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Results & Lemmas (6)

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Theorem 1.1. Theorem 1.1. Let f(eiθ) be defined in (1.2), |||β||| < 1, ℜαj > −1/2, αj ± βj ̸= −1, −2,... for j, k = 0, 1,..., m, and let V (z) satisfy…
Theorem 1.1. Let f(eiθ) be defined in (1.2), |||β||| < 1, ℜαj > −1/2, αj ± βj ̸= −1, −2, . . . for j, k = 0, 1, . . . , m, and let V (z) satisfy the condition (1.11), (1.12) below. Then as n →∞, (1.10) Dn(f) = exp " nV0 + ∞ X k=1 kVkV−k # m Y j=0 b+(zj)−αj+βjb−(zj)−αj−βj × n
Proposition 3.1. Proposition 3.1. Let V (z) ≡0. Let γ = αk or γ = βk, k = 0, 1,..., m, and Dn(f(z)) ̸= 0 for all n. Then for any n = 1, 2,..., (3.20) ∂ ∂γ…
Proposition 3.1. Let V (z) ≡0. Let γ = αk or γ = βk, k = 0, 1, . . . , m, and Dn(f(z)) ̸= 0 for all n. Then for any n = 1, 2, . . . , (3.20) ∂ ∂γ ln Dn(f(z)) = −2χ−1 n ∂χn ∂γ  n + m X j=0 αj 
Proposition 3.3. Proposition 3.3. Let f(z, t) be given by (3.21) and Dn(f(z, t)) ̸= 0 for all n. Let φk(z, t), bφk(z, t), k = 0, 1,..., be the corresponding…
Proposition 3.3. Let f(z, t) be given by (3.21) and Dn(f(z, t)) ̸= 0 for all n. Let φk(z, t), bφk(z, t), k = 0, 1, . . . , be the corresponding orthogonal polynomials. Then for any n = 1, 2, . . . , (3.26) ∂ ∂t ln Dn(f(z, t)) = 1 2πi Z C z−n  Y11(z, t)∂Y21(z, t) ∂z −Y21(z, t)∂Y11(z, t) ∂z
Proposition 5.1. Proposition 5.1. Let (α0, β0,..., αm, βm) be in a compact subset, denote it Λ, belonging to the subset |||β||| < 1, αj ± βj ̸= −1, −2,...…
Proposition 5.1. Let (α0, β0, . . . , αm, βm) be in a compact subset, denote it Λ, belonging to the subset |||β||| < 1, αj ± βj ̸= −1, −2, . . . of the parameter space P = {(α0, β0, . . . , αm, βm) : αj, βj ∈ C, ℜαj > −1/2} and including the point αj = βj = 0, j = 0, 1, . . . , m. Let βj = 0 if αj = 0, j = 0, 1, . . . , m, δ = n2(|||β|||−1). Then for n →∞, and ν = 0, 1, . . . , m, (5.1) ∂ ∂αν ln Dn(f(z)) = 2αν + (αν + βν)  ∂ ∂αν ln Γ(1 + αν + βν) Γ(1 + 2αν) + ln n 
Lemma 6.1. Lemma 6.1. Let the Riemann-Hilbert problem for Φ be solvable. For any k = 0, 1,..., m, (6.27) ∂ln Dn ∂zk = m X j=0 trace Aj ∂eΦj(zj) ∂zk…
Lemma 6.1. Let the Riemann-Hilbert problem for Φ be solvable. For any k = 0, 1, . . . , m, (6.27) ∂ln Dn ∂zk = m X j=0 trace Aj ∂eΦj(zj) ∂zk eΦ−1 j (zj) + trace B ∂eΦ(0)(0) ∂zk 
Lemma 6.2. Lemma 6.2. Let the Riemann-Hilbert problem for Φ be solvable. Let αj ̸= 0, j = 0,..., m. Then for any k = 0, 1,..., m, (6.28) ∂ln Dn ∂αk =…
Lemma 6.2. Let the Riemann-Hilbert problem for Φ be solvable. Let αj ̸= 0, j = 0, . . . , m. Then for any k = 0, 1, . . . , m, (6.28) ∂ln Dn ∂αk = m X j=0 trace Aj ∂eΦj(zj) ∂αk eΦ−1 j (zj) + trace B ∂eΦ(0)(0) ∂αk

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