Journal of Pure and Applied Algebra

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1 Journal of Pure and Applied Algebra 25 (2) Contents lists available at ScienceDirect Journal of Pure and Applied Algebra journal homepage: wwwelseviercom/locate/jpaa An algebraic certificate for Budan s theorem Daniel Bembé Mathematisches Institut der Universität München, Theresienstr 39, D-8333 München, Germany a r t i c l e i n f o a b s t r a c t Article history: Received 9 March 2 Received in revised form 3 July 2 Available online 5 September 2 Communicated by G Rosolini MSC: 2D; 2J5; 4Q2 The existing algorithms to construct the real closure of an ordered field involve very high complexities These algorithms are based on Sturm s theorem which we suspect to be one reason for the complexities since all known proofs of Sturm s theorem use Rolle s theorem which is problematic in a constructive context Therefore we propose to replace the use of Sturm s theorem by Budan s theorem In this paper we present as a first step in this direction an algebraic certificate for Budan s theorem An algebraic certificate is a certain kind of proof of a statement In particular, it is an algorithm which produces, from an arbitrary data in the premise of the statement, explicit (in)equalities which express the conclusion 2 Elsevier BV All rights reserved Introduction The classical proof for the existence of the real closure of an ordered field uses the axiom of choice [,3,6] In contrast, the constructive proofs show without the axiom of choice the existence of the real closure by demonstrating that calculating in the real closure does not lead to contradictions [8,] As a central argument these proofs use the fact that a polynomial with positive derivative on an interval increases on this interval In the proof of [8] this is based on Rolle s theorem concerning the real closure which makes a complicated induction necessary, while the proof of [] uses a much more elementary argument which simplifies the construction in a considerable way The idea for this paper arose as in the computations of [8,3] Sturm s sequence is used to count the real roots of a polynomial on an interval And all known proofs of Sturm s theorem rely on Rolle s theorem In contrast, Budan s sequence (ie, for a polynomial f the sequence (f, f, f, )) counts real roots in a similar way (in fact it counts even more points; therefore we speak of virtual roots instead of real roots) [5] And since the proof of Budan s theorem uses only combinatoric arguments [4] Budan s theorem could be useful in a constructive aspect In this paper we undertake a first step in the constructive use of Budan s theorem by presenting an algebraic certificate First, we hope that in the case of nondiscrete ordered fields, ie, fields without sign test (as many subfields of R), a good notion of the real closure can be obtained with a corresponding good construction of the real closure Second, in the discrete case, an improved construction of the real closure would lead to better constructive versions of Stengle s Positivstellensätze [7,9,8] The actual constructive versions are presented in [,2] The case of Schmüdgen s Positivstellensätze [9,5] is also of interest For a constructive version see [7] For general references about Positivstellensätze see [,3,4] At this time it is not yet clear if the described idea will succeed in improving the construction of the real closure address: bembe@mathlmude /$ see front matter 2 Elsevier BV All rights reserved doi:6/jjpaa286

2 D Bembé / Journal of Pure and Applied Algebra 25 (2) Preliminaries Budan s theorem claims the following It holds for an ordered field K, f K[X] and a < b K that the number of sign changes in the sequence (f (a), f (a), f (a), ) is not less than the number of sign changes in (f (b), f (b), ) [4,2],2 Here, an algebraic certificate for Budan s theorem is an algorithm which receives as input data n N and two sequences of sign conditions (σ,, σ n ), ( σ,, σ n ) {,, +} n+ such that the second one has more sign changes than the first one And it calculates for i n some coefficients z i, z i Z such that σ i z i, σ i z i () for all i where at least one of the inequalities in () is a strict one and i z i f (i) () i! + i z i f (i) () i! = (2) for every degree-n polynomial f K[X] With the coordinate transformation g(x) := f ((X a)/(b a)), () and (2), the assumption sign(g (i) (a)) = σ i, sign(g (i) (b)) = σ i (3) for all i leads to the contradiction < Hence the certificate proves Budan s theorem We call this algorithm a certificate as well as z i, z i a linear certificate (as (2) displays the contradiction as a linear combination of the inequalities (3); in fact, the algorithm does not use multiplication in K ) For linear incompatibilities and linear certificates consider [7,6] The certificate proves the existence of z i, z i and calculates them The existence follows also from the baby Positivstellensätze [7] together with any proof of Budan s theorem And the calculation can be solved by linear programming in an ordered group [6] Nevertheless, we consider the direct proof presented here as interesting A certain algorithm of linear programming is stated in the Appendix Unlike the certificate, this algorithm does not prove the existence of z i, z i but calculates all of them Here, we consider (2) as a system of linear equalities which is solved by the z i, z i and restrict the vector space of solutions to those which fulfill the sign conditions () (Geometrically this means the intersection of certain half spaces) The solutions can be described by the convex hull of finitely many points which are calculated by the second algorithm 3 An algebraic certificate for Budan s theorem Definition 3 Let be 2 n N Let Ab(a,, a n ) denote the free abelian group generated by a,, a n 2 Let I n := {,, n} {,, n} denote the index set of Budan s matrix; I n := {(i, j) I n i + j n + } the index set of the upper left triangle; I n := {(i, j) I n i = } the index set of the top row; I n := {(i, j) I n i + j = n + } the index set of the diagonal and I n := I n I n In fact, Budan s theorem claims something more Moreover, it says that the difference between the number of sign changes at a and the number of sign changes at b is not less than the number of real roots in ]a, b] 2 It is necessary to cite the (very complicated) proof of Fourier In fact, Budan s counting of roots is today known as Budan Fourier count The reason is that Fourier has known the result some years before Budan, but he published his proof only years after Budan s communication to the Académie des Sciences Budan s communication was examined by Lagrange and Legendre (rapporteur) They considered the paper as essentially correct, but the Académie did not yet publish the result; they waited for Fourier s paper

3 362 D Bembé / Journal of Pure and Applied Algebra 25 (2) Let Budan s matrix of dimension n be the (n + ) n-matrix a a 2 a 3 a n k 2 2 k 3 3 k n n k a k a k a k 2 k 2 3 k n a k a k n k a k 3 k a k 2 2 n k a k n 2 n k a k n a k =: α () I n Ab(a,, a n ) (n+) n 4 For A = I n (resp I n ), let a sign condition σ on A be a map σ : A {,, } Z, (i, j) σ 5 Two pairs (i, j), (ĩ, ȷ) I n are said to lie in the same connected component wrt a sign condition σ on I n (in symbols: (i, j) σ (ĩ, ȷ)) if there exists a path (i, j) = (i, j ), (i, j ),, (i m, j m ) = (ĩ, ȷ) in I n with σ ik,j k = σ for all k and (i k+, j k+ ) {(i k +, j k ), (i k, j k ), (i k, j k + ), (i k, j k ), (i k, j k )} for all k < m 6 For a sign condition σ on I n, let the connected component of (i, j) I n be C(i, j) := {(ĩ, ȷ) I n (ĩ, ȷ) σ (i, j)} 7 For A = I n (resp I n ) and a sign condition σ on A with σ for at least one (i, j) I n, let a linear incompatibility (resp a linear certificate) z wrt σ be a map z : A Z, (i, j) z such that σ z for all (i, j) A, σ z > for at least one (i, j) I n and z α = () A in Ab(a,, a n ) Remark 32 Notice that the top row of Budan s matrix is subscripted (, ),, (, n) 2 For all following considerations, exclusively elements of the upper left triangle of Budan s matrix I n play a role

4 D Bembé / Journal of Pure and Applied Algebra 25 (2) Fig Proof of Lemma 33; connected components cannot cross 3 For example, Budan s matrix for n := 5: a a 2 a 3 a 4 a 5 a a + a 2 a + a 2 + a 3 a + a 2 + a 3 + a 4 a + a 2 + a 3 + a 4 + a 5 a 2a + a 2 3a + 2a 2 + a 3 4a + 3a 2 + 2a 3 + a 4 a 3a + a 2 6a + 3a 2 + a 3 a 4a + a 2 a 4 α () = α () + α () for all (i, j) I n with i and j 2 5 (i, j) σ (ĩ, ȷ) means that there is a path through I n from (i, j) to (ĩ, ȷ) for which every move is either up, down, to the left or to the right (not diagonal) and every visited element has the same sign ( is regarded as a distinct sign) 6 σ is an equivalence relation; therefore the {C(i, j) P (I n ) (i, j) I n } define a partition of I n Let K be an ordered field and h a homomorphism from Ab(a,, a n ) to the additive group of K 7 For every polynomial f (x) := h(a )x n + + h(a n )x + h(a n ) K[X], the sequence f (), f (),, f (n) () = (h(a n ), h(a n ),, h(a ))!! (n )! appears in the top row and the sequence f (), f (),, f (n) () = (h(α,n ), h(α 2,n ),, h(α n, ))!! (n )! in the diagonal of Budan s matrix 8 A linear incompatibility (resp linear certificate) expresses a contradiction to the hypothesis sign(h(α )) = σ for all (i, j) I n (resp I n ) as it claims that < z h(α ) = Lemma 33 Let σ be a sign condition on I n such that the sequence (σ n,, σ n,2,, σ,n ) has more sign changes than (σ,,, σ,n ) (in particular, σ for at least one (i, j) I n ) Then there exists a linear incompatibility z : I n Z wrt σ Proof First, we want to prove the following claim: There exists at least one (ĩ, ȷ) I n with σĩ, ȷ and min () C(ĩ, ȷ) (i) >, ie, there is at least one nonzero connected component which touches the diagonal and does not touch the top row Proof by contradiction Supposed min(i) = for all nonzero (ĩ, ȷ) I n Then for arbitrary (ĩ, ȷ), (î, ˆȷ) I n with opposite nonzero signs and ȷ < ˆȷ and arbitrary (, ȷ ) C(ĩ, ȷ), (, ˆȷ ) C(î, ˆȷ) it holds that ȷ < ˆȷ (Otherwise let (ĩ k, ȷ k ) k C(ĩ, ȷ) resp (î k, ˆȷ k ) k C(î, ˆȷ) be two paths which connect (ĩ, ȷ) with (, ȷ ) resp (î, ˆȷ) with (, ˆȷ ) Wlog ĩ k = î k for all k with ĩ k î and (î k, ˆȷ k ) = (î, ˆȷ) for all k with ĩ k > î otherwise this can be achieved by inserting additional steps since both the connected components contain some (i, j) for every i î (here we use that they both touch the top row) Now let k be the smallest k for which ȷ k > ˆȷ k Then ȷ k = ȷ k = ˆȷ k which is a contradiction since the two connected components have different signs (Fig ))

5 364 D Bembé / Journal of Pure and Applied Algebra 25 (2) Fig 2 Example 34 This means that two connected components cannot cross each other It follows that we get for every pair (ĩ, ȷ), (î, ˆȷ) in the diagonal as above, a pair (, ȷ ), (, ˆȷ ) in the top row as above which means that the top row has at least as many sign changes as the diagonal which was excluded by the assumption, and the first claim is proven Next, we take such a (ĩ, ȷ) I n with σĩ, ȷ and min () C(ĩ, ȷ) (i) > Every α with (i, j) C(ĩ, ȷ) and j 2 can be written as the sum of the element one column to the left and the element one row up: α = α + α ; and every α i, equals the element one row above: α i, = α i, It follows by induction that every α with (i, j) C(ĩ, ȷ) in particular αĩ, ȷ itself can be written as sum of direct neighbors of C(ĩ, ȷ): αĩ, ȷ = n α (4) () D(ĩ, ȷ) with n N (including ) and D(ĩ, ȷ) := {(i, j) I n \ C(ĩ, ȷ) (i, j + ) C(ĩ, ȷ) (i +, j) C(ĩ, ȷ)} And from the connectedness of C(ĩ, ȷ) it follows that σ σĩ, ȷ for (i, j) D(ĩ, ȷ) Therefore (4) leads with to zĩ, ȷ := σĩ, ȷ, z := σ n z := for (i, j) D(ĩ, ȷ), otherwise () I n z α = Since σ z for all (i, j) I n, σĩ, ȷ zĩ, ȷ = and (ĩ, ȷ) I n we are done Example 34 For n := 7, Fig 2 shows a sign condition σ on I 7 with four sign changes in the top row and five in the diagonal The colors mean: σ = ; σ = ; σ = + It holds for (ĩ, ȷ) := (4, 4) that σ 4,4 = and min () C(4,4) (i) = 3 > Therefore starting at (4, 4) we can construct the linear incompatibility (2a + b + 4c + d) (a + 6b + 3c + d) (4a + b) (3a + b) (3a + 2b + c) =, which leads to the contradiction < since < (2a + b + 4c + d) and the other summands are nonnegative Lemma 35 Let σ be a sign condition on I n, and for every sign condition σ on I n with σ In = σ let a linear incompatibility z σ : I n Z be given Then there exists a linear certificate z : I n Z wrt σ Proof Let Σ denote the sign conditions σ on I n with σ In = σ, and let o : {,, (n )n/2} I n \ I n be an arbitrary order on I n \ I n By induction on k we will show for every σ Σ, the existence of a linear incompatibility z σ,k with z σ,k o(l) = for all l k In the base case (k := ) let z σ, := z σ In the inductive step (k k), for every σ Σ, a linear incompatibility z σ,k with z σ,k o(l) provided = for all l < k is

6 D Bembé / Journal of Pure and Applied Algebra 25 (2) Now let σ Σ be arbitrary and define σ (resp σ + ) by σ (resp σ + (resp + ) ) := σ if (i, j) = o(k), otherwise Now consider z σ,k and z σ +,k If z σ,k o(l) z σ +,k o(l) = we are done Otherwise we define z σ,k := z σ +,k o(k) z σ,k z σ,k o(k) z σ +,k (5) for all (i, j) I n Since z σ +,k o(l) z σ,k o(k) = by (5), z σ,k o(l) = as well as z σ,k o(l) for l < k by induction hypothesis, is positive it holds that σ z σ,k for all (i, j) I n by induction hypothesis, σ z σ,k > for at least one (i, j) I n by induction hypothesis End of the inductive step Finally we define z := zσ,(n)n/2 for (i, j) I n and an arbitrary sign condition σ Σ Keep in mind that z σ,(n)n/2 = for (i, j) / I n Therefore z is a linear certificate as desired Lemmas 33 and 35 together lead to Theorem 36 Let σ be a sign condition on I n such that the sequence (σ n,, σ n,2,, σ,n ) has more sign changes than (σ,,, σ,n ) Then there exists a linear certificate z : I n Z wrt σ Example 37 For the sign condition σ on I 7 (σ,,, σ,7 ) := (+,,, +, +,, ), (σ 7,,, σ,7 ) := (+, +,, +,, +, ) which is displayed in Fig 2, is (z,,, z,7, z 7,,, z,7 ) := (5,, 2,,,,,,, 4, 3,,, ) a linear certificate since σ z for all (i, j), σ 5,3 z 5,3 > and < 5(a) 2(c) + (e) 4(5a + 5b + c) + 3(2a + b + 4c + d) (5a + b + 6c + 3d + e) = Acknowledgements I would like to thank Henri Lombardi and Peter Schuster for interesting discussions, the reviewers for their helpful comments as well as Hannah Henker and Nathan Swerling for language support Appendix A second algorithm for the certificate Definition A Let 2 n N For x Q, let x denote its absolute value; 2 for a matrix a, let a t denote its transposed; 3 for a = (a,, a n ) t, b = (b,, b n ) t Q n, let a, b := n i= a ib i denote the scalar product; 4 for a Q n, let { a, x = } := {x Q n a, x = } denote the hyperplane with normal a and { a, x } := { a, x } := {x Q n a, x }, { a, x } := { a, x } := {x Q n a, x } the corresponding half spaces (resp <, > );

7 366 D Bembé / Journal of Pure and Applied Algebra 25 (2) for X Q n, let l xk conv(x) := a k x k X, a k Q, l a k = denote the convex hull of X; 6 for b, c Q n, let bc := conv(b, c) denote the line between a and b; 7 for B, C Q n, let BC := (b,c) B C bc; 8 let B : {,, 2n } I n \ {(, )} denote the following correspondence to the indices of Budan s matrix B(),B(2) B(n),B(n + ) B(2n ) := (n, ),(, 2) (, n),(n, 2) (, n) Lemma A2 Let σ be a sign condition on I n with σ, = σ n,, n n n n 2 n 2 n 2 n 2 n 2 U n := Q n (2n), and let V be the set of v := (v,, v 2n ) t Z 2n for which U n v = and which fulfill the sign condition σ, ie, σ B(k) v k (A) for all k Then it holds for every linear certificate z wrt σ that (z B() + z,, z B(2),, z B(2n) ) t V, (A2) and for every v V with σ B(i) v k > for at least one k {, n +,, 2n } is z := v B () for (i, j) = (, ) and otherwise a linear certificate wrt σ Proof Consider the monomorphism h from Ab(a,, a n ) to the additive group of Q n which throws a (,,, ) t,, a n (,,, ) t Then the columns of U n correspond to the elements of the top row and the diagonal of Budan s matrix, ie, U n = h(α B() ),, h(α B(2n) ) = h(α, ) = h(α n, ), h(α,2 ),, h(α,n ), h(α n,2 ),, h(α,n ) Now the assertions follow directly from the definition of linear certificate

8 D Bembé / Journal of Pure and Applied Algebra 25 (2) Lemma A3 Under the conditions of Lemma A2 let n n n n 2 v n 2 n 2 n 2 n 2 v V n n := v n := Q (2n) (n) v n+ v 2n and W := {x Z n σ B(k) v t k, x for all k} (A3) the intersection of all (n )-dimensional half spaces, whose normals are the rows v k, for which σ B(k), and which have the orientation demanded by σ B(k) Then it holds for the set V in Lemma A2 that V = {V n w w W} Z 2n Proof It holds that {v Z 2n U n v = } = {V n w w Z n } (for w to be an integer consider the lower half of V n ) It remains to prove the equivalence between the conditions (A) and (A3), which is clear since v k = v t k, x for every v = V n x {v Z 2n U n v = } V and every k Theorem A4 Let σ be a sign condition on I n with σ for all (i, j) and σ, = σ n, With v k as in Lemma A3 define the sets W,, W n Q n inductively σ B(n+) W :=,, ; σ B(2n) W := W {σ B() v t, x } B C { v t, x = } ; W n := W n {σ B(n) v t, x } n B n C n { v t n, x = } with B k := W k { v t, x > } k+ and C k := W k { v t k+, x < } for all k Then with W = {r w Q n r >, w conv(w n )} Z n (A4) and (A2) define a one-to-one relation between w W and the linear certificates z wrt σ with z, = Remark A5 The general case with σ B(k) = for some k can be done in the following way: For k n skip the calculation of W k and let W k := W k and go on with calculating W k+ For k > n calculate both W n for σ B(k) = and σ B(k) = and take the union of the solutions Proof For k > n, (A3) defines a restriction to a certain octant of the Z n Since the length of the considered vectors does not matter, we take only the vectors with sum of absolute coordinates one It holds with W norm := (w,, w n ) t Q n w k = that k conv(w ) = {(w,, w n ) t W norm σ B(k) w kn for n + k 2n } = W norm {x Q n σ B(k) v t k, x for n + k 2n } (A4)

9 368 D Bembé / Journal of Pure and Applied Algebra 25 (2) Fig A3 Proof of Lemma A6 The next Lemma A6 shows that it holds for k n conv(w k ) = conv(w k ) {σ B(k) v t k, x }, which says that conv(w n ) fulfills all sign conditions of (A3) To verify the one-to-one relation take a linear certificate z wrt σ with z, = (A4) leads to a unique v V and (A2) to a unique w W To see that w W write it as w w k w w = = k w k w k n w n w k k For the other direction take a w W which leads to a unique v V and to a unique linear certificate z with z, = ; since w at least for one n + k 2n, z B(k) = v k and therefore σ B(k) z B(k) > It remains to prove Lemma A6 Let Y Q n, a Q n, B := Y { a, x > } and C := Y { a, x < } Then it holds that conv(y) { a, x } = conv (Y { a, x }) (BC { a, x = }) Proof is clear since conv(y) { a, x } (Y { a, x }) (BC { a, x = }) and u, v { a, x } (λu+(λ)v) { a, x } for λ since a, (λu+(λ)v) = λ a, u +(λ) a, v For let d be an arbitrary point of conv(b C) { a, x }, ie, d := β k b k + γ l c l with b k B, c l C, β k, γ l with β k + γ l = and a, d Furthermore with β := β k and γ := γ l let b := β k β b k, c := γ l γ c l, d := b c { a, x = } Then d b c, and it holds with d = λ b +(λ )c and d = λ b +(λ )c that λ λ, since a, λb +(λ)c = λ a, b + ( λ) a, c and a, d = a, d That is, d = λ λ b λ + ( λ λ ) d λ b d (Fig A3) The next argument shows that d conv (Y { a, x }) (BC { a, x = }) =: RHS; according to Lemma A7, b k c { a, x = } RHS for every k and therefore again with A7 d = b c { a, x = } RHS It follows from d b d that d RHS

10 D Bembé / Journal of Pure and Applied Algebra 25 (2) Fig A4 Proof of Lemma A7 And arbitrary d conv(y) { a, x } can be written as d = α j a j + β k b k + γ l c l with a j Y { a, x = } RHS, b k B, c l C and therefore it also holds that d RHS It remains to prove Lemma A7 With the notions of Lemma A6, let b B, c l C and γ l with γ l = for l m and c := γ l c l Then it holds that bc { a, x = } conv {b}{c,, c m } { a, x = } Proof First let m = 2 Let (λ b + ( λ )c ) = bc { a, x = } and (λ 2 b + ( λ 2 )c 2 ) = bc 2 { a, x = } Then it holds with that γ := γ ( λ 2) γ ( λ 2) + γ 2 ( λ ) and γ := γ 2 ( λ ) 2 γ ( λ 2) + γ 2 ( λ ) γ (λ b + ( λ )c ) + γ (λ 2 2b + ( λ 2 )c 2 ) = bc { a, x = } conv {b}{c, c 2 } { a, x = } as γ, γ 2 (Consider Fig A4; the gray triangles are similar) For greater m the claim follows by induction since c = m l= γ l c l = m l= γ l m γ l c l l= m l= γ l + γ m c m and ( m l= γ l c l)/( m l= γ l ) conv({c,, c m }) References [] J Bochnak, M Coste, M-F Roy, Real Algebraic Geometry, in: Erg Math Grenzgeb (3), vol 36, Springer, Berlin, 998 [2] J Borowczyk, Sur la vie et l œuvre de François Budan (76 84), Historia Mathematica 8 (2) (99) [3] S Basu, R Pollack, M-F Roy, Algorithms in Real Algebraic Geometry, in: Algo Comp in Math (), Springer, Berlin, 26 [4] Budan de Boislaurent, Nouvelle méthode pour la résolution des équations numériques d un degré quelconque, Paris, 822 Contains in the appendix a proof of Budan s theorem edited by the Académie des Sciences (8) [5] M Coste, T Lajous, H Lombardi, M-F Roy, Generalized Budan Fourier theorem and virtual roots, Journal of Complexity 2 (25)

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