expo poly

ON THE ZEROS OF EXPONENTIAL POLYNOMIALS* BY J. F. RITT By an exponential polynomial, we shall mean a function (1) a0eao* + • • • + ame"mZ with constant a's and with constant a's distinct from one another. The distribution of the zeros of such functions, and of more general functions in which the a's are polynomials in z, rather than constants, has been in- vestigated by Tamarkin, Pólya and Schwenglert. The very elegant results secured by them will be described, to some extent, below. The present writer has treated the question of factorizing an exponential polynomial into a product of exponential polynomials!. We present here two results. In §1, we prove that if every zero of one exponential polynomial is also a zero of a second exponential polynomial, the quotient of the second function by the first is an exponential polynomial. In §2, we study the function 1 + aie°y +-h amef", with real a's such that 0 < ai < • • • < am. We consider, any horizontal strip of the complex plane, and derive an ex- pression for the sum of the real parts of those zeros of the exponential polynomial which are situated in the strip. The result obtained is analogous to the theorem that the product of the zeros of 1 + aiz + a2z2 + • • • + amzm is (-l)m/am. * Presented to the Society, October 27,1928. Received by the editor of the Bulletin in Novem- ber, 1928, accepted for publication in the Bulletin, and subsequently transferred to these Trans- actions. t Tamarkin, Mathematische Zeitschrift, vol. 27 (1927), p. 1, and earlier papers there referred to; Pólya, Münchner Berichte, 1920; Schwengler, Geometrisches ueber die Verteilung der Nullstellen etc., Dissertation, Zurich, 1925. X These Transactions, vol. 29 (1927), p. 584. 680 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use ZEROS OF EXPONENTIAL POLYNOMIALS 681 1. Division. Theorem. Let (2) Aiz) = floe««1 +-h amea">% Biz) = foe»'' +-\- bn^'. Suppose that Biz) ¿¿0, and that A iz)/Biz) is an integral function. Then there exists a Ciz) = coe-"' + • ■ • + cpew such that Aiz) =5(z) Ciz). We begin our proof by describing a result of Tamarkin, Pólya and Schwengler. Let the exponents a in A (z) be plotted in the complex plane, and let the smallest convex polygon 31 which contains them be constructed. Let the sides of 21 be designated by vi, ■ ■ ■ ,<n. Let di, * = 1, • • • , /, represent a ray which is the image, with respect to the real axis, of a perpendicular to o\ erected exterior to 31. It is proved by the above-named writers that there exist / half-strips,* each parallel to, and ex- tending in the same direction as, one of the rays di, which contain all of the zeros of A. If s< is the length of o-„ the number of zeros in the half-strip parallel to di whose moduli are less than r is asymptotically equivalent to rsi/i2ir). Consider now the convex polygon 93 corresponding to Biz). As every zero of B is also a zero of A, it is clear, from the asymptotic formula for the number of zeros in a half-strip, that to every side t of 33 there corresponds a side of 21 which is parallel to t, at least as long as t, and whose outward perpendicular has the same direction as the outward perpendicular to t. We shall suppose that the a's in (1) are so ordered that a,- comes before a¡ if the real part of a< is less than that of a¡, or if the real parts are equal but the coefficient of ( —1)1/2 in a¡ is less than that in a,-. When the a's are real and non-negative, and am^0, we shall call am the degree of the function (1). Lemma. If Aiz), and Biz)^0, are two exponential polynomials with real non-negative exponents, we have (3) A=QB + R, where Q and R are two exponential polynomials with real non-negative ex- ponents, and where R, if not zero, is of lower degree than B. If, in (2), we have am<ßn, we have (3) with Q = 0, R=A. Suppose then that amtßn- * By a half-strip is meant an infinite region comprised between two half-lines and a line perpen- dicular to both of them. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 682 J. F. RITT [October Let am, • • -, am-i be those a's which are at least as great as /3„. Consider the exponential polynomial (4) C = A - ia^ + • • • + «-"-"V whose exponents are non-negative. Suppose first that B consists of one term. Then C is either zero, or of smaller degree than B, so that we have, in (4), a representation (3) with Q the fraction in the second member of (4), and R = C. Suppose now that B has at least two terms. If C is not zero, its degree is either less than ßn or equal to am— (/3„—ß„_i). If C is zero, or of degree less than ßn, we have, in (4), a representation (3). Otherwise, we repeat the process just described, subjecting C to the treatment received by A. As ßn—/3„_i is a fixed quantity ,we arrive in a finite number of steps at a repre- sentation (3). It is a simple consequence of the asymptotic formula for the distribution of the zeros of exponential polynomials that the representation (3) is unique. We return now to the A and B of our theorem, whose exponents may, of course, be complex. We assume that A?£0. Grouping together those terms of A whose exponents have like real parts, we write (5) A = Fie""2 + • • • + P,eu>', where the u's are real, increasing with their subscripts, and where the P's are of the type (6) gie»'" + ■ • • + gp«"»", with real v's which increase with their subscripts. As it does not disturb the zeros of A, or affect the divisibility problem which we are studying, to multiply A by an exponential, we suppose that Mi, and the smallest v in P, are both zero. Similarly, we suppose that (7) B = Qië°* + ■■■ + Qhew»z, with stipulations identical with those made above for A. The quantity u¡ is the difference between the abscissas of any rightmost point and any leftmost point of 21. Then because to every side of 93 there corresponds a side of 21 at least as long, and having the same direction, it is clear that uj = Wh- Let us suppose, for the present, that B consists of at least two terms, that is, in (7), h^2, Qi, Qn¿¿0. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1929] ZEROS OF EXPONENTIAL POLYNOMIALS 683 Let Uj, ■ • • , u,-r be those u's which exceed w3 — (wn—wn-i). We shall prove that the quotients of P„ • • • , P,_r by Qh are exponential polynomials of the type (6). If Qh is a constant, this is certainly so. Suppose that Qh is not a constant. Then 93 has a right-hand vertical side whose length is the greatest v in Qh. Then 31 must have a right-hand side of at least the same length. That is, the greatest v in P, is not less than that of Qh. By the lemma above, P¡ = SQh + R where S and R are of type (6), with non-negative »'s, and where, if 72=^0, the greatest v in R is less than that of Qh. We say that R is zero. Suppose that this is not so. Then (8) A - SBe^i-™^' = • • • + Re»>'. The terms which precede the last in the second member of (8) are products of polynomials (6) by exponentials edz with every d ^ 0 and less than Uj. Now the first member of (8) has every zero of B. But the right hand vertical side of the polygon for the first member of (8) is shorter than the corresponding side of 93. This shows that R=0. If u,-i>Uj— (wh—Wh-i), we have A - SBe<u>-™»)* = • • • + TVie"»--'«, and it follows as above the P,_i is the product of Qh by a polynomial of type (6)*. Similarly, P,_2, • • • , Pj-r are such products. Consider now TV«* + • • • + P,-reu»-'* QhC"' If it is not identically zero, it is of the form D = Si«««' + • • ■ + S*'», where the S's are of type (6), where the t's are non-negative and increasing, and where tk^u¡—(wh — Wh-i). If D is not zero, we can, since D has every zero of B, repeat the above procedure. As Wh—Wh-i is a fixed positive quan- tity, we can repeat our process only a finite number of times, so that, at some stage, we must reach a function like D above, which is zero. When that happens, we have A expressed as the product of B by an exponential poly- nomial. When A = l, in (7), we have B = Qh. It follows, as above, that every P * The presence of negative v's in P,_i would be of no significance. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 684 J. F. RITT [October is the product of F by a function of type (7). Our theorem is thus com- pletely proved. 2. Real exponents. We deal with functions of the type /(z) = 1 + aie«'' + ■ ■ • + ame°*>z, where the a's are any constants with am^0, and thea's are any real numbers such that 0 < ai < ■ ■ ■ < am. Because/(z) is close to unity when x iz=x+yi) is large and negative, and and close to oo when x is large and positive, there exist two vertical lines be- tween which all of the zeros of /(z) are comprised. Let F(m, v) be the sum of the real parts of those zeros of/(z) for which u <y <v, where u and v are any real numbers with v >u. We shall prove that (o — u) loe I am I(9) Riu,v) = - -i-'—^-L + 0(1). 2x Let A be such that (10) | fiz) - 11 < 1 for x=A, and let B > A be such that fiz)(IDamec">'- 1< 1 for x - B. For any zero oí fiz), we have A <x <B. Let S represent the sum of those zeros oí fiz) for which u<y<v. We assume that no zero oí fiz) lies on the lines y = u or y = v. This assumption does not affect our results. We have /fiz)z-r^dz,fiz) the integration being performed in the positive sense around the rectangle of sides x=A, x = B, y = u, y = v. Now fis)(13) fzf-^-dz = z log fiz)- flogfiz)dzJ fiz) J fiz) As RÇu, v) is the real part of S, we have to determine the imaginary part of the second member of (13). We shall use that determination of log /(z) for which the coefficient of ¿ at the point (.4, v) is greater than — x but not greater than it. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1929] ZEROS OF EXPONENTIAL POLYNOMIALS 685 Let us determine first the variation of z log/(z) as z makes a circuit of the rectangle, starting from and returning to the point (A, v). Evidently z log f(z) is increased by (4 +vi)Ci, where C is the variation in the amplitude of/00. Because of (10), the variation of amp/(z) along the side x =A is less than it. To get an idea of the variation along y=u and y = v, we consider that* Y amp f(z) = arctan — > A where X and F are respectively the real and imaginary parts of f(z). As z travels along a segment of the line y = u, for instance, amp/(z) cannot under- go a variation as great as it unless X is zero at some point on the segment. Hence the variation of amp f(z) along either horizontal side of our rectangle cannot exceed ir(p+l) where p is the number of zeros of X on such a side. On the line y = u, for instance, X = 1 + ôie0"1 + • • • + ôme""*, where the b's are real numbers depending upon u. It is known that a func- tion like X cannot have more than n real zeros, t Hence the total variation of amp/(z) along y = u, y = v, x=A, is less than (2w+3) it. In virtue of (11), the variation of amp/(z) along x=B differs from the variation of the amplitude of ame"m* by less than tt. The variation of the amplitude of ameamZ along x = B is am(v—u). Hence the variation of amp f(z) as z goes around the rectangle differs from am(v—u) by less than (2m+4)7t. The change in z log z is thus of the form (A + vi)[am(v - u) + 0(l)]i. The coefficient of i in this variation is, since the 0(1) is real, (14) A[am(v-u) +0(1)]. We shall estimate the imaginary part of the integral of log/(z). We put upon A the further condition that log f(z) have, for x ^ A, an expression as an absolutely convergent Dirichlet series log/(z) = cie"* + c2e"'z + • • • , where the p's are positive and increase indefinitely. We see immediately that * Tamarkin, loc. cit., p. 27-28, or Wilder, Expansion problems etc., these Transactions, vol. 1& (1917), pp. 415-447; pp. 420-427. t Pólya and Szegö, Aufgaben und Lehrsätze aus der Analysis, vol. 2, p. 49. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 686 J. F. RITT (15) fA\ogfiz)dz = Oil). J v+Ai We now take the side y = «. The amplitude oí fiz) at iA,u), differing by less than w from the amplitude at iA,v), does not exceed 27r in absolute value. As the variation of the amplitude along y = u is less than in + l)ir the absolute value of amp/(z) is less than («+3)7r on y = u. Hence the im- aginary part of the integral along y = u is less in absolute value than (16) in + 3MB-A). We need the integral along x = B of the real part of log/(z). We put upon B the restriction that, for x = B, log/(z) admit an absolutely convergent development log fiz) = amz + log am + diC'z + d2e"«z + • • • , where the o-'s are negative and decrease indefinitely. Thus the coefficient of i in the integral along x = B is (17) amBiv- u) + iv - u)\og\ am\ +0(1). Finally, we must have the integral along y = v of the imaginary part of log/(z). Along y = v, we have | amp/(z) — amÇo — u) \ < (2» + 4)ir. Hence Ip v+Ai amp/(z)dz - amÇu - u)iA - B) J v+Bi< (2w + 4)iriB - A). Understanding now that A and B are fixed, we have A =0(1), B—A =0(1), and find, from (14), (15), (16), (17) and (18), the expression (9) for RÇu,v). Columbia University, New York, N. Y. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use