Convex Optimization. (EE227A: UC Berkeley) Lecture 28. Suvrit Sra. (Algebra + Optimization) 02 May, 2013

Transcription

1 Convex Optimization (EE227A: UC Berkeley) Lecture 28 (Algebra + Optimization) 02 May, 2013 Suvrit Sra

2 Admin Poster presentation on 10th May mandatory HW, Midterm, Quiz to be reweighted Project final report on 16th May upload to easychair Any questions / concerns: me! me if you need to meet 2 / 25

3 Convex sets: geometry vs algebra Geometry of convex sets is very rich and well-understood (we didn t cover much of it) 3 / 25

4 Convex sets: geometry vs algebra Geometry of convex sets is very rich and well-understood (we didn t cover much of it) But what about (efficient) representation of these geometric objects? 3 / 25

5 Convex sets: geometry vs algebra Geometry of convex sets is very rich and well-understood (we didn t cover much of it) But what about (efficient) representation of these geometric objects? How do algebraic, geometric, computational aspects interact? 3 / 25

6 Convex sets: geometry vs algebra Geometry of convex sets is very rich and well-understood (we didn t cover much of it) But what about (efficient) representation of these geometric objects? How do algebraic, geometric, computational aspects interact? Semidefinite programming plays a major role! 3 / 25

7 Convex sets: geometry vs algebra Geometry of convex sets is very rich and well-understood (we didn t cover much of it) But what about (efficient) representation of these geometric objects? How do algebraic, geometric, computational aspects interact? Semidefinite programming plays a major role! A nice book for detailed development of these ideas: G. Blekherman, P. Parrilo, R. R. Thomas. Semidefinite optimization and convex algebraic geometry (2012). 3 / 25

8 Polyhedral sets Recall (convex) polyhedra, described by finitely many half-spaces { x R n a T i x b i, i = 1,..., m }. 4 / 25

9 Polyhedral sets Recall (convex) polyhedra, described by finitely many half-spaces { x R n a T i x b i, i = 1,..., m }. Convex polyhedra have many nice properties: Remain preserved under projection (Fourier-Motzkin elimination) 4 / 25

10 Polyhedral sets Recall (convex) polyhedra, described by finitely many half-spaces { x R n a T i x b i, i = 1,..., m }. Convex polyhedra have many nice properties: Remain preserved under projection (Fourier-Motzkin elimination) Farkas lemma / duality theory gives emptiness test 4 / 25

11 Polyhedral sets Recall (convex) polyhedra, described by finitely many half-spaces { x R n a T i x b i, i = 1,..., m }. Convex polyhedra have many nice properties: Remain preserved under projection (Fourier-Motzkin elimination) Farkas lemma / duality theory gives emptiness test Optimization over cvx polyhedra is linear programming. 4 / 25

12 Polyhedral sets Recall (convex) polyhedra, described by finitely many half-spaces { x R n a T i x b i, i = 1,..., m }. Convex polyhedra have many nice properties: Remain preserved under projection (Fourier-Motzkin elimination) Farkas lemma / duality theory gives emptiness test Optimization over cvx polyhedra is linear programming. But getting away from linearity... 4 / 25

13 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. 5 / 25

14 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. Semidefinite representations 5 / 25

15 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. Semidefinite representations Which sets can be represented via SDPs? 5 / 25

16 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. Semidefinite representations Which sets can be represented via SDPs? LP case well-understood: if a set is polyhedral (i.e., finite number of extreme points / rays) 5 / 25

17 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. Semidefinite representations Which sets can be represented via SDPs? LP case well-understood: if a set is polyhedral (i.e., finite number of extreme points / rays) Do we have a similar nice characterization in SDP case? 5 / 25

18 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. Semidefinite representations Which sets can be represented via SDPs? LP case well-understood: if a set is polyhedral (i.e., finite number of extreme points / rays) Do we have a similar nice characterization in SDP case? We ve seen a few SDRs in Lecture 6 (polyhedra, matrix norms, second order cones, etc.) 5 / 25

19 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. Semidefinite representations Which sets can be represented via SDPs? LP case well-understood: if a set is polyhedral (i.e., finite number of extreme points / rays) Do we have a similar nice characterization in SDP case? We ve seen a few SDRs in Lecture 6 (polyhedra, matrix norms, second order cones, etc.) Preserved under standard convex algebra : affine transformations, convex hulls, taking polars, etc. 5 / 25

20 SDP, LMIs We ve seen SOCPs, SDPs as substantial generalization. Semidefinite representations Which sets can be represented via SDPs? LP case well-understood: if a set is polyhedral (i.e., finite number of extreme points / rays) Do we have a similar nice characterization in SDP case? We ve seen a few SDRs in Lecture 6 (polyhedra, matrix norms, second order cones, etc.) Preserved under standard convex algebra : affine transformations, convex hulls, taking polars, etc. See lecture notes by A. Nemirovski for SDR (and conic) calculus 5 / 25

21 Semidefinite representations Can S be represented via SDPs? S must be convex and semialgebraic 6 / 25

22 Semidefinite representations Can S be represented via SDPs? S must be convex and semialgebraic S can be defined using a finite number of polynomial inequalities. 6 / 25

23 Semidefinite representations Can S be represented via SDPs? S must be convex and semialgebraic S can be defined using a finite number of polynomial inequalities. Exact or approx. representations (also, relaxing nonconvex S) 6 / 25

24 Semidefinite representations Can S be represented via SDPs? S must be convex and semialgebraic S can be defined using a finite number of polynomial inequalities. Exact or approx. representations (also, relaxing nonconvex S) Example ( direct representation) x S A 0 + i x ia i 0 6 / 25

25 Semidefinite representations Can S be represented via SDPs? S must be convex and semialgebraic S can be defined using a finite number of polynomial inequalities. Exact or approx. representations (also, relaxing nonconvex S) Example ( direct representation) x S A 0 + i x ia i 0 Lifted representation (recall HW2), can use extra variables x S y s.t. A(x) + B(y) 0. This projection / lifting technique can be very useful. 6 / 25

26 Lifting / projection Classic example 7 / 25

27 Lifting / projection Classic example n-dimensional l 1 -unit ball (crosspolytope). Requires 2 n inequalities of the form ±x 1 ± x 2 ± x n 1. 7 / 25

28 Lifting / projection Classic example n-dimensional l 1 -unit ball (crosspolytope). Requires 2 n inequalities of the form ±x 1 ± x 2 ± x n 1. But we can efficiently represent it as a projection: { (x, y) R 2n } y i = 1, y i x i y i, i = 1,..., n. i Just 2n variables and 2n + 1 constraints 7 / 25

29 Lifting / projection Classic example n-dimensional l 1 -unit ball (crosspolytope). Requires 2 n inequalities of the form ±x 1 ± x 2 ± x n 1. But we can efficiently represent it as a projection: { (x, y) R 2n } y i = 1, y i x i y i, i = 1,..., n. i Just 2n variables and 2n + 1 constraints Moral: When playing with convexity, rather than eliminating variables, often nicer to add new variables with which description of set can become simpler! 7 / 25

30 SDRs and LMIs Does every convex semialgebraic set S have a direct SDR? 8 / 25

31 SDRs and LMIs Does every convex semialgebraic set S have a direct SDR? (answer known in 2-dimensions) Does ever basic convex semialgebraic set have a lifted SDR? 8 / 25

32 SDRs and LMIs Does every convex semialgebraic set S have a direct SDR? (answer known in 2-dimensions) Does ever basic convex semialgebraic set have a lifted SDR? Answers to both are unknown as of now 8 / 25

33 SDRs and LMIs Does every convex semialgebraic set S have a direct SDR? (answer known in 2-dimensions) Does ever basic convex semialgebraic set have a lifted SDR? Answers to both are unknown as of now Some partial results known. See references 8 / 25

34 SDRs and LMIs Does every convex semialgebraic set S have a direct SDR? (answer known in 2-dimensions) Does ever basic convex semialgebraic set have a lifted SDR? Answers to both are unknown as of now Some partial results known. See references Let us look at SDR and approx SDR for polynomials 8 / 25

35 Polynomials Def. (Polynomial). Let K be a field and x 1,..., x n be indeterminates. A polynomial f with coefficients in a field K is a finite linear combination of monomials: f = α c α x α = α x α 1 1 xαn n, c α K; we sum over finite n-tuples α = (α 1,..., α n ), each α i N 0. Degree: d = i α i (largest such sum over all α) 9 / 25

36 Polynomials Def. (Polynomial). Let K be a field and x 1,..., x n be indeterminates. A polynomial f with coefficients in a field K is a finite linear combination of monomials: f = α c α x α = α x α 1 1 xαn n, c α K; we sum over finite n-tuples α = (α 1,..., α n ), each α i N 0. Degree: d = i α i (largest such sum over all α) Def. Ring of all polynomials K[x 1,..., x n ] 9 / 25

37 Polynomials Def. (Polynomial). Let K be a field and x 1,..., x n be indeterminates. A polynomial f with coefficients in a field K is a finite linear combination of monomials: f = α c α x α = α x α 1 1 xαn n, c α K; we sum over finite n-tuples α = (α 1,..., α n ), each α i N 0. Degree: d = i α i (largest such sum over all α) Def. Ring of all polynomials K[x 1,..., x n ] Eg: Univariate polynomials with real coefficients R[x] 9 / 25

38 Nonnegativity We care about whether p(x) 0 for all x 10 / 25

39 Nonnegativity We care about whether p(x) 0 for all x Question equivalent to SDR for univariate polynomials 10 / 25

40 Nonnegativity We care about whether p(x) 0 for all x Question equivalent to SDR for univariate polynomials For multivariate polynomials, question remains very important 10 / 25

41 Nonnegativity We care about whether p(x) 0 for all x Question equivalent to SDR for univariate polynomials For multivariate polynomials, question remains very important (Nonnegativity intimately tied to convexity (formally real fields, algebraic closure, ordered property etc.)) 10 / 25

42 Nonnegativity We care about whether p(x) 0 for all x Question equivalent to SDR for univariate polynomials For multivariate polynomials, question remains very important (Nonnegativity intimately tied to convexity (formally real fields, algebraic closure, ordered property etc.)) If p(x) 0, then degree of p must be even Set of nonnegative polynomials quite interesting. Theorem Let P n denote the set of all nonnegative univariate polynomials of degree n. Identifying a polynomial with its n + 1 coefficients (p n,..., p 0 ), the set P n is a closed, convex, pointed cone in R n+1 10 / 25

43 Testing nonnegativity Def. (SOS). A univariate polynomial p(x) is a sum of squares (SOS) if there exist q 1,..., q m R[x] such that p(x) = m k=1 q2 k (x). 11 / 25

44 Testing nonnegativity Def. (SOS). A univariate polynomial p(x) is a sum of squares (SOS) if there exist q 1,..., q m R[x] such that p(x) = m k=1 q2 k (x). Theorem A univariate polynomial is nonneg if and only if it is SOS Proof: Obviously, if p(x) is SOS, then p(x) / 25

45 Testing nonnegativity Def. (SOS). A univariate polynomial p(x) is a sum of squares (SOS) if there exist q 1,..., q m R[x] such that p(x) = m k=1 q2 k (x). Theorem A univariate polynomial is nonneg if and only if it is SOS Proof: Obviously, if p(x) is SOS, then p(x) 0. For converse, recall by the fundamental theorem of algebra, we can factorize p(x) = p n (x r j) nj (x z k) m k (x z k ) m k, j k where r j and z k are real and complex roots, respectively. 11 / 25

46 Testing nonnegativity Def. (SOS). A univariate polynomial p(x) is a sum of squares (SOS) if there exist q 1,..., q m R[x] such that p(x) = m k=1 q2 k (x). Theorem A univariate polynomial is nonneg if and only if it is SOS Proof: Obviously, if p(x) is SOS, then p(x) 0. For converse, recall by the fundamental theorem of algebra, we can factorize p(x) = p n (x r j) nj (x z k) m k (x z k ) m k, j k where r j and z k are real and complex roots, respectively. Since p(x) 0, p n > 0, multiplicities n j of real roots are even. Also, note (x z)(x z) = (x a) 2 + b 2, if z = a + ib. 11 / 25

47 Testing nonnegativity Def. (SOS). A univariate polynomial p(x) is a sum of squares (SOS) if there exist q 1,..., q m R[x] such that p(x) = m k=1 q2 k (x). Theorem A univariate polynomial is nonneg if and only if it is SOS Proof: Obviously, if p(x) is SOS, then p(x) 0. For converse, recall by the fundamental theorem of algebra, we can factorize p(x) = p n (x r j) nj (x z k) m k (x z k ) m k, j k where r j and z k are real and complex roots, respectively. Since p(x) 0, p n > 0, multiplicities n j of real roots are even. Also, note (x z)(x z) = (x a) 2 + b 2, if z = a + ib. Thus, we have p(x) = j (x r j) 2sj k [(x a k) 2 + b 2 k] m k. Expand out above product of SOS into a sum to see that p(x) is SOS. 11 / 25

48 SOS Exercise: Show that in fact if p(x) 0, then it can be written as a sum of just two squares, i.e., p(x) = q1 2(x) + q2 2 (x). (Hint: It may help to notice (a 2 + b 2 )(c 2 + d 2 ) = (ac bd) 2 + (ac + bd) 2 ) 12 / 25

49 SOS Exercise: Show that in fact if p(x) 0, then it can be written as a sum of just two squares, i.e., p(x) = q1 2(x) + q2 2 (x). (Hint: It may help to notice (a 2 + b 2 )(c 2 + d 2 ) = (ac bd) 2 + (ac + bd) 2 ) Unfortunately, for multivariate polynomials SOS not equivalent to p(x 1,..., x m ) 0 (Motzkin polynomial) M(x, y) := x 4 y 2 +x 2 y x 2 y 2 nonneg but not SOS. 12 / 25

50 SOS and SDP Theorem Let p(x) be of degree 2d. Then, p(x) 0 (or SOS) if and only if there exists a Q S+ d+1 that satisfies p(x) = z T Qz, where z = [1, x,..., x d ] T. 13 / 25

51 SOS and SDP Theorem Let p(x) be of degree 2d. Then, p(x) 0 (or SOS) if and only if there exists a Q S+ d+1 that satisfies p(x) = z T Qz, where z = [1, x,..., x d ] T. If p(x) 0, then we have p(x) = m i qi 2(x) 13 / 25

52 SOS and SDP Theorem Let p(x) be of degree 2d. Then, p(x) 0 (or SOS) if and only if there exists a Q S+ d+1 that satisfies p(x) = z T Qz, where z = [1, x,..., x d ] T. If p(x) 0, then we have p(x) = m i qi 2(x) Obviously, degree of any q i at most d 13 / 25

53 SOS and SDP Theorem Let p(x) be of degree 2d. Then, p(x) 0 (or SOS) if and only if there exists a Q S+ d+1 that satisfies p(x) = z T Qz, where z = [1, x,..., x d ] T. If p(x) 0, then we have p(x) = m i qi 2(x) Obviously, degree of any q i at most d Write a vector of polynomials q 1 (x) x 0 q 2 (x). = V x 1. q m (x) x d where row i of V R m (d+1) contains coefficients of the q i. 13 / 25

54 SOS and SDP Denote [x] d := [x 0, x 1,..., x d ] T 14 / 25

55 SOS and SDP Denote [x] d := [x 0, x 1,..., x d ] T Then, since q = V [x] d, we have i q2 i (x) = (V [x] d) T (V [x] d ) which is nothing but [x] T d Q[x] d, where Q = V T V / 25

56 SOS and SDP Denote [x] d := [x 0, x 1,..., x d ] T Then, since q = V [x] d, we have i q2 i (x) = (V [x] d) T (V [x] d ) which is nothing but [x] T d Q[x] d, where Q = V T V 0. Conversely, if there is a Q such that p(x) = [x] T d Q[x] d, just take Cholesky factorization Q = R T R, to obtain SOS decomp. of p 14 / 25

57 SOS and SDP Denote [x] d := [x 0, x 1,..., x d ] T Then, since q = V [x] d, we have i q2 i (x) = (V [x] d) T (V [x] d ) which is nothing but [x] T d Q[x] d, where Q = V T V 0. Conversely, if there is a Q such that p(x) = [x] T d Q[x] d, just take Cholesky factorization Q = R T R, to obtain SOS decomp. of p If we are given p, how to find SOS decomp / matrix Q? Remark: N. Z. Shor (inventor of subgradient method), seems to be first to establish connection between SOS decompositions and convexity. 14 / 25

58 SOS and SDP ` SOSTOOLS package automatically translates between SOS polynomial and its SDP representation. 15 / 25

59 SOS and SDP ` SOSTOOLS package automatically translates between SOS polynomial and its SDP representation. Suppose p(x) 0, then p(x) = [x] T d Q[x] d. 15 / 25

60 SOS and SDP ` SOSTOOLS package automatically translates between SOS polynomial and its SDP representation. Suppose p(x) 0, then p(x) = [x] T d Q[x] d. We need to find Q. 15 / 25

61 SOS and SDP ` SOSTOOLS package automatically translates between SOS polynomial and its SDP representation. Suppose p(x) 0, then p(x) = [x] T d Q[x] d. We need to find Q. Expanding out the product above we have d j,k=0 q jkx j+k = 2d i=0 ( j+k=i q jk ) x i. Since p(x) = p n x n p 1 x + p 0. Thus, matching coeffts 15 / 25

62 SOS and SDP ` SOSTOOLS package automatically translates between SOS polynomial and its SDP representation. Suppose p(x) 0, then p(x) = [x] T d Q[x] d. We need to find Q. Expanding out the product above we have d j,k=0 q jkx j+k = 2d i=0 ( j+k=i q jk ) x i. Since p(x) = p n x n p 1 x + p 0. Thus, matching coeffts p i = j+k=i q jk, i = 0,..., 2d. These are 2d + 1 linear constraints on Q We also have Q 0 15 / 25

63 SOS and SDP ` SOSTOOLS package automatically translates between SOS polynomial and its SDP representation. Suppose p(x) 0, then p(x) = [x] T d Q[x] d. We need to find Q. Expanding out the product above we have d j,k=0 q jkx j+k = 2d i=0 ( j+k=i q jk ) x i. Since p(x) = p n x n p 1 x + p 0. Thus, matching coeffts p i = j+k=i q jk, i = 0,..., 2d. These are 2d + 1 linear constraints on Q We also have Q 0 Thus, finding feasible Q is an SDP 15 / 25

64 Mini-challenge Exercise: Prove that for 1 n m, the polynomial p(x) = 1 ( 2m 2 2n) (1 + x) 2m 2n + 1 2q(x) is nonnegative, where q(x) = ( ) m 2m (1 x) 2m 2j ( 4x) j n. j=n 2j 16 / 25

65 Mini-challenge Exercise: Prove that for 1 n m, the polynomial p(x) = 1 ( 2m 2 2n) (1 + x) 2m 2n + 1 2q(x) is nonnegative, where q(x) = ( ) m 2m (1 x) 2m 2j ( 4x) j n. j=n 2j Other computational tricks may be more suitable? 16 / 25

66 Mini-challenge Exercise: Prove that for 1 n m, the polynomial p(x) = 1 ( 2m 2 2n) (1 + x) 2m 2n + 1 2q(x) is nonnegative, where q(x) = ( ) m 2m (1 x) 2m 2j ( 4x) j n. j=n 2j Other computational tricks may be more suitable? Remark: We note that testing nonnegativity of multivariate polynomials (of degree 4 or higher) is NP-Hard. 16 / 25

67 Application Global optimization of a univariate polynomial p(x) 17 / 25

68 Application Global optimization of a univariate polynomial p(x) Instead of seeking x argmin p(x), first attempt to find a good lower bound on optimal value p(x ) 17 / 25

69 Application Global optimization of a univariate polynomial p(x) Instead of seeking x argmin p(x), first attempt to find a good lower bound on optimal value p(x ) A number γ is a global lower bound on p(x), iff p(x) γ x p(x) γ 0, x. 17 / 25

70 Application Global optimization of a univariate polynomial p(x) Instead of seeking x argmin p(x), first attempt to find a good lower bound on optimal value p(x ) A number γ is a global lower bound on p(x), iff p(x) γ x p(x) γ 0, x. Now optimize to get tightest bound, so max γ s.t. p(x) γ is SOS. Turn this into SDP for SOS; solve SDP to obtain γ 17 / 25

71 Application Global optimization of a univariate polynomial p(x) Instead of seeking x argmin p(x), first attempt to find a good lower bound on optimal value p(x ) A number γ is a global lower bound on p(x), iff p(x) γ x p(x) γ 0, x. Now optimize to get tightest bound, so max γ s.t. p(x) γ is SOS. Turn this into SDP for SOS; solve SDP to obtain γ Note, optimal γ gives global minimum of polynomial, even though p may be highly nonconvex! 17 / 25

72 Applications Polynomials nonnegative only over an interval 18 / 25

73 Applications Polynomials nonnegative only over an interval Minimizing ratio of two polynomials (where q(x) > 0) p(x) γ p(x) γq(x) 0. q(x) 18 / 25

74 Applications Polynomials nonnegative only over an interval Minimizing ratio of two polynomials (where q(x) > 0) p(x) γ p(x) γq(x) 0. q(x) Several others (in nonlinear control, etc.) 18 / 25

75 Applications Polynomials nonnegative only over an interval Minimizing ratio of two polynomials (where q(x) > 0) p(x) γ p(x) γq(x) 0. q(x) Several others (in nonlinear control, etc.) (Lower bounds for minima of multivariate polynomials) 18 / 25

76 References P. Parrilo. Algebraic techniques and semidefinite optimization. MIT course, P. Parrilo s website. G. Blekherman, P. Parrilo, R. R. Thomas. Semidefinite optimization and convex algebraic geometry (2012). 19 / 25

77 What we did not cover? See Springer Encyclopedia on Optimization (over 4500 pages!) Convex relaxations of nonconvex problems in greater detail Algorithms (trust-region methods, cutting plane techniques, bundle methods, active-set methods, and 100s of others) Applications of our techniques Software, systems ideas techniques, implementation details Theory: convex analysis, geometry, probability Noncommutative polynomial optimization (where often we might just care for just a feasibility test) Convex optimization in inf-dimensional Hilbert, Banach spaces Semi-infinite and infinite programming Multi-stage stochastic programming, chance constraints, robust optimization, tractable approximations of hard problems Optimization on manifolds, on matrix manifolds And 100s of other things! 20 / 25

78 Thanks! Hope you learned something new!! 21 / 25

79 Thanks! Hope you learned something new!! 21 / 25

80 Some algebra Ideals and cones Given a set of multivariate polynomials {f 1,..., f m }, define { ideal(f 1,..., f m ) := f f = } t if i, t i R[x]. i cone(f 1,..., f m ) := g g = s 0 + s i f i + s ij f i f j +..., {i} {i,j} where each term is a squarefree product of f i, with a coefficient s α R[x] that is a sum of squares. The sum is finite, with a total of 2 m 1 terms, corresponding to the nonempty subsets of {f 1,..., f m }. 22 / 25

81 Algebraic connections Note: Every polynomial in ideal(f i ) vanishes in the solution set of f i (x) = 0. Note: Every element of cone(f i ) is nonnegative on the feasible set f i (x) 0. Example Ax = b is infeasible there exists a µ, such that A T µ = 0 and b T µ = 1. Theorem Hilbert s Nullstellensatz: Let f 1 (z),..., f m (z) be polynomials in complex variables z 1,..., z n. Then, f i (z) = 0, (i = 1,..., m) is infeasible in C n 1 ideal(f 1,..., f m ). Exercise: Verify the easy direction of the above theorems. 23 / 25

82 Semialgebraic connections Farkas lemma and Positivstellensatz Theorem (Farkas lemma). Ax + b = 0 and Cx + d 0 is infeasible is equivalent to { A T µ + C T λ = 0 λ 0, µ s.t. b T µ + d T λ = 1. Theorem (Positivstellensatz). The system f i (x) = 0 for i = 1,..., m and g i (x) 0 for i = 1,..., p is infeasible in R n is equivalent to F (x) + G(x) = 1 F (x), G(x) R[x] s.t. F (x) ideal(f 1,..., f m ) G(x) cone(g 1,..., g p ). 24 / 25

83 What it means? For every infeasible system of polynomial equations and inequalities, there exists a simple algebraic identity that directly certifies non-existence of real solutions. Evaluation of polynomial F (x) + G(x) at any feasible point should produce a nonnegative number. But this expression is identically equal to 1, a contradiction. Degree of F (x) and G(x) can be exponential. These cones and ideals are always convex sets (regardless of original polynomial); similar to dual function being always concave, regardless of primal. 25 / 25