On the distribution of polynomials with bounded roots, I. Polynomials with real coefficients

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1 c 2014 The Mathematical Society of Japan J. Math. Soc. Japan Vol. 66, No. 3 (2014) pp oi: /jmsj/ On the istribution of polynomials with boune roots, I. Polynomials with real coefficients By Shigeki Akiyama an Attila Pethő (Receive Aug. 18, 2012) Abstract. Let v (s) enote the set of coefficient vectors of contractive polynomials of egree with 2s non-real zeros. We prove that v (s) can be compute by a multiple integral, which is relate to the Selberg integral an its generalizations. We show that the bounary of the above set is the union of finitely many algebraic surfaces. We investigate arithmetical properties of v (s) an prove among others that they are rational numbers. We will show that within contractive polynomials, the probability of picking a totally real polynomial ecreases rapily when its egree becomes large. 1. Introuction. Let P (X) = X + p 1 X p 0 R[X]. In this note, we often have to switch between P (X) an the vector of its coefficients v P = (p 0,..., p 1 ) R. To simplify the notation, we ientify P an v P from now on. Let be a positive integer, B 1 a real number. Denote by E (B) the set of - imensional vectors whose roots (by the above ientification) lie within the ball of raius B centere at the origin. In this Part I of our paper we are ealing solely with the case B = 1 therefore we will use the abbreviation E instea of E (1). The elements of E are calle contractive polynomials. This set was stuie by several authors. I. Schur [12] prove a necessary an sufficient conition for v E, which implies that the bounary of E is the union of finitely many algebraic surfaces. A. T. Fam an J. S. Meitsch [6] improve this result by proving that the bounary of E is the union of two hyperplanes an one hypersurface. The two hyperplanes correspons to roots 1 an 1 respectively. You fin a thorough stuy of the bounary in Kirschenhofer et al. [10]. Later A. T. Fam [7] compute the volume of E : m (j 1)! 4 2 2m2, if = 2m, (2j 1)! 2 v = λ (E ) = m 2 2m2 +2m+1 j! 2 (j 1)! 2, if = 2m + 1, (2j 1)!(2j + 1)! where λ (.) enotes the -imensional Lebesgue measure. (1) 2010 Mathematics Subject Classification. Primary 33B20; Seconary 11B65, 93D10. Key Wors an Phrases. Selberg integral, polynomials with boune roots.

2 928 S. Akiyama an A. Pethő Given a polynomial P in R[x], the non-real roots of P appear in complex conjugate pairs. Thus = r + 2s, where r enotes the number of real an s the number of nonreal pairs of roots. The pair (r, s) is calle the signature of the polynomial. However in this paper, we erive asymptotic formulas with a fixe, we call s the signature for simplicity, because r = 2s. The set E splits naturally into /2 + 1 isjoint subsets accoring to the signature s. In the sequel E (s) enotes the subset of E whose elements have signature s. This paper is organize as follows. First we show in Theorem 2.1 that λ (E (s) ) can be compute by a certain multiple integral. It turns out that for s = 0 this is a simple variant of S (1, 1, 1/2), where S (α, β, γ) enotes the Selberg integral, which is a generalization of the beta integral stuie from many ifferent points of view, see e.g. [13], [3], [8]. By using a generalization of it, ue to K. Aomoto [5], we prove in Theorem 4.2 an expression for s = 1, which makes it possible to compute v (1) for large values of. Here we turn to the investigation of the arithmetic nature of v (s). We prove in Theorem 5.1, that they are rational numbers. We express S (1, 1, 1/2) as a prouct of binomial coefficients, which enables us to show that they are reciprocals of integers in Corollary 5.1. After this we summarize our observations on our computations on v (s) We conjecture that v (s) /v(0) is always an integer. In the case even an s = /2 our Conjecture 5.2 is completely explicit. Theorem 5.2 supports our conjectures. The quotient p (s) = v (s) /v may be viewe as the probability of picking an element v E of the signature s. By Theorem 5.1 these probabilities are rational numbers. It might be surprising to observe that totally real polynomials are very rare. If the sets E (s) woul have approximately the same volume then p (s) 2/. However using the explicit formulae for an v we show p (0) 2 2 /2, which is much smaller than we expecte. In the last section we prove a generalization of the results of I. Schur [12] as well as of A. T. Fam an J. S. Meitsch [6], that the bounary of E (s) is the union of finitely many algebraic surfaces in Theorem 7.1. In Part II [1] we apply our results to estimate the istribution of polynomials with integer coefficients with given egree an signature. We are able to complete this program with respect to three natural parameters, which we will call measures. 2. The volume of E (s). The aim of this section is to prove that the volume of E (s) can be expresse by a multiple integral. We enote by Res X (P (X), Q(X)) the resultant of the polynomials P (X), Q(X) R[X]. Theorem 2.1. Let 1 an r, s non-negative integers such that r +2s =. Then the set E (s) is Joran measurable. Let R j (X) = X 2 y j X + z j, where 0 z j 1 an the iscriminant of R j (X) is negative, j = 1,..., s. Put Then we have D r,s = [ 1, 1] r [ 2 z 1, 2 z 1 ] [0, 1] [ 2 z s, 2 z s ] [0, 1]..

3 On the istribution of polynomials with boune roots, I 929 v (s) ( (s)) 1 = λ E = r s r,s X, r!s! D r,s where r = (x j x k ), s = r,s = 1 j<k r 1 j<k s r s R k (x j ) k=1 Res X (R j (X), R k (X)), an X = x 1... x r y 1 z 1... y s z s. To prove this theorem we nee some preparation. Let the signature of P (X) = X +p 1 X 1 + +p 0 R[X] be 0 s /2. Assume that its zeroes are x 1,..., x. Assume further that they are orere such that x 1,..., x r R an the others belong to C \ R. Moreover x r+2j = x r+2j 1, j = 1,..., s, where x enotes the complex conjugate of x. Denote S j (x 1,..., x ), j = 1,..., the j-th elementary symmetric polynomial of x 1,..., x, i.e., let S j (x 1,..., x ) = 1 i 1< <i j x i1... x ij, where the sum is extene to all possible values of the inices i 1,..., i j. For later use we efine S 0 (x 1,..., x ) = 1. The classical Viéta s formulae connect the roots an coefficients of P (X). With the notation p = 1 they are p j = ( 1) j S j, j = 0,...,. (2) The last system of equations efines a mapping R r C s R. To compute the volume of E (s) we nee a mapping R R. Write P (X) = = = (X x j ) r (X x j ) r (X x j ) s ((X x r+2j 1 )(X x r+2j )) s (X 2 (x r+2j 1 + x r+2j )X + x r+2j 1 x r+2j ). As the coefficients of the quaratic factors are real numbers, this form together with

4 930 S. Akiyama an A. Pethő (2) gives the esire relations. Therefore we introuce the following new variables y j = x j (j = 1,..., r) an y r+2j 1 = x r+2j 1 + x r+2j, y r+2j = x r+2j 1 x r+2j (j = 1,..., s). Uner this notation, we prove a Lemma 2.1. Let = r + 2s, R j (X) = X 2 y r+2j 1 X + y r+2j for j = 1,..., s an J = ( S i (x 1,..., x )/ y j ) 1 i,j. Then et(j) = 1 j<k r (y j y k ) r k=1 s R k (y j ) 1 j<k s Res X (R j (X), R k (X)). Proof. Let J 1 = ( S i (x 1,..., x )/ x j ) 1 i,j. We easily see et(j) = et(j 1 ) s k=1 (x r+2k 1 x r+2k ). (3) by the Jacobian computation to transform variables from x j to y j. In the secon step we prove et(j 1 ) = k=j+1 (x j x k ). (4) Let H be a subset of {1,..., }. If t H then enote S t,h the t-th elementary symmetric polynomial of the variables {x 1,..., x }\{x j : j H}, otherwise let S t,h = 0. For 1 t, k we have which implies S t, = y k S t 1,{k} + S t,{k}, S t, x k = S t 1,{k}. Subtract the 1-st column of J 1 from the k-th column, where 2 k. All the entries of the first row will be zero, except the north west entry. Let t 2. Then the t-th entry of the k-th column is S t 1,{k} S t 1,{1}, which we can rewrite as follows S t 1,{k} S t 1,{1} = x 1 S t 2,{k,1} + S t 1,{k,1} x k S t 2,{k,1} S t 1,{k,1} = (x 1 x k )S t 2,{k,1}. Thus all entries of the k-th column are ivisible by (x 1 x k ). Factoring out all these factors from the eterminant we get et(j 1 ) = et(j 2 ) (x 1 x k ), k=2

5 On the istribution of polynomials with boune roots, I 931 where J 2 is a ( 1) ( 1) matrix, which has the same structure as J 1, but without the variable x 1. With inuction we get (4). Summarizing our computation we prove so far that et(j) = 1 j<k / s (x j x k ) (x r+2k 1 x r+2k ). After canceling by the enominator, split the remaining prouct into three factors as follows Π 1 = Π 2 = Π 3 = 1 j<k r 1 j<k s r k=1 (x j x k ), k=1 (x r+2j 1 x r+2k 1 )(x r+2j 1 x r+2k )(x r+2j x r+2k 1 )(x r+2j x r+2k ), s (x k x r+2j 1 )(x k x r+2j ). Obviously, Π 1 = r. We have (x k x r+2j 1 )(x k x r+2j ) = x 2 k y r+2j 1x k + y r+2j = R j (x k ), thus Π 3 = r,s. Finally as the roots of R j (X) are x r+2j 1, x r+2j we get Res X (R j (X), R k (X)) = (x r+2j 1 x r+2k 1 )(x r+2j 1 x r+2k ) (x r+2j x r+2k 1 )(x r+2j x r+2k ), which means Π 2 = s an the lemma is prove. Remark 2.1. By equation (3) Lemma 2.1 can be written in the form et(j) = Disc(x 1,..., x ) s (2iIx r+2j), where Disc(x 1,..., x ) enotes the iscriminant of P (X) = (X x j). Thus, if s = 0 we obtain exactly the iscriminant of P (X). Now we are in the position to prove Theorem 2.1. In the proof of Lemma 2.1 it was convenient to use the same name for the variables. To continue this notation woul make our presentation unnecessarily complicate. Therefore in the sequel we use the notation: x i = y i, i = 1,..., r, y i = y r+2i 1, z i = y r+2i, i = 1,..., s introuce in the theorem. Proof of Theorem 2.1. Let v E (s). The polynomial P v(x) can be expresse by its coefficients an by its roots, moreover we have the following relation between these representations

6 932 S. Akiyama an A. Pethő r s P (X) = X + p 1 X p 0 = (X x j ) (X 2 y k X + z k ). k=1 It is clear that ( (s)) λ E = p 0... p 1. E (s) Now we change the variables p 0,..., p 1 to x 1,..., x r, y 1, z 1,..., y s, z s. This gives rise to one to r!s! corresponence, up to measure zero exceptions. It is clear that x j [ 1, 1], j = 1..., r. The variable z k is the absolute value of a complex number lying in the close unit circle, thus z k [0, 1], k = 1,..., s. The polynomial R k (X) has two non-real roots, thus its iscriminant y 2 k 4z k 0, hence y k [ 2 z k, 2 z k ]. Moreover the Jacobian of the variable change was compute in Lemma 2.1. Thus we obtain ( (s)) 1 λ E = et(j) X. r!s! D r,s As the polynomials R k (X) are positive efinite, R k (x j ) 0 an we also have Res X (R j (X), R k (X)) 0, thus we may omit the absolute value sign aroun s an r,s. 3. On the Selberg integral an its generalization by Aomoto. After expressing v (s) in Theorem 2.1 by a multiple integral, the main question is how to compute it. We will show in the next section, that it can be expresse by Selberg s integral, if s = 0 an by a generalization of Selberg s integral ue to K. Aomoto, if s = 1. To prepare our results we summarize the necessary knowlege about these integrals in this section. Let n be a positive integer, C n = [0, 1] n an = (t 1,..., t n ) = 1 j<k n In 1944 A. Selberg [13] prove the beautiful formula S n (α, β, γ) = = n C n n 1 j=0 (t j t k ). t α 1 j (1 t j ) β 1 2γ t 1... t n Γ(α + jγ)γ(β + jγ)γ(1 + (j + 1)γ) Γ(α + β + (n + j 1)γ)Γ(1 + γ), which is vali for complex parameters α, β, γ such that R(α) > 0, R(β) > 0, R(γ) > min{1/n, R(α)/(n 1), R(β)/(n 1)}.

7 On the istribution of polynomials with boune roots, I 933 Besies the original proof of Selberg, there are at least two more essentially ifferent proofs of this formula, ue to G. W. Anerson [2] an K. Aomoto [5]. You fin a goo overview of the history, the generalizations an applications of the Selberg integral in the book of G. E. Anrews an R. Askey an R. Roy [3] as well as in the survey paper [8]. We formulate here one generalization of the Selberg integral, which we will nee later. Let w(t) = w(t 1,..., t n ) = n (1 t j ) β 1 2γ an B n (j, k, l) = tα 1 j j C n i=1 t i j+k l i=j+1 l (1 t i )w(t) t 1... t n. Thus j represents the number of extra t i factors, k the number of extra 1 t i factors, an l the number of variables that overlap among the extra factors. Assuming l j, k n an j + k l n then by Theorem of [3] we have B n (j, k, l) = l i=1 α + β + (n i 1)γ α + β (2n i 1)γ j i=1 (α + (n i)γ) k i=1 (α + (n i)γ) j+k i=1 (α + β + (2n i 1)γ) S n (α, β, γ) for all complex numbers α, β, γ satisfying the former conitions. We nee only the special case k = l, α = β = 1, γ = 1/2. Then, as is pointe out on p. 408 of [3], the integral efining B n (j, k, k) can be written in the form B n (j, k, k) = j k t i C n i=1 i=1 (1 t i )w(t) t 1... t n. Setting (α, β, γ) = (1, 1, 1/2) an B n (j, k) = B n (j, k, k) we obtain for n j k. B n (j, k) = k i=1 2 + (n i 1)/2 3 + (2n i 1)/2 j i=1 (1 + (n i)/2) k i=1 (1 + (n i)/2) j+k i=1 (2 + (2n i 1)/2) S n (1, 1, 1/2) 4. Computation of an v (1). In this section we prove expressions for v (s) for s = 0, 1. In Remark 2.1 an explicit formula for et(j) is given in the case s = 0. Then our multiple integral simplifies to a transforme Selberg integral [13], [8] an we obtain

8 934 S. Akiyama an A. Pethő Theorem 4.1. Let be a positive integer. Then = 2(+1)/2 S (1, 1, 1/2)! = 2(+1)/2! 1 j=0 (Γ(1 + j/2)) 2 Γ(1 + (j + 1)/2) Γ(2 + ( + j 1)/2)Γ(3/2). Proof. Using the notations of Theorem 2.1 we have r =, s = 0, hence = 1! [ 1,1] 1 j<k (x j x k ) x 1... x. Rearranging x 1,..., x in ecreasing orer all factors are non-negative an we may omit the absolute value. Taking in account that x 1,..., x have! ifferent orerings we obtain =... 1 x 1 x 1 1 j<k (x j x k ) x 1... x. Now we change the variables by x i = 2X i 1, i = 1,..., an get = 2 (+1)/ X X 1 1 j<k Finally we perform the first step backwars an obtain = 2(+1)/2! j<k (X j X k ) X 1... X. (X j X k ) X 1... X = 2(+1)/2 S (1, 1, 1/2).! This relation implicitly appears in G. W. Anerson [2]. For s > 0 we were not able to fin a similar simple relation between the Selberg integral or its generalizations an our expressions for v (s), although their form an numerical investigations suggest strong connections. For s = 1, using Aomoto s generalization [5], more precisely its variant in the book [3], we were able to erive a bit more complicate formula, which we present now.

9 On the istribution of polynomials with boune roots, I 935 Theorem 4.2. Let be an integer. Then v (1) 2 = 2 ( 1)( 2)/2 j=0 2 j k=0 B 2 ( 2 k, 2 k j) ( 1) k k j j!k!( 2 j k)! 1 2 z 0 2 z y j (y + z + 1) k y z. Proof. As in the proof 2 appears often an this makes the formulae more complicate, we use the abbreviation δ = 2. By Theorem 2.1 we have where an = δ,1 = v (1) = 1 δ (x) Xy z, δ! D δ,1 δ (x 2 i yx i + z), δ (x) = i=1 X = x 1... x δ. 1 j<k δ (x j x k ) Remark that for simplicity we replace y 1, z 1 by y, z. We transform the range of integration for x 1,..., x δ to the range use by Selberg an Aomoto by performing the substitutions x i = 2X i 1, i = 1,..., δ. (These early substitutions results much simpler final formulae, than we woul o them later.) After some simple calculations we obtain with v (1) = 2(δ+1)δ/2 δ! P (X, y, z) = = [0,1] δ δ (X) 1 2 z 0 2 z δ ((2X i 1) 2 (2X i 1)y + z) i=1 P (X, y, z) y z X, (5) δ ( 4X i (1 X i ) 2X i y + (y + z + 1)). i=1 By performing the multiplications we can separate the variables as follows where P (X, y, z) = δ j=0 ( 2) j δ j ( 4) δ k j y j (y + z + 1) k Σ 1 Σ 2, k=0

10 936 S. Akiyama an A. Pethő Σ 1 = Σ 2 = (i 1,...,i j) I 1 X i1 X ij, (i j+1,...,i δ k ) I 2 X ij+1 (1 X ij+1 ) X iδ k (1 X iδ k ) an I 1, I 2 run through all orere isjoint subsets of {1,..., δ} with size j an δ j k respectively. Inserting these expressions into (5) we can separate the variables. Moreover it is obvious that [0,1] δ δ (X) X i1 X ij X ij+1 (1 X ij+1 ) X iδ k (1 X iδ k ) X oes not epen on the actual values of (i 1,..., i j ), (i j+1,..., i δ k ) but only on the size of the (orere) sets I 1, I 2 to which they belong. It is also clear that this integral is equal to B δ (δ k, δ j k). With these observations we can consierably simplify our integral, an obtain v (1) = 2(δ+1)δ/2 δ! δ ( δ ( 2) j j j=0 1 2 z 0 2 z ) δ j ( ) δ j ( 4) δ k j B δ (δ k, δ j k) k k=0 y j (y + z + 1) k y z. After some obvious simplification we obtain the final formula v (1) = 2 (δ+1)δ/2 δ δ j j=0 k=0 ( 1) δ k 2 2δ 2k j j!k!((δ j k)! B δ(δ k, δ j k) 1 2 z 0 2 z y j (y + z + 1) k y z. 5. Arithmetical properties of v (s). After expressing v (s) in Theorem 2.1 by a multiple integral, which was simplifie in the cases s = 0, 1 in the last section, we investigate arithmetical properties of these numbers. We start with a general fact. Theorem 5.1. The numbers v an v (s) are rational. Moreover v = /2 s=0 v (s). (6)

11 On the istribution of polynomials with boune roots, I 937 Proof. The first an the last statements are obvious by the formula (1) an by the efinitions of the volumes respectively. We inclue them only for completeness. The secon assertion is true for s = 0, 1 by Theorems 4.1 an 4.2 respectively. Thus we may assume s 2. By Theorem 2.1 we have v (s) = 1 r s r,s X r!s! D r,s with the notations explaine there. In the first step we prove v (s) = 1 W (x 1,..., x r ) r x, (7) r!s! D r,0 where x = x 1... x r an W (x 1,..., x r ) Q[x 1,..., x r ] is symmetric. As r oes not epen on y 1, z 1,..., y s, z s we may take W (x 1,..., x r ) = s r,s Y, F s where F s = [ 2 z 1, 2 z 1 ] [0, 1] [ 2 z s, 2 z s ] [0, 1] an Y = y 1 z 1... y s z s. Remark that s r,s = 1 j<k s Res X (R j (X), R k (X)) s k=1 r R j (x k ) is obviously a symmetric polynomial with rational coefficients in x 1,..., x r. Further we have where W 1 (y 1, z 1 ) = s r,s = s 1 r,s 1 W 1 (y 1, z 1 ), s Res X (R j (X), R 1 (X)) j=2 r R 1 (x k ) is a polynomial with coefficients in Q = Q[x 1,..., x r, y 2,... y s, z 2,..., z s ], which is again symmetric in x 1,..., x r. Now we can rewrite the formula for W as follows W = s 1 r,s 1 F s 1 with Y 1 = y 2 z 2... y s z s. It is clear that 1 2 z1 0 k=1 2 z 1 W 1 (y 1, z 1 )y 1 z 1 Y 1 2 z1 2 z 1 W 1 (y 1, z 1 ) y 1

12 938 S. Akiyama an A. Pethő is a polynomial in z 1 with coefficients from Q. Thus the same is true for 1 2 z1 0 2 z 1 W 1 (y 1, z 1 ) y 1 z 1. As W 1 is symmetric in x 1,..., x r, this property remains unaffecte after the two integrations. Now we can continue the above escribe process with the pairs of variables (y 2, z 2 ),..., (y s, z s ), which finally leas to the proof of (7). Performing in (7) the variable change x i = 2X i 1, i = 1,..., r we obtain v (s) = 2r(r+1)/2 r!s! C r W (X 1,..., X r ) r (X 1,..., X r ) X with C r = [0, 1] r an X = X 1... X r. Moreover we have W (X 1,..., X r ) = W (2X 1 1,..., 2X r 1). As W is a symmetric polynomial with rational coefficients, the same is true for W. It follows from a very general result of K. Aomoto [4, p. 177] (see also [5, p. 545]), that the last integral ivie by S r (1, 1, 1/2) is a rational number, which implies the assertion immeiately. In our simple situation the proof of rationality can be complete irectly. Inee, as in the proof of Theorem 4.1 we can rearrange the variables x 1,..., x r in (7) in ecreasing orer. Then r > 0 an we may omit the absolute value. Thus we obtain v (s) = W (x 1,..., x r ) (x 1,..., x r ) x r... x 1. s! 1 x 1 x r 1 As the integran is a polynomial with rational coefficients, this property is not affecte uring the successive integration by x r,..., x 1. In the last step we obtain a polynomial with rational coefficients without variables, i.e., a rational number. The theorem is prove. In the sequel we concentrate mainly on the case r = 0. simpler formula for S (1, 1, 1/2) as given in Theorem 4.1. First we prove a much Lemma 5.1. If is a positive integer then S (1, 1, 1/2) =! i=1 (i 1)! 2 (2i 1)!. (8) Proof. In this proof we use the abbreviation S = S (1, 1, 1/2). By Theorem 4.1 we have

13 On the istribution of polynomials with boune roots, I 939 S +1 = Γ(1 + /2)2 1 Γ(1 + ( + 1)/2) Γ(2 + ( + j 1)/2) S Γ( + 1)Γ(3/2) Γ(2 + ( + j)/2) j=0 = Γ(1 + /2)2 Γ(3/2 + /2) 2 Γ( + 2)Γ( + 3/2)Γ(3/2). Using the functional equation Γ(x + 1) = xγ(x) we can consierably simplify the last formula This implies S +1 = Γ(1 + /2)2 Γ(1/2 + /2) 2 (1/2 + /2) 2 S Γ( + 1)Γ( + 1/2)Γ(3/2) ( + 1)( + 1/2) = S S (2 + 1). S +1 S = ( after a short computation. The quotient of consecutive values of the sequence! i=1 (i 1)! (i 1)!/(2i 1)! satisfy the same relation. As the starting values of both sequences coincie we prove the statement. ) 1 Corollary 5.1. For 1 we have 1 1 S (1, 1, 1/2) = j=0 ( 2j + 1 i.e. the number S (1, 1, 1/2) is the reciprocal of an integer. Proof. The statement is true for = 1. Further 1 S +1 (1, 1, 1/2) = 1 S (1, 1, 1/2) j ), ( ) S (1, 1, 1/2) S +1 (1, 1, 1/2) = 1 S (1, 1, 1/2), which proves the assertion. Theorem 2.1 allows the numerical computation of v (s). We i it by using the computer algebra software Mathematica. It was able to compute v (s) for all possible signatures for 8 an check the formula (6). For = 9 we faile for s = 2 by time constraint. We compute v (2) 9 inirectly by formula (6), which is actually v (2) 9 = v 9 ( 9 + v (1) 9 + v (3) 9 + v (4) ) 9. In Table 1, you fin these compute values for 9.

14 940 S. Akiyama an A. Pethő v (1) v (2) v (3) v (4) Table 1.

15 On the istribution of polynomials with boune roots, I 941 We also compute the value v (5) 10 = Unfortunately our metho is not generalizable, therefore we o not present it here. In Table 2 we isplay the quotients v / an v (s) /v(0) for 2 8, 1 s /2. v / v (1) /v(0) v (2) /v(0) v (3) /v(0) v (4) /v(0) Our numerical investigations lea to a Table 2. Conjecture 5.1. The quotient v (s) /v(0) is an integer. The conjecture is true for the compute values, i.e. for any signatures for 9. It is also true for = 10, s = 5. The formula of Theorem 4.2 makes the computation of v (1) much more efficient. Using it we compute v (1) for 100. This range coul easily be extene, but we coul not expect new information, hence we stoppe there. Our computation confirme Conjecture Conjecture 5.1 together with relation (6) implies that v / is an integer for 1. In this case the formulae (1) an (8) imply m 2 v m (j 1)! 4 2m (2j 1)! (2j 1)! 2, if = 2m, (j 1)! 2 = m 2 m j! 2 (j 1)! 2 2m+1 (2j 1)!, if = 2m + 1, (2j 1)!(2j + 1)! (j 1)! 2 which can easily be transforme to a simpler form 1 Recently P. Kirschenhofer an M. Weitzer, A number theoretic problem on the istribution of polynomials with boune roots, manuscript, 2014, confirme Conjecture 5.1 in case s = 1. They prove that v (1) /v(0) = (P (3) 2 1)/4, where P (x) are the Legenre polynomials. As a consequence they get log p (1) = (log 2/2) 2 + log( ) + O(log ), c.f. Section 6.

16 942 S. Akiyama an A. Pethő 2m ( ) m 1 2j 2 v m j j=m+1 = 2m+1 ( ) 2j m 2 m 1 j j=m+1 ( ) 1 2j, if = 2m, j ( ) 1 2j, if = 2m + 1. j (9) We are able to prove that these numbers are inee integers. Theorem 5.2. The numbers v / are o integers. To prove this theorem we nee a simple but important lemma. Lemma 5.2. Let a n, n = 0, 1,... be a purely perioic sequence of real numbers with perio length l. Assume that a i a i+1, 0 i l 2. Let c, be integers such that 0, c > 0, gc(c, l) = 1. If R is a non-negative integer then R a i i=0 R a ci+. i=0 Proof. There exist integers q, t such that R = ql + t, 0 t < l. Then R t 1 R a i = a i + i=0 i=0 i=t a i t 1 q 1 l 1 = a i + i=0 i=0 j=0 i=0 t 1 l 1 = a i + q a i. i=0 a jl+i Similarly R t 1 q 1 l 1 a ci+ = a ci+ + i=0 i=0 i=0 j=0 i=0 t 1 l 1 = a ci+ + q a ci+. i=0 a cjl+ci+ However, as c an l are coprime, if i runs through a complete resiue system moulo l, then ci + o the same. Hence l 1 i=0 a ci+ = l 1 i=0 a i an we only have to compare t 1 i=0 a i an t 1 i=0 a ci+. Since (c, l) = 1 the mapping i mo l ci+ mo l is injective for 0 i < l. Thus the monotonicity assumption implies that t 1 i=0 a i attains the minimum of all the sums of the shape i K a i for any subsets K Z/lZ of carinality t.

17 On the istribution of polynomials with boune roots, I 943 Proof of Theorem 5.2. Let f(x) = 2x 2 x. It is a perioic function with perio one. Moreover f(x) = 0 if x [0, 1/2) an f(x) = 1 if x [1/2, 1) thus f(x) is increasing in the interval [0, 1). Let p be an o prime an enote ν p (x) the largest exponent e such that p e ivies x. Let R be a positive integer. Then by Legenre s formula ν p ( R n=0 ( )) 2n = n = R n=0 R n=0 ( ) 2n n p j 2 p j ( ) n f p j. For j 1 set a j,n = f(n/p j ). The sequence a j,n is purely perioic with perio length p j. The increasing property of f in [0, 1) implies that a j,n is increasing for 0 n < p j. Using this notation (9) implies ( ) v ν p = ( m 1 k=0 m 1 a j,m+k+1 k=0 a j,k ) for = 2m. To get the same number of factors in numerator an enominator we extene the prouct in the enominator with the trivial factor ( 0 0) = 1. With the choice c = 1, = m + 1 the assumptions of Lemma 5.2 hol, thus all ifferences in the brackets are non-negative. Hence the enominator of v / has no o ivisors. Finally we prove that v / is o. This is true for = 2. Assume that it is true for = 2m. By (9) we have ( ) v+2 ν 2 +2 ( ) (( )) (( )) v 4m + 4 4m + 2 = ν ν 2 + ν 2 2m + 2 2m + 1 )) ν 2 (( 2m + 2 m + 1 ν 2 (( 2m m It follows from a classical (( result of E. E. Kummer [11], (see also the expository paper of Granville [9]) that ν 2n )) 2 n is exactly the number of ones in the binary expansion of n. As 2m + 2 = 2(m + 1) the number of one s in the binary expansions of 2m + 2 an m + 1 is equal, thus ν 2 (( 4m+4 2m+2 )). )) (( = 2m+2 )) ν2 m+1. Further 2m + 1 has exactly one more one s in its binary expansion than m, thus ν 2 (( 4m+2 2m+1)) = ν2 (( 2m m )) + 1. Hence ( ) v+2 ν 2 +2 ( ) v = ν 2 hols for even. As this number is zero for = 2, it is zero for all even. The proof of the case o is similar, therefore we left it for the reaer.

18 944 S. Akiyama an A. Pethő We finish this part with a conjecture, which is true for 5. Conjecture 5.2. We have v () 2 2 = 2 2( 1) ( 2 ). (10) Combining Theorems 2.1 an 4.1 this conjecture can be written in the form Conjecture 5.3. Let 1, 0 z j 1, j = 1,...,, D 0, = [ 2 z 1, 2 z 1 ] [0, 1] [ 2 z, 2 z ] [0, 1], R k (X) = X 2 y j X + z j, j = 1,..., an X = y 1 z 1... y z. Then we have 1! D 0, 1 j<k Res X (R j (X), R k (X)) X = 242 (2)! ( ) 2 S 2 (1, 1, 1/2). All our attempts to prove this conjecture faile. It seems unlikely to have a simple formula like (10) for other ratios v (s) 2 /v(0) 2 or v(s) 2+1 /v(0) 2+1 with 0 < s <, because we fin large prime factors. 6. Probability results. It is natural to ask: what is the probability p (0) that picking v E the corresponing polynomial P v is totally real, i.e. has only real roots? Let = r + 2s, where r, s are nonnegative integers. More generally we can ask the probability p (s) that picking v E such that the corresponing polynomial P v has signature (r, s)? Notice that in this setting we pick the coefficients of the polynomial! Of course we can express these probabilities with our former notations as p (s) = v(s). v By Theorem 5.1 these probabilities are rational numbers. The next natural question is the behavior of these numbers. Are they of similar size or is there significant ifference between them? We have the complicate, but exact formula (1) for v, but not for v (s) for s 0, thus not for p (s) either. However, in the case s = 0 Theorem 4.1 makes it possible to prove an accurate estimate for the size of p (0) estimate for p (/2), provie is even.. We exten this with a hypothetical Theorem 6.1. Let 2 be an integer. Then log ( p (0) ) log 2 = log + O(1). 8

19 On the istribution of polynomials with boune roots, I 945 Moreover, if is even an Conjecture 5.2 is true then log ( p (/2) ) 3 = log + O(1). (11) 8 Proof. We give a non-computer assiste proof 2. Using ieas of A. T. Fam [7], we have ( ) 2n = (2n)! n n! 2 = n k=1 (2k)(2k 1) k 2 = 2 2n for all positive integers n. First we prove (11). Let be even, say = 2m. As n k=1 2k 1 2k (12) we get p (m) 2m p (m) 2m = 22m2 m = v (m) p(0) 2m 2m 2m 2m j=m+1 ( 2j j ) 1 m ( ) 2j j by using (9) an Conjecture 5.2. Combining this with (12) we obtain p (m) 2m = 22m2 m = 2 m m = 2 m m Taking logarithms we get log ( p (m) 2m 2m j=m+1 2 2j j k=1 ( 2j ) m 2m 2j 1 2k 2k 1 j=m+1 ( ) j 4m 2j + 2 m 4m 2j + 1 ) m = m log 2 (j 1) log = m log m = 3 log m + O(1). 8 m j 2 2j k=1 2k 1 2k ( 2j ) 2m+1 j m 2j 1 ( ) j+1 2j 1. 2j ( 1 1 ) + 2j m ( 1 2m + 1 ) + m 2m + 1 j ( 2j 1 2j ) m+1 j ( ) 1 j log 1 + 4m 2j + 1 m 1 j + O(1) 2 The computation is not easy even by symbolic computation programs like Mathematica, because we have to fix properly the branches of complex functions.

20 946 S. Akiyama an A. Pethő Here we use the three term Taylor expansion of log(1 + x). Now we compute an asymptotic estimate for p (0) 2m. We start with p (0) 2m = p(m) 2m 2 2m(m 1) ( 2m m ) 1. Using (12) we get p (0) 2m = p(m) 2m 2 2m(m 1) 2 2m m k=1 ( 1 1 ) 1. 2k Taking logarithms, using (11) an the first term of the Taylor expansion for log(1 x) we obtain log ( p (0) ) = 2m 2 log m k=1 1 k 3 log m + O(1), 8 which proves the first assertion for even. Finally we are ealing with the case is o, say = 2m + 1. By (9) we have By applying (12) this implies p (0) 2m+1 = p(0) 2m 2 ( 2m m p (0) 2m+1 = p(0) 2m 2 2m 1 )( ) 1 4m m + 1 2m+1 k=m+1 Taking logarithms, using the expression for log p (0) 2m result. ( 1 1 ) 1. 2k an Taylor s formula we obtain the Theorem 6.1 means that the probability of picking a totally real polynomial of egree is asymptotically 2 2 /2. On the other han there are only [/2]+1 ifferent signatures, thus the polynomials are istribute among the ifferent signatures not equally, the totally real signature for example is a very rare one. Using the formula in Theorem 4.2 we compute v (1) for 100. After having these values we also compute v (1) /v(0), which happen to be integers. Moreover we foun that the growth rate of this integer sequence is monotonically increasing for 5 an lies in the interval [ , ]. We are not able to prove, but it seems that the growth rate is boune above, say by q. (We expect that q 6.) If this is true then q. Combining this with Theorem 6.1 we get v (1) /v(0) log p (1) log log q,

21 On the istribution of polynomials with boune roots, I 947 i.e., the signature s = 1 has a bit larger probability than the case s = 0, but it is still very rare comparing to the uniform istribution. On the other han we have some information on the totally complex case, which is the other extremum. For even we propose Conjecture 5.2, which is supporte by our computations. Assuming it we obtaine that in this case the frequency of totally complex polynomials is about 3/8. This number is much bigger than 2/, which woul be the probability if the polynomials were uniformly istribute among the ifferent signature classes. As alreay for = 6 the most frequent signature is (4, 1) one may expect that the peak of the frequency curve will move from the totally complex case. As we have no theoretical an only few computational support, we o not continue the speculation. 7. On the bounary of E (s). For the sake of completeness we now turn to the bounary of E (s) an prove a generalization of the result of I. Schur [12] as well as of A. T. Fam an J. S. Meitsch [6]. Although this escription is not as explicit as the cite ones, we inclue it, because we nee it in Part II. Theorem 7.1. Let 1 an r, s be non-negative integers such that r + 2s =. Then the bounary of the set E (s) is the union of finitely many algebraic surfaces. Proof. Let v 0 R be a bounary point of E (s). If one of its coorinates is 1 or 1 then v 0 satisfies the linear equation v 0 (1,..., 1) = 0 or v 0 (1, 1, 1,..., ( 1) ) = 0 respectively, where vu enotes the inner prouct of the vectors v an u. If the polynomial corresponing to v 0 has a non-real root lying on the unit circle then by [6] v 0 lies on a hypersurface. Assume in the sequel that v 0 is a bounary point of E (s), such that the roots of P v0 (X) lie insie the unit circle. It will be calle an inner bounary point. Then for any δ > 0 there exist v 1, v 2 R such that v 0 v 1, v 0 v 2 < δ, v 1 E (s) an v 2 E (s1), s 1 s. (13) We may assume without loss of generality that s > s 1, whence r < r 1 an it is fixe. Denote by α i, i = 0, 1, 2 the vectors of roots of P vi (X) orere such that the real roots come first, then the non-real ones such that the complex conjugates follow each other. Then there exists a coorinate, say 1 j s, such that α 1,r+2j 1 C \ R an α 2,r+2j 1 R (α ij enotes the j-t coorinate of the vector α i ). As v 1 is a real vector an the corresponing polynomial has a non-real root then its conjugate is a ifferent root of the polynomial. Thus ᾱ 1,r+2j 1 is a root of P v1 (x) an because of the orering of the roots ᾱ 1,r+2j 1 = α 1,r+2j. Notice that α 2,r+2j R hols too. The roots are continuous functions of the coefficients, thus for any ε > 0 there exists δ > 0 such that if for v 1, v 2 R the inequality (13) hols then α 1,r+2j 1 α 2,r+2j 1, ᾱ 1,r+2j 1 α 2,r+2j < ε.

22 948 S. Akiyama an A. Pethő This can only happen, if α 0,r+2j 1 = α 0,r+2j, i.e., P v0 (X) has a multiple real root. Thus the inner bounary points of E (s) lie on the surface 1 j<k r (x j x k ) r k=1 s (x 2 j y r+2k 1 x j + y r+2k ) = 0, where the y s has to be chosen such that 0 < y r+2k < 1 an y r+2k 1 < 2 y r+2k. These are obviously algebraic relations. In the opposite irection we prove that polynomials with multiple real roots lie on the inner bounary of ifferent signature boies. Inee, assume that P (X) has a multiple real root α < 1 an signature (r, s). Then for any ε > 0 we can fin real numbers 0 < δ 1, δ 2 < ε such that Consier the polynomials an α 2 + δ 2 2 < 1, α + δ 1, α + δ 2 < 1. P 1 (X) = P (X)(X (α + δ 1 ))(X (α + δ 1 ))/(X α) 2 P 2 (X) = P (X)(X (α + iδ 2 ))(X (α iδ 2 ))/(X α) 2, i = 1. Obviously P 1 (X), P 2 (X) E, they have ifferent signature an are arbitrary near to each other. This proves the claim an finishes the proof of the statement. Acknowlegments. We are very inebte to Professor Masanobu Kaneko, who rew our attention to Selberg integeral an suggeste us the formula (8). The first author move from Niigata University to the current aress in August The paper was written, when the secon author was visiting Niigata University as a long term research fellow of JSPS. Both of us wish to express our eep gratitue to all the staffs in Department of Mathematics, Niigata University an the support from JSPS, Grant in ai an Invitation Fellowship Program FY2011, L The secon author was partially supporte by the OTKA grant K an by the TÁMOP C-11/1/KONV project. The project is implemente through the New Hungary Development Plan, co-finance by the European Social Fun an the European Regional Development Fun. References [ 1 ] S. Akiyama an A. Pethő, On the istribution of polynomials with boune roots II. Polynomials with integer coefficients, submitte. [ 2 ] G. W. Anerson, A short proof of Selberg s generalize beta formula, Forum Math., 3 (1991), [ 3 ] G. E. Anrews, R. Askey an R. Roy, Special Functions, Encyclopeia Math. Appl., 71, Cambrige University Press, 1999.

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