Uniform asymptotic expansions for hyper geometric functions with large

Document Sample
Uniform asymptotic expansions for hyper geometric functions with large Powered By Docstoc
					        Uniform asymptotic expansions for
            hypergeometric functions
                 with large parameters


                       A. B. Olde Daalhuis

                       University of Edinburgh


Abstract. We will present asymptotic expansions for the Gauss hy-
                              a + e1 λ, b + e2 λ
pergeometric function 2 F1                       ; z , as |λ| → ∞, where
                                  c + e3 λ
ej = 0, ±1. The expansions hold for fixed values of a, b, c, and are
uniformly valid for z in large domains.




                                   1
                                 1. Hypergeometric functions

The Gauss function
                                ∞
                  a, b                (a)s (b)s s   ab  a(a + 1)b(b + 1) 2
        2 F1           ;z   =                  z =1+ z+                 z + ···
                   c            s=0
                                       (c)s s!      c      c(c + 1)2!

for |z| < 1 and c = 0, −1, −2, · · ·.
Linear transformations

       a, b                             a, c − b    z                           c − a, c − b
2 F1        ;z     = (1−z)−a 2 F1                ;       = (1−z)c−a−b 2 F1                   ;z .
        c                                   c      z−1                               c

Contiguous Relations

                 a − 1, b                                a, b                    a + 1, b
(c−a)2 F1                 ; z + 2a−c+(b−a)z 2 F1              ; z +a(z−1)2 F1             ;z   = 0,
                    c                                     c                         c

                         a, b                                       a, b                               a, b
c(c−1)(z−1)2 F1               ; z +c c−1−(2c−a−b−1)z 2 F1                ; z +(c−a)(c−b)z 2 F1              ;z   = 0.
                        c−1                                          c                                c+1




                                                  2
             2. Kummer’s 24 solutions and connection formulas
The hypergeometric differential eq.: z(1 − z)w + c − (a + b + 1)z w − abw = 0.

                a, b                                c − a, c − b                           a, c − b    z
w1 (z) = 2 F1        ;z   = (1 − z)c−a−b 2 F1                    ;z   = (1 − z)−a 2 F1              ;       = ···
                 c                                       c                                     c      z−1


                      a − c + 1, b − c + 1                                      a − c + 1, 1 − b    z
w2 (z) = z 1−c 2 F1                        ;z       = z 1−c (1 − z)c−a−1 2 F1                    ;          = ···
                             2−c                                                     2−c           z−1


                   a, b                                   a, a − c + 1      1
w3 (z) = 2 F1           ;1 − z             = z −a 2 F1                 ;1 −       = ···
                a+b−c+1                                   a+b−c+1           z


                                c − a, c − b                                           1 − a, c − a      1
w4 (z) = (1 − z)c−a−b 2 F1                   ;1 − z       = z a−c (1 − z)c−a−b 2 F1                 ;1 −       = ···
                               c−a−b+1                                                c−a−b+1            z


                          a, a − c + 1 1                           a, c − b   1
w5 (z) = (−z)−a 2 F1                  ;      = (1 − z)−a 2 F1               ;            = ···
                           a−b+1 z                                a−b+1 1−z


                          b, b − c + 1 1                          b, c − a   1
w6 (z) = (−z)−b 2 F1                  ;     = (1 − z)−b 2 F1               ;          = ···.
                           b−a+1 z                               b−a+1 1−z
                                                3
Connection Formulae
               Γ(1 − c)Γ(a + b − c + 1)          Γ(c − 1)Γ(a + b − c + 1)
    w3 (z) =                            w1 (z) +                          w2 (z).
               Γ(a − c + 1)Γ(b − c + 1)                  Γ(a)Γ(b)


               Γ(1 − c)Γ(c − a − b + 1)          Γ(c − 1)Γ(c − a − b + 1)
    w4 (z) =                            w1 (z) +                          w2 (z).
                   Γ(1 − a)Γ(1 − b)                  Γ(c − a)Γ(c − b)

               Γ(1 − c)Γ(a − b + 1)                   Γ(c − 1)Γ(a − b + 1)
    w5 (z) =                        w1 (z) + e(c−1)πi                      w2 (z).
               Γ(a − c + 1)Γ(1 − b)                       Γ(a)Γ(c − b)


               Γ(1 − c)Γ(b − a + 1)                   Γ(c − 1)Γ(b − a + 1)
    w6 (z) =                        w1 (z) + e(c−1)πi                      w2 (z).
               Γ(b − c + 1)Γ(1 − a)                       Γ(b)Γ(c − a)


               Γ(c)Γ(c − a − b)          Γ(c)Γ(a + b − c)
    w1 (z) =                    w3 (z) +                  w4 (z).
               Γ(c − a)Γ(c − b)              Γ(a)Γ(b)


               Γ(2 − c)Γ(c − a − b)            Γ(2 − c)Γ(a + b − c)
    w2 (z) =                        w3 (z) +                          w4 (z).
                Γ(1 − a)Γ(1 − b)             Γ(a − c + 1)Γ(b − c + 1)
         .
         .
         .

                                           4
                              3. From 27 cases to 4


                   a + e1 λ, b + e2 λ
            2 F1                      ;z ,       where    ej = 0, ±1.
                       c + e3 λ
Watson noted in 1918 that the 27 possible cases can be reduced to the 5 cases, and
these can be reduced to 4 cases.

  e1       e2        e3
   0       0        +1                                            e1       e2        e3
  +1      −1         0                                             0       0         +1
   0      −1        +1                                            +1      −1          0
  +1      +1        −1                                             0      −1         +1
  −1      −1        +1           Final two cases reduce to →      +1      +2          0




                                             5
                                            4. Case I

Wagner (1988) showed that: When a, b, z ∈ C are fixed such that one of the following
cases holds
  (i) a or b ∈ {0, −1, −2, · · ·},
 (ii) z < 1 and |c + n| δ > 0 for all n ∈ {0, 1, 2, · · ·},
             2
             1
(iii) z = 2 and |ph c| π − δ < π,
             1                                                           ph (1−z)−ph z+π
(iv)    z>   2
                         1
                 and −β− 2 π+δ ph c −α+ 1 π−δ, where α = arctan
                                        2                                            1
                                                                              ln |1− z |
                                                                                           ,
                           ph (1−z)−ph z−π                                          1
       and β = arctan                  1
                                ln |1− z |
                                             , where z is restricted so that α ∈ (− 2 π, 0]
       and β ∈ [0, 1 π),
                   2
then
                                            m−1
                              a, b                (a)s (b)s s
                       2 F1        ;z   ∼                  z + O c−m
                               c            s=0
                                                   (c)s s!

as |c| → ∞.




                                                  6
For fixed a, b, c and z
                                                       ∞
                        a, b           Γ(c + λ)
                2 F1         ;z    ∼                        qs (z) (b)s λ−s−b ,            (∗)
                       c+λ           Γ(c − b + λ)     s=0


as |λ| → ∞, where q0 (z) = 1 and
                                                                  ∞
                  et − 1   b−1
                                                      −t −a
                                 et(1−c) 1 − z + ze           =         qs (z)ts .
                     t                                            s=0

                                                            1          1                    1
When |ph (1 − z)| < π then (∗) holds for |ph λ|             2π   − δ < 2 π, and when   z    2
then (∗) holds for |ph λ| π − δ < π.




                                             7
                                     5. Case II

For fixed a, b, c ∈ C and |ph (z − 1)| < π

                                                            (c−a−b−1)/2                           1−c
       a + λ, b − λ 1 1                           (z + 1)                              a−b
2 F1               ; 2 − 2z   ∼ 2(a+b−1)/2 Γ(c)                 c/2
                                                                          ζ sinh ζ λ +
            c                                         (z − 1)                           2
                          a−b          Ic−2 (λ + a−b )ζ
                                                  2
          × Ic−1     (λ +     )ζ     +                          (c − 1 )(c − 3 )(1/ζ − coth(ζ))
                                                                     2       2
                           2              2λ + a − b
                              + 1 (2c − a − b − 1)(a + b − 1) tanh(ζ/2)
                                2                                             ,

as λ → ∞, in |ph λ| π − δ < π, where ζ = arccosh(z). For the corresponding
asymptotic expansion and error bounds see Jones (2001)




                                            8
                                          6. Case III

For fixed a, b, c ∈ C and |ph z| < π

                                                λ                                                                √
       a, b − λ                       z+1                                       α        1−a
2 F1            ; −z     ∼ 2b−c+1/2    √             (1 + z)c−a−b z 1−c                        λa/2 U (a − 1 , −α λ)
                                                                                                           2
        c+λ                           2 z                                      z−1
                                          1−a             1
                                                                    a
                        c−a−b 1−c    α               c−b− 2    α
              (1 + z)        z      z−1         −2            z−1                                       √
                                                                            (a−1)/2            3
          +                                                             λ             U (a −   2 , −α       λ) ,
                                      α

                                            1                               z−1 2
as λ → ∞, in |ph λ|        π − δ < π, where 2 α2 = − ln 1 −                 z+1       , such that

                                                       z−1
                                    α > 0 ⇐⇒               > 0.
                                                       z+1

U (a, z) is the parabolic cylinder function.




                                                 9
                                           7. Case IV

For fixed a, b, c ∈ C and |ph z| < π

       a + λ, b + 2λ                                              1−c
2 F1                 ; −z      ∼ Γ(c)(z + 1)−3λ/2 (2λ)
             c
                                           1                      2/3                      1                  2/3
             ×    a0 λ−1/3 eπi(a−c+λ+ 3 ) Ai             e−πi λ         χ + e−πi(a−c+λ+ 3 ) Ai        eπi λ         χ

                                               2                      2/3                         2                 2/3
                  +a1 λ−2/3 eπi(a−c+λ+ 3 ) Ai               e−πi λ          χ + e−πi(a−c+λ+ 3 ) Ai       eπi λ            χ   ,


                          1          1
as λ → ∞, in |ph λ|       2π   − δ < 2 π, where ζ = arccosh(z/4 − 1),

 4 3/2                      2 + eζ
 3χ        = −2ζ + 3 ln                        such that                z > 8 ⇐⇒ ζ > 0 ⇐⇒ χ > 0,
                           2 + e−ζ

                                                                                        1/4
                    √         ±(a− 1 c)ζ            ±ζ c−a−b         1     −c/2    zχ
                 g(± χ) = e        2       2+e                         z                      ,
                                                                     2            z−8
       1     √        √                            1       √        √
a0 =   2   g( χ) + g(− χ) , and a1 =               √
                                                   2 χ   g( χ) − g(− χ) .



                                                       10
        0.2

         0.1
               2             4               6             8         10
           0
                                                                 z
       –0.1

       –0.2

       –0.3

       –0.4


                                          a + λ, b + 2λ
               (z + 1)3λ/2 λc−2/3 2 F1                  ; −z ,
                                                c
with λ = 20.

                                     11
                                     8. Case III, ‘proof ’

We start with the integral representation

                                                                   (0+) b−1−λ
         a, b − λ              (λ−b)πi Γ(c + λ)Γ(1 − b + λ)             t           (1 + t)a−c−λ
 2 F1             ; −z    =e                                                                     dt,
          c+λ                             Γ(c − b + 2λ)2πi        ∞             (t + zt + 1)a

where     (c − b + 2λ) > 0, b − λ = 1, 2, 3, · · ·, and |ph (1 + z)| < π.
                                    z−1                                         1
       For the moment take          z+1      < 0, |λ| is large with |ph λ|      2 π−δ   and t = τ − 1 .
                                                                                                    2


                                                                   i∞       1        b−1   1        a−c       1          −λ
        a, b − λ           (1 + z)−a Γ(c + λ)Γ(1 − b + λ)                   2   −τ         2   +τ             4   − τ2
2 F1             ; −z    =                                                                                a                   dτ.
         c+λ                      Γ(c − b + 2λ)2πi                −i∞                      τ−    z−1
                                                                                                2(z+1)


The integrand has a saddle point at τ = 0. The branch points at τ = ± 1 are at
                                                                      2
                                                                         z−1
fixed positions and not important. However, the branch point at τ = τc ≡ 2(z+1)
can coalesce with the saddle point.




                                                   12
The simplest integral with a saddle point near a branch point is
           i∞
                 1   2           −a
                                            √    1         1   2            √
                e 2 λu (u − α)        du = i 2πλ 2 (a−1) e 4 λα U a − 1 , −α λ .
                                                                      2
          −i∞

Hence, we use the transformation

                                      1 2
                                      2u    = − ln 1 − 4τ 2 ,

where we choose the branch of the logarithm so that the τ -imaginary axis is mapped
onto the u-imaginary axis.




                                                 13
                       A                     τ               A
                           B
       u
                               C



                                                                        C
                                                   1
                                                  −2         B



0
Figure 1. The three lines (‘steepest descents’ paths) in the u-plane (left) are
given by ph A = π + ε, ph B = π , and ph C = π − ε, and are mapped onto the
                  4              4               4
curves in the τ -plane (right). In the τ -plane point A spirals to infinity, point
             1
C spirals to 2 , and B moves on a periodic orbit through the origin.




                                       14
                                                                 i∞
        a, b − λ          (1 + z)−a Γ(c + λ)Γ(1 − b + λ)4λ                  1    2             −a
                 ; −z                                                       2 λu     (u − α)
2 F1                    =                                               e                           G0 (u) du,
         c+λ                      Γ(c − b + 2λ)2πi              −i∞

where
                                    1 2
                                    2u    = − ln 1 − 4τ 2 ,
                                     b−1            a−c                      a
                             1               1               u−α                 dτ
                 G0 (u) =      −τ              +τ               z−1                 ,
                             2               2             τ − 2(z+1)            du

                                     dτ   (1 − 4τ 2 )u
                                        =              ,
                                     du       8τ
and
                                                   (z + 1)2
                                    α=−       2 ln
                                                      4z
is the u-image of τc . Since, for the moment, τc < 0, we want α < 0. This explains
the minus sign in front of the square root.
     The function G0 (u) has singularities in the u-plane at the points
     √              √
  ±2 kπe   πi/4
                , ±2 kπe−πi/4 k = 1, 2, 3, · · · , which are mapped onto τ = 0, and
                           √
at half of the points ± α2 + 4kπi k = ±1, ±2, ±3, · · · , which are mapped onto
τ = τc .

                                               15
                           9. The uniform asymptotic expansion

To obtain the uniform asymptotic expansion we use Bleistein’s method, that is, we
substitute
                   Gn (u) = γ0,n + γ1,n (u − α) + u(u − α)Hn (u)               (∗)
and use one integration by parts. We obtain
  i∞
        1   2           −a
       e 2 λu (u − α)        G0 (u) du
 −i∞
            √     1    2             1                          √                1                         √
                  4 λα               2 (a−1)   U a−    1                         2 (a−2)          3
        = i 2πe             γ0,0 λ                     2 , −α       λ + γ1,0 λ             U a−   2 , −α       λ
                     i∞
              1              1   2             −a
            +              e 2 λu (u − α)           G1 (u) du,
              λ    −i∞

where
                                  d
  Gn+1 (u) = −(u − α)a              (u − α)1−a Hn (u) = (a − 1)Hn (u) − (u − α)Hn (u).
                                 du
The coefficients in (∗) are given by

                                                                      Gn (α) − Gn (0)
                    γ0,n = Gn (α)               and         γ1,n =                    .
                                                                             α


                                                       16
We can repeat this process and obtain
  i∞
        1   2               −a
       e 2 λu (u − α)            G0 (u) du
 −i∞
                                                                √       n−1                               √      n−1
            √      1    2            1                                        γ0,s    1                                γ1,s
                   4 λα              2 (a−1)   U a−   1                               2 (a−2) U     3
       = i 2πe                   λ                    2 , −α        λ
                                                                               λs
                                                                                   +λ           a − 2 , −α λ
                                                                                                                        λs
                                                                        s=0                                      s=0
                            i∞
              1                      1   2           −a
            + n                  e 2 λu (u − α)           Gn (u) du.
             λ          −i∞

We also give an integral representation for the coefficients:

                                                                          1        b−1   1        a−c
                  n!                               (u − α)a−1 u1−j        2   −τ         2   +τ
     γj,n =             2                                                        n+1                    du dτ,
                (2πi)                                 1 2
                             {0,τc }     {u(τ )}
                                                      2u    + ln(1 −      4τ 2 )       (τ − τc   )a

where the τ -contour of integration is a simple loop that encircles 0 and τc once in
the positive direction. The u-contour of integration is a small loop that encircles
the point u(τ ).




                                                           17
                                    10. Case IV, ‘proof ’

We start with the integral representation

                                                                                1                        a−c+λ
        a + λ, b + 2λ               (a−c+λ)πi Γ(c)Γ(1  − c + b + 2λ)                 tc−b−1−2λ (t − 1)
2 F1                  ; −z     =e                                                                                dt
              c                                    Γ(b + 2λ)2πi                −i∞       (z + 1 − t)a+λ
                                                                                         i∞ c−b−1−2λ        a−c+λ
                                          (c−a−λ)πi Γ(c)Γ(1 − c + b + 2λ)                  t          (t − 1)
                                    +e                                                                                dt.
                                                        Γ(b + 2λ)2πi                 1          (z + 1 − t)a+λ

For the moment we will assume that z > 8 and λ is large and positive, and continue
with the asymptotics of the first integral:

                                                           1                               a−c+λ
                      (a−c+λ)πi Γ(c)Γ(1 − c + b + 2λ)            tc−b−1−2λ (t − 1)
       F (λ, z) = e                                                                                dt
                                    Γ(b + 2λ)2πi          −i∞        (z + 1 − t)a+λ
                                                           1  c−b−1                  a−c
                      (a−c+λ)πi Γ(c)Γ(1 − c + b + 2λ)            t           (t − 1)
               =e                                                                          eλf (t) dt,
                                    Γ(b + 2λ)2πi          −i∞            (z + 1 − t)a

where
                                                     t−1
                                  f (t) = ln                         .
                                                t2 (z + 1 − t)




                                                 18
                                                          1                  a−c
                   (a−c+λ)πi Γ(c)Γ(1− c + b + 2λ)              tc−b−1 (t − 1)
    F (λ, z) = e                                                                   eλf (t) dt,
                                Γ(b + 2λ)2πi             −i∞      (z + 1 − t)a

where
                                                 t−1
                               f (t) = ln                       .
                                            t2 (z + 1 − t)
Differentiating this function we obtain
                         1
                2 t2 − 2 z + 2 t + 1 + z   2 t − 2 − eζ t − 2 − e−ζ
        f (t) =                          =                          ,
                    t(t − 1)(z + 1 − t)        t(t − 1)(z + 1 − t)

where
                                                  1
                                  ζ = arccosh     4z   −1 .
Hence, the saddle points are

                         sp+ = 2 + eζ ,     and        sp− = 2 + e−ζ .

These saddle points coalesce when ζ = 0, that is, when z = 8.




                                             19
To obtain an uniform asymptotic expansion we use the Chester, Friedman and Ursell
ansatz, that is the transformation

                              f (t) = 1 u3 − xu + γ.
                                      3

The saddle points of the left-hand side of this transformation should correspond to
                                             √
those of the right-hand side, that are u = ± x. Thus

                                            2 2/3
                              f (sp± ) =    3x      + γ.

We obtain
                  γ = − 3 ln (2 + eζ )(2 + e−ζ ) = − 2 ln(z + 1),
                        2
                                                     3


and
                         4 3/2                   2 + eζ
                         3x      = −2ζ + 3 ln              .
                                                2 + e−ζ




                                           20
 D = eπi ∞          C = −1 B = 0                    A=8      E = +∞
D = e−πi ∞             C B



                                                2
    B = πi      C = πi + ln(2)            C = e 3 πi ∞
                             D



         A                    E = +∞                            D
                                           B        A           E = +∞
                                           B                    D


        B
                C             D
                                      C
            Figure 2. The z-plane |ph (z−8)| < π (top) is mapped
            onto the semi-strip ζ > 0, | ζ| < π (bottom left),
            which is mapped onto the shaded region (bottom right)
            in the x-plane.

                                     21
We obtain integral representation
                                                                             −∞
               (a−c+λ)πi             −3λ   Γ(c)Γ(1 − c + b + 2λ)                         1      3
                                                                                                    −xu)λ
F (λ, z) = e               (z + 1)    2
                                                                                       e( 3 u               G0 (u) du,
                                               Γ(b + 2λ)2πi                 e−πi/3 ∞

where
                                                                    dt
               G0 (u) = tc−b−1 (t − 1)a−c (z + 1 − t)−a
                                                                    du
                           1 c−b             a−c+1                 1−a         u2 − x
                     =     2t    (t   − 1)           (z + 1 − t)                              .
                                                                         (t − sp− )(t − sp+ )
The only singularities in the bounded u-plane are points u, such that t, as a function
                                    √
of u is equal to sp± , but not u = ± x. These points are solutions of

        1 3
        3u    − xu + γ = f (sp± ) + 2kπi,                where      k = 1, −1, 2, −2, 3, −3, · · · .




                                                       22
                                11. The uniform asymptotic expansion

To obtain the uniform asymptotic expansion we use Bleistein’s method again, but
omit the details.

            −∞
 1                    1     3
                                −xu)λ
                   e( 3 u               G0 (u) du
2πi     e−πi/3 ∞
                                                            n−1                                                             n−1
                            1                       2                                                               2
                                                                                 −s− 1         2                                               2
                   =e       3 πi   Ai     e−πi λ    3
                                                        x         (−)s αs λ          3   −e    3 πi   Ai   e−πi λ   3
                                                                                                                        x         (−)s βs λ−s− 3
                                                            s=0                                                             s=0
                                                 −∞
                                  (−λ)−n                      1      3
                                                                         −xu)λ
                                +                           e( 3 u               Gn (u) du,
                                    2πi         e−πi/3 ∞

We also give an integral representation for the coefficients:

(−)n n!                                 tc−b−1 (t − 1)a−c (z + 1 − t)−a uj                                   αn ,   when j = 1,
        2                                                                                n+1   du dτ =
(2πi)        {sp± }   {u(t)}            1 3                               t−1                                βn ,   when j = 0,
                                        3u    − xu + γ − ln          t 2 (z+1−t)



where the t-contour of integration is a simple loop that encircles the saddle points
t = sp± once in the positive direction. The u-contour of integration is a small loop
that encircles the point u(t).


                                                               23