Solution to Problem 88-11: A generalized hypergeometrictype identity
Lossers, O.P.
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SIAM Review
DOI:
10.1137/1031101
Published: 01/01/1989
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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. DOI: 10.1137/1031101
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Download date: 14. Jun. 2017
496
PROBLEMS AND SOLUTIONS
SOLUTIONS
A Generalized Hypergeometric-Type Identity
Problem 88-11, by C. C. GnosJEAN (State University of Ghent, Belgium).
Let the rational function f(a, b, c, d; k) be defined as follows:
f(a,b,c,d;k)=my=o 2k+-m
=o (a+c-Cm+
2k+l
(-2k- )m+r (2a),,(b)m (2c)n(d)n
m!(2b)m n!(2d),
with the usual notation
(Z)o 1, (z).=z(z+ 1)... (z +j- 1) VzeC, Vjo,
whereby k e J and a, b, c, and d represent arbitrary real or complex numbers, except
that b and d t {0,--5,-1, ...,-k} as well as a + c t {0, +1, +2,
+k}, solely in
in the
indeterminacies
to
denominators
zero
leading
of
appearance
order to avoid the
above-mentioned definition.
Prove that f(a, b, c, d; k) is identically equal to zero.
This problem arose from some work on generalized hypergeometric functions
which I was led to from a solution of a problem in electrostatics.
...,
Solution by O. P. LossEns (Eindhoven University of Technology, Eindhoven, the
Netherlands).
We define
(bh-); (C/)
Fh.k.t{(ah);
(dr); (e,.); (f)
[(a)]p+q[(b)]p[(C)]xPyq
r.s.,
p=Oq=O
where (a) represents the array of symbols a, a2,
product (al )p+q
(ah )p+q. Then
..., ah, and [(ah)]p+q means the
1; b, 2a; d, 2c
f (a, b, c, d; k) F] ’2’2{-2k1,1\ a+c-k;2b;2d
Choosing in formula (23) of[l]
b,
,
2a, k2-- d
"y2=2C,
we obtain
-(’r+,2-k)=
(2a+2e-2h)=a+c-k, 2)=2b, 2X2=2d.
We find that f(a, b, c, d; k)
O.
REFERENCE
1] C. C. GROSJEAN AND R. K. SHARMA, Transformation formulae for hypergeometric series in two
variables II, Simon Stevin, 62 (1988), pp. 97-125.
PROBLEMS AND SOLUTIONS
497
Also solved by ELIZABETH A. MILLER and H. M. SRIVASTAVA (University of Victoria,
British Columbia, Canada) and by the proposer.
A Laplace Transform
Problem 88-13, by M. L. GLASSER (Clarkson University, Potsdam, New York).
The Laplace transform
p(p)
1-1 -tz/4" dt
which is needed extensively in field theory and statistical mechanics, is tabulated in
[1 ], for example. However, this (and the value listed elsewhere) is only valid for real
p _-> 0. Show that for p real 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 of the result for p > 0, it is possible that
erroneous results may have appeared due to the uncritical use of the tabulated
formula.
REFERENCE
[1] A. ERDELYI, ED., Tables of Integral Transforms, Vol. 1, McGraw-Hill, New York, 1954, p. 146.
Solution by NORBERT ORTNER and PETER WAGNER (University of Innsbruck).
The Laplace transform of the function f(t)= t-1/:e -’2/4" is given 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 absolutely convergent integral (p) ff(t)e-P’dt is
a holomorphic function of p in the whole complex plane. On the other hand, the
function ,I,(p) can be analytically continued to the Riemannian surface S of log p
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 S which lies above z on the next leaf. Since (e’z) /= -z /, we can
infer that (p) is actually an analytic function in C\0 and, by continuity, even in C.
Using the uniqueness of the analytic continuation we conclude that ,I, and ,I must
coincide everywhere, i.e.,
-’-/4"
ot-/Ze-P’e
dt= x/a IPl exp [(i arg p+ ap:)/2]K/4(a IPl2e:iargP/2)
for all complex p different from 0. Finally, we want to point out that, in contrast to
the assertion of the proposer, the value of (p) for negative real p can be deduced by
the method of analytic continuation. Indeed, applying again [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,
we obtain
K/a(e2itxp2/2) -igl/4(op2/2)- xTriI1/a(Op2/2)