Interpolatory bivariate refinable functions and subdivision
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(2) De laration By submitting this thesis ele troni ally, I de lare that the entirety of the work ontained therein is my own, original work, that I am the owner of the opyright thereof (unless to the extend expli itly otherwise stated) and that I have not previously in its entirety or in part submitted it for obtaining any quali ation.. Signature: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.F. Rabarison. Date: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Copyright. ©. 2008 Stellenbos h University. All rights reserved.. i.
(3) Summary In this thesis, we introdu e bivariate re nable fun tions whi h are fun tions that are expressible as linear ombinations of the shifts of their own dilation by a fa tor of a dilation matrix. For the orresponding re nement masks, we de ne the mask symbols as the Laurent polynomials whose oe ients are the elements of the re nement masks. Of parti ular interest are interpolatory re nable fun tions, that is, re nable fun tions whi h vanish at all integers ex ept the origin at whi h they take the value. 1.. We present simple. hara terization of the orresponding interpolatory masks in terms of both the delta sequen e and the determinant of the dilation matrix. The orresponding interpolatory mask symbols are hara terized by some polynomial identities.. An important tool for our work is the Eu lidean algorithm, whi h, in asso iation with the Bezout theorem, helps us to provide an expli it omputational algorithm to nd the general solution for some polynomial identities. Using the algorithm thus presented, we introdu e the general form of an interpolatory mask symbol asso iated with the dilation matrix. 2I ,. and the result thus obtained is applied to the mask symbols orresponding to. the box splines.. The on epts of interpolatory subdivision s hemes and as ade algorithms are also investigated. Subdivision s hemes, as usually used to generate urves and surfa es, are interpolatory when the initial data points are preserved at all the steps of the subdivision pro ess. We show that interpolatory subdivision s hemes and the as ade algorithm are. ii.
(4) iii. SUMMARY. strongly linked to ea h other. For a well- hosen dilation matrix and interpolatory re nement mask, we nd that the asso iated as ade algorithm preserves ertain properties of the initial fun tions, allowing us to prove that as ade algorithm onvergen e implies the existen e of a orresponding interpolatory re nable fun tion, whi h in turn implies subdivision s heme onvergen e.. Spe ializing only to the ase where the dilation matrix is. M = 2I ,. we present some. workable methods applied for both non-negative interpolatory masks and interpolatory masks obtained by tensor produ ts in order to investigate the existen e of orresponding interpolatory re nable fun tions. For interpolatory masks onstru ted to satisfy the sum rules, we provide numeri al proofs towards investigating the existen e of orresponding interpolatory re nable fun tions by using the as ade algorithm with an appropriate initial fun tion. Numeri al illustrations by means of subdivision graphs are also provided..
(5) Opsomming In hierdie tesis beskou ons tweeveranderlike verfynbare funksies, oftewel funksies wat uitdrukbaar is as lineêre kombinasies van die skuiwe van hulle eie dilasie deur die faktor van die dilasiematriks.. Vir die ooreenkomstige verfyningsmaskers de nieer ons die. maskersimbole as Laurent polinome waarvan die koë siënte die elemente van die verfyningsmaskers is.. Van besondere belang is interpolerende verfynbare funksies, dit wil sê. verfynbare funksies wat gelyk aan nul is by alle heelgetalle behalwe die oorsprong waar hulle die waarde 1 aanneem. Ons gee 'n eenvoudige karakterisering van die ooreenstemmende interpolerende maskers, beide in terme van die delta ry en die determinant van die dilasiematriks. Die ooreenstemmende interpolerende maskersimbole word gekarakteriseer deur sekere polinoom identiteite.. 'n Belangrike stuk gereedskap vir ons werk is die Euklidiese algoritme, wat, tesame met die Bezout stelling, ons help om 'n eksplisiete algoritme te bepaal vir die algemene oplossing van sekere polinoom identiteite. Met behulp van hierdie algoritme stel ons dan bekend die algemene vorm van 'n interpolerende maskersimbool wat ooreenstem met die dilasiematriks. 2I , en die resultaat wat sodanig verkry is word dan toegepas op die masker-. simbole wat ooreenstem met 'n sekere klas tweeveranderlike latfunksies (box splines).. Die konsepte van interpolerende subdivisie skemas en kaskade algoritmes word ook ondersoek.. Subdivisieskemas, soos gewoonlik gebruik om krommes en oppervlakke te. genereer, is interpolerend indien die begin-datapunte gepreserveer word by elke stap van. iv.
(6) v. OPSOMMING. die subdivisie proses. Ons toon aan dat interpolerende skemas en die kaskade algoritme sterk aanmekaar verbind is. Vir 'n goedgekose dilasiematriks en interpolerende verfyningsmasker vind ons dat die ooreenstemmende kaskade algoritme sekere eienskappe van die beginfunksie preserveer, met behulp waarvan ons dan kan bewys dat kaskade algoritme konvergensie die bestaan van 'n ooreenstemmende interpolerende verfynbare funksie impliseer, en wat op die beurt dan die konvergensie van die subdivisieskema impliseer.. Deur te spesialiseer na die geval waar die dilasiematriks. M = 2I , verskaf ons werkbare. metodes vir toepassing op beide nie-negatiewe interpolerende maskers en interpolerende maskers soos verkry met behulp van tensor produkte met die doel om die bestaan van ooreenstemmende interpolerende verfynbare funksies te ondersoek.. Vir interpolerende. maskers wat die somreëls bevredig, gee ons numeriese bewyse ten opsigte van die ondersoek na die bestaan van ooreenstemmende verfynbare funksies, deur die kaskade algoritme met 'n gepaste beginfunksie te gebruik. Numeriese illustrasies deur middel van subdivisie gra eke word ook verskaf..
(7) A knowledgements I thank God for everything He have done for me, strength, ourage and faith I have re eived from Him in order to a omplish this thesis. My deepest a knowledgment goes to my supervisor Prof Johan de Villiers, working with him was a pleasant and unforgettable experien e. I am ex eptionally grateful to Gaelle Andriamaro for the heartful help and supports she has given without any hesitation while I was working on this thesis.. My. thanks also go to all of my friends, my o emates, with a spe ial appre iation to my parents and family for their thoughtful prayers. Finally, I am very grateful to AIMS, the Afri an Institute for Mathemati al S ien es, and to the University of Stellenbos h for the nan ial supports permitting me to nish this thesis. God bless and thank you all.. vi.
(8) Contents De laration. i. Summary. ii. Opsomming. iv. Contents. vii. List of symbols. ix. Introdu tion. 1. 1 Interpolatory bivariate re nable fun tions. 4. 1.1. Notation and general on epts . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 1.2. Interpolatory re nement masks. . . . . . . . . . . . . . . . . . . . . . . . .. 6. 1.3. Box splines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 9. 2 The interpolatory mask symbols for M = 2I. 19. 2.1. Simple hara terization . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 19. 2.2. General form. 23. 2.3. Appli ation to box splines interpolatory mask symbols. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3 Interpolatory subdivision s hemes 3.1. Preliminaries. 3.2. Subdivision s hemes onvergen e. 43. 47. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. vii. 47 51.
(9) viii. CONTENTS. 3.3. Property preservation in the as ade algorithm . . . . . . . . . . . . . . . .. 4 Existen e of interpolatory re nable fun tions. 54. 60. 4.1. For non-negative masks . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 61. 4.2. Tensor produ ts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 63. 4.3. Mask onstru tion based on sum rules. 69. Bibliography. . . . . . . . . . . . . . . . . . . . .. 80.
(10) List of symbols Symbol. De nition. N. the set of natural numbers. Z, Z+. the sets of integers and non-negative integers. Z2 , Z2+. the sets of integer pairs and non-negative integer pairs. Q, Q2. the sets of rational numbers and rational pairs. R, R2. the sets of real numbers and real pairs. C, C2. the sets of omplex numbers and omplex pairs. M(Z). the linear spa e of bi-in nite real-valued sequen es in. Z,. i.e.. c ∈ M(Z) ⇐⇒ c = {cj : j ∈ Z} ⊂ R M(Z2 ). the linear spa e of bi-in nite real-valued sequen es in. Z2 ,. c ∈ M(Z2 ) ⇐⇒ c = {cj : j ∈ Z2 } ⊂ R2 M(R). the linear spa e of real-valued fun tions in. R,. i.e. the set. {f : R → R} M(R2 ). the linear spa e of real-valued fun tions in. R2 ,. i.e. the set. {f : R2 → R} M0 (Z). the subset of nitely supported sequen es in. M(Z). M0 (Z2 ). the subset of nitely supported sequen es in. M(Z2 ). M0 (R). the subset of nitely supported fun tions in. M(R). M0 (R2 ). the subset of nitely supported fun tions in. M(R2 ). ix. i.e..
(11) x. LIST OF SYMBOLS. supp(c). the support of the sequen e. c ∈ M0 (Z2 ),. supp(f ). the support of the fun tion. f ∈ M0 (R2 ),. ontaining. i.e. the set. {j ∈ Z2 : cj 6= 0}. i.e. the smallest losed set. {x ∈ R2 : f (x) 6= 0}. C(R). the subset of ontinuous fun tions in. M(R). C(R2 ). the subset of ontinuous fun tions in. M(R2 ). C0 (R). the subset of nitely supported fun tions in. C(R). C0 (R2 ). the subset of nitely supported fun tions in. C(R2 ). C α (R). the subset of. α-times. ontinuously dierentiable fun tions in. C(R). C α (R2 ). the subset of. α-times. ontinuously dierentiable fun tions in. C(R2 ). C0α (R). the subset of nitely supported fun tions in. C0 (R). C0α (R2 ) X X. the subset of nitely supported fun tions in. C0 (R2 ). and. the summations. j. j. X j. and. the summation. X. j∈Z2. j∈Z. i,j. sup. X. X. (i,j)∈Z2 and. sup x. the suprema over all. 2×2. j ∈ Z2 and over all x ∈ R2. I. the. M. dilation matrix, i.e. a. a. re nement mask in. Π. the spa e of all polynomials with omplex variables. Πk. the subspa e of. A. mask symbol asso iated with the re nement mask. identity matrix. Π. 2×2. invertible matrix with integer entries. M0 (Z2 ). onsisting of polynomials of degree at most. Laurent polynomial. X. k ∈ Z+. a ∈ M0 (Z2 ),. i.e. the. ai,j z1i z2j. i,j. φ. re nable fun tion, i.e. a fun tion satisfying the re nement equation. φ=. X j. aj φ(M · −j) δ0 = 1 and δj = 0 for j = 6 0 i jT = j for j = (i, j). δ. the delta sequen e de ned by. jT. the transpose of. j ∈ Z2 , i.e..
(12) xi. LIST OF SYMBOLS. Sa. the subdivision operator mapping. (Sa c)j = Sar. X k. the sequen e. k · k∞. the uniform norm in. c ∈ M(Z2 ),. applied. Sar c,. r -times,. with the onvention that. and. c ∈ M(Z2 ). where. M(Z2 ). and in. kf k∞ = sup|f (x)| x. M(R2 ),. for. i.e.. kck∞ = sup|cj | j. X j. f ∈ M(R2 ) Sa. with initial. f ∈ M(R2 ). to. Ta f ∈ M(R2 ),. with. aj f (M · −j). the as ade operator. Ta0. for. c ∈ M(Z2 ). the as ade operator mapping. Ta f = Tar. Sa. the limit fun tion of a onvergent subdivision s heme sequen e. Ta. with. is the identity operator. c(r). Sa∞ c. Sa c ∈ M(Z2 ),. to. aj−M kT ck , j ∈ Z2. the subdivision operator. Sa0. c ∈ M(Z2 ). Ta. applied. r -times,. with the onvention that. is the identity operator. M(R2 ). g. an initial fun tion in. fr. the fun tion. Ta∞ g. the limit fun tion of a onvergent as ade algorithm. for the as ade algorithm. Tar g , r ∈ Z+. g ∈ C0 (R2 ) −r T. dyadi set M j : j ∈ Z2 , r ∈ Z+. Ta. with initial. fun tion. D. the. φ˜ · ψ˜. the tensor produ t of the univariate fun tions bivariate fun tion. whi h is dense in. φ˜ and ψ˜,. ˜ ψ(y) ˜ , (x, y) ∈ R2 (x, y) 7→ φ(x). R2. i.e. the.
(13) Introdu tion A. re nable fun tion,. or a fun tion expressible as a linear ombination of the shifts of its. own dilations by a fa tor of a. dilation matrix, i.e.. is always linked to a ertain sequen e alled the orresponds to a Laurent polynomial alled the. an invertible matrix with integer entries,. re nement mask. mask symbol,. are the elements of the re nement mask. The ardinal. The re nement mask. the oe ients of whi h. B -spline. fun tions presented in. [dV07℄ are among the rst examples of univariate re nable fun tions whi h have enormous appli ations in wavelet analysis and approximation theory.. In general, it is hard to investigate whether a given fun tion is re nable, sin e both the asso iated re nement mask, as well as the orresponding the dilation matrix have to be found. It is thus better to start with a given dilation matrix and a nitely supported sequen e, and investigate the existen e of a orresponding re nable fun tion.. Based on a given dilation matrix and a nitely supported sequen e, the asso iated. subdivision s heme is de ned as an operator whi h re ursively produ es denser and denser data points by means of linear ombinations of the previous ones.. as ade algorithm. The orresponding. is also de ned as a fun tional operator whi h iteratively produ es a. sequen e of fun tions by means of linear ombinations of the previous ones.. Subdivision methods, as initialy introdu ed by de Rham (1956) and later by Chaikin (1974), play important roles in omputer aided geometri design (CAGD) by generating urves and surfa es in omputer graphi s (see e.g.. 1. [Dyn92℄).. Cas ade algorithms, on.
(14) 2. INTRODUCTION. the other hand, are useful in the sense that as ade algorithm onvergen e implies the re nability of the limit fun tion.. Spe ializing only to the ase where the dilation matrix is. M = 2I ,. our goal in this. thesis is to give a purely algebrai method for the study of both bivariate re nable fun tions and their asso iated subdivision s hemes, in ontrast to methods based on Fourier transforms as mostly en ountered in the literature. sis is that of. interpolatory. take the value. 1. A fundamental theme in this the-. bivariate re nable fun tions, that is, re nable fun tions that. at the origin and. 0. at all other integers.. We pro eed to introdu e in. Chapter 1 a brief overview of interpolatory re nable fun tions. The orresponding re nement masks, alled. interpolatory masks,. and the asso iated. are respe tively hara terized by (1.8) and (1.10).. interpolatory mask symbols. We refer to the Dubu -Deslauriers. interpolatory re nable fun tion, as investigated in [VGH03℄ (see also [Hun05, Goo00℄) for the univariate setting, and to the interpolatory re nable fun tions onstru ted in [RS97℄ (see also [Jia00℄) for the multivariate ase.. Several studies of re nement masks have been developed by using the asso iated mask symbols, whi h often help to prove the onvergen e of the subdivision s hemes to whi h they are asso iated (e.g. [DL02, pages 37-70℄, [CDM91℄). Motivated by this perspe tive, we take a spe ial interest in interpolatory mask symbols for the spe ial ase where the dilation matrix is. 2I .. In Chapter 2, an alternative riterion to interpolatory mask symbols. whi h is easier to use than (1.10) is given. In Theorem 2.2.3, we dedu e the general form of an interpolatory mask symbol by using some polynomial identities and the. algorithm.. Eu lidean. The results thus obtained are then applied to the mask symbols orresponding. to the well-known box splines.. An interpolatory re nement mask generates an. interpolatory subdivision s heme, that. is, a subdivision s heme for whi h the initial data points are preserved at all the steps of the re ursive pro ess (see [Dyn92℄). This is extremely relevant in ertain appli ation.
(15) INTRODUCTION. 3. areas in CAGD, where the initial data are required to be preserved while applying the subdivision pro ess. In Chapter 3, we dis uss the onvergen e of interpolatory subdivision s hemes, and we investigate in Se tion 3.3 the issue of property preservation with respe t to the as ade algorithm.. Though remarkable progress by mathemati ians in the area have been made, omputationally ine ient onditions are still often applied to re nement masks in order to ensure the onvergen e of the asso iated subdivision s hemes. For instan e, the hara terization by using the joint spe tral radius for subdivision s hemes investigated in [HJ98a℄ an take impra ti ally long to test omputationally, whereas the alternative method based on ontra tivity onditions, as introdu ed in [DL02℄ (see also [Dyn02℄), an also be a formidable omputational task to perform. Under ertain restri tions, we therefore develop in Chapter 4 three feasible methods to examine the existen e of interpolatory re nable fun tions from a pra ti al point of view. The presented methods are applied on interpolatory mask symbols, and are based on the results of Mi helli in [Mi 96℄ and on tensor produ ts.. Unfortunately, for the general setting, the existing methods investigating the existen e of interpolatory re nable fun tions are not always feasible to implement. By using the above-mentioned general form of an interpolatory mask symbol, an interesting ontinuation of this thesis thus in lude nding easily he kable su ient onditions on interpolatory mask symbols for them to omply with the onditions of the existing methods..
(16) Chapter 1 Interpolatory bivariate re nable fun tions We rst give in this hapter a brief introdu tion to interpolatory bivariate re nable fun tions and the orresponding interpolatory masks. Then, we elaborate a simple riterion in (1.8) and in (1.10) to re ognize simultaneously an interpolatory mask and the asso iated interpolatory mask symbol. We end the hapter by presenting the box splines as examples of interpolatory bivariate re nable fun tions.. 1.1 Notation and general on epts We shall denote the set of natural numbers by integers respe tively by numbers by. C.. Z. and. Z+ ,. N,. the set of integers and non-negative. the set of real numbers by. Similarly, the symbols. Z2 , R2. and. C2. R. and the set of omplex. denote the set of ordered pairs with. respe tively integer, real number and omplex number entries.. For the linear spa e. M(Z2 ). support is denoted by supp(c). of all real-valued sequen es. := {j ∈ Z2 : cj 6= 0},. 4. c = {cj ∈ R :. j ∈ Z2 } whi h. the subspa e of nitely supported.
(17) CHAPTER 1.. 5. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. sequen es, i.e. whose supports are nite, onstitute a linear subspa e denoted by In the same way, for the linear spa e. M(R2 ) of all real-valued bivariate fun tions f. whi h support supp(f ) is the smallest losed set ontaining. {x ∈ R2 : f (x) 6= 0},. of nitely supported fun tions onstitute a linear subspa e denoted by the subspa es of ontinuous fun tions respe tively in by. C(R2 ). and. M(R2 ). and in. M0 (R2 ).. M0 (R2 ). on. R2. the set. Moreover,. are denoted. C0 (R2 ).. For a given termed. M0 (Z2 ).. 2×2. M -re nable. invertible matrix. M. if there exists a sequen e. φ=. X j. We shall refer to. M. with integer entries, a fun tion. as the. a = {aj : j ∈ Z2 } ∈ M0 (Z2 ). φ ∈ M0 (R2 ). is. su h that. aj φ(M · −j).. (1.1). dilation matrix, whereas the sequen e a is alled the re nement. mask (or simply the mask), and the equation (1.1) is referred to as the re nement equation. Note that an. M -re nable. fun tion is therefore expressible as a linear ombinations of. the shifts of its own dilations with the fa tor of the dilation matrix re nement mask. a.. M,. as spe i ed by the. For onvenien e, we shall often simplify M -re nable to re nable.. The problem of existen e of re nable fun tions by using re nement masks is fundamental, but most importantly in this thesis, is that our study is fo ussed on. interpolatory. re nable fun tions, that is, re nable fun tions that satisfy. φ(j) = δj ,. where the delta fun tion. δ. j ∈ Z2 ,. (1.2). (also alled the delta sequen e) is de ned by. δj =. 1, 0,. j = 0, j 6= 0,. , j ∈ Z2 .. (1.3).
(18) CHAPTER 1.. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. 6. In other words, a re nable fun tion is interpolatory if it vanishes at all integers ex ept at the origin. 0 ∈ Z2. 1.. where it takes the value. interpolatory re nement masks. We pro eed to hara terize the so- alled. asso iated with interpolatory re nable fun tions.. 1.2 Interpolatory re nement masks We present in this se tion a hara terization theory for re nement masks asso iated with interpolatory re nable fun tions. Thereafter we introdu e the on ept of re nement mask symbols and then spe ialize to the ase. M = 2I ,. with some examples of bivariate inter-. polatory re nable fun tions.. By using the symbol. jT. for the transpose of the integer pair. j ∈ Z2 , we ome rst to. the following result.. Proposition 1.2.1. For a given dilation matrix M and a mask a ∈ M0 (Z2 ), suppose the re nement equation (1.1) holds for a re nable fun tion φ. If φ is interpolatory, then a satis es aM jT = δj ,. Proof.. From (1.2) and (1.1), we have that, for. δj = φ(j) =. X k. j. ∈ Z2 .. j ∈ Z2 ,. ak φ(M jT − k) =. Our next result was proved for the ase. (1.4). X k. M = 2I. ak δM jT −k = aM jT .. in [CDM91℄. Our general proof is. based on a suggestion in [HJ98a℄.. Proposition 1.2.2. For a given dilation matrix M and a mask a ∈ M0 (Z2 ), suppose the re nement equation (1.1) holds for a re nable fun tion φ. If φ is nitely supported and.
(19) CHAPTER 1.. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. 7. integrable with non-zero integral over R2 , then a satis es X. aj = |det(M)|.. (1.5). j. Proof.. Suppose that the dilation matrix has the form. . . c d M = . e f Writing. ai,j = aj ,. we an now integrate the re nement equation (1.1) to obtain. Z Z. φ(x, y)dxdy = R2. X. ai,j.
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(27). φ(M(x, y)T − (i, j))dxdy.. (1.6). R2. i,j. Sin e the variable transformation. Z Z. (X, Y )T = M(x, y)T. ∂X ∂x. ∂X ∂y. ∂Y ∂x. ∂Y ∂y.
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(42). has Ja obian.
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(47) = det(M),
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(49). it follows from standard multivariate integration theorems in analysis that. Z Z. T. φ(M(x, y) − (i, j))|det(M)|dxdy = R2. Z Z. φ((X, Y ) − (i, j))dX dY. R2. =. Z Z. φ(X, Y )dX dY .. R2. We then dedu e from (1.6) and (1.7) that. Z Z. φ(x, y)dxdy = R2. X i,j. 1 ai,j |det(M)|. Z Z. φ(x, y)dxdy. R2. (1.7).
(50) CHAPTER 1.. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. Moreover, sin e we assume the integral of. X. ai,j. i,j. φ. to be non-zero over. R2 ,. 8. we obtain. 1 = 1, |det(M)|. from whi h the result (1.5) follows.. Therefore, given a dilation matrix latory re nable fun tion mask. a. φ. M,. the existen e of a ompa tly supported interpo-. with non-zero integral over. R2. requires for a given re nement. to satisfy the onditions. 2 aM jT = δj , j ∈ Z , X aj = |det(M)|. . (1.8). j. Now, onsidering a re nement mask. a = {aj } = {ai,j },. we de ne the orresponding. re nement mask symbol, or simply the mask symbol, as the bivariate Laurent polynomial A. given by. A(z1 , z2 ) =. X. ai,j z1i z2j , z1 , z2 ∈ C \ {0}.. (1.9). i,j. Also, we say that a re nement mask brevity, we all. a is interpolatory. a an interpolatory mask.. if it satis es (1.8). In that ase, for. Moreover, its symbol. A is alled an interpolatory. mask symbol. Sin e, a ording to (1.9), re nement masks and their symbols are bije tively linked, the restri tions (1.8) on a mask symbol. A . a. an equivalently be expressed in terms of the mask. as follows:. The onstant term in. A(z1 , z2 ). is. 1,. and. (α1 , α2 ) = M(i, j)T 6= (0, 0) X A(1, 1) = ai,j = |det(M)|. su h that. i,j. A. has no term in. for some. (i, j) ∈ Z2 ;. z1α1 z2α2 also,. (1.10).
(51) CHAPTER 1.. 9. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. It is often onvenient to use re nement mask symbols instead of their orresponding re nement masks.. Indeed, as presented in [CDM91, Mi 96, Der99℄, some properties of. masks symbols lead to the existen e of ompa tly supported re nable fun tions.. The following se tion presents some examples of interpolatory re nable fun tions with dilation matrix. M = 2I .. 1.3 Box splines In this se tion, we x the dilation matrix tory mask. a. M = 2I .. The onditions (1.8) on an interpola-. an then be re-written as. . a2i,2j = δ(i,j) , (i, j) ∈ Z2 , X ai,j = 4, i,j. whereas the onditions (1.10) on an interpolatory mask symbol. . The onstant term in. A(z1 , z2 ). z12α1 z22α2 , for X A(1, 1) = ai,j = 4. no term in. any. is. 1,. and. A. (1.11). A. be ome. has. (α1 , α2 ) ∈ Z2 \ {(0, 0)};. also,. (1.12). i,j. The box spline N1 The box spline fun tion. N1. is de ned by. N1 (x, y) =. The graph of. N1. 1, (x, y) ∈ [0, 1)2 ,. / [0, 1)2 . 0, (x, y) ∈. is shown in Figure 1.1 (b), from whi h we see that. (1.13). N1. is nitely.
(52) CHAPTER 1.. 10. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. 1. 1. 0.8 0.6. 0.5. 0.4 0.2 0 0. 1 1 0. (a) Support of. a(1). 0. (b) Graph of Figure 1.1:. N1. The box spline N1. supported, and though it is not ontinuous, we laim that fun tion with respe t to the interpolatory mask. a(1). N1. is an interpolatory re nable. whi h support is delimitated by the. dotted lines in Figure 1.1 (a), as given by. (1). (1). (1). (1). (1). a0,0 = a0,1 = a1,0 = a1,1 = 1; ai,j = 0. To prove this, observe rst that, for. N1 (2x, 2y) =. otherwise.. x, y ∈ R,. 1, (x, y) ∈ [0, 21 )2 ,. / [0, 1)2 ; 0, (x, y) ∈ 1, (x, y) ∈ [ 21 , 1) × [0, 21 ), N1 (2x − 1, 2y) = / [0, 1)2; 0, (x, y) ∈ 1, (x, y) ∈ [0, 21 ) × [ 21 , 1), N1 (2x, 2y − 1) = / [0, 1)2; 0, (x, y) ∈. (1.14).
(53) CHAPTER 1.. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. N1 (2x − 1, 2y − 1) =. the unit square. [0, 1)2 ,. 1, (x, y) ∈ [ 21 , 1)2, / [0, 1)2 . 0, (x, y) ∈. [0, 21 )2 , [ 21 , 1) × [0, 21 ), [0, 12 ) × [ 21 , 1). Then, sin e the squares. we obtain, for. and. [ 12 , 1)2. form a partition of. (x, y) ∈ R2 ,. N1 (x, y) = N1 (2x, 2y) + N1 (2x − 1, 2y) + N1 (2x, 2y − 1) + N1 (2x − 1, 2y − 1),. thereby proving that. N1. 11. is re nable with orresponding mask. a(1). a ording to (1.14) and (1.9), the orresponding mask symbol. A1. (1.15). given in (1.14). Hen e, is given by. A1 (z1 , z2 ) = 1 + z1 + z2 + z1 z2 = (1 + z1 )(1 + z2 ), z1 , z2 ∈ C.. (1.16). Note that the onditions (1.11) and (1.12) are respe tively ful lled by the re nement mask. a(1) N1. and its symbol. A1 .. Moreover, (1.13) shows that. N1 (j) = δj , j ∈ Z2 ,. whi h means that. is an interpolatory re nable fun tion.. The box spline N2 Using the box spline. N1. given in (1.13), the box spline fun tion. N2 (x, y) =. Z. N2. is de ned by. 1. N1 (x − t, y − t)dt, x, y ∈ R.. (1.17). 0. Let us rst prove that. N2. end, observe that, for. t ∈ (0, 1). is a ontinuous fun tion by nding its expli it formula. To this and. x, y ∈ R,. N1 (x − t, y − t) 6= 0 ⇐⇒ x − t ∈ [0, 1) =⇒ 0 < x < 2. and. and. y − t ∈ [0, 1). 0 < y < 2.. (1.18).
(54) CHAPTER 1.. 12. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. Hen e, from (1.18) and (1.17), we dedu e that. Moreover, for. x, y ∈ [0, 2),. N2 (x) = 0,. x ∈/ [0, 2]2 .. we have. 0 ≤ x − t < 1 ⇐⇒ x − 1 < t ≤ x. and. 0 ≤ y − t < 1 ⇐⇒ y − 1 < t ≤ y,. whi h, together with (1.17), yields. N1 (x − t, y − t) 6= 0 ⇐⇒ t ∈ (0, 1) ∩ (x − 1, x] ∩ (y − 1, y], x, y ∈ [0, 2).. (1.19). We then have the following result.. Proposition 1.3.1. The box spline N2 , as de ned in (1.17), is expli itly given by. N2 (x, y) =. min{x, y}, 2 − max{x, y},. 1 + min{x, y} − max{x, y}, 0. if (x, y) ∈ [0, 1)2 , if (x, y) ∈ [1, 2)2 , (1.20). if (x, y) ∈ ∆, otherwise,. where ∆ is the set de ned by. ∆ = {(x, y) : min{x, y} ∈ [0, 1); max{x, y} ∈ [1, 2); 1 + min{x, y} ≥ max{x, y}},. (1.21). i.e., ∆ = B ∪ E,. with B and E as in Figure 1.2. Consequently, the support of N2 is the polygon A ∪ B ∪ C ∪ D ∪ E ∪ F = [0, 1]2 ∪ ∆ ∪ [1, 2]2 in Figure 1.2.. Proof.. Observe from Figure 1.2 that. [0, 1)2 = A ∪ F , [1, 2)2 = C ∪ D. and. ∆ = B ∪ E..
(55) CHAPTER 1.. Figure 1.2:. Support of the box spline N2 .. Therefore, from (1.19), we have that, for. •. If. y ∈ [0, 1). that. If. N2 (x, y) =. y ∈ [1, 2), ◦ ◦. If. If. y ≤ x (resp. y ≥ x), then t ∈ [0, y] Z x y dt = y resp. N2 (x, y) = dt = x .. Z. If. ◦. If. t ∈ [0, x]),. so. two ases o ur:. y − 1 > x,. then. y − 1 ≤ x,. y ∈ [0, 1),. (resp.. 0. then. t∈∅. and. N2 (x, y) = 0;. t ∈ (y − 1, x]. Similarly, from (1.19), we have that, for. •. x ∈ [0, 1):. is su h that. 0. •. 13. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. and therefore. N2 (x, y) =. Z. x dt. = 1 + x − y.. dt. = 1 + y − x;. y−1. x ∈ [1, 2):. two ases o ur:. y > x − 1,. then. t ∈ (x − 1, y]. and therefore. N2 (x, y) =. Z. y. x−1. ◦ •. If. If. y ≤ x − 1,. y ∈ [1, 2) is su h. so that. N2 (x, y) =. then. that. Z. 1. x−1. t∈∅. and. N2 (x, y) = 0.. y ≤ x (resp. y ≥ x), then t ∈ (x − 1, 1] (resp. t ∈ (y − 1, 1]), Z 1 dt = 2 − x resp. N2 (x, y) = dt = 2 − y . y−1. By taking the appropriate ombination of the four ases above, we obtain the desired result (1.20)..
(56) CHAPTER 1.. 14. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. Next, by using Proposition 1.3.1 and Figure 1.2, we dedu e that the restri tions of to the respe tive regions. A, B, C, D, E. and. F. N2. are given as follows:. ⋄. In the region. A: x, y ∈ [0, 1),. with. y ≥ x,. we have. N2 |A (x, y) = x;. ⋄. In the region. F : x, y ∈ [0, 1),. with. y ≤ x,. we have. N2 |F (x, y) = y ;. ⋄. In the region. B : x ∈ [0, 1). and. y ∈ [1, 2),. with. x ≥ y − 1,. we have. N2 |B (x, y) =. E : x ∈ [1, 2). and. y ∈ [0, 1),. with. y ≥ x − 1,. we have. N2 |E (x, y) =. 1 + x − y; ⋄. In the region. 1 + y − x; ⋄. In the region. C : x, y ∈ [1, 2),. with. y ≥ x,. we have. N2 |C (x, y) = 2 − y ;. ⋄. In the region. D : x, y ∈ [1, 2),. with. x ≥ y,. we have. N2 |D (x, y) = 2 − x.. Hen e,. F.. N2. de nes a dierent plane in ea h of the respe tive regions. It will therefore su e to prove the ontinuity of. along the lines. x = 0, x = 1, x = 2,. y = x, y = x + 1. and. the lines. y → 2),. we also have that. (resp.. have that. x → 1. (resp.. N2 |B (x, y) → 2 − y. N2 |D (x, y) → 2 − x),. (resp.. D. F ),. when. x→0. C ),. (resp.. (resp.. when. y → 0),. x → 2. (resp.. N2 (x, y) → 0. (resp.. N2 |A (x, y) → x. F ∪E. A. Similarly, for the region. Next, observe that, when. ous in the region. as well as the lines. y = x − 1.. N2 (x, y) → 0.. N2 |E (x, y) → y. and. at the edges of these regions, i.e. y = 0, y = 1, y = 2,. To this end, observe rst that, for the region we have that. N2. A, B, C, D, E. so that. A ∪ B ).. and. N2. and. y → 1),. we have. N2 |B (x, y) → x),. Similarly, when. N2 |C (x, y) → 2 − y. N2 |F (x, y) → y. so that. x → 1. (resp.. is also ontinuous in the region. N2. (resp.. is ontinu-. y → 1),. N2 |E (x, y) → 2 − x B∪C. (resp.. and. we and. E ∪ D )..
(57) CHAPTER 1.. Finally, along the line. N2 |D (x, y), y = x+1. 15. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. so that. N2. y = x,. N2. we on lude that. we have that. is ontinuous on. We pro eed now to prove that. N2. A∪F. N2 |B (x, y) = 0. and. C ∪ D.. and. (resp.. N2 |C (x, y) =. Along the line. N2 |E (x, y) = 0).. Thus,. R2 . is re nable. From the re nement equation (1.15),. x, y ∈ R,. we have that, for. N2 (x, y) =. N2 |A (x, y) = N2 |F (x, y). is ontinuous in the regions. y = x − 1),. (resp.. we have that. Z. 1. N1 (x − t, y − t)dt. 0. =. Z. 1. [N1 (2x − 2t, 2y − 2t) + N1 (2x − 2t − 1, 2y − 2t). 0. + N1 (2x − 2t, 2y − 2t − 1) + N1 (2x − 2t − 1, 2y − 2t − 1)] dt.. Using the transformations. t˜ = 2t for t ∈ [0, 21 ] and t˜ = 2t−1 for t ∈ [ 21 , 1], the rst integral x, y ∈ R,. in (1.22) an be re-written, for. Z. 1. N1 (2x − 2t, 2y − 2t)dt =. Z. 1 2. as. N1 (2x − 2t, 2y − 2t)dt +. 0. 0. (1.22). Z. 1 1 2. N1 (2x − 2t, 2y − 2t)dt. Z Z 1 1 1 1 N1 (2x − t, 2y − t)dt + N1 (2x − t − 1, 2y − t − 1)dt = 2 0 2 0 1 1 = N2 (2x, 2y) + N2 (2x − 1, 2y − 1), (1.23) 2 2. by virtue of the de nition of. N2. in (1.17). Similarly, we get, for. x, y ∈ R,. 1. 1 1 N1 (2x − 2t − 1, 2y − 2t)dt = N2 (2x − 1, 2y) + N2 (2x − 2, 2y − 1), 2 2 0 Z 1 1 1 N1 (2x − 2t, 2y − 2t − 1)dt = N2 (2x, 2y − 1) + N2 (2x − 1, 2y − 2), 2 2 0 Z 1 1 1 N1 (2x − 2t − 1, 2y − 2t − 1)dt = N2 (2x − 1, 2y − 1) + N2 (2x − 2, 2y − 2). 2 2 0 Z. (1.24). (1.25). (1.26).
(58) CHAPTER 1.. 16. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. Consequently, from (1.22), (1.23), (1.24), (1.25) and (1.26), we obtain. 1 N2 (x, y) = {N2 (2x, 2y) + N2 (2x − 1, 2y) + N2 (2x, 2y − 1) + 2N2 (2x − 1, 2y − 1) 2 + N2 (2x − 1, 2y − 2) + N2 (2x − 2, 2y − 1) + N2 (2x − 2, 2y − 2)} ,. whi h shows that. N2. is re nable with orresponding mask. a(2). given by. (2) (2) (2) (2) (2) (2) 1 a(2) 1,1 = 1, a0,0 = a0,1 = a1,0 = a2,1 = a1,2 = a2,2 = 2 ,. (2) / {(0, 0), (0, 1), (1, 0), (1, 1), (1, 2), (2, 1), (2, 2)}, ai,j = 0, (i, j) ∈ A2. a ording to whi h the orresponding mask symbol. A2 (z1 , z2 ) = (1 + z1 )(1 + z2 ). However, observe from (1.28) that term in. A2 (z1 , z2 ). is not. 1. . (2). (1.27). (1.28). is given by. 1 + z1 z2 2. . , z1 , z2 ∈ C.. (1.29). (2). a0,0 6= 1 and a2,2 6= 0 (or, equivalently, the onstant. and it has a term in. z12 z22 ),. that is,. N2. is not interpolatory.. The shifted box spline N˜2 Using the box spline. N2. de ned in (1.17), we de ne the shifted box spline fun tion. ˜2 (x, y) = N2 (x + 1, y + 1), x, y ∈ R. N. We laim that the fun tion. ˜2 N. by. (1.30). ˜2 , as drawn in Figure 1.3 (b), is an interpolatory re nable N. fun tion asso iated with the interpolatory mask. a ˜(2). whi h support is delimitated by the.
(59) CHAPTER 1.. 17. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0. 1. 0.5 1 0 0. -1. -1. 0 1. (a) Support of. a ˜(2). (b) Graph of Figure 1.3:. ˜2 N. The shifted box spline N˜2. dotted lines in Figure 1.3 (a), as given by. (2) (2) (2) (2) (2) (2) a˜(2) ˜1,1 = a˜0,1 = a ˜1,0 = a ˜−1,0 = a˜0,−1 = a ˜−1,−1 = 21 , 0,0 = 1, a. (2) / {(0, 0), (0, 1), (1, 0), (−1, 0), (0, −1), (1, 1), (−1, −1)}, a˜i,j = 0, (i, j) ∈ with orresponding mask symbol. A˜2. A˜2 (z1 , z2 ) = (1 + z1 )(1 + z2 ). (1.31). given by. . 1 + z1 z2 2. . z1−1 z2−1 , z1 , z2 ∈ C \ {0}.. To prove this, we use (1.30) and (1.27) to dedu e that, for. (1.32). x, y ∈ R,. ˜2 (x, y) =N2 (x + 1, y + 1) N 1 = {N2 (2x + 2, 2y + 2) + N2 (2x + 1, 2y + 2) + N2 (2x + 2, 2y + 1) 2 + 2N2 (2x + 1, 2y + 1) + N2 (2x + 1, 2y) + N2 (2x, 2y + 1) + N2 (2x, 2y)}.
(60) CHAPTER 1.. 18. INTERPOLATORY BIVARIATE REFINABLE FUNCTIONS. 1n˜ ˜2 (2x, 2y + 1) + N ˜2 (2x + 1, 2y) = N2 (2x + 1, 2y + 1) + N 2. o ˜2 (2x, 2y) + N ˜2 (2x, 2y − 1) + N ˜2 (2x − 1, 2y) + N ˜2 (2x − 1, 2y − 1) , + 2N. whi h implies that. ˜2 N. is a re nable fun tion with re nement mask. a ˜(2). (1.33). given by (1.31).. Moreover, by using (1.31) and (1.9), we nd that the orresponding mask symbol given by (1.32). It an now be veri ed from (1.31) and (1.32) that. a˜(2). and. A˜2. A˜2. is. satisfy. respe tively the interpolatory onditions (1.11) and (1.12).. To prove that. ˜2 N. is interpolatory, we use (1.30) and (1.17) to obtain, for. ˜2 (x, y) = N2 (x + 1, y + 1) = N. Z. 1. N1 (x + 1 − t, y + 1 − t)dt.. whereas, for. (i, j) 6= (0, 0),. Z. (1.34). 0. Taking into a ount the de nition of the box spline. ˜2 (0, 0) = N. x, y ∈ R,. N1. 1. N1 (1 − t, 1 − t)dt = 0. in (1.13), we dedu e that. Z. 1. 1dt = 1,. (1.35). 0. we have that. ˜2 (i, j) = N. Z. 1. N1 (i + 1 − t, j + 1 − t)dt = 0,. (1.36). 0. for if. i 6= 0 (resp. j 6= 0) then i+ 1 −t ∈ / [0, 1) (resp. j + 1 −t ∈ / [0, 1)), for any t ∈ (0, 1).. It. follows from (1.35) and (1.36) that the interpolatory ondition (1.2) is satis ed, thereby showing that the shifted box spline. ˜2 N. is an interpolatory re nable fun tion.. Note in parti ular from Figure 1.3 (b) that. ˜2 N. belongs to. C0 (R2 ) \ C01 (R2 )..
(61) Chapter 2 The interpolatory mask symbols for M = 2I We x the dilation matrix. M = 2I. in this hapter. In Se tion 2.1 below, we produ e the. alternative riterion (2.9) for interpolatory mask symbols. In Se tion 2.2, after solving some polynomial identities by means of the well-known Bezout identity and the Eu lidean algorithm, we provide in Theorem 2.2.3 a useful hara terization result for interpolatory mask symbols. In Se tion 2.3, we spe ialise to the ase of box splines interpolatory mask symbols.. 2.1 Simple hara terization We pro eed to establish an alternative hara terization to interpolatory mask symbols whi h is simpler to use than (1.12), and whi h will be used in Se tion 2.2. Re all from Chapter 1 that the lass of interpolatory mask symbols onsists of all Laurent polynomials. 19.
(62) CHAPTER 2.. A. THE INTERPOLATORY MASK SYMBOLS FOR. 20. M = 2I. satisfying the onditions (1.12), i.e.. . where. a. The onstant term in no term in. z12α1 z22α2 ,. A(z1 , z2 ) for any. is. 1,. and. A. has. (α1 , α2 ) ∈ Z2 \ (0, 0);. also,. (2.1). A(1, 1) = 4, is the orresponding interpolatory mask, as de ned by (1.9), and satisfying the. onditions (1.11), i.e.. = δ(i,j) , (i, j) ∈ Z2 , a2i,2j X ai,j = 4. . (2.2). i,j. Let us denote by. F ∩G. F ⊔G. EE , EO , OE. is empty, whereas. respe tively integers. Z2. the union of two sets and. even-even, even-odd, odd-even. OO. and. F. and. G. for whi h the interse tion. stand for the sets of integer pairs with. odd-odd. entries. Observe that the set of. onsists of the union of the four disjoint subsets. EE , EO , OE. and. Z2 = EE ⊔ EO ⊔ OE ⊔ OO.. Given a mask symbol and (1.9) that, for. A. with orresponding mask. OO ,. i.e.. (2.3). a ∈ M0 (Z2 ),. we obtain from (2.3). z1 , z2 ∈ C \ {0},. A(z1 , z2 ) =. X. a2i,2j z12i z22j +. i,j. +. X i,j. X. a2i,2j+1 z12i z22j+1 +. i,j. a2i+1,2j+1 z12i+1 z22j+1 ,. X. a2i+1,2j z12i+1 z22j. i,j. (2.4).
(63) CHAPTER 2.. whereas also, by repla ing. A(−z1 , z2 ) =. X. z1. by. −z1. in (2.4), we have, for. a2i,2j z12i z22j +. i,j. −. X. z1 , z2 ∈ C \ {0},. a2i,2j+1 z12j z22j+1 −. i,j. X. 21. M = 2I. THE INTERPOLATORY MASK SYMBOLS FOR. X. a2i+1,2j z12i+1 z22j. i,j. a2i+1,2j+1 z12i+1 z22j+1 .. (2.5). i,j. Combining (2.4) and (2.5), we obtain, for. A(z1 , z2 ) + A(−z1 , z2 ) = 2. X. z1 , z2 ∈ C \ {0},. a2i,2j z12i z22j + 2. i,j. Now repla e. z1. by. −z1. and. z2. by. −z2. X. a2i,2j+1 z12i z22j+1 .. in (2.6) to dedu e that, for. A(−z1 , −z2 ) + A(z1 , −z2 ) = 2. X. (2.6). i,j. a2i,2j z12i z22j − 2. i,j. X. z1 , z2 ∈ C \ {0},. a2i,2j+1 z12i z22j+1 .. (2.7). i,j. By adding (2.6) and (2.7), we obtain the identity. A(z1 , z2 ) + A(−z1 , z2 ) + A(z1 , −z2 ) + A(−z1 , −z2 ) = 4. X. a2i,2j z12i z22j , z1 , z2 ∈ C \ {0},. i,j. (2.8). whi h we an now use to prove the following hara terization result.. Theorem 2.1.1. Suppose that a is a re nement mask su h that. X. aj = 4. Then a is. j. interpolatory if and only if the orresponding mask symbol A, as de ned by (1.9), satis es the identity. A(z1 , z2 ) + A(−z1 , z2 ) + A(z1 , −z2 ) + A(−z1 , −z2 ) = 4, z1 , z2 ∈ C \ {0}.. Proof.. Suppose rst that. a. is interpolatory. From (2.2), sin e. X i,j. a2i,2j z12i z22j = 1,. a2i,2j = δi,j ,. (2.9). we have that.
(64) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 22. M = 2I. whi h, together with (2.8), implies that (2.9) holds.. Conversely, if (2.9) holds, we obtain from (2.8) that. X. a2i,2j z12i z22j = 1,. i,j. whi h proves that. a2i,2j = δi,j .. Therefore, (2.2) holds and. Note that, for a given re nement mask a hieved if the orresponding mask symbol positive integers. k1 , k2. a,. a. is interpolatory.. the ondition in the se ond line of (2.2) is. A satis es the identity (2.9), and if there exist. and a Laurent polynomial. B. su h that. A(z1 , z2 ) = (1 + z1 )k1 (1 + z2 )k2 B(z1 , z2 ), z1 , z2 ∈ C \ {0},. sin e then. A(−1, z2 ) = A(z1 , −1) = 0. A(1, 1) = 4. and thus the mask symbol. A. for any. z1 , z2 ∈ C \ {0},. (2.10). so that (2.9) yields. is interpolatory. Hen e the following result.. Corollary 2.1.2. For a Laurent polynomial A satisfying the identity (2.9), if there exists a Laurent polynomial B su h that (2.10) holds, then A is an interpolatory mask symbol. A. is an inter-. A(1, 1) = 4,. we only get. Note that the onverse of Corollary 2.1.2 is not ne essarily true, for if polatory mask symbol that sati es the identity (2.9), then sin e that. A(−1, 1) + A(1, −1) + A(−1, −1) = 0,. whi h does not ne essarily imply that. A. is of. the fa torized form (2.10).. Motivated by the result of Corollary 2.1.2, we pro eed to hara terize in Se tion 2.2 below the interpolatory mask symbols whi h are in the fa torized form (2.10)..
(65) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 23. M = 2I. 2.2 General form We pro eed to give the general form of interpolatory mask symbols that are fa torizable in the sense of (2.10). More pre isely, we start by solving for the Laurent polynomial. A. in the identity (2.9) with the help of the Bezout theorem, to nally establish a general formulation of interpolatory mask symbols.. To fa ilitate our investigation, we hen eforth assume that the mask symbol. A has. the. fa torized form. A(z1 , z2 ) = 22−k1 −k2 (1 + z1 )k1 (1 + z2 )k2 B(z1 , z2 ), z1 , z2 ∈ C \ {0},. for some integers. B(−1, z2 ) 6= 0 that. k1 , k2 ∈ N. and. A(1, 1) = 4.. and some Laurent polynomial. B(z1 , −1) 6= 0. for all. z1 , z2 ∈ C \ {0},. Also, we shall assume that. a ording to Corollary 2.1.2,. A. A. B. su h that. (2.11). B(1, 1) = 1,. so that, from (2.11), it holds. satis es the identity (2.9), in whi h ase,. is an interpolatory mask symbol.. Polynomial identities To hara terize the mask symbol. A,. we rst prove the following result.. Lemma 2.2.1. Let k1 , k2 ∈ N and suppose α1 , α2 are two odd integers in N. Then: (a) if α1 < 2k1 , there exists a polynomial S1 whi h is odd in z2 , with degree α2 in z2 , and degree less than k1 in z1 , su h that the general Laurent polynomial solution K1 of the identity (1 + z1 )k1 K1 (z1 , z2 ) − (1 − z1 )k1 K1 (−z1 , z2 ) = z1α1 z2α2 , z1 , z2 ∈ C \ {0},. (2.12).
(66) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 24. M = 2I. is the Laurent polynomial given by K1 (z1 , z2 ) = S1 (z1 , z2 ) + T1 (z1 , z2 )(1 − z1 )k1 , z1 , z2 ∈ C \ {0},. (2.13). with T1 denoting an arbitrary even Laurent polynomial in z1 ; also, K1 is odd in z2 if and only if T1 is odd in z2 . (b) if α2 < 2k2 , there exists a polynomial S2 whi h is odd in z1 , with degree α1 in z1 , and degree less than k2 in z2 , su h that the general Laurent polynomial solution K2 of the identity (1 + z2 )k2 K2 (z1 , z2 ) − (1 − z2 )k2 K2 (z1 , −z2 ) = z1α1 z2α2 , z1 , z2 ∈ C \ {0},. (2.14). is the Laurent polynomial given by K2 (z1 , z2 ) = S2 (z1 , z2 ) + T2 (z1 , z2 )(1 − z2 )k2 , z1 , z2 ∈ C \ {0},. (2.15). with T2 denoting an arbitrary even Laurent polynomial in z2 ; also, K2 is odd in z1 if and only if T2 is odd in z1 . Proof.. (a) Sin e the two univariate polynomials. (1 + z1 )k1. and. (1 − z1 )k1. fa tor, there exist by the Bezout theorem two univariate polynomials. U1. have no ommon and. V1. (1 + z1 )k1 U1 (z1 ) + (1 − z1 )k1 V1 (z1 ) = 1, z1 ∈ C.. Multiplying both sides of (2.16) by. z1α1 z2α2. yields, for. su h that. (2.16). z1 , z2 ∈ C,. (1 + z1 )k1 [z1α1 z2α2 U1 (z1 )] + (1 − z1 )k1 [z1α1 z2α2 V1 (z1 )] = z1α1 z2α2 , z1 , z2 ∈ C.. (2.17).
(67) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 25. M = 2I. Using the polynomial division theorem, we dedu e the existen e of two polynomials and. R1. satisfying. z1α1 V1 (z1 ) = Q1 (z1 )(1 + z1 )k1 + R1 (z1 ), z1 ∈ C,. su h that the degree of by. α1. and. V1 .. R1. is less than. k1 ,. and where. Q1. and. R1. (2.18). are uniquely determined. It then follows from (2.17) that. ˜ 1 (z1 , z2 ) = z1α1 z2α2 , z1 , z2 ∈ C, (1 + z1 )k1 S1 (z1 , z2 ) + (1 − z1 )k1 R. where. ˜1 R. Q1. S1. is the polynomial de ned by. is the polynomial given by. z1. the degree in. of. S1. S1 (z1 , z2 ) = z1α1 z2α2 U1 (z1 ) + (1 − z1 )k1 z2α2 Q1 (z1 ), and. ˜ 1 (z1 , z2 ) = z α2 R1 (z1 ), R 2. is less than. k1 .. (2.19). z1 , z2 ∈ C.. for all. We laim that. To prove this, we rst note from (2.19) that. ˜ 1 (z1 , z2 ), z1 , z2 ∈ C, (1 + z1 )k1 S1 (z1 , z2 ) = z1α1 z2α2 − (1 − z1 )k1 R. a ording to whi h, sin e the degree of ne essarily have that the degree in. Repla ing. k1. z1. by. −z1. z1. ˜1 R. of. in. S1. z1. is less than. is less than. (1 − z1 ) [−S1 (−z1 , z2 )] + (1 + z1 ). h. and sin e. α1 < 2k1 ,. we. k1 .. in (2.19), and using the fa t that. k1. k1 ,. α1. is odd, we dedu e that. i ˜ −R1 (−z1 , z2 ) = z1α1 z2α2 , z1 , z2 ∈ C.. (2.20). Substra ting the identities (2.19) and (2.20) now yields. ˜ 1 (−z1 , z2 )] = −(1 − z1 )k1 [S1 (−z1 , z2 ) + R ˜ 1 (z1 , z2 )], z1 , z2 ∈ C, (1 + z1 )k1 [S1 (z1 , z2 ) + R. and thus. ˜ 1 (−z1 , z2 ) = M1 (z1 , z2 )(1 − z1 )k1 , z1 , z2 ∈ C, S1 (z1 , z2 ) + R. (2.21).
(68) CHAPTER 2.. for some polynomial (2.21) is less than. M1 .. k1 ,. Sin e the degree in. we ne essarily have. z1. 26. M = 2I. THE INTERPOLATORY MASK SYMBOLS FOR. of the polynomial in the left-hand-side of. M1 = 0. in (2.21), or, equivalently,. ˜ 1 (−z1 , z2 ), z1 , z2 ∈ C, S1 (z1 , z2 ) = −R. (2.22). ˜ 1 (z1 , z2 ) = −S1 (−z1 , z2 ), z1 , z2 ∈ C. R. (2.23). Using (2.19), (2.22) and (2.23), we nd that the polynomial. S1. satis es. (1 + z1 )k1 S1 (z1 , z2 ) − (1 − z1 )k1 S1 (−z1 , z2 ) = z1α1 z2α2 , z1 , z2 ∈ C,. whi h means that degree in Sin e. α2. z1. S1. less than. is a parti ular polynomial solution of the identity (2.12) with a. k1 .. Moreover, from (2.22), we see that. is odd, we on lude that. Now, let. K1. (2.24). S1. is odd in. z2 ,. S1 (z1 , z2 ) = −z2α2 R1 (−z1 ).. and that its degree in. z2. is. α2 .. denote the general Laurent polynomial solution of (2.12). Substra ting. (2.12) from (2.24), we obtain, for. z1 , z2 ∈ C \ {0},. (1 + z1 )k1 [K1 (z1 , z2 ) − S1 (z1 , z2 )] = (1 − z1 )k1 [K1 (−z1 , z2 ) − S1 (−z1 , z2 )] .. Sin e. (1 + z1 )k1. and. (1 − z1 )k1. exists a Laurent polynomial. T1. have no ommon fa tor, it follows from (2.25) that there satisfying. K1 (z1 , z2 ) − S1 (z1 , z2 ) = T1 (z1 , z2 )(1 − z1 )k1 , z1 , z2 ∈ C \ {0}.. Substituting (2.26) into (2.25) yields that. T1. is even in. z1 .. (2.25). T1 (z1 , z2 ) = T1 (−z1 , z2 ). Thus, we dedu e from (2.26) that. K1. for. (2.26). z1 , z2 ∈ C \ {0},. i.e. is given by. K1 (z1 , z2 ) = S1 (z1 , z2 ) + T1 (z1 , z2 )(1 − z1 )k1 , z1 , z2 ∈ C \ {0},. (2.27).
(69) CHAPTER 2.. where. T1. THE INTERPOLATORY MASK SYMBOLS FOR. is an arbitrary even Laurent polynomial in. Also, sin e. S1. is odd in. z2 ,. 27. M = 2I. z1 .. we get from (2.27) that, for. z1 , z2 ∈ C \ {0},. K1 (z1 , −z2 ) = S1 (z1 , −z2 ) + T1 (z1 , −z2 )(1 − z1 )k1 = −S1 (z1 , z2 ) + T1 (z1 , −z2 )(1 − z1 )k1 ,. whereas also, for. (2.28). z1 , z2 ∈ C \ {0},. −K1 (z1 , z2 ) = −S1 (z1 , z2 ) − T1 (z1 , z2 )(1 − z1 )k1 .. Substra ting the identities (2.28) and (2.29) gives, for. (2.29). z1 , z2 ∈ C \ {0},. K1 (z1 , −z2 ) + K1 (z1 , z2 ) = (1 − z1 )k1 [T1 (z1 , −z2 ) + T1 (z1 , z2 )],. from whi h it then immediately follows that. K1. is odd in. z2. if and only if. T1. is odd in. z2 .. (b) The proof is similar to (a).. The Eu lidean algorithm We present here a detailed method to ompute the polynomials. S1. and. S2. in Lemma 2.2.1. by using the Eu lidean algorithm.. Under the onditions of Lemma 2.2.1, with integers su h that also and. V1. α1 < 2k1 ,. k 1 , k 2 ∈ N,. and where. α1 , α2 ∈ N. are odd. we rst pro eed to nd the univariate polynomials. U1. su h that (2.16) holds.. From the polynomial division theorem, there exist univariate polynomials. q0 , q1. and.
(70) CHAPTER 2.. r 1 , r2. THE INTERPOLATORY MASK SYMBOLS FOR. su h that, for. 28. M = 2I. z1 ∈ C,. (1 + z1 )k1 =q0 (z1 )(1 − z1 )k1 + r1 (z1 ), (1 − z1 )k1 =q1 (z1 )r1 (z1 ) + r2 (z1 ),. deg(r1 ). deg(r2 ). <. < k1 ,. (2.30). deg(r1 ).. (2.31). Repeated appli ations of polynomial division then yield the existen e of univariate polynomials. qj , j = 2, . . . , n + 1. r1 (z1 ) = q2 (z1 )r2 (z1 ) + r3 (z1 ),. and. rj , j = 3, . . . , n + 2,. deg(r3 ). <. su h that, for. . deg(r2 ),. . ... rn−1 (z1 ) = qn (z1 )rn (z1 ) + rn+1 (z1 ),. deg(rn+1 ). ≥ 1,. rn (z1 ) = qn+1 (z1 )rn+1 (z1 ) + rn+2 (z1 ), rn+2 (z1 ) = c,. so that, by ba k substitution, it holds that, for. z1 ∈ C,. n ∈ N. a. r−1 (z1 ) = (1 + z1 )k1. and. r0 (z1 ) = (1 − z1 )k1 , z1 ∈ C. (1 + z1 )k1. (2.32). by. for. j = 0, . . . , n + 1. T3,−1 (z1 ) = T3,n+2 (z1 ) = 0,. otherwise,. {Ti,j (z1 ) : i = 0, 1, 2, 3; j = −1, 0, . . . , n + 2}. Ti,j+1 (z1 ) = Ti,j−1 (z1 ) − qj (z1 )Ti,j (z1 ), for. (1 − z1 )k1. c 6= 0,. as. and. a ommon fa tor, whi h is impossible sin e deg(rn+1 )≥. T3,j (z1 ) = qj (z1 ),. Observe that. (2.33). rn+1 (z1 ). by ba k substitution and by using (2.33),. Now de ne the polynomial sequen e. z1 ∈ C,. onstant, . rj+1 (z1 ) = rj−1(z1 ) − qj (z1 )rj (z1 ), j = 0, . . . , n + 1,. with. and. would have. 1.. i = 0, 1, 2. and. j = 1, . . . , n + 1, . (2.34).
(71) CHAPTER 2.. 29. M = 2I. THE INTERPOLATORY MASK SYMBOLS FOR. with also. T0,−1 (z1 ) = (1 + z1 ) , T1,−1 (z1 ) = 0, T2,−1 (z1 ) = 1, k1. (2.35). T0,0 (z1 ) = (1 − z1 )k1 , T1,0 (z1 ) = 1, T2,0 (z1 ) = 0. . (2.36). Observe from (2.34), (2.33) and the rst lines of (2.35) and (2.36) that then. T0,j (z1 ) = rj (z1 ), j = 1, 2, . . . , n + 2.. It follows that the matrix. −1 ≤ j ≤ n + 2,. T. onsisting of the polynomials. [ Ti,j (z1 ) ],. (2.37). for. 0≤i≤3. and. is given by. . k k r2 (z1 ) (1 + z1 ) 1 (1 − z1 ) 1 r1 (z1 ) 0 1 −q0 (z1 ) 1 + q1 (z1 )q0 (z1 ) T = 1 0 1 −q1 (z1 ) 0 q0 (z1 ) q1 (z1 ) q2 (z1 ) We laim that, for. .... rn+1 (z1 ). . rn+2 (z1 ). . . . T1,n+1 (z1 ) T1,n+2 (z1 ) . . . T2,n+1 (z1 ) T2,n+2 (z1 ) .... qn+1 (z1 ). 0. . . j = 1, . . . , n + 2,. (1 + z1 )k1 T2,j (z1 ) + (1 − z1 )k1 T1,j (z1 ) = rj (z1 ), z1 ∈ C.. We prove this by indu tion on. j.. (2.38). Observe rst from (2.34) (2.30) that (2.38) holds for.
(72) CHAPTER 2.. j = 1.. THE INTERPOLATORY MASK SYMBOLS FOR. Also, from (2.31), (2.30) and (2.34), we obtain, for. 30. M = 2I. z1 ∈ C,. r2 (z1 ) =(1 − z1 )k1 − q1 (z1 )r1 (z1 ) =(1 − z1 )k1 − q1 (z1 )[(1 + z1 )k1 − q0 (z1 )(1 − z1 )k1 ] =[−q1 (z1 )](1 + z1 )k1 + [1 + q1 (z1 )q0 (z1 )](1 − z1 )k1 =T2,2 (z1 )(1 + z1 )k1 + T1,2 (z1 )(1 − z1 )k1 ,. thereby proving that (2.38) holds for. j = 2.. Suppose now that (2.38) is true for both sides of (2.38) by. −qj (z1 ). j−1. and. j. with. j ∈ {2, . . . , n + 1}.. Multiplying. yields. (1 + z1 )k1 [−qj (z1 )T2,j (z1 )] + (1 − z1 )k1 [−qj (z1 )T1,j (z1 )] = −qj (z1 )rj (z1 ), z1 ∈ C.. (2.39). From the indu tive assumption, re all that. (1 + z1 )k1 T2,j−1 (z1 ) + (1 − z1 )k1 T1,j−1 (z1 ) = rj−1(z1 ), z1 ∈ C.. (2.40). Addition of equations (2.39) and (2.40), and using also (2.34) and (2.33), then yield. (1 + z1 )k1 T2,j+1 (z1 ) + (1 − z1 )k1 T1,j+1 (z1 ) = rj+1(z1 ), z1 ∈ C,. thereby ompleting our indu tive proof of (2.38).. In parti ular, by hoosing. j = n+2. in (2.38), and sin e. rn+2 (z1 ) = c 6= 0,. we dedu e. that. (1 + z1 )k1 U1 (z1 ) + (1 − z1 )k1 V1 (z1 ) = 1, z1 ∈ C,. (2.41).
(73) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. where the polynomials. U1. U1 (z1 ) =. and. V1. 31. M = 2I. are given by. T2,n+2 (z1 ) c. and. V1 (z1 ) =. T1,n+2 (z1 ) , z1 ∈ C. c. (2.42). Next, from the polynomial division theorem, there exist univariate polynomials. R1. su h that (2.18) holds, that is, for. Q1. and. z1 ∈ C,. z1α1 V1 (z1 ) = Q1 (z1 )(1 + z1 )k1 + R1 (z1 ),. with deg(R1 )<k1 ,. so that, from the proof of Lemma 2.2.1 (a), by hoosing the polynomial. S1. (2.43). as. S1 (z1 , z2 ) = −z2α2 R1 (−z1 ), z1 , z2 ∈ C,. (2.44). it follows that (2.24) holds. In other words, we have the identity. (1 + z1 )k1 S1 (z1 , z2 ) − (1 − z1 )k1 S1 (−z1 , z2 ) = z1α1 z2α2 , z1 , z2 ∈ C.. Moreover, we know from Lemma 2.2.1 (a) that and that its degree in. z1. is less than. S1. is odd in. z2 ,. (2.45). that its degree in. z2. is. α2 ,. k1 .. We have now proved the following algorithm for the expli it omputation of the polynomial. S1. of Lemma 2.2.1 (a). Algorithm for the omputation of S1 : 1. Use polynomial division to obtain the polynomials. {rj (z1 ) : j = 1, . . . , n + 2},. with. 2. De ne the polynomial sequen e. rn+2 (z1 ) = c 6= 0. {qj (z1 ) : j = 0, . . . , n + 1}. as in (2.32).. {Ti,j (z1 ) : i = 0, 1, 2; j = −1, . . . , n + 2}. by means of (2.34), (2.35) and (2.36).. and. re ursively.
(74) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 3. De ne the polynomials. U1. and. V1. by (2.42);. 4. Use the polynomial division theorem to nd. 5. The polynomial. S1. Q1. and. R1. su h that (2.43) holds;. is then given by (2.44).. The onstru tion of the polynomial that of. 32. M = 2I. S2 ,. under the onstraint. α2 < 2k2 ,. is analogous to. S1 .. We pro eed to give an example by nding the polynomial. k1 = 1 will be presented. S1. for. k 1 = 2.. The ase. in Se tion 2.3, and will be used to hara terize the mask symbols. of the box spline fun tions from Chapter 1. Under the onditions of Lemma 2.2.1 and the above algorithm, let that, for. k1 = 2, α1 ∈ {1, 3},. and let. α2 ∈ N. be any odd integer. Observe. z1 ∈ C,. (1 + z1 )2 =q0 (z1 )(1 − z1 )2 + r1 (z1 ), (1 − z1 )2 =q1 (z1 )r1 (z1 ) + r2 (z1 ),. It follows that the matrix. . T. with. with. q0 (z1 ) = 1. r1 (z1 ) = 4z1 ,. and. 1 1 q1 (z1 ) = z1 − 4 2. and. r2 (z1 ) = 1.. is given by. (1 + z1 ) 0 T = 1 0. 2. (1 − z1 ). 2. 4z1. 1. −1. 0. 1. 1. 1 z 4 1. −. . 1 1 z 4 1. +. 1 2. − 14 z1 + 1 2. whi h, together with (2.42), yields that the polynomials. 1 2. 0. U1. and. V1. 1 1 1 U1 (z1 ) = − z1 , V1 (z1 ) = z1 + , z1 ∈ C. 4 4 2. , are given by. (2.46).
(75) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 33. M = 2I. Two ases o ur:. •. if. α1 = 1:. we dedu e from (2.43) that, for. z1 ∈ C,. 1 1 z1 V1 (z1 ) = z12 + z1 = Q1 (z1 )(1 + z1 )2 + R1 (z1 ), 4 2 with. S1. Q1 (z1 ) =. 1 4. and. R1 (z1 ) = −. 1 , 4. and it follows from (2.44) that the polynomial. is given by. 1 S1 (z1 , z2 ) = z2α2 , z1 , z2 ∈ C. 4 •. if. α1 = 3:. we dedu e from (2.43) that, for. (2.47). z1 ∈ C,. 1 1 z13 V1 (z1 ) = z14 + z13 = Q1 (z1 )(1 + z1 )2 + R1 (z1 ), 4 2 with. Q1 (z1 ) =. polynomial. S1. 1 2 1 z − 4 1 4. and. R1 (z1 ) =. 1 1 z1 + , 2 4. and it follows from (2.44) that the. is given by. 1 S1 (z1 , z2 ) = (2z1 − 1)z2α2 , z1 , z2 ∈ C. 4. Observe in parti ular from (2.47) and (2.48) that the degree of and that. S1. is odd in. z2. with degree. α2. in. S1. in. (2.48). z1. is less than. k 1 = 2,. z2 .. First fa torization of mask symbols With the help of Lemma 2.2.1, we an prove the following formula.. Lemma 2.2.2. For an interpolatory mask symbol A, suppose there exist integers k1 , k2 ∈ N and a Laurent polynomial B su h that (2.11) holds, and let α1 and α2 be any pair of. odd integers su h that α1 < 2k1 and α2 < 2k2 . Then both the following results hold:.
(76) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 34. M = 2I. (a) There exist Laurent polynomials K1 , K2 and T3 su h that the Laurent polynomial B has, for z1 , z2 ∈ C \ {0}, the form. B(z1 , z2 ) = 2k1 +k2 z1−2α1 z2−2α2 [K1 (z1 , z2 )K2 (z1 , z2 ) + T3 (z1 , z2 )(1 − z2 )k2 ],. (2.49). where the Laurent polynomial T3 is odd in z2 , and with K1 , K2 satisfying the respe tive identities (1 + z1 )k1 K1 (z1 , z2 ) − (1 − z1 )k1 K1 (−z1 , z2 ) = z1α1 z2α2 ,. α α (1 + z2 )k2 K2 (z1 , z2 ) − (1 − z2 )k2 K2 (z1 , −z2 ) = z1 1 z2 2 ,. , z1 , z2 ∈ C \ {0}.. (2.50). Moreover, K1 and K2 are formulated expli itly by the expressions (2.13), (2.15), with S1 , T1 , S2 and T2 as des ribed in Lemma 2.2.1, and where both K1 and T1 are odd in z2 .. (b) There exist Laurent polynomials L1 , L2 and T˜3 su h that the Laurent polynomial B has, for z1 , z2 ∈ C \ {0}, the form. B(z1 , z2 ) = 2k1 +k2 z1−2α1 z2−2α2 [L1 (z1 , z2 )L2 (z1 , z2 ) + T˜3 (z1 , z2 )(1 − z1 )k1 ],. (2.51). where the Laurent polynomial T˜3 is odd in z1 , and with L1 , L2 satisfying respe tive identities (1 + z1 )k1 L1 (z1 , z2 ) − (1 − z1 )k1 L1 (−z1 , z2 ) = z1α1 z2α2 ,. α α (1 + z2 )k2 L2 (z1 , z2 ) − (1 − z2 )k2 L2 (z1 , −z2 ) = z1 1 z2 2 ,. , z1 , z2 ∈ C \ {0}.. (2.52). Moreover, L1 and L2 are formulated expli itly by the expressions (2.13), (2.15), with S1 , T1 , S2 and T2 as des ribed in Lemma 2.2.1, and where both L2 and T2 are odd in z1 ..
(77) CHAPTER 2.. Proof.. THE INTERPOLATORY MASK SYMBOLS FOR. (a) By de ning the Laurent polynomial. H. 35. M = 2I. as. H(z1 , z2 ) = A(z1 , z2 ) + A(z1 , −z2 ), z1 , z2 ∈ C \ {0},. (2.53). we observe that the identity (2.9) is equivalent to. H(z1 , z2 ) + H(−z1 , z2 ) = 4, z1 , z2 ∈ C \ {0}.. (2.54). Also, by using (2.11) and (2.53), we have that. H(z1 , z2 ) = 22−k1 −k2 (1 + z1 )k1 G(z1 , z2 ), z1 , z2 ∈ C \ {0},. where the Laurent polynomial. G. (2.55). is de ned by. G(z1 , z2 ) = (1 + z2 )k2 B(z1 , z2 ) + (1 − z2 )k2 B(z1 , −z2 ), z1 , z2 ∈ C \ {0},. with. B. (2.56). denoting the Laurent polynomial for whi h (2.11) is satis ed.. It then follows from (2.54) and (2.55) that. G. satis es the identity. 2−k1 −k2 (1 + z1 )k1 G(z1 , z2 ) + 2−k1 −k2 (1 − z1 )k1 G(−z1 , z2 ) = 1, z1 , z2 ∈ C \ {0}.. Now, hoose any pair of odd integers for the Laurent polynomial. G. α1 , α2 ∈ N. su h that. α1 < 2k1. and. α2 < 2k2 .. given by (2.56), we de ne the Laurent polynomial. G(z1 , z2 ) = 2k1 +k2 z1−α1 z2−α2 K1 (z1 , z2 ), z1 , z2 ∈ C \ {0}.. It follows from (2.58) and (2.57) that. K1. (2.57). Then,. K1. by. (2.58). satis es the identity. (1 + z1 )k1 z1−α1 z2−α2 K1 (z1 , z2 ) − (1 − z1 )k1 z1−α1 z2−α2 K1 (−z1 , z2 ) = 1, z1 , z2 ∈ C \ {0},.
(78) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 36. M = 2I. or, equivalently,. (1 + z1 )k1 K1 (z1 , z2 ) − (1 − z1 )k1 K1 (−z1 , z2 ) = z1α1 z2α2 , z1 , z2 ∈ C \ {0}.. Hen e, a ording to Lemma 2.2.1 (a), there exist a polynomial mial. T1. S1. (2.59). and a Laurent polyno-. su h that. K1 (z1 , z2 ) = S1 (z1 , z2 ) + (1 − z1 )k1 T1 (z1 , z2 ), z1 , z2 ∈ C \ {0},. with the polynomial. S1. and the Laurent polynomial. T1. satisfying the properties as stated. in Lemma 2.2.1 (a). Besides, (2.55) and (2.58) yield. H(z1 , z2 ) = 4(1 + z1 )k1 z1−α1 z2−α2 K1 (z1 , z2 ), z1 , z2 ∈ C \ {0},. a ording to whi h, sin e the Laurent polynomial dedu e that. K1. is odd in. z2 ,. H. and hen e also, from Lemma 2.2.1 (a),. Next, we de ne the Laurent polynomial. ˜ B. z2 ,. we. is also odd in. z2 .. de ned by (2.53) is even in. T1. by. ˜ 1 , z2 ), z1 , z2 ∈ C \ {0}. B(z1 , z2 ) = 2k1 +k2 z1−2α1 z2−2α2 B(z. (2.60). From (2.58) and (2.56) we then obtain. (1 + z2 )k2 B(z1 , z2 ) + (1 − z2 )k2 B(z1 , −z2 ) = 2k1 +k2 z1−α1 z2−α2 K1 (z1 , z2 ), z1 , z2 ∈ C \ {0}, (2.61) whi h, together with (2.60), shows that. ˜ B. satis es the identity. ˜ 1 , z2 ) + (1 − z2 )k2 B(z ˜ 1 , −z2 ) = z α1 z α2 K1 (z1 , z2 ), z1 , z2 ∈ C \ {0}. (1 + z2 )k2 B(z 1 2. (2.62).
(79) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. It now remains to nd. ˜. B. 37. M = 2I. To this end, we rst obtain a parti ular solution of (2.62) by. onsidering the Laurent polynomial. B1. de ned by. B1 (z1 , z2 ) = K1 (z1 , z2 )K2 (z1 , z2 ), z1 , z2 ∈ C \ {0},. for some arbitrary appropriate Laurent polynomial. K2. su h that. B1. (2.63). satis es (2.62), i.e.. (1 + z2 )k2 B1 (z1 , z2 ) + (1 − z2 )k2 B1 (z1 , −z2 ) = z1α1 z2α2 K1 (z1 , z2 ), z1 , z2 ∈ C \ {0}.. Sin e. K1. is odd in. z2 ,. we have from (2.63) that, for. (2.64). z1 , z2 ∈ C \ {0},. B1 (z1 , −z2 ) = K1 (z1 , −z2 )K2 (z1 , −z2 ) = −K1 (z1 , z2 )K2 (z1 , −z2 ),. so that, from (2.64) and (2.63), and after dividing by Laurent polynomial. K2. K1 (z1 , z2 ),. we dedu e that, if the. is hosen to satisfy the identity. (1 + z2 )k2 K2 (z1 , z2 ) − (1 − z2 )k2 K2 (z1 , −z2 ) = z1α1 z2α2 , z1 , z2 ∈ C \ {0},. then the Laurent polynomial. B1. (2.65). de ned by (2.63) satis es the identity (2.64). But a -. ording to Lemma 2.2.1 (b), the general Laurent polynomial solution. K2. of the identity. (2.65) is given by. K2 (z1 , z2 ) = S2 (z1 , z2 ) + (1 − z2 )k2 T2 (z1 , z2 ), z1 , z2 ∈ C \ {0},. with the polynomial. S2. and the Laurent polynomial. T2. satisfying the properties as stated. in Lemma 2.2.1 (b). Substra ting the equations (2.62) and (2.64) now yields, for. z1 , z2 ∈ C \ {0},. ˜ 1 , z2 ) − B1 (z1 , z2 )] = −(1 − z2 )k2 [B(z ˜ 1 , −z2 ) − B1 (z1 , −z2 )], (1 + z2 )k2 [B(z. (2.66).
(80) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. and, sin e the univariate polynomials there exists a Laurent polynomial. T3. (1 + z2 )k2. and. su h that, for. 38. M = 2I. (1 − z2 )k2. have no ommon fa tor,. z1 , z2 ∈ C \ {0},. ˜ 1 , z2 ) − B1 (z1 , z2 ) = (1 − z2 )k2 T3 (z1 , z2 ). B(z. Substituting the expressions in (2.67) into (2.66), we obtain, for. (2.67). z1 , z2 ∈ C \ {0},. (1 + z2 )k2 (1 − z2 )k2 T3 (z1 , z2 ) = −(1 − z2 )k2 (1 + z2 )k2 T3 (z1 , −z2 ),. from whi h we dedu e that. T3. is odd in. z2 .. Also, we dedu e from (2.67) that. ˜ 1 , z2 ) = B1 (z1 , z2 ) + T3 (z1 , z2 )(1 − z2 )k2 , z1 , z2 ∈ C \ {0}, B(z. whi h, together with (2.60) and (2.63), shows that. (b) By de ning the Laurent polynomial. J. B. is indeed given by (2.49).. as. J(z1 , z2 ) = A(z1 , z2 ) + A(−z1 , z2 ), z1 , z2 ∈ C \ {0},. observe that the identity (2.9) is equivalent to. (2.68). J(z1 , z2 ) + J(z1 , −z2 ) = 4, z1 , z2 ∈ C \ {0}.. The rest of proof then uses a similar argument as in (a).. The hara terization result Note that (2.49) and (2.51) yield two dierent formulae for the Laurent polynomial in Lemma 2.2.2. We pro eed here to give an alternative expression for simultaneously (2.49) and (2.51).. B. B. whi h veri es.
(81) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 39. M = 2I. Using Lemmas 2.2.1 and 2.2.2, we prove the following result whi h yields an important hara terization for interpolatory mask symbols.. Theorem 2.2.3. For a Laurent polynomial A, suppose that there exist integers k1 , k2 ∈ N and a Laurent polynomial B su h that (2.11) holds. Then A de nes an interpolatory mask symbol if and only if for any pair of odd integers α1 and α2 su h that α1 < 2k1 and α2 < 2k2 , the Laurent polynomial B has, for z1 , z2 ∈ C \ {0}, the form. B(z1 , z2 ) =2k1 +k2 z1−2α1 z2−2α2 T (z1 , z2 )(1 − z1 )k1 (1 − z2 )k2. (2.69). + S1 (z1 , z2 ) + T1 (z1 , z2 )(1 − z1 )k1 S2 (z1 , z2 ) + T2 (z1 , z2 )(1 − z2 )k2 ,. where the polynomials S1 and S2 are as in Lemma 2.2.1, i.e. S1 and S2 are respe tively odd in z2 and odd in z1 , they satisfy the respe tive identities (1 + z1 )k1 S1 (z1 , z2 ) − (1 − z1 )k1 S1 (−z1 , z2 ) = z1α1 z2α2 ,. α α (1 + z2 )k2 S2 (z1 , z2 ) − (1 − z2 )k2 S2 (z1 , −z2 ) = z1 1 z2 2 ,. , z1 , z2 ∈ C,. (2.70). where also S1 has a degree less than k1 in z1 , and S2 has a degree less than k2 in z2 . Besides, the Laurent polynomials T1 , T2 and T are respe tively even in z1 but odd in z2 , even in z2 but odd in z1 , and odd in both z1 and z2 . Proof.. We show that the proof in the ne essary dire tion an be obtained either by starting. with the formula given by (2.49) with an appropriate hoi e for the polynomial. L1 ,. or by. starting with the formula given by (2.51) with an appropriate hoi e for the polynomial. K2 .. We then prove the theorem in the su ient dire tion by using Theorem 2.1.1.. To prove the theorem in the ne essary dire tion, we suppose that latory mask symbol and onsider any pair of odd integers and. α2 < 2k2 .. A de nes an interpo-. α1 , α2 ∈ N. su h that. α1 < 2k1.
(82) CHAPTER 2.. B. A ording to Lemma 2.2.2, the Laurent polynomial. 40. M = 2I. THE INTERPOLATORY MASK SYMBOLS FOR. for whi h (2.11) is satis ed, has. the forms given by (2.49) and (2.51), where the Laurent polynomials. K2. in (2.49) and. L1. in (2.51) are to be hosen as spe i ed in Lemma 2.2.2.. We see from Lemma 2.2.1 and 2.2.2 that we may hoose it then holds that both. K1. and. L1. are even in. z1. L1 = K1 ,. and odd in. z2 .. a ording to whi h. It follows that, from. (2.11) and (2.51), it holds that. A(z1 , −z2 ) =4(1 + z1 )k1 (1 − z2 )k2 z1−2α1 z2−2α2 [−L1 (z1 , z2 )L2 (z1 , −z2 ) + T˜3 (z1 , −z2 )(1 − z1 )k1 ], z1 , z2 ∈ C \ {0},. whi h, together with (2.11), (2.51) and the se ond line of (2.52), shows that, for. z1 , z2 ∈. C \ {0},. A(z1 , z2 ) + A(z1 , −z2 ) =4(1 + z1 )k1 z1−2α1 z2−2α2 [z1α1 z2α2 L1 (z1 , z2 ) +(1 − z1 )k1 {(1 + z2 )k2 T˜3 (z1 , z2 ) + (1 − z2 )k2 T˜3 (z1 , −z2 )}].. Next, we note that, sin e the Laurent polynomials Lemma 2.2.2, odd in. z2 ,. T3. and. K1. (2.71). in (2.49) are, a ording to. we have from (2.11) and (2.49) that. A(z1 , −z2 ) =4(1 + z1 )k1 (1 − z2 )k2 z1−2α1 z2−2α2 [−K1 (z1 , z2 )K2 (z1 , −z2 ) − T3 (z1 , z2 )(1 + z2 )k2 ], z1 , z2 ∈ C \ {0},. whi h, together with (2.11), (2.49) and the rst line of (2.50), shows that, for. z1 , z2 ∈. C \ {0},. A(z1 , z2 ) + A(z1 , −z2 ) = 4(1 + z1 )k1 z1−2α1 z2−2α2 [z1α1 z2α2 K1 (z1 , z2 )].. (2.72).
(83) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. It then follows from (2.71) and (2.72) that, sin e also we have hosen polynomial. T˜3. 41. M = 2I. L1 = K1 , the Laurent. satis es. (1 + z2 )k2 T˜3 (z1 , z2 ) + (1 − z2 )k2 T˜3 (z1 , −z2 ) = 0, z1 , z2 ∈ C \ {0},. or, equivalently,. (1 + z2 )k2 T˜3 (z1 , z2 ) = −(1 − z2 )k2 T˜3 (z1 , −z2 ), z1 , z2 ∈ C \ {0}.. Sin e the univariate polynomials. (1 + z2 )k2. and. (1 − z2 )k2. dedu e from (2.73) the existen e of a Laurent polynomial. have no ommon fa tor, we. T˜4. satisfying. T˜3 (z1 , z2 ) = T˜4 (z1 , z2 )(1 − z2 )k2 , z1 , z2 ∈ C \ {0},. so that, sin e. T˜3. is odd in. expression in (2.74) of. T˜3. z1 ,. we nd that. T˜4. is odd in. (2.73). z1 .. (2.74). Also, by substituting the. into (2.73), we obtain. (1 + z2 )k2 (1 − z2 )k2 T˜4 (z1 , z2 ) = −(1 − z2 )k2 (1 + z2 )k2 T˜4 (z1 , −z2 ), z1 , z2 ∈ C \ {0},. showing that. T˜4. z1 , z2 ∈ C \ {0},. is also odd in. z2 .. Combining (2.51) with (2.74), we dedu e that, for. the Laurent polynomial. B. is of the form. B(z1 , z2 ) = 2k1 +k2 z1−2α1 z2−2α2 [L1 (z1 , z2 )L2 (z1 , z2 ) + T (z1 , z2 )(1 − z1 )k1 (1 − z2 )k2 ],. where. T = T˜4. is a Laurent polynomial whi h is odd in both. z1. and. (2.75). z2 .. Our proof in the ne essary dire tion is now ompleted by appealing to Lemma 2.2.1 and 2.2.2, and using (2.75), with spe i ally the Laurent polynomial T2 in Lemma 2.2.1 (b) hosen to also be odd in. z1 ..
(84) CHAPTER 2.. THE INTERPOLATORY MASK SYMBOLS FOR. 42. M = 2I. Note from Lemmas 2.2.1 and 2.2.2 that the result (2.69) an similarly be a hieved by means of the hoi e. K2 = L2. in (2.49).. Next, we prove the theorem in the su ient dire tion. for any pair of odd integers polynomial. B. α1. and. α2. su h that. To this end, suppose that,. α1 < 2k1. and. α2 < 2k2,. the Laurent. has the form given by (2.69). To show that the Laurent polynomial. an interpolatory mask symbol, it will su e to prove that. A satis es. A. is. the identity (2.9) in. Theorem 2.1.1.. To this end, sin e by assumption (2.69) that, for. S2 , T2. and. T. are odd in. z1 ,. observe from (2.11) and. z1 , z2 ∈ C \ {0},. A(z1 , z2 )+A(−z1 , z2 ) =4z1−2α1 z2−2α2 (1 + z1 )k1 (1 + z2 )k2 T (z1 , z2 )(1 − z1 )k1 (1 − z2 )k2. + S1 (z1 , z2 ) + T1 (z1 , z2 )(1 − z1 )k1 S2 (z1 , z2 ) + T2 (z1 , z2 )(1 − z2 )k2. +4z1−2α1 z2−2α2 (1 − z1 )k1 (1 + z2 )k2 −T (z1 , z2 )(1 + z1 )k1 (1 − z2 )k2. + S1 (−z1 , z2 ) + T1 (z1 , z2 )(1 + z1 )k1 −S2 (z1 , z2 ) − T2 (z1 , z2 )(1 − z2 )k2 ,. whi h, together with (2.70), yields, for. z1 , z2 ∈ C \ {0},. A(z1 , z2 )+A(−z1 , z2 ) =4z1−2α1 z2−2α2 (1 + z2 )k2 z1α1 z2α2 S2 (z1 , z2 ) + T2 (z1 , z2 )(1 − z2 )k2 . Repla ing. z2. by. −z2. in (2.76), and using the fa t that. T2. is even in. z2 ,. (2.76). we obtain, for. z1 , z2 ∈ C \ {0},. A(z1 , −z2 )+A(−z1 , −z2 ) =4z1−2α1 z2−2α2 (1 − z2 )k2 −z1α1 z2α2 S2 (z1 , −z2 ) + T2 (z1 , z2 )(1 + z2 )k2 .. (2.77).
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