,`,s)6=(1,0,1) where Q16 ( j, n) = O (log2n/n2 ) as n → ∞, uniformly with respect to j ∈ Jn . Eigenvalues of Hessenberg Toeplitz matrices generated by symbols with several singularities 39 −3 7 x 10 6 5 4 3 2 1 0 0 50 100 150 200 250 ¡ ¢ Figure 6. The absolute value of the difference between the eigenvalues of T256 t −1 (1 − t)0.6 (1 + t)0.9 obtained with Matlab and formula (6.2). The red, blue, and green dots correspond to the approximations of (6.2) with 2, 3, and 4 terms, respectively. 6 Examples In this Section we consider two particular situations for symbols with two and three singularities. In these situations we employ our formulas for tλ j,n and λ j,n , and with the aid of Matlab, we calculate the corresponding numerical errors. Example 6.1. Consider the symbol a(t) = t −1 (1 −t)0.6 (1 +t)0.9 with two singularities. In this case equations (1.2) and (1.3) become à à ! ! nA 2 (ω j ) (−1) 1 a n 2,0,1 tλ j,n = ωnj n1.6/n 1 + log − + R2 ( j, n) (6.1) 2j j n A1,0,1 n1.3 A1,0,1 ωn a0 (ωn ) and log n ωn a0 (ωn ) λ j,n = a(ωnj ) + 1.6ωnj a0 (ωnj ) + log n n j à ! j a2 (ωn ) A1,0,1 ωn a0 (ωn ) 2j j (−1)n A2,0,1 ωn a0 (ωn ) + R3 ( j, n), A1,0,1 n1.3 j − j j (6.2) respectively. Here A1,0,1 = 20.9 sin(0.6π)Γ(1.6)/π, A2,0,1 = 20.6 sin(0.9π)Γ(1.9)/π, and R2 , R3 are O (1/n1.6 ) as n → ∞, uniformly with respect to j ∈ Jn . Table 1 shows the data, see also Figures 2 and 6. Example 6.2. Consider now the symbol a(t) = t −1 (1 − t)0.4 (1 − t/e2i )0.6 (1 − t/e4i )0.7 40 J. M. Bogoya, A. Böttcher, S. M. Grudsky, and E. A. Maksimenko n (6.1) with 1 term (6.1) with 2 terms (6.1) with 3 terms (6.2) with 2 term (6.2) with 3 terms (6.2) with 4 terms 256 512 1024 2048 4096 1.1×10−2 6.8×10−3 3.3×10−3 1.7×10−3 2.6×10−3 2.5×10−3 1.4×10−2 1.6×10−3 1.4×10−3 7.9×10−4 7.9×10−4 7.1×10−3 5.8×10−4 4.4×10−4 2.3×10−4 2.2×10−4 3.5×10−3 2.2×10−4 1.8×10−4 7.1×10−5 6.6×10−5 1.7×10−3 7.5×10−5 6.0×10−5 8.4×10−4 2.2×10−5 1.9×10−5 8.5×10−4 2.6×10−5 2.0×10−5 Table shows¢the maximum error obtained with formulas (6.1) and (6.2) for the eigenvalues of the matrices ¡ −11. The table 0.6 Tn t (1 − t) (1 + t)0.9 for different values of n. The data was obtained by comparison with the solutions given by Matlab, taking into account only the 90% best approximated eigenvalues. with three singularities. In this case equations (1.2) and (1.3) read à à ! 2 (ω j ) a 1 n tλ j,n = ωnj n1.4/n 1 + log 2j j n A1,0,1 ωn a0 (ωn ) ! A2,0,1 e−2ni A3,0,1 e−4ni − − + R2 ( j, n) A1,0,1 n1.2 A1,0,1 n1.3 and ωn a0 (ωn ) + log n n log n λ j,n = a(ωnj ) + 1.4ωnj a0 (ωnj ) j à ! j a2 (ωn ) A1,0,1 ωn a0 (ωn ) 2j j A2,0,1 e−2ni ωn a0 (ωn ) A3,0,1 e−4ni ωn a0 (ωn ) − + R3 ( j, n) A1,0,1 n1.2 A1,0,1 n1.3 j − j (6.3) j j j (6.4) respectively. Here A1,0,1 = sin(0.4π)Γ(1.4)(1 − e−2i )0.6 (1 − e−4i )0.7 /π, A2,0,1 = sin(0.6π)Γ(1.6)(1 − e2i )0.4 (1 − e−2i )0.7 /(πe4i ), A3,0,1 = sin(0.7π)Γ(1.7)(1 − e4i )0.4 (1 − e2i )0.6 /(πe8i ), and R2 , R3 are O (1/n1.4 ) as n → ∞, uniformly with respect to j ∈ Jn . Table 2 shows the data, see also Figure 2. n (6.3) with 1 term (6.3) with 2 terms (6.3) with 4 terms (6.4) with 2 terms (6.4) with 3 terms (6.4) with 5 terms 256 512 1024 2048 4096 2.5×10−2 1.1×10−2 6.2×10−3 3.1×10−3 1.0×10−2 7.8×10−3 2.6×10−2 9.2×10−3 5.7×10−3 3.0×10−3 2.4×10−3 1.2×10−2 2.0×10−3 1.8×10−3 9.0×10−4 6.8×10−4 6.4×10−3 6.3×10−4 5.2×10−4 2.8×10−4 2.3×10−4 3.2×10−3 2.1×10−4 1.9×10−4 1.6×10−3 9.5×10−5 7.8×10−5 1.6×10−3 7.8×10−5 7.0×10−5 Table shows the maximum error ¢obtained with formulas (6.3) and (6.4) for the eigenvalues of the matrices ¡ −12. The table 2i 0.4 Tn t (1 − t/e ) (1 − t/e4i )0.6 (1 − t/e6i )0.7 for different values of n. The data was obtained by comparison with the solutions given by Matlab, taking into account only the 90% best approximated eigenvalues. 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