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Solution to Problem 88-11: A generalized hypergeometric-type

identity

Citation for published version (APA):

Lossers, O. P. (1989). Solution to Problem 88-11: A generalized hypergeometric-type identity. SIAM Review, 31(3), 496-497. https://doi.org/10.1137/1031101

DOI:

10.1137/1031101

Document status and date: Published: 01/01/1989 Document Version:

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496 PROBLEMS ANDSOLUTIONS

SOLUTIONS

A

Generalized Hypergeometric-Type Identity

Problem 88-11, by

C. C. GnosJEAN (State

Universityof Ghent, Belgium).

Let

the rational function

f(a, b,

c,

d;

k)

be defined asfollows:

2k+l

2k+-m

(-2k- )m+r (2a),,(b)m (2c)n(d)n

f(a,b,c,d;k)=m

y=o

=o

(a+c-Cm+

m!(2b)m

n!(2d),

withthe usual notation

(Z)o

1,

(z).=z(z+

1)... (z

+j-

1)

VzeC,

Vjo,

whereby ke

J

and a,

b,

c, anddrepresent arbitrary real or complex numbers,except thatb and d

t

{0,--5,-1, ...,-k}

aswell as a

+

c

t

{0,

+1, +2,

...,

+k},

solely in order to avoid the appearanceofzerodenominatorsleadingtoindeterminacies in the above-mentioned definition.

Prove

that

f(a,

b,

c,

d; k)

isidenticallyequal to zero.

This problem arose from some work on generalized hypergeometric functions which

I

wasledtofromasolution ofa problem in electrostatics.

Solution by

O. P. LossEns

(Eindhoven University of Technology, Eindhoven, the

Netherlands).

We

define

Fh.k.t{(ah);

(bh-); (C/)

r.s.,

(dr);

(e,.);

(f)

[(a)]p+q[(b)]p[(C)]xPyq

p=Oq=O

where

(a)

represents the array ofsymbols a, a2,

...,

ah, and

[(ah)]p+q

means the

product

(al

)p+q

(ah )p+q.

Then

f

(a, b,

c,

d;

k)

F]

’2’2{-2k-

1;

b,

2a;

d,

2c

1,1\

a+c-k;2b;2d

Choosinginformula

(23) of[l]

b,

,

2a,

k2--

d "y2=2C, weobtain

-(’r+,2-2k)=-

(2a+2e-2h)=a+c-k,

2)=2b,

2X2=2d.

We

findthat

f

(a, b,

c,

d;

k)

O.

REFERENCE

1] C. C. GROSJEAN AND R. K. SHARMA, Transformationformulae forhypergeometric series in two

(3)

PROBLEMSAND SOLUTIONS 497

Also

solved by

ELIZABETH A. MILLER

and

H. M. SRIVASTAVA

(University of Victoria, BritishColumbia,

Canada)

andby theproposer.

A

LaplaceTransform

Problem

88-13,

by

M. L. GLASSER (Clarkson

University, Potsdam,

New York).

TheLaplacetransform

p(p)

1-1

-tz/4"dt

which isneeded extensively in fieldtheory and statisticalmechanics, istabulated in

[1 ],

forexample.

However,

this

(and

the value listed

elsewhere)

isonly valid forreal p_->0. Showthat forpreal and

Re

a

=>

0

(p)=r

2

e"p2/2

(1-sgnp)I1/4(ap2/2)+X/K/4(PZ/2)

Since this is not the analytic continuation ofthe result forp

>

0, it is possible that erroneous results may have appeared due to the uncritical use of the tabulated formula.

REFERENCE

[1] A. ERDELYI,ED.,TablesofIntegralTransforms,Vol. 1, McGraw-Hill,NewYork,1954,p. 146.

Solution by

NORBERT ORTNER

and

PETER

WAGNER (University of

Innsbruck).

The Laplacetransform of the function

f(t)=

t-1/:e-’2/4" isgiven for

Re

a

>

0 in

[1,

p. 43,

5.44]

by

(p):=

x/e"P2/:K,/a(ap:/2),

p>0.

By [2,

p. 125, Thm.

13.8.6]

the absolutelyconvergent integral (p)

ff(t)e-P’dt

is a holomorphic function ofp in the whole complex plane.

On

the other hand, the function ,I,(p) can be analytically continued to the Riemannian surface

S

of logp since the same holds true for the square root and the function

K,(z).

Taking into

account

[3, Form. 8.475.5]

we obtain

K/4((eziz)2)

---K/4(z:),

interpreting

eeiz

as the point on Swhich lies above z onthe next leaf. Since

(e’z)

/= -z

/,

we can

inferthat (p)isactuallyan analytic function inC\0and, by continuity, even in

C.

Using the uniqueness ofthe analytic continuation we conclude that ,I, and ,I must

coincide everywhere, i.e.,

ot-/Ze-P’e

-’-/4"dt=

x/a

IPl

exp [(iarg

p+

ap:)/2]K/4(a

IPl2e:iargP/2)

forall complex p different from 0. Finally,we want to point outthat, in contrast to theassertionofthe proposer, the valueof(p) for negative realp can bededuced by the method of analytic continuation. Indeed, applyingagain

[3, Form. 8.475.5],

i.e.,

K/4(e’"iz)

e-mi/4K/4(z)

x/ri

sin

(mTr/4)I/a(Z),

with z

ap2/2,

p

elPl

< O,

weobtain

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