This is a preprint of: “Around some extensions of Casas-Alvero conjecture for non-polynomial functions”, Anna Cima, Armengol Gasull, Francesc Mañosas, Extracta Math., vol. 35(2), 221–228, 2020. DOI: [10.17398/2605-5686.35.2.221] Around some extensions of Casas-Alvero conjecture for non-polynomial functions A. Cima, A. Gasull, and F. Mañosas Departament de Matemàtiques Universitat Autònoma de Barcelona, Barcelona, Spain {cima,gasull,manyosas}@mat.uab.cat Abstract. We show that two natural extensions of the real Casas-Alvero conjecture in the non-polynomial setting do not hold. 2010 MSC: Primary: 30C15. Secondary: 12D10; 13P15; 26C10. Keywords: polynomial, Casas-Alvero conjecture, zeroes of functions 1 Introduction and main results The Casas-Alvero conjecture affirms that if a complex polynomial P of degree n > 1 shares roots with all its derivatives, P (k) , k = 1, 2 . . . , n − 1, then there exist two complex numbers, a and b 6= 0, such that P (z) = b(z − a)n . Notice that, in principle, the common root between P and each P (k) might depend on k. Casas-Alvero arrived to this problem at the turn of this century, when he was working in his paper [1] trying to obtain an irreducibility criterion for two variable power series with complex coefficients. See [2] for an explanation of the problem in his own words. Although several authors have got partial answers, to the best of our knowledge the conjecture remains open. For n ≤ 4 the conjecture is a simple consequence of the wonderful Gauss-Lucas Theorem ([6]). In 2006 it was proved in [5], by using Maple, that it is true for n ≤ 8. Afterwards in [6, 7] it was proved that it holds when n is pm , 2pm , 3pm or 4pm , for some prime number p and m ∈ N. The first cases left open are those where n = 24, 28 or 30. See again [6] for a very interesting survey or [3, 8] for some recent contributions on this question. Adding the hypotheses that P is a real polynomial and all its n roots, taking into account their multiplicities, are real, the conjecture has a real counterpart, that also remains open. It says that P (x) = b(x − a)n for some real numbers a and b 6= 0. For this real case, the conjecture can be proved easily for n ≤ 4, simply by using Rolle’s Theorem. This tool does not suffice for n ≥ 5, see for instance [4] for more details, or next section. Also in the real case, in [6] it is proved that if the condition for one of the derivatives of P is removed, then there exist polynomials satisfying the remaining n − 2 conditions, different from b(x − a)n . The construction of some of these polynomials presented in that paper is very nice and is a consequence of the Brouwer’s fixed point Theorem in a suitable context. 1 Finally, it is known that if the conjecture holds in C, then it is true over all fields of characteris- tic 0. On the other hand, it is not true over all fields of characteristic p, see again [7]. For instance, consider P (x) = x2 (x2 + 1) in characteristic 5 with roots 0, 0, 2 and 3. Then P 0 (x) = 2x(2x2 + 1), P 00 (x) = 12x2 + 2 = 2(x2 + 1) and P 000 (x) = 4x and all them share roots with P. The aim of this note is to present two natural extensions of the real Casas-Alvero conjecture to smooth functions and show that none of them holds. Question 1. Fix 1 < n ∈ N. Let F be a class C n real function such that F (n) (x) 6= 0 for all x ∈ R, and has n real zeroes, taking into account their multiplicities. Assume that F shares zeroes with all its derivatives, F (k) , k = 1, 2 . . . , n − 1. Is it true that F (x) = b(f (x))n for some 0 6= b ∈ R and some f, a class C n real function, that has exactly one simple zero? Notice that one of the hypotheses of the real Casas-Alvero conjecture can be reformulated as follows: The polynomial F shares roots with all its derivatives but one, precisely the one corresponding to its degree. Trivially, this is so, because all the derivatives of order higher than n are identically zero. The second question that we consider is: Question 2. Fix 1 < n ∈ N. Let F be a real analytic function that shares zeroes with all its derivatives but one, say F (n) . Is it true that F (x) = b(f (x))n for some 0 6= b ∈ R and some real analytic function f, that has exactly one simple zero? Theorem A. (i) The answer to the Question 1 is “yes” for n ≤ 4 and “no” for n = 5. (ii) The answer to the Question 2 is already “no” for n = 2. Our result reinforces the intuitive idea that Casas-Alvero conjecture is mainly a question related with the rigid structure of the polynomials. 2 Proof of Theorem A (i) The answer to Question 1 is “yes” for n = 2, 3, 4 because the proof of the real Casas-Alvero conjecture for the same values of n, based on the Rolle’s Theorem and given in [4], does not uses at all that P is a polynomial. Let us adapt it to our setting. Since F (n) does not vanish we know that F has exactly n real zeroes, taking into account their multiplicites. Moreover we know that F has to have at least a double zero, that without loss of generality can be taken as 0. Next we can do a case by case study to discard all situations except that F has only a zero and it is of multiplicity n. For the sake of brevity, we give all the details only in the most difficult case, n = 4. Assume, to arrive to a contradiction, that n = 4, F is under the hypotheses of Question 1 and x = 0 is not a zero of multiplicity four. Notice that by Rolle’s theorem, for k = 1, 2, 3, each F (k) has exactly 4 − k zeroes, taking into account their multiplicities. Moreover, the only zero of F 000 must be one of the zeroes of F. If F 00 (0) = 0 and F 000 (0) 6= 0 then F has only another zero at x = a and, without loss of generality, we can assume that a > 0. Applying three times Rolle’s theorem we get that F 000 (b) = 0 for some b ∈ (0, a) which is a contradiction with the hypotheses. If F 00 (0) 6= 0 then F has two more zeroes counting multiplicities. There are three possibilities. The first one is that there is a > 0 such that F (a) = F 0 (a) = 0. In this case, applying two times Rolle’s theorem we obtain that there 2 exist b, c ∈ (0, a) with F 00 (b) = F 00 (c) = 0 and they are the only zeroes of F 00 . This fact gives again a contradiction because none of them is a zero of F. The second one is that there exist a1 , a2 ∈ R with 0 ∈ (a1 , a2 ) such that F (a1 ) = F (a2 ) = 0. Also in this case, by applying two times Rolle’s theorem we obtain that there exist b, c ∈ (a1 , a2 ) such that 0 ∈ (b, c) and F 00 (b) = F 00 (c) = 0 giving us the desired contradiction. Lastly, assume that the other two zeroes of F are a1 and a2 , with 0 < a1 < a2 . By Rolle’s Theorem the zeroes of F 0 are 0, b1 and b2 and satisfy 0 < b1 < a1 < b2 < a2 . Then, since F 00 has to have two zeroes, say c1 , c2 , and they satisfy 0 < c1 < b1 < c2 < b2 , the hypotheses force that c2 = a1 . Hence the zero of F 000 has to be between c1 and c2 = a1 , that is in particular in (0, a1 ), interval that contains no zero of F, arriving once more to the desired contradiction. In short, we have proved for n ≤ 4, that F (x) = xn G(x), for some class C n function G, that does not vanish. Hence s !n G(x) F (x) = sign(G(0)) x n = b(f (x))n , sign(G(0)) where f has only one zero, x = 0, that is simple, as we wanted to prove. To find a map F for which the answer to Question 1 is “no” we consider n = 5 and a configu- ration of zeroes of F and its derivatives proposed in [4] as the simplest one, compatible with the hypotheses of the Casas-Alvero conjecture and Rolle’s Theorem. Specifically, we will search for a function F, of class at least C 5 , with the five zeroes 0, 0, 1, c, d, to be fixed, satisfying 0 < 1 < c < d, and moreover F 0 (0) = 0, F 00 (1) = 0, F 000 (c) = 0, F (4) (1) = 0, (1) and such that F (5) does not vanish. Notice that F 0 (0) = 0 is not a new restriction. We start assuming that F (5) (x) = r − sin(x), for some r > 1 to be determined. By imposing that conditions (1) hold, together with F (0) = 0, we get that Z xZ uZ wZ zZ y  F (x) = r − sin(t) dt dy dz dw du. 0 0 1 c 1 Some straightforward computations give that r 5 r + cos(1) 4 2rc − 2 sin(c) + 2 cos(1)c − rc2 3 F (x) = x − x + x 120 12 12 6 sin(c) + 2r + 9 cos(1) − 6rc + 3rc2 − 6 cos(1)c 2 + x − 1 + cos(x). 12 Imposing now that F (1) = 0 we obtain that  5 8 cos(1)c − 41 cos(1) − 8 sin(c) + 24 r= = R(c). 4(5c2 − 10c + 4) Next we have to impose that F (c) = 0. By replacing the above expression of r in F we obtain that G(c) F (c) = , 96(5c2 − 10c + 4) 3 where  G(c) = − c2 12 c4 − 369 c3 + 1437 c2 − 1708 c + 532 cos (1) − 8 c2 (c − 1) (c − 2)2 sin (c)   + 480 c2 − 960 c + 384 cos (c) − 24 (c − 1) 9 c4 − 36 c3 + 24 c2 + 24 c − 16 . A carefully study shows that G has exactly one real zero c1 ∈ (17/10, 19/10) = I, with c1 ≈ 1.79343096. To prove its existence it suffices to show that       17 99211099 18207 17 696 17 1583211 G =− cos (1) − sin + cos + > 0, 10 500000 12500 10 5 10 12500       19 180110481 3249 19 1464 19 3616677 G =− cos (1) − sin + cos + < 0. 10 500000 12500 10 5 10 12500 By using Taylor’s formula we know that for any c > 0, S − (c) < sin(c) < S + (c) and C − (c) < cos(c) < C + (c) where c3 c5 c7 c9 c11 c2 c4 c6 c8 c10 S ± (c) = c − + − + ± and C ± (c) = 1 − + − + ± . 3! 5! 7! 9! 11! 2! 4! 6! 8! 10! Hence we can replace the values of the trigonometric functions in G by rational numbers to have upper or lower bounds of this function evaluated at 1, 17/10 or 19/10. For instance, 1960649 280093 0.5403023 ≈ = C − (1) < cos(1) < C + (1) = ≈ 0.5403028. 3628800 518400 We obtain that       17 99211099 + 18207 + 17 696 − 17 1583211 G >− C (1) − S + C + 10 500000 12500 10 5 10 12500 3444600099561969856969 = >0 49896000000000000000000 and       19 180110481 − 3249 − 19 1464 + 19 3616677 G <− C (1) − S + C + 10 500000 12500 10 5 10 12500 1689627895469649855823 =− < 0. 16632000000000000000000 To show the uniqueness of the zero in I, we will prove that G is strictly decreasing in this interval. It holds that G0 (c) = T (c) cos (1) + U (c) sin (c) + V (c) cos (c) + W (c), with    U (c) = − 8 5 c2 − 10 c + 4 c2 − 2 c + 12 , V (c) = −8 (c − 1) c4 − 4 c3 + 4 c2 − 120 ,  T (c) = − c 72 c4 − 1845 c3 + 5748 c2 − 5124 c + 1064 , W (c) = −120(9c4 − 36c3 + 36c2 − 8). 4 Figure 1: Plot of a map F for which the answer to Question 1 for n = 5 is “no”. By computing the Sturm sequences of T, U and V we can prove that T (c) < 0, U (c) < 0 and V (c) > 0 for all c ∈ I. Hence, for c ∈ I, G0 (c) < T (c)C − (c) + U (c)S − (c) + V (c)C + (c) + W (c) = Q(c), where 72469 669211 2 18852329 3 8854991 4 4732471 5 532 6 8 7 191 8 Q(c) = c− c + c − c + c − c + c + c 64800 43200 302400 80640 50400 15 7 70 797 9 34 10 1651 11 3533 12 193 13 1 1 − c − c + c + c − c + c14 − c15 . 1890 405 103950 2494800 623700 142560 831600 The Sturm sequence of Q shows that it has no zeroes in I. Moreover, it is negative in this interval, and as a consequence, G0 is also negative, as we wanted to prove. We fix c = c1 . Then, r = R(c1 ) and F is also totally fixed. Moreover, by using the same techniques we get that r = R(c1 ) > R(19/10) > 1 and as a consequence F (5) does not vanish. In fact, r = R(c1 ) ≈ 1.04591089. Finally, F has one more real zero d ∈ (33/10, 34/10). In fact, d ≈ 3.32178369. This F gives our desired example, see Figure 1. (ii) Consider F (x) = 4x2 + π 2 (cos(x) − 1) that has a double zero at 0 and also vanishes at ±π/2. Moreover, F 0 (x) = 8x − π 2 sin(x) vanishes at x = 0, F 00 (x) = 8 − π 2 cos(x) has no common zeroes with F and, for any k > 1, |F (2k) (x)| = π 2 | cos(x)| vanishes at x = π/2 and |F (2k−1) (x)| = π 2 | sin(x)| vanishes at x = 0. A similar example for n = 3 is F (x) = 4x3 −6πx2 +π 3 (1−cos(x)), that vanishes at 0, π (double zeroes) and π/2. Acknowledgements The authors are supported by Ministerio de Ciencia, Innovación y Universidades of the Spanish Government through grants MTM2016-77278-P (MINECO/AEI/FEDER, UE, first and second 5 authors) and (MTM2017-86795-C3-1-P third author). The three authors are also supported by the grant 2017-SGR-1617 from AGAUR, Generalitat de Catalunya. References [1] E. Casas-Alvero. Higher order polar germs. J. Algebra 240 (2001), 326–337. [2] Interview to E. Casas-Alvero in Spanish. https://www.gaussianos.com/la-conjetura-de-casas- alvero-contada-por-eduardo-casas-alvero/ [3] W. Castryck, R. Laterveer, M. Ounaı̈es. 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