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On the quotient of two integral functions
JOURNAL
OF MATHEMATICAL
ANALYSIS
AND
On the Quotient
54, 408-418 (1976)
APPLICATIONS
of Two Integral Functions
CHARLES F. OSGOOD AND CHUNG-CHUN YANG Mathematical
Sciences Staff, Naval Research Laboratory Washington, D.C. 20375 Submitted by R. P. Boas
I. INTRODUCTION An exponential polynomial is a function of the form:
where the aj(j = 0, 1, 2 ,..., n) are constants and the ai(i = 0, 1, 2 ,..., n) are distinct constants. Let us denote the set of all exponential polynomials by H. The following result is proved in [4]. THEOREM 1. The quotient of two exponential polynomials, an exponential polynomial.
;f entire, is itself
Remark. Rubel suggested the more general conjecture that any function algebraic over H, if entire, belongs to H. The proof of Theorem 1 has something to do with the distribution of the zeros of an exponential polynomial and the convex hull of the 0~~‘s.Thus, the methods used in [4] by their nature will fail to deal with the situation that some or all of the q’s are replaced by polynomials. In this paper, we shall treat a special case of the above-mentioned situation, As an application of the study, we shall present a result concerning the zero-one set problem (see [5]) for the class of all integral functions of finite order. Let us denote by G the class of all integral functions of the form: a
ebo(z)
0
h(z) +
+
. . . +
anebzi(z),
ale
where the ai(i = 0, 1, 2 ,..., n) are constants, and the b,(z)(j = 0, 1, 2 ,..., n) are polynomials differing from one another by a nonconstant polynomial. A statement analogous to that of Theorem 1 for the class G would be the following: 408 Copyright All rights
Q 1976 by Academic Press, Inc. of reproduction in any form reserved.
QUOTIENT
OF TWO INTEGRAL FUNCTIONS
409
Conjecture. The quotient of two elements of G, if entire, is an element of G. Ozawa pointed out that the conjecture is not always true by exhibiting the quotient: f(z) == (Piz* - l)/(e2niz - 1), which is entire but not a member of G. However, in this paper, we shall prove the following: THEOREM
degree.
2.
Let P(z), Q(z) be two nonconstant pol~~nomials of the same
Jf f(x) = (&-) - I)/(@‘) - 1)
is entire then P(z) =
m@(z)) + 2n7ri,
where m, 12are integers. Remark.
It follows
from this that f~ G.
II. APPLICATION As we know, there are integral functions f such that any other integral function g that has the same zeros and one-points with the same multiplicities as f must be f. For instance, take f = sin z [5]. There are also pairs of distinct integral functions f and g having the same zeros and one-points with the same multiplicities, for instance, f = eZ and 2 = e-“. Kate here that f and 0” have the same rate of growth. It would not be an unreasonable guess that if two integral functions f and g have the same zeros and one-points with the same multiplicities, then T(r, f) y T(r, g), where T(r, h) denotes the Nevanlinna characteristic function for the function h (for the details of the notation see, e.g., [2]). It has been shown by Nevanlinna [3, p. 1251 that if two meromorphic functions fi and f2 having the same zeros and one-points, and poles with the same multiplicities, then fi and f2 can be expressed as
and 1 _ ee&) f2@>
=
em,(Z)
_
ee,(z)
’
where or and v1 are two integral functions. (Here we would like to point out that the form of fi in [3] was incorrect.) There was nothing said about the
410
OSGOODAND YANG
relation between vi and ‘pa. We have not been able to settle this question completely. However, for the class F of all nonconstant integral functions of finite order, we can obtain the following result. Particularly, it will tell us the relation between pr and vz . THEOREM 3. Let f, g E F. Suppose that f andg have the same zeros and onepoints Edith the same multiplicities. Then either f = g, or
f(z) = 1 + ey + ... + etn-ljY = (1 - eny)/(l - eY), and ... +
g(z) = 1 + eeYf
,-tn-lh
=
(1 _
e--"')/(l
_
e-Y),
or more generally f(z)
=
(1
-
e2niP(Y))/(
_
e2aiP(v))l(e2ni(p(v)-v)
1 _
e2niY)
and g(z)
=
(1
_
&hiP(Y)))
where y is a polynomial in z and p(z) is a polynomial in z with rational numbers as the coe#icients. It follows that q2 = 2rrip(1/2G((p2 - w)) in Nevanlinna’s results, which we just mentioned above. Furthermore, we always have T(r, f) v, g>asr+ 00.
III. PROOFS OF THEOREMS 2 AND 3 To prove our theorems, we shall first have to prove a lemma, which is interesting in itself. LEMMA 1. Let P(z) and Q(z) be two nonconstant polynomials. Suppose that wherever Q(z) assumes an integral value, so also does P(z). Then (deg P)/(degQ) = d is an integer and P(z) = L(Q(x)) where L(z) = CIdZd$- cid++l
+ *.. + 010,
and the aj(j = 0, 1, 2 ,..., d) are rationals with C+ # 0. Proof. It is clear that the degree of P must be larger than or equal to that of Q, since otherwise, there would be more points in some finite part of the plane causing Q(z) to be an integer than points causing P(z) to be an integer. We shall see that n = deg Q and m = deg P cannot have their greatest common divisor less than n. The roots of the equations
Q(z) = fN(N
= 1, 2,...)
QUOTIENT
OF TWO
INTEGRAL
FVNCTIONS
411
lie asymptotically near differing radii each of the form pf+y1 = 0, 1, 2 ,...) 2n), where t& = e, + 27r
and
I e, - e,_., ) = 7+z.
All the roots of the equations P(z) = &m(m = 1,2,...) lie asymptotically near differing radii each of the form p+j(j
= 0, I, 2 ,..., 3m),
where km = $0 + 2i-r
and
j #j - $J-~ i = n/m.
It is impossible to have any inclusion relation between these two sets of roots. Hence, we conclude m 3 n and m/n is an integer. Let us now set d = m/n and we shall prove the statement by induction on d. For d = 1 we have the following statement. LEMMA 2. Let P(z), Q(z) be two nonconstant polynomials of the same degree, sa-v deg P = deg Q = k. Suppose that whenever Q(z) assumes an integral value or zero at some point z, , so does P(z). Then P(z) = -4Q(z) + B, where both A and B are integers. Proof. Let n be a given integer and let the wj(n)(j = 1, 2,..., k) be the algebraic functions satisfying Q(m) = n. Let us also assume that
and id
First, we will show that po/qo is a rational number. Let us consider the algebraic function of n q(n) = t
P(wj(n)).
(1)
j=~l
Sow p)(n) is symmetric in the wj(n); hence, it is a polynomial in ~8-14~1,..., n) 4;’ over C (see [6, p. 791). F or each integer n, F(n) is an integer. By the nonvanishing of the determinant 1: : /, say, we see that the coefficients of q(n) are rational numbers. By growth arguments deg, q(n) < 1. For large values of n, we have
qlqkl, (qO -
wj(n) = pjp;“kfll’k 409:54/2-g
f o(l) (n”“)
(2)
412
OSGOODAND YANG
where p is a primitive kth root of unity; hence, P(w(n)) = (p&&f 3
+ o(l)(nl-l’k).
(3)
Substituting this into (I), we have:
&> = k(Pol~o)n+ C? for some rational number c, so p,,/q,, = integers. Next consider B(z) 3 sP(z) -
Y/S
is rational, where
(4) Y
and s are
(5)
YQ(Z).
Now B(z) is a polynomial of degree I < k, and B(z) is an integer whenever Q(z) is an integer. If B(z) is not a constant, then we have a contradiction, since previously we showed that under these hypotheses one must have deg P(z) 3 degQ(z). Then B(z) = Br and (r/s)t + B,/s is an integer for all integers t. Setting t = 0, 1 we see that Y/S and B/s are each integers. Let d = r/s = p&a. Then P(z) = /IQ(z) + B, and we have proved Lemma 2. Now we are going to complete our induction proof of Lemma 1. Assume that for all 1 < dr < d the statement of Lemma 1 holds. First we see from the construction above, for d > 1, that 9(n) is a polynomial of degree d with rational coefficients (use the nonvanishing of the Vandermonde determinant) and that v(n) = k(pO/q,,)nd + .... Thus, ps/q,, = Y/S is rational. Now B(z) zs
- s(&(z))~
rP(z)
(6)
is a polynomial of degree m, < 7n,which is an integer at all points where Q(z) is an integer. Hence, (deg BWl(degQM)
= 4 < 4
and we have, by the induction assumption, B(z) = L,(Q(z)), where L,(z) is a polynomial of degree d, with rational coefficients. Lemma 1 now follows immediately. LEMMA 3.
(See, e.g., [2, p. 71.)
Let P(z)
be a poZynomia1 of degree n
having the form P(z) = i
pizF,
i=O
then T(Y,
ePtr)) = (l/n) 1PO[ P(l + 0(1/r)).
QUOTIENT
OF TWO
INTEGRAL
FI’NCTIONS
413
Proof of Theorem 2. First we rewrite P(z) = 2nip(z) and Q(z) = 27+(z). Since f (z) is an integral function, we conclude that whenever Q(Z) assumes an integral value, so does?(z). By Lemma 1 it follows thatp(z) -= ms(z) + n; hence, P(z) = mQ(z) + 2~2 for some integers n and m. The theorem is thus proved. Proof of Theorem 3.
By assumption, we have f/g = e”(“),
(7)
l)/(g - 1) =I eBtz).
(8)
and (f -
It is clear that both S(X) and p(z) have to be polynomials, because otherwise at least one off and g would be of infinite order, which is a contradiction. Solving for g in the two equations, we obtain
It follows that the quotient: (1 _ escz,)/l _ e”‘Z’-“‘“’ has to be an integral function, Thus, by Lemma 1 we have PC4 = wc4
- 44)J
(10)
where P(z) is a polynomial with rational coefficients. Then, it follows from this (7) and (8) that 1 _ e8’z’ 1 - edlz) g(z) =: p(2) _ @cc,= @xlZ)1 _ eE(2)-~Y(z) I’ [ and
Now we shall treat two cases separately: (i) (ii)
P is a linear polynomial. P is a nonlinear polynomial.
For Case (i) there are two situations that may arise: (ia)
P is a constant, in which case we conclude easily that f = g.
414
OSGOOD
AND
YANG
(ib) P is a nonconstant linear polynomial, in which case, according to Theorem 2, we have /3(z) = n@(z) - cd(z)) + 2m7ri
(13)
where m, n are integers. Thus
@(a)- a(x)= T
- $ Tri,
(14)
and a(z) = ?/3(z)
+ 2
(15)
ni.
Substituting (14) and (15) into (1 l), we obtain 1 - @(a &4 =
e”n-l)/n)6(z)+(2?n/,r)ni 1 _
=
(
1 _
e(6(z)/n~-~2?~1/71)ni
) (16)
en[(6(zljnI-(2?1~/,l)ni]
e((n-l)/n)8(z)-(n-l)(2,n!n)Ri
(
1 _
&3(z)/nk(Pmmni
1
*
Setting
pc4- -
$7ri
n
= y(z)
in the above equation, we have 1 + id4
=
,(,-:,,“’
=
1 +
p
+
e2Y +
eY) = e-Y +
e-2v
... +
&l-lb
e’“-“Y +
... +
(17)
e-(n-lh.
Similarly, by noting the relation that (n - 1)~ = (Y- 2mi, one can derive f(z) = 1 + ey + e2y+ ..* + fP-l)v.
(18)
Therefore, by Lemma 3, we conclude immediately that
w, j-1- qy,g)-
(19)
Finally, we deal with case (ii). In this case, we obviously must have deg OL= deg g and that the deg(/3 - CX)< deg /3. It follows that O(T(y, e6(r))) as
Y
-
00.
(20)
QUOTIENT
Now clearly if (27ri)-1(&) integer, hence so is (2ai)-’ that
OF TWO
INTEGRAL
415
FUNCTIONS
- a(z)) is an integer, then (2+l/3(z) is also an a(z). Th ere f ore we have, according to Lemma 1,
(2%-i)-1/3(z) = p[(27$1(/3(z)
- a(z))]
(21)
(2xi)-’
-- a(z))],
(22)
and a(2) =: q[(2&-‘(P(z)
where p and q are two polynomials It follows that (2ni)-y/3
with rational numbers as their coefficients.
- a) = p[(27ri)-y/3
- a)] - q[(2xi)-yp
-
a)].
Hence p(z) z q(2) + 2.
(23)
From this one can derive easily the forms f(=)
=
(1
--
e2siP(Yl)/y
1 -
g(2)
=
(1
_
e2rriP(v))!(e2niP(P(v)-v)
$“lY)
and _
e2niPbl),
(24)
where y -== ljZrri@ - a). Comparing these forms with that of fi and &: obtained by Nevanlinna (which were mentioned in section IT) one can prove easily that y+ +z 2+(y) mod 2~ri and v1 z= 29~i(p(y) - y) mod 27~i. Thus qJf .~- 2Gp( l/27-++ - qJ1)) mod 2372’is the most general relation one can obtain for ‘pr and ~a . Finally, the relation T(r, g) ,- T(r,f)
as
Y --f cc
(25)
holds since both functionals are w to T(r, eY(v)). The theorem is thus proved completely. The following generalization of our Lemmas 1 and 2 is interesting in its own right: THEOREM 4. Let P(w) and Q(w) be two relative& Q(w) not equal to a power of w. Suppose
prime po[vnomials
with
is an entire function, p(z) and q(z) are polynomials, and deg(q(a)) > 1. Then there exist complex numbers c1 and c2 and a polynomial L(w) with rational coeficients such that L(q(n) + CJ = p(a) t c., .
416
OSGOOD
AND
YANG
Proof There must exist a factor of e2ai(~(z)+cl)- 1 in the denominator off(z) for some constant cr . Suppose thatf(z) $ 0 and that the numerator of f(z) factors into fi
(e2ni(dz)+K,)-
1)
times a function that does not vanish. We must have degp(z) > 0. By the same argument as in the proof of Lemma 1, we must have that degp(z) = d(deg q(z)), for some positive integer d. Let wl(n),..., wk(n) denote the distinct algebraic functions that are solutions of q(w) + c, = n. For each integer rz and each integer 1 < t < K, there exists 1 < j, < k, such that p(w,(n)) + Ki, is an integer. Hence, one may choosej, ,...,I; so that
idn) = i Mwt(~Jl>> + Kjt) t=1
is an integer infinitely often. Now $(n) is a polynomial in 1zsince it is symmetric in w,(n),..., wk(n). By the nonvanishing of the Vandermonde determinant 4(n) is seen to have rational coefficients. We may write each wj(n) as p$ik-*nk-l + O(l), where p = e2aijk-1and q(w) = q,,wk + .a.. Then Jl($ =P,q,-d~d + ...I wherep(w) = pt,wdk + .... It follows that pa& is a rational number. Hence, we see that there exist two integers, M and N, such that JM4
+ wz(4
+ Cl) = %.I(4
has degree less than dk. Now e2ni(rlW+Kj)
_
1 =
(e2niUdpW+Kj) _
(e2niUdp(z)+Kj
_
l)(e2niN(aLz)+cl) _
1) _
_
(e2niN(4(zI+c1)
1) _
I),
for each 1 < j < 12, so for every complex number a and each positive integer s if (z - a)g divides both e2ni(*(zJ+cl)- 1 and Z’(esniPtz)),then (z - a)* also divides P(e2ni71(s)).Thus, fl@)
=
p(e2nirl(z))/(e2ni~s(z,+el,
_
1)
is entire. One may now repeat the argument above, if jr(z) $ 0, replacing p(.z) by ylW f(z) by f&4, and Q(e2ni~(r)) by e2ni(*(z)+cl)- 1. After 7 < d steps, we arrive at fT 5 0. Then r,(z) is a constant, and we have proved Theorem 4.
QUOTIENT
417
OF TWO INTEGRAL FUNCTIONS
IV. CONCLUDING
REMARKS AND A CONJECTURE
1. It is easy to see that in Lemma 1, if one assumes that P(z) assumes an integral value whenever Q(z) does, then the conclusion will become P(z) = fQ(,n) A- s. where t, 9 are rational numbers. -.7 In the hypotheses of Lemmas 1 and 2, we have asked that P(a) assume an integral value whenever Q(x) does. From the proofs of these two Lemmas, it follows that this requirement can be replaced by the requirement that P(z) assume an integral value whenever Q(s) does with at most finitel! many exceptions, and the two results still hold. ? Because of Lemma 2, one might suspect that the coefficients of the _. polynomial L(z) in the statement actually have to be integers. This is, however, not the case in general. Consider, for instance,
P(,) = 064 (Q(4 - 1) 2
.
4. It is natural to ask what can one say about two functions f and g that have the property that whenever integral value, so doesf(a). One thing that we can say right will be, in general, no result analogous to Lemma 1. As f and g can be related very wildly. Consider. (a)
f(z)
(b)
f@) z @g(r),
arbitrary integral g(a) assumes an away is that there a matter of fact,
== fzg(z)(?l an integer),
and
(4 !I4 =
nreanig(z) f *yeanig2(z) (n, , no are integers).
One notes that of all the four cases, only in case (a) is the growth between .f and g restricted by
relation
(26) This is a relation that always holds for two nonconstant thus suggest the following conjecture: Conjecture. condition:
Letf
and g be two integral functions
W,-==k
polynomials.
Tl’e
that satisfy the following
418
OSGOODAND YANG
Furthermore, suppose that f assumes an integral value whenever g(z) does. Then f(z) = J%!i+4), (28) where L(z) is a polynomial of degree k with rational coefficients. Hence, lim To f”= W,g)
= A.
We believe that the problem is much harder if one tries to deal with two meromorphic functions, because in this case, one cannot use the argument that f may be expressed in the form f = filf2 , where whenever f2 vanishes then so does fi and to at least the same order. 5. As a corollary of proof of Lemmas I and 2, we may obtain the following result: Let f(z) and g( z ) each be entire and each have an infinite number of zeros, all but a finite number of which are contained in a subfield F of the real numbers. Then if we ever have f(p(z)) = g(q(z)), for polynomials p(z) and q(z), it f o11ows that p(z) = IQ(z) + Ks , where Kl and K2 both belong to F. One may use the proof of Lemma 1 to see that p(z) = L@(z)), where 44 E 04 and degp(z) 3 deg d z ) , so L(w) is linear and, by symmetry, we are through. One may compare this result with that in [I].
REFERENCES 1. L. FLATTO, A theorem on level curves of harmonic functions, J. London Math. Sot. 1 (1969), 410-472. 2. W. K. HAYMAN, “Meromorphic Functions,” Oxford Univ. Press, London/New York, 1964. 3. R. NE~ANLINNA, “ThtorCme de Picard-Bore1 et la ThCorie des Fonctions Meromorphes,” Gauthier-Villars, Paris, 1929. 4. J. F. RITT, On the zeros of exponential polynomials, Trans. Amer. Math. Sot. 29 (1927), 681. 5. LEE A. RUBEL AND C. C. YANG, Interpolation and unavoidable families of meromorphic functions, Michigan Math. J. 20 (1973). 6. VAN DER WAERDEN, “Modern Algebm,” Vol. 1, Ungar, New York, 1949.
OF MATHEMATICAL
ANALYSIS
AND
On the Quotient
54, 408-418 (1976)
APPLICATIONS
of Two Integral Functions
CHARLES F. OSGOOD AND CHUNG-CHUN YANG Mathematical
Sciences Staff, Naval Research Laboratory Washington, D.C. 20375 Submitted by R. P. Boas
I. INTRODUCTION An exponential polynomial is a function of the form:
where the aj(j = 0, 1, 2 ,..., n) are constants and the ai(i = 0, 1, 2 ,..., n) are distinct constants. Let us denote the set of all exponential polynomials by H. The following result is proved in [4]. THEOREM 1. The quotient of two exponential polynomials, an exponential polynomial.
;f entire, is itself
Remark. Rubel suggested the more general conjecture that any function algebraic over H, if entire, belongs to H. The proof of Theorem 1 has something to do with the distribution of the zeros of an exponential polynomial and the convex hull of the 0~~‘s.Thus, the methods used in [4] by their nature will fail to deal with the situation that some or all of the q’s are replaced by polynomials. In this paper, we shall treat a special case of the above-mentioned situation, As an application of the study, we shall present a result concerning the zero-one set problem (see [5]) for the class of all integral functions of finite order. Let us denote by G the class of all integral functions of the form: a
ebo(z)
0
h(z) +
+
. . . +
anebzi(z),
ale
where the ai(i = 0, 1, 2 ,..., n) are constants, and the b,(z)(j = 0, 1, 2 ,..., n) are polynomials differing from one another by a nonconstant polynomial. A statement analogous to that of Theorem 1 for the class G would be the following: 408 Copyright All rights
Q 1976 by Academic Press, Inc. of reproduction in any form reserved.
QUOTIENT
OF TWO INTEGRAL FUNCTIONS
409
Conjecture. The quotient of two elements of G, if entire, is an element of G. Ozawa pointed out that the conjecture is not always true by exhibiting the quotient: f(z) == (Piz* - l)/(e2niz - 1), which is entire but not a member of G. However, in this paper, we shall prove the following: THEOREM
degree.
2.
Let P(z), Q(z) be two nonconstant pol~~nomials of the same
Jf f(x) = (&-) - I)/(@‘) - 1)
is entire then P(z) =
m@(z)) + 2n7ri,
where m, 12are integers. Remark.
It follows
from this that f~ G.
II. APPLICATION As we know, there are integral functions f such that any other integral function g that has the same zeros and one-points with the same multiplicities as f must be f. For instance, take f = sin z [5]. There are also pairs of distinct integral functions f and g having the same zeros and one-points with the same multiplicities, for instance, f = eZ and 2 = e-“. Kate here that f and 0” have the same rate of growth. It would not be an unreasonable guess that if two integral functions f and g have the same zeros and one-points with the same multiplicities, then T(r, f) y T(r, g), where T(r, h) denotes the Nevanlinna characteristic function for the function h (for the details of the notation see, e.g., [2]). It has been shown by Nevanlinna [3, p. 1251 that if two meromorphic functions fi and f2 having the same zeros and one-points, and poles with the same multiplicities, then fi and f2 can be expressed as
and 1 _ ee&) f2@>
=
em,(Z)
_
ee,(z)
’
where or and v1 are two integral functions. (Here we would like to point out that the form of fi in [3] was incorrect.) There was nothing said about the
410
OSGOODAND YANG
relation between vi and ‘pa. We have not been able to settle this question completely. However, for the class F of all nonconstant integral functions of finite order, we can obtain the following result. Particularly, it will tell us the relation between pr and vz . THEOREM 3. Let f, g E F. Suppose that f andg have the same zeros and onepoints Edith the same multiplicities. Then either f = g, or
f(z) = 1 + ey + ... + etn-ljY = (1 - eny)/(l - eY), and ... +
g(z) = 1 + eeYf
,-tn-lh
=
(1 _
e--"')/(l
_
e-Y),
or more generally f(z)
=
(1
-
e2niP(Y))/(
_
e2aiP(v))l(e2ni(p(v)-v)
1 _
e2niY)
and g(z)
=
(1
_
&hiP(Y)))
where y is a polynomial in z and p(z) is a polynomial in z with rational numbers as the coe#icients. It follows that q2 = 2rrip(1/2G((p2 - w)) in Nevanlinna’s results, which we just mentioned above. Furthermore, we always have T(r, f) v, g>asr+ 00.
III. PROOFS OF THEOREMS 2 AND 3 To prove our theorems, we shall first have to prove a lemma, which is interesting in itself. LEMMA 1. Let P(z) and Q(z) be two nonconstant polynomials. Suppose that wherever Q(z) assumes an integral value, so also does P(z). Then (deg P)/(degQ) = d is an integer and P(z) = L(Q(x)) where L(z) = CIdZd$- cid++l
+ *.. + 010,
and the aj(j = 0, 1, 2 ,..., d) are rationals with C+ # 0. Proof. It is clear that the degree of P must be larger than or equal to that of Q, since otherwise, there would be more points in some finite part of the plane causing Q(z) to be an integer than points causing P(z) to be an integer. We shall see that n = deg Q and m = deg P cannot have their greatest common divisor less than n. The roots of the equations
Q(z) = fN(N
= 1, 2,...)
QUOTIENT
OF TWO
INTEGRAL
FVNCTIONS
411
lie asymptotically near differing radii each of the form pf+y1 = 0, 1, 2 ,...) 2n), where t& = e, + 27r
and
I e, - e,_., ) = 7+z.
All the roots of the equations P(z) = &m(m = 1,2,...) lie asymptotically near differing radii each of the form p+j(j
= 0, I, 2 ,..., 3m),
where km = $0 + 2i-r
and
j #j - $J-~ i = n/m.
It is impossible to have any inclusion relation between these two sets of roots. Hence, we conclude m 3 n and m/n is an integer. Let us now set d = m/n and we shall prove the statement by induction on d. For d = 1 we have the following statement. LEMMA 2. Let P(z), Q(z) be two nonconstant polynomials of the same degree, sa-v deg P = deg Q = k. Suppose that whenever Q(z) assumes an integral value or zero at some point z, , so does P(z). Then P(z) = -4Q(z) + B, where both A and B are integers. Proof. Let n be a given integer and let the wj(n)(j = 1, 2,..., k) be the algebraic functions satisfying Q(m) = n. Let us also assume that
and id
First, we will show that po/qo is a rational number. Let us consider the algebraic function of n q(n) = t
P(wj(n)).
(1)
j=~l
Sow p)(n) is symmetric in the wj(n); hence, it is a polynomial in ~8-14~1,..., n) 4;’ over C (see [6, p. 791). F or each integer n, F(n) is an integer. By the nonvanishing of the determinant 1: : /, say, we see that the coefficients of q(n) are rational numbers. By growth arguments deg, q(n) < 1. For large values of n, we have
qlqkl, (qO -
wj(n) = pjp;“kfll’k 409:54/2-g
f o(l) (n”“)
(2)
412
OSGOODAND YANG
where p is a primitive kth root of unity; hence, P(w(n)) = (p&&f 3
+ o(l)(nl-l’k).
(3)
Substituting this into (I), we have:
&> = k(Pol~o)n+ C? for some rational number c, so p,,/q,, = integers. Next consider B(z) 3 sP(z) -
Y/S
is rational, where
(4) Y
and s are
(5)
YQ(Z).
Now B(z) is a polynomial of degree I < k, and B(z) is an integer whenever Q(z) is an integer. If B(z) is not a constant, then we have a contradiction, since previously we showed that under these hypotheses one must have deg P(z) 3 degQ(z). Then B(z) = Br and (r/s)t + B,/s is an integer for all integers t. Setting t = 0, 1 we see that Y/S and B/s are each integers. Let d = r/s = p&a. Then P(z) = /IQ(z) + B, and we have proved Lemma 2. Now we are going to complete our induction proof of Lemma 1. Assume that for all 1 < dr < d the statement of Lemma 1 holds. First we see from the construction above, for d > 1, that 9(n) is a polynomial of degree d with rational coefficients (use the nonvanishing of the Vandermonde determinant) and that v(n) = k(pO/q,,)nd + .... Thus, ps/q,, = Y/S is rational. Now B(z) zs
- s(&(z))~
rP(z)
(6)
is a polynomial of degree m, < 7n,which is an integer at all points where Q(z) is an integer. Hence, (deg BWl(degQM)
= 4 < 4
and we have, by the induction assumption, B(z) = L,(Q(z)), where L,(z) is a polynomial of degree d, with rational coefficients. Lemma 1 now follows immediately. LEMMA 3.
(See, e.g., [2, p. 71.)
Let P(z)
be a poZynomia1 of degree n
having the form P(z) = i
pizF,
i=O
then T(Y,
ePtr)) = (l/n) 1PO[ P(l + 0(1/r)).
QUOTIENT
OF TWO
INTEGRAL
FI’NCTIONS
413
Proof of Theorem 2. First we rewrite P(z) = 2nip(z) and Q(z) = 27+(z). Since f (z) is an integral function, we conclude that whenever Q(Z) assumes an integral value, so does?(z). By Lemma 1 it follows thatp(z) -= ms(z) + n; hence, P(z) = mQ(z) + 2~2 for some integers n and m. The theorem is thus proved. Proof of Theorem 3.
By assumption, we have f/g = e”(“),
(7)
l)/(g - 1) =I eBtz).
(8)
and (f -
It is clear that both S(X) and p(z) have to be polynomials, because otherwise at least one off and g would be of infinite order, which is a contradiction. Solving for g in the two equations, we obtain
It follows that the quotient: (1 _ escz,)/l _ e”‘Z’-“‘“’ has to be an integral function, Thus, by Lemma 1 we have PC4 = wc4
- 44)J
(10)
where P(z) is a polynomial with rational coefficients. Then, it follows from this (7) and (8) that 1 _ e8’z’ 1 - edlz) g(z) =: p(2) _ @cc,= @xlZ)1 _ eE(2)-~Y(z) I’ [ and
Now we shall treat two cases separately: (i) (ii)
P is a linear polynomial. P is a nonlinear polynomial.
For Case (i) there are two situations that may arise: (ia)
P is a constant, in which case we conclude easily that f = g.
414
OSGOOD
AND
YANG
(ib) P is a nonconstant linear polynomial, in which case, according to Theorem 2, we have /3(z) = n@(z) - cd(z)) + 2m7ri
(13)
where m, n are integers. Thus
@(a)- a(x)= T
- $ Tri,
(14)
and a(z) = ?/3(z)
+ 2
(15)
ni.
Substituting (14) and (15) into (1 l), we obtain 1 - @(a &4 =
e”n-l)/n)6(z)+(2?n/,r)ni 1 _
=
(
1 _
e(6(z)/n~-~2?~1/71)ni
) (16)
en[(6(zljnI-(2?1~/,l)ni]
e((n-l)/n)8(z)-(n-l)(2,n!n)Ri
(
1 _
&3(z)/nk(Pmmni
1
*
Setting
pc4- -
$7ri
n
= y(z)
in the above equation, we have 1 + id4
=
,(,-:,,“’
=
1 +
p
+
e2Y +
eY) = e-Y +
e-2v
... +
&l-lb
e’“-“Y +
... +
(17)
e-(n-lh.
Similarly, by noting the relation that (n - 1)~ = (Y- 2mi, one can derive f(z) = 1 + ey + e2y+ ..* + fP-l)v.
(18)
Therefore, by Lemma 3, we conclude immediately that
w, j-1- qy,g)-
(19)
Finally, we deal with case (ii). In this case, we obviously must have deg OL= deg g and that the deg(/3 - CX)< deg /3. It follows that O(T(y, e6(r))) as
Y
-
00.
(20)
QUOTIENT
Now clearly if (27ri)-1(&) integer, hence so is (2ai)-’ that
OF TWO
INTEGRAL
415
FUNCTIONS
- a(z)) is an integer, then (2+l/3(z) is also an a(z). Th ere f ore we have, according to Lemma 1,
(2%-i)-1/3(z) = p[(27$1(/3(z)
- a(z))]
(21)
(2xi)-’
-- a(z))],
(22)
and a(2) =: q[(2&-‘(P(z)
where p and q are two polynomials It follows that (2ni)-y/3
with rational numbers as their coefficients.
- a) = p[(27ri)-y/3
- a)] - q[(2xi)-yp
-
a)].
Hence p(z) z q(2) + 2.
(23)
From this one can derive easily the forms f(=)
=
(1
--
e2siP(Yl)/y
1 -
g(2)
=
(1
_
e2rriP(v))!(e2niP(P(v)-v)
$“lY)
and _
e2niPbl),
(24)
where y -== ljZrri@ - a). Comparing these forms with that of fi and &: obtained by Nevanlinna (which were mentioned in section IT) one can prove easily that y+ +z 2+(y) mod 2~ri and v1 z= 29~i(p(y) - y) mod 27~i. Thus qJf .~- 2Gp( l/27-++ - qJ1)) mod 2372’is the most general relation one can obtain for ‘pr and ~a . Finally, the relation T(r, g) ,- T(r,f)
as
Y --f cc
(25)
holds since both functionals are w to T(r, eY(v)). The theorem is thus proved completely. The following generalization of our Lemmas 1 and 2 is interesting in its own right: THEOREM 4. Let P(w) and Q(w) be two relative& Q(w) not equal to a power of w. Suppose
prime po[vnomials
with
is an entire function, p(z) and q(z) are polynomials, and deg(q(a)) > 1. Then there exist complex numbers c1 and c2 and a polynomial L(w) with rational coeficients such that L(q(n) + CJ = p(a) t c., .
416
OSGOOD
AND
YANG
Proof There must exist a factor of e2ai(~(z)+cl)- 1 in the denominator off(z) for some constant cr . Suppose thatf(z) $ 0 and that the numerator of f(z) factors into fi
(e2ni(dz)+K,)-
1)
times a function that does not vanish. We must have degp(z) > 0. By the same argument as in the proof of Lemma 1, we must have that degp(z) = d(deg q(z)), for some positive integer d. Let wl(n),..., wk(n) denote the distinct algebraic functions that are solutions of q(w) + c, = n. For each integer rz and each integer 1 < t < K, there exists 1 < j, < k, such that p(w,(n)) + Ki, is an integer. Hence, one may choosej, ,...,I; so that
idn) = i Mwt(~Jl>> + Kjt) t=1
is an integer infinitely often. Now $(n) is a polynomial in 1zsince it is symmetric in w,(n),..., wk(n). By the nonvanishing of the Vandermonde determinant 4(n) is seen to have rational coefficients. We may write each wj(n) as p$ik-*nk-l + O(l), where p = e2aijk-1and q(w) = q,,wk + .a.. Then Jl($ =P,q,-d~d + ...I wherep(w) = pt,wdk + .... It follows that pa& is a rational number. Hence, we see that there exist two integers, M and N, such that JM4
+ wz(4
+ Cl) = %.I(4
has degree less than dk. Now e2ni(rlW+Kj)
_
1 =
(e2niUdpW+Kj) _
(e2niUdp(z)+Kj
_
l)(e2niN(aLz)+cl) _
1) _
_
(e2niN(4(zI+c1)
1) _
I),
for each 1 < j < 12, so for every complex number a and each positive integer s if (z - a)g divides both e2ni(*(zJ+cl)- 1 and Z’(esniPtz)),then (z - a)* also divides P(e2ni71(s)).Thus, fl@)
=
p(e2nirl(z))/(e2ni~s(z,+el,
_
1)
is entire. One may now repeat the argument above, if jr(z) $ 0, replacing p(.z) by ylW f(z) by f&4, and Q(e2ni~(r)) by e2ni(*(z)+cl)- 1. After 7 < d steps, we arrive at fT 5 0. Then r,(z) is a constant, and we have proved Theorem 4.
QUOTIENT
417
OF TWO INTEGRAL FUNCTIONS
IV. CONCLUDING
REMARKS AND A CONJECTURE
1. It is easy to see that in Lemma 1, if one assumes that P(z) assumes an integral value whenever Q(z) does, then the conclusion will become P(z) = fQ(,n) A- s. where t, 9 are rational numbers. -.7 In the hypotheses of Lemmas 1 and 2, we have asked that P(a) assume an integral value whenever Q(x) does. From the proofs of these two Lemmas, it follows that this requirement can be replaced by the requirement that P(z) assume an integral value whenever Q(s) does with at most finitel! many exceptions, and the two results still hold. ? Because of Lemma 2, one might suspect that the coefficients of the _. polynomial L(z) in the statement actually have to be integers. This is, however, not the case in general. Consider, for instance,
P(,) = 064 (Q(4 - 1) 2
.
4. It is natural to ask what can one say about two functions f and g that have the property that whenever integral value, so doesf(a). One thing that we can say right will be, in general, no result analogous to Lemma 1. As f and g can be related very wildly. Consider. (a)
f(z)
(b)
f@) z @g(r),
arbitrary integral g(a) assumes an away is that there a matter of fact,
== fzg(z)(?l an integer),
and
(4 !I4 =
nreanig(z) f *yeanig2(z) (n, , no are integers).
One notes that of all the four cases, only in case (a) is the growth between .f and g restricted by
relation
(26) This is a relation that always holds for two nonconstant thus suggest the following conjecture: Conjecture. condition:
Letf
and g be two integral functions
W,-==k
polynomials.
Tl’e
that satisfy the following
418
OSGOODAND YANG
Furthermore, suppose that f assumes an integral value whenever g(z) does. Then f(z) = J%!i+4), (28) where L(z) is a polynomial of degree k with rational coefficients. Hence, lim To f”= W,g)
= A.
We believe that the problem is much harder if one tries to deal with two meromorphic functions, because in this case, one cannot use the argument that f may be expressed in the form f = filf2 , where whenever f2 vanishes then so does fi and to at least the same order. 5. As a corollary of proof of Lemmas I and 2, we may obtain the following result: Let f(z) and g( z ) each be entire and each have an infinite number of zeros, all but a finite number of which are contained in a subfield F of the real numbers. Then if we ever have f(p(z)) = g(q(z)), for polynomials p(z) and q(z), it f o11ows that p(z) = IQ(z) + Ks , where Kl and K2 both belong to F. One may use the proof of Lemma 1 to see that p(z) = L@(z)), where 44 E 04 and degp(z) 3 deg d z ) , so L(w) is linear and, by symmetry, we are through. One may compare this result with that in [I].
REFERENCES 1. L. FLATTO, A theorem on level curves of harmonic functions, J. London Math. Sot. 1 (1969), 410-472. 2. W. K. HAYMAN, “Meromorphic Functions,” Oxford Univ. Press, London/New York, 1964. 3. R. NE~ANLINNA, “ThtorCme de Picard-Bore1 et la ThCorie des Fonctions Meromorphes,” Gauthier-Villars, Paris, 1929. 4. J. F. RITT, On the zeros of exponential polynomials, Trans. Amer. Math. Sot. 29 (1927), 681. 5. LEE A. RUBEL AND C. C. YANG, Interpolation and unavoidable families of meromorphic functions, Michigan Math. J. 20 (1973). 6. VAN DER WAERDEN, “Modern Algebm,” Vol. 1, Ungar, New York, 1949.