Plancherel–Rotach asymptotics

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The Plancherel–Rotach asymptotics are asymptotic results for orthogonal polynomials. They are named after the Swiss mathematicians Michel Plancherel and his PhD student Walter Rotach, who first derived the asymptotics for the Hermite polynomial and Laguerre polynomial. Nowadays asymptotic expansions of this kind for orthogonal polynomials are referred to as Plancherel–Rotach asymptotics or of Plancherel–Rotach type.[1]

The case for the associated Laguerre polynomial was derived by the Swiss mathematician Egon Möcklin, another PhD student of Plancherel and George Pólya at ETH Zurich.[2]

Hermite polynomials

Let Hn(x) denote the n-th Hermite polynomial. Let ϵ and ω be positive and fixed, then

  • for x=(2n+1)1/2cos⁡φ and ϵ≤φ≤π−ϵ
e−x2/2Hn(x)=2n/2+1/4(n!)1/2(πn)−1/4(sin⁡φ)−1/2{sin⁡[(n2+14)(sin⁡2φ−2φ)+3π4]+𝒪(n−1)}
  • for x=(2n+1)1/2cosh⁡φ and ϵ≤φ≤ω
e−x2/2Hn(x)=2n/2−3/4(n!)1/2(πn)−1/4(sinh⁡φ)−1/2exp⁡[(n2+14)(2φ−sinh⁡2φ)]{1+𝒪(n−1)}
  • for x=(2n+1)1/2−2−1/23−1/3n−1/6t and t complex and bounded
e−x2/2Hn(x)=31/3π−3/42n/2+1/4(n!)1/2n−1/12{A(t)+𝒪(n−2/3)}

where A(t)=3−1/3πAi⁡(−3−1/3t) and Ai denotes the Airy function.[3]

Probabalist's Hermite polynomials

Let Hen(x) denote the n-th probabalist's Hermite polynomial. These are related to the (physicist's) Hermite polynomials Hn(x) by the relations Hn(x)=2n2Hen(2x),Hen(x)=2−n2Hn(x2).

Three approximations for the probabilists Hermite polynomials illustrated when n=25

The Plancherel–Rotach asymptotics for Hen(x) can be written as[4]:

  • [Small x regime] For fixed α∈(0,π/2), set x=2cos⁡(α)n and then as n→∞ one has

Hen(x)=2nnsin⁡(α)en(cos2α−12)cos⁡(n(sin⁡αcos⁡α−α)−12(α+π2))(1+o(1))

  • [Large x regime] For fixed γ>0, set x=2cosh⁡(γ)n. Then as n→∞ one has Hen(x)=nn/2en(γ+12e−2γ)eγ/22sinh⁡(γ)(1+o(1)).
  • [Edge regime] For fixed u∈ℝ, set x=2n+n−1/6u. Then as n→∞ one has

Hen(x)=2πen/2nn/2n1/6eun1/3Ai⁡(u)(1+o(1)).

(Associated) Laguerre polynomials

Let Ln(α)(x) denote the n-th associate Laguerre polynomial. Let α be arbitrary and real, ϵ and ω be positive and fixed, then

  • for x=(4n+2α+2)cos2φ and ϵ≤φ≤π2−ϵn−1/2
e−x/2Ln(α)(x)=(−1)n(πsin⁡φ)−1/2x−α/2−1/4nα/2−1/4{sin⁡[(n+α+12)(sin⁡2φ−2φ)+3π/4]+(nx)−1/2𝒪(1)}
  • for x=(4n+2α+2)cosh2φ and ϵ≤φ≤ω
e−x/2Ln(α)(x)=12(−1)n(πsinh⁡φ)−1/2x−α/2−1/4nα/2−1/4exp⁡[(n+α+12)(2φ−sinh⁡2φ)]{1+𝒪(n−1)}
  • for x=4n+2α+2−2(2n/3)1/3t and t complex and bounded
e−x/2Ln(α)(x)=(−1)nπ−12−α−1/331/3n−1/3{A(t)+𝒪(n−2/3)}

where A(t)=πAi⁡(−3−1/3t) and Ai denotes the Airy function.[3]

Literature

References