Level Sets of Arbitrary Dimension Polynomials with Positive Coefficients and Real Exponents

Transcription

1 Level Sets of Arbitrary Dimension Polynomials with Positive Coefficients and Real Exponents Spencer Greenberg April 20, 2006 Abstract In this paper we consider the set of positive points at which a polynomial with positive coefficients, arbitrary dimension n, and real exponents is equal to a fixed positive constant c. We find that when this set is nonempty and is bestowed with the relative Euclidean topology coming from R n, it is homeomorphic to a codimension one piecewise linear set that depends only on the polynomial s exponents. This piecewise linear set can in a certain sense be interpreted as a bijectively mapped version of the original set as the constant c approaches infinity. In addition to this result, we provide a condition on the polynomial exponents for testing if the solution space is homeomorphic to the n-1 dimensional sphere S n 1, and derive piecewise linear inner and outer bounds for our solution set. Each point in our solution set that lies on a fixed ray originating at the origin is trapped between a unique inner and outer bound point also on that ray. While this paper provides insight into the level sets of only a specific type of polynomial, an appropriate generalization of these observations might one day lead to improved techniques for analyzing level sets of high dimension polynomials in general, objects which appear frequently throughout mathematics. Acnowledgments I would lie to than Professor Charles Doran for acting as my advisor on this paper, and Professor Stanislav Jabua for his time and advice. 1

2 Let T c be the set of positive real points where a multivariate polynomial in n variables, with positive coefficients c and real exponent vectors p, equals a positive constant c. That is, T c {t > 0 c t p = c} R n (1) where we require that c > 0, c > 0, p (0, 0,..., 0), m > 0 and use the following conventions: t (t 1, t 2,..., t n ) R n p (p (,1), p (,2),..., p (,n) ) R n t p t p (,1) 1 t p (,2) 2 t p (,n) n R t > 0 t 1 > 0, t 2 > 0,..., t n > 0. We will also assume, without loss of generality, that c > 1. This can always be achieved without changing our set T c if we multiply both our constant c and each of our coefficients c by the same sufficiently large value. It is worth noting that the polynomials referred to in this paper are not true polynomials, but rather generalizations of them which allow for real valued exponents. The focus of this paper will be the set T c when it is thought of as a topological space under the relative Euclidean topology induced by R n. To facilitate our later proofs, we will start out by applying a homeomorphic transformation to our space, converting T c into a slightly simpler space Z c. When this is complete, we will define a new topological space Z which is piecewise linear and can be thought of as a certain limit of Z c as c approaches infinity. Theorem 1 on page 5 will then demonstrate that whenever Z c is not empty, it is homeomorphic to Z. Next, theorem 2 on page 13 will clarify when Z c (and therefore Z and T c ) are topologically equivalent to a sphere. Finally, theorem 3 on page 14 will establish two piecewise linear spaces between which every point in a special translation of Z c is trapped. This provides a subset of Euclidean space in which Z c is completely contained. You may find it useful at this point to examine the figures on pages 18, 19 and 20 which should give you some idea of what the spaces Z c can loo lie in dimension n = 2. The sets In c and Out c shown in these plots are examples of the piecewise linear spaces which we will use to bound Z c. To avoid studying the properties of T c directly, let us now formally introduce the simpler to analyze but topologically equivalent space Z c log c T c = {z c c p z = c} R n (2) 2

3 where is defined as log c : R n + R n log c t = (log c t 1, log c t 2,..., log c t n ) = (z 1, z 2,..., z n ) = z and we use the dot ( ) to denote the dot product so, p z = p (,1) z 1 + p (,2) z p (,n) z n. Since we are only considering positive t, log c is clearly continuous and bijective with a continuous inverse, and thus is a homeomorphism, even when restricted to log c : T c log c (T c ) = Z c. Let us now define z argmax p z (3) by which we mean that, for z R n fixed, z is any predetermined choice of that maximizes p z. Notice that Z c {z c c p z = c} = {z log c ( c c p z ) = log c (c)} = {z log c ( c c p z c p z z c p z z ) = 1} = {z p z z = 1 log c ( c c (p p z ) z )} = {z p z z = 1 log c (c z + z c c (p pz ) z )} (4) where (p p z ) z 0 by the construction of z. This non-positivity implies that 0 lim c c (p p z ) z c c and thus that log(c z ) lim c log(c z + z c c (p pz ) z ) log( c ) (5) where you should notice that in the above inequalities the base of the logarithms used does not depend on c. These inequalities will come into play again later on 3

4 when we attempt to bound our solution set, but for now they just demonstrate to us that Z {z p z z = 1 lim log c( c c (p p z ) z )} c log( m = {z p z z = 1 lim c c (p p z ) z ) } c log(c) = {z p z z = 1} = {z max p z = 1}. (6) It is worth noting that in some cases lim c Z c = {z p z z = 1 lim c log c( c c (p p z ) z )} and when that is so, we can use the elegant definition Z lim c Z c. (7) Not only is Z a piecewise linear set, but it also satisfies the rather strong condition that it is the boundary of the convex region R {z max p z 1}. (8) To see that R is convex, we need only observe that for z µ z 1 + (1 µ) z 2 where z 1 and z 2 are points in R and 0 µ 1, we have: max p z max µ max p (µ z 1 + (1 µ) z 2 ) p z 1 + (1 µ) max µ (1) + (1 µ) (1) = 1 p z 2 implying that z R. In other words, given any two points in R, we now that all points on the line segment between these two points are also in R. As it turns out, even for finite c, Z c is always the boundary of the convex region R c {z c c p z c}. (9) To demonstrate that R c is convex, we again define z µ z 1 + (1 µ) z 2 with 0 µ 1, where this time z 1 and z 2 are points in R c. We now need only show that c c p z c c µ p z 1 c (1 µ) p z 2 c 4

5 to demonstrate that z R c. But oddly enough, this is just a consequence of Hölder s famous inequality for sums, which states that for real numbers a, b 0, with 0 µ 1, ( m ) µ ( m ) 1 µ (a ) µ (b ) 1 µ a b. (10) Using a c c p z 1 and b c c p z 2 gives us: c c p z c c µ p z 1 c (1 µ) p z 2 ( m ) µ ( m ) 1 µ c c p z 1 c c p z 2. But now, since z 1 and z 2 are elements of R c, we have (by the definition of R c ) that: ( m ) µ ( m ) 1 µ c c p z 1 c c p z 2 (c) µ (c) 1 µ = c completing our proof that Z c is the boundary of the convex region R c. Theorem 1 If Z c does not equal the empty set then Z c is homeomorphic to the piecewise linear set Z {z max p z = 1} where both Z c and Z are considered topological spaces under the induced Euclidean topologies coming from R n. The five lemmas that follow will play an integral part in proving this theorem. Instead of woring with Z c as it stands, however, we will wor with the topological space Z s c, which is just Z c translated in R n by s = (s 1, s 2,..., s n ) R n units. Thus, Z s c {z c c p (z+s) = c} = {z c c p s c p z = c}. Zc s and Z c are homeomorphic, so anything that we prove about the homeomorphism type of Zc s for any s R n will also be true of Z c and T c. We will identify a special translation vector s = s 0 such that Zc s0 contains at most one point lying on each ray emanating from the origin, which, as we will see, is a very desirable property. 5

6 To address this problem more formally, consider the intersection of Z s c with a ray which starts at the origin and is directed at an angle = ( 1, 2,..., n 1 ) [0, 2π) n 1. Any z Z s c on this ray can be written z = r α(), where r is z s Euclidean distance from the origin (r = z ), and α is a smooth, n-valued function of satisfying α() = 1. Thus, for example, in dimension n = 2, and α( 1 ) = (cos( 1 ), sin( 1 )) α( 1, 2 ) = (cos( 1 ) sin( 2 ), sin( 1 ) sin( 2 ), cos( 2 )) in dimension n = 3. Dimension one is something of a special case since there we require that is of dimension zero, taing on the two values and + where α( ) = 1 and α( + ) = 1. To continue with our definitions, for fixed s and let f s : (0, ) (0, ) be the smooth function of r f s (r) c c p s c p z = c c p s c r(p α()) (11) which is found in the definition of Zc s. Notice that if we define Zc s, to be the points in Zc s which lie on the ray starting at the origin and directed at angle, then we have and Z s, c {z = r α() f s (r) = c, r > 0} (12) Z s c = [0,2π) n 1 Z s, c (13) so, the points r > 0 where f s (r) = c for each completely characterize the set Zc s. The proof of theorem 1 will proceed as follows: In lemma 1 we will identify and prove the existence of our special translation vector s 0, by which we will translate Z c, producing Zc s0. In lemma 2 we will show that on each ray emanating from the origin of Euclidean space there can exist no more than one point of Zc s0. The translation vector s 0 is essential in this argument because, if necessary, it shifts Z c so that Z c surrounds the origin. For example, when Z c is roughly spherical, Zc s0 will have the origin of Euclidean space trapped within the volume it encloses, even if Z c does not. In lemma 3 we will demonstrate that on each ray emanating from the origin Zc s0 contains the same number of points as Z. On all rays where Zc s0 and Z contain exactly one point, we can then treat the parameterized sets Zc s0, and as functions, each taing a ray angle and producing the point where the ray at that angle intersects Zc s0 or Z. Z 0, 6

7 Lemma 4 and lemma 5 will go on to show that Zc s0, and Z 0, are continuous functions of. At that point, we will combine our lemmas by constructing a bijection from Zc s0 to Z which explicitly taes Zc s0, to Z 0, for each. The continuity of the functions Zc s0, and Z 0, in will lend continuity to our bijection in both directions, implying that it is a homeomorphism, and completing the proof of our theorem. Lemma 1 If Z c is not the empty set, then there exists a translation vector s = s 0 such that (0) lim (r) < c for all, where f s0 f s (r) f s0 r 0 c c p s c r(p α()) is the function arising in the construction Zc s {r α() f s (r) = c, r > 0}. [0,2π) n 1 Proof: Notice that lim f s (r) = r 0 c c p s so f s0 (0) is less than or equal to c for all if and only if c c p s c. (14) To begin with, lets find an s = s 0 that satisfies the above non-strict version of the inequality. Well, clearly s 0 R n such that c c p s 0 c z R n such that c c p z c and so, since m c c p z = c for all z Z c by the definition of Z c, we will always be able to find our required s 0 by choosing any s 0 Z c. We then need only use the assumption that Z c to be sure that some such s 0 exists. But now, the question arises as to whether we can complete the more difficult tas of finding an s 0 such that f s0 (0) is strictly less than c for all. Well, suppose not, and choose an s 0 Z c so that c c p s 0 = c 7

8 and so f s0 m (r) = c c p s 0 c r p α(). Now, if f s0 (r) < c for some r = r 0 and = 0 then the translation vector s 1 = s 0 + r 0 α( 0 ) satisfies f s1 (0) < c contradicting our assumption that no translation vector with this property exists. Thus, we must have that f s0 (r) c for all and r. Observe though that since each c > 0 by definition, our function f s for any s satisfies d 2 dr 2 f s (r) = c c p s (log(c) p α()) 2 c r(p α()) 0 (15) and so f s(r) is convex in the variable r. In addition, since f s (r) is analytic and non-constant by construction (it is just a sum of exponentials in r), it cannot be constant on any open interval in the variable r. This, together with convexity, tells us that if f s (r) ever begins to increase as r > 0 increases, then it can never again decrease. Therefore, f s (r) must either increase forever, decrease forever, or initially decrease and then increase forever after. In our case, however, since f s0 s0 (0) = c and f (r) c (as demonstrated above), our function must initially increase and therefore must simply increase forever as r increases on the region where r > 0. But this implies that f s0 (r) can never attain the value c for any and any r > 0, so Zc s0 (by its definition) must be empty, producing a contradiction when combined with the fact that the set Z c which is homeomorphic to Zc s0 was assumed to be non-empty. This guarantees that a translation vector s 0 with f s0 (0) < c exists, so our lemma is proved. Note in particular that if m c < c we need only choose s 0 = (0, 0,..., 0) to satisfy our condition f s0 (0) < c. This will come into play later on when we consider bounding our solution set Z c. Lemma 2 If Z c and we choose an s 0 such that f s0 (0) < c, then for all fixed, the set Zc s0, {r α() f s0 (r) = c, r > 0} {r α() c c p s 0 c r p α() = c, r > 0} contains one element if p α() > 0 for some, and zero elements otherwise. 8

9 Proof: As we have mentioned before, the convexity and non-constantness of f s0 s0 (r) in the variable r place tight restrictions on its behavior. Since f (0) < c, we must have one of the following scenarios as r > 0 increases: 1. f s0 (r) increases forever and thus attains c exactly once. 2. f s0 (r) decreases initially but then levels out and increases forever after, again attaining c exactly once. 3. f s0 (r) is non-increasing forever, and thus never attains the value c. The number of times that c is reached therefore depends only on the long run behavior of f s0 (r). In particular, if p α() > 0 for any then our function will have at least one term with a positive exponent and thus will eventually increase forever and attain c exactly once. On the other hand, if this inequality is not satisfied for any, then our function will have all non-positive exponents and therefore will never increase and never reach c. This completes the proof of lemma 2. Lemma 3 If we choose an s 0 such that f s0 (0) < c then, for all fixed, the set, contains the same number of elements as Z s0, c Z 0, = {z = r α() max p z = 1, r > 0}. Proof: Suppose that Zc s0, contains one element for some fixed. As we have seen in lemma 2, this implies that there exists some = such that p α() > 0. But that means that z = r α() satisfies p z > 0 for all r > 0. Note however that max p z p z > 0. Thus, since max p z = r max p α() is positive for all positive r, and since max p α() is constant due to our choice of a constant, that means that there is a unique r > 0 such that max p z = r max p α() = 1. In particular, it is just r = 1 max p α(). But this implies that the set Z 0, contains exactly one element. 9

10 On the other hand, suppose that Zc s0, has no elements (which we now, by lemma 2 is the only other possibility for Zc s0, ). In this case, lemma 2 implies that there is no such that p α() > 0. But, in particular, this means that max p α() 0. Specifically, this tells us that there is no z = r α() for r > 0 satisfying r max p α() = 1, so Z 0, is an empty set. Thus the proof of lemma 3 is complete. Lemma 4 For fixed c and any choice of s 0 such that f s0 (0) < c, the parameterized set Zc s0, = {r α() f s0 (r) = c, r > 0} is a continuous function of at all points 0 where Zc s0,0. Proof: From lemma 2 we now that Zc s0, always has zero or one element, and thus can be thought of as a function of so long as we restrict ourselves to those points 0 where it is not empty. Then, all we need to show is that Zc s0, is continuous in at these special points. But this is equivalent to demonstrating that for all directions at which can approach 0. lim Zc s0, = Zc s0,0 (16) 0 Consider the graphs of f s0 s0 (r) and f 0 (r) as functions of r. By the continuity, convexity and non-constantness of f s0 0 (r), and our choice of s 0 so that f s0 0 (0) < c, it must be the case that f s0 0 (r) is increasing on some open interval around the unique point r 0 satisfying f s0 0 (r 0 ) = c. Now, choose an ɛ > 0 small enough so that r 0 + ɛ and r 0 ɛ lie in this interval. Since f s0 (r) is continuous in we must have that s0 lim (r) = f 0 (r) r > 0. (17) 0 f s0 Thus, in particular, we can choose our vector to be close enough to 0 (component wise) so that f s0 s0 (r + ɛ) f 0 (r + ɛ) and f s0 s0 (r ɛ) f 0 (r ɛ) are as close to zero as we lie. By the increasingness of f s0 0 (r) on our interval, we have that f s0 0 (r 0 ɛ) < c and f s0 0 (r 0 + ɛ) > c and thus, by maing our vector sufficiently close to 0, we can ensure that f s0 (r 0 ɛ) < c and f s0 (r 0 + ɛ) > c. By the mean value theorem f s0 (r) must then attain the value c at some point r 1 in the interval [r 0 ɛ, r 0 + ɛ]. But now, since we can choose any ɛ > 0 that we lie merely by maing close enough to 0, it is clear that as 0 we have r 1 r 0 and so lim Zc s0, = lim {r α() f s0 (r) = c, r > 0 } 0 0 = {r α( 0 ) f s0 0 (r) = c, r > 0 } = Z s0,0 c (18) 10

11 by the continuity of α. This holds regardless of the direction from which approaches 0, thus completing the proof of lemma 4. Lemma 5 The parameterized set Z 0, 0 where Z 0,0. is a continuous function of at all points Proof: Since Z 0, has at most one element (as demonstrated by lemma 3 in conjunction with lemma 2), it can be thought of as a function at those points 0 where it is non-empty. To prove continuity, we must show that lim Z 0, = lim {z = r α() r max p α() = 1} = Z 0,0 0 0 for all directions at which can approach 0. However, for any fixed, we can express Z 0, succinctly as { Z 0, α() { = max p }, max α() p α() > 0., otherwise Thus, we need only show that lim 0 α() max p α() = α( 0 ) max p α( 0 ). But note that α() is bounded since, by construction, α() = 1. This, together with the continuity of α, implies that it suffices to show that lim max p α() = max p α( 0 ). 0 Well, for r α(0) argmax p r α( 0 ) consider any 0 satisfying (p r α(0 ) p 0 ) α( 0 ) > 0. Then, since α varies continuously in, any 1 sufficiently close to 0 (component wise), will satisfy (p r α(0 ) p 0 ) α( 1 ) > 0. But that implies that in our limit as approaches 0, the vector p r α() eventually satisfies p r α() α( 0 ) = p α( r α(0 ) 0) giving us that = ( lim 0 lim max p α() lim 0 0 p r α() ) ( lim 0 p r α() α() α()) = ( lim 0 p r α() ) α( 0 ) = lim 0 p r α() α( 0 ) = p r α(0 ) α( 0) 11

12 completing the proof of lemma 5. = max p α( 0 ) At long last we are in a position to prove theorem 1, which states that Z and Z c are homeomorphic whenever Z c. By lemma 1 we now that there will always be some translation vector s 0 satisfying f s0 (0) < c. Lemma 2 and 3 then tell us that Z 0, and Zc s0, always have the same number of elements (zero or one). Finally, lemma 4 and 5 tell us that Z 0, and Zc s0, vary continuously (when thought of as functions) in the variable at all points 0 where they are non-empty. Now, to prove our theorem, we consider the map ψ : Z c Z ψ(z s0,0 c ) = Z 0,0 (19) The inverse of ψ clearly also exists, and is just given by ψ 1 : Z Z c ψ 1 (Z 0,0 ) = Zc s0,0 (20) so ψ is bijective. To prove that ψ is a homeomorphism, all that remains to be shown is that it and its inverse are continuous at all points 0 where they are defined. Thus we need to demonstrate that and lim Z s 0, c Z s 0, 0 c ψ(zc s0, ) = ψ(zc s0,0 ) lim ψ 1 (Z Z 0, Z 0, 0, ) = ψ 1 (Z 0,0 ) 0 Note, however, that points in Z c or Z which have differing lie on different rays shooting out from the origin. In particular, since each angle is associated with at most one unique point in each of our two sets, the only possible way that Zc s0, Thus we have and can approach Z s0,0 c or Z 0, can approach Z 0,0 lim Z s 0, c Z s 0, 0 c ψ(zc s0, ) = lim ψ(zc s0, ) 0 = lim Z 0, = Z 0,0 = ψ(zc s0,0 ) 0 lim ψ 1 (Z Z 0, Z 0, 0, ) = lim ψ 1 (Z 0, ) 0 0 = lim Zc s0, = Zc s0,0 = ψ 1 (Z 0,0 ) 0 is if approaches 0. 12

13 due to the continuity of Z 0, and Zc s0, in. Therefore, we have that ψ is a homeomorphism from Z c to Z, completing the proof of theorem 1. There is an important special case of the homeomorphism type of Z c that is worth addressing, namely when Z c is homeomorphic to S n 1, the n-1 dimensional sphere. To investigate the circumstances under which this occurs, we define m = ConvexHull( {p } ) (21) by which we mean that is the minimal convex polytope (with its interior filled) that contains all the points in m {p }. When the exponent vectors p are integer valued, is commonly nown as the Newton polytope of the polynomial m c t p. As it turns out, the set Z = {z max p z = 1} has a very similar construction to the boundary of what is nown as the dual polytope associated with the Newton polytope. Theorem 2 Z c is homeomorphic to S n 1 if and only if contains the origin (0, 0,..., 0) R n in its interior. Proof: If Zc s0, is non-empty for all, then due to lemmas 1 and 2 there is exactly one element of Zc s0, lying on each ray protruding outward from the origin. But then, since Zc s0, varies continuously in, and since two values Zc s0,a and Z s0, b c are close to each other if and only if a is close to b, Zc s0, interpreted as a function of is itself a homeomorphism between Zc s0 and [0,2π) = n 1 Sn 1. If, on the other hand, Zc s0, is empty for any, then Z c is homeomorphic to the n-1 sphere with some points removed, and thus can never be homeomorphic to the sphere itself. But when will this condition that Zc s0, is always non-empty be satisfied? Lemma 2 tells us that Zc s0, will be empty if and only if α() satisfies p α() 0 for all. In other words, Zc s0, will be empty for some if and only if there exists some vector z = r α() (0, 0,..., 0) with r > 0 and p z 0 for all. Assume that we can find such a vector z. Well, the fact that p z 0 for all means in particular that the minimal angle measured between the vector z and each of the vectors p is not in the interval ( π 2, π 2 ). But this means that we can construct a codimension one plane in R n passing through the origin such that all of the points p lie on the opposite side of the plane as z, where we consider any points lying on the plane itself to be on the opposite side of the plane as m all other points. However, this means that the convex hull of {p } cannot cross through our plane passing through the origin, which means that no points lying on this plane can be in the interior of our convex hull. In particular, this implies that the origin is not in = ConvexHull( m {p } ). On the other hand, assume that contains the origin. By the definition of the convex hull, the points d are just those that can be written d = 13

14 m w p for some weights w R with m w = 1 and w 0. Then, since contains the origin by assumption, there must be some set of weights w satisfying w p = (0, 0,..., 0) But that means that and so, for any vector z p 1 = 1 w 1 p 1 z = 1 w 1 m =2 m =2 w p w p z (22) Assume that there exists a vector z satisfying p z 0 for all. This implies that p 1 z is non-positive and 1 m w 1 =2 w p z is non-negative. But due to equation 22 above, both of these conditions can only be satisfied if p z = 0 for all. But this implies that the vectors p all lie on the same side of an n-1 dimensional plane that passes through the origin (and is perpendicular to z), prohibiting the convex hull of the p from containing the origin in its interior, which contradicts our assumption. Thus we are guaranteed that for every vector z = r α(), there exists some so that p z > 0, which, due to lemma 2, implies that there is exactly one point in the set Zc s0, homeomorphic to S n 1. This completes the proof of theorem 2. for each, giving us that Z s0 c As it turns out, there is more that we can say about Zc s0 than just its homeomorphism type. Consider our final theorem which places restrictions on how far the points of Zc s0, can and must lie from the origin. is Theorem 3 For fixed c and any choice of s 0 such that f s0 (0) < c for all fixed, we have that each non-empty set Z s0, c {r α() {r α() f s0 (r) = c, r > 0} c c p s 0 c r p α() = c, r > 0} consists of a single point that lies between the unique points in the sets and In c { 1 log c( m c c p s 0 ) α()} max p α() Out c { 1 log c(c z c p z s0 ) max p α() on a line passing through the origin. α()} 14

15 Proof: Consider equation 4 on page 3, which tells us that Z c = {z p z z = 1 log c (c z + z c c (p pz ) z )}. Extending this result to Z s0 c Z s0 c we have = {z p z z = 1 log c (c z c p z s0 + z c c p s0 c (p pz ) z )}. (23) Now, since (p p z ) z 0 by the construction of z, and we can always guarantee that c > 1, we have that 0 < c (p p z ) z 1. Since log c is a strictly increasing function for positive inputs and each term in our sum is positive, this tells us that: log c (c z c p z s0 ) log c ( c c p s 0 c (p p z ) z ) log c ( c c p s 0 ). (24) However, our choice of s 0 was such that m c c p s 0 < c. This implies that log c ( m c c p s 0 ) < 1, which gives us the inequalities 0 < 1 log c ( c c p s 0 ) 1 log c ( c c p s 0 c (p p z ) z ) 1 log c (c z c p z s0 ). (25) Equation 23 then tells us that for all z Zc s0 0 < 1 log c ( c c p s 0 ) p z z 1 log c (c z c p z s0 ). (26) And thus, if we fix to some value 0 such that z = r α( 0 ) Z c for some r > 0, then 0 < 1 log c( m c c p s 0 ) p z α( 0 ) This motivates us to construct the sets, and Out c = {z max In c = {z max r 1 log c(c z c p z s0 ). (27) p z α( 0 ) p z = 1 log c (c z c p z s0 )} (28) p z = 1 log c ( c c p s 0 )} (29) which naturally provide an outer and inner bound for Z c. Consider Out c = {r α() r max p α() = 1 log c (c z c p z s0 ), r > 0 } 15

16 and = { 1 log c(c z c p z s0 ) α() 1 log c(c z c p z s0 ) max p α() max p α() In c = {r α() r max > 0} p α() = 1 log c ( c c p s 0 ), r > 0 } = { 1 log c( m c c p s 0 ) max p α() α() 1 log c( m c c p s 0 ) > 0}. max p α() By the same argument given in lemma 3 on page 9 which explains why Z and Zc must have the same number of elements, we can see that Out c and In c have exactly one element if and only if Zc has exactly one element, and zero elements otherwise. Thus, if for all sets A R n containing just one element we define A to be the euclidean distance of the (unique) point in the set A from the origin, then at angles 0 where Zc 0 we have (by equation 27) In 0 c = 1 log c( m c c p s 0 ) α( 0 ) = 1 log c( m c c p s 0 ) max p α( 0 ) max p α( 0 ) Zc 0 1 log c(c z c p z s0 ) = 1 log c(c z c p z s0 ) α() = Out 0 c. max p α() max p α() Therefore, a ray projecting outwards from the origin along any such angle 0 will first hit the unique point in the set In 0 c, then the unique point in the set Zc s0,0, and finally the unique point in the set Out 0 c. In this sense, Z c, when shifted appropriately by a value s 0, lies between In c and Out c. This completes the proof of theorem 3. It is worth reiterating that when the sum of our coefficients c is less than c, we can choose s 0 = (0, 0,..., 0), so in this case no translation of our initial set is necessary. However, even when the sum of our coefficients is larger than c, computing a suitable s 0 should not be terribly difficult. In fact, it seems that an uncountable infinity of them lie in every open neighborhood of each point in Z c. Thus, as long as we can compute at least a single point z 0 Z c we now where to loo. We might consider a tiny codimension 1 ball around z 0, and choose values s 1 at random from this ball until one satisfying m c c p s 1 < c is found. This algorithm will probably halt rapidly in practice, so long as our ball is chosen to have a sufficiently small radius, and our value z 0 is computed with sufficient accuracy. We might even try shrining the diameter of our ball after each failed iteration of the algorithm to cause it to halt faster. 16

17 Conclusions We have demonstrated that the polynomial level set T c is homeomorphic to c Z and that it is trapped between the curves c Inc and c Outc. In addition, we have seen that T c is homeomorphic to S n 1 if and only if contains the origin. There are a number of directions in which this research could be extended. For example, we might consider the case where our c are allowed to be negative or even complex. In addition, we might relax the restriction that we are only considering our polynomial slice for positive real inputs. Other possibilities would be to search for tighter piecewise linear bounds on our solution set T c than c Inc and c Outc provide, or to study the points of T c that stretch out towards infinity. It is the author s hope that if the properties discovered in this paper generalize elegantly to include negative coefficients and negative points of polynomials then they might lead to improved techniques for studying level sets of general high dimensional polynomials, and perhaps even assist in solving systems of such polynomials. References [1] Weisstein, Eric W. MathWorld, A Wolfram Web Resource. 14 September 2005 < [2] Wiipedia: The Free Encyclopedia. 14 September 2005 < 17

18 Figure 1: Sphere-lie example curve with shifting. 18

19 Figure 2: Sphere-lie example curve without shifting. 19

20 Figure 3: Line-lie example curve without shifting. 20