Biquaternion Z Transform Is Illustrated Via Several Exam- Explanation

Home , Biquaternion, Definable real number

25 15 20 10 arXiv:2108.02975v15 [math.CA] 6 Aug 2021 F ifrnetm oanmto 4.Dgtlcmestr ca compensators Digital [4]. method domain fini time frequency-dependent difference of model mathematical the tablish estrt upeste”idn siltos n[] Rab [8], In oscillations. ”hidden” be- the the c suppress digital examine 7 a to design to pensator [6, to transform and Jury transients Z and sampling modified between system [5] havior the data Schroeder use sampled and to the Jury proposed of particular, response In the [5]-[7]. improve to used be uaintm.I 9,Ln n oge l ple Z in del CZT uncertain applied of problem the al. solved et orthogona and modulation, coherent Song plitude digital and of com Leng estimation the offset [9], frequency reducing In use greatly to thus time. is size, putation dis any idea the of main of transform transform The Fourier Fourier fast the transform. realize Z to convolution classical the (CZT on transform Z based chirp the defined together Rader and Schafer ucinof function a n upto h ape-aasses h rnfr of transform Z input The the systems. of sequence sampled-data analysis a the work an of includes The output [3] and transforms University. Zadeh Columbia Z and of called Ragazzini the Zadeh was of of and and problem Ragazzini This [2], function by Hurewicz transfer [1]. by the filter equation to difference impulse applied the fir was solve was to method transform 1947 Z in The proper proposed [1]-[13]. interesting applications its to potential due and engineers from attention more Introduction 1. biquate the of comple convergence of the region of pr the extension is analyze natural to method an employed transformation is Z transform biquaternion Z queternion the work, this In Abstract engvn ti hw httepooe rprisi helpf is properties proposed the that shown is It given. been rpitsbitdt ora fL of Journal to submitted Preprint aebe ie oilsrt h fetvns ftepropo Keywords: the of effectiveness the illustrate to given been have ( ∗ z mi address: Email author Corresponding uigteps eae,Ztasomhscpue oeand more captured has transform Z decades, past the During ) ol ecmlxvle.Oeo t plctosi oes- to is applications its of One complex-valued. be could utrinagba,rcrec eain,tase func transfer relations, recurrence algebras, Quaternion f n z sdfie as defined is [email protected] ncmlxdmi.I eea,both general, In domain. complex in eateto ahmtc,Fclyo cec n Technolo and Science of Faculty Mathematics, of Department eateto ahmtc,Fclyo cec n Technolo and Science of Faculty Mathematics, of Department A T E Templates X F colo cec,HbiUiest fTcnlg,Whn H Wuhan, Technology, of University Hubei Science, of School ( z KtInKou) Ian (Kit = ) P n ∞ =0 f iutrinZTransform Z Biquaternion n z − n hc is which , f n hnFn Cai Zhen-Feng am- l e einmethod. design sed crete iner, i a Kou Ian Kit om- lt nesadtebqaeno rnfr.Fnly sev Finally, transform. Z biquaternion the understand to ul esa Bi Wenshan and ties ay. te- st n ] ) - rnfr.I h einpoes pca ompresentat norm special process, design the In transform. Z x no emtysqec.I diin oeueu properti useful some addition, In sequence. geometry rnion psdt ov ls fbqaeno eurnerelation recurrence biquaternion of class a solve to oposed 40 30 55 50 35 45 in transform Z tion, ini ope-aud hc sdfcl oaayeiscon biquater its analyze of to difficult norm is the which complex-valued, speaking, is nion Generally before. discussed method was transform Laplace wi It by equations solved transform. be differential can Z quaternion domain the continuous the using that by [14] domain in Z shown tran discrete the a with into in transformed system function is fer a model sequence with the dealing time, sampling When fo function analysis. transfer oretical domain ordin the the into transform model, to equation domain used differential usually time is continuous transform Laplace the the in ti continuous Specifically, in transform Laplace systems. the by acted is which role 13]. [12, frequencies low at accura even enables detection CZT environment, fading in compensation nel 1,1] eie,i h hne rdcinpoeso chan of process prediction channel the in Besides, sys distribution 11]. in effectiv the [10, component in flicker role voltage important of an plays estimation CZT hand, other the On utrini xlie,abqaeno rnfr method transform bi Z of biquaternion presented. norm a bette real-valued exploited, that special fact is a the quaternion when by obtained motivated be paper, can results this In domain. gence .Fo h ento,w aclt h rnfr fsome of transform Z the calculate we definition, the From 2. rapid made has transform Z the of theory the Although 1. ∗ h rnfr ftebqaeno oanhsrrl been rarely has domain biquaternion the of transform Z The the acts transform Z the data sampled with working When h otiuin fti okaesmaie sfollows. as summarized are work this of contributions The pca ucin,gtmn neetn rpris The properties. interesting many get functions, special 2. Section suit- in the transform give Z biquaternion we of paper definition in this able discussed In been domain. not biquaternion has the issue this However, progress. y nvriyo aa,Mco China Macao, Macau, of University gy, y nvriyo aa,Mco China Macao, Macau, of University gy, bi400,P.R.China 430000, ubei rlexamples eral uut9 2021 9, August shave es .Bi- s. o is ion the- r ver- ha th tem ary me s- te is e - r - - . basic properties and special function’s Z transform are Besides, Gürlebeck give the real-valued norm in H(C). Let given in Table 1 and Table 2. x ∈ H(C), the real-valued norm kxk has the following form

4 2 2 3. The difficulty process of solving a class of biquaternion kxk = |kxkC| . (4) 60 recurrencerelations is transformed into an algebraic quaternion problem. This method makes solving a class of bi- It’s worth noting that the real-valued norm satisfies kxyk = quaternion recurrence relations and the related initial val- kxkkyk. In this paper, we use k·k to represent real-valued ues problem much convenient. Finally, the validity of the norm of biquaternion as above defined unless we give a special biquaternion Z transform is illustrated via several exam- explanation. 65 ples. 90 Accordingto [15, 16], the overall descriptionof high-dimensional data by using biquaternion can effectively represent its inherent The rest of this paper is organized as follows. The defi- properties. In this paper, in order to better distinguish domains, nitions about biquaternion sequence, such as convergence of we use plain fonts to represent real or complex numbers, and biquaternion sequence and biquaternion Z transform are given use bold to represent real quaternions or biquaternions, unless in section 2. In the following, we investigate its main prop- 95 we give a special explanation. In addition, we use i, j, k to rep- 70 erties and special biquaternion valued function’s Z transform resent the imaginary unit of a quaternion number, noting that are given in section 3. The method of using biquaternion z they are commutative with the complex unit I. For more de- transform to solve biquaternion recurrence relations (60) is pre- tails, please refer to [17, 18]. sented in section 4. Finally, we conclude this paper in section 5. 2.2. Biquaternion geometry sequence

100 Before disscussing the definition of the Z transform in the 75 2. Preliminary H(C) domain, we first introduce the convergence of biquaternion sequence. 2.1. Biquaternions An infinite sequence The concept of biquaternion, a generalization of complex numbers, was first proposed by Hamilton in 1853. In general, f0, f1, ··· , fn, ··· (5) we adopt the symbol H(C) to denote the set of biquaternionand of biquaternion has a limit f, f ∈ H(C), if, for each positive H(R) to denotethe set of real quaternion. For each biquaternion q, it can be written as the following form number ε, there exists a positive integer n0 such that

kfn − fk < ε, (6) q = q0 + q1i + q2j + q3k, (1) whenever n>n . The sequence (5) can have at most one limit. where q ∈ C,n =0, 1, 2, 3. The basis elements {i, j, k} obey 0 n f the Hamilton’s multiplication rules That is, a limit is unique if it exists. When that limit exists, the sequence is said to converge to f ; and we write i2 = j2 = k2 = ijk = −1; lim fn = f. (7) ij = k, jk = i, ki = j; (2) n→∞ ji k kj i ik j = − , = − , = − . Let fn ∈ H(C), (n =1, 2,...). If the sequence

For any biquaternion p, q ∈ H(C), we write p = p0 + N p1i + p2j + p3k, q = q0 + q1i + q2j + q3k, where pn, qn ∈ SN = fn = f1 + f2 + ··· + fN (8) C,n = 0, 1, 2, 3. Let p = p + p, we use p := Sc(p) be n=0 0 0 X 80 the scalar part of p, p =: V ec(p) be the vector part of p, and of patital sums convergence to S, S ∈ H(C). Then we say the p =: p0 −p be the quaternion conjugate of p. Define (p, q) := infinite sequence p1q1+p2q2+p3q3 and pq := p0q0−(p, q)+p0q+q0p+p×q. In [18], Gürlebeck give the complex-valued norm in H(C). ∞ H C H R R Let x ∈ ( ), a, b ∈ ( ), an,bn ∈ (n = 0, 1, 2, 3), fn = f1 + f2 + ··· + fn + ··· (9) x = a + Ib, where a = a + a i + a j + a k = a + a, n=0 0 1 2 3 0 X b = b0 + b1i + b2j + b3k = b0 + b, and I be the complex unit. convergence to the sum S, and we write Then we adopt the symbol kxkC to denote the complex-valued norm of the element x, which has the following form ∞ f = S. (10) 2 2 2 n kxkC := xx = |a| − |b| +2I[a0b0 + (a, b)], (3) n=0 X 2 2 where |a| := aa, |b| := bb, a =: a0 − a, b =: b0 − b. In When a sequence does not convergence,we say that it diverges. −1 x general, we use x := 2 to denote the inverse of x, where By direct calculate, we have the following proposition. kxkC 85 kxkC 6=0.

2 ∞ H C ∞ Proposition 2.1. Suppose f = {fn}n=0, fn, S ∈ ( ), and Deﬁnition 2.1. (Biquaternion Z transform) Let f = {fn}n=0, fn = mn + iwn + jxn + kyn, S = M + iW + jX + kY , then fn, x ∈ H(C). The Z transform X [f](x) of f is deﬁned by

∞ ∞ −1 −n −n fn = S. (11) X [f](x) := f0 +f1x +···+fnx +··· = fnx . (21) n=0 n=0 X X If and only if 115 The region of convergence (ROC) is the set of all x values that ∞ ∞ ∞ ∞ make the defined sequence converge, that is kxk > σf . Among mn = M, wn = W, xn = X, yn = Y. (12) them, σf (0 ≤ kσf k < ∞) is defined as the radius of conver- n=0 n=0 n=0 n=0 gence of the Z transform of f. X X X X 105 2.3. Biquaternion Z transform Remark 2.1. (i) If fn = 1 for all n ≥ 0, the Z transform ∞ H C In this section, we define the Z transform in the H(C) do- X [f](x) of sequence {fn}n=0 in ( ) is defined by main. Firstly, we introducethe biquaternion geometric sequence ∞ as follows. X [f](x) := x−n = (1 − x−1)−1, (22) n=0 Lemma 2.1. Let y ∈ H(C), we have X ∞ which is convergent for all x such that kxk > 1. More yn = (1 − y)−1, kyk < 1. (13)120 precise, the ROC of X [f](x) is kxk > 1. n=0 n X (ii) When fn ∈ H(C)(n ≥ 0), x ∈ C. Let fn = p , where p is Proof. Firstly, let SN ∈ H(C). We use SN (y) to denote the a nonzero biquaternion number. The Z transform X [f](x) ∞ n ∞ partial sums of n=0y , that is of sequence {fn}n=0 is defined by

N−1 ∞ P −n −1 −1 S (y)= yn =1+ y + y2 + ··· + yN−1. (14) X [f](x) := fnx = (1 − px ) , N (23) n=0 n=0 X X |x| > kpk, Then we multiply Eq. (14) by y on the left-hand side,

N−1 X [f](x) is convergence for all x such that |x| > kpk. n 2 N f p ySN (y)= y y = y + y + ··· + y . (15) More precise, the ROC of X [ ](x) is |x| > k k. Define n=0 σf to be the radius of convergence of the Z transform of X fn. Here σf := kpk. Using Eq. (14)-(15) C H C n N (iii) When fn ∈ (n ≥ 0), x ∈ ( ). Let fn = p , where (1 − y)SN (y)=1 − y . (16) p is a nonzero complex number. The Z transform X [f](x) ∞ Thus we know of sequence {fn}n=0 is defined by ∞ −1 N SN (y) =(1 − y) (1 − y ). (17) −n −1 −1 X [f](x) := fnx = (1 − px ) , −1 n=0 (24) When kyk < 1, (1 − y) 6=0, (1 − y) is well defined. Define X kxk > |p|, S(y) := (1 − y)−1, (18) 125 X [f](x) is convergence for all x such that kxk > |p|. using Eq. (17)-(18) More precise, the ROC of X [f](x) is kxk > |p|.

ρN (y) :=S(y) − SN (y) (iv) When x, p ∈ C, and p 6= 0. The Z transform X [f](x) of n =(1 − y)−1 − (1 − y)−1(1 − yN ) (19) the sequence fn = p (n ≥ 0) is deﬁned by =(1 − y)−1yN . ∞ −n −1 −1 X [f](x) := fnx = (1 − px ) , Thus n=0 (25) X −1 N −1 N |x| > |p|, kρN (y)k = k(1 − y) y k = k(1 − y) kkyk , (20) where |x| denote the absolute value of x. and it is clear from this that the remainders ρN (y) tends to zero 110 when kyk < 1 but not when kyk ≥ 1. Therefore, the proof of equation (13) is completed. 3. Properties

As an immediate consequenceof Proposition 2.1 and Lemma Similar to complex Z transform, the biquaternion Z trans- 2.1, we present the deﬁnition of biquaternion Z transform in the130 form has many useful properties. First of all, the following the- following. orem gives some calculation rules for Z transformation. For 3 the sake of simplicity, we have introduced some symbols for For (II) we ﬁnd numerical sequence operations. Here, σ represents the radius f ∞ of convergence of the Z transform of fn, which indicates that −n X [g](x)= gnx 135 the sequence is convergent for x > σ and divergent for k k f n=0 kxk < σf . X∞ n −n = fnq x ∞ ∞ Theorem 3.1. Let f = {fn}n=0, g = {gn}n=0, fn, gn ∈ n=0 (33) H(C). X [f] and X [g] denote the biquaternion Z transform of f X∞ −1 −n and g, respectively. = fn(q x) n=0 (I) For all constants c , c ∈ H(C), we have X 1 2 =X [f](q−1x).

X [c1f + c2g](x)=c1X [f](x)+ c2X [g](x), −1 (26) Since kq xk > σf , therefore kxk > σf kqk. kxk > max(σf , σg). For (III),

∞ X [fc1 + gc2](x)=X [f](x)c1 + X [g](x)c2, −n (27) X [g](x)= gnx kxk > max(σ , σ ). f g n=0 X∞ X [c1f + gc2](x)=c1X [f](x)+ X [g]c2(x), −n (28) = fn+kx kxk > max(σf , σg). n=0 X∞ (II) Let q be a nonzero biquaternion number and g f qn, −(n+k) k n = n = fn+kx x (34) n =0, 1, 2, ··· . If qx = xq, then n=0 X −1 −1 =(X [f](x) − f0 − f1x X [g](x)= X [f](q x), kxk > σf kqk. (29) −2 −(k−1) k − f2x −···− fk−1x )x k (III) Let k > 0 be a ﬁxed integer and gn = fn+k, n = −1 k k−n 0, 1, 2, ··· , then =X [f](x)x − fnx , kxk > σf . n=0 X X [g](x) =(X [f](x) − f − f x−1 0 1 For (IV), −2 −(k−1) k − f2x −···− fk−1x )x ∞ k k −n =X [f](x)x − f0x (30) X [g](x)= gnx k−1 n=0 − f1x −···− fk−1x, X∞ kxk > σf . −n = fn−kx n=0 X 140 (IV) Conversely, if k is a positive integer and gn = fn−k ∞ −(n−k) −k (35) for n ≥ k and gn = 0 for n < k, then X [g](x) = = fn−kx x X [f](x)x−k . n=0 X∞ −(n−k) −k (V) Let x be a complex-valued number and gn = nfn,n = = fn−kx x 0, 1, 2, ··· ,then n=k X =X [f](x)x−k , kxk > σ . d f X [g](x)= −x [X [f](x)], |x| > σf . (31) dx For (V), the right-hand side of equation (31) can write

Proof. For (I), we only give the proof of Eq.(26). Eq.(27) and d d ∞ −x [X [f](x)] = − x ( f x−n) Eq.(28) can be derived by the similar manner. Since X [f](x)= dx dx n ∞ −n ∞ −n n=0 fnx and X [g](x)= gnx , we have n=0 n=0 ∞ X −n−1 P ∞ P ∞ = − x fn(−n)x −n −n (36) X [c1f + c2g](x)= c1fnx + c2gnx n=0 ∞ X n=0 n=0 −n X∞ X∞ = (nfn)x −n −n (32) n=0 =c1 fnx + c2 gnx X n=0 n=0 =X [g](x). X X =c1X [f](x)+ c2X [g](x), where |x| > σf . kxk > max(σf , σg). 4 Table 1: Properties of z transform Property Function Ztransform ROC

Left linearity c1f + c2g c1X [f](x)+ c2X [g](x) kxk > max(σf , σg) Right linearity fc1 + gc2 X [f](x)c1 + X [g](x)c2 kxk > max(σf , σg) Two-side linearity c1f + gc2 c1X [f](x)+ X [g](x)c2 kxk > max(σf , σg) n n −1 q -scaling fnq X [f](q x) kxk > σf kqk ′ n-scaling nfn −xX [f] (x) |x| > σf , x ∈ C k k−1 k−n Shifting I fn+k X [f](x)x − n=0fnx kxk > σf −k Shifting II fn−k X [f](x)x kxk > σf n P C Convolution k=0fn−kgk X [f](x)X [g](x) |x| > max(σf , σg), x ∈ P Theorem 3.2. Let x ∈ C. If f and g are two biquaternion- Example 3.1. valued number sequence, w, called the convolution of f and g, −1 −1 by writing X [1](x) =(1 − x ) , kxk > 1, x ∈ H(C), X [n](x)=x(x − 1)−2, kxk > 1, x ∈ H(C), w =(f ∗ g) n n 2 2 −3 H C n X [n ](x) =(x + x)(x − 1) , kxk > 1, x ∈ ( ), n −1 −1 = fn−kgk X [p ](x) =(1 − px ) , |x| > kpk, x ∈ C, p ∈ H(C), (37) k=0 n −1 −1 C H C Xn X [np ](x)=p(1 − px ) , |x| > kpk, x ∈ , p ∈ ( ). (42) = fkgn−k, n =0, 1, 2, ··· . k=0 X Example 3.2. Let q ∈ H(C), and q = q0 + q = q0 + q1i + j k The z transform X [f](x) is convergence in |x| > σf and X [g](x) q2 + q3 . According to [14], we know

is convergence in |x| > σg, where x is complex valued. Then q q 1 |q| qn − |q| qn the convolution of f and g has Z transform in |x| > max(σf , σg) e + e , |q| 6=0 cos(qn)= 2 (43) and  cos( q0n) − qn sin(q0n), |q| =0, X [w](x)= X [f](x)X [g](x). (38) 

q q  q qn − qn Proof. From the definition, we have − 1 e |q| − e |q| , |q| 6=0 sin(qn)= 2 |q| (44)  ∞ n sin(q0n)+ qn cos(q0n), |q| =0, −n  X [w](x)= fn−kgk x ! and n=0 k=0 (39)  X∞ ∞X q0 −n e (cos(|q|)+ sgn(q) sin(|q|)), |q| 6=0 = fn−kgkx . eq = (45) q0 k n k e (1 + q), |q| =0. X=0X= ( Let m = n − k, we can rewrite Then, we have ∞ ∞ X [cos(qn)](x) w x f g x−(n−k)x−k X [ ]( )= n−k k q q k=0n=k 1 |q| −1 −1 2 (1 − e x ) XX (40) q (46) ∞ ∞ − q −m −k = + 1 (1 − e |q| x−1)−1, |q| 6=0 = fmx gkx . 22  x −x cos(q) k=0 m=0 !  2 , |q| =0. X X x −2x cos(q)+1 Because f and g both have z transform, then we get  m k X[sin(qn)](x) q ∞ ∞ q q −m −k 1 e |q| x−1 −1 X [w](x)= fmx gkx = X [f](x)X [g](x). (41) − 2 |q| (1 − ) q q (47) m=0 k=0  1 − q q X X = + (1 − e | | x−1)−1, |q| 6=0  2 |q| It is clear that since both sequences needs to converge in this  x sin(q) x2−2x cos(q)+1 , |q| =0. 145 domain, the final ROC is |x| > max(σf , σg).  Proof. Without loss of generality, we only give the proof of the The properties of z transform are summarized in Table 1. In X [cos qn](x) here. Nowwe see that, when |q| =0, X [cos qn](x) the following, we give the z transformation of several special functions. 5 −1 −1 is actually doing the classical Z transformation. As for |q| 6=0, and S0 = (1 − qx ) , then taking Z transform on both sides of the Eq. (43), we have m n X [Cn+mq ](x) ∞ q q −1 −m −1 −1 1 q qn − q qn −n (54) X [cos(qn)](x)= e | | + e | | x =(1 − x q) (1 − qx ) , 2 m=0 kxk > kqk. X ∞ q 1 qn −n = e |q| x (48) ∞ m n −n 2 (II) Let Sm = n=m Cn q x , then m=0 X ∞ q −1 1 − qn (xq P− 1)Sm + e |q| x−n, 2 ∞ ∞ m=0 m n−1 −n+1 m n −n (55) X = Cn q x − Cn q x , q q n=m n=m and since X [pn](x)=(1 − px−1)−1, let p = e |q| , therefore, X X since the first part of the above formula is equal to q m−1 −m+1 ∞ m n−1 −n+1 1 |q| q −1 −1 q x C q x X [cos(qn)](x)= (1 − e x ) + n=m+1 n , then 2 q (49) −1 m−1 −m+1 1 − q (xq − 1)S P=q x + (1 − e |q| x−1)−1, kxk > kpk. m 2 ∞ m m n −n + (Cn+1 − Cn )q x n=m ∞X (56) m−1 n −n m n m n = C q x Example 3.3. Let an = Cn mq , bn = Cn q , and cn = n n + q ∞ ∞ n=m−1 n! . Z transform of the sequences a = {an}n=0, b = {bn}n=0, X ∞ =Sm−1. c = {cn}n=0 are the following: −1 −1 −1 −2 X [a](x) =(1 − x−1q)−m(1 − qx−1)−1, kxk > kqk, Then, Sm = (xq −1) Sm−1 = (xq −1) Sm−2, −1 −1 and S0 = (1 − qx ) . Therefore, X [b](x) =(xq−1 − 1)−m(1 − qx−1)−1, kxk > kqk, 1 m n −1 −m −1 −1 qx− X [C q ](x) =(xq − 1) (1 − qx ) , X [c](x)=e , kxk > kqk, n (57) (50) kxk >kqk.

150 Proof. 1 (III) First, let dn = n! , then ∞ m n −n (I) Let Sm = n=0 Cn+mq x , then ∞ 1 −1 X [d ](x)= x−n = ex , kxk > 0. (58) P ∞ n n! −1 −1 m n −n n=0 (1 − x q)Sm =(1 − x q) Cn+mq x X n=0 By rule (ii) of Theorem 3.1, ∞ X m n −n n = C q x (51) q 1 −1 n+m X [ ](x)= X [ ](q−1x)= eqx . (59) n=0 n! n! X∞ m n −n − Cn+m−1q x , n=1 X since the ﬁrst part of the aforementionedequation is equal ∞ m n −n 4. Examples to 1+ n=1 Cn+mq x , then we have

P ∞ Suppose sequence fn is unknown and some initial values −1 m m n −n (1 − x q)Sm =1 + (Cn+m − Cn+m−1)q x of fn are given. Consider a class of biquaternion recurrence n=1 relations as follows X∞ m−1 n −n M =1 + Cn+m−1q x fn mpm n=1 + X m=0 S . X (60) = m−1 M T L (52) = g q + h u + ··· + y d , Therefore, n+k k n+t t n+l l k t=0 l X=0 X X=0 −1 −1 Sm =(1 − x q) Sm−1 where sequence f , g , h , , y , and coefﬁcients p , q , (53) n n n ··· n n n −1 −2 =(1 − x q) Sm−2, un, ··· , dn are all known. If gn = hn = ··· = yn = 0 or 6 Table 2: Some sequence gn and their biquaternion z transform X [g](x)

gn X [g] 1 1 (1 − x−1)−1 2 n x(x − 1)−2 3 n2 (x2 + x)(x − 1)−3 4 pn (1 − px−1)−1 5 npn p(1 − px−1)−1 q q 1 |q| q −1 −1 1 − |q| q −1 −1 6 cos qn 2 (1 − e x ) + 2 (1 − e x ) q q q q 1 |q| q −1 −1 1 − |q| q −1 −1 7 sin qn − 2 |q| (1 − e x ) + 2 |q| (1 − e x ) m n −1 −m −1 −1 8 Cn+mq (1 − x q) (1 − qx ) 9 Cmqn (xq−1 − 1)−m(1 − qx−1)−1 n n q qx−1 10 n! e

155 qn = un = ··· = dn = 0, then we call the equation (60) Thus the equation (62) can be written as homogeneous, otherwise it becomes inhomogeneous. Next, we X [f](x) − 1 − (i + j)x−1 x2 will give several examples of the right-side coefficients homo- (65) geneous linear biquaternion equation and the right-side coef- = (X [f](x) − 1) x(i + j − 1)− X [f](x)(i + j), ficients inhomogeneous linear biquaternion equation, respec- 160 tively, to further explain how to solve the equations with the from which X [a](x) can be solved. After simplification we biquaternion z transform. have X [f](x) =(1 − (i + j)x−1)−1. (66) Example 4.1. If we know that f =1, f = i + j and 0 1 We found in the preceding Table 2 that this is the z transformation of the sequence fn+2 = fn+1(i + j − 1) + fn(i + j), n =0, 1, 2, ··· , (61) n fn = (i + j) , n =0, 1, 2, ··· . (67) find a formular for fn. Next, we can check the result by returning to the initial condi- Let X [f](x)= ∞ f x−n. It is clear that z transformationon n=0 n tions of the problem: f = 1, f = i + j are all right; and if both sides of the aforementioned equation yields an algebraic 0 1 f = (i + j)n and f = (i + j)n+1, then equation of X [fP](x), which is n n+1

∞ ∞ fn+1(i + j − 1)+ fn(i + j) −n −n n+1 n fn+2x = fn+1x (i + j − 1) =(i + j) (i + j − 1)+(i + j) (i + j) (68) n=0 n=0 n+2 X X ∞ (62) =(i + j) , + f x−n(i + j). n which is also right. n=0 X We notice that, firstly, Example 4.2. Determine the numbers fn, n =0, 1, 2, ··· , if f0 = 1, f1 = Ij, and for n =0, 1, 2, ··· ∞ ∞ −n −k fn+1x = fkx x fn+2 =2fn + fn+1(Ij). (69) n=0 k=1 ! X X F ∞ f −n ∞ (63) Let (x) = n=0 nx . It is clear that z transformation = f x−k − f x on both sides of equation (69) yields an algebraic equation of k 0 P k ! F(x), which is X=0 = (X [f](x) − 1) x, ∞ ∞ ∞ f x−n =2 f x−n + f x−n(Ij). (70) and n+2 n n+1 n=0 n=0 n=0 X X X ∞ ∞ −n −k 2 Using (30) form the Theorem 3.1, we get fn+2x = fkx x n=0 k=2 ! −1 2 X X [F(x) − f0 − f1x ]x =2F(x) + [F(x) − f0]x, (71) ∞ (64) = f x−k − f − f x−1 x2 k 0 1 where f0 =1, f1 = Ij and after computation we have k ! X=0 = X [f](x) − 1 − (i + j)x−1 x2. F(x) = [1 − (Ij)x−1]−1. (72) 7 Then we get the expression of fn, that is It is clear that if we taking z transforms for both sides of above equation, it will gives f Ij n. (73) n = ( ) ∞ ∞ ∞ f x−n − 2 f x−n + f x−n When we check the result, we find that it satisfies the initial n+2 n+1 n n=0 n=0 n=0 165 conditions and equation (69). X X X (79) =2x(x − 1)−1 + 2(1 − Ik)[1 − (Ik)x−1]−1 Example 4.3. Find the numbers fn, n = 0, 1, 2, ··· , such that − [1 − (Ik + 1)x−1]−1. −1 f0 =1, f1 = Ii + I and fn+2 = fn+1(−2I )+2fn. ∞ −n Let F(x) = n=0fnx . Using (30) from the Theorem 3.1, ∞ −n Let F(x) = n=0fnx . It is clear that z transformation on we get both sides of the aforementioned equation yields an algebraic P x2F(x) − 2xF(x)+ F(x) equation of FP(x), which is =2x(x − 1)−1 + 2(1 − Ik)[1 − (Ik)x−1]−1 (80) ∞ ∞ ∞ −n −n −1 −n − [1 − (Ik + 1)x−1]−1. fn+2x = fn+1x (−2I )+2 fnx . (74) n=0 n=0 n=0 X X X That is Using (30) from the Theorem 3.1, we get F(x) =2(1 − Ik)(x − 1)−2[1 − (Ik)x−1]−1 −3 −2 −1 −1 [F(x)−f −f x−1]x2 = [F(x)−f ]x(−2I−1)+2F(x), (75) 2x(x − 1) − (x − 1) [1 − (Ik + 1)x ] . 0 1 0 (81) Noting that where f0 =1, f1 = Ii + I. Then ∞ F(x)(x2 +2I−1x − 2) = x(x + Ii + I +2I−1). (76) (x − 1)−1 = x−n−1, n=0 That is X∞ F(x) = [1 − (Ii + I)x−1]−1. (77) (x − 1)−2 = (n + 1)x−n−2, (82) n=0 Using the equation (24) from definition 2.1, the final expression X ∞ (n + 1)(n + 2) can be rewritten as (x − 1)−3 = x−n−3, 2 n n=0 fn = (Ii + I) . (78) X and Obviously, it satisfies the initial conditions. ∞ [1 − (Ik)x−1]−1 = (Ik)nx−n, Remark 4.1. The solving process of the right-side coefficient n=0 X 170 linear homogeneous biquaternion recurrence relations is as fol- ∞ lows: [1 − (Ik + 1)x−1]−1 = (Ik + 1)nx−n (83) n=0 Step 1. Apply the biquaternion z transform (24) on both X∞ sides of the given equation. = 2n−1(Ik + 1)x−n, n=0 Step 2. Use the basic properties in Theorem 3.1 to sim- X which provide that 175 plify the result of step 1. ∞ (n + 1)(n + 2) Step 3. Using quaternion algebra, the result of step 2 is 2x(x − 1)−3 =2x x−n−3 reduced to a monadic equation about the z transforma- 2 n=0 (84) tion of the sequence. ∞X = (n2 − n)x−n, Step 4. The biquaternion sequence is obtained by com- n=0 180 paring Table 2 and step 3. X 2(1 − Ik)(x − 1)−2[1 − (Ik)x−1]−1 The three homogeneous biquaternion recurrence relations ∞ ∞ with right-side coefficients have been solved by the biquater- =2(1 − Ik) (n + 1)x−n−2 (Ik)nx−n nion z transformation method. In the following, let’s focus on n=0 n=0 X X the inhomogeneous biquaternion recurrence relations with the ∞ 2 n−1 185 right-side coefficients. = {2n − 2 − 2[(Ik) + (Ik) + ··· + (Ik) ]} n=0 X ∞ −n Example 4.4. Find fn, n = 0, 1, 2, ··· , such that f0 = f1 = 0 n=0{2n − 2 − (n − 1)(1 + Ik)}x , when n is odd; n n = ∞ −n and fn+2 − 2fn+1 + fn =2+(2 − 2Ik)(Ik) − (Ik + 1) . n=0{2n − 2 − (n − 2)(1 + Ik) − Ik}x , when n is even, P (85) P 8 and It is clear that the left hand side is the convolution of x and the −2 −1 −1 (x − 1) [1 − (Ik + 1)x ] function t → (3j)t, so that taking z transform of both members ∞ ∞ gives −n−2 n−1 −n = (n + 1)x 2 (Ik + 1)x (1 − 3jx−1)−1X [f](x)=(1 − 2ix−1)−1. (92) n=0 n=0 (86) X∞ X Use the result of Eq. (24). We get = {n − 1+(2n−1 − n)(Ik + 1)}x−n. n=0 X [f](x) X Thus, when n is odd, the equation (81) can be rewritten as =(1 − 3jx−1)(1 − 2ix−1)−1 (93) −1 −1 −1 −1 −1 ∞ =(1 − 2ix ) − 3jx (1 − 2ix ) , F(x)= {n2 − n +2n − 2 − (n − 1)(1 + Ik) and using Eq. (24) and rule (iv) of Theorem 3.1, we see that n=0 X n−1 −n + n − 1+(2 − n)(Ik + 1)}x (87) f(t)=(2i)t − 3j(2i)t−1,t =0, 1, 2, ··· . (94) ∞ = {n2 − (Ik + 1)n + Ik}x−n, n=0 5. Conclusion X and when n is even, the equation (81) can be rewritten as The solution of a class of biquaternion recurrence relations

∞ has been solved in this paper. After introducing the z transfor- F(x)= {n2 − n +2n − 2 − (n − 2)(1 + Ik) 200 mation of the biquaternion sequence, we give its basic proper- n=0 ties. Besides, we discussed the relationship between a biquaternion- X valued number sequence and a complex-valuednumber sequence: − Ik + n − 1+(2n−1 − n)(Ik + 1)}x−n (88) the convolution theorem. Moreover, applying the biquaternion ∞ z transform to the class of biquaternion recurrence relations can = {n2 − (Ik + 1)n +1}x−n. 205 indicate the capability and efﬁciency of the method. n=0 X From the above equations (87) and (88), the ﬁnal expression References can be rewritten as

2 n n References fn = n − (Ik + 1) + (Ik) . (89) [1] E. R. Kanasewich, Time sequence analysis in geophysics. University of We can also check the result by returningto the statement of Alberta, 1981. 2 210 the question: f0 =0 and f1 =0 are all right; and if fn = n − [2] H. M. James, W. B. Nichols, R. S. Phillips. Theory of Esrvomechanisms. n n 2 n+1 n+1 McGraw- Hill Book Company, Inc., New York, N. Y., chapter 5, 1947. (Ik+1) +(Ik) , fn+1 = (n+1) −(Ik+1) +(Ik) , 2 n+2 n+2 [3] J. R. Ragazzini, L. A. Zadeh. The analysis of sampled-data systems. Trans- and fn+2 = (n + 2) − (Ik + 1) + (Ik) , then actions of the American Institute of Electrical Engineers, Part II: Applica- tions and Industry, 1952, 71(5): 225-234. fn+2 − 2fn+1 + fn 215 [4] D. M. Sullivan. Frequency-dependent FDTD methods using Z transforms. 2 n+2 n+2 IEEE Transactions on Antennas and Propagation, 1992, 40(10): 1223-1230. =(n + 2) − (Ik + 1) + (Ik) [5] I. E. Jury, W. Schroeder. Discrete compensation of sampled-data and con- − 2(n + 1)2 + 2(Ik + 1)n+1 + n2 (90) tinuous control systems. Transactions of the American Institute of Electrical Engineers, Part II: Applications and Industry, 1957, 75(6): 317-325. n+1 n n − 2(Ik) − (Ik + 1) + (Ik) 220 [6] H. Freeman, O. Lowenschuss. Bibliography of sampled-data control sys- n n tems and Z-transform applications. IRE Transactions on Automatic Control, =2 + (2 − 2Ik)(Ik) − (Ik + 1) , 1958, 4(1): 28-30. [7] I. E. Jury. Synthesis and critical study of sampled-data control systems. which is also right. Transactions of the American Institute of Electrical Engineers, Part II: Ap- 225 plications and Industry, 1956, 75(3): 141-151. Remark 4.2. Different from solving the right-side coefficient [8] L. Rabiner, R. W. Schafer, C. Rader. The chirp z-transform algorithm. IEEE transactions on audio and electroacoustics, 1969, 17(2): 86-92. 190 linear inhomogeneous biquaternion recurrence relations, the [9] H. Leng, S. Yu, X. Li, et al. Frequency offset estimation for optical coher- inhomogeneous cases are often more complicated, which in- ent M-QAM detection using chirp z-transform. IEEE Photonics Technology creases the difficulty of calculation of quaternion algebras. In230 Letters, 2012, 24(9): 787-789. this case, we can consider using the Taylor expansion in the [10] F. Li, Y. Gao, Y. Cao, et al. Improved teager energy operator and im- third step, and finally find the sequence by referring to the def- proved chirp-Z transform for parameter estimation of voltage flicker. IEEE Transactions on Power Delivery, 2015, 31(1): 245-253. 195 inition (24) and properties of the z transformation of the bi- [11] W. Kang, J. Guo, H. Li, et al. Voltage flicker detection based on Chirp- quaternion. 235 z transform. 2010 Asia-Pacific Power and Energy Engineering Conference. IEEE, 2010: 1-4. Example 4.5. Find f(t), t =0, 1, 2, ··· , from the equation [12] T. Ding, A. Hirose. Fading channel prediction based on combination of complex-valued neural networks and chirp Z-transform. IEEE Transactions t on Neural Networks and Learning Systems, 2014, 25(9): 1686-1695. (3j)nf(t − n)=(2i)n,t =0, 1, 2, ··· . (91)240 [13] S. Tan, A. Hirose. Low-calculation-cost fading channel prediction using chirp z-transform. Electronics letters, 2009, 45(8): 418-420. n=0 X 9 [14] Z. F. Cai, K. I. Kou. Laplace transform: a new approach in solving linear quaternion differential equations. Mathematical Methods in the Applied Sciences, 2018, 41(11): 4033-4048. 245 [15] T. Bülow, G. Sommer, ”Hypercomplex signals-a novel extension of the analytic signal to the multidimensional case,” IEEE Trans. Signal Process., 2001, 49(11): 2844-2852. [16] J. P. Ward. Quaternions and Cayley numbers: Algebra and applications. Springer Science and Business Media, 2012. 250 [17] S. J. Sangwine, T. A. Ell, N. Le Bihan. Fundamental representations and algebraic properties of biquaternions or complexified quaternions. Advances in Applied Clifford Algebras, 2011, 21(3): 607-636. [18] K. Gürlebeck, W. Sprössig. Quaternionic and Clifford calculus for physi- cists and engineers. Wiley, 1997.

10 25 15 20 10 arXiv:2108.02975v15 [math.CA] 6 Aug 2021 F ifrnetm oanmto 4.Dgtlcmestr ca compensators Digital [4]. method domain fini time frequency-dependent difference of model mathematical the tablish estrt upeste”idn siltos n[] Rab [8], In oscillations. ”hidden” be- the the c suppress digital examine 7 a to design to pensator [6, to transform and Jury transients Z and sampling modified between system [5] havior the data Schroeder use sampled and to the Jury proposed of particular, response In the [5]-[7]. improve to used be uaintm.I 9,Ln n oge l ple Z in del CZT uncertain applied of problem the al. solved et orthogona and modulation, coherent Song plitude digital and of com Leng estimation the offset [9], frequency reducing In use greatly to thus time. is size, putation dis any idea the of main of transform transform The Fourier Fourier fast the transform. realize Z to convolution classical the (CZT on transform Z based chirp the defined together Rader and Schafer ucinof function a n upto h ape-aasses h rnfr of transform Z input The the systems. of sequence sampled-data analysis a the work an of includes The output [3] and transforms University. Zadeh Columbia Z and of called Ragazzini the Zadeh was of of and and problem Ragazzini This [2], function by Hurewicz transfer [1]. by the filter equation to difference impulse applied the fir was solve was to method transform 1947 Z in The proper proposed [1]-[13]. interesting applications its to potential due and engineers from attention more Introduction 1. biquate the of comple convergence of the region of pr the extension is analyze natural to method an employed transformation is Z transform biquaternion Z queternion the work, this In Abstract engvn ti hw httepooe rprisi helpf is properties proposed the that shown is It given. been rpitsbitdt ora fL of Journal to submitted Preprint aebe ie oilsrt h fetvns ftepropo Keywords: the of effectiveness the illustrate to given been have ( ∗ z mi address: Email author Corresponding uigteps eae,Ztasomhscpue oeand more captured has transform Z decades, past the During ) ol ecmlxvle.Oeo t plctosi oes- to is applications its of One complex-valued. be could utrinagba,rcrec eain,tase func transfer relations, recurrence algebras, Quaternion f n z sdfie as defined is [email protected] ncmlxdmi.I eea,both general, In domain. complex in eateto ahmtc,Fclyo cec n Technolo and Science of Faculty Mathematics, of Department eateto ahmtc,Fclyo cec n Technolo and Science of Faculty Mathematics, of Department A T E Templates X F colo cec,HbiUiest fTcnlg,Whn H Wuhan, Technology, of University Hubei Science, of School ( z KtInKou) Ian (Kit = ) P n ∞ =0 f iutrinZTransform Z Biquaternion n z − n hc is which , f n hnFn Cai Zhen-Feng am- l e einmethod. design sed crete iner, i a Kou Ian Kit om- lt nesadtebqaeno rnfr.Fnly sev Finally, transform. Z biquaternion the understand to ul esa Bi Wenshan and ties ay. te- st n ] ) - rnfr.I h einpoes pca ompresentat norm special process, design the In transform. Z x no emtysqec.I diin oeueu properti useful some addition, In sequence. geometry rnion psdt ov ls fbqaeno eurnerelation recurrence biquaternion of class a solve to oposed 40 30 55 50 35 45 in transform Z tion, ini ope-aud hc sdfcl oaayeiscon biquater its analyze of to difficult norm is the which complex-valued, speaking, is nion Generally before. discussed method was transform Laplace wi It by equations solved transform. be differential can Z quaternion domain the continuous the using that by [14] domain in Z shown tran discrete the a with into in transformed system function is fer a model sequence with the dealing time, sampling When fo function analysis. transfer oretical domain ordin the the into transform model, to equation domain used differential usually time is continuous transform Laplace the the in ti continuous Specifically, in transform Laplace systems. the by acted is which role 13]. [12, frequencies low at accura even enables detection CZT environment, fading in compensation nel 1,1] eie,i h hne rdcinpoeso chan of process prediction channel the in Besides, sys distribution 11]. in effectiv the [10, component in flicker role voltage important of an plays estimation CZT hand, other the On utrini xlie,abqaeno rnfr method transform bi Z of biquaternion presented. norm a bette real-valued exploited, that special fact is a the quaternion when by obtained motivated be paper, can results this In domain. gence .Fo h ento,w aclt h rnfr fsome of transform Z the calculate we definition, the From 2. rapid made has transform Z the of theory the Although 1. ∗ h rnfr ftebqaeno oanhsrrl been rarely has domain biquaternion the of transform Z The the acts transform Z the data sampled with working When h otiuin fti okaesmaie sfollows. as summarized are work this of contributions The pca ucin,gtmn neetn rpris The properties. interesting many get functions, special 2. Section suit- in the transform give Z biquaternion we of paper definition in this able discussed In been domain. not biquaternion has the issue this However, progress. y nvriyo aa,Mco China Macao, Macau, of University gy, y nvriyo aa,Mco China Macao, Macau, of University gy, bi400,P.R.China 430000, ubei rlexamples eral uut9 2021 9, August shave es .Bi- s. o is ion the- r ver- ha th tem ary me s- te is e - r - - . basic properties and special function’s Z transform are Besides, Gürlebeck give the real-valued norm in H(C). Let given in Table 1 and Table 2. x ∈ H(C), the real-valued norm kxk has the following form

N−1 ∞ P −n −1 −1 S (y)= yn =1+ y + y2 + ··· + yN−1. (14) X [f](x) := fnx = (1 − px ) , N (23) n=0 n=0 X X |x| > kpk, Then we multiply Eq. (14) by y on the left-hand side,

X [c1f + c2g](x)=c1X [f](x)+ c2X [g](x), −1 (26) Since kq xk > σf , therefore kxk > σf kqk. kxk > max(σf , σg). For (III),

155 qn = un = ··· = dn = 0, then we call the equation (60) Thus the equation (62) can be written as homogeneous, otherwise it becomes inhomogeneous. Next, we X [f](x) − 1 − (i + j)x−1 x2 will give several examples of the right-side coefficients homo- (65) geneous linear biquaternion equation and the right-side coef- = (X [f](x) − 1) x(i + j − 1)− X [f](x)(i + j), ficients inhomogeneous linear biquaternion equation, respec- 160 tively, to further explain how to solve the equations with the from which X [a](x) can be solved. After simplification we biquaternion z transform. have X [f](x) =(1 − (i + j)x−1)−1. (66) Example 4.1. If we know that f =1, f = i + j and 0 1 We found in the preceding table 2 that this is the z transformation of the sequence fn+2 = fn+1(i + j − 1) + fn(i + j), n =0, 1, 2, ··· , (61) n fn = (i + j) , n =0, 1, 2, ··· . (67) find a formular for fn. Next, we can check the result by returning to the initial condi- Let X [f](x)= ∞ f x−n. It is clear that z transformationon n=0 n tions of the problem: f = 1, f = i + j are all right; and if both sides of the aforementioned equation yields an algebraic 0 1 f = (i + j)n and f = (i + j)n+1, then equation of X [fP](x), which is n n+1

∞ ∞ fn+1(i + j − 1)+ fn(i + j) −n −n n+1 n fn+2x = fn+1x (i + j − 1) =(i + j) (i + j − 1)+(i + j) (i + j) (68) n=0 n=0 n+2 X X ∞ (62) =(i + j) , + f x−n(i + j). n which is also right. n=0 X We notice that, firstly, Example 4.2. Determine the numbers fn, n =0, 1, 2, ··· , if f0 = 1, f1 = Ij, and for n =0, 1, 2, ··· ∞ ∞ −n −k fn+1x = fkx x fn+2 =2fn + fn+1(Ij). (69) n=0 k=1 ! X X F ∞ f −n ∞ (63) Let (x) = n=0 nx . It is clear that z transformation = f x−k − f x on both sides of equation (69) yields an algebraic equation of k 0 P k ! F(x), which is X=0 = (X [f](x) − 1) x, ∞ ∞ ∞ f x−n =2 f x−n + f x−n(Ij). (70) and n+2 n n+1 n=0 n=0 n=0 X X X ∞ ∞ −n −k 2 Using (30) form the Theorem 3.1, we get fn+2x = fkx x n=0 k=2 ! −1 2 X X [F(x) − f0 − f1x ]x =2F(x) + [F(x) − f0]x, (71) ∞ (64) = f x−k − f − f x−1 x2 k 0 1 where f0 =1, f1 = Ij and after computation we have k ! X=0 = X [f](x) − 1 − (i + j)x−1 x2. F(x) = [1 − (Ij)x−1]−1. (72) 7 Then we get the expression of fn, that is It is clear that if we taking z transforms for both sides of above equation, it will gives f Ij n. (73) n = ( ) ∞ ∞ ∞ f x−n − 2 f x−n + f x−n When we check the result, we find that it satisfies the initial n+2 n+1 n n=0 n=0 n=0 165 conditions and equation (69). X X X (79) =2x(x − 1)−1 + 2(1 − Ik)[1 − (Ik)x−1]−1 Example 4.3. Find the numbers fn, n = 0, 1, 2, ··· , such that − [1 − (Ik + 1)x−1]−1. −1 f0 =1, f1 = Ii + I and fn+2 = fn+1(−2I )+2fn. ∞ −n Let F(x) = n=0fnx . Using (30) from the Theorem 3.1, ∞ −n Let F(x) = n=0fnx . It is clear that z transformation on we get both sides of the aforementioned equation yields an algebraic P x2F(x) − 2xF(x)+ F(x) equation of FP(x), which is =2x(x − 1)−1 + 2(1 − Ik)[1 − (Ik)x−1]−1 (80) ∞ ∞ ∞ −n −n −1 −n − [1 − (Ik + 1)x−1]−1. fn+2x = fn+1x (−2I )+2 fnx . (74) n=0 n=0 n=0 X X X That is Using (30) from the Theorem 3.1, we get F(x) =2(1 − Ik)(x − 1)−2[1 − (Ik)x−1]−1 −3 −2 −1 −1 [F(x)−f −f x−1]x2 = [F(x)−f ]x(−2I−1)+2F(x), (75) 2x(x − 1) − (x − 1) [1 − (Ik + 1)x ] . 0 1 0 (81) Noting that where f0 =1, f1 = Ii + I. Then ∞ F(x)(x2 +2I−1x − 2) = x(x + Ii + I +2I−1). (76) (x − 1)−1 = x−n−1, n=0 That is X∞ F(x) = [1 − (Ii + I)x−1]−1. (77) (x − 1)−2 = (n + 1)x−n−2, (82) n=0 Using the equation (24) from definition 2.1, the final expression X ∞ (n + 1)(n + 2) can be rewritten as (x − 1)−3 = x−n−3, 2 n n=0 fn = (Ii + I) . (78) X and Obviously, it satisfies the initial conditions. ∞ [1 − (Ik)x−1]−1 = (Ik)nx−n, Remark 4.1. The solving process of the right-side coefficient n=0 X 170 linear homogeneous biquaternion recurrence relations is as fol- ∞ lows: [1 − (Ik + 1)x−1]−1 = (Ik + 1)nx−n (83) n=0 Step 1. Apply the biquaternion z transform (24) on both X∞ sides of the given equation. = 2n−1(Ik + 1)x−n, n=0 Step 2. Use the basic properties in Theorem 3.1 to sim- X which provide that 175 plify the result of step 1. ∞ (n + 1)(n + 2) Step 3. Using quaternion algebra, the result of step 2 is 2x(x − 1)−3 =2x x−n−3 reduced to a monadic equation about the z transforma- 2 n=0 (84) tion of the sequence. ∞X = (n2 − n)x−n, Step 4. The biquaternion sequence is obtained by com- n=0 180 paring table 2 and step 3. X 2(1 − Ik)(x − 1)−2[1 − (Ik)x−1]−1 The three homogeneous biquaternion recurrence relations ∞ ∞ with right-side coefficients have been solved by the biquater- =2(1 − Ik) (n + 1)x−n−2 (Ik)nx−n nion z transformation method. In the following, let’s focus on n=0 n=0 X X the inhomogeneous biquaternion recurrence relations with the ∞ 2 n−1 185 right-side coefficients. = {2n − 2 − 2[(Ik) + (Ik) + ··· + (Ik) ]} n=0 X ∞ −n Example 4.4. Find fn, n = 0, 1, 2, ··· , such that f0 = f1 = 0 n=0{2n − 2 − (n − 1)(1 + Ik)}x , when n is odd; n n = ∞ −n and fn+2 − 2fn+1 + fn =2+(2 − 2Ik)(Ik) − (Ik + 1) . n=0{2n − 2 − (n − 2)(1 + Ik) − Ik}x , when n is even, P (85) P 8 and It is clear that the left hand side is the convolution of x and the −2 −1 −1 (x − 1) [1 − (Ik + 1)x ] function t → (3j)t, so that taking z transform of both members ∞ ∞ gives −n−2 n−1 −n = (n + 1)x 2 (Ik + 1)x (1 − 3jx−1)−1X [f](x)=(1 − 2ix−1)−1. (92) n=0 n=0 (86) X∞ X Use the result of Eq. (24). We get = {n − 1+(2n−1 − n)(Ik + 1)}x−n. n=0 X [f](x) X Thus, when n is odd, the equation (81) can be rewritten as =(1 − 3jx−1)(1 − 2ix−1)−1 (93) −1 −1 −1 −1 −1 ∞ =(1 − 2ix ) − 3jx (1 − 2ix ) , F(x)= {n2 − n +2n − 2 − (n − 1)(1 + Ik) and using Eq. (24) and rule (iv) of Theorem 3.1, we see that n=0 X n−1 −n + n − 1+(2 − n)(Ik + 1)}x (87) f(t)=(2i)t − 3j(2i)t−1,t =0, 1, 2, ··· . (94) ∞ = {n2 − (Ik + 1)n + Ik}x−n, n=0 5. Conclusion X and when n is even, the equation (81) can be rewritten as The solution of a class of biquaternion recurrence relations