Finding positive instances of parametric polynomials Salvador Lucas - - PowerPoint PPT Presentation

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Finding positive instances of parametric polynomials

Salvador Lucas Departamento de Sistemas Inform´ aticos y Computaci´

• n (DSIC)

Universidad Polit´ ecnica de Valencia, Spain

Salvador Lucas – MAP’10 Finding positive instances of parametric polynomials

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Motivation: automatic proofs of termination

Salvador Lucas – MAP’10 Finding positive instances of parametric polynomials

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Motivation: automatic proofs of termination

Example

Consider the following Term Rewriting System (TRS) R (cf. [Der95]): fact(0) → s(0) fact(s(x)) → s(x) × fact(p(s(x)))) 0 × y → s(x) × y → (x × y) + y x + 0 → x x + s(y) → s(x + y) p(s(x)) → x together with the following dependency pairs [AG00] associated to R:

FACT(s(x)) → s(x) ×♯ fact(p(s(x)))) FACT(s(x)) → FACT(p(s(x))) FACT(s(x)) → P(s(x)) s(x) ×♯ y → x ×♯ y s(x) ×♯ y → (x × y) +♯ y x +♯ s(y) → x +♯ y

During the proof of termination, we have to show that p(s(x)) x and FACT(s(x)) ❂ FACT(p(s(x))) for some (appropriate) quasi-ordering and well-founded ordering ❂ (on terms).

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Motivation: automatic proofs of termination

A fruitful approach: polynomial interpretations, where the k-ary symbols f are given polynomial functions [f ] : Ak → A (for some numeric domain A) which are defined by some polynomial [f ] ∈ R[X1, . . . , Xk]. Terms are interpreted inductively.

Example

The following polynomial interpretation (where A = [0, +∞)) [p](x) =

1 2x

[s](x) = 2x + 1

2

[FACT](x) = 2x can be used to prove the previous constraints (for all x ∈ A): [p(s(x))] = x + 1

4

≥ x = [x] [FACT(s(x))] = 4x + 1 > 2x + 3

4

= [FACT(p(s(x)))]

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Motivation: automatic proofs of termination

Relevant issues

1 The shape of the polynomials in the interpretation. Linear

polynomials lead to linear constraints (easy to check), but polynomials of bigger degree are often required in practice.

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Motivation: automatic proofs of termination

Relevant issues

1 The shape of the polynomials in the interpretation. Linear

polynomials lead to linear constraints (easy to check), but polynomials of bigger degree are often required in practice.

2 We can obtain arbitrary polynomials Ps,t = [s] − [t] when dealing

with constraints s t or s ❂ t as Ps,t ≥ 0 and Ps,t > 0 (for variables ranging in A), respectively.

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Motivation: automatic proofs of termination

Relevant issues

1 The shape of the polynomials in the interpretation. Linear

polynomials lead to linear constraints (easy to check), but polynomials of bigger degree are often required in practice.

2 We can obtain arbitrary polynomials Ps,t = [s] − [t] when dealing

with constraints s t or s ❂ t as Ps,t ≥ 0 and Ps,t > 0 (for variables ranging in A), respectively.

3 We face the problem of testing (semidefinite) positiveness of arbitrary

polynomials (on a restricted domain A).

Salvador Lucas – MAP’10 Finding positive instances of parametric polynomials

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Motivation: automatic proofs of termination

Relevant issues

1 The shape of the polynomials in the interpretation. Linear

polynomials lead to linear constraints (easy to check), but polynomials of bigger degree are often required in practice.

2 We can obtain arbitrary polynomials Ps,t = [s] − [t] when dealing

with constraints s t or s ❂ t as Ps,t ≥ 0 and Ps,t > 0 (for variables ranging in A), respectively.

3 We face the problem of testing (semidefinite) positiveness of arbitrary

polynomials (on a restricted domain A).

4 We have to generate the interpretations. In practice, we use

parametric polynomials whose coefficients are parameters.

Salvador Lucas – MAP’10 Finding positive instances of parametric polynomials

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Motivation: automatic proofs of termination

Relevant issues

1 The shape of the polynomials in the interpretation. Linear

polynomials lead to linear constraints (easy to check), but polynomials of bigger degree are often required in practice.

2 We can obtain arbitrary polynomials Ps,t = [s] − [t] when dealing

with constraints s t or s ❂ t as Ps,t ≥ 0 and Ps,t > 0 (for variables ranging in A), respectively.

3 We face the problem of testing (semidefinite) positiveness of arbitrary

polynomials (on a restricted domain A).

4 We have to generate the interpretations. In practice, we use

parametric polynomials whose coefficients are parameters.

5 Time is important: all should be done within a fraction of second.

Salvador Lucas – MAP’10 Finding positive instances of parametric polynomials

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Motivation: automatic proofs of termination

Relevant issues

1 The shape of the polynomials in the interpretation. Linear

polynomials lead to linear constraints (easy to check), but polynomials of bigger degree are often required in practice.

2 We can obtain arbitrary polynomials Ps,t = [s] − [t] when dealing

with constraints s t or s ❂ t as Ps,t ≥ 0 and Ps,t > 0 (for variables ranging in A), respectively.

3 We face the problem of testing (semidefinite) positiveness of arbitrary

polynomials (on a restricted domain A).

4 We have to generate the interpretations. In practice, we use

parametric polynomials whose coefficients are parameters.

5 Time is important: all should be done within a fraction of second. 6 We have to provide certificates of the proofs. The users should be

able to check the proofs either by hand or by using automatic tools (based on theorem provers like Coq or Isabelle).

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Motivation: automatic proofs of termination

Some history

1 In his IPL’79 paper, Dershowitz pointed out that the use of Tarski’s

results would ’circumvent’ the practical difficulties of solving polynomial interpretations over the naturals (after Matjasevich).

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Motivation: automatic proofs of termination

Some history

1 In his IPL’79 paper, Dershowitz pointed out that the use of Tarski’s

results would ’circumvent’ the practical difficulties of solving polynomial interpretations over the naturals (after Matjasevich).

2 Giesl, in his RTA’95 paper, says the following

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Motivation: automatic proofs of termination

Some history

1 In his IPL’79 paper, Dershowitz pointed out that the use of Tarski’s

results would ’circumvent’ the practical difficulties of solving polynomial interpretations over the naturals (after Matjasevich).

2 Giesl, in his RTA’95 paper, says the following 3 Since 2005 [Luc05], polynomials over the reals are really used (and

useful) in proofs of termination. But we can hardly say that the specific theory of the area is used in the implementations.

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Motivation: automatic proofs of termination

Definition (Parametric polynomial interpretation)

Let F be a signature. A parametric polynomial interpretation for F is a pair A = (A, FA) where A is a numeric domain and FA consists of parametric polynomials, i.e., for all k-ary symbols f ∈ F, ⌊f ⌋ is a polynomial ⌊f ⌋ ∈ Z[C1, . . . , CMf , X1, . . . , Xk]. Here, C1, . . . , CMf are variables, often called parameters, and X1, . . . , Xk are the usual domain variables.

State-of-the-art

1 Tests of (semidefinite) positiveness proceed as follows: 1 P ∈ Z[C1, . . . , CM][X1, . . . , Xn] is PSD on [0, +∞)n if all coefficients in

P are non-negative.

2 P ∈ Z[C1, . . . , CM][X1, . . . , Xn] is PD on [0, +∞)n if all coefficients in

P are non-negative and the constant coefficient is positive.

2 Dealing with parametric polynomials, polynomial constraints are

translated into constraint solving problems by using this rule

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Summary

Our goal:

1 Review the existing literature about testing P(S)Dness of polynomials

and devise how to apply it in our setting.

2 We contribute with a new technique.

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Summary

Our goal:

1 Review the existing literature about testing P(S)Dness of polynomials

and devise how to apply it in our setting.

2 We contribute with a new technique.

Summary:

1 A parametric vector basis for positive polynomials in one variable 2 Correctness and completeness 3 Cost analysis 4 Related work 5 Conclusions

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A parametric basis for positive polynomials in one variable

Observation

The current approach to PSDness corresponds to checking (or imposing, in the parametric case) that the coordinates [P]Sd ∈ Rd+1 of P are non-negative (w.r.t. the standard vector base Sd = (1, x, . . . , xd) for polynomials P ∈ Pd of degree d).

Idea:

If we change the vector base B = (v0, v1, . . . , vd), then some polynomials P with negative coefficients could get non-negative coordinates [P]B. If the vi are PSD on [0, +∞), then P is PSD on [0, +∞) as well.

Example

Consider Q(x) = x3 − 4x2 + 6x + 1 and B = {1, x, x2, x(x − 2)2}. Since [Q]B = (1, 2, 0, 1)T ≥ 0, we conclude that Q is PSD on [0, +∞). Moreover, since the first coordinate is 1 > 0, Q is actually PD on [0, +∞).

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A parametric basis for positive polynomials in one variable

For d ∈ N, and 2 ≤ i ≤ d, consider the parametric univariate polynomials: P0(x) = 1 P1(x) = x Pi(x) = i

2

j=1(x − γij)2

if i is even Pi(x) = x i−1

2

j=1(x − γij)2

if i is odd where the γij are parameters. This sequence Pd of d + 1 polynomials is a vector base for Pd, the set of polynomials P ∈ R[X] whose degree does not exceed d. If the coordinates [P]Pd ∈ Rd+1 of P are non-negative, then P is non-negative on [0, +∞).

Theorem

Let P ∈ Pn. If [P]Pn ≥ 0 (resp. [P]Pn > 0), then P(x) ≥ 0 (resp. P(x) > 0) for all x ≥ 0.

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A parametric basis for positive polynomials in one variable

On the other hand, for all P ∈ Pn which is non-negative on [0, +∞), there is an instance of Pd for which P has non-negative coordinates only.

Theorem

Let P ∈ R[X] be a polynomial of degree n. If P(x) ≥ 0 (P(x) > 0) for all x ≥ 0, then [P]Pn ≥ 0 (resp. [P]Pn > 0) for some assignment of values γij ≥ 0 to the parameters in Pn.

Example

With γ21 = 0 and γ31 = 2 we obtain the required instance of Pn: B = {1, x, x2, x(x − 2)2}.

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A parametric basis for positive polynomials in one variable

In order to prove (semidefinite) positiveness of P ∈ Pd we compute a (parametric) cb-matrix M = MSd→Pd, where Sd = (1, x, . . . , xd) is the standard basis for Pd. The matrix-vector product M [P]Sd yields the coordinates of P w.r.t. Pd. Then, M [P]Sd > 0 (resp. M [P]Sd ≥ 0) yields a set of d + 1 constraints which we have to solve w.r.t. the parameters γij.

Theorem (Incremental computation of the cb-matrix)

We have M0 = I1 and for all n > 1, Mn =

• Mn−1

−Mn−1[Pn(x)]1,...,n

Sn

I1

• where [Pn(x)]1,··· ,n

Sn

is the n-dimensional vector containing the first n coordinates of [Pn(x)]Sn (the last one is 1).

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A parametric basis for positive polynomials in one variable

Example

Q is PD on [0, +∞). Since [Q]S3 = (1, 6, −4, 1)T, we impose

[Q]P3 = M3[Q]S3 = B B @ 1 −γ2

21

−2γ2

21γ31

1 2γ21 4γ21γ31 − γ2

31

1 2γ31 1 1 C C A B B @ 1 6 −4 1 1 C C A = B B @ 1 + 4γ2

21 − 2γ2 21γ31

6 − 8γ21 + 4γ21γ31 − γ2

31

−4 + 2γ31 1 1 C C A > B B @ 1 C C A

Thus, we have to solve the following constraints: 1 + 4γ2

21 − 2γ2 21γ31 > 0

6 − 8γ21 + 4γ21γ31 − γ2

31 ≥ 0

2γ31 − 4 ≥ 0 With γ21 = 0 and γ31 = 2, we prove Q positive on [0, +∞) through the representation [Q]P3 = (1, 2, 0, 1)T where P3 = {1, x, x2, x(x − 2)2}.

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Cost analysis

1 The number V (d) of auxiliary variables γij which are needed is

V (d) =

• d2

4

if d is even

d2−1 4

• therwise

2 The number I(d) of inequalities to be solved is I(d) = d. 3 The degree of the polynomial in the inequalities goes from 1 to d. A

constraint of degree n contains terms of degree 0, 1, . . . , n.

Note

The method is able to deal with parametric polynomials.

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Related work

Our approach can be compared with other well-known techniques for proving (semidefinite) positiveness of polynomials on [0, +∞):

1 P´

• lya and Szeg¨
• representation (PSD).

2 Karlin and Studden (PD). 3 Goursat’s transformation and Bernstein base. 4 PSD as SOS representation through the change of variables X → X 2

(e.g., use of Gram matrices).

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P´

• lya and Szeg¨
• representation (for PSD polynomials)

Proposition (P´

• lya-Szeg¨
• )

If P is PSD on [0, +∞), then there are SOS polynomials f , g such that P(x) = f (x) + xg(x) and deg(f ), deg(xg) ≤ deg(p). According to Hilbert, we can assume that both f and g consists of a sum

• f two squares f 2

1 , f 2 2 and g2 1 , g2 2 , respectively.

If deg(P) = d, the two fi have at most degree d1 = ⌊ d

2 ⌋, and the two gi

have at most degree d2 = ⌊ d−1

2 ⌋.

Write fi = ai,d1xd1 + · · · + ai,1x + ai,0 and gi = bi,d2xd2 + · · · + bi,1x + bi,0 for i = 1, 2. Try to match this representation against the target polynomial P through appropriate equalities between the coefficients of the monomials.

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P´

• lya and Szeg¨
• representation (for PSD polynomials)

Example

Consider the polynomial Q(X) = X 3 − 4X 2 + 6X + 1 = f1(X) + f2(X) + X(g1(X) + g2(X)) where fi(X) = (aiX + bi)2 and gi(X) = (ciX + di)2 for i = 1, 2.

Q(X) = (c2

1 +c2 2)X 3 +(a2 1 +a2 2 +2c1d1 +2c2d2)X 2 +(2a1b1 +2a2b2 +d2 1 +d2 2 )X +b2 1 +b2 2

which amounts at solving the following equalities:

c2

1 + c2 2

= 1 a2

1 + a2 2 + 2c1d1 + 2c2d2

= −4 2a1b1 + 2a2b2 + d2

1 + d2 2

= 6 b2

1 + b2 2

= 1

The following assignment solves the system:

a1 = a2 = b1 = 1 b2 = c1 = 1 c2 = d1 = −2 d2 = √ 2

and proves that Q is PSD on [0, +∞).

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P´

• lya and Szeg¨
• representation (variant 1)

The Pythagoras number of Z and Q is 4 (for instance, there is no x, y ∈ Z such that x2 + y2 = 3, but 12 + 12 + 12 = 3). If we want to restrict the attention to Z or Q when solving these quadratic equalities, we have to use four additive components for f and g.

Example

Now we have

Q(X) = (

4

X

i=1

c2

i )X 3 + ( 4

X

i=1

a2

i + 2cidi)X 2 + ( 4

X

i=1

2aibi + d2

i )X + 4

X

i=1

b2

i

which amounts at solving the following equalities:

P4

i=1 c2 i = 1

P4

i=1 a2 i + 2cidi = −4

P4

i=1 2aibi + d2 i = 6

P4

i=1 b2 i = 1

The following assignment (without irrationals) solves the system:

a1 = 0 a2 = 0 a3 = 0 a4 = 0 b1 = 1 b2 = 0 b3 = 0 b4 = 0 c1 = 1 c2 = 0 c3 = 0 c4 = 0 d1 = −2 d2 = 1 d3 = 1 d4 = 0

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P´

• lya and Szeg¨
• representation (variant 2)

Proposition

Let P, Q ∈ R[X1, . . . , Xn] be polynomials P =

α aαX α and

Q =

α bαX α. If aα ≥ bα for all α ∈ Nn and Q(x1, . . . , xn) ≥ 0 for all

x1, . . . , xn ≥ 0, then P(x1, . . . , xn) ≥ 0 for all x1, . . . , xn ≥ 0.

Example

Instead of equalities, we solve now the following inequalities:

c2

1 + c2 2

≤ 1 a2

1 + a2 2 + 2c1d1 + 2c2d2

≤ −4 2a1b1 + 2a2b2 + d2

1 + d2 2

≤ 6 b2

1 + b2 2

≤ 1

which now is solved with the following assignment:

a1 = a2 = b1 = 1 b2 = c1 = 1 c2 = d1 = −2 d2 = 1

which does not require irrational numbers.

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Karlin and Studden representation (for PD polynomials)

Theorem (Karlin and Studden)

Let P2m be a polynomial of degree 2m for some m ≥ 0 with leading coefficient a2m > 0. If P2m > 0 on [0, +∞), then there exists a unique representation P2m(X) = a2m

m

• j=1

(X − αj)2 + βX

m

• j=2

(X − γj)2 where β > 0 and 0 = γ1 < α1 < γ2 < · · · < γm < αm < ∞. Similarly, if P2m+1 is a polynomial of degree 2m + 1 for some m ≥ 0, with leading coefficient a2m+1 > 0 and P2m+1 > 0 on [0, +∞), then there exists a unique representation P2m+1(X) = a2m+1X

m+1

• j=2

(X − αj)2 + β

m

• j=1

(X − γj)2 where β > 0 and 0 = α1 < γ1 < α2 < γ2 < · · · < γm < αm+1 < ∞.

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Karlin and Studden representation (for PD polynomials)

Example

Since the degree of Q is odd, we let KQ(X) = X(X−α2)2+β(X−γ1)2 = X 3+(β−2α2)X 2+(α2

2−2βγ1)X+βγ2 1

Thus, we have the following constraints: −4 ≥ β − 2α2 1 > 0 6 ≥ α2

2 − 2βγ1

β > 0 1 ≥ βγ2

1

0 < γ1 < α2 The satisfaction of the 1 ≥ βγ2

1 requires the use of positive numbers below

1 (note that βγ2

1 > 0). The following assignment solves the system:

α2 = 9

4

β = 1

4

γ1 = 1

2

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Goursat’s transformation and Bernstein basis

Theorem (Goursat transform)

Given P ∈ R[X] of degree d, we let P(X) = (1 + X)dP(1−X

1+X ). Then, P is

P(S)D on [−1, 1] if and only if P is P(S)D on [0, +∞).

Theorem (Use of Bernstein’s basis)

If [P]Bp ≥ 0 (for the Bernstein basis Bd which consists of polynomials of degree d only) then P is positive on [−1, 1].

Theorem (Bernstein’s Theorem)

If P is PD on [−1, 1], there is p ≥ d (Bernstein degree) s.t. [P]Bp ≥ 0.

Problems

1 p can be much higher than n. Furthermore, 2 We have to (over)estimate p as q ≥ p and use Bq.

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Goursat’s transformation and Bernstein basis

Example

For Q(X) = X 3 − 4X 2 + 6X + 1, we get Q(X) = −10X 3 + 4X 2 + 10X + 4. For B3 = {1

8(1 − 3x + 3x2 − x3), 3 8(1 − x − x2 + x3), 3 8(1 + x − x2 − x3), 1 8(1 + 3x + 3x2 + x3)}

we obtain [ Q]B3 = SS3→B3[ Q]S3 =     1 −1 1 −1 1 −1

3

−1

3

1 1

1 3

−1

3

−1 1 1 1 1         4 10 4 −10     =     8 −32

3

16 8     which does not prove PDness of Q on [0, +∞). By using available estimations for the Bernstein degree of Q [BCR08], we

• btain q = 40, which amounts at computing a cb-matrix of 40×40 entries.

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Goursat’s transformation and Bernstein basis

Note

The method is not well-suited to deal with parametric polynomials: current estimations of Bernstein’s degree require lower bounds λ on the value which is reached by P on [−1, 1]. Then, q ∝ 1

λ.

With parametric polynomials we would obtain λ = 0 and q = +∞.

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P(S)D as SOS representation - Hilbert

Since we have the following

Proposition

Let P ∈ R[X1, . . . , Xn] and Q(X1, . . . , Xn) = P(X 2

1 , . . . , X 2 n ). Then, P is

P(S)D on [0, +∞)n if and only if Q is P(S)D on Rn. we can use all available techniques for proving P(S)Dness on R in proofs

• f P(S)Dness on [0, +∞). In particular:

Proposition (Hilbert)

If P is PSD on R, then P is a sum of two squares of polynomials.

Note

This transformation duplicates the degree of P.

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P(S)D as SOS representation - Hilbert

Example

Consider the polynomial Z(X) = Q(X 2) = X 6 − 4X 4 + 6X 2 + 1 = f1(X) + f2(X) where fi(X) = (aiX 3 + biX 2 + ciX + di)2 for i = 1, 2. Z(X) = 2

i=1 a2 i X 6 + 2aibiX 5 + (b2 i + 2aici)X 4 + 2(bici + aidi)X 3

+(2bidi + c2

i )X 2 + 2cidiX + d2 i

which amounts at solving the following equalities:

P2

i=1 a2 i

= 1 P2

i=1 aibi

= P2

i=1 b2 i + 2aici

= −4 P2

i=1 bici + aidi

= P2

i=1 2bidi + c2 i

= 6 P2

i=1 cidi

= P2

i=1 d2 i

= 1

A solution to this set of equations can be obtained by using Mathematica, but it yields (quite involved) irrational numbers.

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P(S)D as SOS representation - Gram matrices

Theorem ([CLR95])

Let P be a polynomial of degree 2m and z(X) be the vector of all monomials X α such that |α| ≤ m. Then, P is a sum of squares in R[X] if and only if there exists a real, symmetric, psd matrix B such that P = z(X)TBz(X). The problem of proving that P is SOS is translated into the problem of proving whether a matrix B is positive semidefinite. This can be done (in practice) in several ways:

1 Compute the characteristic polynomial of B and require that all its

roots are non-negative [PW98]. Powers and W¨

• rmann show that this

can be translated into a constraint solving problem involving d2−d

2

new parameters λi.

2 Use semidefinite programming techniques [PP08].

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Quantitative comparison

In the following table, V (d) is the number of auxiliary variables which are introduced by the method and I(d) is the number of (in)equalities which are obtained.

Method: Vector P&S P&S (1) P&S (2) K&S Hilbert Gram V(d):

d2 4

2d + 2 4d + 4 2d + 2 d + 1 2d + 2

d2−d 2

I(d): d d + 1 d + 1 d + 1 2d + 1 2d + 1 d

Note that P&S (1) and Hilbert yield equations.

Summary

The most critical aspect is the number of variables V (d) which are

• introduced. In this sense, we have the following:

1 For proving PSDness on [0, +∞), Vector is the best choice for

1 ≤ d ≤ 9 and P&S (2) is the best choice for d ≥ 10.

2 For proving PDness on [0, +∞), Vector is the best choice for

1 ≤ d ≤ 5 and K&S is the best choice for d ≥ 6.

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Conclusions and future work

1 We have introduced a new method for proving P(S)Dness of

univariate polynomials.

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Conclusions and future work

1 We have introduced a new method for proving P(S)Dness of

univariate polynomials.

2 The method is sound and complete and well-suited for dealing with

parametric polynomials.

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Conclusions and future work

1 We have introduced a new method for proving P(S)Dness of

univariate polynomials.

2 The method is sound and complete and well-suited for dealing with

parametric polynomials.

3 The quantitative analysis of the method shows that it can work better

than other methods for polynomials of small degree (below 10).

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Conclusions and future work

1 We have introduced a new method for proving P(S)Dness of

univariate polynomials.

2 The method is sound and complete and well-suited for dealing with

parametric polynomials.

3 The quantitative analysis of the method shows that it can work better

than other methods for polynomials of small degree (below 10).

Future work

1 The method can be extended to multivariate polynomials. A thorough

analysis and comparison with existing techniques (Positivstellensatz, P´

• lya’s theorem, (S)DS, SOS representation, etc., should be

addressed).

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Conclusions and future work

1 We have introduced a new method for proving P(S)Dness of

univariate polynomials.

2 The method is sound and complete and well-suited for dealing with

parametric polynomials.

3 The quantitative analysis of the method shows that it can work better

than other methods for polynomials of small degree (below 10).

Future work

1 The method can be extended to multivariate polynomials. A thorough

analysis and comparison with existing techniques (Positivstellensatz, P´

• lya’s theorem, (S)DS, SOS representation, etc., should be

addressed).

2 The possibility of restricting the attention to consider rational and

even integer solutions for the constraints (as in P&S (1)) should be investigated for Vector.

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Conclusions and future work

1 We have introduced a new method for proving P(S)Dness of

univariate polynomials.

2 The method is sound and complete and well-suited for dealing with

parametric polynomials.

3 The quantitative analysis of the method shows that it can work better

than other methods for polynomials of small degree (below 10).

Future work

1 The method can be extended to multivariate polynomials. A thorough

analysis and comparison with existing techniques (Positivstellensatz, P´

• lya’s theorem, (S)DS, SOS representation, etc., should be

addressed).

2 The possibility of restricting the attention to consider rational and

even integer solutions for the constraints (as in P&S (1)) should be investigated for Vector.

3 Implementation and benchmarks.

Salvador Lucas – MAP’10 Finding positive instances of parametric polynomials