NON-NEGATIVE POLYNOMIALS

CARLOS ANGLAS AND DWAYNE WOOLWARD

Contents

Introduction 21. Non-Negative Polynomials on the Whole Real Line 32. Non-Negative Polynomials on the Half-Line 113. Non-Negative Polynomials on [0, 1] 154. Non-Negative Polynomials in Two-Variables 25References 30Appendix A. Factorization Theorem 31Appendix B. Section 1 Examples 34Appendix C. Section 2 Examples 43Appendix D. Section 3 Examples 52Appendix E. Section 4 Examples 77

Research was conducted during the Spring, 2010 and Summer I, 2010 semesters with ProfessorLawrence A. Fialkow as the mentor. During Spring, 2010 the project was sponsored by SUNY/NSFAlliance for Minority Participation at SUNY New Paltz. The New Paltz AMP Director is ProfessorStacie Nunes from the Department of Physics. During Summer I, 2010 the project was sponsoredunder National Science Foundation Grant DMS0758378. Special thanks to the sponsors for theirgenerous help and support of this project.

1

Introduction

The use of positive polynomials in one or more variables is essential in severalareas of mathematics, including optimization and algebraic geometry [4][3]. As forpolynomials that are non-negative on the real line, half-line, or on an interval, thispaper derives a “weighted sums of squares” description. These results are knownfrom Polya and Szego [5], where the proofs are based on complex analysis. In thispaper, starting from the Factorization Theorem for real polynomials, we prove thesetheorems using only basic algebra. These proofs explain how to construct the weightedsums of squares expressions, and we illustrate these constructions with examples in thetext and in the Mathematica appendices. In section 1 (Theorem 1.1), we discuss howevery polynomial which is non-negative on the whole real line is a sum of two squaresof polynomials. In section 2 (Theorem 2.1), we use Theorem 1.1 to demonstrate thatevery non-negative polynomial on the half line [0,+∞) is of the form f 2+g2+x(h2+k2)for certain polynomials f , g, h, and k. In section 3 (Theorem 3.1), we use Theorems1.1 and 2.1 to demonstrate that every non-negative polynomial on the interval [0, 1]can be represented in the form f 2 + g2 + x(1 − x)(h2 + k2) for certain polynomialsf , g, h, and k. In Section 4, we look at non-negative polynomials in two variables.First, by studying the example of Motzkin, we prove that not every non-negativepolynomial of deg 6 can be represented as a sum of squares. We also study a methodfor identifying when a polynomial is a sum of squares.

2

1. Non-Negative Polynomials on the Whole Real Line

1.1. Section A. In this section we look at non-negative polynomials on the real lineand introduce our first and most important theorem. For polynomials f(x) and g(x),the expression f(x)2+g(x)2 is known as a sum of two squares. We know that if p(x) isa sum of two squares, then p(x) ≥ 0 for all x. But if p(x) ≥ 0 for all x (a non-negativepolynomial), then is p(x) necessarily a sum of two squares? The following theorem,from Polya and Szego, answers this question.

Theorem 1.1. (See Polya and Szego [5, Part VI, Sec. 6, 44, page 77].) If we have apolynomial p(x) ≥ 0 for all x, then p(x) admits polynomials f(x) and g(x) such thatp(x) = f(x)2 + g(x)2 for all x.

In an attempt to prove Theorem 1.1, we first turn to the quadratic q(x) =ax2 + bx + c, assuming a 6= 0 and q(x) ≥ 0 for all x . Consider the three possiblecases:

(i) q(x) has no real roots(ii) q(x) has one repeated, real root

(iii) q(x) has two distinct, real roots

1.1.1. Case 1: No Real Roots. In order to compute the roots of q(x) = 0, we use thequadratic formula

(1.1) x =−b±

√b2 − 4ac

2a.

From this we see there are no real roots when the discriminant of the quadratic,b2 − 4ac, is negative. Suppose q(x) > 0 for all x. We write q(x) as

(1.2) q(x) = a

(x2 +

b

ax+

c

a

).

Since limx→+∞

x2 +b

ax+

c

a= +∞, it follows that a > 0. We can now express q(x) as a

sum of two squares by completing the squares, as follows:

q(x) = a

(x2 +

b

ax+

c

a

)=

= a

[x2 +

bx

a+

(b

2a

)2

−(b

2a

)2

+c

a

]

= a

[(x+

b

2a

)2

−(b

2a

)2

+c

a

]

= a

(x+

b

2a

)2

+ a

[−(b

2a

)2

+c

a

]

=

[√a

(x+

b

2a

)]2+

[−a(b

2a

)2

+ c)

]3

=

[(√a)(

x+b

2a

)]2+

4ac− b2

4a.

From the final expression, we let

f(x) =√a

(x+

b

2a

)and let

α =

√4ac− b2

4a.

So now we have q(x) = f(x)2 + (α)2, a sum of two squares.We illustrate this case with the following example.

Example 1.2. We start with a polynomial, not in its sum of squares form.p = x2 + 5x+ 1010 + 5x+ x2

Factor[p]10 + 5x+ x2

Now we check to see if p has any roots using the discriminant, b2 − 4ac < 0a = 1b = 5c = 10b2 − 4ac = −15-15 is less than zero so p has no real roots.Plot[p, x,−10, 10]

-10 -5 5 10

50

100

150

The plot shows us that p is positive for all x. Now we can convert p into its sum oftwo squares form by completing the square.(√

a

(x+

b

2a

))2

+

(√4ac− b2

4a

)2

=15

4+

(5

2+ x

)2

f =5

2+ x

α =

√15

4

f 2 + α2 =15

4+

(5

2+ x

)2

4

This our new sum of squares form.Expand[%] = 10 + 5x+ x2

Expanded, we get the polynomial we started off with.

1.1.2. Case 2: One Repeated, Real Root. Suppose q(x) = 0 has one root, xo, repeatedtwice, so that q(x) = a(x− x0)2, with a 6= 0. Here a must be positive since q(x) ≥ 0for all x. To make q(x) a sum of two squares we simply take f(x) =

√a(x− xo) and

g(x) = 0, i.e., q(x) = f(x)2 + 02. Here is an example of a polynomial with repeatedroots.

Example 1.3. We have a polynomial with a repeated root, a(x− x0)2P = (x− 3)2

Expand[P ] = 9− 6x+ x2

Factor[P ] = (−3 + x)2

Plot[P, x,−5, 10]

-4 -2 2 4 6 8 10

10

20

30

40

50

60

Plot shows p is non-negative for all xf = (x− 3)g = 0Expand[f 2 + g2] = 9− 6x+ x2

Expand[P − (f 2 + g2)] = 0So our sum of two squares is simply (x− 3)2 + 02

1.1.3. Case 3: Two Distinct, Real Roots. For there to be two distinct real roots, thenq(x) = a(x − xo)(x − x1), with a 6= 0 and xo 6= x1 But then q(x) changes sign atx = xo and x = x1, so this case cannot occur in Theorem 1.1. Here is an example ofa polynomial with two distinct roots.

Example 1.4. A polynomial of the form a(x−xo)(x−x1) with a = 3, xo = −3, x1 = 4P = −36 + 3x+ 3x2

Factor[P ] = 3(−3 + x)(4 + x)Plot[P, x,−10, 10]

5

-10 -5 5 10

50

100

150

200

250

The plot clearly shows that the function is negative as it goes below the x - axis, itchanges sign at the roots +3,−4. This is a contradiction of the rule p ≥ 0 for all x,so this case cannot occur.

1.2. Section B. In the proof of Theorem 1.1, we use the following theorem.

Theorem 1.5. (Factorization Theorem) If p(x) is a polynomial, then

p(x) = Aq1(x) . . . qk(x)(x− x1)n1 . . . (x− xr)nr(x− y1)m1 . . . (x− ys)ms ,

where A is a constant, each of q1(x) . . . qk(x) are strictly non-zero quadratics, x1 . . . xr,y1 . . . ys are all distinct, n1 . . . nr are even exponents, and m1 . . .ms are odd exponents,(Note that any of these groups may be missing in the expression).

For the proof, see Appendix A. Here is an example of the factorization theorem.

Example 1.6. We start with our polynomial pp = 11150950400+1944616960x−94936064x2+12993152x3−4111322x4−650222x5+150728x6 − 8798x7 + 166x8

Factor[p] = 166(−16 + x)4(5 + x)2(41 + x+ x2)p is now in its factored form given by the Factorization Theorem. Notice A = 166, q1 =x2 + x+ 41 with no real roots, and then there are some more linear terms with evenexponents.Plot[p, x,−10, 20]

-10 -5 5 10 15 20

1.´1010

2.´1010

3.´1010

4.´1010

Plot tells us p is non-negative for all x. But what happens when we add a term withan odd exponent? Observe r.r = 166(x2 + x+ 41)(x− 16)4(x+ 5)2(x+ 5)3

Factor[r] = 166(−16 + x)4(5 + x)5(41 + x+ x2)Plot[r, x,−10, 3]

6

-10 -8 -6 -4 -2 2

-4.´1012

-2.´1012

2.´1012

4.´1012

This plot tells us that the linear term with the odd exponent made p negative forx < −5.

1.3. Section C. In this section we introduce a lemma which shows that a product oftwo terms, each a sum of two squares, can also be expressed as a sum of two squares.The lemma can be applied as many times as needed, depending on the number ofterms in a product. For example, if you have three terms, each sums of two squares,you will use the lemma twice, once to combine the first two terms into a sum of twosquares, and once more to combine with the last term. In the end, this will result ina polynomial of the form f(x)2 + g(x)2.

Lemma 1.7. If A(x) = f(x)2 + g(x)2(Type I) and B(x) = h(x)2 + k(x)2(Type I),then A(x) × B(x) can be expressed as R(x)2 + S(x)2(Type I), where R(x) and S(x)are polynomials; in brief, Type I×Type I = Type I.

Proof. We can expand the product A(x)B(x) :

A(x)B(x) = f 2h2 + f 2k2 + g2h2 + g2k2

=(f 2h2 + 2fhgk + g2k2

)+(f 2k2 − 2fhgk + g2h2

).

Factoring the sub-expressions further will give us two sums of squares:

f 2h2 + 2fghk + g2k2 = (fh+ gk)2

g2h2 − 2fghk + f 2k2 = (fk − gh)2.

So A(x)B(x) = R(x)2 + S(x)2 with R(x) = fh+ gk and S(x) = fk − gh.�

The following example shows how Lemma 1.7 is applied.

Example 1.8. We start with a polynomial of several termsp = 3360 + 224x+ 1002x2 − 353x3 + 101x4 − 71x5 + 12x6

Factor[p] = (−4 + x)2(15 + x+ 3x2)(14 + 7x+ 4x2)Plot[p, x,−10, 10]

7

-10 -5 5 10

2.´106

4.´106

6.´106

8.´106

Plot shows polynomial is non-negative for all x.Expand[(15 + x+ 3x2)] = 15 + x+ 3x2

a = 3b = 1c = 15((√

a)(

x+b

2a

))2

+

(√4ac− b2

4a

)2

=179

12+ 3

(1

6+ x

)2

Prove that the completed square expression and the original are equal. Expand [(15 + x+ 3x2)−%] =0Now combine (−4 + x)2(15 + x+ 3x2) using the Lemma.X = Expand[(−4 + x)2(15 + x+ 3x2)] = 240− 104x+ 55x2 − 23x3 + 3x4

X represents the first two terms.f = (−4 + x)g = 0

h =

√179

12

k =√

3

(1

6+ x

)We’ll let A = (fh+ gk) and B = (fk − gh)

A = fh+ gk =1

2

√179

3(−4 + x)

B = fk − gh =√

3(−4 + x)(1

6+ x)

Expand[X − (A2 +B2)] = 0There is the product of our two sum of squares. Double-checking, we get a zero toshow we haven’ t changed anything.Convert polynomial to sum of squares form.Expand[14 + 7x+ 4x2] = 14 + 7x+ 4x2

a = 4b = 7c = 14((√

a)(

x+b

2a

))2

+

(√4ac− b2

4a

)2

=175

16+ 4

(7

8+ x

)2

8

Now lets combine the third term in sums of two squares form using the Lemma again.

f = A =1

2

√179

3(−4 + x)

g = B =√

3(−4 + x)

(1

6+ x

)h =

√175

16

k = 2

(7

8+ x

)AA = fh+ gk =

5

8

√1253

3(−4 + x) + 2

√3(−4 + x)

(1

6+ x

)(7

8+ x

)AA = Factor[AA] =

((−4 + x)(7 + 5√

1253 + 50x+ 48x2))

(8√

3)

Expand[AA] = − 72√3− 5√

12533

2− 193x

8√3

+ 58

√12533x− 71x2

4√3

+ 2√

3x3

BB = fk − gh = −5

4

√21(−4 + x)

(1

6+ x

)+

√179

3(−4 + x)

(7

8+ x

)BB = Factor[BB] = −(((−4 + x)(5

√7− 7

√179 + 30

√7x− 8

√179x)))

(8√

3)

Expand[BB] =5√

73

2− 7√

1793

2+ 115

8

√73x− 25

8

√1793x− 5

√21x2

4+√

1793x2

Expand[p− (AA2 +BB2)] = 0Now we have one big sum of squares that equals our original polynomial p.

Proof of Theorem 1.1

Proof. Suppose p(x) ≥ 0 for every real number x. By the Factorization Theorem,we have p(x) = Aq1(x) . . . qk(x) (x− x1) n1 . . . (x− xr)nr (x− y1) m1 . . . (x− ys) ms ,where the qi are quadratics with no real roots, x1, . . . , xr, y1 . . . , ys are distinct,n1, . . . , nr are even, say ni = 2di, and m1, . . . ,ms are odd. We can also assume A 6= 0.Since p(x) ≥ 0 for all x, and mi is odd, we cannot have any factors (x− yi)mi , sincep(x) will change sign at the root yi. So p(x) = Aq1(x) . . . qk(x)(x−x1)n1 . . . (x−xr)nr .If some qi(x) < 0 for all x, we can write

p(x) = (-1)Aq1(x) . . . (-qi(x)) . . . (qk(x))(x− x1)n1 . . . (x− xr)nr

(notice the −1’s cancel, leaving the value of p(x) unchanged). We now have p(x) =

Aqi(x) . . . qk(x)(x−x1)n1 . . . (x−xr)nr , where each qi(x) is a strictly positive quadratic.Since qi(x) > 0 for all x, each factor (x−xi)ni ≥ 0 for all x, and p(x) ≥ 0 for all x, we

must have A > 0. So A is a square of the form√A

2

, each qi(x) is a sum of 2 squares

(by section A, case 1), and each (x − xi)ni is a square, (x − xi)

ni =[(x− xi)di

]2.

The proof is now completed by applying Lemma 1.7 to the factors in p(x), one afteranother. �

The following corollary generalizes Lemma 1.7.9

Corollary 1.9. If A(x) =n∑

i=1

fi2(x) and B(x) =

n∑i=1

gi2(x), then A(x) × B(x) can

be expressed as R(x)2 + S(x)2, a sum of two squares.

Proof. Since A(x)×B(x) is non-negative for every x, the Type I representation followsfrom Theorem 1.1. �

10

2. Non-Negative Polynomials on the Half-Line

In this section we look at non-negative polynomials on the half-line and intro-duce our second theorem.

Theorem 2.1. (See Polya and Szego [5, Part VI, Sec. 6, 45, page 78].) If we havea polynomial p(x) that satisfies p(x) ≥ 0 for every x ≥ 0, then there are polynomialsf(x), g(x), h(x), and k(x) such that p(x) = f(x)2 + g(x)2 + x[h(x)2 + k(x)2] for allx.

We can use Theorem 1.1 to prove Theorem 2.1. But before we begin we mustdefine a couple of things to make the process easier.

• A Type I polynomial is a polynomial p(x) with a representation as in Theorem1.1, a sum of two squares, i.e. f(x)2 + g(x)2.• A Type II polynomial is a polynomial p(x) with a representation as in the form

of Theorem 2.1, ([Type I] + x[Type I]), i.e. f(x)2 + g(x)2 + x[h(x)2 + k(x)2].

Now that we have defined the two different types, we can continue. In Theorem 1.1,we used Lemma 1.7, which shows that Type I × Type I = Type I. Similarly, to proveTheorem 2.1, we need the following result:

Lemma 2.2. If xi ≤ 0, then Type II × (x− xi) is also of the form Type II.

Proof. Suppose p(x) is Type II: p = f 2 + g2 + x(h2 + k2), where f , g, h, and k areall polynomials. Let q(x) := p(x − xi) = x(f 2 + g2) − xi(f 2 + g2) + x2(h2 + k2) −(x)(xi)(h

2 + k2)

=[x2(h2 + k2)− xi(f 2 + g2)

]︸ ︷︷ ︸P (x)

+x[(f 2 + g2)− xi(h2 + k2)

]︸ ︷︷ ︸Q(x)

.

Since −xi ≥ 0, we can see that P (x) ≥ 0 and Q(x) ≥ 0 for each x, so by Theorem1.1, P (x) and Q(x) have Type I representations. At the same time, we see thatq(x) ≡ P (x) + xQ(x) has a Type II representation. �

Example 2.3. Suppose P is a non-negative polynomial for x ≥ 0.P = 7 + 10x+ 4x2 + x3

Factor[P ] = (1 + x)(7 + 3x+ x2)xo = −5Q = Expand[P (x− xo)] = 35 + 57x+ 30x2 + 9x3 + x4

Plot[P, x,−6, 5]

-6 -4 -2 2 4

500

1000

1500

11

From the plot, we can clearly see that P > 0 when x > 0.Factor[Q] = (1 + x)(5 + x) (7 + 3x+ x2)Convert to sum of squares form using Lemma 1.7 from Theorem 1.1.Z=Expand[7 + 3x+ x2] = 7 + 3x+ x2

a = 1b = 3c = 7((√a)(x+ b

2a

))2+(√

4ac−b∧24a

)2= 19

4+(32

+ x)2

Expand[Z −%] = 0Prove P has Type II.

f =

√19

2

g =

(3

2+ x

)h =

√19

2

k =

(3

2+ x

)P has a Type II representation.Expand[P − (f 2 + g2 + x(h2 + k2))] = 0Apply Lemma 2.2 from Theorem 2.1 with xo = −5: TypeII × (x − xo) = TypeII.We show that Q is Type II.

R = x2(h2 + k2)− xo(f 2 + g2) = 5(

194

+(32

+ x)2)

+ x2(

194

+(32

+ x)2)

Factor[R] = (5 + x2)(7 + 3x+ x2)((√a)(x+ b

2a

))2+

(√4ac−b2

4a

)2

= 194

+(32

+ x)2

f = xg =√

5

h =√192

k = 32

+ x

A = fk − gh = −√952

+ x(32

+ x)

B = fh+ gk =√19x2

+√

5(32

+ x)

Expand[R− (A2 +B2)] = 0R has Type I representation

S = (f 2 + g2)− xo(h2 + k2) = 194

+(32

+ x)2

+ 5(

194

+(32

+ x)2)

Factor[S] = 6(7 + 3x+ x2)f = 0g =√

6

h =√192

k =(32

+ x)

12

J = fk − gh = −√

572

K = fh+ gk =√

6(32

+ x)

Expand[S − (J2 +K2)] = 0S has a Type I representation.Expand[Q− (A2 +B2 + x(J2 +K2))] = 0Polynomial Q has a Type II representation.

Proof of Theorem 2.1

Proof. We use the Factorization Theorem to write p from Theorem 2.1 as:

p(x) = A [q1 . . . qk(x− y1)n1 . . . (x− ym)nm(x− x1)s1 . . . (x− xp)sp ] ,

where A is a constant, the qi’s are quadratics that cannot be factored (i.e., theyhave no real roots), the exponents ni are even, the exponents si are odd, and all ofx1,. . .xp,y1,. . . ym are distinct. (Note some of these factors may be missing in anyexpression p.) Since p(x) ≥ 0 for all x ≥ 0, we can assume that each qi is positive forall x, each xi ≤ 0 (since p(x) changes sign at xi), and A > 0. Let’s say si = 2ti + 1,with ti ≥ 0, so that (x− xi)si = (x− xi)2ti (x− xi). Let

r(x) := Aq1 . . . qk(x− y1)n1 . . . (x− ym)nm(x− x1)2t1 . . . (x− xp)2tp ,

so that p(x) = r(x)(x− xi) . . . (x− xp).Clearly, r(x) ≥ 0 for every x, so r(x) has a Type I representation by Theorem 1.1,and therefore a Type II representation:

r(x) = f(x)2 + g(x)2 = f(x)2 + g(x)2 + x(02 + 02).

If there are no xi’s, then we just have p(x) = r(x), so p(x) has a Type II representation.If there are some xi’s, they must be ≤ 0, since p(x) changes sign at xi.Now, we can apply Lemma 2.2 to

p(x) = r(x)(x− x1)(x− x2) . . . (x− xp)︸ ︷︷ ︸R(x)

Since each xi ≤ 0, Lemma 2.2 implies that r(x)(x − x1), and each partial productr(x)(x− x1) . . . (r − xi) is Type II, so p(x) ≡ R(x) is Type II. �

Before continuing to the next section we will introduce three facts concerningdegrees of polynomials. We will use these facts in Theorem 3.1.

2.1. Fact 1. First we compute the degree of a sum of squares of two polynomials.Let q(x) = a0 + aix+ . . .+ anx

n, where an 6= 0 (so the deg q = n).Let r(x) = b0 + bix+ . . .+ bmx

m, where bm 6= 0 (so the deg r = m).Then s(x) := q(x)2 + r(x)2

= ao2 + . . .+ 2an−1anx

2n−1 + an2x2n + bo

2 + . . .+ 2bm−1bmx2m−1 + b2mx

2m.13

But what is the degree of s(x)? Let’s consider the degrees n and m.

deg s(x) =

2n n > m2m m > n

2n = 2m n = m.According to this, the degree of the sum is twice n if n > m, twice m if m > n, ortwice n or m if n = m. So we can say the degree of the sum is twice the highestdegree,

(2.1) deg (q(x)2 + r(x)2) = 2max(deg q(x), deg r(x))

2.2. Fact 2. For any polynomials a(x), b(x), the degree of their sum, a(x) + b(x),satisfies

(2.2) deg (a(x) + b(x)) ≤ max (deg a(x), deg b(x)) .

Also, if the highest degree coefficients in a(x) and b(x) are both positive, then

(2.3) deg (a(x) + b(x)) = max(deg a(x), deg b(x)).

2.3. Fact 3. Consider a Type II polynomial:

p(x) = f(x)2 + g(x)2 + x[h(x)2 + g(x)2

]Then by Fact 1, deg (f(x)2 + g(x)2) = 2max(deg f(x), deg g(x)). Also by Fact 1,deg (h(x)2 + k(x)2) = 2max(deg h(x), deg k(x)). Then deg (x(h(x)2 + k(x)2)) =

1 + 2max(deg h(x), deg k(x)) = max(1 + 2deg h(x), 1 + 2deg k(x)).

So, by Fact 2,deg

(f(x)2 + g(x)2 + x

[h(x)2 + g(x)2

])=

max(deg (f(x)2 + g(x)2), deg x(h(x)2 + k(x)2)) =

max(2max(deg f, deg g)), 1 + 2max(deg h(x), deg k(x)).

So we discover that

(2.4) deg (f 2 + g2 + x(h2 + k2)) = max (2deg f, 2deg g, 1 + 2deg h, 1 + 2deg k) .

We will use Fact 3, mainly equation 2.4, in the proof of the next theorem.

14

3. Non-Negative Polynomials on [0, 1]

In this section, we look at polynomials that are positive in the interval [0, 1].The next theorem is due to M. Fekete (1935) (see [3], Theorem 2.7 and page 49).

Theorem 3.1. The polynomial p(x) satisfies p(x) ≥ 0 for all x on the interval [0, 1]if and only if p(x) is a Type III polynomial, i.e. there are polynomials f(x), g(x),h(x), and k(x) such that p(x) = f(x)2 + g(x)2 + x(1− x)[h(x)2 + k(x)2].

In the proof of this theorem we will utilize more resources, which we will discussnow. The following corollaries concern the degree of a Type II polynomial.

Corollary 3.2. Consider the Type II polynomial,p(x) = f(x)2 + g(x)2 + x(h(x)2 + k(x)2) where f, g, h, and k are polynomials. Ifdeg p(x) = 2m, then deg f(x), deg g(x) ≤ m and deg h(x), deg k(x) ≤ m− 1.

Proof. By equation 2.4, we can say the following:2m = deg p(x) ≥ deg f(x)2 = 2× deg f(x), or more simply deg f(x) ≤ m. Similarly,deg g(x) ≤ m. Also, 2m = deg p(x) ≥ deg xh(x)2 = 1 + 2 × deg h(x), so m ≥12

+ deg h(x). Finally, we get deg h(x) ≤ m− 12, but since deg h(x) is an integer, we

must have deg h(x) ≤ m− 1. Similarly, deg k(x) ≤ m− 1. �

Corollary 3.3. Consider the Type II polynomial,p(x) = f(x)2 + g(x)2 + x(h(x)2 + k(x)2) where f, g, h, and k are polynomials. Ifdeg p(x) = 2m+ 1, then deg f(x), deg g(x), deg h(x), deg k(x) ≤ m.

Proof. By equation 2.4, we can say the following:2m+ 1 = deg p(x) ≥ deg f(x)2 ≥ 2× deg f(x), so m+ 1

2≥ deg f(x), or more simply,

deg f(x) ≤ m. Similarly, deg g(x) ≤ m. Also 2m + 1 = deg p(x) ≥ deg xh(x)2 =1 + 2× deg h(x), or more simply, deg h(x) ≤ m. Similarly, deg k(x) ≤ m. �

The following Lemma will be utilized to help us create Type III polynomials.

Lemma 3.4. If r(x) and s(x) are both Type III polynomials, then the product of r(x)and s(x) is also a Type III polynomial.

Proof. Suppose r(x) = f(x)2 + g(x)2 + x(1− x)[h(x)2 + k(x)2]and s(x) = a(x)2 + b(x)2 + x(1 − x)[c(x)2 + d(x)2] where f, g, h, k, a, b, c, and d arepolynomials.Then r(x)× s(x) =

[f(x)2 + g(x)2][a(x)2 + b(x)2] + x2(1− x)2[h(x)2 + k(x)2][c(x)2 + d(x)2]︸ ︷︷ ︸P (x)

+x(1− x)[f(x)2 + g(x)2

] [c(x)2 + d(x)2

]+ [a(x)2 + b(x)2][h(x)2 + k(x)2]︸ ︷︷ ︸

Q(x)

So r(x)× s(x) = P (x) + x(1− x)Q(x)Clearly, both P (x) and Q(x) are non-negative for all x, so, by Theorem 1.1, eachpolynomial is a Type I. So r(x)× s(x) is a Type III polynomial. �

15

Proof of Theorem 3.1To prove Theorem 3.1, we now consider the two possible cases:

(i) polynomials of even degree(ii) polynomials of odd degree

3.1. Case 1: Polynomials of Even Degree. Suppose p(x) ≥ 0 for 0 ≤ x ≤ 1 andsuppose p(x) has even degree, deg p(x) = 2k, so that

p(x) = a0 + a1x+ a2x2 + . . .+ a2kx

2k.

For x 6= 0,

p

(1

x

)= a0 + a1

1

x+ a2

1

x+ . . .+ a2k

1

x2k,

so we can define

q(x) := x2kp

(1

x

)= a0x

2k + a1x2k−1 + . . .+ a2k.

If x ≥ 1, then 0 <1

x≤ 1, so p

(1

x

)≥ 0, and so q(x) = x2kp

(1

x

)≥ 0.

Let r(x) := q(x + 1). If x ≥ 0, then x + 1 ≥ 1, so q(x + 1) ≥ 0, and we find thatr(x) ≥ 0 for x ≥ 0. By Theorem 2.1, there exists polynomials A(x), B(x), C(x), andD(x) such that

r(x) = A(x)2 +B(x)2 + x[C(x)2 +D(x)2].

We have r(x) = q(x + 1), so deg r = deg q = 2k. Corollary 3.2 shows that deg A,deg B ≤ k and deg C, deg D ≤ k − 1. Now,

r(x− 1) = A(x− 1)2 +B(x− 1)2 + (x− 1)[(C(x− 1)2 +D(x− 1)2

]and

r(x− 1) = q(x− 1 + 1) = q(x) = x2kp

(1

x

).

For x 6= 0, let u =1

xand in turn x =

1

u. Then,

1

u2k× p(u) = r

(1

u− 1

)

= A

(1

u− 1

)2

+B

(1

u− 1

)2

+

(1

u− 1

)[C

(1

u− 1

)2

+D

(1

u− 1

)2],

so, p(u) =

u2k ×

[A

(1

u− 1

)2

+B

(1

u− 1

)2

+

(1

u− 1

)[C

(1

u− 1

)2

+D

(1

u− 1

)2]]

.

Next, distribute the u2k term to each polynomial A,B,C, and D. For example, sincedeg A ≤ k, we can write A(x) = co + c1x+ . . .+ ckx

k. Now,

A

(1− uu

)= co + c1

(1− uu

)+ . . .+ ck

((1− u)k

uk

),

16

so

u2k ×

[A

(1

u− 1

)2]

= u2k × A(

1− uu

)2

.

Consider

(3.1) A := ukA

(1− uu

).

Since, deg A ≤ k, then A(u)

= ukco + c1 (1− u)uk−1 + . . .+ ck(1− u)k.

Now we see that A(u) is a polynomial in u. Similarly, B(u):= ukB

(1

u− 1

)is a

polynomial, where B(x) = do + d1x+ . . . dkxk (from Corollary 3.2, deg B ≤ k).

Next,

u2k(

1− uu

)[C

(1− uu

)2

+D

(1− uu

)2]

= u2(

1− uu

)[u2k−2C

(1− uu

)2

+ u2k−2D

(1− uu

)2]

= u(1− u)[C(u)2 + D(u)2],

where C(u):=uk−1C

(1− uu

)and D(u):=uk−1D

(1− uu

). Since, from Corollary 3.2,

deg C, deg D ≤ k − 1, we see that C(u) and D(u) are polynomials. So,

p(u) = A(u)2 + B(u)2 + u(1− u)[C(u)2 + D(u)2]

and therefore p is Type III. This concludes the proof of Case 1.

Here is an example illustrating equation 3.1.

Example 3.5. Let A(x) = 1014x7 + 366x6 + 99x5 + 32x4 + 678x3 + 44x2 + 4x+ 14.Then A(x) := x7A

(1−xx

).

A(x) =(

1014(1−x)7x7 + 366(1−x)6

x6 + 99(1−x)5x5 + 32(1−x)4

x4 + 678(1−x)3x3 + 44(1−x)2

x2 + 4(1−x)x

+ 14)x7

Expand[A(x)] = 1014− 6732x+ 19197x2 − 30463x3 + 29710x4 − 18592x5 + 7219x6 −1339x7

So we see that A(x) is a polynomial in x as we claimed in Equation 3.1.

3.2. Case 2: Polynomials of Odd Degree. Suppose p(x) ≥ 0 for 0 ≤ x ≤ 1 andsuppose that p(x) has odd degree, deg p(x) = 2m+1. By the Factorization Theorem,

p(x) = Aq1 . . . qr(x− y1)s1 . . . (x− yt)st(x− x1)t1 . . . (x− xu)tu ,

where q1 . . . qr are quadratics with no real roots, x1 . . . xu, y1 . . . yt are distinct, si iseven, and ti is odd. We let ti = 2vi + 1, so

(x− xi)ti = (x− xi)2vi(x− xi).17

Now we have p(x) =

Aq1 . . . qr(x− y1)s1 . . . (x− yt)st(x− x1)2v1 . . . (x− xu)2vu︸ ︷︷ ︸Q(x)

(x− xi) . . . (x− xu).

We can assume that each qi > 0 for all x, so Q(x) ≥ 0 for all x. Since p(x) ≥ 0 for0 ≤ x ≤ 1, it is clear that the xi’s are not in the interval [0, 1] (otherwise, p(x) wouldchange sign in [0, 1].)To continue further we need two Lemmas.

Lemma 3.6. If x0 ≤ 0, then x− x0 is of Type III.

Proof. We want a Type III representation for x− x0 of the form

(3.2) p(x) := (x− x0) = (Ax+B)2 + x(1− x)C2,

where A,B, and C are constants.Let α = −x0 ≥ 0, so we can now sayx+ α ≡ x− x0 = A2x2 + 2ABx+B2 + xC2 − x2C2.Now we compare coefficients on each side of the equation:B2 = α ≥ 0, so B =

√α.

Next,

(3.3) 2AB + C2 = 1.

Also, A2 − C2 = 0, so A2 = C2, or | A |=| C |.So we can substitute A for C in equation 3.3 to get2AB + A2 = 1, so finally we have

(3.4) A2 + 2√αA− 1 = 0.

For the discriminant of this quadratic in A we get

b2 − 4ac = 4α− 4(1)(−1) = 4(α + 1) > 0.

This tells us that we can solve equation 3.4 for A, and get the Type III representationsuch as in equation 3.2. �

The following is an example of Lemma 3.6

Example 3.7. We start with P , a (x− xo) term where xo ≤ 0.P = (x+ 5)Plot[P, x, 0, 1]

18

0.2 0.4 0.6 0.8 1.0

5.2

5.4

5.6

5.8

6.0

Think of the number 5 as a square, (√α)2 and since we know h and k cannot be 0 or

else there will be no x term.x0 = −5α = −x0 = 5B =

√α =√

5H = AUse equation 3.4 from Lemma 3.60 == 2AB + A2 − 1 = −1 + 2

√5A+ A2

a = 1b = 2B = 2

√5

c = −1Apply the quadratic formula(−b+√b2−4ac2a

)= 1

2

(−2√

5 + 2√

6)

Expand[%] = −√

5 +√

6(−b−√b2−4ac2a

)= 1

2

(−2√

5− 2√

6)

Expand[%] = −√

5−√

6

A =(−b+√b2−4ac2a

)= 1

2

(−2√

5 + 2√

6)

H = A = 12

(−2√

5 + 2√

6)

B =√α =√

5Verify that the expression is equal to 1

A2 + 2BA =√

5(−2√

5 + 2√

6)

+ 14

(−2√

5 + 2√

6)2

FullSimplify[%] = 1A2 + 2BA = 1 so the above expression should be zero.Expand[((x− xo))− ((A2 + 2BA)x− xo)] = 0Type III representation below.Expand[(x− xo)− ((Ax+B)2 + x(1− x)H2)] = 0It is concluded that original linear term x+ 5 has a Type III representation.

Lemma 3.8. If xo ≥ 1, then −(x− xo) is of Type III.

Proof. We try for a Type III representation of the form

(3.5) p(x) = −(x− x0) = (Ax+B)2 + x(1− x)C2.19

Now, −x+ x0 = A2x2 + 2ABx+B2 + xC2 − x2C2. To solve 3.5 we needx0 = B2, so B =

√x0. We also need

(3.6) 2AB + C2 = −1

andA2 − C2 = 0, so A2 = C2, or | A |=| C |.So we can substitute A for C in equation 3.6 and get

(3.7) 2AB + A2 = −1.

So finally we need A2 + 2√x0A+ 1 = 0.

From the discriminant, we getb2 − 4ac = 4(x0 − 1) ≥ 0. Therefore we can find A,B, and C that satisfy equation3.5, so p(x) is Type III.

�

The following example illustrates Lemma 3.8.

Example 3.9. Here is an example of why −(x− xo), where xo ≥ 1, is of Type III.P = −(x− 4)Plot[P, x, 0,−1]

0.2 0.4 0.6 0.8 1.0

3.2

3.4

3.6

3.8

4.0

Plot shows P is non-negative for [0, 1].xo = 4B =

√xo = 2

H = ASolve using equation 3.40 == 2AB + A2 + 1 = 1 + 4A+ A2

a = 1b = 2B = 4c = 1Apply the quadratic formula.(−b+√b2−4ac2a

)= 1

2

(−4 + 2

√3)

Expand[%] = −2 +√

3(−b−√b2−4ac2a

)= 1

2

(−4− 2

√3)

20

Expand[%] = −2−√

3

A =(−b+√b2−4ac2a

)= 1

2

(−4 + 2

√3)

H = A = 12

(−4 + 2

√3)

B =√xo = 2

Verify the expression is - 1

A2 + 2BA = 2(−4 + 2

√3)

+ 14

(−4 + 2

√3)2

FullSimplify[%] = −1Expand[(− (x− xo))− ((A2 + 2BA)x+ xo)] = 0Expand[− (x− xo)− ((Ax+B)2 + x(1− x)H2)] = 0−(x− 4) has a Type III representation.

For the proof of Theorem 3.1 with deg p odd, we will now look at two differentcases depending on the sign of A from the factorization theorem.

3.2.1. Case i. When A > 0, p(x) =

Aq1 . . . qr(x− y1)s1 . . . (x− yt)st(x− x1)2v1

. . . (x− xm)2vm(x− x1) . . . (x− xi)(x− xi+1) . . . (x− xm),

where x1 . . . xi ≤ 0 and xi+1 . . . xm ≥ 1. Let

Q(x) := Aq1 . . . qr(x− y1)s1 . . . (x− yt)st(x− x1)2v1 . . . (x− xm)2vm .

Q(x) ≥ 0 because A > 0 and the qi’s are strictly positive quadratics. These pa-rameters make Q(x) a Type I polynomial, which can be expressed as a Type IIIpolynomial. So now we have

p(x) = Q(x) (x− x1) . . . (x− xi)︸ ︷︷ ︸O(x)

(x− xi+1) . . . (x− xm)︸ ︷︷ ︸E(x)

,

where x1 . . . xi are ≤ 0 and xi+1, . . . , xm are ≥ 1. By Lemma 3.6, each of the factorsin (x−x1) . . . (x−xi) are Type III so by Lemma 3.4, O(x) is Type III. For 0 < x < 1,each factor in E(x) is negative, so there must be an even number of these factorsin E(x). We can rewrite E(x) as (−(x − xi+1)) . . . (−(x − xm))(since the number offactors is even, the value does not change). Since each −(x − xi+j) is Type III byLemma 3.6, then E(x) is Type III by Lemma 3.4. Now, p(x) = Q(x)×O(x)×E(x),so by Lemma 3.4, p(x) is Type III.

3.2.2. Case ii. : When A < 0, p(x) =

−Aq1q1 . . . qr(x− y1)s1 . . . (x− yt)st(x− x1)2v1

. . . (x− xm)2vm(x− x1) . . . (x− xi)(x− xi+1) . . . (x− xm),

where A > 0, each qi is strictly positive, x1 . . . xi ≤ 0, and xi+1 . . . xm ≥ 1. Let21

Q(x) := Aq1 . . . qr(x − y1)s1 . . . (x − yt)

st(x − x1)2v1 . . . (x − xm)2vm . Then Q(x) is

non-negative for all x and we can write

p(x) = (−1)Q(x) (x− x1) . . . (x− xi)︸ ︷︷ ︸E(x)

(x− xi+1) . . . (x− xm)︸ ︷︷ ︸O(x)

For 0 < x < 1, each factor (x− xj) with xj ≥ 1 is negative, so there must be an oddnumber of factors in O(x) (since otherwise p(x) < 0). So we can write

p(x) = (−1)Q(x)E(x) ((−1)(x− xi+1) . . . (−1)(x− xm))︸ ︷︷ ︸R(x)

(−1).

(The number of negative signs inserted in O(x) is odd, so in order for them to canceleach other out we compensate with an extra negative at the end). Let F (x) =Q(x) × E(x). Since each factor in E(x) is Type III by Lemma 3.6, then E(x) isType III by Lemma 3.4. So F (x) is also Type III by Lemma 3.4. Since each factor−(x − xj) in R(x) is Type III by Lemma 3.8, then R(x) is Type III by Lemma 3.4,so now we have

p(x) = F (x)R(x).

(The leading coefficient(-1) and the very last (-1) at the end cancel out.) Since F (x)and R(x) are Type III, by Lemma 3.4, p(x) = F (x) × R(x) is also Type III. Thiscompletes the proof of case 3.2, polynomials with odd degree.

The following corollary tells us that Theorem 3.1 works for any interval [a, b].

Corollary 3.10. Suppose p(x) is non-negative on the interval [a, b], where a < b.Then there are polynomials R(x), S(x), T (x), and U(x) such that

p(x) = R(x)2 + S(x)2 + (x− a)(b− x)[T (x)2 + U(x)2].

Proof. Consider the function, q(x) := p (u(x)) where u ≡ u(x) := a + x(b − a). q(x)is a polynomial with deg q = deg p, and if 0 ≤ x ≤ 1, then a ≤ u(x) ≤ b, so q(x) ≥ 0.According to Theorem 3.1, there are polynomials A(x), B(x), C(x), D(x) such that

q(x) = A(x)2 +B(x)2 + x(1− x)[C(x)2 +D(x)2

].

We know that u = a+ x(b− a); if we solve for x, we get the expression:

x =u− ab− a

.

Now replace x in the function q(x) withu− ab− a

,

p (u(x)) = A(x)2 +B(x)2 + x(1− x)[C(x)2 +D(x)2

]= A

(u− ab− a

)2

+B

(u− ab− a

)2

+

(u− ab− a

)(1− u− a

b− a

)[C

(u− ab− a

)2

+D

(u− ab− a

)2].

22

Now simplify 1− u− ab− a

to get

b− ab− a

− u− ab− a

=b− ub− a

,

then replace in the expression for p(u(x)):

p (u(x)) = A(u)2

+ B(u)2

+

((u− a)

(b− a)2(b− u)

)[C(u)

2+ D(u)

2],

where

A(u) := A

(u− ab− a

), B(u) := B

(u− ab− a

), C(u) := C

(u− ab− a

), D(u) := D

(u− ab− a

).

Note that A, B, C, and D are polynomials in u.

The final expression is derived by distributing1

b− a2,

p (u) = A(u)2

+ B(u)2

+ (u− a)(b− u)[C ′(u)

2+D′(u)

2],

where C ′(u) :=C(u)

b− a, D′(u) :=

D(u)

b− aare polynomials in u. �

The following example illustrates Corollary 3.10.

Example 3.11. Take P (u) = u3, positive on [2, 4] and give a Type III representation.P = u3

a = 2b = 4Plot[P, u,−5, 5]

-4 -2 2 4

-100

-50

50

100

u = a+ x(b− a) = 2 + 2xP = (2 + 2x)3

Factoring the polynomial shows an odd degree, so we use Theorem 3.1 Case 2.Factor[P ] = 8(1 + x)3

Expand[P − (8(1 + x)2(1 + x))] = 0Next we get the Type III representation of the polynomial P .

23

(x+ 1) =

(1 +

1

2

(−2 + 2

√2)x

)2

+ x(1− x)

(1

2

(−2 + 2

√2))2

8 ((1 + x)2) =(√

8(1 + x))2

+ 02 + x(1− x) ((0)2 + (0)2)Now use Lemma 3.4 regarding multiplication of Type III polynomials.Using Lemma 3.4, we can express P as F + x(1 − x)G, where F and G are the fol-lowing sums of two squares:

F = X2 +W 2 = 02 +(2√

2(1 + x)(1− x+√

2x))2

G = V 2 + U2 = 02 +(√

2(−2 + 2√

2)(1 + x))2

.Now solve for x in the equation u = a+ x(b− a)

x =u− ab− a

=1

2(−2 + u)

(X2 +W 2 + x(1− x) (U2 + V 2)) =

(8

(1 +

1

2(−2 + u)

)2(1 +

2− u2

+−2 + u√

2

)2

+(−2 + 2

√2)2(

1 +2− u

2

)(1 +

1

2(−2 + u)

)2

(−2+

u)

RR =√

8

(1 +

1

2(−2 + u)

)(1 +

2− u2

+−2 + u√

2

)SS = X = 0TT = V = 0

UU =1

(b− a)U =

(−2 + 2

√2) (

1 + 12(−2 + u)

)√

2ExpandAll[P − (RR2 + SS2 + (u− a)(b− u)(TT 2 + UU2))] = 0So the original P = u3, has a Type III representation on [2, 4].(Refer to Appendix D for the full problem implementation.)

24

4. Non-Negative Polynomials in Two-Variables

In this section we will discuss non-negative polynomials in two variables. In1888, David Hilbert proved that there are polynomials p(x, y) that are non-negativeon R2 but cannot be expressed as a sum of squares. The first specific example ofsuch a polynomial was discovered by T. Motzkin in 1967. In this section we studyMotzkin’s example. We then illustrate a general method for indicating whether ornot a polynomial p(x, y) is a sum of squares.

Theorem 4.1. Given the Motzkin polynomial

(4.1) M(x, y) = x4y2 + x2y4 − 3x2y2 + 1,

(i) M(x, y) ≥ 0 for every x, y and(ii) M cannot be expressed as a sum of squares.

In an attempt to prove this two-part theorem we will use two methods. Thefirst will utilize the following theorem.

Theorem 4.2. (Inequality of Arithmetic and Geometric Means) When we have thenumbers a1 . . . an > 0, we can say that

(4.2)a1 + . . .+ an

n≥ (a1 . . . an)

1n .

Proof of Theorem 4.1

Proof. Here we try to prove M(x, y) ≥ 0 for every x, y with the Motzkin polynomial

M(x, y) = x2 + y2(x2 + y2 − 3) + 1.

If x2 + y2 − 3 ≥ 0, then clearly M(x, y) ≥ 0.Suppose

x2 + y2 < 3,

letw := 3− x2 − y2 > 0,

and letz :=

√w.

Then

(4.3) z2 = 3− x2 − y2,or

(4.4) 3 = x2 + y2 + z2.

By equation 4.2, we can say the following:

x2 + y2 + z2

3≥ (x2y2z2)

13 .

If we substitute in equation 4.4 we get,

1 =3

3=x2 + y2 + z2

3≥ (x2y2z2)

13 .

25

Since1 ≥ (x2y2z2)

13 ,

then1 ≥ x2y2z2.

Substitute z2 in from equation 4.3:

1 ≥ x2y2(3− x2 − y2).Distribute the −1

−1 ≤ x2y2(x2 + y2 − 3).

Bring the 1 to the other side,

0 ≤ x2y2(x2 + y2 − 3) + 1 = M(x, y),

and we discover that M(x, y) ≥ 0 for every x, y.

Now we will try to prove the second part of Theorem 4.1 by attempting to expressM(x, y) as a sum of squares. Suppose M(x, y) is a sum of squares of polynomials ofdegree 3:

M(x, y) = s(x, y) :=p∑

i=1

(ai + bix+ ciy + dix2 + eixy + fiy

2 + gix3 + hix

2y + jixy2 + kiy

3)2,

where ai, bi, ci, di, ei, fi, gi, hi, ji, and ki are constants. Comparing to equation 4.1, theMotzkin polynomial does not contain most of the terms in s(x, y) so some of theseconstants may equal 0. But let’s look at certain terms to see if we can really expressM(x, y) as a sum of squares. There is no x2 term in equation 4.1, so the coefficient

of x2 in s(x, y) must be 0. Therefore,

p∑i=1

b2i = 0, so each bi = 0. Similarly, there is

no y2 term in Equation 4.1, so

p∑i=1

c2i = 0, and therefore ci = 0. Most noticeably we

see that the coefficient of the x2y2 term in equation 4.1 is −3, so we must have

(4.5)

p∑i=1

2fidi + e2i + 2biji + 2cihi =

p∑i=1

2fidi + e2i = −3.

But let’s look at di. Since there is no x4 term in equation 4.1, then

p∑i=1

d2i + 2bigi = 0.

But we discovered above that bi = 0, so

p∑i=1

d2i = 0, and therefore each di = 0.

When we go back to equation 4.5 we see that we now have

p∑i=1

e2i = −3. This is

a contradiction and is not possible. (The sum of all the e2i terms cannot equal anegative number.) Therefore M cannot be expressed as a sum of squares. �

26

We now continue by studying some consequences of the following theorem ofHilbert:

Theorem 4.3. Every polynomial p(x1, x2, · · ·xn) of degree 2k that is non-negative isa sum of squares if and only if

(i) n = 1, k ≥ 0 (Theorem 1.1)(ii) n > 1, k = 1

(iii) n = 2, k = 2

Hilbert’s theorem shows for n = 2 and deg p = 6 there are polynomials that arenon-negative but cannot be expressed as sums of squares. Next we study a methodfor determining if a polynomial p(x, y) is a sum of squares.

Proposition 4.4. A polynomial p(x, y) with deg p = 2k can be expressed as a sumof squares if and only if there is a matrix A such that

p(x, y) = (A v)TA v,

wherev ≡ vk :=

(1, x, y, x2, x y, y2, · · · , xk, xk−1 y, · · · , yk

).

Proof. The entries of Av are polynomials, so (A v)T (A v) is a sum of squares ofpolynomials. So if

p(x, y) = (A v)T (A v),

then p is a sum of squares of polynomials. Conversely, if p =k∑

i=1

fi2 (fi is a poly-

nomial), define a matrix A so that row i of A has the coefficients of fi. Thenp(x, y) = (A v)T (A v). �

To decide if a given p(x, y) is a sum of squares, we use the following steps.

(i) Find a matrix M with M = MT such that p(x, y) = vTMv.(ii) Find a matrix M that satisfies step(i), and M � 0 (M is positive, i.e., where

all eigenvalues of M are non-negative). Then let P = M so, p(x, y) = vT P v.(iii) Use Cholesky Decomposition to find a matrix A such that

P = ATA.

(iv) Then p(x, y) = vT P v = vTATAv = (A v)T (A v), giving a sum of squaresrepresentation for p.

Example 4.5. p = 7 + 2x+ 14x2 + 14y + 28xy + 49y2

The vector v, must count up to half of the total degree.v = (1, x, y)The matrix is self-adjoint for simplicity so there is easy variable comparison.P = {{a, b, c}, {b, d, e}, {c, e, f}}Solve p = v.P.v for P.

MatrixForm[P ] =

a b cb d ec e f

27

Together[p−v.P.v] = 7−a+2x−2bx+14x2−dx2+14y−2cy+28xy−2exy+49y2−fy2a = 7b = 1c = 7d = 14e = 14f = 49p− v.P.v = x+ 14x2 + 7y + 28xy + 49y2 − x(1 + 14x+ 14y)− y(7 + 14x+ 49y)Factor[%] = 0P= v.M.v

P = {7, 1, 7}, {1, 14, 14}, {7, 14, 49} =

7 1 71 14 147 14 49

Check for positivity, determinant, cholesky decomposition, eigenvalues.Det[P ] = 2891

CholeskyDecomposition[P ] ={{√

7, 1√7,√

7},{

0,√

977, 13√

797

},{

0, 0, 7√

5997

}}A={{√

7, 1√7,√

7},{

0,√

977, 13√

797

},{

0, 0, 7√

5997

}}MatrixForm[A]=

√

7 1√7

√7

0√

977

13√

797

0 0 7√

5997

tA=Transpose[A]=

{{√7, 0, 0

},{

1√7,√

977, 0},{√

7, 13√

797, 7√

5997

}}MatrixForm[tA]=

√

7 0 01√7

√977

0√

7 13√

797

7√

5997

Double check that P = Transpose[A].A:P − tA.A = {0, 0, 0}, {0, 0, 0}, {0, 0, 0}To express p as a sum of squares, we now use (v.tA).(A.v)

(v.tA).(A.v) = 2891y2

97+(√

977x+ 13

√797y)2

+(√

7 + x√7

+√

7y)2

This is a sum of squares expression; We double check that it equals pq = (v.tA).(A.v)Factor[p− q] = 0Another way to check that a real symmetric matrix is positive is to verify that all ofits eigenvalues are nonnegative :Eigenvalues[P ] = Root

[−2891 + 881#1− 70#12 + #13&, 3

],

Root[−2891 + 881#1− 70#12 + #13&, 2

],

Root[−2891 + 881#1− 70#12 + #13&, 1

]N [%] = 54.9159, 9.60073, 5.48334This means that there are vectors a, b, c such that P.a = 54.91592601638876‘a, P.b =

28

9.600729026932235‘b, Pc = 5.483344956679007‘c. Further, any vector w can be ex-pressed as w = ra+sb+tc, so Pw = 54.91592601638876‘(ra)+9.600729026932235‘(sb)+5.483344956679007‘(tc). This shows that P is like stretching by positive multipliersin three directions.

29

References

[1] Ahlfors, L. Complex Analysis. McGraw-Hill Book Company, 2nd Edition, 1966.[2] Hilbert, D., Uber die Darstellung definiten formen als Sumen von Formenquadraten.

Math.Ann.32. 32(1888),342-350.[3] Lasserre, J. Moments, Positive Polynomials, and Their Applications. Imperial College Press

Optimization Series Vol. 1, Imperial College Press, London, 2010.[4] Laurent, M. Sums of squares, moment matrices and optimization over polynomials. Emerging

Applications of Algebraic Geometry, Vol. 149 (Springer, New York, NY), 157-270.[5] Polya G. and Szego, G. Problems and Theorems in Analysis II (Springer-Verlag), 1976.

30

Appendix A. Factorization Theorem

In the previous sections we used the Factorization Theorem:If p(x) is a polynomial, then p(x) = Aq1(x) . . . qk(x)(x − x1)

n1 . . . (x − xr)nr(x −

y1)m1 . . . (x−ys)ms , where A is a constant, q1(x) . . . qk(x) are strictly non-zero quadrat-

ics, x1 . . . xr, y1 . . . ys are all distinct, n1 . . . nr are even exponents, and m1 . . .ms areodd exponents, (Note that any of these groups may be missing in the expression).Here we explain the theorem’s origin.Suppose z is a complex number,

z = x+ iy,

where x, y are real numbers and i is imaginary, with i2 = −1. Let z be the complexconjugate of z,

z := x− iy.Consider a complex polynomial

p(z) = a0 + a1z + . . .+ akzk

(where the ai’s are complex).We state without proof the following theorem (see [1]).

Theorem A.1. (Fundamental Theorem of Algebra) There is a complex number z0such that p(z0) = 0.

Next, we recall the Division Theorem.

Theorem A.2. (Division Theorem) If f(z) and g(z) are polynomials, then there arepolynomials, q(z) and r(z), with deg r < deg g, such that f(z) = g(z)q(x) + r(x).

Now if we havef(z) = p(z)

andg(z) = z − z0,

where p(z0) = 0, thenp(z) = (z − z0)q(z) + r(z),

wheredeg r < deg z − z0 = 1.

Sodeg r = 0,

and therefore r is a constant. Then

0 = p(z0) = 0 + r,

sor = 0.

So we see that if z0 is a root of p(z) = 0, then z − z0 is a factor of p(z), i.e.,

p(z) = (z − z0)q(z),31

anddeg p = 1 + deg q.

Corollary A.3. A non-constant polynomial p(z) factors completely as A(z−z0) . . . (z−zk−1). Also, if p(z) has real coefficients and if zi is a root, then so is zi.

Proof. The Corollary is true if deg p = 1:

az + b = a(z +b

a) where a 6= 0.

Assume by induction that the Corollary is true for all complex polynomials up todeg k − 1. Suppose deg p = k. From above,

p = (z − zo)q, deg q = k − 1.

By induction, q = A(z − z1) . . . (z − zk). So

p = (z − zo)q = A(z − zo)(z − z1) . . . (z − zk).

Suppose p has real coefficients and w is a root.

p(z) =k∑

i=o

aizi.

So

0 = p(w) =k∑

i=o

aiwi.

Then

0 = p(w) =k∑

i=o

aiwi =

k∑i=o

aiwi =k∑

i=o

aiwi =k∑

i=o

aiwi = p(w)

(because wi = wi). �

Suppose p(z) is a real polynomial. So p(z) factors completely as a complexpolynomial and if zo is a non-real root, then so is zo. So now we have

p(z) = (x− zo)(x− zo)q(x).

We can express f(x) as a real quadratic with no real roots. Let zo = a + ib whereb 6= 0 and zo = a− ib. So

f(x) := (x− zo)(x− zo) = x2 − 2ax+ a2 + b2.

Checking the discriminant,

4a2 − 4(a2 + b2) = 4a2 − 4a2 − 4b2,

we end up with−4b2 < 0

(since b 6= 0). Therefore f(x) has no real roots. If we apply this argument to thefactorization in Corollary A.3, we get:

(A.1) p(z) = (x− z0) . . . (x− zm)q1(x) . . . qk(x),32

where z0, . . . , zm are real roots and q1(x) . . . qk(x) are real quadratics with no realroots. Notice that the z0, . . . , zm are not necessarily distinct, but the repeated factorsin equation A.1 can be grouped together, and thus factors with powers of even andodd degrees are created. So finally we have our Factorization Theorem.

33

Appendix B. Section 1 Examples

H* Example 1.1-Find a Type I representation for the

product of two polynomials that are non-negative for all x *L

P = Hx^2 + 9L Hx^2 + x + 1L

I9 + x2M I1 + x + x2M

Plot@P, 8x, -10, 10<D

-10 -5 5 10

2000

4000

6000

8000

10 000

12 000

H* make H1+x+x^2L a sum of squares by using the formula derived from a quadratic*L

H* we find h and k so that we can use the product formula*L

h = x + 1 � 2

1

2+ x

k = Sqrt@3 � 4D

3

2

H*from the first question we find f and g*L

f = x

x

g = 3

3

H*now the product formula*L

Hf^2 + g^2L * Hh^2 + k^2L

I9 + x2M3

4+

1

2+ x

2

34

Hf * h + g * kL^2 + Hf * k - g * hL^2

3 x

2- 3

1

2+ x

2

+

3 3

2+ x

1

2+ x

2

p =3 x

2- 3

1

2+ x

2

+

3 3

2+ x

1

2+ x

2

3 x

2- 3

1

2+ x

2

+

3 3

2+ x

1

2+ x

2

Expand@pD

9 + 9 x + 10 x2 + x3 + x4

Factor@pD

I9 + x2M I1 + x + x2M

H*notice p is equivalent to P at the beginning of the page*L

P - p

I9 + x2M I1 + x + x2M -3 x

2- 3

1

2+ x

2

-

3 3

2+ x

1

2+ x

2

Expand@%D

0

35

H*Example 1.2~ Find a sum of two squares representation for a polynomial of degree 12 *L

P = 7 517�52 1131�13 x4 +7 51�4 x5

2 2

+ 543�52 1131�13 x5 +49

1251�4 x6 +

53�4 x6

2 2

+ 4 517�52 1131�13 x6 +

2 51�4 x7 +7

1253�4 x7 +

7

351�4 x8 + 7 543�52 1131�13 x8 +

7 53�4 x9

2 2

+ 5 517�52 1131�13 x9 +

5 51�4 x10

2 2

+

49

1253�4 x10 + 4 543�52 1131�13 x10 +

35

1251�4 x11 + 2 53�4 x11 +

7

353�4 x12

7 517�52 1131�13 x4 +7 51�4 x5

2 2+ 543�52 1131�13 x5 +

49

1251�4 x6 +

53�4 x6

2 2+ 4 517�52 1131�13 x6 +

2 51�4 x7 +7

1253�4 x7 +

7

351�4 x8 + 7 543�52 1131�13 x8 +

7 53�4 x9

2 2+ 5 517�52 1131�13 x9 +

5 51�4 x10

2 2+

49

1253�4 x10 + 4 543�52 1131�13 x10 +

35

1251�4 x11 + 2 53�4 x11 +

7

353�4 x12

Factor@PD

1

1251�4 x4 J12 5651�13 + 3 2 x + 7 x2N J7 + 5 x + 4 x2 + 7 5 x4 + 5 x5 + 4 5 x6N

H*Factor PHxL to see its roots where the polynomials can be condensed,

combined, or easily differeniatable

q = Factor@PD

1

1251�4 x4 J7 + 5 x + 4 x2N J12 5651�13 + 3 2 x + 7 x2N J1 + 5 x4N

H*Expand the polynomial into its expanded form.

Expand@PD

7 517�52 1131�13 x4 +7 51�4 x5

2 2+ 543�52 1131�13 x5 +

49

1251�4 x6 +

53�4 x6

2 2+ 4 517�52 1131�13 x6 +

2 51�4 x7 +7

1253�4 x7 +

7

351�4 x8 + 7 543�52 1131�13 x8 +

7 53�4 x9

2 2+ 5 517�52 1131�13 x9 +

5 51�4 x10

2 2+

49

1253�4 x10 + 4 543�52 1131�13 x10 +

35

1251�4 x11 + 2 53�4 x11 +

7

353�4 x12

H*Plot PHxL to make sure it is > 0 for all x.

36

Plot @ P, 8x, -100, 100 <D

-100 -50 50 100

5.´1023

1.´1024

1.5´1024

H*Break down the first polynomial into its SOS form, by completing the square.

A = Expand B J 5 x + 4 x^2 + 7NF

7 + 5 x + 4 x2

a = 4

4

b = 5

5

c = 7

7

a x +

b

2 a^2 +

4 a c - b^2

4 a

2

107

16+ 4

5

8+ x

2

H*Break down into SOS form, by completing the square,

first considering it's a square of the fourth degree we

take the square root of the both terms then complete the square,

then when we apply the Lemma to the quadractic form of the polynomial. Note that 1 2= 14.

1i =Expand B 5 Ix4M + 1 F

1+51�4 x4

1ii =Expand B 1 + 51

4 x2F1+51�4 x2

Expand @1i - 1ii D-51�4 x2

+51�4 x4= 0*L

37

B = ExpandBJ 5 x^4 + 1NF

1 + 5 x4

a = 5

5

b = 0

0

c = 1

1

J a N x +

b

2 a^2 +

4 a c - b^2

4 a

2

1 + 5 x2

Z = Expand@A HBLD

7 + 5 x + 4 x2 + 7 5 x4 + 5 x5 + 4 5 x6

H*Apply the Lemma*L

f =

107

4

107

4

g = 45

8+ x

25

8+ x

h = 1

38

k = 5^H1 � 4L x^2

51�4 x2

T = f h - g k

107

4- 2 51�4 x2

5

8+ x

U = f k + g h

1

451�4 107 x2 + 2

5

8+ x

H*Z = T2 + U2= AHBL*L

Expand@Z - HT^2 + U^2LD

0

H*Note we group the first set of polynomials which are in SOS form f2+g2 and combine them

together with the text term which is a square to get something like this: x2If2+g2M*L

ZZ = T^2 + U^2

1

451�4 107 x2 + 2

5

8+ x

2

+

107

4- 2 51�4 x2

5

8+ x

2

Expand@ZZD

7 + 5 x + 4 x2 + 7 5 x4 + 5 x5 + 4 5 x6

H*CC is a square of the fourth degree I51�8 x2M2=

I51�8M2 Ix2M2 so we use the terms without the squares when we apply the lemma *L

CC = Expand@H5^H1 � 4L x^4LD

51�4 x4

ExpandB CC F

51�8 x4

H*Apply the Lemma, where f = T, g =U, h = I51�8M, k= 0 *L

f = T

107

4- 2 51�4 x2

5

8+ x

39

g = U

1

451�4 107 x2 + 2

5

8+ x

h = 5^H1 � 8L x^2

51�8 x2

k = 0

0

TT = f h - g k

51�8 x2107

4- 2 51�4 x2

5

8+ x

UU = f k + g h

51�8 x21

451�4 107 x2 + 2

5

8+ x

H*x2If2+g2M = LHxL2+ KHxL2 = CC*ZZ = TT2+UU2, where x is a square, f2+g2 are SOS. *L

Expand@CC HZZL - HTT^2 + UU^2LD0

H*Now the first three sets of polynomials are in SOS form If2+g2M and combine them

together with the next term which is a mixture of fractions and radicals.*L

ZZZ = TT^2 + UU^2

51�4 x41

451�4 107 x2 + 2

5

8+ x

2

+ 51�4 x4107

4- 2 51�4 x2

5

8+ x

2

Expand@ZZZD

7 51�4 x4 + 53�4 x5 + 4 51�4 x6 + 7 53�4 x8 + 5 51�4 x9 + 4 53�4 x10

H*Expand the equation and complete the square*L

EE = ExpandBJH7 � 12L x^2 + J1 � 8 N x + H565^H1 � 13LLNF

5651�13 +x

2 2+

7 x2

12

40

a = 7 � 12

7

12

b =

1

2 2

1

2 2

c = 5651�13

5651�13

J a N x +

b

2 a^2 +

4 a c - b^2

4 a

2

3

7-

1

8+

7 5651�13

3+

7

12

3

7 2+ x

2

H*Apply the Lemma once again, taking the previous set of SOS and

multiplying them by the final polynomial that is now in SOS form *L

f = TT

51�8 x2107

4- 2 51�4 x2

5

8+ x

g = UU

51�8 x21

451�4 107 x2 + 2

5

8+ x

h =

3

7-

1

8+

7 5651�13

3

3

7-

1

8+

7 5651�13

3

k =

7

12

3

7 2

+ x

1

2

7

3

3

7 2+ x

41

TTT = f h - g k

-

51�8 7 x2 K 3

7 2+ xO K 1

451�4 107 x2 + 2 K 5

8+ xOO

2 3+

51�83

7-

1

8+

7 5651�13

3x2

107

4- 2 51�4 x2

5

8+ x

UUU = f k + g h

51�83

7-

1

8+

7 5651�13

3x2

1

451�4 107 x2 + 2

5

8+ x +

51�8 7 x2 K 3

7 2+ xO K 107

4- 2 51�4 x2 K 5

8+ xOO

2 3

Expand@P - HTTT^2 + UUU^2LD

0

H*As stated in Section D a product of two sums of squares is another

sums of squares so P = TTT2+ UUU2= Hf h - g kL2+ Hf k + g hL2= F2+*G2M

42

Appendix C. Section 2 Examples

H* Example 2.1- Given a Polynomial of degree 3, find a type II representation for P.*L

P = 3 + 4 x + 4 x2 + x3

3 + 4 x + 4 x2 + x3

Factor@PD

H3 + xL I1 + x + x2M

Plot@P, 8x, -5, 5<D

-4 -2 2 4

50

100

150

200

250

H*Step One: Get the quadratic in SOS formL

Expand@Hx^2 + x + 1LD

1 + x + x2

a = 1

1

b = 1

1

c = 1

43

J a N x +

b

2 a

2

+

4 a c - b^2

4 a

2

3

4

+

1

2

+ x

2

H* The trick is to get the P in form of the lemma f 2+ g2

+x Ih2+k2M M

H* So we provide some manipulation to do so taking the original problem

P = H3+xL I1+x+x2M which is the same as 3 I1+x+x2M + xI1+x+x2M so,

below I verified directly verified P =

f 2+ g2

+x Ih2+k2M however it isn't too easy to see so I'll explain in the next few steps.

Expand AP - I3 I1 + x + x2M + x I1 + x + x2MM E0

H*In order to to figure out the x Ih2+ k2M portion we subtract out If 2

+ g2M= 3I1+x+x2M,

which in turn will give us x Ih2+ k2M

Expand AP - If 2+ g2ME

x + x2 + x3

H* Factoring this portion that is left order

it immediately gives us the x Ih2+k2M portion of the lemma.

Factor @%D

x I1 + x + x2M

H*Below is the 3 I1+x+x2M in SOS form. M

f =

3

4

3

2

g =

1

2+ x

2

1

2+ x

2

h = 3

3

44

k = 0

0

A = f h - g k

3

2

B = f k + g h

31

2+ x

2

H*Below we verify that P = f2+ g2 +x Ih2+k2MM

f = A

3

2

g = B

3

1

2

+ x

2

h =1

2+ x

2

1

2

+ x

2

k =3

4

3

2

Pi = ExpandAf2 + g2 + x Ih2 + k2ME

3 + 4 x + 4 x2+ x

3

Expand@P - PiD

0

45

H* Example 2.2 Given two polynomials P and Q, which are given as

the following. Find the Type II x Type II expression of their product. *L

P = Hx - 3L2 + x Ix2 + 4M

H-3 + xL2 + x I4 + x2M

Q = Ix2 + 1M + x Ix2 + 8 x + 17M

1 + x2 + x I17 + 8 x + x2M

H*Convert to SOS form b2-4a c < 0.*L

ExpandAI17 + 8 x + x2ME

17 + 8 x + x2

a = 1

1

b = 8

8

c = 17

17

J a N x +

b

2 a

2

+

4 a c - b^2

4 a

2

1 + H4 + xL2

f = x - 3

-3 + x

g = 0

0

h = x

x

k = 2

46

ExpandAP - If2 + g2 + x Ih2 + k2MME

0

H*P is of type II*L

a = x

x

b = 1

1

c = 1

1

d = 4 + x

4 + x

ExpandAQ - Ia2 + b2 + x Ic2 + d2MME

0

H*Q is of type II*L

Pi = IIf2 + g2M Ia2 + b2MM + Ix2 Ih2 + k2M Ic2 + d2MM

H-3 + xL2 I1 + x2M + x2 I4 + x2M I1 + H4 + xL2M

Qi = IIf2 + g2M Ic2 + d2M + Ia2 + b2M Ih2 + k2MM

I1 + x2M I4 + x2M + H-3 + xL2 I1 + H4 + xL2M

Expand@HHQL HPLL - HPi + x QiLD

0

47

H*Example 2.3- Given a polynomial of degree 7,find a type II representation for P.

P = 36 + 84 x + 49 x2 - 3 x3 - 2 x4 - 2 x5 - 3 x6 + x7

36 + 84 x + 49 x2 - 3 x3 - 2 x4 - 2 x5 - 3 x6 + x7

Factor@PD

H-3 + xL2 H1 + xL3 I4 + x2M

Plot@P, 8x, -10, 10<D

-10 -5 5 10

-2.´106

-1.5´106

-1.´106

-500 000

500 000

1.´106

1.5´106

H*Group the first two polynomials that have even

powers or are SOS form and combine them together into one SOS *L

Z = Expand@Hx^2 + 4L Hx - 3L^2D

36 - 24 x + 13 x2 - 6 x3 + x4

f = H-3 + xL

-3 + x

g = 0

0

h = 4

2

k = x

x

A = f h - g k

2 H-3 + xL

48

B = f k + g h

H-3 + xL x

H*Here you verify that Z = A2+B2 *L

ExpandAZ - IA2 + B2ME0

H*Subtract out the portion f2+ g2 leaving you with x Ih2+k2M*L

X = ExpandAP - IA2 + B2ME

108 x + 36 x2 + 3 x3 - 3 x4 - 2 x5 - 3 x6 + x7

H*Factor the polynominal we see that its a bunch of polynomials

multiplied by x: xII4+x2MI3+3 x+x2MH-3+xL2M = x Ih2+k2M *L

Factor@XD

H-3 + xL2 x I4 + x2M I3 + 3 x + x2M

H*Convert the polynomial with no real roots*L

FullSimplifyA3 + 3 x + x2E3 + x H3 + xL

a = 1

1

b = 3

3

c = 3

3

J a N x +

b

2 a

2

+

4 a c - b^2

4 a

2

3

4+

3

2+ x

2

H*Group in the form of h2+k2M

f = A

2 H-3 + xL

49

g = B

H-3 + xL x

h =

3

4

3

2

k =

3

2+ x

3

2+ x

AA = f h - g k

3 H-3 + xL - H-3 + xL x3

2+ x

BB = f k + g h

1

23 H-3 + xL x + 2 H-3 + xL

3

2+ x

H*h = AA, k = BB, X = xIAA2+BB2M = xIh2+k2M

ExpandAX - x IAA2 + BB2ME

0

H*Now the final part is to apply the lemma f2+

g2 +x Ih2+k2M verfying that P = f2+ g2 +x Ih2+k2M

f = A

2 H-3 + xL

g = B

H-3 + xL x

h = AA

3 H-3 + xL - H-3 + xL x3

2+ x

k = BB

1

23 H-3 + xL x + 2 H-3 + xL

3

2+ x

50

Pi = ExpandAf2 + g2 + x Ih2 + k2ME

36 + 84 x + 49 x2 - 3 x3 - 2 x4 - 2 x5 - 3 x6 + x7

Expand@P - PiD

0

H*So P= f2+ g2 +x Ih2+k2

51

Appendix D. Section 3 Examples

H* Example 3.1 -The polynomial is of this

form: A Aq1... q kHx-y i Lni ... Hx-ymLnm Hx-x i Ls i ... Ix-xpMspEIx-x j M

s i Hx-xkLni ,

Where A is a constant >0. The q i 's are quadratics that cannot be factored,

so they have no real roots. The exponents n i are even and the exponents s i are odd. Notice

that in this case as long as the constant A is positive x j £ 0 and each x k³ 0. *L

P = 32 400 - 7200 x + 303 684 x 2- 69 048 x 3

+ 109 848 x 4- 38 796 x 5

+ 9852 x 6- 5064 x 7

+ 216 x 8+ 108 x 9

32 400 - 7200 x + 303 684 x2 - 69 048 x3 + 109 848 x4 - 38 796 x5 + 9852 x6 - 5064 x7 + 216 x8 + 108 x9

Factor @PD

12 H-5 + xL H-3 + xL H9 + xL I4 + x2M I5 + x + x2M I1 + 9 x2M

H*Plotting the Polynomial shows that on the interval @0,1 D it is strictly positive. *L

Plot @P, 8x, -5, 5 <D

-4 -2 2 4

5.´107

1.´108

1.5´108

Plot @P, 8x, 0, 1 <D

0.2 0.4 0.6 0.8 1.0

50 000

100 000

150 000

200 000

250 000

300 000

H*First group the first polynomials into a sum of squares,

here we have polynomials that are already in sum of squares form

so we combine them into one sum of squares and group them together. *L

52

Z = ExpandAIx2 + 4M I9 x2 + 1ME

4 + 37 x2 + 9 x4

f = x

x

g = 2

2

h = 9 x

3 x

k = 1

1

A = f h - g k

-2 + 3 x2

B = f k + g h

7 x

H*Verifty that Z and A2 + B2*L

ExpandAZ - IA2 + B2ME0

H*Convert into sum of squares form, b2-4 a c < 0 *L

ExpandAI5 + x + x2ME

5 + x + x2

a = 1

1

b = 1

1

c = 5

5

J a N x +

b

2 a

2

+

4 a c - b^2

4 a

2

19

4+

1

2+ x

2

53

ExpandAI5 + x + x2M - %E

0

H*Apply the Lemma from theorem 1*L

X = ExpandAI4 + x2M I5 + x + x2M I1 + 9 x2ME

20 + 4 x + 189 x2 + 37 x3 + 82 x4 + 9 x5 + 9 x6

f = A

-2 + 3 x2

g = B

7 x

h =19

4

19

2

k =1

2+ x

1

2+ x

AA = f h - g k

-7 x1

2+ x +

1

219 I-2 + 3 x2M

BB = f k + g h

7 19 x

2+

1

2+ x I-2 + 3 x2M

H*Verify that AA2+BB2=X*L

ExpandAX - IAA2 + BB2ME

0

H*Now apply the Lemma from Theorem 3 that states that Hx-xoL, where xo£ 0*L

B0 = x + 9

9 + x

54

xo = -9

-9

Α = -xo

9

B = Α

3

H = Y

Y

a = 1

1

b = 2 B

6

c = -1

-1

-b + b2 - 4 a c

2 a

1

2J-6 + 2 10 N

Expand@%D

-3 + 10

-b - b2 - 4 a c

2 a

1

2J-6 - 2 10 N

Expand@%D

-3 - 10

Y = %

-3 - 10

55

H = Y

-3 - 10

B = Α

3

Y2 + 6 Y

6 J-3 - 10 N + J-3 - 10 N2

FullSimplify@%D

1

ExpandAHHB0LL - IIY2 + 6 YM x + 9ME0

Expand@HB0L - HHY x + BL^2 + x H1 - xL H^2LD

0

H*So x+9= 6J-3- 10 N+J-3- 10 N2x+J 9 N

2*L

H*So now combine Type III with Type III since now x +9 is in Type III form.*L

H*Bi is in the form Hx-xoL,Ci is a constant, and X are all now Type III*L

CC = 12

12

ZZ = Expand@B0 CC HXLD

2160 + 672 x + 20 460 x2 + 6264 x3 + 9300 x4 + 1956 x5 + 1080 x6 + 108 x7

l = CC AA

2 3 -7 x1

2+ x +

1

219 I-2 + 3 x2M

m = CC BB

2 37 19 x

2+

1

2+ x I-2 + 3 x2M

n = 0

0

56

o = 0

0

ExpandACC X - Il2 + m2 + x H1 - xL In2 + o2MME

0

a = Y x + B

3 + J-3 - 10 N x

b = 0

0

c = 0

0

d = H

-3 - 10

ExpandAB0 - IIa2 + b2M + x H1 - xL Ic2 + d2MME

0

RR = IIl2 + m2M Ia2 + b2MM + Ix2 H1 - xL2 In2 + o2M Ic2 + d2MM

J3 + J-3 - 10 N xN2

12 -7 x1

2+ x +

1

219 I-2 + 3 x2M

2

+ 127 19 x

2+

1

2+ x I-2 + 3 x2M

2

Factor@RRD

12 J-3 + 3 x + 10 xN2I4 + x2M I5 + x + x2M I1 + 9 x2M

QQ = IIl2 + m2M Ic2 + d2M + Ia2 + b2M In2 + o2MM

J-3 - 10 N2

12 -7 x1

2+ x +

1

219 I-2 + 3 x2M

2

+ 127 19 x

2+

1

2+ x I-2 + 3 x2M

2

Factor@QQD

12 J3 + 10 N2I4 + x2M I5 + x + x2M I1 + 9 x2M

57

H*Find the type I representation of RR and QQ.*L

f = AA

-7 x1

2+ x +

1

219 I-2 + 3 x2M

g = BB

7 19 x

2+

1

2+ x I-2 + 3 x2M

h = CC J-3 + 3 x + 10 xN

2 3 J-3 + 3 x + 10 xN

k = 0

0

A4 = f k - g h

-2 3 J-3 + 3 x + 10 xN7 19 x

2+

1

2+ x I-2 + 3 x2M

B4 = f h + g k

2 3 J-3 + 3 x + 10 xN -7 x1

2+ x +

1

219 I-2 + 3 x2M

ExpandARR - IA42 + B42ME0

f = AA

-7 x1

2+ x +

1

219 I-2 + 3 x2M

g = BB

7 19 x

2+

1

2+ x I-2 + 3 x2M

h = CC J3 + 10 N

2 3 J3 + 10 N

58

k = 0

0

F0 = f k - g h

-2 3 J3 + 10 N7 19 x

2+

1

2+ x I-2 + 3 x2M

G0 = f h + g k

2 3 J3 + 10 N -7 x1

2+ x +

1

219 I-2 + 3 x2M

ExpandAQQ - IF02 + G02ME0

ExpandAZZ - IA42 + B42 + x H1 - xL IF02 + G02MME0

A0 = Expand@H-5 + xL H-3 + xLD

15 - 8 x + x2

h = A0 �. x ®

1

x

15 +1

x2-

8

x

g = x2 h

15 +1

x2-

8

xx2

q = g �. x ® s + 1

H1 + sL2 15 +1

H1 + sL2-

8

1 + s

Factor@qD

H2 + 3 sL H4 + 5 sL

Expand@%D

8 + 22 s + 15 s2

FactorBq - 15 - -

2

3+ s - -

4

5+ s F

59

f = 15

15

g = 0

0

h = 0

0

k = 0

0

xo =

-2

3

-

2

3

IIs2 Ih2 + k2M - xo If2 + g2MM + s IIf2 + g2M - xo Ih2 + k2MMM

10 + 15 s

f = 10

10

g = 0

0

h = 15

15

k = 0

0

xo =

-4

5

-

4

5

IIs2 Ih2 + k2M - xo If2 + g2MM + s IIf2 + g2M - xo Ih2 + k2MMM

8 + 22 s + 15

60

R = 15 s2

15 s2

S = 8

2 2

U = 22

22

V = 0

0

q = ExpandAR2 + S2 + s IU2 + V2ME

8 + 22 s + 15 s2

q = q �. s ® Hx - 1L

8 + 22 H-1 + xL + 15 H-1 + xL2

P0 = w2 q �. x ®

1

w

8 + 22 -1 +1

w+ 15 -1 +

1

w

2

w2

P0 - A0 �. x ® w

-15 + 8 w - w2 + 8 + 22 -1 +1

w+ 15 -1 +

1

w

2

w2

Expand@%D

0

H*So the final form of the Ai in Type III form is 152H1-xL2+Jx 8 N

2+xH1-xLJ 22 N

2*L

l = 15 H1 - xL

15 H1 - xL

m = x 8

2 2 x

61

n = 22

22

o = 0

0

ExpandAA0 - Il2 + m2 + x H1 - xL In2 + o2MME

0

H*Apply the formula once more Type III * Type III = Type III*L

a = A4

-2 3 J-3 + 3 x + 10 xN7 19 x

2+

1

2+ x I-2 + 3 x2M

b = B4

2 3 J-3 + 3 x + 10 xN -7 x1

2+ x +

1

219 I-2 + 3 x2M

c = F0

-2 3 J3 + 10 N7 19 x

2+

1

2+ x I-2 + 3 x2M

d = G0

2 3 J3 + 10 N -7 x1

2+ x +

1

219 I-2 + 3 x2M

RRR = IIl2 + m2M Ia2 + b2MM + Ix2 H1 - xL2 In2 + o2M Ic2 + d2MM

22 H1 - xL2 x2

12 J3 + 10 N2-7 x

1

2+ x +

1

219 I-2 + 3 x2M

2

+ 12 J3 + 10 N2 7 19 x

2+

1

2+ x I-2 + 3 x2M

2

+

I15 H1 - xL2 + 8 x2M 12 J-3 + 3 x + 10 xN2-7 x

1

2+ x +

1

219 I-2 + 3 x2M

2

+

12 J-3 + 3 x + 10 xN2 7 19 x

2+

1

2+ x I-2 + 3 x2M

2

Factor@RRRD

12 I4 + x2M I5 + x + x2M I1 + 9 x2MJ135 - 540 x - 90 10 x + 1450 x2 + 402 10 x2 - 1820 x3 - 582 10 x3 + 855 x4 + 270 10 x4N

62

QQQ = IIl2 + m2M Ic2 + d2M + Ia2 + b2M In2 + o2MM

I15 H1 - xL2 + 8 x2M

12 J3 + 10 N2-7 x

1

2+ x +

1

219 I-2 + 3 x2M

2

+ 12 J3 + 10 N2 7 19 x

2+

1

2+ x I-2 + 3 x2M

2

+

22 12 J-3 + 3 x + 10 xN2-7 x

1

2+ x +

1

219 I-2 + 3 x2M

2

+

12 J-3 + 3 x + 10 xN2 7 19 x

2+

1

2+ x I-2 + 3 x2M

2

Factor@QQQD

36 I4 + x2M I5 + x + x2M I1 + 9 x2M J161 + 30 10 - 322 x - 104 10 x + 285 x2 + 90 10 x2N

Expand@P - HHRRR + QQQ H1 - xL xLLD

0

H*P has a type III expression;

RRR and QQQ are not in SOS form but positive everywhere for all x,

further implementation is needed to derive those

expressions which will not be done due to time restrictions.*L

63

H* Example 3.2 -The polynomial is of this

form: A Aq1... q kHx-y i Lni ... Hx-ymLnm Hx-x i Ls i ... Ix-xpMspEIx-x j M

ni Hx-xkLs i ,

Where A is a constant <0. The q i 's are quadratics that cannot be factored,

so they have no real roots. The exponents n i are even and the exponents s i

are odd. Notice that in this case as long as the constant A is positive x j £

0 and each x k³ 0. Note that some of these factors are missing. *L

A = -1

-1

P = -22 + 13 x - x2

-22 + 13 x - x2

Factor @PD

-H-11 + xL H-2 + xL

Expand @P - HH-Hx - 2LL H-Hx - 11LL H-1LLD

0

Factor @PD

-H-11 + xL H-2 + xL

Plot @HPL, 8x, 0, 1 <D

0.2 0.4 0.6 0.8 1.0

-20

-18

-16

-14

-12

-10

A1 = -H-2 + xL

2 - x

Plot @A1, 8x, 0, 1 <D

0.2 0.4 0.6 0.8 1.0

1.2

1.4

1.6

1.8

2.0

64

xo = 2

2

B = xo

2

A =.

H = A

A

H*Apply Lemma 3.6 equation H3.5L*L

0 == 2 A B + A2 + 1

0 � 1 + 2 2 A + A2

a = 1

1

b = 2 B

2 2

c = 1

1

-b + b2 - 4 a c

2 a

1

2J2 - 2 2 N

Expand@%D

1 - 2

-b - b2 - 4 a c

2 a

1

2J-2 - 2 2 N

Expand@%D

-1 - 2

A =-b + b2 - 4 a c

2 a

1

2J2 - 2 2 N 65

H = A

1

2J2 - 2 2 N

B = xo

2

H*Verify the expression *L

A2 + 2 B A

2 J2 - 2 2 N +1

4J2 - 2 2 N2

FullSimplify@%D

-1

ExpandAH-Hx - xoLL - IIA2 + 2 B AM x + xoME

0

Expand@-Hx - xoL - HHA x + BL^2 + x H1 - xL H^2LD

0

Expand@A1 - HHHA x + BL^2 + x H1 - xL H^2LLD

0

A0 = A x + B

2 +1

2J2 - 2 2 N x

A01 = H

1

2J2 - 2 2 N

H********************************************************************L

B1 = -H-11 + xL

11 - x

Plot@B1, 8x, 0, 1<D

0.2 0.4 0.6 0.8 1.0

10.2

10.4

10.6

10.8

11.0

66

xo = 11

11

B = xo

11

A =.

H = A

A

H*Apply Lemma 3.6 equation H3.5L*L

0 == 2 A B + A2 + 1

0 � 1 + 2 11 A + A2

a = 1

1

b = 2 B

2 11

c = 1

1

-b + b2 - 4 a c

2 a

1

2J2 10 - 2 11 N

Expand@%D

10 - 11

-b - b2 - 4 a c

2 a

1

2J-2 10 - 2 11 N

Expand@%D

- 10 - 11

A =-b + b2 - 4 a c

2 a

1

2J2 10 - 2 11 N 67

H = A

1

2J2 10 - 2 11 N

B = xo

11

H*Verify the expression is equal to 1 *L

A2 + 2 B A

11 J2 10 - 2 11 N +1

4J2 10 - 2 11 N2

FullSimplify@%D

-1

ExpandAH-Hx - xoLL - IIA2 + 2 B AM x + xoME

0

Expand@-Hx - xoL - HHA x + BL^2 + x H1 - xL H^2LD

0

Expand@B1 - HHHA x + BL^2 + x H1 - xL H^2LLD

0

B0 = A x + B

11 +1

2J2 10 - 2 11 N x

B01 = H

1

2J2 10 - 2 11 N

H*Apply Multiplication Lemma For Type III polynomials*L

a = A0

2 +1

2J2 - 2 2 N x

b = 0

0

c = A01

1

2J2 - 2 2 N

68

d = 0

0

f = B0

11 +1

2J2 10 - 2 11 N x

g = 0

0

h = 0

0

k = B01

1

2J2 10 - 2 11 N

R = IIf 2+ g2M Ia2

+ b2MM + Ix2 H1 - xL2 Ih2+ k2M Ic2

+ d2MM

1

16J2 - 2 2 N2 J2 10 - 2 11 N2 H1 - xL2 x2 + 2 +

1

2J2 - 2 2 N x

2

11 +1

2J2 10 - 2 11 N x

2

Factor @RD

-2 J-11 + 44 x - 11 2 x - 2 110 x - 113 x2 + 54 2 x2 - 8 55 x2 + 9 110 x2 +

138 x3 - 85 2 x3 + 16 55 x3 - 13 110 x3 - 63 x4 + 42 2 x4 - 8 55 x4 + 6 110 x4N

Q = IIf 2+ g2M Ic2

+ d2M + Ia2+ b2M Ih2

+ k2MM

1

4J2 10 - 2 11 N2 2 +

1

2J2 - 2 2 N x

2

+

1

4J2 - 2 2 N2 11 +

1

2J2 10 - 2 11 N x

2

Factor @QD

75 - 22 2 - 4 110 - 150 x + 86 2 x - 16 55 x +

14 110 x + 126 x2 - 84 2 x2 + 16 55 x2 - 12 110 x2

H*Mathematica doesn't factor Q and R,

however they are positive for all x. The polynomials Q and R have type I representation ,

due to time restrictions this will not be explored. *L

H*Remember to include the last -

1 at the end into the equation to balance since theres one -

1 in P and now one -1 in the Type III representation of P =HHR+Q xH1-xLLH-1LL*L

Expand @P- HHR+ Q x H1 - xLL H-1LLD

0

69

H* Example 3.3-Corollary 3.10 take pHuL= u3,

positive @2,4D and give a Type III representation.*L

P = u3

u3

a = 2

2

b = 4

4

Plot@P, 8u, -5, 5<D

-4 -2 2 4

-100

-50

50

100

H*Apply the formula from Corollary 3.10*L

u = a + x Hb - aL

2 + 2 x

H*Factoring the polynomial shows a odd degree, we use theorem 3.1 case 2.*L

Factor@QD

8 H1 + xL3

ExpandAQ - I8 H1 + xL2 H1 + xLME

0

H*Next we get the type III representation of the polynomial Q*L

xo = -1

-1

70

Α = -xo

1

B = Α

1

A =.

H = A

A

H*Use equation H3.4L from Lemma 3.6*L

0 == 2 A B + A2 - 1

0 � -1 + 2 A + A2

a = 1

1

b = 2 B

2

c = -1

-1

H*Apply the quadratic formula.*L

-b + b2 - 4 a c

2 a

1

2J-2 + 2 2 N

Expand@%D

-1 + 2

-b - b2 - 4 a c

2 a

1

2J-2 - 2 2 N

71

Expand@%D

-1 - 2

A =

-b + b2 - 4 a c

2 a

1

2J-2 + 2 2 N

H = A

1

2J-2 + 2 2 N

B = Α

1

H*Verify that the expression is equal to 1*L

A2 + 2 B A

-2 + 2 2 +1

4J-2 + 2 2 N

2

FullSimplify@%D

1

H*A2+2B A =1 so the above expression should be zero.*L

ExpandAHHx - xoLL - IIA2 + 2 B AM x - xoME

0

H*Type III representation below.*L

Expand@Hx - xoL - HHA x + BL^2 + x H1 - xL H^2LD

0

H*It is concluded that original linear term x+1 has a type III representation.*L

f = 0

0

g = A x + B

1 +1

2J-2 + 2 2 N x

72

h = H

1

2J-2 + 2 2 N

k = 0

0

ExpandAHx + 1L - If2 + g2 + x H1 - xL Ih2 + k2MME

0

l = 8 H1 + xL

2 2 H1 + xL

m = 0

0

n = 0

0

o = 0

0

H*Now apply Lemma 3.4 regarding multiplication of Type III polynomials.*L

F = IIf2 + g2M Il2 + m2MM + Ix2 H1 - xL2 Ih2 + k2M In2 + o2MM

8 H1 + xL2 1 +1

2J-2 + 2 2 N x

2

Factor@FD

8 H1 + xL2 J1 - x + 2 xN2

G = If2 + g2M In2 + o2M + Il2 + m2M Ih2 + k2M

2 J-2 + 2 2 N2H1 + xL2

Expand@P - HF + G x H1 - xLLD

0

H*Use Lemma 1.7 to get two sums of squares. *L

f = 8 H1 + xL

2 2 H1 +

73

g = 0

0

h = J1 - x + 2 xN

1 - x + 2 x

k = 0

0

X = f k - g h

0

W = f h + g k

2 2 H1 + xL J1 - x + 2 xN

ExpandAF - IX2 + W2ME0

H*****************************************************L

f = 2 J-2 + 2 2 N

2 J-2 + 2 2 N

g = 0

0

h = H1 + xL

1 + x

k = 0

0

V = f k - g h

0

U = f h + g k

2 J-2 + 2 2 N H1 + xL

74

Expand@UD

4 - 2 2 + 4 x - 2 2 x

ExpandAG - IV2 + U2ME

0

ExpandAQ - IX2 + W2 + x H1 - xL IV2 + U2MME

0

H*Now with the type III expression for Q, on @0,1D,we continue using corollary 3.10 to find the type III representation on@2,4D*L

a = 2

2

b = 4

4

u =.

x =

u - a

b - a

1

2H-2 + uL

ZZ = X2 + W2 + x H1 - xL IV2 + U2M

8 1 +1

2H-2 + uL

2

1 +2 - u

2+

-2 + u

2

2

+ J-2 + 2 2 N2

1 +2 - u

21 +

1

2H-2 + uL

2

H-2 + uL

RR = 8 1 +

1

2H-2 + uL 1 +

2 - u

2+

-2 + u

2

2 2 1 +1

2H-2 + uL 1 +

2 - u

2+

-2 + u

2

SS = X

0

TT = V

75

H*The 1

Hb-aLgets absorbed into each polynomial TT and UU*L

UU =

1

Hb - aL U

J-2 + 2 2 N I1 + 1

2H-2 + uLM

2

H*Prove that ZZ = Q = P is equal to the expression below*L

ExpandAllAZZ - IRR2 + SS2 + Hu - aL Hb - uL ITT2 + UU2MME

0

ExpandAllAP - IRR2 + SS2 + Hu - aL Hb - uL ITT2 + UU2MME

0

H*So the orginal P = u3, has a type III representation on @2,4D.*L

76

Appendix E. Section 4 Examples

H*Example 4.1- sum of squares in 2 variables, show that pHx,yL is a sum of squares*L

p = 7 + 2 x + 14 x2 + 14 y + 28 x y + 49 y2

7 + 2 x + 14 x2 + 14 y + 28 x y + 49 y2

v = 81, x, y<

81, x, y<

P = 88a, b, c<, 8b, d, e<, 8c, e, f<<

88a, b, c<, 8b, d, e<, 8c, e, f<<

MatrixForm@PD

a b c

b d e

c e f

Together@p - v.P.vD

7 - a + 2 x - 2 b x + 14 x2 - d x2 + 14 y - 2 c y + 28 x y - 2 e x y + 49 y2 - f y2

a = 7

7

b = 1

1

c = 7

7

d = 14

14

e = 14

14

f = 49

49

p - v.P.v

x + 14 x2 + 7 y + 28 x y + 49 y2 - x H1 + 14 x + 14 yL - y H7 + 14 x + 49 yL

Factor@%D

77

P

887, 1, 7<, 81, 14, 14<, 87, 14, 49<<

MatrixForm@%D

7 1 7

1 14 14

7 14 49

Det@PD

2891

CholeskyDecomposition@PD

:: 7 ,1

7, 7 >, :0,

97

7, 13

7

97>, :0, 0, 7

59

97>>

A = %

:: 7 ,1

7, 7 >, :0,

97

7, 13

7

97>, :0, 0, 7

59

97>>

MatrixForm@AD

7 1

77

0 97

713 7

97

0 0 7 59

97

tA = Transpose@AD

:: 7 , 0, 0>, :1

7,

97

7, 0>, : 7 , 13

7

97, 7

59

97>>

MatrixForm@tAD

7 0 0

1

7

97

70

7 13 7

977 59

97

78

H*Double check that P = Transpose @AD.A: *L

P- tA.A

880, 0, 0<, 80, 0, 0<, 80, 0, 0<<

H* to express p as a sum of squares, we now use Hv.tA L. HA.v L *L

Hv.tA L. HA.v L

2891 y2

97+

97

7x + 13

7

97y

2

+ 7 +x

7+ 7 y

2

H*this is a sum of squares expression;

we double check that it equals p *L

q = %

2891 y2

97+

97

7x + 13

7

97y

2

+ 7 +x

7+ 7 y

2

Factor @p - qD

0

H*Another way to check that a real

symmetric matrix is positive is to

verify that all of its eigenvalues are nonnegative: *L

Eigenvalues @PD

9RootA-2891 + 881 ð1 - 70 ð12 + ð13 &, 3E,

RootA-2891 + 881 ð1 - 70 ð12 + ð13 &, 2E, RootA-2891 + 881 ð1 - 70 ð12 + ð13 &, 1E=

N@%D

854.9159, 9.60073, 5.48334<

H*This means that there are

vectors a,b,c such that P.a = 54.91592601638876` a,

P.b = 9.600729026932235` b, Pc = 5.483344956679007` < c. Further,

any vector w can be expressed

as w = ra + sb + tc,

so Pw = 54.91592601638876` Hra L + 9.600729026932235` HsbL + 5.483344956679007` Htc L.

This shows that P is like

stretching by positive multipliers

in three directions

79

H* Example 4.2 - Consider a polynomial of two variables x,

y of degree 4. We wish to express P Hx,y L as a sum of squares. *L

P = 6 + 8 x + 16 x 2+ 16 x 3

+ 16 x 4+ 8 y + 28 x y + 24 x 2 y + 10 y 2

+ 9 x 2 y2

6 + 8 x + 16 x2 + 16 x3 + 16 x4 + 8 y + 28 x y + 24 x2 y + 10 y2 + 9 x2 y2

H*Notice mathematica doesn't know how to factor this polynomial. *L

Factor @PD

6 + 8 x + 16 x2 + 16 x3 + 16 x4 + 8 y + 28 x y + 24 x2 y + 10 y2 + 9 x2 y2

H*The vector v, must count up to half of the total degree. *L

v = 91, x, y, x 2, x y, y 2=

91, x, y, x2, x y, y2=

H*The table is created to generated to compare to coefficients of variables *L

M= Table @g@i, j D, 8i, 6 <, 8j, 6 <D

88g@1, 1D, g@1, 2D, g@1, 3D, g@1, 4D, g@1, 5D, g@1, 6D<,8g@2, 1D, g@2, 2D, g@2, 3D, g@2, 4D, g@2, 5D, g@2, 6D<,8g@3, 1D, g@3, 2D, g@3, 3D, g@3, 4D, g@3, 5D, g@3, 6D<,8g@4, 1D, g@4, 2D, g@4, 3D, g@4, 4D, g@4, 5D, g@4, 6D<,8g@5, 1D, g@5, 2D, g@5, 3D, g@5, 4D, g@5, 5D, g@5, 6D<,8g@6, 1D, g@6, 2D, g@6, 3D, g@6, 4D, g@6, 5D, g@6, 6D<<

H*The code below makes the matrix M its own transpose M = MT*L

For @i = 1, i £ 6, i = i + 1,

For @j = 1, j £ i, j = j + 1,

g@i, j D = g@j, i DDD

MatrixForm @MD

g@1, 1D g@1, 2D g@1, 3D g@1, 4D g@1, 5D g@1, 6Dg@1, 2D g@2, 2D g@2, 3D g@2, 4D g@2, 5D g@2, 6Dg@1, 3D g@2, 3D g@3, 3D g@3, 4D g@3, 5D g@3, 6Dg@1, 4D g@2, 4D g@3, 4D g@4, 4D g@4, 5D g@4, 6Dg@1, 5D g@2, 5D g@3, 5D g@4, 5D g@5, 5D g@5, 6Dg@1, 6D g@2, 6D g@3, 6D g@4, 6D g@5, 6D g@6, 6D

H*Solve p = v.M.v for M *L

Factor @v.M.v D

g@1, 1D + 2 x g@1, 2D + 2 y g@1, 3D + 2 x2 g@1, 4D + 2 x y g@1, 5D + 2 y2 g@1, 6D + x2 g@2, 2D + 2 x y g@2, 3D +2 x3 g@2, 4D + 2 x2 y g@2, 5D + 2 x y2 g@2, 6D + y2 g@3, 3D + 2 x2 y g@3, 4D + 2 x y2 g@3, 5D +2 y3 g@3, 6D + x4 g@4, 4D + 2 x3 y g@4, 5D + 2 x2 y2 g@4, 6D + x2 y2 g@5, 5D + 2 x y3 g@5, 6D + y4 g@6, 6D

Expand @P- %D

6 + 8 x + 16 x2 + 16 x3 + 16 x4 + 8 y + 28 x y + 24 x2 y + 10 y2 + 9 x2 y2 - g@1, 1D - 2 x g@1, 2D -2 y g@1, 3D - 2 x2 g@1, 4D - 2 x y g@1, 5D - 2 y2 g@1, 6D - x2 g@2, 2D - 2 x y g@2, 3D - 2 x3 g@2, 4D -2 x2 y g@2, 5D - 2 x y2 g@2, 6D - y2 g@3, 3D - 2 x2 y g@3, 4D - 2 x y2 g@3, 5D - 2 y3 g@3, 6D -x4 g@4, 4D - 2 x3 y g@4, 5D - 2 x2 y2 g@4, 6D - x2 y2 g@5, 5D - 2 x y3 g@5, 6D - y4 g@

80

H*Now try to guess the coefficients for the terms starting with constants, x, x2,etc.*L

g@1, 1D = 6

6

g@1, 2D = 4

4

g@1, 3D = 4

4

g@2, 2D = 16 - 2 g@1, 4D

16 - 2 g@1, 4D

g@2, 3D = H14 - g@1, 5DL

14 - g@1, 5D

g@3, 3D = 10 - 2 g@1, 6D

10 - 2 g@1, 6D

g@2, 4D = 8

8

g@3, 4D = 12 - g@2, 5D

12 - g@2, 5D

g@3, 5D = -g@2, 6D

-g@2, 6D

g@3, 6D = 0

0

g@4, 4D = 16

16

g@4, 5D = 0

0

g@5, 5D = -2 g@4, 6D + 9

9 - 2 g@4, 6D

g@5, 6D = 0

0

81

g@6, 6D = 0

0

MatrixForm@MD

6 4 4 g@1, 4D g@1, 5D g@1, 6D

4 16 - 2 g@1, 4D 14 - g@1, 5D 8 g@2, 5D g@2, 6D

4 14 - g@1, 5D 10 - 2 g@1, 6D 12 - g@2, 5D -g@2, 6D 0

g@1, 4D 8 12 - g@2, 5D 16 0 g@4, 6D

g@1, 5D g@2, 5D -g@2, 6D 0 9 - 2 g@4, 6D 0

g@1, 6D g@2, 6D 0 g@4, 6D 0 0

ExpandAll@P - v.M.vD

0

H*Notice again that the lower corner of the matrix M has

a zero. This means that the entire row and column must be zero.*L

MatrixForm@MD

6 4 4 g@1, 4D g@1, 5D g@1, 6D

4 16 - 2 g@1, 4D 14 - g@1, 5D 8 g@2, 5D g@2, 6D

4 14 - g@1, 5D 10 - 2 g@1, 6D 12 - g@2, 5D -g@2, 6D 0

g@1, 4D 8 12 - g@2, 5D 16 0 g@4, 6D

g@1, 5D g@2, 5D -g@2, 6D 0 9 - 2 g@4, 6D 0

g@1, 6D g@2, 6D 0 g@4, 6D 0 0

g@1, 6D = g@2, 6D = g@4, 6D = 0

0

MatrixForm@MD

6 4 4 g@1, 4D g@1, 5D 0

4 16 - 2 g@1, 4D 14 - g@1, 5D 8 g@2, 5D 0

4 14 - g@1, 5D 10 12 - g@2, 5D 0 0

g@1, 4D 8 12 - g@2, 5D 16 0 0

g@1, 5D g@2, 5D 0 0 9 0

0 0 0 0 0 0

H*Define the different determinants of each corners.*L

T1 = Det@M@@81, 2<, 81, 2<DDD

80 - 12 g@1, 4D

T2 = Det@M@@81, 2, 3<, 81, 2, 3<DDD

-184 - 88 g@1, 4D + 136 g@1, 5D - 6 g@1, 5D2

T3 = Det@M@@81, 2, 3, 4<, 81, 2, 3, 4<DDD

-4224 + 256 g@1, 4D - 156 g@1, 4D2 + 20 g@1, 4D3 + 1024 g@1, 5D + 160 g@1, 4D g@1, 5D -

28 g@1, 4D2 g@1, 5D - 96 g@1, 5D2 + g@1, 4D2 g@1, 5D2 + 832 g@2, 5D - 304 g@1, 4D g@2, 5D +

16 g@1, 4D2 g@2, 5D + 96 g@1, 5D g@2, 5D - 8 g@1, 4D g@1, 5D g@2, 5D - 80 g@2, 5D2 + 12 g@1, 4D g@2, 5

82

T4 = Det@M@@81, 2, 3, 4, 5<, 81, 2, 3, 4, 5<DDD

-38 016 + 2304 g@1, 4D - 1404 g@1, 4D2 + 180 g@1, 4D3 + 9216 g@1, 5D +1440 g@1, 4D g@1, 5D - 252 g@1, 4D2 g@1, 5D - 32 g@1, 5D2 + 32 g@1, 4D g@1, 5D2 +9 g@1, 4D2 g@1, 5D2 - 256 g@1, 5D3 + 16 g@1, 5D4 + 7488 g@2, 5D - 2736 g@1, 4D g@2, 5D +144 g@1, 4D2 g@2, 5D - 32 g@1, 5D g@2, 5D + 104 g@1, 4D g@1, 5D g@2, 5D - 32 g@1, 5D2 g@2, 5D +24 g@1, 4D g@1, 5D2 g@2, 5D - 16 g@1, 5D3 g@2, 5D - 560 g@2, 5D2 + 12 g@1, 4D g@2, 5D2 +10 g@1, 4D2 g@2, 5D2 + 128 g@1, 5D g@2, 5D2 - 28 g@1, 4D g@1, 5D g@2, 5D2 +16 g@1, 5D2 g@2, 5D2 - 144 g@2, 5D3 + 8 g@1, 4D g@2, 5D3 - 8 g@1, 5D g@2, 5D3 + 6 g@2, 5D4

H*Since there is only three variables left define them as s, t, and u.*L

g@1, 4D = s

s

g@1, 5D = t

t

g@2, 5D = u

u

Factor@T1D

-4 H-20 + 3 sL

Factor@T2D

-2 I92 + 44 s - 68 t + 3 t2M

Factor@T3D

-4224 + 256 s - 156 s2 + 20 s3 + 1024 t + 160 s t - 28 s2 t -

96 t2 + s2 t2 + 832 u - 304 s u + 16 s2 u + 96 t u - 8 s t u - 80 u2 + 12 s u2

Factor@T4D

-38 016 + 2304 s - 1404 s2 + 180 s3 + 9216 t + 1440 s t - 252 s2 t - 32 t2 + 32 s t2 + 9 s2 t2 -

256 t3 + 16 t4 + 7488 u - 2736 s u + 144 s2 u - 32 t u + 104 s t u - 32 t2 u + 24 s t2 u - 16 t3 u -

560 u2 + 12 s u2 + 10 s2 u2 + 128 t u2 - 28 s t u2 + 16 t2 u2 - 144 u3 + 8 s u3 - 8 t u3 + 6 u4

Solve@T1 � 0, sD

::s ®20

3>>

N@%D

88s ® 6.66667<<

H*Solving for the determinant of T1 shows that s,

must be less then 6.6667, in this case we chose s =4. *L

s = 4

83

MatrixForm@MD

6 4 4 4 t 0

4 8 14 - t 8 u 0

4 14 - t 10 12 - u 0 0

4 8 12 - u 16 0 0

t u 0 0 9 0

0 0 0 0 0 0

T1

32

H*T1 is now positive, now continue the process, we solve for t and u

using the same method as above in order to have a positive matrix M. *L

Solve@T2 � 0, tD

::t ®2

3J17 - 2 22 N>, :t ®

2

3J17 + 2 22 N>>

N@%D

88t ® 5.07945<, 8t ® 17.5872<<

H*t must be between the intervals of @5,17.5D, so we pick t =6.*L

t = 6

6

MatrixForm@MD

6 4 4 4 6 0

4 8 8 8 u 0

4 8 10 12 - u 0 0

4 8 12 - u 16 0 0

6 u 0 0 9 0

0 0 0 0 0 0

T2

64

H*T2 is positive, now move on the next determinant T3 and solve for u.*L

Solve@T3 � 0, uD

88u ® 0<, 8u ® 8<<

H*T3 must be in the interval of @0,8D so choose u =0*L

u = 0

84

MatrixForm @MD

6 4 4 4 6 0

4 8 8 8 0 0

4 8 10 12 0 0

4 8 12 16 0 0

6 0 0 0 9 0

0 0 0 0 0 0

T3

0

ExpandAll @P- Hv.M.v LD

0

H*Now verify that P is positive using eigenvalues, and cholesky decomposition. *L

Eigenvalues @MD

9RootA-1428 + 576 ð1 - 49 ð12 + ð13 &, 3E,

RootA-1428 + 576 ð1 - 49 ð12 + ð13 &, 2E, RootA-1428 + 576 ð1 - 49 ð12 + ð13 &, 1E, 0, 0, 0=

N@%D

832.7382, 12.8735, 3.38826, 0., 0., 0.<

CholeskyDecomposition @MD

CholeskyDecomposition::posdef:

Thematrix886,4,4,4,6,0<,84,8,8,8,0,0<,84,8,10,12,0,0<,84,8,12,16,0,0<,86,0,0,0,9,0<,80,0,0,0,0,0<< isnot

sufficientlypositivedefinitetocompletetheCholeskydecompositiontoreasonableaccuracy.�

CholeskyDecomposition@886, 4, 4, 4, 6, 0<, 84, 8, 8, 8, 0, 0<,84, 8, 10, 12, 0, 0<, 84, 8, 12, 16, 0, 0<, 86, 0, 0, 0, 9, 0<, 80, 0, 0, 0, 0, 0<<D

H*The choleskydecomposition results in a error message however M is definitely positive,

the only problem is the Matrix M has zeros in some of its

rows and columns and mathematica simply can't produce a outcome. *L

H*The next best step is to break up the matrix M into 4 sections of 3x3 matricies. *L

R = M@@81, 2, 3 <, 81, 2, 3 <DD

886, 4, 4<, 84, 8, 8<, 84, 8, 10<<

Eigenvalues @RD

9RootA-64 + 92 ð1 - 24 ð12 + ð13 &, 3E,

RootA-64 + 92 ð1 - 24 ð12 + ð13 &, 2E, RootA-64 + 92 ð1 - 24 ð12 + ð13 &, 1E=

N@%D

819.4359

85

H*Eigenvalues shows that R is positive.*L

MatrixForm@RD

6 4 4

4 8 8

4 8 10

Det@RD

64

CholeskyDecomposition@RD

:: 6 , 22

3, 2

2

3>, :0,

4

3,

4

3>, :0, 0, 2 >>

X = :: 6 , 22

3, 2

2

3>, :0,

4

3

,4

3

>, :0, 0, 2 >>

:: 6 , 22

3, 2

2

3>, :0,

4

3,

4

3>, :0, 0, 2 >>

MatrixForm@CDD

CD

S = M@@81, 2, 3<, 84, 5, 6<DD

884, 6, 0<, 88, 0, 0<, 812, 0, 0<<

W = Inverse@RD.S

::0,3

2, 0>, :-1, -

3

4, 0>, 82, 0, 0<>

H*S= R.W*L

S - R.W

880, 0, 0<, 80, 0, 0<, 80, 0, 0<<

U = M@@84, 5, 6<, 84, 5, 6<DD

8816, 0, 0<, 80, 9, 0<, 80, 0, 0<<

U - Transpose@SD.W

880, 0, 0<, 80, 0, 0<, 80, 0, 0<<

Transpose@XD.X - R

880, 0, 0<, 80, 0, 0<, 80, 0, 0<<

86

MatrixForm@XD

6 2 2

32 2

3

0 4

3

4

3

0 0 2

X.W

::22

3, 6 , 0>, :

4

3, - 3 , 0>, :2 2 , 0, 0>>

MatrixForm@%D

2 2

36 0

4

3- 3 0

2 2 0 0

A = :: 6 , 22

3, 2

2

3, 2

2

3, 6 , 0>, :0,

4

3

,4

3

,4

3

, - 3 , 0>,

:0, 0, 2 , 2 2 , 0, 0>, 80, 0, 0, 0, 0, 0<, 80, 0, 0, 0, 0, 0<>

:: 6 , 22

3, 2

2

3, 2

2

3, 6 , 0>, :0,

4

3,

4

3,

4

3, - 3 , 0>,

:0, 0, 2 , 2 2 , 0, 0>, 80, 0, 0, 0, 0, 0<, 80, 0, 0, 0, 0, 0<>

MatrixForm@AD

6 2 2

32 2

32 2

36 0

0 4

3

4

3

4

3- 3 0

0 0 2 2 2 0 0

0 0 0 0 0 0

0 0 0 0 0 0

A0 = Transpose@AD

:: 6 , 0, 0, 0, 0>, :22

3,

4

3, 0, 0, 0>, :2

2

3,

4

3, 2 , 0, 0>,

:22

3,

4

3, 2 2 , 0, 0>, : 6 , - 3 , 0, 0, 0>, 80, 0

87

H*M = AT.A*L

MatrixForm@A0D

6 0 0 0 0

2 2

3

4

30 0 0

2 2

3

4

32 0 0

2 2

3

4

32 2 0 0

6 - 3 0 0 0

0 0 0 0 0

M - A0.A

880, 0, 0, 0, 0, 0<, 80, 0, 0, 0, 0, 0<, 80, 0, 0, 0, 0, 0<,80, 0, 0, 0, 0, 0<, 80, 0, 0, 0, 0, 0<, 80, 0, 0, 0, 0, 0<<

Factor@Hv.A0L HA.vLD

:2

3I3 + 2 x + 2 x2 + 2 y + 3 x yM2,

1

3I4 x + 4 x2 + 4 y - 3 x yM2, 2 I2 x2 + yM2, 0, 0>

2

3I3 + 2 x + 2 x2 + 2 y + 3 x yM

2+

1

3I4 x + 4 x2 + 4 y - 3 x yM

2+ 2 I2 x2 + yM

2

2 I2 x2 + yM2 +1

3I4 x + 4 x2 + 4 y - 3 x yM2 +

2

3I3 + 2 x + 2 x2 + 2 y + 3 x yM2

ExpandAll@P - %D

0

H*P is a sum of squares and positive for each x,

88