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On computing of arbitrary positive integer powers for one type of odd order tridiagonal matrices with zero row—II
Applied Mathematics and Computation 174 (2006) 490–499 www.elsevier.com/locate/amc
On computing of arbitrary positive integer powers for one type of odd order tridiagonal matrices with zero row—II Jonas Rimas Department of Applied Mathematics, Faculty of Fundamental Sciences, Kaunas University of Technology, Kaunas 51368, Lithuania
Abstract This article is an extension of the work [J. Rimas, On computing of arbitrary positive integer powers for one type of odd order diagonal matrices with zero row—I, Appl. Math. Comput., in press], in which the general expression of the lth power (l 2 N) for one type of odd order tridiagonal matrices is given. In this new paper we present the complete derivation of this general expression. Expressions of eigenvectors and JordanÕs form of the matrix and of the transforming matrix and its inverse are given, too. Ó 2005 Elsevier Inc. All rights reserved. Keywords: Tridiagonal matrices; Eigenvalues; Eigenvectors; Chebyshev polynomials
1. Introduction Solving some difference, differential equations and delay differential equations we meet the necessity to compute the arbitrary positive integer powers
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.04.100
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
491
of square matrix [1,2]. In the work [3] the general expression of the lth power (l 2 N) for one type of odd order tridiagonal matrices with zeros in the first row is presented. In this new paper we give the complete derivation of the general expression, presented in [3].
2. Formulation of the problem Consider the 0 0 B1 B B B B B¼B B B B @
nth order (n = 2p + 1, p 2 N) matrix B of the following type: 1 0 C 0 1 0 C C C 1 0 1 C ð1Þ C. .. C . C C 0 1 0 1A 2
0
We will derive expression of the lth power (l 2 N) of the matrix (1) (see formula (14) in [3]) applying the expression Bl = TJlT1 [4], where J is the JordanÕs form of B, T is the transforming matrix. Matrices J and T can be found provided eigenvalues and eigenvectors of the matrix B are known. The eigenvalues of B are defined by the characteristic equation jB kEj ¼ 0.
ð2Þ
In the paper [3] it is shown that the roots of the characteristic equation (2) (the eigenvalues of the matrix B) are defined by the following expression: 8 ð2k1Þp n1 > < 2 cos 2n2 ; k ¼ 1; 2; . . . ; 2 ; k ¼ nþ1 ; kk ¼ 0; ð3Þ 2 > : ð2k3Þp nþ3 nþ5 2 cos 2n2 ; k ¼ 2 ; 2 ; . . . ; n. Since all the eigenvalues kk ðk ¼ 1; nÞ are simple, for eigenvalue kk corresponds single Jordan cell J1(kk) in the matrix J. Taking this into account we write down the JordanÕs form of the matrix B: J ¼ diagðk1 k2 kn Þ.
ð4Þ
3. Eigenvectors of matrix B and transforming matrix T Consider the relation J = T1BT (BT = TJ); here B is the nth order matrix (1) (n = 2p + 1, p 2 N), J is the JordanÕs form of B, T is the transforming matrix. Since all the eigenvalues of B are simple, columns of the transforming
492
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
matrix T are the eigenvectors of the matrix B [4]. Denoting jth column of T by T j ðj ¼ 1; nÞ, we have T = (T1 T2 Tn) and (BT1 BT2 BTn) = (T1k1 T2k2 Tnkn). The latter expression gives BT j ¼ T j kj ;
j ¼ 1; n.
ð5Þ
Solving the set of systems (5), we find the eigenvectors of the matrix B: 1 0 k U 1 2j B C C B B U 0 k2j C C B B C nþ1 C kj ð6Þ Tj ¼ B B U 1 2 C; j ¼ 1; n and j 6¼ 2 ; C B C B .. C B A @ . k
U n2 ð 2j Þ 0
T nþ1 2
sin p2
1
0
1
1
B sin 2p C B 0 C C B B 2 C C B C B C C B sin 3p B 2 C B B 1 C B . C B . C C B . C B ¼ B .. C ¼ B . C; C B C B B sin ip C B sin ip2 C 2 C C B B B . C B . C B . C B . C @ . A @ . A sin np sin np 2 2
ð7Þ
here Uk(x) is the kth degree Chebyshev polynomial of the second kind [5]: U k ðxÞ ¼
sinðk þ 1Þ arccos x ; sin arccos x
1 6 x 6 1.
ð8Þ
Taking into account (6) and (7) we write down the transforming matrix T: 0
k 1 U 1 k22 B U 1 2 B B B B U 0 k21 U 0 k22 B B B T ¼B B U 1 k21 U 1 k22 B B B .. .. B B . . B @ k U n2 21 U n2 k22
kn1
U 1
2
k 1
U 1
nþ3 2
U 1
kn1
U 1
kn
1
C C C nþ3 kn1 kn C U 0 22 U0 2 C 0 U 0 22 U0 2 C C k kn1 nþ3 k k C C U 1 22 U 1 2n C. 1 U 1 22 U 1 n1 2 C C C .. .. .. .. .. C C . . . . . C k kn1 A nþ3 U n2 22 U n2 k21 sin np U n2 kn1 U n2 22 2 2 2
kn1
2
2
2
k
ð9Þ
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
493
Denoting jth column of the inverse matrix T1 by sj (T1 = (s1 s2 sn)) and implementing the necessary transformations, we obtain 1 0 2t1 0 1 U 0 k21 U n2 k21 t1 k1 B B C C B 2t2 B t U 0 k22 C U n2 k22 C C C B k2 B 2 C C B B C C B B .. .. C C B B C C B B . . C C B B B 2tn1 B kn1 C kn1 C B 2 U0 2 C B tn1 U n2 2 C C C B kn1 B 2 2 2 C C B 2 B C C B B C C; B B ; sn ¼ B ð10Þ s1 ¼ B 1 0 C k C k C C B 2tnþ3 B nþ3 nþ3 C C B 2 B U n2 22 C B knþ3 U 0 22 C B tnþ3 C C B 2 B 2 C C B B C C B B .. .. C C B B C C B B . . C B B kn1 C kn1 C C B 2tn1 B Bk B tn1 U n2 2 C U0 2 C A A @ n1 @ kn k 2tn n t U U0 2 n n2 2 kn 0
2t1
B B 2t2 B B B B B B B 2tn1 B 2 B B sj ¼ B B B B 2tnþ3 B 2 B B B B B B B 2t @ n1 2tn
U j2
k 1 1 2
C U j2 k22 C C C C .. C . C kn1 C U j2 22 C C C C C; 0 k C C nþ3 U j2 22 C C C C C .. C . C kn1 C U j2 2 C A k n U j2 2
j ¼ 2; n 1;
ð11Þ
here
tk ¼
8 k2nþ1k > > 2 n1 > > < 2ð2n 2Þ ; k ¼ 1; 2; . . . ; 2 ; > k23ðnþ1Þk > > > 2 : ; k ¼ nþ3 ; nþ5 ; . . . ; n. 2 2 2ð2n 2Þ
ð12Þ
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J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
Taking these expressions into account we write down the matrix T1:
0
2t1 U 0 k21 k1
2t1 U 0
k1
2t1 U 1
2
k1
2
B B B 2t2 U k2 2t2 U 0 k22 2t2 U 1 k22 B k2 0 2 B B B .. .. .. B . . . B B k k k B 2t n1 n1 n1 B n1 B k 2 U 0 22 2tn1 2tn1 U 0 22 U 1 22 B n1 2 2 2 B B B 1 0 0 T 1 ¼ B B k k k B 2t nþ3 nþ3 nþ3 B nþ3 2 2 B k 2 U 0 22 2t 2t nþ3 U 0 nþ3 U 1 2 2 B nþ3 2 2 B 2 B B B .. .. .. B . . . B B kn1 B 2tn1 kn1 B k U0 2 2tn1 U 0 2 2tn1 U 1 kn1 2 B n1 @ kn kn kn 2tn 2tn U 0 2 2tn U 1 2 U0 2 kn
2t1 U n3
k1
t1 U n2
2
k1 1 2
C C 2t2 U n3 2 t2 U n2 k22 C C C C C .. .. C . . C k k C C n1 n1 C 2tn1 tn1 U n3 22 U n2 22 C C 2 2 C C C 0 0 C. C k k C nþ3 nþ3 C 2tnþ3 U n3 22 tnþ3 U n2 22 C C 2 2 C C C C .. .. C . . C kn1 kn1 C C 2tn1 U n3 2 tn1 U n2 2 C C A kn kn 2tn U n3 2 tn U n2 2 k2
ð13Þ Applying (9) and (13) we can find the transforming matrix T and its inverse for the matrix (1) of arbitrary odd order n (n = 2p + 1, p 2 N). For example, if n = 3, we would have k1 ¼ a;
k2 ¼ 0;
k3 ¼ a;
k1 k3 U 1 ¼ U 1 ¼ 0; 2 2 ki U1 ¼ ki ; i ¼ 1; 3 2
a¼
pffiffiffi 2;
t1 ¼
k21 ; 8
t3 ¼
ki U0 ¼ 1; i ¼ 1; 3; 2
k23 ; 8
and U 1 k21 B k 1 T ¼B @ U0 2 k U 1 21 0
0 2t T 1
B ¼B @
1
k1
U0 1
2t3 k3
U0
1 0 1
k 1
2
k 3
2
1 0 U 1 k23 0 k C B U 0 23 C A¼@ 1 k a U 1 23 2t1 U 0 0 2t3 U 0
k 1
2
k 3
2
k 1
1
1
0
0
C 1 A;
1
a
0 a C 1B C¼ @ 4 0 k A 4 t3 U 1 23 a t1 U 1
1
2
2
a
1
0
C 0 A.
2
a
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
495
If n = 5, we would obtain p k1 ¼ a; k2 ¼ c; k3 ¼ 0; k4 ¼ c; k5 ¼ a; a ¼ 2 cos ; 8 3p k22 k21 k25 k24 ki c ¼ 2 cos ; t1 ¼ ; t2 ¼ ; t4 ¼ ; t5 ¼ ; U 1 ¼ 0; 8 16 16 16 16 2 ki ki i ¼ 1; 2; 4; 5; U 0 ¼ 1; i ¼ 1; 2; 4; 5; U 1 ¼ ki ; 2 2 k1 k5 a k2 k4 c i ¼ 1; 2; 4; 5; U 2 ¼ U2 ¼ ; U2 ¼ U2 ¼ ; c a 2 2 2 2 k1 2 k2 2 k4 2 k5 2 U3 ¼ ; U3 ¼ ; U3 ¼ ; U3 ¼ c a a c 2 2 2 2 and
U 1 k21 B U k1 B 0 2 B k 1 T ¼B B U 1 2 B k1 @ U2 2 U 3 k21 0
0 2t
1
k1
T 1
U0
U 1 k22 U 0 k22 U 1 k22 U 2 k22 U 3 k22
k 1
2
1 U 1 k24 0 U 0 k24 1 U 1 k24 0 U 2 k24 1 U 3 k24
2t1 U 0
k 1
2
1 0 U 1 k25 0 U 0 k25 C 1 C B k5 C B B a C U1 2 C ¼ B C B a U 2 k25 A @ c 2c U 3 k25
2t1 U 1
k 1
2
2t1 U 2
k 1
2
B B 2t2 U k2 2t U k2 2t U k2 2t U k2 2 0 2 2 1 2 2 2 2 B k2 0 2 B 1 0 0 0 ¼B B k k B 2t4 k4 4 4 B k4 U 0 2 2t4 U 0 2 2t4 U 1 2 2t4 U 2 k21 @ 2t5 2t5 U 0 k25 2t5 U 1 k25 2t5 U 2 k25 U 0 k25 k5 1 0 c2 a c2 c2 a ac c C B a2 B a2 a2 c ac a C c C B 1B C ¼ B 8 0 0 0 0 C. C 8B 2 B a a2 a2 c ac a C A @ c c2 a
c2
c2 a
ac
0 1 c ac 2 a
t1 U 3
1 0 0 1 1C C c aC C; C ac ac A 2a 2c
k 1 1
2
C t2 U 3 k22 C C C C 0 k C C 4 t4 U 3 2 C A t5 U 3 k25
c
Derive expression of Bl. Denoting ith row of T by Li ði ¼ 1; nÞ, we have (see (9)): 0 1 L1 BL C B 2C C T ¼B B .. C @ . A Ln
1 0 1 0 1
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J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
and 0
U i2
k 1 1 2
C B B U k2 C i2 2 C B C B C B .. C B C B . B C B kn1 C C B B U i2 22 C C B C B C B ip T C; B sin Li ¼ B 2 C C B knþ3 C B B U i2 2 C C B 2 C B C B C B .. C B C B . C B kn1 C B B U i2 2 C A @ kn U i2 2
i ¼ 1; n.
ð14Þ
Since Bl = TJlT1 and J l ¼ diagðkl1 kl2 kl3 kln Þ, we can write 0
2t1 kl1 1 U0
k 1
1
2
B C C B C B 2t2 kl1 U 0 k22 2 C B C B C B . .. C B C B B C B kn1 C B 2tn1 kl1 U 0 22 C C B 2 n1 2 C B C B C B 0 fBl gi1 ¼ fTJ l T 1 gi1 ¼ Li J l s1 ¼ Li B C B k C C B nþ3 C B 2 kl1 nþ3 U 0 C B 2tnþ3 2 C B 2 2 C B C B C B .. C B . C B B kn1 C C B l1 B 2tn1 kn1 U 0 2 C A @ k l1 n 2tn kn U 0 2 n X kk ¼2 tk kl1 U ; i ¼ 1; n; i2 k 2 k¼1
ð15Þ
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
0
t1 kl1 U n2
k 1
1
2
C B B t2 kl U n2 k2 C C B 2 2 C B C B .. C B . B C C B kn1 C B B tn1 kln1 U n2 22 C C B 2 2 C B C B l l 1 l 0 fB gin ¼ fTJ T gin ¼ Li J sn ¼ Li B C k C B nþ3 C B l B tnþ3 knþ3 U n2 2 C 2 C B 2 2 C B C B C B .. C B . C B C B B tn1 kln1 U n2 kn1 C 2 A @ k l n tn kn U n2 2 n X l kk kk ¼ tk kk U i2 U n2 ; i ¼ 1; n; 2 2 k¼1 0
2t1 kl1 U j2
k 1
497
ð16Þ
1
2
B C C B 2t2 kl U j2 k2 2 C B 2 C B C B . .. C B B C C B kn1 C B B 2tn1 kln1 U j2 22 C C B 2 2 C B C B l l 0 fB gij ¼ Li J sj ¼ Li B C k C B nþ3 C B B 2tnþ3 klnþ3 U j2 2 C 2 C B 2 2 C B C B C B .. C B . C B C B B 2tn1 kl U j2 kn1 C n1 2 A @ 2tn kln U j2 k2n n X kk kk l ¼2 tk kk U i2 U j2 ; i ¼ 1; n; j ¼ 2; n 1 2 2 k¼1 (we have evaluated that knþ1 ¼ 0 and U 0 k2k ¼ 1; k ¼ 1; n). 2 Expressions (15)–(17) can be written in the following unique form: n X kk kk la tk kk j U i2 fBl gij ¼ 2pij U j2þaj ; i; j ¼ 1; n; 2 2 k¼1
ð17Þ
ð18Þ
498
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
here tk is defined by (12), kk ðk ¼ 1; nÞ is the eigenvalue of the matrix B (see (3)), n is the order of the matrix B (n = 2p + 1, p 2 N), Un(x) is the kth degree Chebyshev polynomial of the second kind (see (8)), 8 > < 0; if i ¼ 1; ð19Þ pij ¼ 12 ; if i 6¼ 1 and j ¼ n; > : 1; if i 6¼ 1 and j 6¼ n; ai ¼
if j 6¼ 1; if j ¼ 1.
0; 1;
ð20Þ
Expression (18) can be written as follows: fBl gij ¼ S 1 þ S 2 ;
ð21Þ
here n1
S 1 ¼ 2pij
2 X
la tk kk j U i2
k¼1
S 2 ¼ 2pij
n X
laj
t k kk
kk kk U j2þaj ; 2 2
i; j ¼ 1; n;
ð22Þ
kk kk U j2þaj ; 2 2
i; j ¼ 1; n.
ð23Þ
U i2
k¼nþ3 2
The second sum can be written by applying the substitution k ¼ nþ2mþ1 : 2 ! ! n1 2 X knþ2mþ1 knþ2mþ1 laj 2 2 S 2 ¼ 2pij tnþ2mþ1 knþ2mþ1 U i2 U j2þaj ; i; j ¼ 1; n. 2 2 2 2 m¼1 ð24Þ Changing m to k and changing direction of summing in (24), we obtain n1 2 X knkþ1 knkþ1 laj tnkþ1 knkþ1 U i2 S 2 ¼ 2pij U j2þaj ; i; j ¼ 1; n. 2 2 k¼1 Taking into account the relations kk = knk+1 and tk ¼ tnkþ1 k ¼ 1; n1 2 (see (3) and (12)), we get kk kk U i2 U j2þaj 2 2 k¼1 n1 2 X kk kk laj lþiþj ¼ ð1Þ 2pij tk kk U i2 U j2þaj ; 2 2 k¼1 n1
S 2 ¼ 2pij
2 X
tk ðkk Þ
laj
(we have evaluated that Un(x) = (1)kUn(x) [5]).
i; j ¼ 1; n
ð25Þ
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
Substituting (22) and (25) into (18), we obtain n1 2 X kk kk la lþiþj l fB gij ¼ ð1 þ ð1Þ Þ2pij tk kk j U i2 U j2þaj ; 2 2 k¼1
499
i; j ¼ 1; n.
Taking into account (12) we get the final result n1 lþiþj 2 X 1 þ ð1Þ kk kk la l fB gij ¼ pij kk j k2nþ1k U i2 U j2þaj ; i; j ¼ 1; n; 2 2n 2 2 2 k¼1 which coincides with the expression of the lth power of B, presented in [3].
References [1] R.P. Agarwal, Difference Equations and Inequalities, Marcel Dekker, NY, 1992. [2] J. Rimas, Investigation of the dynamics of mutually synchronized systems, Telecommun. Radio Eng. 32 (1977) 68–79. [3] J. Rimas, On computing of arbitrary positive integer powers for one type of odd order tridiagonal matrices with zero row—I, Appl. Math. Comput., in press. [4] P. Horn, Ch. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1986. [5] L. Fox, J.B. Parke, Chebyshev Polynomials in Numerical Analysis, Oxford University Press, London, 1968.
On computing of arbitrary positive integer powers for one type of odd order tridiagonal matrices with zero row—II Jonas Rimas Department of Applied Mathematics, Faculty of Fundamental Sciences, Kaunas University of Technology, Kaunas 51368, Lithuania
Abstract This article is an extension of the work [J. Rimas, On computing of arbitrary positive integer powers for one type of odd order diagonal matrices with zero row—I, Appl. Math. Comput., in press], in which the general expression of the lth power (l 2 N) for one type of odd order tridiagonal matrices is given. In this new paper we present the complete derivation of this general expression. Expressions of eigenvectors and JordanÕs form of the matrix and of the transforming matrix and its inverse are given, too. Ó 2005 Elsevier Inc. All rights reserved. Keywords: Tridiagonal matrices; Eigenvalues; Eigenvectors; Chebyshev polynomials
1. Introduction Solving some difference, differential equations and delay differential equations we meet the necessity to compute the arbitrary positive integer powers
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.04.100
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
491
of square matrix [1,2]. In the work [3] the general expression of the lth power (l 2 N) for one type of odd order tridiagonal matrices with zeros in the first row is presented. In this new paper we give the complete derivation of the general expression, presented in [3].
2. Formulation of the problem Consider the 0 0 B1 B B B B B¼B B B B @
nth order (n = 2p + 1, p 2 N) matrix B of the following type: 1 0 C 0 1 0 C C C 1 0 1 C ð1Þ C. .. C . C C 0 1 0 1A 2
0
We will derive expression of the lth power (l 2 N) of the matrix (1) (see formula (14) in [3]) applying the expression Bl = TJlT1 [4], where J is the JordanÕs form of B, T is the transforming matrix. Matrices J and T can be found provided eigenvalues and eigenvectors of the matrix B are known. The eigenvalues of B are defined by the characteristic equation jB kEj ¼ 0.
ð2Þ
In the paper [3] it is shown that the roots of the characteristic equation (2) (the eigenvalues of the matrix B) are defined by the following expression: 8 ð2k1Þp n1 > < 2 cos 2n2 ; k ¼ 1; 2; . . . ; 2 ; k ¼ nþ1 ; kk ¼ 0; ð3Þ 2 > : ð2k3Þp nþ3 nþ5 2 cos 2n2 ; k ¼ 2 ; 2 ; . . . ; n. Since all the eigenvalues kk ðk ¼ 1; nÞ are simple, for eigenvalue kk corresponds single Jordan cell J1(kk) in the matrix J. Taking this into account we write down the JordanÕs form of the matrix B: J ¼ diagðk1 k2 kn Þ.
ð4Þ
3. Eigenvectors of matrix B and transforming matrix T Consider the relation J = T1BT (BT = TJ); here B is the nth order matrix (1) (n = 2p + 1, p 2 N), J is the JordanÕs form of B, T is the transforming matrix. Since all the eigenvalues of B are simple, columns of the transforming
492
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
matrix T are the eigenvectors of the matrix B [4]. Denoting jth column of T by T j ðj ¼ 1; nÞ, we have T = (T1 T2 Tn) and (BT1 BT2 BTn) = (T1k1 T2k2 Tnkn). The latter expression gives BT j ¼ T j kj ;
j ¼ 1; n.
ð5Þ
Solving the set of systems (5), we find the eigenvectors of the matrix B: 1 0 k U 1 2j B C C B B U 0 k2j C C B B C nþ1 C kj ð6Þ Tj ¼ B B U 1 2 C; j ¼ 1; n and j 6¼ 2 ; C B C B .. C B A @ . k
U n2 ð 2j Þ 0
T nþ1 2
sin p2
1
0
1
1
B sin 2p C B 0 C C B B 2 C C B C B C C B sin 3p B 2 C B B 1 C B . C B . C C B . C B ¼ B .. C ¼ B . C; C B C B B sin ip C B sin ip2 C 2 C C B B B . C B . C B . C B . C @ . A @ . A sin np sin np 2 2
ð7Þ
here Uk(x) is the kth degree Chebyshev polynomial of the second kind [5]: U k ðxÞ ¼
sinðk þ 1Þ arccos x ; sin arccos x
1 6 x 6 1.
ð8Þ
Taking into account (6) and (7) we write down the transforming matrix T: 0
k 1 U 1 k22 B U 1 2 B B B B U 0 k21 U 0 k22 B B B T ¼B B U 1 k21 U 1 k22 B B B .. .. B B . . B @ k U n2 21 U n2 k22
kn1
U 1
2
k 1
U 1
nþ3 2
U 1
kn1
U 1
kn
1
C C C nþ3 kn1 kn C U 0 22 U0 2 C 0 U 0 22 U0 2 C C k kn1 nþ3 k k C C U 1 22 U 1 2n C. 1 U 1 22 U 1 n1 2 C C C .. .. .. .. .. C C . . . . . C k kn1 A nþ3 U n2 22 U n2 k21 sin np U n2 kn1 U n2 22 2 2 2
kn1
2
2
2
k
ð9Þ
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
493
Denoting jth column of the inverse matrix T1 by sj (T1 = (s1 s2 sn)) and implementing the necessary transformations, we obtain 1 0 2t1 0 1 U 0 k21 U n2 k21 t1 k1 B B C C B 2t2 B t U 0 k22 C U n2 k22 C C C B k2 B 2 C C B B C C B B .. .. C C B B C C B B . . C C B B B 2tn1 B kn1 C kn1 C B 2 U0 2 C B tn1 U n2 2 C C C B kn1 B 2 2 2 C C B 2 B C C B B C C; B B ; sn ¼ B ð10Þ s1 ¼ B 1 0 C k C k C C B 2tnþ3 B nþ3 nþ3 C C B 2 B U n2 22 C B knþ3 U 0 22 C B tnþ3 C C B 2 B 2 C C B B C C B B .. .. C C B B C C B B . . C B B kn1 C kn1 C C B 2tn1 B Bk B tn1 U n2 2 C U0 2 C A A @ n1 @ kn k 2tn n t U U0 2 n n2 2 kn 0
2t1
B B 2t2 B B B B B B B 2tn1 B 2 B B sj ¼ B B B B 2tnþ3 B 2 B B B B B B B 2t @ n1 2tn
U j2
k 1 1 2
C U j2 k22 C C C C .. C . C kn1 C U j2 22 C C C C C; 0 k C C nþ3 U j2 22 C C C C C .. C . C kn1 C U j2 2 C A k n U j2 2
j ¼ 2; n 1;
ð11Þ
here
tk ¼
8 k2nþ1k > > 2 n1 > > < 2ð2n 2Þ ; k ¼ 1; 2; . . . ; 2 ; > k23ðnþ1Þk > > > 2 : ; k ¼ nþ3 ; nþ5 ; . . . ; n. 2 2 2ð2n 2Þ
ð12Þ
494
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
Taking these expressions into account we write down the matrix T1:
0
2t1 U 0 k21 k1
2t1 U 0
k1
2t1 U 1
2
k1
2
B B B 2t2 U k2 2t2 U 0 k22 2t2 U 1 k22 B k2 0 2 B B B .. .. .. B . . . B B k k k B 2t n1 n1 n1 B n1 B k 2 U 0 22 2tn1 2tn1 U 0 22 U 1 22 B n1 2 2 2 B B B 1 0 0 T 1 ¼ B B k k k B 2t nþ3 nþ3 nþ3 B nþ3 2 2 B k 2 U 0 22 2t 2t nþ3 U 0 nþ3 U 1 2 2 B nþ3 2 2 B 2 B B B .. .. .. B . . . B B kn1 B 2tn1 kn1 B k U0 2 2tn1 U 0 2 2tn1 U 1 kn1 2 B n1 @ kn kn kn 2tn 2tn U 0 2 2tn U 1 2 U0 2 kn
2t1 U n3
k1
t1 U n2
2
k1 1 2
C C 2t2 U n3 2 t2 U n2 k22 C C C C C .. .. C . . C k k C C n1 n1 C 2tn1 tn1 U n3 22 U n2 22 C C 2 2 C C C 0 0 C. C k k C nþ3 nþ3 C 2tnþ3 U n3 22 tnþ3 U n2 22 C C 2 2 C C C C .. .. C . . C kn1 kn1 C C 2tn1 U n3 2 tn1 U n2 2 C C A kn kn 2tn U n3 2 tn U n2 2 k2
ð13Þ Applying (9) and (13) we can find the transforming matrix T and its inverse for the matrix (1) of arbitrary odd order n (n = 2p + 1, p 2 N). For example, if n = 3, we would have k1 ¼ a;
k2 ¼ 0;
k3 ¼ a;
k1 k3 U 1 ¼ U 1 ¼ 0; 2 2 ki U1 ¼ ki ; i ¼ 1; 3 2
a¼
pffiffiffi 2;
t1 ¼
k21 ; 8
t3 ¼
ki U0 ¼ 1; i ¼ 1; 3; 2
k23 ; 8
and U 1 k21 B k 1 T ¼B @ U0 2 k U 1 21 0
0 2t T 1
B ¼B @
1
k1
U0 1
2t3 k3
U0
1 0 1
k 1
2
k 3
2
1 0 U 1 k23 0 k C B U 0 23 C A¼@ 1 k a U 1 23 2t1 U 0 0 2t3 U 0
k 1
2
k 3
2
k 1
1
1
0
0
C 1 A;
1
a
0 a C 1B C¼ @ 4 0 k A 4 t3 U 1 23 a t1 U 1
1
2
2
a
1
0
C 0 A.
2
a
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
495
If n = 5, we would obtain p k1 ¼ a; k2 ¼ c; k3 ¼ 0; k4 ¼ c; k5 ¼ a; a ¼ 2 cos ; 8 3p k22 k21 k25 k24 ki c ¼ 2 cos ; t1 ¼ ; t2 ¼ ; t4 ¼ ; t5 ¼ ; U 1 ¼ 0; 8 16 16 16 16 2 ki ki i ¼ 1; 2; 4; 5; U 0 ¼ 1; i ¼ 1; 2; 4; 5; U 1 ¼ ki ; 2 2 k1 k5 a k2 k4 c i ¼ 1; 2; 4; 5; U 2 ¼ U2 ¼ ; U2 ¼ U2 ¼ ; c a 2 2 2 2 k1 2 k2 2 k4 2 k5 2 U3 ¼ ; U3 ¼ ; U3 ¼ ; U3 ¼ c a a c 2 2 2 2 and
U 1 k21 B U k1 B 0 2 B k 1 T ¼B B U 1 2 B k1 @ U2 2 U 3 k21 0
0 2t
1
k1
T 1
U0
U 1 k22 U 0 k22 U 1 k22 U 2 k22 U 3 k22
k 1
2
1 U 1 k24 0 U 0 k24 1 U 1 k24 0 U 2 k24 1 U 3 k24
2t1 U 0
k 1
2
1 0 U 1 k25 0 U 0 k25 C 1 C B k5 C B B a C U1 2 C ¼ B C B a U 2 k25 A @ c 2c U 3 k25
2t1 U 1
k 1
2
2t1 U 2
k 1
2
B B 2t2 U k2 2t U k2 2t U k2 2t U k2 2 0 2 2 1 2 2 2 2 B k2 0 2 B 1 0 0 0 ¼B B k k B 2t4 k4 4 4 B k4 U 0 2 2t4 U 0 2 2t4 U 1 2 2t4 U 2 k21 @ 2t5 2t5 U 0 k25 2t5 U 1 k25 2t5 U 2 k25 U 0 k25 k5 1 0 c2 a c2 c2 a ac c C B a2 B a2 a2 c ac a C c C B 1B C ¼ B 8 0 0 0 0 C. C 8B 2 B a a2 a2 c ac a C A @ c c2 a
c2
c2 a
ac
0 1 c ac 2 a
t1 U 3
1 0 0 1 1C C c aC C; C ac ac A 2a 2c
k 1 1
2
C t2 U 3 k22 C C C C 0 k C C 4 t4 U 3 2 C A t5 U 3 k25
c
Derive expression of Bl. Denoting ith row of T by Li ði ¼ 1; nÞ, we have (see (9)): 0 1 L1 BL C B 2C C T ¼B B .. C @ . A Ln
1 0 1 0 1
496
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
and 0
U i2
k 1 1 2
C B B U k2 C i2 2 C B C B C B .. C B C B . B C B kn1 C C B B U i2 22 C C B C B C B ip T C; B sin Li ¼ B 2 C C B knþ3 C B B U i2 2 C C B 2 C B C B C B .. C B C B . C B kn1 C B B U i2 2 C A @ kn U i2 2
i ¼ 1; n.
ð14Þ
Since Bl = TJlT1 and J l ¼ diagðkl1 kl2 kl3 kln Þ, we can write 0
2t1 kl1 1 U0
k 1
1
2
B C C B C B 2t2 kl1 U 0 k22 2 C B C B C B . .. C B C B B C B kn1 C B 2tn1 kl1 U 0 22 C C B 2 n1 2 C B C B C B 0 fBl gi1 ¼ fTJ l T 1 gi1 ¼ Li J l s1 ¼ Li B C B k C C B nþ3 C B 2 kl1 nþ3 U 0 C B 2tnþ3 2 C B 2 2 C B C B C B .. C B . C B B kn1 C C B l1 B 2tn1 kn1 U 0 2 C A @ k l1 n 2tn kn U 0 2 n X kk ¼2 tk kl1 U ; i ¼ 1; n; i2 k 2 k¼1
ð15Þ
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
0
t1 kl1 U n2
k 1
1
2
C B B t2 kl U n2 k2 C C B 2 2 C B C B .. C B . B C C B kn1 C B B tn1 kln1 U n2 22 C C B 2 2 C B C B l l 1 l 0 fB gin ¼ fTJ T gin ¼ Li J sn ¼ Li B C k C B nþ3 C B l B tnþ3 knþ3 U n2 2 C 2 C B 2 2 C B C B C B .. C B . C B C B B tn1 kln1 U n2 kn1 C 2 A @ k l n tn kn U n2 2 n X l kk kk ¼ tk kk U i2 U n2 ; i ¼ 1; n; 2 2 k¼1 0
2t1 kl1 U j2
k 1
497
ð16Þ
1
2
B C C B 2t2 kl U j2 k2 2 C B 2 C B C B . .. C B B C C B kn1 C B B 2tn1 kln1 U j2 22 C C B 2 2 C B C B l l 0 fB gij ¼ Li J sj ¼ Li B C k C B nþ3 C B B 2tnþ3 klnþ3 U j2 2 C 2 C B 2 2 C B C B C B .. C B . C B C B B 2tn1 kl U j2 kn1 C n1 2 A @ 2tn kln U j2 k2n n X kk kk l ¼2 tk kk U i2 U j2 ; i ¼ 1; n; j ¼ 2; n 1 2 2 k¼1 (we have evaluated that knþ1 ¼ 0 and U 0 k2k ¼ 1; k ¼ 1; n). 2 Expressions (15)–(17) can be written in the following unique form: n X kk kk la tk kk j U i2 fBl gij ¼ 2pij U j2þaj ; i; j ¼ 1; n; 2 2 k¼1
ð17Þ
ð18Þ
498
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
here tk is defined by (12), kk ðk ¼ 1; nÞ is the eigenvalue of the matrix B (see (3)), n is the order of the matrix B (n = 2p + 1, p 2 N), Un(x) is the kth degree Chebyshev polynomial of the second kind (see (8)), 8 > < 0; if i ¼ 1; ð19Þ pij ¼ 12 ; if i 6¼ 1 and j ¼ n; > : 1; if i 6¼ 1 and j 6¼ n; ai ¼
if j 6¼ 1; if j ¼ 1.
0; 1;
ð20Þ
Expression (18) can be written as follows: fBl gij ¼ S 1 þ S 2 ;
ð21Þ
here n1
S 1 ¼ 2pij
2 X
la tk kk j U i2
k¼1
S 2 ¼ 2pij
n X
laj
t k kk
kk kk U j2þaj ; 2 2
i; j ¼ 1; n;
ð22Þ
kk kk U j2þaj ; 2 2
i; j ¼ 1; n.
ð23Þ
U i2
k¼nþ3 2
The second sum can be written by applying the substitution k ¼ nþ2mþ1 : 2 ! ! n1 2 X knþ2mþ1 knþ2mþ1 laj 2 2 S 2 ¼ 2pij tnþ2mþ1 knþ2mþ1 U i2 U j2þaj ; i; j ¼ 1; n. 2 2 2 2 m¼1 ð24Þ Changing m to k and changing direction of summing in (24), we obtain n1 2 X knkþ1 knkþ1 laj tnkþ1 knkþ1 U i2 S 2 ¼ 2pij U j2þaj ; i; j ¼ 1; n. 2 2 k¼1 Taking into account the relations kk = knk+1 and tk ¼ tnkþ1 k ¼ 1; n1 2 (see (3) and (12)), we get kk kk U i2 U j2þaj 2 2 k¼1 n1 2 X kk kk laj lþiþj ¼ ð1Þ 2pij tk kk U i2 U j2þaj ; 2 2 k¼1 n1
S 2 ¼ 2pij
2 X
tk ðkk Þ
laj
(we have evaluated that Un(x) = (1)kUn(x) [5]).
i; j ¼ 1; n
ð25Þ
J. Rimas / Appl. Math. Comput. 174 (2006) 490–499
Substituting (22) and (25) into (18), we obtain n1 2 X kk kk la lþiþj l fB gij ¼ ð1 þ ð1Þ Þ2pij tk kk j U i2 U j2þaj ; 2 2 k¼1
499
i; j ¼ 1; n.
Taking into account (12) we get the final result n1 lþiþj 2 X 1 þ ð1Þ kk kk la l fB gij ¼ pij kk j k2nþ1k U i2 U j2þaj ; i; j ¼ 1; n; 2 2n 2 2 2 k¼1 which coincides with the expression of the lth power of B, presented in [3].
References [1] R.P. Agarwal, Difference Equations and Inequalities, Marcel Dekker, NY, 1992. [2] J. Rimas, Investigation of the dynamics of mutually synchronized systems, Telecommun. Radio Eng. 32 (1977) 68–79. [3] J. Rimas, On computing of arbitrary positive integer powers for one type of odd order tridiagonal matrices with zero row—I, Appl. Math. Comput., in press. [4] P. Horn, Ch. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1986. [5] L. Fox, J.B. Parke, Chebyshev Polynomials in Numerical Analysis, Oxford University Press, London, 1968.