I give a short and completely elementary proof of Takagi's 1921 theorem on the zeros of a composite polynomial f(d/dz)g(z)

Sokal, AD

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Full Terms & Conditions of access and use can be found athttps://www.tandfonline.com/action/journalInformation?journalCode=uamm20The American Mathematical MonthlyISSN: (Print) (Online) Journal homepage: https://www.tandfonline.com/loi/uamm20An Elementary Proof of Takagi’s Theorem on theDifferential Composition of PolynomialsAlan D. SokalTo cite this article: Alan D. Sokal (2022) An Elementary Proof of Takagi’s Theorem on theDifferential Composition of Polynomials, The American Mathematical Monthly, 129:4, 381-384, DOI:10.1080/00029890.2022.2027719To link to this article:  https://doi.org/10.1080/00029890.2022.2027719© 2022 The Author(s). Published withlicense by Taylor & Francis Group, LLC.Published online: 01 Apr 2022.Submit your article to this journal Article views: 655View related articles View Crossmark dataAn Elementary Proof of Takagi’s Theoremon the Differential Composition ofPolynomialsAlan D. SokalAbstract. I give a short and completely elementary proof of Takagi’s 1921 theorem on thezeros of a composite polynomial f (d/dz) g(z).Many theorems in the analytic theory of polynomials [2, 8, 10, 11] are concernedwith locating the zeros of composite polynomials. More specifically, let f and g bepolynomials (with complex coefficients) and let h be a polynomial formed in someway from f and g; under the assumption that the zeros of f (respectively, g) lie in asubset S (respectively, T ) of the complex plane, we wish to deduce that the zeros ofh lie in some subset U . The theorems are distinguished by the nature of the operationdefining h, and the nature of the subsets S, T , U under consideration.Here we shall be concerned with differential composition: h(z) = f (d/dz) g(z), orh = f (D) g for short. In detail, if f (z) =m∑i=1aizi and g(z) =n∑j=1bjzj , then h(z) =m∑i=1ai g(i)(z); and D denotes the differentiation operator, i.e., Dg = g′. The followingimportant result was found by Takagi [13] in 1921, subsuming many earlier results:1Theorem 1 (Takagi). Let f and g be polynomials with complex coefficients, withdeg f = m and deg g = n. Let f have an r-fold zero at the origin (0 ≤ r ≤ m), andlet the remaining zeros (with multiplicity) be α1, . . . , αm−r = 0. Let K be the convexhull of the zeros of g. Then either f (D) g is identically zero, or its zeros lie in the setK +m−r∑i=1[0, n− r]α−1i .Here we have used the notations A + B = {a + b : a ∈ A and b ∈ B} and AB ={ab : a ∈ A and b ∈ B}.Takagi’s proof was based on Grace’s apolarity theorem [3], a fundamental but some-what enigmatic result in the analytic theory of polynomials.2 This proof is also givenin the books of Marden [8, Section 18], Obrechkoff [10, pp. 135–136], and Rahman1See Honda [4], Iyanaga [5,6], Kaplan [7], and Miyake [9] for biographies of Teiji Takagi ( , Tak-agi Teiji, 1875–1960). Takagi’s papers published in languages other than Japanese (namely, English, German,and French) have been collected in [14].2For discussion of Grace’s apolarity theorem and its equivalents—notably Walsh’s coincidence theoremand the Schur–Szegő composition theorem—see Marden [8, Chapter IV], Obrechkoff [10, Chapter VII], andespecially Rahman and Schmeisser [11, Chapter 3].doi.org/10.1080/00029890.2022.2027719MSC: Primary 30C15, Primary 30C10(c) 2022 The Author(s). Published with license by Taylor & Francis Group, LLC.This is an Open Access article distributed under the terms of the Creative Commons Attribution License(http://creativecommons.org/licenses/by/4.0/ ), which permits unrestricted use, distribution, and reproductionin any medium, provided the original work is properly cited.April 2022] NOTES 381and Schmeisser [11, Sections 5.3 and 5.4]. Here I give a short and completely elemen-tary proof of Takagi’s theorem.The key step—as Takagi [13] observed—is to understand the case of a degree-1polynomial f (z) = z− α:Proposition 2 (Takagi). Let g be a polynomial of degree n, and let K be the convexhull of the zeros of g. Let α ∈ C, and define h = g′ − αg. Then either h is identicallyzero, or all the zeros of h are contained in K if α = 0, and in K + [0, n]α−1 if α = 0.The case α = 0 is the celebrated theorem of Gauss and Lucas [8, Section 6], [10,Chapter V], and [11, Section 2.1], which is the starting point of the modern analytictheory of polynomials. My proof for general α will be modeled on Cesàro’s [1] 1885proof of the Gauss–Lucas theorem [11, pp. 72–73], with a slight twist to handle thecase α = 0.Proof of Proposition 2. Clearly, h is identically zero if and only if either (a) g ≡ 0 or(b) g is a nonzero constant and α = 0. Moreover, if g is a nonzero constant and α = 0,then the zero set of h is empty. So we can assume that n ≥ 1.Let β1, . . . , βn be the zeros of g (with multiplicity), so that g(z) = bnn∏i=1(z− βi)with bn = 0. If z /∈ K , then g(z) = 0, and we can considerh(z)g(z)= g′(z) − αg(z)g(z)=n∑i=11z− βi − α.If this equals zero, then by taking complex conjugates we obtain0 =n∑i=11z̄− β̄i− ᾱ =n∑i=1z− βi|z− βi |2 − ᾱ,which can be rewritten as z =n∑i=1λiβi + κᾱ whereλi = |z− βi |−2n∑j=1|z− βj |−2, κ = 1n∑j=1|z− βj |−2.Then λi > 0 andn∑i=1λi = 1, son∑i=1λiβi ∈ K; and of course κ > 0. Moreover, by theSchwarz inequality we have|α|2 =∣∣∣∣∣n∑i=11z− βi∣∣∣∣∣2≤ nn∑i=1|z− βi |−2 = nκ,so κ ≤ n|α|−2. This implies that κᾱ ∈ [0, n]α−1 and hence that z ∈ K + [0, n]α−1.We can now handle polynomials f of arbitrary degree by iterating Proposition 2:Proof of Theorem 1. From f (z) = am(m−r∏i=1(z − αi))zr it is easy to see that f (D) =am(m−r∏i=1(D − αi))Dr . We first apply Dr to g, yielding a polynomial of degree n − rwhose zeros also lie in K (by the Gauss–Lucas theorem); then we repeatedly apply (inany order) the factors D − αi , using Proposition 2.Remark. When α = 0, the zeros of h = g′ lie in K; so one might expect that when αis small, the zeros of h = g′ − αg should lie near K . But when α is small and nonzero,the set K + [0, n]α−1 arising in Proposition 2 is in fact very large. What is going onhere?382 c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 129Here is the answer: Suppose that deg g = n. When α = 0, the polynomial h = g′has degree n− 1; but when α = 0, the polynomial h = g′ − αg has degree n. So, inorder to make a proper comparison of their zeros, we should consider the polynomialg′ corresponding to the case α = 0 as also having a zero “at infinity.” This zero thenmoves to a value of order α−1 when α is small and nonzero.This behavior is easily seen by considering the example of a quadratic polynomialg(z) = z2 − β2. Then the zeros of g′ − αg arez = 1±√1+ α2β2α= −β22α + O(α3), 2α−1 + O(α).So there really is a zero of order α−1, as Takagi’s theorem recognizes.In the context of Proposition 2, one expects that g′ − αg has one zero of order α−1and n− 1 zeros near K (within a distance of order α). More generally, in the context ofTheorem 1, one would expect that h has m− r zeros of order α−1, with the remainingzeros near K . It is a very interesting problem — and one that is open, as far as I know— to find strengthenings of Takagi’s theorem that exhibit these properties. There is anold result that goes in this direction [8, Corollary 18.1], [11, Corollary 5.4.1(ii)], but itis based on a disc D containing the zeros of g, which might in general be much largerthan the convex hull K of the zeros.Postscript. A few days after finding this proof of Proposition 2, I discovered thatan essentially identical argument is buried in a 1961 paper of Shisha and Walsh [12,pp. 127–128 and 147–148] on the zeros of infrapolynomials. I was led to the Shisha–Walsh paper by a brief citation in Marden’s book [8, pp. 87–88, Exercise 11]. So theproof given here is not new; but it deserves to be better known.ACKNOWLEDGMENTS. This research was supported in part by U.K. Engineering and Physical SciencesResearch Council grant EP/N025636/1.REFERENCES[1] Cesàro, E. (1885). Solution de la question 1338. Nouvelles Annales de Mathématiques (3e série). 4:328–330. www.numdam.org/article/NAM 1885 3 4 328 0.pdf[2] Dieudonné, J. (1938). La Théorie Analytique des Polynômes d’une Variable (à Coefficients Quelcon-ques). Mémorial des Sciences Mathématiques, fascicule 93. Paris: Gauthier-Villars. www.numdam.org/issue/MSM 1938 93 1 0.pdf[3] Grace, J. H. (1902). The zeros of a polynomial. Proc. Cambridge Philos. Soc. 11: 352–357.[4] Honda, K. (1975). Teiji Takagi: A biography — on the 100th anniversary of his birth. Comment. Math.Univ. St. Paul. 24(2): 141–167. doi.org/10.14992/00010342[5] Iyanaga, S. (1990). On the life and works of Teiji Takagi, in [14, pp. 354–376].[6] Iyanaga, S. (2001). Memories of Professor Teiji Takagi. In: Miyake, K., ed. Class Field Theory — ItsCentenary and Prospect. Advanced Studies in Pure Mathematics, Vol. 30. Tokyo: Mathematical Societyof Japan, pp. 1–11.[7] Kaplan, P. (1997). Takagi Teiji et la découverte de la théorie du corps de classes. Ebisu — ÉtudesJaponaises. 16: 5–11. www.persee.fr/doc/ebisu 1340-3656 1997 num 16 1 973[8] Marden, M. (1966). Geometry of Polynomials, 2nd ed. Providence, RI: American Mathematical Society.(First edition 1949.)[9] Miyake, K. (2007). Teiji Takagi, founder of the Japanese school of modern mathematics. Japanese J.Math. 2(1): 151–164.[10] Obrechkoff, N. (2003). Zeros of Polynomials. Sofia: Marin Drinov Academic Publishing House. (Orig-inally published in Bulgarian: Obreškov, N. (1963). Nuli na Polinomite. Sofia: Izdat. Bǔlgar. Akad.Nauk.)[11] Rahman, Q. I., Schmeisser, G. (2002). Analytic Theory of Polynomials. Oxford: Clarendon Press.[12] Shisha, O., Walsh, J. L. (1961). The zeros of infrapolynomials with some prescribed coefficients. J.Analyse Math. 9: 111–160.April 2022] NOTES 383[13] Takagi, T. (1921). Note on the algebraic equations. Proc. Phys.-Math. Soc. Japan. 3(11): 175–179.doi.org/10.11429/ppmsj1919.3.11 175 (Reprinted in [14, pp. 175–178].)[14] Takagi, T. (1990). Collected Papers, 2nd ed. Edited and with a preface by S. Iyanaga, K. Iwasawa,K. Kodaira, and K. Yosida. Tokyo: Springer-Verlag. (Reprinted by Springer-Verlag, Heidelberg, 2014.)Department of Mathematics, University College London, London WC1E 6BT, UKand Department of Physics, New York University, New York, NY 10003, USAsokal@nyu.eduA Generalization of Euler’s LimitEuler’s limit is defined as limn→∞(n+1n)n = e. We establish a generalization of thislimit in the following proposition.Proposition. Let An be a strictly increasing sequence of positive numbers satisfyingthe asymptotic formula An+1 ∼ An, and let dn = An+1 − An. Thenlimn→∞(An+1An)Andn = e. (1)Proof. Let us consider the function ln x on the interval [An,An+1] for all n ∈ N. Bythe mean value theorem, we have ln An+1 − ln An = 1c (An+1 −An) for some c withAn < c < An+1. Hence (since 1An+1 <1c< 1An)An+1 − AnAn+1< ln An+1 − ln An < An+1 − AnAn.Since An+1 ∼ An, we have1← AnAn+1<ln An+1 − ln AnAn+1−AnAn< 1;that is,limn→∞ ln(An+1An) AnAn+1−An = 1.This completes the proof.It can be seen that generalization (1) gives Euler’s limit when An = n.—Submitted by Reza Farhadian, Razi University, Irandoi.org/10.1080/00029890.2022.2027718MSC: Primary 11Y60; 40A05, Secondary 11B83; 11B05384 c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 129

An elementary proof of Takagi's theorem on the differential composition of polynomials

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An elementary proof of Takagi's theorem on the differential composition of polynomials

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