Synthesis of the Theorems of Hadamard and Hurwitz on Composition of Singularities

568

MA THEMA TICS: W. J. TRJITZINSK Y

PROC. N. A. S.

A SYNTHESIS OF TIHE THEOREMIS OF HADAMARD AND HURWITZ ON COMPOSITION OF SINGULARITIES By W. J. TRJITZINSKY* DEPARTMENT OF MATHEMATICS, HARVARD UNIVERSITY

Communicated September 10, 1931

The two well-known classical theorems on composition of singularities due to Hadamard' and Hurwitz2 one would expect to be interrelated in some way. In fact, Professor Hille has drawn my attention to this possibility in connection with my earlier paper on composition of singularities.3 The purpose of this paper is to prove the following theorem which contains, as special cases, the two theorems referred to above. THEOREM. Let the function q(x, u) = (au + bx + c)/(a'u + b'x + c') depend both on x and u. Assume the following inequalities

a, I

Ialroj< c I

Ic

|b Ibi> I > r2r2I(l b'b'j± aIrOro ++ c )+ |a|r

I

(ro > ri;

r >

0)

Let p be zero or be a negative integer. Let the series

E

an/x'+l, with

an isolated set of singularities ai ai blx I < ri (* 0)], with an isolated set of singularities (3 [ |i | < r2 (* 0)1 represent the uniform functions f(x) and g(x), respectively. Moreover, let C denote a circle with center at the origin and with a radius R(ri < R < ro). It will follow then that the function

F(x)

=

2- ,f f(u)g(g(x, u)uPdu,

(2)

is uniform and is representable, for x > r, by the series co

E cn/x +

nz=-1

(2a)'

.

The singularities of F(x) will be all included among the expressions

b';- baxr + b'#i - (c'i - a'c) (b'a - a'b)

b'

(3) 3

(3a

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- MATHEMATICS: W. J. TRJITZINSKY

569

C if p . -i).

(3b)

I I

IThere will be no singularitiesfor x > r. Furthermore, when c'b -b'c the expressions (3a), (3b) may be deleted. lIadamard's theorem will result when

=

0

b = a' = -p = 1, a =b' = c = c' =O,

while IIurwitz's theorem will result when

ao=bo= p=c=a'=b'=0, -a=b=c'=1.

I I

Proof.-Let u be within a Laurent ring L(ro) (O < ri < u < ro) and let x be in a region Ro( x > r > 0). Then, writing u' = q(x, u), we shall have

+ c)/b Iu'I=i(b+au lu | = |(b)/(b' + that b > well as the inequality

provided

(| a

b Ic I )/r a'u + c') | > b'i aIa ro+ + (a' I ro + l c'l)/r ro + c |)/r. Now the latter inequality, as +

u' > r2

(u in L(ro);

x

in Ro),

will be seen to hold in virtue of (1). Consider now F(x), as given by (2); C is obviously in L(ro). The series representing f(u) and g(u') are uniformly convergent in u for x in Ro, and it can be shown that F(x) is analytic in Ro. Hence F(x) will be represented by a series (2a). However, F(x) is analytic in the more extensive region Rc, defined by the condition that none of the u-points given by u

= ai,

q(x, u) =

k, U

=

0

(ifp

< -1)

(A)

shall lie within a distance 5( > 0) from C; here C is a contour not necessarily in L(ro). This is due to the fact that for u on C the integrand is analytic in x, for x in Rc, and that this integrand is a continuous function of x and u when x is in Rc and u is on C. If C can be deformed into C1 in a continuous way and without passing over any of the u-points of the set (A) we apparently have

Jfc f(u)g(q(x, u))uldu = fc. f(u)g(q(x, u))u"du. F(x) exists in Rc. This region has with Ro a neighborhood of x = o in common. As x approaches the boundary of Rc some of the singular points in the u-plane may approach the boundary of C. We replace C

570

570CHEMISTRY: BANCROFT AND RUTZLER, JR. PROC. N. A. S.

by Ci as specified above. The integral along C1 is analytic for x in Rc, that is, in a region including a neighborhood of x = o. Hence this integral is an analytic continuation of the series (2a). This process of continuation may fail only when points on opposite sides of C tend to coincide. Thus F(x) is analytic (uniform) in every region such that for no point of the region do any two points of the set (A) coincide. The possible singularities of F(x) are then given by (3), (3a), (3b). Consider the situation when c'b - b'c = 0. Each of the expressions in (3a), (3b) will give x = -c/b. For this value of x we have q(x, u) = a/ai. If now a/a1 $ ,Bi (for all i) we can see directly from (2) that x = - c/b is not a singularity of F(x). If, however, a/ai = #j3 (for some i) the value of x given by (3) (for this value of i) will be (-'c/b). This proves the assertion made in the Theorem to the effect that when c'b - b'c = 0 the expressions (3a), (3b) may be deleted. It is a matter of direct verification to show that Hadamard's and Hurwitz's theorems come out as special cases as stated in the Theorem. It is essential to note that in these special cases the coefficients in (2a) will be desired functions of the an and the bn. Hadamard's theorem will be seen to come out in a formulation different from, though equivalent to, the usual formulation. * NATIONAL RESEARCH FELLOW. 1 Hadamard, Acta Mathematica, 22, p. 55 (1898). 2 Hurwitz, Compt. Rend., t. 128, p. 350 (1899). Trjitzinsky, Trans. Am. Math. Soc., 32, 196-215 (1930).

RE VERSIBLE COAGULATION IN LIVING TISSUE. By WILDER D. BANCROFT AND J. E. RUTZLER, JR.** BAKER CHEMICAL LABORATORY, CORNELL UNIVERSITY

VI*

Communicated August 28, 1931

From experiments on the antagonistic action of sodium rhodanate to histamine with rabbits, Bancroft and Rutzlerl deduced that histamine caused reversible coagulation of the proteins of the sympathetic nervous system. It is now proposed to show that histamine acts as a coagulating agent both on the proteins of the sympathetic nervous system and on those of the central nervous system. Dale and Laidlaw2 report that small doses of histamine produce narcosis in frogs, rabbits and cats, which means that in low concentrations histamine acts more strongly on the central nervous system than on the sympathetic nervous system. With the cat under the influence of a regular