Results & Lemmas (14)
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Lemma 2.1.
Lemma 2.1. We have An(z) = 2n + 1 + 2α 1 −z2 + Rn(t) (1 −z2)2, (2.8) Bn(z) = nz 1 −z2 + z rn(t) (1 −z2)2, (2.9) where Rn(t):= 2t
Lemma 2.1. We have An(z) = 2n + 1 + 2α 1 −z2 + Rn(t) (1 −z2)2, (2.8) Bn(z) = nz 1 −z2 + z rn(t) (1 −z2)2, (2.9) where Rn(t) := 2t
Theorem 2.2.
Theorem 2.2. The recurrence coefficient βn satisfies the following second-order nonlinear difference equation: n 68(n + α)2 −9 β3 n + 12…
Theorem 2.2. The recurrence coefficient βn satisfies the following second-order nonlinear difference equation: n 68(n + α)2 −9 β3 n + 12 −80(n + α)2 + (14n + 5 + 14α)˜βn−1 + (14n −5 + 14α)˜βn+1 β2 n + 24(n + α)2 + 4t(t −2α) −3 −2(2n + 1 + 2α)˜βn−1 −2(2n −1 + 2α)˜βn+1 + ˜βn−1 ˜βn+1
Theorem 2.3.
Theorem 2.3. The sub-leading coefficient of the monic orthogonal polynomials, p(n):= p(n, t), 10
Theorem 2.3. The sub-leading coefficient of the monic orthogonal polynomials, p(n) := p(n, t), 10
Theorem 2.4.
Theorem 2.4. The monic orthogonal polynomials Pn(z), n = 0, 1, 2,..., satisfy the following second-order differential equation: P ′′ n(z) −…
Theorem 2.4. The monic orthogonal polynomials Pn(z), n = 0, 1, 2, . . ., satisfy the following second-order differential equation: P ′′ n(z) − v′(z) + A′ n(z) An(z) P ′ n(z) +
Lemma 3.1.
Lemma 3.1. The auxiliary quantities rn(t) and Rn(t) satisfy the coupled Riccati equations: 2t r′ n(t) = 2nt −r2 n(t) + 2(n + α + 1 −t)rn(t)…
Lemma 3.1. The auxiliary quantities rn(t) and Rn(t) satisfy the coupled Riccati equations: 2t r′ n(t) = 2nt −r2 n(t) + 2(n + α + 1 −t)rn(t) −2(2n + 1 + 2α)(r2 n(t) + 2t rn(t)) Rn(t) , (3.4) 2tR′ n(t) = R2 n(t) + 2(n + α + 1 −t)Rn(t) −2rn(t)(2n + 1 + 2α + Rn(t)) −2t(2n + 1 + 2α). (3.5)
Theorem 3.2.
Theorem 3.2. The auxiliary quantities Rn(t) and rn(t) satisfy the following second-order nonlinear ordinary differential equations: 8t2Rn(2n…
Theorem 3.2. The auxiliary quantities Rn(t) and rn(t) satisfy the following second-order nonlinear ordinary differential equations: 8t2Rn(2n + 1 + 2α + Rn)R′′ n −4t2(4n + 2 + 4α + 3Rn)(R′ n)2 + 8tRn(2n + 1 + 2α + Rn)R′ n − R5 n −2(2n + 1 + 2α)R4 n −4 [(n + α)(n + 1 + α) −t(t −2α)] R3 n + 16t(2n + 1 + 2α)(t −α)R2 n + 4t(2n + 1 + 2α)2(5t −2α)Rn + 8t2(2n + 1 + 2α)3 = 0, (3.6)
Theorem 3.3.
Theorem 3.3. The quantity σn(t) satisfies the second-order nonlinear differential equation n t4(σ′′ n)2 + 2t2 (n + α)2 + t(α −σ′ n) σ′′ n…
Theorem 3.3. The quantity σn(t) satisfies the second-order nonlinear differential equation n t4(σ′′ n)2 + 2t2 (n + α)2 + t(α −σ′ n) σ′′ n + 2t3(σ′ n)3 −t2 (t + α)2 −3n(n + 2α) −1 + 4σn (σ′ n)2 −
Theorem 3.4.
Theorem 3.4. The quantity σn(t) can be expressed in terms of the σ-function of a Painlev´e V as follows: σ2n(t) = 2 ˜Hn t, α, 1 2 + 2…
Theorem 3.4. The quantity σn(t) can be expressed in terms of the σ-function of a Painlev´e V as follows: σ2n(t) = 2 ˜Hn t, α, 1 2 + 2 ˜Hn t, α, −1 2 + 4n(n + α), σ2n+1(t) = 2 ˜Hn
Lemma 4.1.
Lemma 4.1. We have A = t √ 1 −b2 −2n ln b 2 + 2α ln 2 1 + √ 1 −b2. (4.10)
Lemma 4.1. We have A = t √ 1 −b2 −2n ln b 2 + 2α ln 2 1 + √ 1 −b2. (4.10)
Theorem 4.2.
Theorem 4.2. The recurrence coefficient βn(t) has the following large n expansion: βn(t) = 1 4 + ∞ X j=1 aj nj/3, n →∞, (4.20) where the first…
Theorem 4.2. The recurrence coefficient βn(t) has the following large n expansion: βn(t) = 1 4 + ∞ X j=1 aj nj/3, n →∞, (4.20) where the first few terms of expansion coefficients are a1 = 0, a2 = − t2/3 4 × 22/3,
Theorem 4.3.
Theorem 4.3. The sub-leading coefficient p(n, t) has the following expansion as n →∞: p(n, t) = b−3n + b−2n2/3 + b−1n1/3 + b0 + ∞ X j=1 bj…
Theorem 4.3. The sub-leading coefficient p(n, t) has the following expansion as n →∞: p(n, t) = b−3n + b−2n2/3 + b−1n1/3 + b0 + ∞ X j=1 bj nj/3, (4.23) where b−3 = −1 4, b−2 = 0, b−1 = 3 t2/3 4 × 22/3,
Lemma 4.4.
Lemma 4.4. The quantity σn(t) can be expressed in terms of p(n, t) and p(n + 1, t) as follows: σn(t) = −n(n + 2t) −(2n −1 + 2α)p(n, t) −(2n…
Lemma 4.4. The quantity σn(t) can be expressed in terms of p(n, t) and p(n + 1, t) as follows: σn(t) = −n(n + 2t) −(2n −1 + 2α)p(n, t) −(2n + 1 + 2α)p(n + 1, t). (4.24)
Theorem 4.5.
Theorem 4.5. The quantity σn(t) = 2t d dt ln Dn(t) has the following large n asymptotic expansion: σn(t) = −3t2/3n4/3 22/3 − 3√ t(t…
Theorem 4.5. The quantity σn(t) = 2t d dt ln Dn(t) has the following large n asymptotic expansion: σn(t) = −3t2/3n4/3 22/3 − 3√ t(t −2α)n2/3 3√ 2 −24/3t2/3α 3√n + 3t2 + 60tα −24α2 + 4 36 −
Theorem 4.6.
Theorem 4.6. The Hankel determinant Dn(t) has the following expansion as n →∞: ln Dn(t) = −n2 ln 2 −9 t2/3n4/3 4 × 22/3 −˜c1(α)n −3 3√ t (t…
Theorem 4.6. The Hankel determinant Dn(t) has the following expansion as n →∞: ln Dn(t) = −n2 ln 2 −9 t2/3n4/3 4 × 22/3 −˜c1(α)n −3 3√ t (t −8α)n2/3 8 3√ 2 −3t2/3α 3√n 22/3 + (12α2 −5) ln n 36