Hindawi Publishing Corporation ISRN Applied Mathematics Volume 2014, Article ID 498135, 8 pages http://dx.doi.org/10.1155/2014/498135
Research Article Generalizing Krawtchouk Polynomials Using Hadamard Matrices
Peter S. Chami,1 Bernd Sing,1 and Norris Sookoo2
1 Department of Computer Science, Mathematics and Physics, Faculty of Science and Technology, The University of the West Indies, Cave Hill, St. Michael, Barbados 2 The University of Trinidad and Tobago, OβMeara Campus, Arima, Trinidad and Tobago
Correspondence should be addressed to Norris Sookoo; [email protected]
Received 15 November 2013; Accepted 22 December 2013; Published 4 March 2014
Academic Editors: F. Ding and X. Liu
Copyright Β© 2014 Peter S. Chami et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We investigate polynomials, called m-polynomials, whose generator polynomial has coefficients that can be arranged in a square matrix; in particular, the case where this matrix is a Hadamard matrix is considered. Orthogonality relations and recurrence relations are established, and coefficients for the expansion of any polynomial in terms of m-polynomials are obtained. We conclude this paper by an implementation of m-polynomials and some of the results obtained for them in Mathematica.
1. Introduction where π is a natural number and π₯ β {0,1,...,π}.The generator polynomial is Matrices have been the subject of much study, and large bodies of results have been obtained about them. We study π (1+π§)πβπ₯(1βπ§)π₯ = βπΎ (π₯) π§π. the interplay between the theory of matrices and the theory π (2) of orthogonal polynomials. For Krawtchouk polynomials, π=0 introduced in [1], interesting results have been obtained in The generalized Krawtchouk polynomial πΎπ(π ) is obtained by [2β4]; also see the review article [5]andcompare[6]for generalizing the above generator polynomial as follows: generalized Krawtchouk polynomials. More recently, condi- π tions for the existence of integral zeros of binary Krawtchouk πβ1 πβ1 π π polynomials have been obtained in [7], while properties for β(βπ(ππππ)π§π) = β πΎπ (π ) π§ , (3) generalized Krawtchouk polynomials can be found in [8]. π=0 π=0 πβπ(π,π) Other generalizations of binary Krawtchouk polynomials have also been considered; for example, some properties of where βπ π =βππ =π, π is a prime power, the π§π are binary Krawtchouk polynomials have been generalised to π- indeterminate, the field πΊπΉ(π) with π elements is {0, π1, Krawtchouk polynomials in [9]. Orthogonality relations for π2,...,ππβ1},andπ is a character. quantum and π-Krawtchouk polynomials have been derived The above information about Krawtchouk polynomials in [10], and it has been shown that affine π-Krawtchouk poly- and generalized Krawtchouk polynomials was taken from [6]. nomials are dual to quantum π-Krawtchouk polynomials. In If we replace the π(ππππ) by arbitrary scalars in the this paper, we define and study generalizations of Krawtchouk lastequation,weobtainthegeneratorpolynomialofπ- polynomials, namely, π-polynomials. polynomials ππΊ(π, π );seeDefinition 2 below. These π- The Krawtchouk polynomial πΎπ(π₯) is given by polynomialsarethesubjectofstudyinthispaper. In Section 2, we present relevant notations and defini- π π₯ πβπ₯ tions. In Section 3, we introduce the generator polynomial. πΎ (π₯) = β(β1)π ( )( ) , π π πβπ (1) The associated matrix of coefficients πΊ can be any square π=0 matrix, and so the question that immediately arises is how 2 ISRN Applied Mathematics the properties of the π-polynomials are related to the respect to πΊ having parameter π is denoted by ππΊ(π; π ) or properties of πΊ.Wewillestablishthat,ifπΊ is a generalized ππΊ(π0,π1,...,ππβ1;π 0,π 1,...,π πβ1) and given by Hadamard matrix, then the associated π-polynomials satisfy πβ1 πβ1 orthogonality conditions. In Section 4,weestablishrecur- π ! π ππΊ (π; π )= β ββπ π,π rence relations for π-polynomials.Afterwards,weobtain ππ π β π ! coefficients for the expansion of a polynomial in terms of π- π,π π,π π,π π=0 π=0 (5) polynomials in Section 5.Finally,inSection 6,weimplement π πβ1 πβ1 π π,π the results obtained here in Mathematica, so the reader may =π !βββ ππ , π πΊ π ! easily derive and explore -polynomials for any matrix . ππ,π π=0 π=0 π,π
where the summation is taken over all sets of nonnegative π π,π=0,1,...,πβ1 2. Definitions and Notations integers π,π (with )satisfying
In this paper, N0 denotes {0,1,2,3,...}.Weusetheconven- πβ1 π tion that if πβN0,thenπ has components (π0,π1,...,ππβ1). βππ,π =π π, π=0,1,...,πβ1, π π=0 So, if π(π) β N then π(π) denotes (ππ,0,ππ,1,...,ππ,πβ1).Wewill 0 (6) also use the π elementary unit vectors π0 = (1, 0, 0, . . . , 0), πβ1 π =(0,1,0,...,0),...,π =(0,0,...,0,1) 1 πβ1 . βππ,π =ππ, π=0,1,...,πβ1. 1 We use the β -norm (the βtaxicab-metricβ) to measure the π=0 π πβ1 length |π| of πβN0,thatis,|π| = βπ=0 ππ. We define the set of weak compositions of π into π numbers by π(π, π) = 3. The Generator Polynomial and {π β Nπ :|π|=π} π(π, π) 0 ;inotherwords, is the subset Orthogonality Relations of π-dimensional nonnegative vectors of length π.Wenote π+πβ1 π that the set π(π, π) has cardinality ( πβ1 ).ForπβN0,we The values of the π-polynomial with respect to a matrix πΊ πβ1 can be derived from a generator polynomial. use the multi-index notation π! = βπ=0 ππ!.Similarly,fora π§=(π§,π§ ,...,π§ )βCπ πβNπ variable 0 1 πβ1 and 0,wewrite Theorem 3 π =(π ,π ,...,π )β πβ1 π (generator polynomial). Let 0 1 πβ1 π§π =β π§ π 00 =1 π=0 π (where the convention is used). We note π(π, π) and let πΊ=(πππ ) be a πΓπmatrix. Then, that the multinomial theorem reads π πβ1 πβ1 π π β(βπππ π§π) = β ππΊ (π; π )π§ , (7) π π=1 π=0 πβπ(π,π) πβ1 π (βπ§ ) = β ( )β π§π π π π=(π,π ,...,π ) π=0 πβπ(π,π) where 0 1 πβ1 . (4)
π! π ππβ1 Proof. This is an application of the multinomial theorem 0 π = β β π§0 β β β π§πβ1 . π !β β β π ! (recall that for π(π) β N0,wehaveπ(π) =(ππ,0,ππ,1,...,ππ,πβ1)): πβπ(π,π) 0 πβ1 π πβ1 πβ1 π πβ1 πβ1 { π ! π } π [ π,π ] β(βπππ π§π) = β { β β(πππ π§π) } π=0 π=0 π=0 π(π)! π=0 In the following, πΊ denotes an arbitrary πΓπmatrix. We {π(π)βπ(π π,π) [ ]} use the following convention to refer to the entries of a πΓπ πβ1 πβ1 π π π ! π matrix [11]. The entry in the th row and th column is called = β (ββπ π,π ) the (πβ1,πβ1)entry,whereπ,π = 1,2,...,π.Thus,the(π, π) β π ! ππ π βπ π ,π , π,π π,π π=1 π=0 πΊ π π,π=0,1,...,πβ1 (π) ( π ) entry of is denoted by ππ ,where .Given π=0,1,...,πβ1 (8) amatrixπΊ, the matrix that remains when the first row and thefirstcolumnofπΊ are removed is called the core of πΊ. πβ1 π0,π+π1,π+β β β +ππβ1,π The next definition is well known. Γ βπ§π π=0 π πΓπ Definition 1. For an integer greater than one, a matrix π π = β ππΊ (π; π )π§ , π» is called a generalized Hadamard matrix if π»π» =ππΌπ, πβπ π,π π ( ) where π» is the complex conjugate transpose of π» and πΌπ is the πΓπidentity matrix. where ππ =π0,π +π1,π +β β β +ππβ1,π. We now define the π-polynomial with respect to a matrix πΊ. As an immediate consequence, we can recover the entries πππ of the matrix πΊ as the π-polynomials of minimum order. Definition 2. Let π =(π 0,π 1,,...,π πβ1) and π=(π0,π1,..., Corollary 4. ππΊ(π ;π )=π ππβ1) be elements of π(π,. π) The π-polynomial in π with We have β π πβ. ISRN Applied Mathematics 3
Proof. For π =ππ, the left-hand side of (7)becomes On the other hand, the multinomial theorem yields π π§ +π π§ +β β β +π π§ , π,0 0 π,1 1 π,πβ1 πβ1 (9) π π π π! πβ1 πβ1 πβ1 π πβπ(1,π) π ,...,π π½π = β β(βπ π§ ) (βπ π¦ ) and since and thus runs through 0 πβ1 in π ! ππ π ππ π this case, the rest follows by comparing coefficients of π§π. π βπ(π,π) π=0 π=0 π=0
Remark 5. Theorem 3 canalsobeusedforsummationresults π! π π(π, π) π§=(1,...,1) = β [ β ππΊ (π; π )π§π] of -polynomials over :using ,weobtain π ! (16) π πβ1 πβ1 π π βπ(π,π) [πβπ(π,π) ] β ππΊ (π; π )= β(βπππ ) , (10) πβπ(π,π) π=0 π=0 Γ [ β ππΊ (π‘; π )π¦π‘] . that is, the product of the π πth power of the πth column sum [π‘βπ(π,π) ] of πΊ. π π‘ As an immediate result of this remark, we can establish Equating coefficients of π§ π¦ in the two above expressions π the following corollary. for π½ , we obtain the desired result. Corollary 6. βπβ1 π =0 0β€πβ€πβ1 If π=0 ππ for one If πΊ is a generalized Hadamard matrix and also satisfies (i.e., one of the row-sums of the matrix πΊ is zero), then certain additional conditions, then it is possible to establish βπβπ(π,π) ππΊ(π; π ). =0 that the corresponding π-polynomials satisfy additional orthogonality conditions. We use the following three results For a generalized Hadamard matrix πΊ,themultinomial in proving this. theorem yields the following orthogonality relation for the π corresponding π-polynomials. Lemma 8. If πΊ is symmetric, that is, πΊ=πΊ ,thenππΊ(π ; π)= (π!/π !)ππΊ(π; π ). Theorem 7 (orthogonality relation). If πΊ is a generalized Hadamard matrix, then the π-polynomials ππΊ(π, π ),with πβ1 Proof. By Definition,wehave(where 2 βπ=0 βπ,π =ππ and π, π β π(π,, π) satisfy the orthogonality relations πβ1 β βπ,π =π π) 1 ππ π=0 β ππΊ (π; π ) ππΊ (π‘; π ) = πΏπ,π‘, π ! π! (11) πβ1 πβ1 βπ,π π βπ(π,π) π ππΊ (π ; π) =π! βββ ππ βπ,π! where βπ,π π=0 π=0
π =(π 0,π 1,...,π πβ1),π=(π0,π1,...,ππβ1), β π! πβ1 πβ1 π π,π = π !βββ ππ π‘=(π‘,π‘ ,...,π‘ ), π ! β ! 0 1 πβ1 β π=0 π=0 π,π (12) π,π (17) πβ1 πβ1 πβ1 π 1, πππ π =π‘π βπ, π! π π,π πΏπ,π‘ = βπΏπ ,π‘ ={ ππ π π 0, ππ‘βπππ€ππ π, = π !βββ π=0 π ! π ! ππ,π π=0 π=0 π,π denotes Kroneckerβs delta. π! π = ππΊ (π; π ), Proof. Let π§=(π§0,π§1,...,π§πβ1), π¦=(π¦0,π¦1,...,π¦πβ1)βC , π ! and define πβ1 πβ1 πβ1 πβ1 πβ1 πβ1 πβ1 where we define ππ,π =βπ,π (and thus βπ=0 ππ,π =π π and π½=β (βπ π§ )(βπ π¦ )=β β βπ π π§ π¦ . πβ1 ππ π ππ π ππ ππ π π βπ=0 ππ,π =ππ), and since πΊ is symmetric (i.e., πππ =πππ). π=0 π=0 π=0 π=0 π=0 π=0 (13) The following results can be obtained in a similar manner Then, on the one hand, we have to Lemma 8. πβ1 Lemma 9. If πΊ has a symmetric core, π0,π =1for π= π½=πβπ§ππ¦π (14) 0,...,πβ1and ππ,0 =β1for π=1,...,πβ1,thenππΊ(π ; π)= π=0 π +π (β1) 0 0 (π!/π !)ππΊ(π; π ). since πΊ is a generalized Hadamard matrix. Consequently, Such symmetry relations yield additional orthogonality { π! π π π } π½π =ππ β (π§ π¦ ) 0 (π§ π¦ ) 1 β β β (π§ π¦ ) πβ1 . relations. { π ! 0 0 1 1 πβ1 πβ1 } π βπ(π,π) { } Theorem 10 (additional orthogonality relation). Let πΊ be a (15) generalized Hadamard matrix. 4 ISRN Applied Mathematics
(i) If in addition πΊ is symmetric, then Proof. Using (7)twice,weobtain(ifπ π >0) β ππΊ (π; π )π§π π β ππΊ (π; π ) ππΊ (π ;) π‘ =ππΏ . πβπ(π,π) π,π‘ (18) π βπ π,π ( ) π πβ1 πβ1 π
= β(βπππ π§π) π=1 π=0
π π (ii) If in addition πΊ has a symmetric core, π0,π =1for π= πβ1 πβ1 0 πβ1 πβ1 0,...,πβ1and ππ,0 =β1for π=1,...,πβ1,then =(βπππ π§π)(βπ0ππ§π) β β β (βππβ1,ππ§π) , π=0 π=0 π=0
π 0 π‘0 π π β1 π π β (β1) ππΊ (π; π ) ππΊ (π ;) π‘ = (β1) π πΏ . πβ1 π πβ1 π+1 πβ1 πβ1 π,π‘ (19) π βπ π,π ( ) (βπππ π§π) (βππ+1,ππ§π) β β β (βππβ1,ππ§π) π=0 π=0 π=0
πβ1 =(βπ π§ ) β ππΊ (π;Μ π βπ )π§πΜ. Proof. This follows immediately from Theorem 7 and Lem- ππ π π π=0 mas 8 and 9. πβπΜ (πβ1,π) (22) Remark 11. It turns out that a slight modification of the above π-polynomials has already been considered in [12, 13]. For Multiplying the right-hand side out and equating coefficients π§π amatrixπΊ and π, π β π(π,,[ π) 13, Equation (2.3)] defines of establish the claim. polynomials πΏπ,π (πΊ) by πΏπ,π (πΊ) = (1/π !)ππΊ(π; π ) in our Theorem 13 (recurrence relations II). Let π, π β π(π, π) with notation. For these polynomials, the following multiplication π >0 πβ{0,...,πβ1} property is proved (see [13, Theorem 3.2]): π for one .Then, πβ1 π ππΊ (π; π )= β π π ππΊ (π βπ ;π βπ), π ππ π π (23) πΏπ,π‘ (πΊ1πΊ2)= β π !πΏπ,π (πΊ1)πΏπ ,π‘ (πΊ2), π=0, π (20) π >0 π βπ(π,π) π where the sum on the right is taken over all π such that the πth π component π π of π βπ(π,π)is positive. for πΓπmatrices πΊ1,πΊ2.WithπΊ1 =πΊ, πΊ2 = πΊ and thus π πΊ1πΊ2 =πΊπΊ =ππΌπ, Theorem 7 becomes a special case of this Proof. We note that we have πΏ (πΊπ)=πΏ (πΊ) π β1 equation by noting that π ,π‘ π‘,π (see [13,Equation π πβ1 π π πβ1 π { (4.5)]) and πΏπ,π‘(ππΌπ)=(π/π!)πΏπ,π‘ (using the notation of π {π π (βπ π§ ) , π >0, (βπ π§ ) = π ππ ππ π if π Definition, 2 the polynomial on the left can only be nonzero ππ π { π=0 ππ§π π=0 { if ππ,π =0whenever π =πΜΈ and ππ =ππ,π =π‘π otherwise). {0, if π π =0. (24) 4. Recurrence Relations and Summation of Thus, on the one hand, the derivative of the left-hand side of Certain Polynomials (7)withrespecttoπ§π is
π We use Theorem 3 to derive a recurrence relation for π- πβ1 πβ1 π πβ1 πβ1 πβ1 π polynomials in π having parameter π where both π , π β [ ] β [π ππππ (βππβπ§β) β(βπππ π§π) ] . (25) π(π, π) by a sum of π-polynomials where both π , π β π(π β π=0, β=0 π=0, π=0 π >0 [ π =πΜΈ ] 1, π). π Using (7), this is equal to Theorem 12 (recurrence relations I). Let π, π β π(π, π) with π π >0for one πβ{0,...,πβ1}.Then, πβ1 β [π π β ππΊ (π;Μ π βπ )π§πΜ] . π ππ π (26) π=0, Μ πβ1 [ πβπ(πβ1,π) ] π π>0 ππΊ (π; π )= β π ππΊ (π βπ ;π βπ), ππ π π (21) π=0, On the other hand, the derivative of the right-hand side of (7) π >0 π with respect to π§π is πβπ β π ππΊ (π; π )π§ π . π π π where the sum on the right is taken over all such that the th πβπ(π,π), (27) component ππ of πβπ(π,π)is positive. ππ>0 ISRN Applied Mathematics 5
Equating the last two expressions and interchanging the order Corollary 15. Let π, π β π(π, π) with π π β₯ππ β₯0for all πβ of summation yield {0,...,πβ1}.Then,
πβ1 π ππΊ (π; π )= β [β ( π )] πβ1 π [ πΜ] (π) π(π)βπ(ππ,π),s.t. π=0 β [ β π ππππππΊ (π;Μ π π βπ )π§ ] πββπβ1 πβπ(πβ|π|,π) πβπ(πβ1,π)Μ π=0, π=0 [π π>0 ] (28) πβ1 πβ1 πβ1 πβπ π = β π ππΊ (π; π )π§ π ; [ ππ ] π Γ ββπππ ππΊ (π β βπ(π);π βπ), πβπ(π,π), π=0 π=0 π=0 π >0 [ ] π (32) πβ1 by comparing coefficients, we must have π=πβπΜ π,and where the sum on the right is taken over all the vectors π βπ(π,π) πββπβ1 π β π(π β |π|, π) therefore (noting that ππ >0) (π) π such that π=0 .
Proof. This follows from repeatedly applying Proposition 14, πβ1 π π π π π ππΊ (π; π )= β π π ππΊ (π βπ ;π βπ ). by subtracting 0 from 0, 1 from 1,andsoon. π ππ π π (29) π=0, π We note that taking π=π in the previous corollary yields π π>0 the statement of Definition 2 (since, by definition, ππΊ(0; 0) = 1). Combining the recurrence relation in Theorem 12 with Theorem 7 yields the following summation formula for π- We note that (21)and(23)canberewrittenas polynomials.
π π π Proposition 16 (summation formula). Let πΊ be a generalized ππΊ (π; π )= β π 0 π 1 β β β π πβ1 π0 π1 π,πβ1 Hadamard matrix and π, π β π(π,. π) Then, for any pair of πβπ(1,π),s.t. πβπβπ(πβ1,π) fixed, arbitrary elements π,πβ{0,...,πβ1},onehas 1 ππ ΓππΊ(πβπ;π βππ), β ππΊ (π +π ;π +π) ππΊ (π; π ) =π . π ! π π ππ π! (33) π π π π βπ(π,π) π 0 π 1 β β β π πβ1 0 1 πβ1 π π π ππΊ (π; π )= β π 0 π 1 β β β π πβ1 π 0π 1π πβ1,π πβπ(1,π),s.t. π 5. Expansion of a Polynomial π βπβπ(πβ1,π) The coefficients for the expansion of a polynomial in terms ΓππΊ(πβπ ;π βπ), π of Krawtchouk polynomials have been obtained (cf. [6]). (30) We obtain a similar result for π-polynomials defined in terms of a generalized Hadamard matrix, using orthogonality respectively. properties. In particular, Theorem 12 canbeeasilyiteratedtoobtain Theorem 17 (polynomial expansion). Let πΊ be a generalized the following statements. Hadamard matrix, and let π₯=(π₯0,π₯1,...,π₯πβ1) be a variable element of π(π,. π) Proposition 14. Let π, π β π(π, π) with π π β₯πβ₯0for one πβ{0,...,πβ1}.Then, (i) If πΊ is symmetric and the expansion of a polynomial πΎ(π₯) in terms of π-polynomials defined with respect to πΊ πΌ β C π π π π is (with coefficients π ) ππΊ (π; π )= β ( )π0 π 1 β β β π πβ1 π π0 π1 π,πβ1 πβπ(π,π),s.t. πΎ (π₯) = β πΌπ β ππΊ(π₯;) π , πβπβπ(πβπ,π) (31) π βπ(π,π) (34)
ΓππΊ(πβπ;π βπππ), then, for all ββπ(π,π), πΌ =πβπ β πΎ (π) β ππΊ (β, π). πβπ(π,π) β (35) where the sum on the right is taken over all such πβπ(π,π) that πβπis in π(π β .π, π) Similarly, if the expansion of a polynomial πΎ(π₯) in terms Proof. This follows by repeatedly applying (21)inTheorem 12. π π½π β C π of -polynomials is (with coefficients ) The multinomial coefficient ( π ) arisesasthenumberofways π canbechangedtoπβπstep-by-step. πΎ (π₯) = β π½π β ππΊ(π ; π₯) , π βπ(π,π) (36) 6 ISRN Applied Mathematics
then, for all ββπ(π,π), {j, 1, q}])Μ(s[[i]]), {i, 1, q}]] π½ =πβπ β πΎ (π) β ππΊ (π, β). > mg[p ,s]:= Coefficient[generator[s], β (37) πβπ(π,π) Inner[Power, z, p, Times]]
(ii) If πΊ has a symmetric core, π0,π =1for π = 0,...,πβ1 π and ππ,0 =β1for π=1,...,πβ1, then the corresponding The values of the -polynomial can be shown using the coefficients can be calculated for all ββπ(π,π)by following command:
β βπ π πΌ = (β1) 0 π β (β1) 0 πΎ (π) β ππΊ (β, π), > TableForm[Table[mg[pall[[i]], β (38) πβπ(π,π) pall[[j]]], {i, Length[pall]}, respectively, {j, Length[pall]}], TableHeadings -> β βπ π π½ = (β1) 0 π β (β1) 0 πΎ (π) β ππΊ (π, β). β (39) πβπ(π,π) {pall, pall}, TableDepth -> 2]
π₯=π ππΊ(β, π) For the above chosen matrix πΊ and length π,theoutput Proof. Let . Multiplying both sides of (34)by ππΊ(π; π ) π π and summing each side over all elements of π(π, π) yield for looks as follows ( denotes the row and the column; e.g., ππΊ((2, 4); (3, 3))): =3 β πΎ (π) ππΊ (β, π) = β β πΌπ ππΊ (β, π) ππΊ (π, π ) πβπ(π,π) πβπ(π,π) π βπ(π,π) {0, 6}{1, 5}{2, 4}{3, 3}{4, 2}{5, 1}{6, 0} {0, 6} 1β11β11β11 = β πΌπ β ππΊ (π,) π ππΊ (π, π ) {1, 5} β64β202β46 π βπ(π,π) πβπ(π,π) {2, 4} 15 β5 β1 3 β1 β5 15 {3, 3} β20 0 4 0 β4 0 20 (41) = β πΌπ β ππΊ (π,) π ππΊ (π, π ) {4, 2} 15 5 β1 β3 β1 5 15 π βπ(π,π) πβπ(π,π) {5, 1} β6 β4 β2 0 2 4 6 {6, 0} 1111111 (β) π = β πΌπ β π β πΏβ,π π βπ(π,π) Thus,weareabletoevaluatetheπ-polynomial ππΊ(π; π ) π for any π, π β π(π, π) (and thus check orthogonality and =πΌβ π , recurrence relations). With the help of the Mathematica (40) function Fit, we can express the m-polynomials as univariant (π βπ ) where step (β) uses Theorem 10; the first result in (i) follows. polynomials in 0 1 as follows (we also recall that |(π ,π )| = π +π =6 Theotherresultsareprovedsimilarly. 0 1 0 1 here): 1 5 ππΊ ( 0, 6) ;(π ,π )) = (π βπ )6 β (π βπ )4 6. Mathematica Code 0 1 720 0 1 72 0 1 π Here,wepresentMathematicacodetoobtain -polynomials. 34 2 πΊ π + (π βπ ) β1, First, we specify the matrix and the length : 45 0 1 > g={{1, 1}, {1, β1}} 1 ππΊ ( 1, 5) ;(π ,π )) = (π βπ )5 > n=6 0 1 120 0 1 From this we can calculate π,thesetπ(π, π) and initialize the 1 3 11 β (π 0 βπ 1) + (π 0 βπ 1), variable π§: 3 5 1 7 > q = Union[Dimensions[g]][[1]] ππΊ ( 2, 4) ;(π ,π )) = (π βπ )4 β (π βπ )2 +3, 0 1 24 0 1 6 0 1 > pall = Sort[Flatten[Map[Permutations, 1 8 IntegerPartitions[n, {q}, ππΊ ( 3, 3) ;(π ,π )) = (π βπ )3 β (π βπ ), 0 1 6 0 1 3 0 1 Range[0, n]]], 1]] 1 > z = Table[f[i], {i, q}] ππΊ ( 4, 2) ;(π ,π )) = (π βπ )2 β3, 0 1 2 0 1 Using Theorem 3,weobtaintheπ-polynomial via its gener- ππΊ ( 5, 1) ;(π ,π )) = (π βπ ), ator: 0 1 0 1
> generator[s ]:= Expand[Product[ ππΊ ( 6, 0) ;(π 0,π 1)) = 1. (Sum[g[[i, j]] z[[j]], (42) ISRN Applied Mathematics 7
Similarly, we may express the π-polynomials as univariant giving the list of coefficients πΌπ ,respectively,π½π for π βπ(π,π). polynomials in (π0 βπ1) as follows: That is, the expansion in terms of π-polynomials here reads as follows: 77 6 203 4 ππΊ ((π0,π1);(0, 6))= (π0 βπ1) β (π0 βπ1) 3 3840 192 π₯ π₯ =β ππΊ ((π₯ ,π₯ );(0, 6)) 0 1 64 0 1 1519 2 + (π0 βπ1) β 20, 39 120 β ππΊ ((π₯ ,π₯ );(2, 4)) 64 0 1 7 5 49 3 ππΊ ((π0,π1);(1, 5))= (π0 βπ1) β (π0 βπ1) 7 640 96 + ππΊ ((π₯ ,π₯ );(4, 2)) 64 0 1 131 (45) + (π0 βπ1), 35 30 + ππΊ ((π₯ ,π₯ );(6, 0)), 64 0 1 7 6 7 4 ππΊ ((π0,π1);(2, 4))=β (π0 βπ1) + (π0 βπ1) 1 3840 64 π₯ π₯ =β ππΊ ( 4, 2) ;(π₯ ,π₯ )) 0 1 2 0 1 199 2 β (π0 βπ1) +4, 15 120 + ππΊ ( 6, 0) ;(π₯ ,π₯ )) . 2 0 1 7 5 ππΊ ((π0,π1);(3, 3))=β (π0 βπ1) 1920 As a check, we have (recall that π₯0 +π₯1 =6) 19 3 67 1 15 + (π0 βπ1) β (π0 βπ1), β ππΊ ( 4, 2) ;(π₯ ,π₯ )) + ππΊ ( 6, 0) ;(π₯ ,π₯ )) 96 30 2 0 1 2 0 1
7 6 25 4 1 1 2 15 ππΊ ((π0,π1);(4, 2))= (π0 βπ1) β (π0 βπ1) =β β ( (π₯ βπ₯ ) β3)+ β 1 (46) 11520 576 2 2 0 1 2
329 2 + (π βπ ) β4, =π₯0 (6 β π₯0)= π₯0π₯1. 360 0 1 1 For a generalized Hadamard matrix with symmetric core, ππΊ ((π ,π );(5, 1))= (π βπ )5 π =1 π = 0,...,πβ1 π =β1 π = 1,...,πβ1 0 1 384 0 1 0,π for and π,0 for ; for example, 17 19 β (π βπ )3 + (π βπ ), 96 0 1 6 0 1 11 1 1ββ3 1+β3 1 6 23 4 πΊ=(β1 ). ππΊ ((π0,π1);(6, 0))=β (π0 βπ1) + (π0 βπ1) 2 2 (47) 2304 576 1+β3 1ββ3 β1 101 2 2 β (π βπ )2 + 20. 72 0 1 (43) The above mathematica code using Theorem 17 has to be modified as follows: By Theorem 17, we can express any polynomial πΎ(π₯) = > Table[1/qΜn Total [Table πΎ(π₯0,...,π₯πβ1) in terms of π-polynomials. Using Mathemat- Μ ica to find the corresponding coefficients πΌπ and π½π in the [(-1) (pall[[i, 1]] + pall[[j, 1]]) expansion in terms of π-polynomials, for example, for the p[pall[[i]]] Conjugate[mg[pall[[j]], polynomial π(π₯0,π₯1)=π₯0π₯1, is achieved as follows: pall[[i]]]], > { } p[ x0 ,x1 ]:= x0 x1 {i, Length[pall]}]], {j, Length[pall]}] > Μ Table[1/q n Total[Table[p[pall[[i]]] > Table[1/qΜn Total[ Conjugate[mg[pall[[j]], pall[[i]]]], Table[(-1)Μ(pall[[i, 1]] + pall[[j, 1]]) { } { } i, Length[pall] ]], j, Length[pall] ] p[pall[[i]]] Conjugate[mg[pall[[i]], > Table[1/qΜn Total[Table[p[pall[[i]]] pall[[j]]]], Conjugate[mg[pall[[i]], pall[[j]]]], {i, Length[pall]}]], {j, Length[pall]}] {i, Length[pall]}]], {j, Length[pall]}] The output of the later two lines is 7. Outlook 3 39 7 35 1 15 {β ,0,β ,0, ,0, }, {0,0,0,0,β ,0, }, Krawtchouk polynomials and their generalisation appear 64 64 64 64 2 2 in many areas of Mathematics, see [5]: harmonic analysis (44) [1, 2, 4], statistics [3], combinatorics and coding theory 8 ISRN Applied Mathematics
[6, 7, 9, 11, 14, 15], probability theory [5], representation [14] V. I. Levenshtein, βKrawtchouk polynomials and universal theory (e.g., of quantum groups) [10, 12, 13], difference bounds for codes and designs in Hamming spaces,β IEEE equations [16], and pattern recognition [17](awebsitecalled Transactions on Information Theory,vol.41,no.5,pp.1303β1321, theβKrawtchoukPolynomialsHomePageβbyV.Zelenkov 1995. at http://orthpol.narod.ru/eng/index.html collecting material [15] F.J.MacWilliamsandN.J.A.Sloane,Theory of Error-Correcting about M. Krawtchouk and the polynomials that bear his Codes,North-Holland,NewYork,NY,USA,1978. name has, unfortunately, not been updated in a while). How- [16] L. Boelen, G. Filipuk, C. Smet, W. Van Assche, and L. ever, our motivation in this paper was primarily driven by Zhang, βThe generalized Krawtchouk polynomials and the generalising the results obtained in [8]βand thus shedding fifth Painleve equation,β Journal of Difference Equations and more light on the qualitative structure of (generalisations Applications,vol.19,pp.1437β1451,2013. of) Krawtchouk polynomialsβand not yet with a specific [17] J. Sheeba Rani and D. Devaraj, βFace recognition using application in mind. By providing the Mathematica code to Krawtchouk moment,β SadhanΒ― aΒ―,vol.37,no.4,pp.441β460, obtain π-polynomials, we also hope that other researchers 2012. areencouragedtoexplorethemandseeiftheycanbeusedin their research.
Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.
References
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