# Uniform asymptotic expansions for hyper geometric functions with large

Document Sample

```					        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 ’

(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 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 ’

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 .

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

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 8 posted: 8/17/2010 language: English pages: 23