On the inequalities in Hermite's theorem for a real polynomial to have real zeros
aa r X i v : . [ m a t h . C V ] S e p On the inequalities in Hermite’s theorem for a real polynomialto have real zeros
Mario DeFrancoSeptember 4, 2019
Abstract
We prove expressions for the inequalities in Hermite’s theorem which are conditionsfor a real polynomial to have real zeros. These expressions generalize the discriminantof a quadratic polynomial and the expression of J. Ma˘r´ık for a cubic polynomial. Weshow that the ( k + 1)-th minor of the Hermite matrix associated a polynomial p ( x )is equal to the k -th minor of another matrix we call E ( n ) times n k − and a simpleinteger. To prove this equivalence, we prove generalizations of the discriminant of apolynomial and analyze certain labeled directed graphs. To define this matrix E ( n )we define functions M ( m , m , n ) which are positive if the zeros of p ( x ) are positive. Let p ( x ) be monic polynomial of degree n with real coefficients p ( x ) = n X i =0 a i x n . The Hermite theorem (see [6]) describes how to determine the number of real zeros of p ( x ) by checking certain inequalities involving the coefficients a i . These inequalities aredefined in terms of the leading principal minors of a certain matrix which we denote by H ( n ). This paper proves that those minors are equal to the leading principal minorsof another matrix E ( n ) which we define in Section 3. The matrix E ( n ) thus providesan alternative way of expressing these inequalities.We now state the results more specifically. Let z , ..., z n denote the zeros of monicpolynomial p ( x ) of degree n with real coefficients. For integer k ≥
0, denote thepower-sum functions by p k ( z , ..., z n ) = n X i =1 z ki . The Hermite matrix H ( n ) associated to p ( x ) is the n × n matrix whose entries H ( n ) i,j are H ( n ) i,j = p i + j − ( z , ..., z n ) . et ∆ k ( H ( n )) denote the determinant of the upper left k × k submatrix of H ( n ). Thisdeterminant is known as the k -th leading principal minor with∆ ( H ( n )) = 1 and ∆ n ( H ( n )) = det( H ( n )) . Hermite’s theorem then says that the zeros z , ..., z n are all real if and only if∆ k ( H ( n )) > ≤ k ≤ n . Since power-sum functions are symmetric functions of the z i , they maybe expressed in terms of the elementary symmetric functions e k = e k ( z , ..., z n ) = X ≤ i <...
For integer k ≥ , denote the power-sum functions by p k = n X i =1 z ki . For integer k ≥ , denote the elementary-symmetric functions by e k = X ≤ i <...
Let λ = ( λ , λ , ..., λ k ) be a k -tuple of integers with ≤ λ i < λ i +1 .Define H ( λ ; n ) to be the k × k matrix with entries H ( λ ; n ) i,j = p λ i + j − . For λ = (0 , , , ..., k − , denote the Hermite matrix H ( n ) by H ( n ) = H ( λ ; n ) . We will prove formulas for the leading principal minors ∆ k of H ( λ ; n ). Definition 3.
Let F be an infinite matrix with entries F i,j . Let F k denote the k × k submatrix with entries ( F k ) i,j for ≤ i, j ≤ k . The denote ∆ k ( F ) = det( F k ) . Definition 4.
Let λ be a k -tuple of integers λ = ( λ , ..., λ k ) . Let x , ..., x k be k indeterminates. Then let V ( x , x , ..., x k ; λ ) denote the k × k matrixwith entries V ( x , x , ..., x k ; λ ) i,j = x λ i j . Also denote D ( x , ..., x k ) = det( V ( x , ..., x k ; (0 , , , ..., k − Y ≤ i Let C ( k, n ) denote the set of all subsets of order k of the set { , , ..., n } .For b ∈ C ( k, n ) b = { b , ..., b k } with b i < b i +1 , let z ( b ) denote the k -tuple z ( b ) = ( z b , ..., z b k ) . For ≤ i ≤ k , let ˆ b i ∈ C ( k − , n ) be ˆ b i = ( b , ..., b i − , b i +1 , ..., b k ) . heorem 1. Let λ be a k -tuple of integers λ = ( λ , ..., λ k ) with ≤ λ i ≤ λ i +1 . Then det( H ( λ ; n )) = X b ∈ C ( k,n ) S ( z ( b ); λ ) D ( z ( b )) Proof. We use induction on k . The statement is true for k = 1 because D ( z i ) = 1 , S ( z i ; ( λ )) = z λ i , and det( H ( λ ; n )) = p λ . Assume it is true for some k ≥ 1. Let λ = ( λ , ..., λ k +1 ). Then we calculatedet( H ( λ ; n ))by expanding along the rightmost column of the matrix H ( λ ; n ). Let ˆ λ i denote the k -tuple ˆ λ i = ( λ , ..., λ i − , λ i +1 , ..., λ k +1 ) . Then det( H ( λ ; n )) = k +1 X i =1 ( − k + i det( H (ˆ λ i ; n )) p λ i + k = X b ∈ C ( k,n ) D ( z ( b )) k +1 X i =1 ( − k + i det( V ( z ( b ); ˆ λ i )) D ( z ( b )) p λ i + k by the induction hypothesis. Using the definition of p λ i + k we re-write the last line ofthe above equation as= X b ∈ C ( k,n ) D ( z ( b )) k +1 X i =1 ( − k + i det( V ( z ( b ); ˆ λ i )) D ( z ( b )) n X j =1 z λ i + kj = X b ∈ C ( k,n ) D ( z ( b )) n X j =1 z kj k +1 X i =1 ( − k + i det( V ( z ( b ); ˆ λ i )) D ( z ( b )) z λ i j = X b ∈ C ( k,n ) D ( z ( b )) n X j =1 z kj det( V (( z ( b ) , z j ); λ )) D ( z ( b ))where ( z ( b ) , z j ) denotes the ( k + 1)-tuple( z ( b ) , z j ) = ( z b , z b , ..., z b k , z j ) . Continuing, we apply the definition of the Schur polynomial and obtain= X b ∈ C ( k,n ) D ( z ( b )) n X j =1 z kj S (( z ( b ) , z j ); λ ) D ( z ( b ) , z j ) D ( z ( b )) . o the above expression we apply D ( z ( b ) , z j ) D ( z ( b )) = k Y i =1 ( z j − z b i ) , which yields X b ∈ C ( k,n ) D ( z ( b )) n X j =1 z kj S (( z ( b ) , z j ); λ ) k Y i =1 ( z j − z b i )= X b ′ ∈ C ( k +1 ,n ) S ( z ( b ′ ); λ ) k +1 X i =1 z kb ′ i D ( z ( ˆ b ′ i )) k +1 Y l =1 , = i ( z b ′ i − z b ′ l ) . Therefore we must show that for b ∈ C ( k + 1 , n ) k +1 X i =1 z kb i D ( z (ˆ b i )) k +1 Y l =1 , = i ( z b i − z b l ) = D ( z ( b )) This follows from Lemma 1 and completes the proof. Lemma 1. For integer k ≥ and indeterminates x , ..., x k +1 , k +1 X l =1 ( − l − x kl Y ≤ i Expand the product on the right side of the lemma statement into monomialterms, treating the x i as non-commuting variables. Each such term m is indexed by aset E ( m ) of k ( k +1)2 ordered pairs: if in the factor( x i − x j )the x i contributes to m , then let ( i, j ) ∈ E ( m ); otherwise ( j, i ) ∈ E ( m ). Thus each m corresponds to a directed graph G ( m ) whose vertex set is V = { , , ..., k + 1 } and whose edge set is E ( m ). We say that ( i, j ) is an outgoing edge from the vertex i and an incoming edge to the vertex j .We claim that for every such G , either there is some vertex in G ( m ) with all outgoingedges or there is a 3-cycle in G ( m ). A “cycle” means a directed cycle and a 3-cycle is acycle with exactly 3 edges. We use induction on k . This statement is true for k = 1 , k ≥ 2. Note in G ( m ) every vertex has exactly k edges.We first show that cycles exist in G ( m ) if there is no vertex with all outgoing edges.Suppose there is no vertex in G ( m ) with all outgoing edges. If there exists a vertex v with all incoming edges, then by the induction hypothesis, the subgraph G ( m ) \ v haseither a vertex v ′ with all outgoing edges or a 3-cycle. If there isa 3-cycle, then we are one. Therefore assume there is such a v ′ with all outgoing edges in G ( m ) \ v . Thensince v has all incoming edges by assumption, we have the edge ( v ′ , v ) in G ( m ). Thus v ′ has all outgoing edges in G ( m ), contradicting the assumption of no such vertex in G ( m ). Therefore if G ( m ) has no vertex with all outgoing edges, then it has no vertexwith all incoming edges.Thus we can assume that no vertex in G ( m ) has all outgoing edges and that everyvertex has at least one incoming and at least one outgoing edge. Therefore there existssome cycle C in G ( m ) with at least 3 edges. Let C consist of the vertices v , ..., v n withedges ( v i , v i +1 ) and ( v n , v ). Then one of the triples { v , v i , v i +1 } for 2 ≤ i ≤ n − v , v i ) for 2 ≤ i ≤ n − 1. But then we havethe triple { v , v n − , v n } which would then be a 3-cycle. This proves the induction step.Now we can prove the lemma. Allowing the x i to commute, the term( − l − x kl Y ≤ i Let P ( k, n ) denote the set of all k -tuples bb = ( b , ..., b k ) such that b i = b j for i = j and b i ∈ { , ..., n } . For a finite non-increasing sequence ofnon-negative integers d = { d , d , ..., d l } , define monomial( d, n ) = X b ∈ P ( l,n ) l Y i =1 z d i b i . Define monomial ( m , m , n ) = monomial( { , , ..., , , , ..., } , n ) m ! m ! where there are m m emma 2. For integers m, k ≥ , e m e m + k = m X i =0 (cid:18) k + 2 ii (cid:19) monomial ( m − i, k + 2 i ; n ) . Proof. As a function of the z i , e m e m + k (2)is a symmetric polynomial. Since each z i appears with exponent at most 1 in eachelementary-symmetric function, any product of two elementary symmetric functionsis then a linear combination of the functions monomial ( m , m ; n ). Expanding theproduct (2) into monomials, we get terms of the form z j ...z j m − i z l ...z k +2 i where 0 ≤ i ≤ m ; to see this, one z j h factor comes from e m and another z j h factorcomes from e m + k . That is, the term from e m and the term from e m + k overlap in m − i indeterminates. Then the remaining indeterminates coming from e m are distinct fromthe remaining ones coming from e m + k ( z ; n ). The total number of these indeterminatesthat do not overlap is i + ( m + k − ( m − i )) = k + 2 i. and i of these indeterminates come from e m . Thus there are (cid:18) k + 2 ii (cid:19) ways to makesuch a product of two terms. This proves the lemma. Lemma 3. For integers m, k ≥ , monomial ( m, k, n ) = m X i =0 ( − i k + 2 ii (cid:18) k + i − i − (cid:19) e m − i e m + k + i Proof. Solving for monomial ( m, k ) using the system of equations given by Lemma 2gives monomial ( m, k, n ) = m X i =0 c i ( k ) e m − i e m + k + i where c ( k ) = 1 c i ( k ) = − i X j =1 (cid:18) k + 2 jj (cid:19) c i − j ( k + 2 j ) for i ≥ . Then for i ≥ c i ( k ) = ( − i k + 2 ii (cid:18) k + i − i − (cid:19) . e prove this by induction on i . It is true for i = 1. Assume it is true for some i ≥ − i +1 k + 2 i + 2 i + 1 (cid:18) k + ii (cid:19) + i +1 X j =1 (cid:18) k + 2 jj (cid:19) c i +1 − j ( k + 2 j )=( k + 2 i + 2) i +1 X j =0 ( − j ( k + 2 i − − j ) i j !( i + 1 − j )!=( k + 2 i + 2) 1 i ! ( ddt ) i t k + i (1 − t ) i +1 | t =1 =0 . This proves the induction step and the lemma.Now we define the functions M ( m , m , n ). Definition 7. Let d denote the sequence d = { , , ..., , , , ..., , } where there are m m d denote the sequence d = { , , ..., , , , ..., } where there are m − m + 2 M ( m , m , n ) = monomial( d , n ) − monomial( d , n ) . Lemma 4. M ( m , m , n ) = m ! m !( n − m − m ) m X i =0 ( − i ( m + 2 i )( m + i − i ! m ! e m − i e m + m + i − ( m − m + 2)! m − X i =0 ( − i ( m + 2 + 2 i )( m + i + 1)! i !( m + 2)! e m − − i e m + m +1+ i Proof. This follows from applying the definitions and Lemma 3. Definition 8. For a finite non-increasing sequence of non-negative integers d = { d , d , ..., d l } ,define monomial inc ( d, i, j ; n ) = X b ∈ P ( l,n ); i,j / ∈ b l Y h =1 z d h b h . Define incomplete elementary-symmetric functions e inc ( k ; i ; n ) = X ≤ l <... Let d denote the sequence d = { , , ..., , , , ..., } where there are m − m M ( m , m , n ) = X ≤ i Applying the definitions we obtain M ( m , m , n ) = X b ∈ P ( m + m +1 ,n ) z b ...z b m z b m ...z b m m z b m m − X b ∈ P ( m + m +1 ,n ) z b ...z b m − z b m z b m ...z b m m z b m m . Now we partition the set P ( m + m + 1 , n ) into pairs { b, b ′ } where for any b ∈ P ( m + m + 1 , n ), we let b ′ be obtained from b by switching the elements b m and b m + m +1 . Then we get M ( m , m , n ) = X { b,b ′ } z b ...z b m − z b m ...z b m m ( z b m + z b m m − z b m z b m m )= X { b,b ′ } z b ...z b m − z b m ...z b m m ( z b m − z b m m ) = X ≤ i Define the infinite matrix E ( n ) with entries E ( n ) r,s E ( n ) r,s = ( r − s − X ≤ i By Lemma 5, it is sufficient to prove m X i =0 (cid:18) mi (cid:19) ( m + k )!( i + k )! ( m − i )!( k + 2 i )!monomial ( m − i, k + 2 i, n ) = ( m !)( m + k )! e m e m + k . By Lemma 2, the coefficient of monomial ( m − i, k + 2 i, n ) in e m e m + k is (cid:18) k + 2 ii (cid:19) .Then m !( m + k )!( m − i )!( k + 2 i )! (cid:18) k + 2 ii (cid:19) = (cid:18) mi (cid:19) ( m + k )!( i + k )! . This completes the proof. Theorem 2. E ( n ) m +1 ,m + k +1 ( n ) = m ! k !( n − m − k ) e m e m + k + n m − X i =0 A i e i e m + k − i for some numbers A i .Proof. Using Lemma 3 and the definition of M ( m, k, n ) we have M ( m, k, n ) = m ! k !( n − m − k ) e m e m + k + m − X i =0 ( m ! k !( n − m − k )( − m − i k + 2 m − im − i (cid:18) k + m − i − k (cid:19) − ( m − k + 2)!( − m − i − k + 2 m − im − i − (cid:18) k + m − ik + 2 (cid:19) ) e m − i e m + k − i Re-indexing i m − i and applying Lemma 6, we get that the coefficient of e m − i e m + k − i in E ( n ) m +1 ,m + k +1 that is constant in n is( k + 2 − i + 2 m ) m X j =0 ( − m − i − j +1 (cid:18) mj (cid:19) ( m + k )!( j + k )! × (( − − j − k − m )( m − j + 1)! ( m + k + j − i )!( m − j − i + 1)! + ( m − j )! ( m + k + j − i + 1)!( m − j − i )! ) . (3)We must show that the above sum is 0 for 0 ≤ i ≤ m − 1. We simplify line (3) toobtain m !( m + k )! j !( j + k )! i ( i − m − k − m + k + j − i ) j + k − . his shows that the sum is 0 for i = 0. We this must prove that m X j =0 ( − j ( m + k + j − i ) j + k − j !( j + k )!has a factor of ( i − h ) for 1 ≤ h ≤ m − 1. This follows from the identity m X j =0 ( − j ( m + k + j − i ) j ( m − i + 1) j j !( j + k )! = ( − m ( Q mh =1 ( i − h )) ( Q mh =1 ( i − (2 m + k ) + h )) m !( m + k )! . To prove this we use induction on m . It is trie for m = 0. Assume it is true for m ≥ − m + k − i + 1)( m − i + 1) j ( j + k ) = ( m − j − i + 1)( m + k + j − i + 1) j ( j + k ) . E ( n ) Definition 10. For integer k ≥ and indeterminates x and x , define f k ( x , x ) f k ( x , x ) = k X j =0 x k − j x j . For a pair of integers b = ( b , b ) , we also use the notation f k ( b ) = f k ( z b , z b ) . Lemma 7. e inc ( k ; i ; n ) = k X h =0 ( − h z hi e k − h (4) e inc ( i, j ; n ) = k X h =0 ( − h f j ( z i , z j ) e k − h (5) Proof. We have e inc ( k ; i ; n ) = e k − z i e inc ( k − i ; n ) . This implies equation (4). We also have e inc ( k ; i, j ; n ) = e k − z i e inc ( k ; i ; n ) − z j e inc ( k ; j ; n ) + z i z j e inc ( k − i, j ; n ) . The above equation combined with (4) implies (5).We will use the following definitions in Theorem 3. efinition 11. Let Pairs( k, n ) denote the set of elements β where each element β isa multi-set of k pairs of integers: β = { β (1) , β (2) , ..., β ( k ) } (6) where β ( i ) = { β ( i, , β ( i, } such that β ( i, < β ( i, and each β ( i, j ) ∈ { , , ..., n } . Note that β is a set: eventhough we have used an ordering of the pairs β (1) , ..., β ( k ) in the notation of (6) ,another ordering would result in the same element β .Let | β | denote | β | = distinct numbers that appear as β ( i, or β ( i, . For β ∈ Pairs( k, n ) , define D ( β ) = k Y i =1 ( z β ( i, − z β ( i, ) Definition 12. For β ∈ Pairs( k, n ) , define the k × k matrix R ( β ) with entries R ( β ) u,v by R ( β ) u,v = e inc ( u ; β ( v, , β ( v, n ) . Define the k × k matrix R ( β ) with entries R ( β ) u,v by R ( β ) u,v = f u ( z β ( v, , z β ( v, ) . Define the k × k matrix R ( β ) with entries R ( β ) u,v by R ( β ) u,v = z uβ ( v, − z uβ ( v, . Theorem 3. ∆ k ( E ( n )) = ( k Y i =1 i !) X β ∈ Pairs( k,n ) det( R ( β )) Proof. We write the definition of E ( n ) u,v as E ( n ) u,v = X β ∈ B (1 ,n ) e inc ( u ; β (1 , , β (1 , e inc ( v ; β (1 , , β (1 , z β (1 , − z β (1 , ) . We first prove that∆ k ( E ( n )) = ( k Y i =1 i !) X β ∈ Pairs( k,n ) D ( β ) det( R ( β )) . (7)When calculating the determinant ∆ k ( E ( n )), we write as a sum over Pairs( k, n ):∆ k ( E ( n )) = ( k Y i =1 i !) X β ∈ Pairs( k,n ) D ( β ) X τ ∈ S k X σ ∈ S k sgn( σ ) k Y u =1 e inc ( u, β ( τ ( u )) e inc ( σ ( u ) , β ( τ ( u )) ! ow X τ ∈ S k X σ ∈ S k sgn( σ ) k Y u =1 e inc ( u, β ( τ ( u )) e inc ( σ ( u ) , β ( τ ( u ))= X τ ∈ S k X σ ∈ S k sgn( σ ) k Y u =1 e inc ( u, β ( τ ( u )) e inc ( u, β ( τ ( σ − ( u )))= X σ ∈ S k X σ ∈ S k sgn( σ )sgn( σ ) k Y u =1 e inc ( u, β ( σ ( u )) e inc ( u, β ( σ ( u ))where σ = τ, σ = τ σ − . Thus continuing we get= X σ ′ ∈ S k sgn( σ ′ ) k Y u =1 e inc ( u, β ( σ ′ ( u ))) ! = det( R ( β )) . This proves equation (7).We next prove that det( R ( β )) = det( R ( β )) . As a sum over S k , each of the k ! terms in det( R ) is of the formsgn( σ ) k Y u =1 e inc ( u, β ( σ ( u ))) (8)for some σ ∈ S k . We apply Lemma 7 to write (8) assgn( σ ) X g h ( g ) k Y u =1 f g u ( β ( σ ( u ))where the sum is over all k -tuples gg = ( g , ..., g k )such that 0 ≤ g i ≤ i − 1; and h ( g ) is some product of the elementary symmetricfunctions e ( i ). We claim that in the sum over S k , each term of the formsgn( σ ) h ( g ) f g u ( β ( σ ( u ))is canceled out by another unless g = (0 , , , ..., k − . (9) e prove the claim now. For a given g , take the smallest pair of indices ( i, j ) in thelexicographic order such that g i = g j and pair the same term arising from the termwith σ ′ , where σ ′ ( i ) = σ ( j ) and σ ′ ( j ) = σ ( i ) and σ ′ ( m ) = σ ( m ) otherwise.Thus the only terms that remain are those with g i all distinct. The only such g is givenby (9) for which h ( g ) = ( − k . Thereforedet( R ( β )) = ( X σ ∈ S k sgn( σ ) k Y u =1 f u − ( β ( σ ( u )))) = det( R ( β )) . Thus we have shown that∆ k ( E ( n )) = X β ∈ Pairs( k,n ) D ( β ) det( R ( β )) . Using f u − ( x , x ) = x u − x u x − x , we write ( D ( β ) det( R ( β ))) = ( X σ ∈ S k k Y u =1 ( z σ ( u ) β ( u, − z σ ( u ) β ( u, )) = det( R ( β )) . This completes the proof. We prove Theorem 4. For integer k ≥ , we have X β ∈ Pairs( k,n ) det( R ( β )) = n k − X b ∈ C ( k +1 ,n ) D ( b ) . We first show how to index the terms on the left side. To do this, we define a setof functions S ( β ). Definition 13. For β ∈ Pairs( k, n ) , let S ( β ) denote the set of k functions s suchthat the domain of each s is the set of k pairs β ( i ) , ≤ i ≤ k and such that the output s ( β ( i )) on the i -th pair of β is s ( β ( i )) ∈ { β ( i, , β ( i, } . hat is, a function s chooses one element in each pair of β . For s ∈ S ( β ) , define sgn( s ) = ( − N where N = { i : s ( β ( i )) = max( β ( i, , β ( i, } . For s , s ∈ S ( β ) , define sgn( s , s ) = ( − N where N = { i : s ( β ( i )) = s ( β ( i )) } . Then sgn( s )sgn( s ) = sgn( s , s ) . Define z ( β ) z ( β ) = (( z β (1 , , z β (1 , ) , ..., ( z β ( k, , z β ( k, )) . Definition 14. Let V = V ( z j , z j , ..., z j k ; { , , ..., k } ) denote the k × k matrix withentries V u,v V u,v = z vj u . Let d ( j , j , .., j k ) denote d ( j , j , .., j k ) = det( V ) . Remark 1. Suppose for a given β we have an ordering β = { β (1) , ..., β ( k ) } . Wedenote d ( s ( β )) d ( s ( β )) = d ( s ( β (1)) , s ( β (2)) , ..., s ( β ( k ))) . Note that the expression d ( s ( β )) depends up to sign on a choice of ordering on β . Butfor s , s ∈ S ( β ), the expression d ( s ( β )) d ( s ( β ))is independent of a choice of ordering, because for σ in S k d ( s ( β ( σ (1))) , ..., s ( β ( σ ( k )))) d ( s ( β ( σ (1))) , ..., s ( β ( σ ( k ))))= sgn( σ ) d ( s ( β (1)) , ..., s ( β ( k ))) d ( s ( β (1)) , ..., s ( β ( k )))= d ( s ( β (1)) , ..., s ( β ( k ))) d ( s ( β (1)) ..., s ( β ( k ))) . (cid:3) Applying the definition of R ( β ), we getdet( R ( β )) = ( X s ∈ S ( β ) sgn( s ) d ( s ( β ))) . Then X β ∈ Pairs( k,n ) det( R ( β )) = X β ∈ Pairs( k,n ) ( X s ∈ S ( β ) sgn( s ) d ( s ( β ))) = X β ∈ Pairs( k,n ) X s ,s ∈ S ( β ) sgn( s , s ) d ( s ( β )) d ( s ( β ))The terms in the sum on the right are thus indexed by ordered triples( β, s , s )for β ∈ Pairs( k, n ) and s , s ∈ S ( β ). As discussed above, each term d ( s ( β )) d ( s ( β ))is well-defined independent of an ordering on β . efinition 15. Given such a triple ( β, s , s ) , choose an ordering on β and let I ( β, s , s ) be the set of indices I ( β, s , s ) = { i : s ( β ( i )) = s ( β ( i )) } . Theorem 5. X β ∈ Pairs( k,n ) det( R ( β )) = X ( β,s ,s ): | I ( β,s ,s ) |≤ sgn( s , s ) d ( s ( β )) d ( s ( β )) . Proof. Consider those terms with triples for which | I ( β, s ( β , s ) | ≥ 2. We define abijection to show that all such terms cancel. Let i , i ∈ I ( β, s , s ) be the indices suchthat s ( β ( i )) and s ( β ( i )) are the two smallest numbers in the set { s ( β ( i )) : i ∈ I ( β, s , s ) } . Note that the elements β ( i ) and β ( i ) of β do not depend on the ordering on β . Let β ′ = { β ′ β ′ (1) , ..., β ′ ( k ) } ∈ Pairs( k, n ) be defined by β ′ ( i ) = β ( i ) if i = i , i { s ( β ( i )) , s ( β ( i )) } if i = i { s ( β ( i )) , s ( β ( i )) } if i = i . Define s ′ , s ′ ∈ S ( β ′ ) by s ′ ( β ′ ( i )) = s ( β ( i ))for all 1 ≤ i ≤ k and s ′ ( β ′ ( i )) = s ( β ( i )) if i = i , i s ( β ( i )) if i = i s ( β ( i )) if i = i . This completes the definition of the bijection ( β, s , s ) ( β ′ , s ′ , s ′ ). We next showthat sgn( s ′ , s ′ ) d ( s ′ ( β ′ )) d ( s ′ ( β ′ )) = − sgn( s , s ) d ( s ( β )) d ( s ( β )) . By construction I ( β, s , s ) = I ( β ′ , s ′ , s ′ )so sgn( s ′ , s ′ ) = sgn( s , s ) . Without loss of generality assume i < i . Then d ( s ′ ( β ′ )) d ( s ′ ( β ′ ))= d ( s ′ ( β ′ (1)) , ..., s ′ ( β ′ ( k ))) d ( s ′ ( β ′ (1)) , ..., s ′ ( β ′ ( k )))= d ( s ( β (1)) , ..., s ( β ( k ))) × d ( s ( β (1)) , ..., s ( β ( i − , s ( β ( i )) , s ( β ( i + 1)) , ..., s ( β ( i − , s ( β ( i )) , s ( β ( i + 1)) , ..., s ( β ( k )))= − d ( s ( β (1)) , ..., s ( β ( k ))) × d ( s ( β (1)) , ..., s ( β ( i − , s ( β ( i )) , s ( β ( i + 1)) , ..., s ( β ( i − , s ( β ( i )) , s ( β ( i + 1)) , ..., s ( β ( k )))= − d ( s ( β )) d ( s ( β )) . e next show how an element β ∈ Pairs( k, n ) corresponds to a graph G ( β ), andhow s ∈ S ( β ) directs the edges of G ( β ). Definition 16. For β ∈ Pairs( k, n ) we define a graph G ( β ) . The set of vertices V ( G ( β )) of G ( β ) is the set of distinct numbers that appear as a β ( i, or β ( i, . Thus | V ( G ( β )) | = | β | The edge set E ( G ( β )) is E ( G ( β )) = { ( β ( i, , β ( i, ≤ i ≤ k } . Given an s ∈ S ( β ) , for each i write β ( i ) = { β i , s ( β i ) } . Then we say that the edge ( s ( β i ) , β i ) is an outgoing edge from s ( β i ) and an incomingedge to β i . Denote the resulting directed graph by G ( β, s ) . For ≤ h ≤ k − , let G ( k, h ) denote the set of graphs G ( β ) such that | V ( G ( β )) | = k + h + 1 . Lemma 8. For integer k ≥ , X β ∈ Pairs( k,n ) det( R ( β )) = X ( β,s ,s ) sgn( s , s ) d ( s ( β )) d ( s ( β )) where the sum on the right is over all triples ( β, s , s ) such that β has no repeatedpairs and:1. | I ( β, s , s ) | ≤ .2. G ( β, s ) and G ( β, s ) each have no vertex with more than one outgoing edge.3. G ( β ) has no cycles.Proof. If β has a repeated pair,that β ( i ) = β ( j ) for some i = j , thendet( R ( β )) . Thus we may assume β has no repeated pairs.Statement 1 was proven above. Statement 2 follows from the fact that if G ( β, s )has a vertex with more than one outgoing edge, then the determinant d ( s ( β )) has arepeated index and thus equals 0. To prove statement 3, suppose that G ( β ) has acycle. By statement 2, for d ( s ( β )) to be non-zero, s must make each cycle in G ( β ) adirected cycle. Thus we take the cycle C of G ( β ) whose vertex set is smallest in thelexicographic ordering and write the edges as( v , v ) , ( v , v ) , ...., ( v m − , v m ) , ( v m , v )for some m ≥ 2. Without loss of generality we may assume s (( v i , v i +1 )) = v i and s (( v m , v )) = v m . We then match the triple ( β, s , s ) to ( β, s ′ , s ) where s ′ is thesame as s except that it reverses the cycle C . Thussgn( s ) = ( − m sgn( s ′ ) hen in s ( β ) we have the subsequence { v , v , ..., v m } and in s ′ ( β ) we have the subsequence { v , v , ..., v m , v } . Thus d ( s ( β )) = ( − m − d ( s ′ ( β )) . Thus the terms from I ( β, s , s ) and I ( β, s ′ , s ) cancel. This proves the statement3. Now we index the terms on the right side of Theorem 4. Lemma 9. For b ∈ C ( k + 1 , n ) , D ( b ) = k +1 X i =1 ( − k +1 − i d (ˆ b i ) ! Proof. This follows from Lemma 1 or by taking the definition of D ( b ) as the determi-nant of a Vandermonde matrix and expanding along the row with all 1’s.Thus X b ∈ C ( k +1 ,n ) D ( b ) = X b ∈ C ( k +1 ,n ) k +1 X i =1 d (ˆ b i ) + k +1 X j =1 k +1 X i =1 , = j ( − i + j d (ˆ b i ) d (ˆ b j ) (10)We consider the sum X b ∈ C ( k +1 ,n ) k +1 X j =1 k +1 X i =1 , = j ( − i + j d (ˆ b i ) d (ˆ b j ) . This sum is indexed by ordered triples ( b, i, j )for 1 ≤ i = j ≤ k + 1. In the sum from (5), we map these terms to sets of terms in thesum X ( β,s ,s ): | I ( β,s ,s ) | =1 sgn( s , s ) d ( s ( β )) d ( s ( β )) . We thus associate to each such triple ( b, i, j ) a set of triples ( β, s , s ): Definition 17. Let b ∈ C ( k + 1 , n ) and ≤ i = j ≤ k + 1 . Given a triple ( b, i, j ) ,define G ( b, i, j ) to be the set of all ( β, s , s ) such that s ( β ) as a set is equal to ˆ b i nd s ( β ) as a set is equal to ˆ b j and d ( s ( β )) d ( s ( β )) = 0 . Define the subset G ( b, i, j ; h ) ⊂ G ( b, i, j ) to consist of all triples ( β, s , s ) such that | β | = k + 1 + h. Thus G ( b, i, j ; h ) = G ( b, i, j ) ∩ G ( k, h ) . Given a pair ( b, i ) , define G ( b, i ) to be the set of all ( β, s ) with s ∈ S ( β ) such thatas sets s ( β ) = ˆ b i and d ( s ( β )) = 0 . Define the subset G ( b, i ; h ) ⊂ G ( b, i ) to consist of those pairs ( β, s ) such that | β | = k + 1 + h. We characterize the possible ( β, s , s ) and ( β, s )that can appear in G ( b, i, j ; h ) and G ( b, i ; h ), respectively. We identify β with the graph G ( β ) in the following lemma. Lemma 10. Let b ∈ C ( k + 1 , n ) . Then G ( b, i, j ; h ) is the set of all ( β, s , s ) such thatfor the graphs G ( β ) :1. b , ..., b k +1 are vertices of G ( β ) , and | V ( G ( β )) | = h + k + 1 .2. There are exactly k edges.3. There are exactly h + 1 components. The vertices b i and b j are in the same com-ponent, and there is exactly one non- b vertex in each of the remaining h components.Here, a “non- b ” vertex means a number not in b .4. Each component has at least two vertices.5. There are no cycles, loops, or multiple edges.6. s and s direct each component of G ( β ) such that the unlabeled vertex corre-sponds to the root vertex of a directed tree, and on the component with b i and b j , s makes b i the root and s makes b j the root.Proof. Statement 1 follows from the definition of h and the fact that as sets s ( β ) ∪ s ( β ) = b. Statement 2 follows from the requirement that β ∈ Pairs( k, n ). That there are nocycles follows from Lemma 8, and there are no loops or multiple edges by construction.Statement 4 also follows by construction. The fact that an s must pick a vertex fromeach edge and that an s cannot pick the same vertex from two edges (or else d ( s ( β )) = 0by Lemma 8) means that s makes any component into a rooted tree, such that theroot is not in d ( s ( β )). Thus a component can have at most one non- b vertex, and s and s must agree on all components that have a non- b vertex. Thus b i and b j mustbe the same component, and s makes b i the root of this component and s makes b j the root. This proves statement 3 and 6. efinition 18. For integer j ≥ let ( x ) j denote the falling factorial ( x ) j = j Y i =1 ( x − i + 1) . Definition 19. Let b = { , , ..., k + 1 } and take a graph G in G ( b, , h ) . Now make a graph G ′ by taking every non- b vertex in G an unlabeled vertex. The setof such G ′ depends only on k and h ; let A ( k, h ) denote this set. Lemma 11. For ≤ h ≤ k − and any b ∈ C ( k + 1 , n ) , |G ( b, i, j ; h ) | = |A ( k, h ) | ( n − k − h . Proof. We can re-name the elements of b to correspond to { , , ..., k + 1 } with b i cor-responding to 1 and b j corresponding to 2. Then any graph in G ( b, i, j ; h ) correspondsto taking a graph in A ( k, h ) and labeling the unlabeled vertices with numbers chosenfrom the set { , , ..., n } − b . There are( n − k − n − k − ... ( n − k − h )ways to do this because each unlabeled vertex appears in a component with a vertexin b and we may order these components. Lemma 12. X β ∈ Pairs( k,n ) , | I ( β,s ,s ) | =1 sgn( s , s ) d ( s ( β )) d ( s ( β ))= k − X h =0 |A ( k, h ) | ( n − k − h ! X b ∈ C ( k +1 ,n ) k +1 X j =1 k +1 X i =1 , = j ( − i + j d (ˆ b i ) d (ˆ b j ) Proof. Each ( β, s , s ) with | I ( β, s , s ) | = 1 is in a unique G ( b, i, j ) with i = j : thesets s ( β ) and s ( β ) determine b , and b i is the only element of b not in s ( β ) and b j isthe only element of b not in s ( β ). Thus X β ∈ Pairs( k,n ) , | I ( β,s ,s ) | =1 | sgn( s , s ) d ( s ( β )) d ( s ( β )) | = X b ∈ C ( k +1 ,n ) k +1 X j =1 k +1 X i =1 , = j | G ( b, i, j ) || ( − i + j d (ˆ b i ) d (ˆ b j ) | = k − X h =0 |A ( k, h ) | ( n − k − h ! X b ∈ C ( k +1 ,n ) k +1 X j =1 k +1 X i =1 , = j | ( − i + j d (ˆ b i ) d (ˆ b j ) | by Lemma 11. All we have to check now is that the signs agree. That is, we show thatfor ( β, s , s ) ∈ G ( b, i, j )sgn( s , s ) d ( s ( β )) d ( s ( β )) = ( − i + j d (ˆ b i ) d (ˆ b j ) . ince G ( β ) has no cycles and vertices b i and b j are in the same component C there isa unique path from b i to b j whose edges we write as( b i , v ) , ( v , v ) , ..., ( v m − , v m ) , ( v m , b j ) . Then s and s agree on all other pairs of β andsgn( s , s ) = ( − m +1 . Order β such that β (1) = { b , v } , β ( m + 1) = { v m , b j } , β ( i + 1) = { v i , v i +1 } for 1 ≤ i ≤ m. Thus d ( s ( β )) = d ( v , v , ..., v m , b j , { l } ) ,d ( s ( β )) = d ( b i , v , v , ..., v m , { l } )= ( − m d ( v , v , ..., v m , b i , { l } ) . where l indicates some sequence. Without loss of generality assume i < j . Then d (ˆ b i ) = d ( b , ..., b i − , b i +1 , ..., b k +1 ) d (ˆ b j ) = d ( b , ..., b j − , b j +1 , ..., b k +1 )= ( − j − i − d ( b , ..., b j − , b i , b j +1 , ..., b k +1 ) . Therefore we may choose some permuations σ, τ ∈ S k such thatsgn( s , s ) d ( s ( β )) d ( s ( β )) = ( − m +1 ( − m sgn( σ ) d ( b i , ˆ b i,j ) d ( b j , ˆ b i,j )and ( − i + j d (ˆ b i ) d (ˆ b j ) = ( − i + j ( − j − i − sgn( τ ) d ( b i , ˆ b i,j ) d ( b j , ˆ b i,j )where ˆ b i,j = ( b , ..., b i − , b i +1 , ..., b j − , b j +1 , ..., b k +1 ) . This completes the proof.Next we prove that k − X h =0 |A ( k, h ) | ( n − k − h = n k − . (11)We prove this by counting the number of graphs described above. The result is Theorem 6. The number of forests such that there is exactly one non-rooted tree thatcontains the vertices 1 and 2; the remaining trees are rooted; there are exactly k + 1 non-root vertices; the non-root vertices chosen from the set { , , ..., k + 1 } ; all non-rootvertices are chosen the set { k + 2 , k + 3 , ..., n } ; and every tree has at least two verticesis n k − . e first determine the relations among the coefficients |A ( k, h ) | implied by (15). Lemma 13. For integer k ≥ , x k − = k − X h =0 B ( k, h )( x − k − h . where the numbers B ( k, h ) are determined by the following relations: B ( k, h ) = B ( k − , h − 1) + (1 + 2 h + k ) B ( k − , h ) + (1 + h )(1 + h + k ) B ( k − , h + 1) B ( k, − 1) = 0 , B (0 , h ) = δ ,h . Proof. We use induction on k . The lemma is true for k = 0. Assume it is true for k ≥ 0. Then take x k = k − X h =0 B ( k, h ) x ( x − k − h and re-express the right side in the basis( x − ( k + 1) − h for 0 ≤ h ≤ k . Computing the coefficients in this basis in terms of B ( k, h ) completesthe proof. Theorem 7. For ≤ h ≤ k − , A ( k, h ) | = B ( k, h ) . Proof. We construct A ( k, h ) from the three sets A ( k − , h − , A ( k, h ), and A ( k − , h + 1) in the following six steps. For a graph G , we let ( u, v ) denote an undirectededge in G between the vertices u and v .1. For G ∈ A ( k − , h − G the component that consists of thevertex ( k + 1) with one edge to an unlabeled vertex. This contributes |A ( k − , h − | graphs to A ( k, h ).2. For G ∈ A ( k − , h ), for each vertex v in G whether labeled or unlabeled, wecreate a graph G ′ by adjoining one edge( k + 1 , v ) . This contributes ( k + h ) |A ( k − , h ) | raphs to A ( k, h ), as each G ∈ A ( k − , h ) has k + h vertices.3. For G ∈ A ( k − , h ), for each unlabeled vertex v in G , we create a graph G ′ bylabeling v as ( k + 1) and then adjoining the edge( k + 1 , v ) . This contributes h |A ( k − , h ) | graphs to A ( k, h ), as each G ∈ A ( k − , h ) has h unlabeled vertices.4. For G ∈ A ( k − , h ), we take the component C that contains the vertices 1and 2 and let e denote the edge e = (2 , v )such that in G \ e the vertices 1 and 2 are in separate components. Then create thegraph G ′ by adjoining the edges( k + 1 , 2) and ( k + 1 , v ) . This contributes |A ( k − , h ) | graphs to A ( k, h ).Thus in total the graphs in A ( k − , h ) contribute(1 + 2 h + k ) |A ( k − , h ) | graphs to A ( k, h ).5. For G ∈ A ( k − , h + 1), let the components be C , C , ..., C h +2 where C is the component containing the vertices 1 and 2. Take the unlabeled vertex v i in the component C i (so i = 0). Label v as ( k + 1), and then for each vertex v ′ notin C i , we create the graph G ′ by adjoining the edge( v, v ′ ) . This contributes k + h + 1 − | V ( C i ) | raphs to A ( k, h ). Doing this for each unlabeled v i in G contributes( k + h + 1)( h + 1) − h +1 X i =1 | V ( C i ) | = ( k + h + 1)( h + 1) − ( k + h + 1) + | V ( C ) | = ( k + h + 1) h + | V ( C ) | (12)graphs to A ( k, h ).6. For the same G as in step 5, we take the component C and let e denote theedge e = (2 , v )such that in G \ e the vertices 1 and 2 are in separate components U and U , respec-tively. Then, in G , for each v ′ in a component C i , i = 0, we label the unlabeled vertexin C i as ( k + 1), remove the edge e , and adjoin the edges( v, k + 1) and (2 , v ′ ) . This contributes h +1 X i =1 | V ( C i ) | (13)graphs to A ( k, h ). Adding (12) and (13) yields( k + h + 1) h + | V ( C ) | + h +1 X i =1 | V ( C i ) | = ( k + h + 1)( h + 1)graphs in A ( k, h ) that come from one G in A ( k − , h + 1). Doing this for each G ∈ A ( k − , h + 1) contributes( k + h + 1)( h + 1) |A ( k − , h + 1) | graphs in A ( k, h ).This accounts for every graph in A ( k, h ): take the vertex v = k + 1 in A ( k, h )and exactly one of the following is true:1. A component consists solely of v and an unlabeled vertex.2. v is a leaf in a component that contains at least three vertices.3. v is adjacent to an unlabeled vertex which is a leaf in a component that containsat least three vertices.4. v is in component C , and v is adjacent to exactly two vertices, one of which is2, and G \ v separates vertices 1 and 2.5. v is in component C , and v is not a leaf, and G \ v does not separate vertices 1and 2.6. v is in component C , G \ v separates vertices 1 and 2, and either v is adjacentto more than two vertices, or v is adjacent to exactly two vertices, neither of which is2. This completes the proof. e now consider the sum from (10) X b ∈ C ( k +1 ,n ) k +1 X i =1 d (ˆ b i ) . This is equal to ( n − k ) X b ∈ C ( k,n ) d ( b ) . And X ( β,s ,s ): | I ( β,s ,s ) | =0 sgn( s , s ) d ( s ( β )) d ( s ( β )) = X ( β,s ) d ( s ( β )) We thus show that X ( β,s ) d ( s ( β )) = n k − ( n − k ) X b ∈ C ( k,n ) d ( b ) . (14) Definition 20. Let b ∈ C ( k, n ) . Define G ( b, i, j ) to be the set of all ( β, s ) ∈ B ∗ ( k, n ) such that s ( β ) as a set is equal to b. Define the subset G ( b ; h ) ⊂ G ( b ) to consist of all ( β, s ) such that | β | = k + 1 + h. We characterize the possible ( β, s )that can appear in G ( b ; h ). We identify β withthe graph G ( β ) in the following lemma. Lemma 14. Let b ∈ C ( k, n ) . Then G ( b ; h ) is the set of all ( β, s ) such that for thegraphs G ( β ) :1. b , ..., b k are vertices of G ( β ) , and | V ( G ( β )) | = h + k + 1 .2. There are exactly k edges.3. There are exactly h + 1 components. There is exactly one non- b vertex in eachcomponent. Here, a “non- b ” vertex means a number not in b .4. Each component has at least two vertices.5. There are no cycles, loops, or multiple edges.6. s makes each component into a directed rooted tree such that the unlabeled vertexis the root.Proof. This follows from the same reasoning in Lemma 10. Definition 21. Let b = { , , ..., k } and take a graph G in G ( b ; h ) . Now make a graph G ′ by taking every non- b vertex in G an unlabeled vertex. The setof such G ′ depends only on k and h ; let A ( k, h ) denote this set. emma 15. For ≤ h ≤ k − and any b ∈ C ( k, n ) , |G ( b ; h ) | = |A ( k, h ) | ( n − k ) h +1 . Proof. We can re-name the elements of b to correspond to { , , ..., k } . Then any graphin G ( b ; h ) corresponds to taking a graph in A ( k, h ) and labeling the unlabeled verticeswith numbers chosen from the set { , , ..., n } − b . There are( n − k )( n − k − ... ( n − k − h )ways to do this because each unlabeled vertex appears in a component with a vertexin b and we may order these components. Lemma 16. X β ∈ Pairs( k,n ) , | I ( β,s ,s ) | =0 d ( s ( β )) = k − X h =0 |A ( k, h ) | ( n − k ) h +1 ! X b ∈ C ( k,n ) d ( b ) Proof. Each ( β, s ) is in a unique G ( b ): the set s ( β ) is b . Thus X β ∈ Pairs( k,n ) , | I ( β,s ,s ) | =0 d ( s ( β )) = X b ∈ C ( k,n ) |G ( b ) | d ( b ) = k − X h =0 |A ( k, h ) | ( n − k ) h +1 ! X b ∈ C ( k,n ) d ( b ) by Lemma 15. This completes the proof.We thus show that k − X h =0 |A ( k, h ) | ( n − k ) h +1 = n k − ( n − k ) . The result is Theorem 8. The number of rooted forests such that there are exactly k non-rootvertices; the non-root vertices chosen from the set { , , ..., k } ; the root vertices arechosen from the set { k + 1 , k + 2 , ..., n } ; and every tree has at least two vertices is n k − ( n − k ) . Now k − X h =0 |A ( k, h ) | ( n − k ) h +1 = ( n − k ) k − X h =0 |A ( k, h ) | ( n − k − h (15) e thus show that k − X h =0 |A ( k, h ) | ( n − k − h = n k − . That is, we prove |A ( k, h ) | = B ( k, h ) Theorem 9. For ≤ h ≤ k − , A ( k, h ) = B ( k, h ) . Proof. We construct A ( k, h ) from the three sets A ( k − , h − , A ( k, h ), and A ( k − , h + 1) in the following six steps. For a graph G , we let ( u, v ) denote an undirectededge in G between the vertices u and v .1. For G ∈ A ( k − , h − G the component that consists of thevertex k with one edge to an unlabeled vertex. This contributes A ( k − , h − A ( k, h ).2. For G ∈ A ( k − , h ), for each vertex v in G whether labeled or unlabeled, wecreate a graph G ′ by adjoining the edge( k, v ) . This contributes ( k + h ) A ( k − , h )graphs to A ( k, h ), as each G ∈ A ( k − , h ) has k + h vertices.3. For G ∈ A ( k − , h ), for each unlabeled vertex v in G , we create a graph G ′ bylabeling v as k and then adjoining the edge( k, v ) . This contributes ( h + 1) A ( k − , h )graphs to A ( k, h ), as each G ∈ A ( k − , h ) has h + 1 unlabeled vertices. Thus intotal the graphs in A ( k − , h ) contribute(1 + 2 h + k ) A ( k − , h )graphs to A ( k, h ). . For G ∈ A ( k − , h + 1), let the components be C , C , ..., C h +2 . Take the unlabeled vertex v i in the component C i . Label v as k , and then for eachvertex v ′ not in C i , we create the graph G ′ by adjoining the edge( v, v ′ ) . This contributes k + h + 1 − | V ( C i ) | graphs to A ( k, h ). Doing this for each i contributes( k + h + 1)( h + 2) − h +2 X i =1 | V ( C i ) | = ( k + h + 1)( h + 2) − ( k + h + 1)= ( k + h + 1)( h + 1) (16)graphs to A ( k, h ).This accounts for every graph in A ( k, h ): take the vertex v = k in A ( k, h ) andexactly one of the following is true:1. A component consists solely of v and an unlabeled vertex.2. v is a leaf in a component that contains at least three vertices.3. v is adjacent to an unlabeled vertex which is a leaf in a component that containsat least three vertices.4. None of the above.This completes the proof. • Analyze the coefficients of e i e j in the entries of E ( n ). • Prove the equivalence of minors using the Newton-Girard identities to expressthe power-sum functions in terms of the elementary symmetric functions, insteadof using the indeterminates z i . • See if there is some family relating the matrix E ( n ) and the Bezoutian matrix,or try to characterize all matrices that have equivalent minors. • See if tensors can be applied instead of just matrices. • See if these expressions for the inequalities can be applied to prove the convergenceof the NRS( m ) algorithms of [2]. • Apply these expressions for the inequalities to the Jensen polynomials of theRiemann xi function, using the integral kernels in [1], [3], or the kernel used inLi’s criterion. • See if the these expressions can be generalized to other root systems. • Use these expressions to directly prove that they determine when a polynomialhas real zeros. eferences [1] G. Csordas; T.S. Norfolk, and R.S. Varga, “The Riemann hypothesis and the Tur´aninequalities”, Trans. Amer. Math. Soc. 296 (1986), no. 2, 521-541.[2] DeFranco, Mario, “On Generalizations of the Newton-Raphson-Simpson Method”,2019, https://arxiv.org/abs/1903.10697[3] DeFranco, Mario, “On properties of the Taylor series coefficients of the Riemannxi function at s = ”, 2019. https://arxiv.org/abs/1907.08984[4] D. K. Dimitrov, F. R. Lucas.,“Higher order Tur´an inequalities for the Riemann ξ function”, Proceedings of the American Mathematical Society 139(3):1013-1022,March 2011.[5] J. Ma˘r´ık, “On polynomials, all of whose zeros are real”, (Czech) ˘Casopis P˘est. Mat.89 1964 5-9.[6] N. Obrechkoff, Zeros of Polynomials, Publ. Bulg. Acad. Sci., Sofia, 1963 (in Bul-garian); English translation (by I. Dimovski and P. Rusev) published by the MarinDrinov Academic Publishing House, Sofia, 2003.function”, Proceedings of the American Mathematical Society 139(3):1013-1022,March 2011.[5] J. Ma˘r´ık, “On polynomials, all of whose zeros are real”, (Czech) ˘Casopis P˘est. Mat.89 1964 5-9.[6] N. Obrechkoff, Zeros of Polynomials, Publ. Bulg. Acad. Sci., Sofia, 1963 (in Bul-garian); English translation (by I. Dimovski and P. Rusev) published by the MarinDrinov Academic Publishing House, Sofia, 2003.