Convex Hulls of Curves Sam Burton

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1 Convex Hull of Curve Sam Burton 1 Introduction Thi paper will primarily be concerned with determining the face of convex hull of curve of the form C = {(t, t a, t b ) t [ 1, 1]}, a < b N in R 3. We hall ee that the vertex point of the hull face can be determined baed only on whether a and b are even or odd. Thi allow u to categorize them with only four different cae which we will explore in later ection. Thi ection will provide a baic introduction to the topic a well a a few fact and exercie which will be ueful later on. 1.1 Definition To begin, we will tate ome of the baic definition in the tudy of convexity. A et S i convex if for any two point 1, 2 S, the line egment connecting them {u 1 +(1 u) 2 u [, 1]} i contained in S. The convex hull of a et S i the mallet convex et containing S or equivalently the et of all point of the form Σ i u i i where the i are point in S and the u i are point in (, 1] uch that Σ i u i = 1. A face of a convex et S in R n i a ubet T S uch that there exit ome non-zero affine map which i zero on T and poitive on S T. An affine map i a function l : R n R of the form l(x 1,..., x n ) = c n x n + c n 1 x n c 1 x 1 + c A vertex i a zero dimenional face, an edge i a one dimenional face, and a facet i a face of dimenion n 1. In a geometric ene, the zero et of a non-contant affine map can be thought of a a hyperplane in R n. One can how that a ubet of a convex body i a face by howing that an affine map i zero on that ubet and poitive on the ret of the convex body. Thi i akin to finding a hyperplane (the affine map zero et) which interect the convex body only at one of it urface (the face). 1.2 An important fact about affine map Theorem 1. For x = (x 1,..., x n ), y = (y 1,..., y n ) R n and u R, and a given affine map l(r 1,..., r n ) = c n r n + c n 1 r n c 1 r 1 + c, l(ux + (1 u)y) = ul(x) + (1 u)l(y). More generally, for any 1,..., m R n and any u 1,..., u m (, 1] uch that m u i = 1, l( m u i i ) = i=1 m u i l( i ) Proof. Thi fact can be hown through ome imple computation. i=1 l(ux + (1 u)y) = l(ux 1 + (1 u)y 1,..., ux n + (1 u)y n ) = c n (ux n + (1 u)y n ) + + c 1 (ux 1 + (1 u)y 1 ) + c = c n (ux n + (1 u)y n ) + + c 1 (ux 1 + (1 u)y 1 ) + c (u + (1 u)) = u(c n x n + c n 1 x n c 1 x 1 + c ) + (1 u)(c n y n + c n 1 y n c 1 y 1 + c ) i=1

2 = ul(x) + (1 u)l(y) The generalization can be obtained from a imple induction which i done in much the ame way. 1.3 Ueful reult which follow from 1.2 Theorem 1 allow u to prove everal important fact about convex hull. Theorem 2. A affine map l which i non-negative on a et S will be non-negative on the convex hull of S. Proof. Every point in the convex hull of S can be written a = Σ i u i i, where i S u i (, 1], and Σ i u i = 1. Therefore, by Theorem 1, l() = l(σ i u i i ) = Σ i (u i )l( i ). A l( i ) and u i are non-negative for each i, we ee that l() mut be non-negative. Theorem 3. All vertice of the convex hull of a et S mut be element of S. Proof. Becaue a vertex i zero dimenional, it i a ingle point. Suppoe that for a given point on the convex hull of a et S, there exit a affine map l which i non-negative on the convex hull and zero at. A can be written a = Σ i u i i, where i S, u i (, 1], and Σ i u i = 1 and Theorem 1 tell u that l() = Σ i u i l( i ). We have that each u i i poitive and l( i ) i non-negative for each i. Hence each term of the um i nonnegative, and o a l() =, we have that l( i ) = for each i, and o we ee that the only way for a affine map to be non-negative on the convex hull and zero at preciely one point on it, i for that point to be in S and hence expreable with a um that ha only one term. 1.4 Convex Hull of curve and polynomial The convex hull of curve of the form C = {(t d 1,..., t dn ) t [ 1, 1]} incline themelve toward tudy uing polynomial, a a given affine map l(x 1,..., x n ) = c n x n + c n 1 x n c 1 x 1 + c correpond directly to the polynomial f(x) = c n x dn + c n 1 x d n c 1 x d 1 + c along the curve, where all vertice of all face of the convex hull mut be. Uing thi fact, we can find the face of the convex hull of the curve, by finding the et of point on [ 1, 1] for which there exit polynomial of the given form which are non-negative on the interval, and have root at thoe point. That i, if there exit a polynomial f of the given form which i non-negative on [ 1, 1] and zero at ome point t 1,..., t k [ 1, 1], then C(t 1 ),..., C(t k ) are the vertice of a k 1 dimenional face of the convex hull of the curve. 1.5 Convex Hull of Curve of the form C = {(t, t 2,..., t d ) t [ 1, 1]} Though lightly different from the curve which we will be focuing on, determining the face of the convex hull of curve of the form C = {(t, t 2,..., t d ) t [ 1, 1]} in R d will erve a an intructive exercie to how how the face of convex hull of curve can be tudied uing polynomial. Theorem 4. Fix d diticnt point t 1,..., t d [ 1, 1] (1) If t 1,..., t n ( 1, 1), then C(t 1 ),..., C(t n ) are the vertice of an n 1 dimenional face of the convex hull of C if and only if d 2n. (2) If one of the t i i at 1 or 1, then C(t 1 ),..., C(t n ) are the vertice of an n 1 dimenional face if and only if d 2n 1. (3) If two of the t i are at 1 and 1, then C(t 1 ),..., C(t n ) are the vertice of an n 1 dimenional face if and only if d 2n 2.

3 Proof. If C(t 1 ),..., C(t n ) are the vertice of a face of the convex hull then there exit a affine map which i zero at C(t 1 ),..., C(t n ), and non-negative on C. Thi i equivalent to there being a non-zero polynomial of degree at mot d which i non-negative on [ 1, 1], and ha root at each of the t i. In order for a given t i (1, 1) to be the root of uch a polynomial, there mut be a multiple root at that t i, a the derivative of the polynomial mut be zero. The lowet degree polynomial which ha a multiple root at each t i and i non-negative on the interval i (x t 1 ) 2 (x t n ) 2, which ha degree 2n, and o we ee that d mut be at leat 2n. In fact, a double root at each t i i ufficient to enure that the polynomial i non-negative on the interval, a (x t i ) 2 i non-negative everywhere for each t i, and o (x t 1 ) 2 (x t n ) 2 i non-negative everywhere. Thu, any d 2n will allow for C(t 1 ),..., C(t n ) to be the vertice of a face of the convex hull of C. The proof for part (2) and (3) are exactly a the proof for (1), except that it i not neceary for there to be a multiple root at an endpoint of the interval, a the polynomial doe not need to be non-negative for value le than 1 or for value greater than 1. In particular, the linear factor (t + 1) and (t 1) are non-negative on [ 1, 1], and o the minimum degree of a polynomial which i non-negative on the interval with root at each t i i reduced by one for each endpoint contained in {t 1,..., t n }. 1.6 Decarte Rule of Sign In many place throughout thi paper, we will make ue of Decarte rule of ign which tate that the number of poitive root i at mot the number of ign change between the coefficient of conecutive term of a polynomial f(x) = c n x an + c n 1 x a n c 1 x a 1, a i < a i+1 for all i, and a a corallary, tate that the number of negative root i at mot the number of ign change between the coefficient of conecutive term of f( x), where the ign of every odd power term of f i changed. In addition, the following Lemma, which i will ee even more frequent ue, follow directly from Decarte rule. Lemma 1. Let a 1,..., a n be non-negative integer uch that a i < a i+1 for all i, and fix c 1,..., c n R. The polynomial f(x) = c n x an + c n 1 x a n c 1 x a 1 can have at mot 2(n 1) P non-zero real root, where P i the number of change of parity in the equence (a 1,..., a n ) (that i, the number of even power term followed by odd power term and vice vera). Proof. Becaue f(x) and f( x) each have n term, each polynomial ha at mot n 1 potential ign change between conecutive term, and o we ee that the number of non-zero root of f i at mot 2(n 1). However, uppoe that for ome i {1,..., n 1}, a i and a i+1 have different parity, that i, one i even and the othe odd. The coefficient of the even power coefficient will have the ame ign in f( x) that it doe in f(x), but the odd power coefficient will have oppoite ign. Hence, if the ign of the coefficient differ in f(x) they will be the ame in f( x), and vice vera, o it i impoible for both polynomial to have a difference of ign between the coefficient of the a i and a i+1 term. Therefore, for each uch change of parity, the upper bound for the number of non-zero root i reduced by one, and o we achieve the deired reult. 1.7 Projection of Convex Bodie Another ueful lemma for finding the face of convex hull ue projection to implify the problem. Lemma 2. Let π : R n+m R n be the projection map defined by (x 1,..., x n+m ) (x i1,..., x in ), where the i j are ditinct poitive integer le than or equal to n+m and let C be the curve {(t a 1,..., t a n+m ) t [ 1, 1]}, where a 1 < < a n+m N. If ome collection of point on π(c) are vertice of a face of the convex hull of

4 π(c), then the preimage of thoe point are vertice of a face of the convex hull of C. (In fact thi tatement i true for any linear projection of any convex body, but we will not require the generalization for our purpoe) Proof. Suppoe that π(c(r 1 )),..., π(c(r l )), r 1,..., r l [ 1, 1], are vertice of a face of π(c). Then there exit ome polynomial f(t) = c i1 t a i c in t a in which i zero at thoe point. Becaue the point on C have t a i j entrie for each ij, thi polynomial alo correpond to a affine map in R n+k which i non-negative on C, and i zero at each C(r) uch that for all a i, r a i = r a i k for ome k {1,..., l}. The et of uch C(r) i the et of preimage point {π 1 (π(c(r k ))) k {1,..., l}}. Thi in turn, allow u to make ue of the following lemma. Lemma 3. For a curve S = {(x, f(x)) x [v, w]} in R 2, where [v, w] i an arbitrary cloed interval and f(x) i a function uch that f (x) > for x (, 1), we have that each point on S i a vertex of the convex hull of S. Proof. Each point on the curve can be written a (t, f(t)) for ome t [v, w]. For a given t, conider the affine map l(x, y) = y f (t)(x t) f(t). For any point on the curve, y = f(x) and o l(x, y) i equal to the function g(x) = f(x) f (t)(x t) f(t). We ee that at x = t, g(x) = f(t) f(t) =, and that for x [v, w], g (x) = f (x) f (t). Now, becaue f (x) > for all x (, 1), f (x) < f (t) for all x [, t) and f (x) > f (t) for all x (t, 1]. Hence, g(x) i decreaing on [, t) and increaing on (t, ], and therefore g(x) > g(t) = for all x [v, w] {t}. Thu l(t, f(t)) = while l(x, f(x)) > for all other x on the interval, and o (t, f(t)) i a vertex of the convex hull for each t [, 1]. 2. Even-Odd Curve The firt of curve of the form C = {(t, t a, t b ) t [ 1, 1]}, a < b N in R 3 which we will examine are the cae where b i odd and a i even. Thi i a natural tarting place, a we upect that the convex hull will reemble the hull of the curve {(t, t 2, t 3 ) t [ 1, 1]} which we can determine to have a vertex at each point on the curve, and edge connecting each point on the curve to both endpoint uing Theorem Vertice Theorem 5. Each point on C i a vertex of the convex hull. Proof. Conider the curve S = {(t, t a ) t [ 1, 1]} in R 2. Becaue a i even, the econd derivative of x a i poitive on R each point on S i a vertex of the convex hull of S by Lemma 3. Moreover, we ee that for the projection π : R 3 R 2, (x 1, x 2, x 3 ) (x 1, x 2 ), a given point (t, t a ) S ha preimage π 1 (t, t a ) = {(t, t a, t b )}. Thu (t, t a, t b ) i a vertex of C for each t [ 1, 1]. Eentially the ame proof will hold for any curve with a t entry and an even power entry, and we will refer back to it later on. 2.2 Edge of the form C(1)C(t) and C( 1)C(t) Theorem 6. For each t [ 1, 1), there exit an edge of the convex hull with vertice at C(t) and C(1) and an edge of the convex hull with vertice at C( t) and C( 1).

5 Proof. It uffice to how that there i a polynomial of the form f(x) = x b + c a x a + c 1 x + c with root at t and 1, which doe not change ign on the interval [ 1, 1]. Becaue t ( 1, 1) thi mean that there mut in fact be at leat double root at t. Firt we mut how that a polynomial of the correct form with a root at 1 and double root at t exit. To achieve thi, we will olve for the coefficient c a, c 1, and c. The root at 1 give u the equation 1 + c a + c 1 + c = c a + c 1 + c = 1. The fact that there i a root at t tell u that t b + c a t a + c 1 t + c = c a t a + c 1 t + c = t b. Latly, becaue there i a double root at t, f (t) =, and thu bt b 1 + ac a t a 1 + c 1 = ac a t a 1 + c 1 = bt b 1. Thi give u the following ytem of equation or equivalently, the following matrix equation t a t 1 at a 1 1 c a + c 1 + c = 1 c a t a + c 1 t + c = t b ac a t a 1 + c 1 = bt b 1 c a c 1 c = 1 t b bt b 1 The determinant of t a t 1 i the polynomial (1 a)t a + at a 1 1, which we ee i equal to zero at a 1 1 for t = 1. We conider thi polynomial derivative t ( 1 + ata 1 + (1 a)t a ) = a(a 1)t a 2 + a(1 a)t a 1 = a(a 1)(t a 2 t a 1 ) = a(a 1)t a 2 (1 t) A a i even, t a 2 i poitive, and o the derivative i poitive for all t ( 1, 1). Hence, the determinant i increaing on ( 1, 1) and mut therefore be negative and in particular non-zero for all t [ 1, 1). Thu a olution to the matrix equation exit for all poible t, and o for any given t [ 1, 1), there exit a polynomial of the form f(x) = x b + c a x a + c 1 x + c with a double root at t and a root at 1.

6 For t, f ha three non-zero real root, and hence cannot have any more by Lemma 1. Moreover, if we aume there i a root at, then we have that c =, and o f(x) = x b + c a x a + c 1 x. Thi would mean that by Lemma 1, f(x) could only have two non-zero real root, which i a contradiction. Thu, we ee that f cannot have any other real root, and hence cannot change ign on [ 1, 1]. If t =, then there i a double root at, and o c and c 1 mut both be zero. Thu, f(x) = x b + c a x a, and o Lemma 1 tell u that there can be only one non-zero real root, the root at 1. Hence, f cannot have any other real root and o cannot change ign on [ 1, 1]. Thu there exit a polynomial f(x) with root at t and 1 which doe not change ign on [ 1, 1] for each t [ 1, 1) and o there i an edge of the convex hull of C with vertice at C(t) and C(1). In addition, we can ee that f( x) will be a polynomial in x with root at 1 and t which doe not change ign on the interval, hence there are alo edge between C( t) and C( 1) for all t [ 1, 1). 2.3 Summary To ummarize, the face of the convex hull of C are a follow. Vertice t [ 1, 1] Edge t-value C(1)C(t) t [ 1, 1) C( 1)C(t) t ( 1, 1] FIGURE 1. The convex hull of C for a = 2 and b = Even-Even Curve The next type of curve C = {(t, t a, t b ) t [ 1, 1]}, a < b N in R 3 whoe convex hull we will examine are thoe in which both a and b are even. Even power term in polynomial are omewhat better behaved than their odd counterpart, which reult in thi cae being omewhat le unuual than thoe we will ee later on. 3.1 Vertice Theorem 7. Each point on C i a vertex of the convex hull.

7 Proof. A C ha a t entry and an even power entry, we can ue exactly the ame proof a in Theorem The Facet C(1)C( 1)C() Theorem 8. There i a two dimenional face of the convex hull with vertice at C(1), C( 1), and C(). Proof. Conider the polynomial f(x) = x a x b. We ee that f ha root at 1, 1, and. For < x < 1, a a < b, x a > x b, o f(x) >. Thu, f i non-negative on the interval with root at 1, 1, and. Thi i ufficient to prove the exitance of uch a face. 3.3 Edge of the form C(t)C( t) Theorem 9. For each t (, 1], there i an edge of the convex hull with vertice at C(t) and C( t). Proof. Conider the curve S = {(t a, t b ) t [ 1, 1]}. We ee that, a a and b are even S = {(t, t b a ) t [, 1]}. The econd derivative of t b a, b a ( b a 1)t b a 2, i poitive for t (, 1). Hence, by Lemma 3, each point on S i a vertex of the convex hull of S. Moreover, we ee that for the projection π : R 3 R 2, (x 1, x 2, x 3 ) (x 2, x 3 ), a given point (t a, t b ) S {(, )} ha preimage π 1 (t a, t b ) = {(t, t a, t b ), ( t, t a, t b )} = {C(t), C( t)} a a and b are even and o ( t) a = t a and ( t) b = t b. Hence, by Lemma 2, C(t) and C( t) are the vertice of an edge of the convex hull of C for all t (, 1]. 3.4 Edge of the form C(t)C(1) and C( t)c( 1) Theorem 1. For each t (, 1), there i an edge of the convex hull with vertice at C(t) and C(1) and an edge of the convex hull with vertice at C( t) and C( 1). Proof. To how that C(1)C(t) i an edge of the convex hull, it uffice to how that there exit a polynomial of the form f(x) = x b + c a x a + c 1 x + c with root at t and 1, which doe not change ign on the interval [ 1, 1]. To begin, we will how the exitance of a polynomial with a double root at t and a root at 1. To accomplih thi, we olve for the coefficient c a, c 1, and c. Firt, we ee that a there i a root at t = 1, we get the equation 1 + c a + c 1 + c = c a + c 1 + c = 1. Next the fact that there i a root at t give u the equation t b + c a a a + c 1 a + c = c a a a + c 1 a + c = t b. Finally, becaue there i a double root at t, f (t) mut alo equal. Hence bt b 1 + ac a t a 1 + c 1 = ac a t a 1 + c 1 = bt b 1.

8 Thi give u the following ytem of equation, or equivalently, the matrix equation t a t at a 1 1 c a + c 1 + c = 1 c a t a + c 1 t + c = t b ac a t a 1 + c 1 = bt b 1 c a c 1 c = 1 t b bt b 1. We proved a part of Theorem 6 that the determinant det t a t 1 = (1 a)t a + at a 1 1 at a 1 1 wa non-zero, and in particular negative for all t [ 1, 1), and a we only ued that a wa even when proving that fact it continue to hold here. Thu a olution exit, and o there exit ome polynomial of uitable form with a root at 1 and a double root at t for all t [ 1, 1). In particular, we compute that olution to be c a c 1 c = t+bt b (b 1)t b+1 t at a +(a 1)t a+1 bt b (t a 1) at a (t b 1) t at a +(a 1)t a+1 (1 a)t a+1 +(a b)t a+b) +(b 1)t b+1 t at a +(a 1)t a+1 We will how that the coefficient i poitive for all t (, 1). The denominator c 1 = btb (t a 1) at a (t b 1) t at a + (a 1)t a+1, t at a + (a 1)t a+1, i equal to the product of t and the determinant. Thu, a the determinant i negative for t [ 1, 1) and t i negative for t (, 1), we have that the denominator i poitive for t (, 1). The numerator bt b (t a 1) at a (t b 1) factor to t a (bt b a (t a 1) a(t b 1)). We can afely ignore the t a factor a it will be poitive for all t (, 1). Thi leave bt b a (t a 1) a(t b 1) = bt b bt b a at b + a = (b a)t b bt b a + a which i equal to zero at t = 1. We now conider the derivative t ((b a)tb bt b a + a)

9 = b(b a)t b 1 b(b a)t b a 1 = b(b a)t b a 1 (t a 1) which i negative for all t (, 1). Hence, the numerator i decreaing on that interval, and therefore mut be greater than for all t (, 1). A the numerator and denominator are both poitive for all t (, 1), we ee that c 1 i potive for all t (, 1). We alo have that 1 + c a + c 1 + c = f(1) =, o f( 1) = 1 + c a c 1 + c = (1 + c a c 1 + c ) (1 + c a + c 1 + c ) = 2c 1 < Now, becaue f i monic and ha even degree, we ee that f(x) a x, o there mut be a root at ome x < 1. Thu, f ha four non-zero real root, and o by Lemma 1 cannot have any more. Moreover, if we aume that f ha a root at, then c =, and o f(x) = x b + c a x a + c 1 x which can only have three non-zero real root by Lemma 1, and o we get a contradiction. Hence, f cannot have any other real root, and o cannot change ign on [ 1, 1]. That i, for each t (, 1), there exit a polynomial f(x) with root at t and 1 which doe not change ign on [ 1, 1], and it follow that f( x) will be a polynomial with root at t and 1. Thu, the convex hull of C ha edge between C(1) and C(t) and between C( 1) and C( t) for all t (, 1). 3.5 Summary To ummarize, the face of the convex hull of C are a follow. FIGURE 2. The convex hull of C for a = 2 and b = 4. Vertice t [ 1, 1] Facet C(1)C( 1)C()

10 FIGURE 3. View of the convex hull from Figure 2 from a different angle. Edge t-value C(t)C( t) t (, 1] C(1)C(t) t (, 1) C( 1)C(t) t ( 1, ) 4 Odd-Even Curve. We now move onto the cae where a i odd and b i even. Thee curve are the mot bizarre yet, a their convex hull are in many way le uniform. Vertice of certain edge change location depending on variable baed on a and b, and ome face do not exit at all in the pecial cae where a + 1 = b. 4.1 Vertice Theorem 11. Each point on C i a vertex of the convex hull. Proof. We can ue exactly the ame proof a in Lemma Interior Edge Theorem 12. There exit ome r ( 1, ), uch that C(t) and C(rt) are vertice of an edge of the convex hull for all t [ 1, 1]. Proof. Fix t, ( 1, 1). Then C(t) and C() are vertice of an edge of the convex hull if and only if there exit a non-negative polynomial f(x) = c b x b +c a x a +c 1 x+c with double root at both t and. In particular, Lemma 1 tell u that thee four root are the only non-zero root poible, and that there can be no root at zero, a that would bound the number of poible non-zero root at three. Hence, the exitence of uch a polynomial i ufficient to how that an edge exit between C(t) and C(). Looking for uitable coefficient for uch a

11 polynomial we get the following ytem of equation c b t b + c a t a + c 1 t + c = bc b t b 1 + ac a t a 1 + c 1 = c b b + c a a + c 1 + c = bc b b 1 + ac a a 1 + c 1 = or equivalently, the following matrix equation t b t a t 1 bt b 1 at a 1 1 b a 1 b b 1 a a 1 1 c b c a c 1 c = A olution to thi matrix equation exit if and only if the determinant, G(, t) = (b a)t a+b 1 + a(1 b)t b a 1 + b(a 1)t b 1 a + b(a 1)t a b 1 + a(1 b)t a 1 b + (b a) a+b 1, i equal to zero. A G i homogeneou of degree a+b 1, t a+b 1 G(1, r) = G(t, tr) for all r R. Hence, we have that if there exit ome r uch that G(1, r) =, then G(t, tr) = t a+b+1 G(1, r) = t a+b+1 = for all t R. So we et t = 1, and get a polynomial g() = (b a) a+b 1 +a(1 b) b +b(a 1) b 1 +b(a 1) a +a(1 b) a 1 +(b a). Decarte rule of ign tell u that g can have at mot four poitive root and two negative root. By examining g and it firt, econd, and third derivative, we can ee that there i a quadruple root at = 1. Becaue g() = b a > we can prove that there i a root of g on ( 1, ) by howing that g i negative at = 1. We have that g( 1) = (b a)( 1) a+b 1 + a(1 b)( 1) b + b(a 1)( 1) b 1 + b(a 1)( 1) a + a(1 b)( 1) a 1 + (b a) = (b a) + a(1 b) b(a 1) b(a 1) + a(1 b) + (b a) = 2((b a) + a(1 b) b(a 1)) = 2(b a + a ab ab + b) = 2(2b 2ab) < hence there exit a root r ( 1, ). Note that becaue G i ymmetric, it other negative root i 1 r < 1. Thu G(t, tr) = for all t R, and a r < 1, tr ( 1, 1) for all t [ 1, 1] o there i a point C(tr) on C. Therefore, there exit an edge between C(t) and C(tr) for all t [ 1, 1]. In particular, all edge between point C(t) and C(),, t ( 1, 1), mut be of thi form, a the only other root of the determinant are 1 and 1. There mut be two vertice in order to have an edge, o we cannot have r an edge between C(t) and C(1 t) = C(t). There i only a point C( t ) on C if t [ 1, 1], which i only the r r cae if t (r, r), in which cae an edge between C(t) and C( t ) can be written a an edge between C() and r C(r), where = t. r 4.3 Facet Theorem 13. If a + 1 < b, then there exit ome (, 1), uch that C(1), C( 1) and C() are vertice of a face of the convex hull. Another facet of the convex hull exit with vertice at C(1), C( 1) and C( ).

12 Proof. Fix t ( 1, 1), C(1), C( 1), and C(t) are vertice of an facet of the convex hull if and only if there exit a non-negative polynomial f(x) = c b x b + c a x a + c 1 x + c with a double root at t and root at both 1 and 1. To find the value of t for which a polynomial with thee root exit, we ue the following ytem of equation c b + c a + c 1 + c = c b c a c 1 + c = c b t b + c a t a + c 1 t + c = bc b t b 1 + ac a t a 1 + c 1 = or equivalently, the matrix equation t b t a t 1 bt b 1 at a 1 1 c b c a c 1 c = A non-zero olution to thi matrix equation exit if and only if the determinant i equal to zero. We have that and that o there i a double root at 1. However h(t) = (a b)t a+b 1 + (b 1)t b at a h(1) = (a b) + (b 1) a + 1 = h (1) = (a + b 1)(a b) + b(b 1) (a 1)a = a 2 a + ab b 2 a + b + b 2 b a 2 + a = h (1) = (a + b 2)(a + b 1)(a b) + b(b 1) 2 (a 2)(a 1)a = b(a 1)(a + 1 b) which i negative when a + 1 < b. Hence, there exit ome point on (, 1) where h i negative, but h() = 1, o thi implie that h mut have a root (, 1). Thu, there exit a polynomial f(x) = c b x b + c a x a + c 1 x + c with root at 1 and 1, and a double root at. By Lemma 1, f cannot have any more non-zero root, and a root at zero would retrict u to at mot three non-zero root. Hence, f cannot change ign anywhere on the interval [ 1, 1]. Thi i ufficient to prove the exitance of a facet with vertice at C(1), C( 1), and C(). It follow trivially that there mut alo be a facet with vertice at C(1), C( 1), and C( ), a f( x) clearly ha root at 1, 1, and, and like f cannot change ign on [ 1, 1]. Note that in the cae where a + 1 = b, h (1) =, o there i a triple root at 1. Becaue each term of h ha an even power h(t) = h( t), and o there i alo a triple root at 1. Decarte Rule of Sign tell u that h can have at mot three poitive and three negative root, o there can be no olution to the matrix equation on ( 1, 1). Moreover, we ee that a the leading coefficient (a b) i negative, h(t) a t, o a h mut change ign at t = 1 and cannot have any other poitive root, it mut be poitive for all t (, 1). For the purpoe of proof in the upcoming ection, we can ay that = 1 in the cae where a + 1 = b.

13 4.3 Edge at Endpoint In determining the exitance of edge we will ue a variant on the matrix equation from ection 1.7. We are looking for a polynomial with a double root at ome t, and a root at 1, though in thi cae we et the contant term equal to 1, giving u the following ytem of equation c b + c a + c = c b t b + c a t a + c 1 t + 1 = bc b t b 1 + ac a t a 1 + c 1 = or equivalently, the matrix equation which ha the olution t b t a t bt b 1 at a 1 1 c b c a c 1 = c b c a c 1 = 1 1 (a 1)t a+1 at a +t (b a)t a+b (b 1)t b+1 +(a 1)t a+1 (b 1)t b+1 bt b +t (b a)t a+b (b 1)t b+1 +(a 1)t a+1 bt b (t a 1) at a (t b 1) (b a)t a+b (b 1)t b+1 +(a 1)t a+1 A the denominator of the olution i non-zero (we will how that it i in fact poitive below) for all t ( 1, 1) {}, the olution exit for thoe t. Lemma 4. For all t (, ) and all t (, 1), c b > 1, where i the value defined in Theorem 13. Proof. To begin, it i neceary to how that for t ( 1, 1) {}, the denominator of c b, i poitive. We ee that (b a)t a+b (b 1)t b+1 + (a 1)t a+1 (b a)t a+b (b 1)t b+1 + (a 1)t a+1 = t a+1 ((b a)t b 1 (b 1)t b a + (a 1)) A a+1 i even, t a+1 > for all t, o we can diregard it and conider the (b a)t b 1 (b 1)t b a +(a 1) factor. At t = 1, (b a)t b 1 (b 1)t b a + (a 1) = (b a) + (b 1) + (a 1) = b + a + b 1 + a 1 = 2a > and at t = 1, (b a)t b 1 (b 1)t b a + (a 1) = (b a) (b 1) + (a 1) = b a b a 1 = We conider the derivative t (b a)tb 1 (b 1)t b a + (a 1) = (b a)(b 1)t b 2 (b a)(b 1)t b a 1,.

14 = t b a 1 (b a)(b 1)(t a+3 1) which i non-poitive for all t ( 1, 1). Hence, (b a)t b 1 (b 1)t b a + (a 1) i between 2a and for all t ( 1, 1), and i therefore poitive. Thu, the denominator i poitive for all t ( 1, 1) {}. Now, we ee that (a 1)t a+1 at a + t c b = > 1 (b a)t a+b (b 1)t b+1 + (a 1)ta+1 ((a 1)t a+1 at a + t) > ((b a)t a+b (b 1)t b+1 + (a 1)t a+1 ) > ((a 1)t a+1 at a + t) ((b a)t a+b (b 1)t b+1 + (a 1)t a+1 ) (b a)t a+b (b 1)t b+1 + at a t > The polynomial (b a)t a+b (b 1)t b+1 +at a t i equal to th(t), where h i the polynomial in Theorem 13. For t (, ), h(t) and t are both poitive, o th(t) i poitive on the interval. Similarly, h(t) and t are both negative for t (, 1), o th(t) i poitive on thoe t a well. Hence for t (, ) and t (, 1), c b > 1. Lemma 5. The derivative of the polynomial f(x) = c b x b +c a x a +c 1 x+1, where c b, c a, and c 1 are the olution to the matrix equation at the beginning of thi ection i negative at x = 1 for all t ( 1, r). Proof. The derivative f i bc b x b + ac a x a + c 1, which i equal to bc b + ac a + c 1 at x = 1. Plugging in the value of c b, c a, and c 1, we get bc b + ac a + c 1 = b((a 1)ta+1 at a + t) a((b 1)t b+1 bt b + t) (bt b (t a 1) at a (t b 1)) (b a)t a+b (b 1)t b+1 + (a 1)t a+1 = (a b)ta+b + a(b 1)t b+1 b(a 1)t b b(a 1)t a+1 + a(b 1)t a + (a b)t (b a)t a+b (b 1)t b+1 + (a 1)t a+1. We have already hown the denominator to be poitive for all t ( 1, 1) {}, o we can diregard it and conider the numerator (a b)t a+b + a(b 1)t b+1 b(a 1)t b b(a 1)t a+1 + a(b 1)t a + (a b)t which we ee i equal to tg(t), where g i the ame polynomial from Theorem 12. We have that g i negative on ( 1, r), while t i poitive on the interval. Thu, the numerator i negative for and therefore f (1) < for all t ( 1, r).

15 Theorem 14. For all t (, r) and all t (, 1), where and r are a defined in Theorem 13 and Lemma 5 repectively, there exit edge with vertice at C(t) and C(1) and edge with vertice at C( 1) and C( t). Proof. Conider the polynomial f(x) = c b x b + c a x a + c 1 x + 1, where the coefficient are a determined by the matrix equation from the beginning of thi ection. We know that f ha a double root at t and a root at 1. We want to how that f i non-negative on [ 1, 1], and can achieve thi by conidering three poible cae. (1) Suppoe that c b >, then f a x, but we know that f(1) = and by Lemma 5 f (1) <. Thi implie that there exit ome x > 1 for which f(x) i negative, which in turn implie that f mut have ome root which i greater than 1 a we know that f(x) i poitive for ufficiently large x. We know that f() = 1, o by Lemma 1, there can be no other root of f, and hence, f(x) for all x [ 1, 1]. (2) Suppoe that c b <, then f a x, but we know that c b > 1 by Lemma 4, and that c b +c a +c 1 +1 = f(1) =. Hence, f( 1) = c b c a c 1 +1 = 2c b +2 >. It follow then, that there mut be ome root of f which i le than 1 a f(x) i negative for a ufficiently negative x. We know that f() = 1, o by Lemma 1, there can be no other root of f, and hence, f(x) for all x [ 1, 1]. (3) Suppoe that c b =, then by Lemma 1, there can only be three non-zero root, and we know that f() = 1. Thu there can be no other root of f, and hence, f(x) for all x [ 1, 1]. Thu, we can ee that we can determine a polynomial f with root at t and 1 which i non-negative on [ 1, 1] for all t (, r) and all t (, 1). We can alo trivially how that there exit edge from C( 1) to C( t) for uch t, a f( x) will be a polynomial with the appropriate root which i non-negative on [ 1, 1]. Note that in the cae where a + 1 = b, = 1, o there are edge from C(1) to all point C(t), t ( 1, r), and edge from C( 1) to all point C(t), t ( r, 1). 4.4 Summary If where a < b 1, then the face of the convex hull of C are a follow. Let be a defined in Theorem 13, and r be a defined in Lemma 5. Vertice t [ 1, 1] Facet C(1)C( 1)C() C(1)C( 1)C( )

16 Edge t-value C(t)C(rt) t [ 1, 1] C(1)C(t) t (, 1) C(1)C(t) t (, r) C( 1)C(t) t ( 1, ) C( 1)C(t) t ( r, ) F IGURE 4. The convex hull of C for a = 3 and b = 6.In thi cae r = 21 ( 3 + 5) and = F IGURE 5. The convex hull of C for a = 3 and b = 4. In thi cae r = 2 + q 1 ( ). 3. However, in the cae where a = b 1, the convex hull ha the following face intead. Note that we can think of a being equal to 1 in thi pecial cae. Vertice t [ 1, 1]

17 Edge t-value C(t)C(rt) t [ 1, 1] C(1)C(t) t ( 1, r) C( 1)C(t) t ( r, 1) 5 Odd-Odd Curve. The final cae which we will examine i the cae in which both a and b are odd. In many way, thi cae i the mot pathological. The curve goe into the interior of it own convex hull, which reult in thi being the only cae in which not every point on the curve i a vertex of the convex hull, a well a ome unuual fact about the location of variou face, and force u to prove the exitance of ome face in new way. 5.1 Vertice Theorem 15. The convex hull of C doe not have a vertex at C(). Proof. Suppoe that C() i a vertex of the convex hull of C. Then there mut exit a non-zero polynomial of the form h(x) = c b x b + c a x a + c 1 x + c which ha a root at x = and i non-negative for x [ 1, 1]. In particular, it mut have a double root at x =. Thu, c = h() = and c 1 = h () =, and o h(x) = c b x b + c a x a. If c a, then h ha a root of multiplicity a at x =. If c a =, then h(x) = c b x b, and o h ha a root of multiplicity b at x =. A a and b are both odd, either cae reult in a contradiction, a there mut be ome point near to x = at which h i negative. Thi complete the proof. It follow that there mut in fact be a neighborhood around uch that C(t) i not a vertex of the convex hull for any t in that neighborhood. A we prove the exitence of the edge and facet of the convex hull, we will alo how that no other facet or edge can exit, which will determine what preciely thi neighborhood i for given value of a and b. 5.2 Facet and Interior Edge Lemma 6. There exit a polynomial g(x) = c b x b + c a x a + c 1 x + 1 with a double root at 1, a double root at ome r ( 1, ), a fifth real root < 1, and no other real root. In particular, g i the unique polynomial of thi form with a double root at 1 and another double root in ( 1, 1) {} up to multiplication by a contant. Proof. We have that a polynomial of the form c b x b + c a x a + c 1 x + c with double root at 1 and ome point t exit when there i a olution to the following ytem of equation or equivalently, the following matrix equation c b + c a + c 1 + c = bc b + ac a + c 1 = c b t b + c a t a + c 1 t + 1 = bc b t b 1 + ac a t a 1 + c 1 = b a 1 t b t a t 1 bt b 1 at a 1 1 c b c a c 1 c =

18 which ha a olution if and only if the determinant, the polynomial f(t) = (b a)t a+b 1 + a(1 b)t b + b(a 1)t b 1 + b(a 1)t a + a(1 b)t a 1 + (b a), i equal to. Thi polynomial i omewhat familiar to u from Theorem 12, and a before we ee that there i a quatruple root at t = 1 and that there can be no other poitive root. Decarte rule of ign tell u that there can be at mot three negative root. We ee that there i a root at t = 1, and examining the derivative f (t) at t = 1 we ee that f ( 1) = (b a)(a+b 1)( 1) a+b 2 +ab(1 b)( 1) b 1 +b(a 1)(b 1)( 1) b 2 +ab(a 1)( 1) a 1 +a(1 b)(a 1)( 1) a 2 = (b a)(a + b 1) + ab(1 b) b(a 1)(b 1) + ab(a 1) a(1 b)(a 1) = (b a)(a + b 1) ab(b a) (b a)(a 1)(b 1) = (b a)(a + b 1 ab ab + a + b + 1) = 2(b a)(a + b ba) and o a a + b < ab for all a, b > 2, and hence all odd a, b > 1, we ee that f ( 1) i negative. It follow that there mut be ome t < 1 at which f(t) >. Now, becaue the leading coefficient (b a)(a + b 1) i poitive and a + b 1 i odd, we ee that f a t, hence, there mut be ome root of f which i le than 1. Becaue f(t) i a ymmetric polynomial of degree a + b 1, we have that f(t) = t a+b 1 f( 1 t ), and o the reciprocal of each root i alo a root. Thu there exit exactly one root r of f on the interval ( 1, ). A r i a root of the determinant, the matrix b a 1 r b r a r 1 br b 1 ar a 1 1 ha determinant and therefore ha a rank of at mot 3. Thu, the fourth row i a linear combination of the firt three row, and o the olution to the matrix equation c b b a 1 c a r b r a c r 1 1 = 1 i alo a olution to the original matrix equation. In particular, it i clear that a r ( 1, ), and more preciely i not equal to 1, no two of the row vector of the 3 4 matrix can pan the third, and o the rank i exactly 3, and it follow that, the olution to the 3 4 matrix equation will be the unique olution to the 4 4 matrix equation uch that c = 1, and o the polynomial c b x b + c a x a + c 1 x + 1 will be the unique polynomial of uch form with a double root at 1 a double root at another point r ( 1, ). More generally, becaue by Lemma 1, there cannot be more than three poitive root, o there cannot exit any of uitable form polynomial with a double root at one and a double root at any t (, 1), o thi polynomial will in fact be the only one of uch form with a double root at 1 and a double root on ( 1, 1) {} up to multiplication by a contant. We compute thi unique olution to be c b c a c 1 = a ar+r a 1 (1 a)r b +(b 1)r a +(a b)r b+br r b +1 (1 a)r b +(b 1)r a +(a b)r b a br a +ar b (1 a)r b +(b 1)r a +(a b)r

19 and conider the polynomial g(x) = c b x b + c a x a + c 1 x + 1. A g(1) =, g() = 1, and all non-contant term have odd power, we have that g( 1) = 2 >. We want to how that the leading coefficient, c b = i poitive. a ar+r a 1 (a 1)r b +(1 b)r a +(a b)r A r ( 1, ), the numerator a ar + r a 1 = (a 1) + (r a ar) i clearly poitive, o we ne ed only conider the denominator (1 a)r b + (b 1)r a + (a b)r. Treating r a a variable, we ee that for r = 1, (1 a)r b 1 + (b 1)r a 1 + (a b) = (1 a) + (b 1) + (a b) = and that r (1 a)rb 1 + (b 1)r a 1 + (a b) = (b 1)(1 a)r b 2 + (a 1)(b 1)r a 2 = (a 1)(b 1)(r a 2 r b 2 ) which i clearly negative for r ( 1, ) and a, b odd. Thu, a we have that r ( 1, ), (1 a)r b 1 + (b 1)r a 1 + (a b) i negative, and hence (1 a)r b + (b 1)r a + (a b)r i poitive. Thu, c b >, and hence, a b i odd g(x) a x. Therefore, there mut be ome root < 1 of the polynomial g(x). Between the double root at 1 and r and the root at, g ha five real root, and we oberve once again that g() = 1. Thu, by Lemma 1, g cannot have any other real root. Thi together with the proof of uniquene above complete the proof FIGURE 6. The graph of g(x) for a = 3 and b = 5.In thi cae r = 1 2 ( 3 + 5) and = 1 5. Theorem 16. The convex hull of C ha edge with vertice at C(t) and C(rt) for all t [ 1, 1 ) and all t ( 1, 1], a facet with vertice at C( 1), C( r), and C(1), and a facet with vertice at C( 1), C( r ), and C( 1), where r and are the root of the polynomial g decribed in Lemma 6. Moreover, thee are the only facet of the convex hull, and the only edge without a vertex at C(1) or C( 1). Proof. Recall the polynomial g(x) from Lemma 6. We ee that for t R, g( x ) i a polynomial in x with double root at t and tr, and a root at t, and that it t cannot have any other real root. Hence, for t [ 1, 1) and all t ( 1, 1], a t > 1, t > 1, we have that t / [ 1, 1], hence g( x ) cannot change ign on the interval, and o we ee that there exit an edge with t vertice at C(t) and C(rt). For t = 1, we get the polynomial g(x), which ha double root at 1 and r and a root at = 1. A there can be no other real root, g(x) doe not change ign on [ 1, 1], and hence there exit a facet with vertice at C( 1), C( r), and C(1). We can prove the exitance of a facet with vertice at C( 1), C( r ), and C( 1) by chooing t = 1.

20 Furthermore, it follow from the uniquene of g that the only way for a polynomial of uitable form to have double root at two different point on ( 1, 1) i for that polynomial to be of the form g( x ). Hence, for t any t ( r, r ) the fifth root of the polynomial will be on ( 1, 1) which will reult in the polynomial being negative at ome point on the interval. Thu, there can be no other face or edge of thi form FIGURE 7. The graph of g(x), the polynomial correponding to the facet C(1)C( 1 )C( r ), for a = 3 and b = Edge at Endpoint The remaining face of the convex hull of C will be edge with a vertex at C(1) or C( 1). Theorem 17. The convex hull of C ha edge with one vertex at C(1) and the other at C(t) for all t ( r, 1), where r and are the root of the polynomial g decribed in Lemma 6. Similarly, the convex hull ha edge with vertice at C( 1) and C(t) for all t ( 1, r ). Proof. We are looking for a polynomial with a double root at t and a root at 1 for any given t. Setting the contant term equal to 1, we get the familiar ytem of equation c b + c a + c = c b t b + c a t a + c 1 t + 1 = bc b t b 1 + ac a t a 1 + c 1 = or equivalently, the matrix equation which ha the olution t b t a t bt b 1 at a 1 1 c b(t) c a (t) c 1 (t) = c b(t) c a (t) c 1 (t) = 1 1 (a 1)t a+1 at a +t (b a)t a+b (b 1)t b+1 +(a 1)t a+1 (b 1)t b+1 bt b +t (b a)t a+b (b 1)t b+1 +(a 1)t a+1 bt b (t a 1) at a (t b 1) (b a)t a+b (b 1)t b+1 +(a 1)t a+1

21 In particular, thi olution i unique, and o there i a unique function f(x, t) of the form c b (t)x b + c a (t)x a + c 1 (t)x + 1 which for a fixed t i a polynomial in x with a root at 1 and a double root at t. Conider the function g(x) from Lemma 6. Becaue f i unique, and g(x) i a polynomial in x of the form c b x b + c a x a + c 1 x + 1 with a root at 1 and double root at r, we ee that g(x) = f(x, r ) for all x. A a, b, and 1 are odd we ee that f(x, t) = (c b (t)( x) b + c a (t)( x) a + c 1 (t)( x) + 1) = c b x b c a x a c 1 x + 1 = f(x, t) + 2 and o it follow that for a fixed t, f(x, t) will be non-negative for x [ 1, 1] if f(x, t) 2, and ha a root at ome x [ 1, ) if and only if f( x, t) = 2. We know that f(x, r ) = g(x) i non-negative for x [ 1, 1] and ha a root at x = 1, hence for t = r the maximum value of f for x (, 1) i 2 at x = 1. We oberve that the partial derivative for q (f(x, t)) = p t m p = (b 1)bt b+1 (t a 1) + a 2 t a (t bt b + (b 1)t b+1 ) at a (t b 2 t b + (b 2 1)t b+1 ) q = t b (x x a ) + t a (x b x) + t(x a x b ) m = t((1 a)t a+1 + (b 1)t b+1 + (a b)t a+b ) 2. Clearly, the denominator, i poitive for all t > a ((1 a)t a+1 + (b 1)t b+1 + (a b)t a+b ) 2 i a quare. We now conider the two factor of the numerator (b 1)bt b+1 (t a 1) + a 2 t a (t bt b + (b 1)t b+1 ) at a (t b 2 t b + (b 2 1)t b+1 ) and t b (x x a ) + t a (x b x) + t(x a x b ). We can compute that (b 1)bt b+1 (t a 1) + a 2 t a (t bt b + (b 1)t b+1 ) at a (t b 2 t b + (b 2 1)t b+1 ) = (a 2 a)t a+1 + (b b 2 )t b+1 + (ab 2 a 2 b)t a+b + (a a 2 b + a 2 b + b 2 ab 2 )t a+b+1 which i a polynomial in t with four non-zero term, and hence can have at mot three poitive root by Decarte rule. Inpection of thi polynomial and it firt and econd derivative with repect to t will how that it ha a triple root at t = 1, hence it cannot change ign on (, 1). In particular, a a a 2 b+a 2 b+b 2 ab 2 = a a 2 +(a 1)(b b 2 ) <, the polynomial tend toward a t, and o it i poitive for all t (, 1). Next, we ee that t b (x x a ) + t a (x b x) + t(x a x b ) = (t a t)x b + (t t b )x a + (t b t a )x which a a polynomial in x can have at mot two poitive root by Decarte rule. We ee that it ha root at x = t and x = 1, o for t > it cannot have any other poitive root. More preciely, for t (, 1), a t a t <, the polynomial tend toward a x, o it i negative for x > 1, and therefore poitive for x (t, 1). Thu, we ee that the both factor of the numerator are poitive, and hence (f(x, t)) i negative for t t (, 1), x (t, 1). Now, we fix t ( r, 1). Taking f(x, t) a a polynomial in x we know that the only poitive root of f(x, t) are a double root at x = t and a root at x = 1. We alo ee that the derivative (f(x, t)) i a polynomial in x x with three non-zero term, and hence it can have at mot two non-zero root by Decarte rule. We know that (f(x, t)) and that it mut have another root between t and 1, o it cannot have any other poitive root. x Therefore, (f(x, t)) ha contant ign for x (, t) and o in particular, 1 = f(, t) f(x, t) > f(t, t) = x

22 for x [, t). For x (t, 1), (f(x, t)) i negative, o a t > r, f(x, t) < f(x, r ) < 2 for all uch x, and t a there are no root of f between t and 1, f(x, t) 2 for x [t, 1]. Thu, f(x, t) 2 for all x [, 1] and o it follow that f(x, t) i non-negative for x [ 1, 1].That i, f(x, t) a a polynomial in x ha root at t and 1 and i non-negative for x [ 1, 1], o there i an edge of the convex hull with vertice at C(t) and C(1) for each t ( r, 1). It alo follow that f( x, t) a a polynomial in x ha root at t and 1 and i non-negative for x [ 1, 1], o there i an edge of the convex hull with vertice at C(t) and C( 1) for each t ( 1, r). To prove the exitence of the final edge, we mut employ different technique from thoe we have ued o far. Lemma 7. Let r and be the value defined in Lemma 6. For a fixed t ( r, 1), let L be the R-vector pace of polynomial {p(x) = c + c 1 x + c a x a + c b x b p(t) =, p (t) = } and let K = {p(x) L p(x) x [ 1, 1]}. We have that K i a pointed convex cone, which contain polynomial other than the zero polynomial. Proof. For any p 1, p 2 K and we can ee that a p 1 (x), p 2 (x) for all x [ 1, 1], given non-negative α, β R, (αp 1 + βp 2 )(x) = αp 1 (x) + βp 2 (x) for all x [ 1, 1], o αp 1 + βp 2 K, and hence K i a convex cone. By the proof of Theorem 17 that there exit ome non-zero polynomial in K. Finally, we can ee that for any non-zero p K, there will exit ome x [ 1, 1] for which p(x) >, o p(x) < and p / K. It follow that K i pointed, and o the proof i complete. Lemma 8. The convex cone K i two-dimenional. Proof. Let α = 1 + ε for ome ɛ >. We conider the matrix equation αb α a α t b t a t c b(α, t) c a (α, t) = 1 1 bt b 1 at a 1 1 c 1 (α, t) From the proof of Theorem 17 we know that for ε =, a olution exit, that i, the determinant of the 3 3 matrix i non-zero. A t i fixed, the determinant i a polynomial in α, and hence mut be non-zero for a ufficiently mall ε, o for uch an ɛ, a olution exit, and the polynomial p ε (x) = c b (α, t)x b + c a (α, t)x a + c 1 (α, t)x + c (α, t) ha a double root at t and a root at α. Becaue it ha a double root at t, p ε L, and in fact, becaue for a fixed t, c b, c a, c 1 are rational function in α with denominator equal to the determinant of the 3 3 matrix above, which we have already een i non-zero. A we ee from the proof in Theorem 17 that p (x) > for x [ 1, 1] apart from t and 1, we can conclude that for a mall enough ε, p ε i alo poitive on thoe x. Hence, a p ε (t) = we can ee that p ε (x) for x [ 1, 1], and o p ε K for a ufficiently mall ε. In particular, a we alo know that p K and can ee that for any ε >, p and p ε are linearly independent over R, K mut be at leat two dimenional. We can alo oberve that L can be identified with the olution pace of the rank 2 matrix equation [ t b t a t 1 bt b 1 at a 1 1 ] c b c a c 1 c = [ o dim(l) = 2. Thu, it mut be exactly two dimenional a K L, o dim(k) dim(l) = 2. Lemma 9. For the value of r and defined in Lemma 6, we have 1 < r. ]

23 Proof. Suppoe that r < 1. Then r ( 1, 1) and o there exit a polynomial of the form f 1(x) = c b x b + c a x a + c 1 x + c with a double root at r and a double root at r 2 which i poitive for all other x [ 1, 1]. Moreover, by Lemma 6 there exit a polynomial f 2 (x) of thi form with a double root at r and a root at 1 which i poitive for all other x [ 1, 1], and by Theorem 17 we ee that a r ( 1, r ) there exit a polynomial f 3 (x) of thi form with a double root at r and a root at 1 which i poitive for all other x [ 1, 1]. Aume for the ake of contradiction that f 1, f 2, and f 3 are linearly dependent over R. Then there exit d 1, d 2, d 3 R, uch that d 1 f 1 (x) + d 2 f 2 (x) + d 3 f 3 (x) = for all x R. There mut be ome i, j {1, 2, 3}, i j uch that d i and d j have the ame ign. Let k {1, 2, 3} be the remaining value not equal to i or j, and let x k be the non-r root of f k in [ 1, 1]. We ee that d 1 f 1 (x k )+d 2 f 2 (x k )+d 3 f 3 (x k ) = d i f i (x k )+d j f j (x k ). Hence, a f i (x k ) and f j (x k ) are both poitive, and o, a d i and d j have the ame ign, d 1 f 1 (x k )+d 2 f 2 (x k )+d 3 f 3 (x k ) and we get a contradiction. Hence, f 1, f 2 and f 3 are linearly independent. However, thi i alo a contradiction, a f 1, f 2, f 3 K, and dim(k) = 2 by Lemma 8. Thu, we conclude that 1 < r. Thu, it make ene to talk about the interval ( 1, r) and ( r, 1 ). Theorem 18. For any t ( r, 1 ), the convex hull of C ha an edge with vertice at C( 1) and C(t). Similarly, for all t ( 1, r), the convex hull of C ha an edge with vertice at C(1) and C(t). Proof. Fix t ( r, 1 ), and let K be the cone of polynomial with double root at t which are non-negative on [ 1, 1] from the above lemma. The boundary of the cone, K, can only contain polynomial with another double root on ( 1, 1) or a root at ±1. A K i a two dimenional pointed cone, K i made up of two ray, one of which we know i generated by the unique polynomial p(x) = c b x b + c a x a + c 1 x + 1 with a double root at t, which we know exit by the proof of Theorem 17. Let q(x) be the polynomial in the other ray with contant term 1, which clearly mut exit. A p and q are linearly independant, we ee by the uniquene of p that q mut have a root at either 1 or ome element in ( 1, 1). However, by Theorem 16 there cannot be a non-negative polynomial with a double root at t and another double root elewhere on ( 1, 1), o we conclude that there mut exit a polynomial q(x) = d b x b + d a x a + d 1 x + 1 with a double root at t and a root at 1, which i non-negative on [ 1, 1], and it follow that q( x) will be a polynomial of imilar form with a double root at t and a root at 1 which i non-negative on the interval. Thi complete the proof. 5.4 Return to Vertice Theorem 19. Let r and be a defined in Lemma 6. If t ( r, r ), then C(t) i not a vertex of the convex hull of C. Proof. Fix t (, r ). We know from the proof of Theorem 16 and Theorem 17 that there cannot be an edge of the convex hull of the form C(t)C(t ) for any t ( 1, 1] and that there cannot be any facet with a vertex at C(t). Moreover, if there were an edge of the form C(t)C( 1) we could ue an argument imilar to the one in Theorem 18 and the lemma preceding it to how that a econd edge with a vertex at C(t) would have to exit, which i clearly a contradiction. Hence there can be no face of the convex hull with a vertex at C(t), and o C(t) i not a vertex of the convex hull. We can imilarly ee that C(t) cannot be a vertex of the convex hull for t ( r, ), and we already know that C() i not a vertex by Theorem 15, o the proof i complete. 5.5 Summary To ummarize, the face of the convex hull of C are a follow. Let r and be a defined in Lemma 6.

24 Vertice t [ 1, r ] t [ r, 1] Facet C( 1 )C( r )C(1) C( 1 )C( r )C( 1) Edge t-value C(t)C(rt) t [ 1, 1 C(t)C(rt) t ( 1, 1] C(1)C(t) t ( r, 1) C(1)C(t) t ( 1, r) C( 1)C(t) t ( 1, r ) C( 1)C(t) t ( r, 1 ) F IGURE 8. The convex hull of C for a = 3 and b = 5.In thi cae r = 12 ( 3 + 5) and = 1 5.