A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID. Pan Liu

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1 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID Pan Liu Department of Mathematical Sciences, Worcester Polytechnic Institute, 100 Institute Road, Worcester MA , USA Abstract We construct a family of Pólya-type volume-filling continuous maps from a rectangle onto a right triangular solid pyramid The family is indexed by a parameter θ 0 between 0 and π/, which is the largest of the smallest angles of the vertical sections of the pyramid We extend to these maps the differentiability results obtained by Peter Lax for the Pólya s map for different values of θ 0 1 Introduction The space-filling curves were well studied by mathematicians since the beginning of the 0th century Examples of such curves are the Peano curve [5], the Hilbert curve [], the Sierpiński curve [10] and the Pólya curve [6] Later on, it was proved that (the coordinate functions of) the Peano, Hilbert, and Sierpiński curves are nowhere differentiable So, it was tacitly assumed that the Pólya curve will be nowhere differentiable too However, in 1973, Peter Lax proved that Pólya s curve is differentiable on a subset of the domain that depends on the smallest angle θ 0 of the triangle, and he further described the properties of this differentiability set in the three regions defined by two special angles, namely 15 and 30, as showed in Fig1 Volume-filling constructions have rarely been studied An example of volume-filling curve was given by Sagan, who extended the Hilbert construction to 3-dimensional space, and this 3D curve is also nowhere differentiable The idea of this paper is to extend Pólya s curve to a map from a rectangle onto a right pyramid preserving, at the same time, Lax differentiability properties The construction of such a map, however, cannot be based on altitude projections, like in the case of Pólya s curve, because there is no geometrical construction of such a kind that partitions the pyramid into smaller similar pyramids We carry out the construction of a continuous map from the rectangle Q = {(t, α) R : 0 t 1, 1

2 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 0 α π/} onto the right pyramid of R 3, with vertices (0, 0, 0), (0, tan θ 0, 0), (0, 0, 1), and (tan θ 0, 0, 0) We observe that now θ 0 (0, π ) is the largest of the smallest angles of the vertical sections of the pyramid R θ 0 We denote this map by L θ 0 = L θ 0 (t, α) and call it Pólya surface We then show that, though not differentiable, the map L θ 0 possesses both partial derivatives α Lθ 0 and t Lθ 0 on a subset of the domain that can be described as a function of the angle θ 0 in a similar way as in Lax result In addition to the two special angles single out by Lax, that is 15 and 30, the fine tuning of the differentiability property brings to light two more relevant angles, namely 036 and 395 In Lax s result, the two angles 15 and 30 distinguish three regions for the parameter θ 0 (0, π ), which are denoted by, and in Fig1 Pólya s curve has the best differentiability behavior in the region and the worse in, as described more precisely in Theorem In the case of Pólya surface, the worse region is still the one corresponding to θ 0 near 5 (flat pyramids) The differentiability behavior, however, improves to the status before the angle θ 0, decreasing from 5, reaches the value 30, more precisely, as soon as θ 0 crosses the new special angle, 395 In fact, for θ 0 strictly less than 395, an intermediate vertical sector appears in the pyramid where the differentiability upgrades to the status The opening of such a sector increases as θ 0 decreases, and the sector fills the whole pyramid when θ 0 reaches the values 30 A similar effect takes place when θ 0 decreases further from 30 to 15 (pointed pyramids) As θ 0 decreases further and crosses the second additional angle 036, again a central section of the pyramid shows upgraded differentiability of status, and this central sector becomes the whole pyramid when θ 0 reaches the value 15 This new phenomenon is described in Theorem 6 and illustrated in Fig As all space-filling constructions, Pólya surface, like Pólya s curve, is self-intersecting However, it is possible to approximate Pólya surface with surfaces with no selfintersections, that fill a pyramid up to a residual small volume of arbitrary size ε These new fractal surfaces are the boundary of an open domain The result presented in this paper is therefore related to the recent literature on boundary value problems in small domains with large boundary, as the ones considered in [] However, we do not develop this aspect of our research in present paper Our paper is organized as follows: In Section we describe the construction of Pólya s curve and the differentiability results of Lax In Section 3 we prove the

3 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 3 self-similarity of Pólya s map, shown recently by Ramsay and Terry in [7], and the representation formulas original given by Sagan in [9] By relying on these properties, we study Pólya s map as a function of both t and θ, and prove, Lemma, the existence of the partial derivative in θ In Section, we give the construction of Pólya surface and our main result, the partial differentiability properties in Theorem 6 Pólya s Function and Lax Result In 1913 Pólya presented an iterative geometric construction of a map, P, mapping the interval [0, 1] onto a non-isosceles right triangle T, [6] In 1973 Lax proved that the Pólya s map is differentiable on a subset of [0, 1] that depends on the smallest angle of T To build Pólya s map, we need to follow the process illustrated below Let t be any number in the unit interval and consider it to be a binary fraction: where the n th digit d n is either 0 or 1 t = d 1 d Definition 1 Let T be a non-isosceles right triangle, we define the Pólya s function P (t), which maps a point t [0, 1] into T, by the following process: i We subdivide T into two smaller and similar triangles in T by drawing the altitude of T Since T is non-isosceles, these two triangles are unequal; call the larger of the two T 0, the smaller T 1 ii We define T 1 to be T 0, if d 1 = 0; T 1 to be T 1, if d 1 = 1; T, T 3 are defined recursively, with T n 1 taking the place of T and d n replacing d 1 iii We denote the n-th triangle assigned to the number t = 0d 1 d d 3 d n by T n (t) The sequence T n (t) is nested, and the hypotenuse of T n (t) goes to 0 as n The triangles T n (t) have exactly one point in common, and we define this point to be P (t) Remark 1 We know that there can be two binary representations of the same real number t Lax, however, has shown that the procedure described in Definition 1 will assign the same point P (t) to different binary representations of t Theorem 1 [6] Pólya s Theorem i The function P maps the interval I = [0, 1] continuously onto the triangle T ii If one chooses the ratio of the shorter side to the hypotenuse to be a transcendental number, then every point in T has, at most, three pre-images in I

4 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID Theorem [3] Lax s Differentiability Theorem Let θ 0 denote the smaller angle of the (non-isoceles) right triangle T Then, the function P (t) has the following properties: If 30 < θ 0 < 5, P is nowhere differentiable If 15 < θ 0 < 30, P is not differentiable on a set of Lebesgue measure 1, but has derivative zero on a non-denumerable set If θ 0 < 15, P = 0 on a set of measure 1 In the picture below (Fig1) we outline the 3 cases Figure 1 The differentiability regions of Pólya s curve 3 Self-Similarity of Pólya s Curve In order to generate Pólya s curve, we divide the triangle T by its altitude and create two similar small triangles In other words, we map the interval [0, 1 ] onto the triangle T 0 and the interval [ 1, 1] onto the triangle T 1 We then iterate this procedure further This suggests that the Pólya s map must possess some self-similarity property This has been shown by Ramsay and Terry, who proved that Pólya s map is indeed self-similar in the sense of the Definition below, [7] Let χ 0 and χ 1 be contractive similitudes in the Euclidian space R and let S be a subset of R We define χ i/n (S) := χ i1 χ i χ in (S), where i/n = i 1 i i n is a sequence of 0 s and 1 s of fixed length n, and then χ i (S) := lim n χ i/n (S) By the contraction principle, χ i (S) is a single point in S The set S is said to be self-similar with respect to χ 0 and χ 1 if for every n S = i/n χ i/n (S) Now we define the concept of self-similar map

5 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 5 Definition Let S be self-similar with respect to χ 0 and χ 1 Let S R be another set which is self-similar with respect to two other contractive similitudes ω 0 and ω 1 Let F be a surjective mapping F : S S such that F (χ i/n (S)) = ω i/n (S ) for every i/n Then we say that the mapping F is self-similar Now, to show that Pólya s function is self-similar, we first define the contractive similitudes that operate on I = [0, 1] in the following way φ 0 (x) = x, φ 1(x) = x + 1 Clearly, the interval I = [0, 1] is self-similar with respect to φ 0 and φ 1 In particular I = i/n φ i/n (I), for every n The length of every interval φ i/n (I) equal to n Now we define the contractive similitudes, ψ 0 and ψ, that operate on a triangle T Assign θ 0 as the smallest angle of T, a = (a 1, a ) as the vertex of T at the angle θ 0, b = (b 1, b ) as the vertex of T at the angle π θ 0: ψ 0 is the similarity which maps T to the larger of the two similar triangles, named as T 0, formed by the altitude of T For an arbitrary point x = (x 1, x ) T : [ ] a1 + cos θ ψ 0 (x) = ψ 0 (x 1, x ) = 0 [(a 1 x 1 ) cos θ 0 + (a x ) sin θ 0 ] a + cos θ 0 [(a 1 x 1 ) sin θ 0 + (x a ) cos θ 0 ] ψ 1 is the similarity which maps T to the smaller of the two similar triangles, named as T 1, formed by the altitude of T For an arbitrary point x = (x 1, x ) T [ ] b1 + sin θ ψ 1 (x) = ψ 1 (x 1, x ) = 0 [(x 1 b 1 ) sin θ 0 + (b x ) cos θ 0 ] b + sin θ 0 [(b 1 x 1 ) cos θ 0 + (b x ) sin θ 0 ] For convenience, we define the triangle T occurring in Pólya construction to be the following triangle of R : The point of smallest angle θ 0 in T is the point (0, 1), that is, a = (a 1, a ) = (0, 1) The point of angle π/ θ 0 is the point (tan θ 0, 0), that is, b = (b 1, b ) = (tan θ 0, 0)

6 6 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID Then, the maps ψ 0 and ψ 1 can be written in the following way: ψ 0 (x) = cos θ 0 O 0 x + sin θ 0 µ with (1) and with () O 0 = O 1 = [ ] [ ] cos θ0 sin θ 0 cos θ0, µ =, sin θ 0 cos θ 0 sin θ 0 ψ 1 (x) = sin θ 0 O 1 x + sin θ 0 µ [ ] [ ] sin θ0 cos θ 0 cos θ0, µ = cos θ 0 sin θ 0 sin θ 0 Note that in this way, the two legs of T lay on the positive x 1 - and x -axis The larger sub-triangle T 0 in Definition 1 is now obtained as T 0 = ψ 0 (T ), and similarly, the smaller sub-triangle T 1 = ψ 1 (T ) At the next subdivision, we obtain the sub-triangles T 00 = ψ 0 ψ 0 (T ), T 01 = ψ 0 ψ 1 (T ), T 10 = ψ 1 ψ 0 (T ), and T 11 = ψ 1 ψ 1 (T ), and we get the decomposition T = T 00 T 01 T 10 T 11 After n iterations, we get the decomposition T = i/n ψ i/n (T ) This shows that the triangle T is self-similar with respect to ψ 0 and ψ 1 Theorem 3 [7] Let P : I T be Pólya s map Then for every i/n, (3) P (φ i/n (I)) = ψ i/n (T ) Moreover, () P (φ i (I)) = ψ i (T ) for every i In particular, P is self-similar Proof Define [ k I i/n = φ i/n ([0, 1]) =, k + 1 ] n n Since every point in I i/n has the same first n digits, these points will all be mapped to the same subtriangle of T We can find which subtriangle of T this is if we take ψ i/n (T ) We note that the action of the maps ψ 0 and ψ 1 reproduces the process carried out by Pólya s function, ie, the division of T by the altitude and the choice

7 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 7 of the smaller or larger triangle Therefore, we have that P (φ i/n (I)) = ψ i/n (T ) For a given i, by taking the limit as n in the previous identity, we obtain by the continuity of P P (φ i (I)) = ψ i (T ) Now that we have established the Self-Similarity of Pólya function, we introduce the Pólya Interpolates, which are suitable Polygonal trajectories based on Pólya construction First of all, we partition the unite interval [0, 1] into n small intervals, [t j, t j+1 ], j = 0, 1,,, n, by letting (5) 0 = t 0 < t 1 < t < < t n 1 < t n = 1, where t j t j+1 = n Clearly, by the construction of [t j, t j+1 ], there exists a i/n such that (6) I j := [t j, t j+1 ] = φ i/n (I), and by using the same i/n, we get the triangle T j := ψ i/n (T ) Lemma 1 For any interval [t j, t j+1 ] of the partition introduced in (5) and (6), there exists a t j [t j, t j+1 ] such that the point (7) G j = P (t j) is the geometric center of triangle T j = T i/n = ψ i/n (T ) Proof By the self-similarity of I = [0, 1], we have (8) [t j, t j+1 ] = φ i/n (I) = φ i/n φ k (I), where k = d 1 d d m is a infinite sequence with d m {0, 1} and K is the collection of all possible sequences k Therefore, (9) P ( φ i/n (I) ) = P ( φ i/n φ k (I) ) By (3), we have k K k K P ( φ i/n (I) ) = ψ i/n (T ) = T i/n,

8 8 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID and by (), Therefore, from (9), we obtain P ( φ i/n φ k (I) ) = ψ i/n ψ k (T ) T i/n = ψ i/n ψ k (T ) k K So, there exists a sequence k K such that ψ i/n ψ k (T ) is the geometric center G j of the triangle T i/n We then define t j to be (10) By (8), we know that (11) Moreover, by the self-similarity of P and this concludes our lemma t j := φ i/n φ k (I) t j [t j, t j+1 ] G j = P (t j), For every partition as in Lemma 1, we obtain the sequence of t j [t j, t j+1 ], such that t 0 < t 0 < t 1 < t 1 < t < < t n 1 < t n 1 < t n for j = 0, 1,,, and we now define the Pólya interpolate, associated with that partition, as the oriented polygonal curve in T that connects each P (t j) to the successive one, as showed in Fig7 Moreover, since (1) t j t j+1 = n 0, as n, we obtain that t j = t j = t j+1 in the limit as n Therefore, the Pólya interpolate converges eventually to the Pólya curve as n, in an obvious sense Theorem [9] The Pólya s mapping, P (t) = P (0d 1 d d 3 d n ), can be represented for any t = 0d 1 d d 3 d n, by (13) P (t) = P (0d 1 d d 3 ) = c Z j s Vj+1 O d1 O d O dj 1 µ, j=1 where c = cos θ 0, s = sin θ 0, and Z i, V i is the number of 0 s, 1 s preceding d i, respectively, and µ is defined in equation (1)

9 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 9 Proof Recall the equation () in Theorem, that is It is clear that P (φ i (I)) = ψ i (T ) φ i (I) = 0d 1 d d 3 d n represents the parameter t [0, 1] in binary form Obviously, for any binary fraction t, we have 0d 1 d d 3 d n = 0d 1 d d 3 d n , and, for convenience, we define ψ n 0 (x) to be ψ 0 applied n times on x ie, ψ n 0 (x) = ψ 0 ψ 0 ψ 0 (x) for n ψ 0 (x) s Taking any x = (x 1, x ) T, we have (1) ψ n 0 (x) =(cos θ 0 O 0 + sin θ 0 µ) (cos θ 0 O 0 x + sin θ 0 µ) = cos n θ 0 O n 0 x + sin θ 0 [cos n 1 θ 0 O n 1 0 µ+ + cos n 3 θ 0 O n 3 0 µ + + cos θ 0 O 0µ + cos θ 0 O 0 µ + µ] Notice that O 0 O 0 = I is a identity matrix, and we obtain { lim n cosn θ 0 O0 n lim n cos n θ 0 I, if n is even = lim n cos n θ 0 O 0, if n is odd = Also, we have [ 0 0] lim [sin θ 0 (cos n 1 θ 0 O0 n 1 µ + + cos θ 0 O d1 µ + µ)] n = sin θ 0 [(1 + cos θ 0 + cos θ 0 + )µ + (cos θ 0 + cos 3 θ 0 + )O 0 µ] [ cos θ0 sin θ 0 = 1 sin θ 0 [ 0 = 1] [ ] cos θ0 + cos θ 0 sin θ 0 sin θ 0 Therefore, by (1), we get (15) Thus (16) sin θ 0 ] [ ] cos θ0 cos θ 0 sin θ 0 lim n ψn 0 (T ) = ψ 0 ψ 0 ψ 0 (T ) = [ 0 1] P (0d 1 d d 3 d n ) = P (0d 1 d d 3 d n 000 ) [ ] 0 = lim ψ d1 d d 3 d n ψ k 0 T = ψ d1 d d 3 d n k 1

10 10 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID By replacing the expression of ψ 0 and ψ 1 into right side of (16), we obtain [ ] P (0d 1 d d 3 d n ) = c Z j s V j 0 n (17) O d1 O d O dn + c 1 Z j s Vj+1 O d1 O dj 1 µ, where as defined in (1) µ = [ ] cos θ0 sin θ 0 To go to the limit as n, we need the property that the Pólya s mapping is continuous We could just accept this property from the original Pólya Theorem 1, or we can show this by taking t 1 t < 1 n Hence, by (17) and the continuity of Pólya map, we obtain (18) P (t) = P (0d 1 d d 3 ) = c Z j s Vj+1 IO d1 O d O dj 1 µ, since Remark j=1 j=1 [ ] [ lim n cz j s V j 0 0 O d1 O d O dn = 1 0] By using a formula similar to equation (), Sagan proved that the Pólya s coordinate functions have the same differentiability shown in Theorem Moreover, Sagan proved that the Pólya s function is nowhere differentiable when the smallest angle θ 0 is 30, and has derivative zero on a non-denumerable set if the smallest angle θ 0 is equal to 15 Definition 3 The General Pólya Function: We define the function P G (θ, t), which maps the rectangle [0, π ] [0, 1] onto a right triangle T θ with vertices (0, 0), (0, 1), and (tan θ, 0), by (19) P G (θ, t) := c Z j s Vj+1 O d1 O d O dj 1 µ, j=1 by treating θ 0, in equation (13), as a variable too Lemma The General Pólya s Function P G (θ, t) is continuous on [0, π ] [0, 1] Moreover, for every t [0, 1], the partial derivative P θ G(θ, t) does exist for all θ (0, π)

11 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 11 Proof Since the equation (19), after we fix t, is only the composition of summations and multiplications of simple functions sin θ and cos θ and these two simple functions are both continuous and differentiable, we know that P (θ, t 0 ) is continuous and differentiable with respect to θ Three Dimensional Constructions and Differentiability The very first and natural idea to extend Pólya s mapping to a pyramid would be trying to partition a triangular pyramid into n small and similar triangular pyramids all similar to the original pyramid But, unfortunately, we know of no construction leading to such kind of partition In this regard, we refer to the recent paper [1], where similar packing constructions are carried out that provide only a partial partition of the initial pyramid up to 8563% of the total volume Now, we introduce our three dimensional Solid Triangular Pyramid Construction The idea is if we take the Pólya interpolate and rotate it around the vertical leg of T, we have a solid cone, as showed in Fig Let α [0, π] denote the rotation angle and t [0, 1] denote the variable in Pólya s Function Clearly, this operation maps a rectangle, [0, π] [0, 1], to a solid cone from R to R 3 But this is not good enough We are looking for a map which can map the rectangle to a solid triangular pyramid Figure The Solid Cone Figure 3 The 3D construction

12 1 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID Notation In the following we will use: Given θ 0 (0, π ), we define O θ 0 p := {(x, y) R, ( x p + y p ) 1 p = tan θ0 }, for p > 0 Given θ 0 (0, π ), we define C θ 0 p := {(x, y, z) : x, y 0, 0 z 1, ( x p + y p ) 1 p tan θ0 (1 z)} For example, C θ 0 is a quarter of the Solid Cone in Fig, and C θ 0 1 is a right triangular pyramid, as showed in Fig3 Given θ 0 (0, π), we define T θ1(α) to be a right triangle generated by the intersection of the vertical plane y = (tan α)x and C θ 0 1 Lemma 3 Given θ 0 (0, π), for each α [0, π], the smallest angle θ 1(α) of the right triangle T θ1(α), has the formula: [ ] tan θ 0 (tan 1 α + 1) θ 1 (α) = arcsin tan θ 0 (tan α + 1) + (1 + tan α) Moreover, the function sin(θ 1 (α)) is continuous for all α [0, π ] Proof Let (x 0, y 0 ) O θ 0 1, that is { x 0 + y 0 = tan θ 0 (0) tan α = x 0 /y 0 Since the height of right triangle T θ1(α) is 1, we have [ ] x sin(θ 1 (α)) = 0 + y0 1 (1), x 0 + y0 + 1 Together with (0), we obtain [ () sin(θ 1 (α)) = tan θ 0 (tan α + 1) tan θ 0 (tan α + 1) + (1 + tan α) ] 1 It is clear that sin(θ 1 (α)) is well definied for all α [0, π/) For α = π/, though the function tan α is not defined at α = π/, the function sin(θ 1 (α)) still makes sense and continuous at α = π/ by setting sin(θ 1 (π/)) = sin(θ 1 (0)), because (3) lim sin(θ 1(α)) = sin(θ 1 (0)) α π Therefore, θ 1 (α) is continuous in an obvious sense

13 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 13 Notation In the following we will use: By sin(θ(α)) and θ(α), we denote sin(θ 1 (α)) and θ 1 (α), respectively By R θ 0, we denote C θ 0 1, a right triangular pyramid with vertices (0, 0, 0), (0, tan θ 0, 0), (0, 0, 1), and (tan θ 0, 0, 0) By T θ(α), we denote the right triangle given by the intersection of the vertical plane y = (tan α)x with R θ 0 Definition The 3D construction map: Given θ 0 (0, π ), we define the function Lθ 0 (α, t), which maps the rectangle Q = [0, π ] [0, 1] onto Rθ 0 by L θ 0 (α, t) = (x, y, z), where, by setting (a, b) = P G (θ(α), t), x = a cos α, y = a sin α and z = b Remark 3 As the definition above of L θ 0 (α, t) is rather abstract and hard to visualize, it is useful to describe the function L θ 0 (α, t) in a geometric way For a better understanding, and to help building a mental picture of L θ 0 (α, t), we can fix the rotation angle α first We then consider the right triangle T θ(α) obtained by intersecting the pyramid R θ 0 with the vertical plane y = (tan α)x Next, we build the Pólya s curve inside T θ(α) Finally, we move the rotation angle The function L θ 0 (α, t) is the function that describes the rotation in α of the point P (t) on the Pólya s curve in T θ(α) Theorem 5 Let 0 < θ 0 < π be fixed, the map Lθ 0 (α, t) is surjective and continuous for all α [0, π ] and t [0, 1] Proof We first prove L θ 0 (α, t) is surjective Taking any point m = (x, y, z) R θ 0, there exists a α 0 [0, π], such that m T θ(α0) Then, since the Pólya map is surjective, there exists a point t 0 [0, 1] such that P G (θ(α 0 ), t 0 ) = (a, b) where a = x/cos(α 0 ) and b = z Therefore, for any point m R θ 0, there exists α 0 and t 0 such that L θ (α 0, t 0 ) = m That is, the map L θ 0 (α, t) is surjective To prove the continuity of L θ 0 (α, t) in t for fixed α, it suffices to note that this map is the map P (t) acting on T θ(α), which we know to be a continuous map from Theorem 1 Next, to prove the continuity of function L θ 0 (α, t) with respect to α, we define an inter-median function () r(α, x, z) := (x cos α, x sin α, z),

14 1 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID which rotates the point (x, y, x) α degrees around z-axis It is clear that the function r(α, x, z) is continuous and we have (5) L θ 0 (α, t) = (r P G )(θ(α), t) We have shown in Lemma 1 that the function P G (θ, t) is continuous for all θ [0, π], and in Lemma 3 that θ(α) is also continuous Therefore, the composite function L θ 0 (α, t) is continuous for all α [0, π] Now, we are ready to introduce our main result, the Solid Triangular Pyramid Differentiability Theorem First of all, for future convenience, we define θ 1 := arcsin 5, θ 1 8 (6) 3 := arcsin , θ(α 1) := π 6, θ(α ) := π 1 Moreover, to be consistent with Lax s paper we convert angles measured in radians into angles measured in degrees Approximately, θ = 395 and θ 036 = 036 Theorem 6 Solid Triangular Pyramid Partial Differentiability Theorem Given θ 0 (0, π), the map Lθ 0 (α, t) defined for α [0, π ], t [0, 1] has the following partial differentiability properties: i Given θ 0 (0, π), α Lθ 0 (α, t) does exist for all α (0, π ) and t [0, 1]; ii Given θ 0 [θ 1, π), for all α [0, π], t Lθ 0 (α, t) has property ; iii Given θ 0 [θ, π), for all α [0, π], 6 t Lθ 0 (α, t) has property ; iv Given θ 0 (0, π ), for all α [0, π], 1 t Lθ 0 (α, t) has property Moreover, v Given θ 0 [ π, θ 6 1), when α [0, α1 ] [ π α 1, π], t Lθ 0 (α, t) has property ; when α (α 1, π α 1), t Lθ 0 (α, t) has property vi Given θ 0 [ π, θ ) 1, when α [0, α ] [ π α, π], t Lθ 0 (α, t) has property ; when α (α, π α ), t Lθ 0 (α, t) has property Figure The 3D partial differentiability properties regions, compared with the D case

15 Proof Case 1: By (5),we have A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 15 L θ 0 (α, t) = (r P G )(θ(α), t) It is clear that the partial derivative r(α, x, z) and P α α G(θ(α), t) both exist for all α (0, π/) Therefore, the partial derivative α Lθ 0 (α, t) does exist for all α (0, π) by Composite function theorem Case : Recall that the Pólya function is nowhere differentiable when θ 0 [ π, π ] Therefore, 6 we want to find a set S such that when the given angle θ 0 S, θ(α) [ π, π ] for all 6 α [0, π] By the properties of θ(α), we have min α [0, π ] θ(α) = θ(0) = θ ( π ) ( π ), max θ(α) = θ α [0, π ] Moreover, since sin x is monotone increasing for x [0, π ], we obtain ( min sin(θ(α)) = sin(θ(0)) = sin θ α [0, π ] ( π )) (, max sin(θ(α)) = sin θ α [0, π ] Therefore, we need to determine the value of θ 0 such that ( ( π )) ( π ) sin θ = sin = 1 6 By () we have Thus (7) ( ( π )) [ ] tan 1 (θ 0 ) ( π ) sin θ = = sin = 1 tan (θ 0 ) + 6 θ 0 = arcsin , ( π )) and we use θ 1, defined in (6), to denote the θ 0 in (7) Therefore, we conclude that given any θ 0 [θ 1, π ], θ(α) [ π 6, π ] for all α [0, π ] Based on Theorem, the partial derivative t Lθ 0 (α, t) does not exist for all θ 0 [θ 1, π ] Case 3: Similar to Case, we would like to find out the set S such that when θ 0 S,

16 16 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID θ(α) [ π, π) for all α [0, π] 1 6 Setting ( ( π )) ( π sin θ = sin, 1) we obtain (8) [ θ 0 = arcsin ] = 036, and we use θ, defined in (6), to denote the θ 0 in (8) Therefore, by Theorem, for θ 0 [θ, π 6 ), the partial derivative t Lθ 0 (α, t) does not exist for all t in a set of measure 1, but it exists when t is in a non-denumerable set Case : Recall that the function θ(α) attains the maximal value at α = 0 So, for any θ 0 (0, π ), θ(α) (0, π ) for all α [0, π] 1 1 Therefore, for all θ 0 (0, π ), the partial derivative 1 t Lθ 0 (α, t) does exist when t on a set of Lebesgue measure 1 Case 5: Based on Case, we know that when θ 0 [θ 1, π) π ( π ) ( π ) 6 min θ(α) = θ < max θ(α) = θ α [0, π ] α [0, π ] = θ(0) π, and for θ 0 = θ 1, ( π ) min θ(α) = θ = π α [0, π ] 6 Thus, θ( π) will be less then π/6 when θ 0 < θ 1 because θ(α) is strictly decreasing for α [0, π] Therefore, for θ 0 [ π, θ 6 1), there exists a α1 [0, π] such that α 1 satisfies the equation (9) (3 tan θ 0 1) tan α 1 tan α 1 + (3 tan θ 0 1) = 0, where α 1 is defined in (6) Clearly, when θ 0 [π/6, θ 1 ), the equation (9) has two solutions and one of it satisfies θ(α 1 ) = θ( π α 1) = π 6, α 1 0 as θ 0 π 6 So, we conclude that for given θ 0 [ π 6, θ 1), these exists a α1 [0, π ] with θ(α 1) = π/6 such that (i) When α [0, α 1 ] [ π α 1, π ], the partial derivative t Lθ 0 (α, t) does not exist for any t [0, 1];

17 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 17 (ii) When α (α 1, π α 1), the partial derivative t Lθ 0 (α, t) does not exist for t in a set of measure 1, but it exists for t in a non-denumerable set of [0, 1] Since α 1 0 as θ 0 π/6, the interval α (α 1, π α 1) will expand and fill the area (0, π ) when θ 0 = π/6 As mentioned in Section 1, we have proved that when θ 0 crosses the new special angle θ 1, an intermediate vertical sector starts to appear in the pyramid where the differentiability property upgrades to the status and this sector will fill the whole pyramid when θ 0 reaches the values π/6 Case 6: Similar as in Case 5, we conclude that for given θ 0 [ π, θ 1 ), there exists an angle α with θ(α ) = π/1 such that (i) When α [0, α ] [ π α, π ], the partial derivative t Lθ 0 (α, t) does not exist for t in a set of measure 1, but it exists for t in a non-denumerable set of [0, 1]; (ii) When α (α, π α ), the partial derivative t Lθ 0 (α, t) does exist for all t in a set of measure 1 in [0, 1] For a better understanding, and to help making a mental picture of the differentiability properties of t Lθ 0 (α, t) in Case 5, as well as Case 6, one can refer to Fig5 and Fig6 Certainly, t Lθ 0 (α, t) becomes better in a vertical central sector, ie, when α belongs to the orange area, as showed in Fig5 Remark As stated in the Theorem 1 part (), if one chooses the ratio of the shorter side to the hypotenuse, ie sin θ, to be a transcendental number, then for every point in T, there are at most three branches of Pólya s curve which intersect at that point This property fails for the map L θ 0 (α, t) as a function of α In fact, for it to hold, sin(θ(α)) must be a transcendental number But the value of θ(α) is changing smoothly with respect to α [0, π ], as also explained in Remark 3 Therefore, it is impossible to have all values of sin(θ(α)) to be transcendental numbers As mentioned in the Introduction, however, it is possible to approximate the Pólya surface with surfaces which have no self-intersections This study of these approximating surfaces will be developed elsewhere ACKNOWLEDGMENTS I would like to express my gratitude to Professor Umberto Mosco, who suggested this research, for his great efforts to explain things clearly and simply This paper would

18 18 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID not have been possible without his support This research was partially supported by NSF grant: DMS References [1] E R Chen, M Engel and S C Glotzer, Dense crystalline dimer packings of regular tetrahedra, Discrete & Computational Geometry, (010), [] D Hilbert, Über die stetige Abbildung einer Linie auf ein Flächenstück, Math Ann, 38 (1891), [3] P D Lax, The differentiability of Pólya s function, Advances in Mathematics, 10 (1973), 56-6 [] U Mosco, Fractal reinforcement of elastic membranes, Arch Rational Mech Anal, 19 (009), 9-7 [5] G Peano, Sur une courbe qui remplit toute une aire plane Math Ann, 36 (1890), [6] G Pólya, Über eine Peanosche kurve, Bull Acad Sci Cracovie, Ser A, (1913), [7] A Ramsay and M Terry, An investigation of Pólya s function, Major Qualifying Project, E-project Worcester Polytechnic Institute (010), 1-18 [8] H Sagan, Space-filling curves, Springer-Verlag New York (199) [9] H Sagan and K Prachar, On the differentiability of the coordinate functions of Pólya s spacefilling curve, Mathematik, 11 (1996), [10] W Sierpiński, Sur une nouvelle courbe continue qui remplit toute une aire plane, Bull Acad Sci Cracovie A, (191), 63-78

19 A TWO-DIMENSIONAL PÓLYA-TYPE MAP FILLING A PYRAMID 19 Figure 5 The differentiability regions in Case v Figure 6 The differentiability regions in Case v Below are multiple images which help provide a basic understanding of L θ 0 (α, t), where θ 0 = arctan(075) Figure 7 Pólya interpolate for n = 5 Figure 8 3D Pólya surface with n = 5