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Convergence of matrix continued fractions
Linear Algebra and its Applications 320 (2000) 115–129 www.elsevier.com/locate/laa
Convergence of matrix continued fractions Mustapha Raissouli a,∗ , Ali Kacha b a Department of Mathematics and Informatic, Sciences Faculty, Moulay Ismail University, AFACS UFR,
AFA Group, B.P. 4010, Meknes, Morocco b Department of Mathematics, Sciences and Technics Faculty, Moulay Ismail University, B.P. 509,
Errachidia, Morocco Received 22 February 2000; accepted 14 June 2000 Submitted by R.A. Brualdi
Abstract The aim of this work is to give some criteria on the convergence of matrix continued fractions. We begin by presenting some new results which generalize the links between the convergent elements of real continued fractions. Secondly, we give necessary and sufficient conditions for the convergence of continued fractions of matrix arguments. This paper will be completed by illustrating the theoretical results with some examples. © 2000 Elsevier Science Inc. All rights reserved. AMS classification: 40A15; 15A60 Keywords: Continued fraction; An m × m matrix; Convergence
1. Definitions, generalities and notations 1.0. Introduction In recent few years, the theory of continued fraction has extensive developments and applications. Since calculations involving matrix valued functions with matrix arguments are feasible with large computers, such functions have been used in solving many difficult problems. In [8], Varga used Padé approximants, equivalent to continued fraction approximants of the matrix exponential to derive and analyze cer∗ Corresponding author.
E-mail address: [email protected] (M. Raissouli). 0024-3795/00/$ - see front matter 2000 Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 0 0 ) 0 0 1 9 6 - 8
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tain discrete approximations to solutions of self-adjoint parabolic differential equations. The matrix continued fraction used by Varga is an example of noncommutative continued fractions about which not much is known. A few convergence theorems on noncommutative continued fractions have been published. Two theorems are stated in [10], where Wynn reviews many aspects of the theory of continued fractions, whose elements do not commute under a multiplication law. In a Banach space, extentions of Worpitsky’s theorem have been proven by Hayden [2] and Negoescu [7]. In this paper, several convergence criteria on the noncommutative continued fractions whose arguments are m × m matrices of the forms K(Bn /An ) are given. 1.1. Convergence of matrices Throughout this paper, we denote by Mm the set of m × m real (or complex) matrices endowed with the classical norm defined by kAxk = sup {kAxk, kxk = 1}. ∀A ∈ Mm , kAk = sup kxk x =0 / This norm satisfies the relationship kABk 6 kAkkBk. Let (An ) be a sequence of matrices in Mm . We say that (An ) converges if there exists a matrix A ∈ Mm , such that kAn − Ak tends to 0 when n tends to +∞. In this case, we write An −→ A
or
lim An = A.
n→+∞
Let A ∈ Mm . We denote by det A the determinant of A. It is well known that if An −→ A in Mm , then det An −→ det A in R. Moreover, if An and A are invertible, −1 in M . then A−1 m n −→ A Let A and B be two matrices with B invertible. We write A/B = B −1 A, in particular, if A = I , where I is the mth order identity matrix, then we write I /B = B −1 . It is clear that for any invertible matrix X, we have XA AX A = = / . B XB BX 1.2. Some results on continued fractions of matrices Let (An )n>0 , (Bn )n>1 be two sequences of Mm . We call K(Bn /An ) the continued fraction if the quantity B1 B1 B2 = A0 ; , ,... . A0 + A1 A2 A 1 + B2 A2 +···
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Sometimes, we use briefly the notation Bn +∞ . A0 ; An n=1 The fractions Bn and An
Pn Bi n = A0 ; Qn Ai i=1
are called, respectively, the nth partial quotient and the nth convergent of the continued fraction K(Bn /An ). When Bn = I for all n > 1, then K(I /An ) is called an ordinary continued fraction. Proposition 1. The elements (Pn )n>−1 and (Qn )n>−1 of the nth convergent of K(Bn /An ) are given by the relationships Pn = An Pn−1 + Bn Pn−2 P−1 = I, P0 = A0 , n > 1. (1) and Q−1 = 0, Q0 = I Qn = An Qn−1 + Bn Qn−2 Proof. The proof of Proposition 1 uses induction on n. The relationships are obviously verified for n = 1. Now, suppose that for all p 6 n Eq. (1) is true. For n + 1, using the definition of K(Bn /An ), to obtain Rn+1 = Pn+1 /Qn+1 from Rn = Pn /Qn , we must replace An by An + (Bn+1 /An+1 ). So, we have Bn+1 An + A Pn−1 + Bn Pn−2 n+1 Rn+1 = Bn+1 An + A Qn−1 + Bn Qn−2 n+1 (An+1 An + Bn+1 )Pn−1 + An+1 Bn Pn−2 (An+1 An + Bn+1 )Qn−1 + An+1 Bn Qn−2 An+1 (An Pn−1 + Bn Pn−2 ) + Bn+1 Pn−1 = An+1 (An Qn−1 + Bn Qn−2 ) + Bn+1 Qn−1 An+1 Pn + Bn+1 Pn−1 = , An+1 Qn + Bn+1 Qn−1 =
and Proposition 1 is proved. Proposition 2. For all ordinary continued fractions K(I /An ), we have for all n > 0, n Y (1 + kAi k), (2) kPn k 6 i=0
kQn k 6
n Y i=0
(1 + kAi k).
(3)
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Proof. Q For n = 0, relation (2) holds because P0 =A0 , suppose that for all l 6 n, kPl k 6 li=0 (1 + kAi k). Then, for n + 1, we have Pn+1 = An+1 Pn + Pn−1 and we obtain kPn+1 k 6 kAn+1 kkPn k + kPn−1 k, which yields kPn+1 k 6kAn+1 k
n Y
(1 + kAi k) +
i=0
6
n−1 Y
n−1 Y
(1 + kAi k)
i=0
(1 + kAi k) (kAn+1 k(1 + kAn k) + 1) .
i=0
Consequently, we find kPn+1 k 6
n−1 Y
(1 + kAi k) (1 + kAn k) (1 + kAn+1 k) ,
i=0
Qn and by this induction, we have for all n > 1, QnkPn k 6 i=0 (1 + kAi k). Similarly, we also prove that kQn k 6 i=0 (1 + kAi k), and the proof is complete. Theorem 1. The convergents of the continued fraction K(Bn /An ) satisfy n Y Pn Pn−1 det = (−1)m(n−1) (det Qn )−1 (det Qn−1 )−1 − det Bi . Qn Qn−1 i=1
Proof. Put Pn Pn−1 −1 − = Q−1 n Pn − Qn−1 Pn−1 . Qn Qn−1 We can write Dn =
−1 Dn = Q−1 n (Pn − Qn Qn−1 Pn−1 ).
It follows that det Dn = (det Qn )−1 det(Pn − Qn Q−1 n−1 Pn−1 ). We now will use the next lemma. Lemma 1 [1]. Let (M)i,j =1,2 be four matrices in Mm , with M22 invertible. Then we have M11 M12 −1 = det M22 det(M11 − M21 M22 det M12 ). (4) M21 M22 Using Lemma 1, one has det Dn = (det Qn )
−1
(det Qn−1 )
−1
P det n Qn
Pn−1 . Qn−1
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If we put that
P δn = det n Qn
Pn−1 , Qn−1
and due to Eq. (1) in Proposition 1, we deduce that Pn−1 An Pn−1 + Bn Pn−2 δn = det . An Qn−1 + Bn Qn−2 Qn−1 If the determinant is multilinear, then Bn Pn−2 An Pn−1 Pn−1 + det δn = det An Qn−1 Qn−1 Bn Qn−2
Pn−1 . Qn−1
Using the previous lemma again, we find that An Pn−1 Pn−1 det =0 An Qn−1 Qn−1 and we deduce that δn = detQn−1 det(Bn Pn−2 − Bn Qn−2 Q−1 n−1 Pn−1 )
= det Qn−1 det Bn det(Pn−2 − Qn−2 Q−1 n−1 Pn−1 ) −1 = det Qn−1 det Bn det Qn−2 Qn−1 (Qn−1 Q−1 n−2 Pn−2 − Pn−1 ) .
Thus, δn = det Bn det Qn−2 det(Qn−1 Q−1 n−2 Pn−2 − Pn−1 )
= (−1)m det Bn det Qn−2 det(Pn−1 − Qn−1 Q−1 n−2 Pn−2 ).
By applying relation (4), we obtain Pn−1 m δn = (−1) det Bn det Qn−1
Pn−2 , Qn−2
which becomes δn = (−1)m (det Bn ) δn−1 . Consequently, we have ! n Y δn = (−1)m(n−1) det Bi δ1 , i=2
where
P1 δ1 = det Q1
P0 I
A1 A0 + B1 = det A1
A0 . I
The lemma gives δ1 = det B1 and Theorem 1 is proved.
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2. Convergence criteria of continued fractions Definition 1. The continued fraction K(Bn /An ) converges in Mm if the sequence (Fn ) = (Pn /Qn ) = (Q−1 n Pn ) converges in Mm in the sense that there exists a matrix F ∈ Mm such that limn→+∞ kFn − F k = 0. In the other case, we say that K(Bn /An ) is divergent. It is clear that n X Pi−1 Pi Pn = A0 + − . Qn Qi Qi−1
(5)
i=1
From (5), we see that the continued fraction K(Bn /An ) converges in Mm if and only if the series +∞ X Pn−1 Pn − Qn Qn−1 n=1
converges in Mm . 2.1. Necessary condition for the convergence of K(I /An ) We now give a necessary condition for the convergence of the matrix continued fraction of the form K(I /An ) . of Mm . If the continued fraction K(I /An ) Theorem 2. Let (An )n>0 be a sequence P converges in Mm , then the series +∞ n=0 kAn k diverges. Proof. We assume that K(I /A P n ) converges and we put M = limn→+∞ Pn /Qn . Suppose We will show that the series n kAn k diverges by an absurd reasoning. Q+∞ P that the series +∞ n=0 kAn k converges, then the infinite product n=0 (1 + kAn k) also converges. P P It follows, by virtue of relationships (2) and (3), that the series n An Pn−1 and n An Qn−1 converge absolutely in Mm . Since Mm is a complete algebra, we conclude that the above series converges in Mm . Using Eq. (1) of Proposition 1, one has Pn − Pn−2 = An Pn−1 and hence, n n X X P2n = A2i P2i−1 and P2n+1 = P1 + A2i+1 P2i . i=1
i=1
Analogous results can be obtained for Qn . Thus, all limits lim P2n+1 = P (1) , lim P2n = P (2) , lim Q2n+1 = Q(1) and lim Q2n = Q(2) exist in Mm . According to Theorem 1, we see that P2n det P2n+1 − Q2n+1 (6) det Q2n = 1, Q2n
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P2n+1 det Q2n − P2n det Q2n+1 = 1. Q2n+1
121
(7)
But, we have lim Pn /Qn = M. Then P2n P2n+1 = lim = M. lim n→+∞ Q2n n→+∞ Q2n+1 Consequently, when n → +∞, relations (6) and (7) give: det(P (1) − Q(1) M) det Q(2) = 1
(8)
and det(Q(2) M − P (2) ) det Q(1) = 1. From (8) and (9), we deduce that det Q(1)
(9) = /
are invertible in Mm . Using (8) and (9) again, we have / 0. det P (1) − Q(1) (Q(2) )−1 P (2) =
0 and det Q(2)
= /
0. Then Q(1)
and Q(2)
Consequently, we obtain det((Q(1) )−1 P (1) − (Q(2) )−1 P (2) ) = / 0, it follows that the two limits lim P2n+1 /Q2n+1 = (Q(1) )−1 P (1) and lim P2n /Q2n = (Q(2) )−1 P (2) are distinct, which is absurd. Finally, we conclude that the continued fraction K(I /An ) diverges. P Corollary 1. If the series +∞ n=0 An converges absolutely in Mm , then the ordinary continued fraction K(I /An ) diverges in Mm .
2.2. Sufficient condition for convergence of K(I /An ) In the following, we give a convergence sufficient condition for continued fraction of the form K(I /An ). −1 Theorem 3. Let (An ) be a sequence of matrices in Mm . If kA−1 2n−1 k 6 α and kA2n k 6 β for all n > 1, where 0 < α < 1, 0 < β < 1 and αβ 6 1/4, then the continued fraction K(I /An ) converges in Mm .
Before proving Theorem 3, we record one lemma dealing with real continued fractions K(an /1). Lemma 2. Let δ be a real number such that 0 < δ 6 1/4. (i) The nth convergent of the continued fraction K(an /1), where a1 = 1 and an = −δ, is less than the nth convergent of K(bn /1), where b1 = 1 and bn = −1/4. (ii) The nth convergent of K(bn /1), where b1 = 1 and bn = −1/4, is fn = 2n/(n + 1).
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Proof. (i) We prove it easily and directly. (ii) We now determine the convergents fn of the real continued fraction K(bn /1) defined in Lemma 1. We have 1 4 1 f2 = = = . f1 = 1, 1 − 1/4 3 3/4 We compute f3 = p3 /q3 , by p3 = p2 − 14 p1 = 1 − 1/4 = 3/4, q3 = q2 − 14 q1 = 3/4 − 1/4 = 2/4, then f3 = 3/2. By an induction argument, we show that for all n > 1, fn = 2n/ (n + 1), and precisely we have: For n odd (n = 2k), 2n n+1 and qn = . k 4 4k For n even (n = 2k + 1), pn =
pn =
n 4k
and qn =
k+1 . 4k
Proof of Theorem 3. (i) We will prove that the sequence (Fn ) = (Pn /Qn ) = (Q−1 n Pn ) is bounded. We can assume that A0 = 0. Then, we have F1 = P1 /Q1 = −1 A−1 1 and kF1 k = kA1 k 6 α. We also have −1 −1 −1 −1 F2 = (A1 + A−1 = (I + A−1 2 ) 1 A2 ) A1 . −1 −1 −1 Since kA−1 1 A2 k 6 αβ < 1. Then I + A1 A2 is invertible and −1 −1 −1 kF2 k = k(I + A−1 1 A2 ) A1 k 6
kA−1 1 k
−1 1 − kA−1 1 kkA2 k
.
Thus, kF2 k 6
α . 1 − (αβ/1)
For n = 3, we find −1 −1 −1 −1 −1 −1 −1 −1 F3 = (A1 + (A2 + A−1 = (I + A−1 3 ) ) 1 (I + A2 A3 ) A2 ) A1 . −1 We note that all the previous inverses exist because kA−1 2 A3 k 6 αβ < 1 and −1 −1 −1 −1 kA−1 1 (I + A2 A3 ) A2 k < 1. So, we deduce that
α αβ αβ . 1− 1− 1 Now, suppose that kF3 k 6
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α αβ αβ , (10) 1− 1 − · · · 1 where the right-hand side of (10) is α times the nth convergent of the real continued fraction K(an /1) with a0 = 0, a1 = 1 and an = −αβ for n > 2. For n + 1, to obtain Fn+1 from Fn , we must replace An by An + A−1 n+1 = An (I + −1 −1 An An+1 ). −1 −1 Therefore, the matrix An + A−1 n+1 is invertible (kAn An+1 k 6 αβ < 1), and it is easy to see that −1 −1 −1 −1 −1 A−1 (11) Fn = I + A−1 1 (· · · (I + An−1 An ) ) · · ·) 1 . kFn k 6
Replacing An by An + A−1 n+1 in (11), we find −1 −1 −1 −1 −1 −1 −1 · · ·)−1 A−1 Fn+1 = I + A−1 1 (· · · (I + An−1 (I + An An+1 ) An ) 1 , (12) with −1 −1 −1 −1 kA−1 n−1 (I + An An+1 ) An k 6 αβ < 1.
(13)
Consequently, using relationships (10) and (12) we conclude that α αβ αβ , kFn+1 k 6 1− 1 − · · · 1 where the right-hand side is α times the (n + 1)th convergent of the real continued fraction K(an /1) defined earlier. Using Lemma 2, we conclude that fn < 2, which yields 0 < kFn k 6 12 fn < 1,
n > 1.
Finally, the sequence (Fn ) = (Q−1 n Pn ) is bounded. (ii) In this section, we show that the series +∞ X Pn Pn+1 − Qn+1 Qn n=0
converges absolutely. We have −1 −1 −1 −1 F1 − F2 =A−1 1 − (I + A1 A2 ) A1 −1 −1 −1 =(I − (I + A−1 1 A2 ) )A1 −1 −1 −1 −1 −1 =(I + A−1 1 A2 ) A1 A2 A1 .
(14)
For n = 2, we have −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 F2 − F3 =(I + A−1 1 A2 ) A1 − (I + A1 (I + A2 A3 ) A2 ) A1 −1 −1 −1 −1 −1 −1 −1 −1 =(I + A−1 − (I + A−1 1 A2 ) 1 (I + A2 A3 ) A2 ) A1 .
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We now use the relation X−1 − Y −1 = Y −1 (Y − X)X−1 ,
(15)
and obtain
−1 −1 −1 −1 −1 −1 (I + A A ) A − A A F2 − F3 =F3 A1 A−1 F2 1 2 3 2 1 2 −1 −1 =F3 (I + A−1 − I A−1 2 A3 ) 2 F2 −1 −1 −1 −1 −1 =−F3 (I + A−1 2 A3 ) A2 A3 A2 F2 .
(16)
In order to formulate Eqs. (14) and (16), we must adopt some news notations. Putting Fi,n =
I I I Ai + Ai+1 + · · · + An
for 1 6 i 6 n,
it becomes I I I , −1 −1 −1 Ai + Ai+1 + · · · + An−2 + (I + A−1 n−1 An ) An−1 −1 −1 −1 −1 −1 −1 −1 −1 (· · · (I + A (I + A A ) A ) · · ·) A−1 = I + A−1 n i n−2 n−1 n−1 i .
Fi,n =
(17) −1 −1 −1 −1 In particular, F1,n = Fn , Fn−1,n = (I + A−1 n−1 An ) An−1 and Fn,n = An . Using notation (17), relationships (14) and (16) yield
F2 − F1 =−F1,2 F2,2 F1,1 , F3 − F2 =F1,3 F2,3 F3,3 F2,2 F1,2 . Since kAi k 6 δ for all i > 1, where δ = α or β and from (17) we deduce that δ αβ αβ , 1− 1 − · · · 1 where the right-hand side is δ times the (n + 1 − i)th convergent of K(an /1). Thus, kFi,n k 6
n+1−i . (18) n+2−i By the same manipulations as in (14) and (16), we show that for all n > 1, we have kFi,n k 6 δfn+1−i = 2δ
Pn+1 Pn − Qn+1 Qn = (−1)n F1,n+1 F2,n+1 · · · Fn,n+1 Fn+1,n+1 Fn,n · · · F2,n F1,n . In (19) use (18) to write
(19)
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kFn+1 − Fn k6 22n−1 α(αβ)n 6
125
21 12 n n+1 n ··· . ··· n+2n+1 32 23 n+1
1 . 2(n + 2)(n + 1)
Since the series +∞ X n=0
1 (n + 2)(n + 1)
converges in IR, this implies that the series +∞ X Pn Pn+1 − Qn+1 Qn n=0
converges absolutely and then converges in Mm . This concludes the proof of Theorem 3. −1 Remark 1. The choice of kA−1 2n−1 k 6 α and kA2n k 6 β with αβ 6 1/4 in Theorem 3 follows from the scalar case, where 1/4 is the best possible. In the scalar continued fraction K(an /1), if a1 = 1, and for n > 2 an = α, where 0 < α 6 1/4, then K(an /1) converges in R. But, if an = α, where α > 1/4, then K(an /1) diverges. Because, we look when we can find a real number x such that
x=
1 1−
α α 1− 1−···
,
which is equivalent to finding a real number y such that y =1−
α 1 = . x 1−y
The previous equation becomes y 2 − y + α = 0, so the discriminant is D = 1 − 4α > 0 if α 6 1/4. 2.3. Convergence of continued fractions of the form K(Bn /An ) 2.3.1. Equivalent continued fractions Let (An ) and (Bn ) be two sequences of Mm , we notice that we can write the first convergents of the continued fractions K(Bn /An ) by −1 −1 F1 = A0 + A−1 1 B1 = A0 + (B1 A1 )
−1 F2 =A0 + A1 + A−1 B1 2 B2 −1 =A0 + B1−1 A1 + (B2−1 A2 B1−1 )−1 .
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If we put, A∗1 = B1−1 A1 and A∗2 = B2−1 A2 B1 , we have F1 = A0 +
I , A∗1
F2 = A0 +
I I A∗1 + A∗2
and generally we prove by a recurrence, that if we put for all k > 1 A∗2k = (B2k · · · B2 )−1 A2k B2k−1 · · · B1 and A∗2k+1 = (B2k+1 · · · B1 )−1 A2k+1 B2k · · · B2 , then the continued fractions A0 + K(Bn /An ) and A0 + K(I /A∗n ) are equivalent. It follows that the convergence of one of these continued fractions implies the convergence of the other continued fraction. 2.3.2. Convergence criteria of K(Bn /An ) In this section, we present a new convergence criteria on the continued fractions of the matrices. Theorem 4. Let (An ), (Bn ) be two sequences in Mm . If k(B2k−2 · · · B2 )−1 A−1 2k−1 B2k−1 · · · B1 k 6 α and k(B2k−1 · · · B1 )−1 A−1 2k B2k · · · B2 k 6 β for all k > 1, where 0 < α < 1, 0 < β < 1 and αβ 6 1/4, then the continued fraction K(Bn /An ) converges in Mm . Corollary 2. If we have for all k > 1, k−1 Y
−1 kB2i kkA−1 2k−1 k
i=1
k Y
kB2i−1 k 6 α
i=1
and k Y i=1
−1 kB2i−1 kkA−1 2k k
k Y
kB2i k 6 β,
i=1
where 0 < α < 1, 0 < β < 1 and αβ 6 1/4, then the continued fraction K(Bn /An ) converges in Mm . Proof of Theorem 4. The continued fractions K(Bn /An ) and K(I /A∗n ) are equivalent and an equivalence transformation does not change the classical approximants of continued fractions. So, from Theorem 3, we deduce the convergence of K(I /A∗n ). And the result of Theorem 4 is proved.
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Remark 2. We remark that we can replace the norm used in this paper by any matricial norm of matrices. The previous results can be generalized if the matrices An and Bn are not invert+ ible by replacing the inverses of An and Bn by their pseudo-inverses A+ n and Bn , respectively.
3. Examples 3.1. The matrix exponential We recall that Euler has proven that the continued fraction expansion of the exponential function ex exists for all real numbers x (see [4]), it was given by x2 2x x 2 x 2 , , ,..., ,... . ex = 1; 2 − x 6 10 4k − 2 Let A be an m × m matrix with real coefficients. We study the convergence of the continued fractions A0 + K(Bk /Ak ), where A0 = I, Ak = (4k − 2)I A1 = 2A and for all k > 2. B1 = 2I − A Bk = A2 We have for k > 2 k(B2k−1 · · · B3 )−1 A−1 2k (B2k · · · B4 )k =
1 8k − 2
and 1 kA2 k. 8k + 2 Let α = β = 1/2. We remark that the first inequality 1/(8k − 2) 6 1/2 is true for all k > 2 and for all matrices A, and when k is a large number, the second inequality k(B2k · · · B4 )−1 A−1 2k+1 (B2k+1 · · · B3 )k =
1 kA2 k 6 1/2 8k + 2 is already verified. According to Theorem 4, we affirm that the following continued fraction +∞ A2 2A , I; 2I − A (4k − 2)I k=2 converges for all m × m matrices A with real coefficients. The limit is the exponential matrix eA , and consequently we have 2A A2 A2 A2 A , , ,..., ,... . (20) e = I; 2I − A 6I 10I (4k − 2)I
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3.2. The function Log(I + A) It is well known that the continued fraction expansion of Log(1 + x), for all real x in ] − 1, 1[ is (see [5]) x x/2 x/6 2x/6 2x/10 , , , ,... , Log(1 + x) = 0; , 1 1 1 1 1 which is equivalent to i h x a x a x an x 2 3 , ,..., ,... , Log(1 + x) = 0; , 1 1 1 1 where k k a2k = and a2k+1 = for all k > 1. 2(2k − 1) 2(2k + 1) Let A be a matrix in Mm such that kAk < 1. We now study the convergence of the following continued fractions A0 + K(Bk /Ak ), where A0 = 0, Ak = I A1 = A and for all k > 2. B1 = I Bk = ak A For this example, we recall that the continued fraction K(I /An ) is equivalent −1 ∗ −1 to the continued fraction K(Bn∗ /I ), where B1∗ = A−1 1 and Bn = An−1 An for all n > 2. So in our proof we will use Theorem 3 and this equivalence. From the above-continued fraction expansion, we see that Bn∗ = an An . Since kAk < 1, then kBn∗ k = an kAk < 1/4 when n is a large number. According to Theorem 3, we affirm that the following continued fraction an A A A/2 A/6 2A/6 2A/10 , , , ,..., ,... 0; , I I I I I I converges for each m × m matrix A such that kAk < 1. Furthermore, the continued fraction expansion of Log(I + A) is A A/2 A/6 2A/6 2A/10 an A Log(I + A) = 0; , , , , ,..., ,... . I I I I I I Remark 3. As is well known (e.g., see [3]) every invertible matrix has a logarithm (not necessary real). Since, in this example kAk < 1, then the matrix I + A is invertible and its logarithm exists.
References [1] P.G. Ciarlet, Introduction à l’analyse numérique matricielle et à l’optimisation, Masson, Paris, 1982. [2] T.L. Hayden, Continued fractions in Banach spaces, Rocky Mtn. J. Math. 4 (1974) 367–369.
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[3] B.W. Helton, Logarithms of matrices, Proc. Amer. Math. Soc. 19 (1968) 733–738. [4] A.N. Khovanski, The Applications of Continued Fractions and their Generalisations to Problems in Approximation Theory, 1963, Noordhoff, Groningen, The Netherlands (Chapter 2). [5] L. Lorentzen, H. Wadeland, Continued Fractions with Applications, Elsevier, Amsterdam, 1992. [6] G.J. Murphy, C ∗ -Algebras and Operators Theory, Academic press, New York, INC Harcourt Brace Jovanovich, publishers, 1990 (Chapter 2). [7] N. Negoescu, Convergence theorems on noncommutative continued fractions, Rev. Anal. Numér. Théorie Approx. 5 (1977) 165–180. [8] R.S. Varga, On higher order stable imlicit methods for solving parabolic partial differential equations, J. Math. Phys. 40 (1961) 220–231. [9] W.B. Jones, W.J. Thron, Continued Fractions: Analytic Theory and Applications, Addison-Wesley, Reading, MA, 1980. [10] P. Wynn, One some recent developments in the theory of continued fractions, SIAM J. Numer. Anal. 1 (1964) 177–197.
Convergence of matrix continued fractions Mustapha Raissouli a,∗ , Ali Kacha b a Department of Mathematics and Informatic, Sciences Faculty, Moulay Ismail University, AFACS UFR,
AFA Group, B.P. 4010, Meknes, Morocco b Department of Mathematics, Sciences and Technics Faculty, Moulay Ismail University, B.P. 509,
Errachidia, Morocco Received 22 February 2000; accepted 14 June 2000 Submitted by R.A. Brualdi
Abstract The aim of this work is to give some criteria on the convergence of matrix continued fractions. We begin by presenting some new results which generalize the links between the convergent elements of real continued fractions. Secondly, we give necessary and sufficient conditions for the convergence of continued fractions of matrix arguments. This paper will be completed by illustrating the theoretical results with some examples. © 2000 Elsevier Science Inc. All rights reserved. AMS classification: 40A15; 15A60 Keywords: Continued fraction; An m × m matrix; Convergence
1. Definitions, generalities and notations 1.0. Introduction In recent few years, the theory of continued fraction has extensive developments and applications. Since calculations involving matrix valued functions with matrix arguments are feasible with large computers, such functions have been used in solving many difficult problems. In [8], Varga used Padé approximants, equivalent to continued fraction approximants of the matrix exponential to derive and analyze cer∗ Corresponding author.
E-mail address: [email protected] (M. Raissouli). 0024-3795/00/$ - see front matter 2000 Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 0 0 ) 0 0 1 9 6 - 8
116
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tain discrete approximations to solutions of self-adjoint parabolic differential equations. The matrix continued fraction used by Varga is an example of noncommutative continued fractions about which not much is known. A few convergence theorems on noncommutative continued fractions have been published. Two theorems are stated in [10], where Wynn reviews many aspects of the theory of continued fractions, whose elements do not commute under a multiplication law. In a Banach space, extentions of Worpitsky’s theorem have been proven by Hayden [2] and Negoescu [7]. In this paper, several convergence criteria on the noncommutative continued fractions whose arguments are m × m matrices of the forms K(Bn /An ) are given. 1.1. Convergence of matrices Throughout this paper, we denote by Mm the set of m × m real (or complex) matrices endowed with the classical norm defined by kAxk = sup {kAxk, kxk = 1}. ∀A ∈ Mm , kAk = sup kxk x =0 / This norm satisfies the relationship kABk 6 kAkkBk. Let (An ) be a sequence of matrices in Mm . We say that (An ) converges if there exists a matrix A ∈ Mm , such that kAn − Ak tends to 0 when n tends to +∞. In this case, we write An −→ A
or
lim An = A.
n→+∞
Let A ∈ Mm . We denote by det A the determinant of A. It is well known that if An −→ A in Mm , then det An −→ det A in R. Moreover, if An and A are invertible, −1 in M . then A−1 m n −→ A Let A and B be two matrices with B invertible. We write A/B = B −1 A, in particular, if A = I , where I is the mth order identity matrix, then we write I /B = B −1 . It is clear that for any invertible matrix X, we have XA AX A = = / . B XB BX 1.2. Some results on continued fractions of matrices Let (An )n>0 , (Bn )n>1 be two sequences of Mm . We call K(Bn /An ) the continued fraction if the quantity B1 B1 B2 = A0 ; , ,... . A0 + A1 A2 A 1 + B2 A2 +···
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Sometimes, we use briefly the notation Bn +∞ . A0 ; An n=1 The fractions Bn and An
Pn Bi n = A0 ; Qn Ai i=1
are called, respectively, the nth partial quotient and the nth convergent of the continued fraction K(Bn /An ). When Bn = I for all n > 1, then K(I /An ) is called an ordinary continued fraction. Proposition 1. The elements (Pn )n>−1 and (Qn )n>−1 of the nth convergent of K(Bn /An ) are given by the relationships Pn = An Pn−1 + Bn Pn−2 P−1 = I, P0 = A0 , n > 1. (1) and Q−1 = 0, Q0 = I Qn = An Qn−1 + Bn Qn−2 Proof. The proof of Proposition 1 uses induction on n. The relationships are obviously verified for n = 1. Now, suppose that for all p 6 n Eq. (1) is true. For n + 1, using the definition of K(Bn /An ), to obtain Rn+1 = Pn+1 /Qn+1 from Rn = Pn /Qn , we must replace An by An + (Bn+1 /An+1 ). So, we have Bn+1 An + A Pn−1 + Bn Pn−2 n+1 Rn+1 = Bn+1 An + A Qn−1 + Bn Qn−2 n+1 (An+1 An + Bn+1 )Pn−1 + An+1 Bn Pn−2 (An+1 An + Bn+1 )Qn−1 + An+1 Bn Qn−2 An+1 (An Pn−1 + Bn Pn−2 ) + Bn+1 Pn−1 = An+1 (An Qn−1 + Bn Qn−2 ) + Bn+1 Qn−1 An+1 Pn + Bn+1 Pn−1 = , An+1 Qn + Bn+1 Qn−1 =
and Proposition 1 is proved. Proposition 2. For all ordinary continued fractions K(I /An ), we have for all n > 0, n Y (1 + kAi k), (2) kPn k 6 i=0
kQn k 6
n Y i=0
(1 + kAi k).
(3)
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Proof. Q For n = 0, relation (2) holds because P0 =A0 , suppose that for all l 6 n, kPl k 6 li=0 (1 + kAi k). Then, for n + 1, we have Pn+1 = An+1 Pn + Pn−1 and we obtain kPn+1 k 6 kAn+1 kkPn k + kPn−1 k, which yields kPn+1 k 6kAn+1 k
n Y
(1 + kAi k) +
i=0
6
n−1 Y
n−1 Y
(1 + kAi k)
i=0
(1 + kAi k) (kAn+1 k(1 + kAn k) + 1) .
i=0
Consequently, we find kPn+1 k 6
n−1 Y
(1 + kAi k) (1 + kAn k) (1 + kAn+1 k) ,
i=0
Qn and by this induction, we have for all n > 1, QnkPn k 6 i=0 (1 + kAi k). Similarly, we also prove that kQn k 6 i=0 (1 + kAi k), and the proof is complete. Theorem 1. The convergents of the continued fraction K(Bn /An ) satisfy n Y Pn Pn−1 det = (−1)m(n−1) (det Qn )−1 (det Qn−1 )−1 − det Bi . Qn Qn−1 i=1
Proof. Put Pn Pn−1 −1 − = Q−1 n Pn − Qn−1 Pn−1 . Qn Qn−1 We can write Dn =
−1 Dn = Q−1 n (Pn − Qn Qn−1 Pn−1 ).
It follows that det Dn = (det Qn )−1 det(Pn − Qn Q−1 n−1 Pn−1 ). We now will use the next lemma. Lemma 1 [1]. Let (M)i,j =1,2 be four matrices in Mm , with M22 invertible. Then we have M11 M12 −1 = det M22 det(M11 − M21 M22 det M12 ). (4) M21 M22 Using Lemma 1, one has det Dn = (det Qn )
−1
(det Qn−1 )
−1
P det n Qn
Pn−1 . Qn−1
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If we put that
P δn = det n Qn
Pn−1 , Qn−1
and due to Eq. (1) in Proposition 1, we deduce that Pn−1 An Pn−1 + Bn Pn−2 δn = det . An Qn−1 + Bn Qn−2 Qn−1 If the determinant is multilinear, then Bn Pn−2 An Pn−1 Pn−1 + det δn = det An Qn−1 Qn−1 Bn Qn−2
Pn−1 . Qn−1
Using the previous lemma again, we find that An Pn−1 Pn−1 det =0 An Qn−1 Qn−1 and we deduce that δn = detQn−1 det(Bn Pn−2 − Bn Qn−2 Q−1 n−1 Pn−1 )
= det Qn−1 det Bn det(Pn−2 − Qn−2 Q−1 n−1 Pn−1 ) −1 = det Qn−1 det Bn det Qn−2 Qn−1 (Qn−1 Q−1 n−2 Pn−2 − Pn−1 ) .
Thus, δn = det Bn det Qn−2 det(Qn−1 Q−1 n−2 Pn−2 − Pn−1 )
= (−1)m det Bn det Qn−2 det(Pn−1 − Qn−1 Q−1 n−2 Pn−2 ).
By applying relation (4), we obtain Pn−1 m δn = (−1) det Bn det Qn−1
Pn−2 , Qn−2
which becomes δn = (−1)m (det Bn ) δn−1 . Consequently, we have ! n Y δn = (−1)m(n−1) det Bi δ1 , i=2
where
P1 δ1 = det Q1
P0 I
A1 A0 + B1 = det A1
A0 . I
The lemma gives δ1 = det B1 and Theorem 1 is proved.
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2. Convergence criteria of continued fractions Definition 1. The continued fraction K(Bn /An ) converges in Mm if the sequence (Fn ) = (Pn /Qn ) = (Q−1 n Pn ) converges in Mm in the sense that there exists a matrix F ∈ Mm such that limn→+∞ kFn − F k = 0. In the other case, we say that K(Bn /An ) is divergent. It is clear that n X Pi−1 Pi Pn = A0 + − . Qn Qi Qi−1
(5)
i=1
From (5), we see that the continued fraction K(Bn /An ) converges in Mm if and only if the series +∞ X Pn−1 Pn − Qn Qn−1 n=1
converges in Mm . 2.1. Necessary condition for the convergence of K(I /An ) We now give a necessary condition for the convergence of the matrix continued fraction of the form K(I /An ) . of Mm . If the continued fraction K(I /An ) Theorem 2. Let (An )n>0 be a sequence P converges in Mm , then the series +∞ n=0 kAn k diverges. Proof. We assume that K(I /A P n ) converges and we put M = limn→+∞ Pn /Qn . Suppose We will show that the series n kAn k diverges by an absurd reasoning. Q+∞ P that the series +∞ n=0 kAn k converges, then the infinite product n=0 (1 + kAn k) also converges. P P It follows, by virtue of relationships (2) and (3), that the series n An Pn−1 and n An Qn−1 converge absolutely in Mm . Since Mm is a complete algebra, we conclude that the above series converges in Mm . Using Eq. (1) of Proposition 1, one has Pn − Pn−2 = An Pn−1 and hence, n n X X P2n = A2i P2i−1 and P2n+1 = P1 + A2i+1 P2i . i=1
i=1
Analogous results can be obtained for Qn . Thus, all limits lim P2n+1 = P (1) , lim P2n = P (2) , lim Q2n+1 = Q(1) and lim Q2n = Q(2) exist in Mm . According to Theorem 1, we see that P2n det P2n+1 − Q2n+1 (6) det Q2n = 1, Q2n
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P2n+1 det Q2n − P2n det Q2n+1 = 1. Q2n+1
121
(7)
But, we have lim Pn /Qn = M. Then P2n P2n+1 = lim = M. lim n→+∞ Q2n n→+∞ Q2n+1 Consequently, when n → +∞, relations (6) and (7) give: det(P (1) − Q(1) M) det Q(2) = 1
(8)
and det(Q(2) M − P (2) ) det Q(1) = 1. From (8) and (9), we deduce that det Q(1)
(9) = /
are invertible in Mm . Using (8) and (9) again, we have / 0. det P (1) − Q(1) (Q(2) )−1 P (2) =
0 and det Q(2)
= /
0. Then Q(1)
and Q(2)
Consequently, we obtain det((Q(1) )−1 P (1) − (Q(2) )−1 P (2) ) = / 0, it follows that the two limits lim P2n+1 /Q2n+1 = (Q(1) )−1 P (1) and lim P2n /Q2n = (Q(2) )−1 P (2) are distinct, which is absurd. Finally, we conclude that the continued fraction K(I /An ) diverges. P Corollary 1. If the series +∞ n=0 An converges absolutely in Mm , then the ordinary continued fraction K(I /An ) diverges in Mm .
2.2. Sufficient condition for convergence of K(I /An ) In the following, we give a convergence sufficient condition for continued fraction of the form K(I /An ). −1 Theorem 3. Let (An ) be a sequence of matrices in Mm . If kA−1 2n−1 k 6 α and kA2n k 6 β for all n > 1, where 0 < α < 1, 0 < β < 1 and αβ 6 1/4, then the continued fraction K(I /An ) converges in Mm .
Before proving Theorem 3, we record one lemma dealing with real continued fractions K(an /1). Lemma 2. Let δ be a real number such that 0 < δ 6 1/4. (i) The nth convergent of the continued fraction K(an /1), where a1 = 1 and an = −δ, is less than the nth convergent of K(bn /1), where b1 = 1 and bn = −1/4. (ii) The nth convergent of K(bn /1), where b1 = 1 and bn = −1/4, is fn = 2n/(n + 1).
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Proof. (i) We prove it easily and directly. (ii) We now determine the convergents fn of the real continued fraction K(bn /1) defined in Lemma 1. We have 1 4 1 f2 = = = . f1 = 1, 1 − 1/4 3 3/4 We compute f3 = p3 /q3 , by p3 = p2 − 14 p1 = 1 − 1/4 = 3/4, q3 = q2 − 14 q1 = 3/4 − 1/4 = 2/4, then f3 = 3/2. By an induction argument, we show that for all n > 1, fn = 2n/ (n + 1), and precisely we have: For n odd (n = 2k), 2n n+1 and qn = . k 4 4k For n even (n = 2k + 1), pn =
pn =
n 4k
and qn =
k+1 . 4k
Proof of Theorem 3. (i) We will prove that the sequence (Fn ) = (Pn /Qn ) = (Q−1 n Pn ) is bounded. We can assume that A0 = 0. Then, we have F1 = P1 /Q1 = −1 A−1 1 and kF1 k = kA1 k 6 α. We also have −1 −1 −1 −1 F2 = (A1 + A−1 = (I + A−1 2 ) 1 A2 ) A1 . −1 −1 −1 Since kA−1 1 A2 k 6 αβ < 1. Then I + A1 A2 is invertible and −1 −1 −1 kF2 k = k(I + A−1 1 A2 ) A1 k 6
kA−1 1 k
−1 1 − kA−1 1 kkA2 k
.
Thus, kF2 k 6
α . 1 − (αβ/1)
For n = 3, we find −1 −1 −1 −1 −1 −1 −1 −1 F3 = (A1 + (A2 + A−1 = (I + A−1 3 ) ) 1 (I + A2 A3 ) A2 ) A1 . −1 We note that all the previous inverses exist because kA−1 2 A3 k 6 αβ < 1 and −1 −1 −1 −1 kA−1 1 (I + A2 A3 ) A2 k < 1. So, we deduce that
α αβ αβ . 1− 1− 1 Now, suppose that kF3 k 6
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123
α αβ αβ , (10) 1− 1 − · · · 1 where the right-hand side of (10) is α times the nth convergent of the real continued fraction K(an /1) with a0 = 0, a1 = 1 and an = −αβ for n > 2. For n + 1, to obtain Fn+1 from Fn , we must replace An by An + A−1 n+1 = An (I + −1 −1 An An+1 ). −1 −1 Therefore, the matrix An + A−1 n+1 is invertible (kAn An+1 k 6 αβ < 1), and it is easy to see that −1 −1 −1 −1 −1 A−1 (11) Fn = I + A−1 1 (· · · (I + An−1 An ) ) · · ·) 1 . kFn k 6
Replacing An by An + A−1 n+1 in (11), we find −1 −1 −1 −1 −1 −1 −1 · · ·)−1 A−1 Fn+1 = I + A−1 1 (· · · (I + An−1 (I + An An+1 ) An ) 1 , (12) with −1 −1 −1 −1 kA−1 n−1 (I + An An+1 ) An k 6 αβ < 1.
(13)
Consequently, using relationships (10) and (12) we conclude that α αβ αβ , kFn+1 k 6 1− 1 − · · · 1 where the right-hand side is α times the (n + 1)th convergent of the real continued fraction K(an /1) defined earlier. Using Lemma 2, we conclude that fn < 2, which yields 0 < kFn k 6 12 fn < 1,
n > 1.
Finally, the sequence (Fn ) = (Q−1 n Pn ) is bounded. (ii) In this section, we show that the series +∞ X Pn Pn+1 − Qn+1 Qn n=0
converges absolutely. We have −1 −1 −1 −1 F1 − F2 =A−1 1 − (I + A1 A2 ) A1 −1 −1 −1 =(I − (I + A−1 1 A2 ) )A1 −1 −1 −1 −1 −1 =(I + A−1 1 A2 ) A1 A2 A1 .
(14)
For n = 2, we have −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 F2 − F3 =(I + A−1 1 A2 ) A1 − (I + A1 (I + A2 A3 ) A2 ) A1 −1 −1 −1 −1 −1 −1 −1 −1 =(I + A−1 − (I + A−1 1 A2 ) 1 (I + A2 A3 ) A2 ) A1 .
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We now use the relation X−1 − Y −1 = Y −1 (Y − X)X−1 ,
(15)
and obtain
−1 −1 −1 −1 −1 −1 (I + A A ) A − A A F2 − F3 =F3 A1 A−1 F2 1 2 3 2 1 2 −1 −1 =F3 (I + A−1 − I A−1 2 A3 ) 2 F2 −1 −1 −1 −1 −1 =−F3 (I + A−1 2 A3 ) A2 A3 A2 F2 .
(16)
In order to formulate Eqs. (14) and (16), we must adopt some news notations. Putting Fi,n =
I I I Ai + Ai+1 + · · · + An
for 1 6 i 6 n,
it becomes I I I , −1 −1 −1 Ai + Ai+1 + · · · + An−2 + (I + A−1 n−1 An ) An−1 −1 −1 −1 −1 −1 −1 −1 −1 (· · · (I + A (I + A A ) A ) · · ·) A−1 = I + A−1 n i n−2 n−1 n−1 i .
Fi,n =
(17) −1 −1 −1 −1 In particular, F1,n = Fn , Fn−1,n = (I + A−1 n−1 An ) An−1 and Fn,n = An . Using notation (17), relationships (14) and (16) yield
F2 − F1 =−F1,2 F2,2 F1,1 , F3 − F2 =F1,3 F2,3 F3,3 F2,2 F1,2 . Since kAi k 6 δ for all i > 1, where δ = α or β and from (17) we deduce that δ αβ αβ , 1− 1 − · · · 1 where the right-hand side is δ times the (n + 1 − i)th convergent of K(an /1). Thus, kFi,n k 6
n+1−i . (18) n+2−i By the same manipulations as in (14) and (16), we show that for all n > 1, we have kFi,n k 6 δfn+1−i = 2δ
Pn+1 Pn − Qn+1 Qn = (−1)n F1,n+1 F2,n+1 · · · Fn,n+1 Fn+1,n+1 Fn,n · · · F2,n F1,n . In (19) use (18) to write
(19)
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kFn+1 − Fn k6 22n−1 α(αβ)n 6
125
21 12 n n+1 n ··· . ··· n+2n+1 32 23 n+1
1 . 2(n + 2)(n + 1)
Since the series +∞ X n=0
1 (n + 2)(n + 1)
converges in IR, this implies that the series +∞ X Pn Pn+1 − Qn+1 Qn n=0
converges absolutely and then converges in Mm . This concludes the proof of Theorem 3. −1 Remark 1. The choice of kA−1 2n−1 k 6 α and kA2n k 6 β with αβ 6 1/4 in Theorem 3 follows from the scalar case, where 1/4 is the best possible. In the scalar continued fraction K(an /1), if a1 = 1, and for n > 2 an = α, where 0 < α 6 1/4, then K(an /1) converges in R. But, if an = α, where α > 1/4, then K(an /1) diverges. Because, we look when we can find a real number x such that
x=
1 1−
α α 1− 1−···
,
which is equivalent to finding a real number y such that y =1−
α 1 = . x 1−y
The previous equation becomes y 2 − y + α = 0, so the discriminant is D = 1 − 4α > 0 if α 6 1/4. 2.3. Convergence of continued fractions of the form K(Bn /An ) 2.3.1. Equivalent continued fractions Let (An ) and (Bn ) be two sequences of Mm , we notice that we can write the first convergents of the continued fractions K(Bn /An ) by −1 −1 F1 = A0 + A−1 1 B1 = A0 + (B1 A1 )
−1 F2 =A0 + A1 + A−1 B1 2 B2 −1 =A0 + B1−1 A1 + (B2−1 A2 B1−1 )−1 .
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If we put, A∗1 = B1−1 A1 and A∗2 = B2−1 A2 B1 , we have F1 = A0 +
I , A∗1
F2 = A0 +
I I A∗1 + A∗2
and generally we prove by a recurrence, that if we put for all k > 1 A∗2k = (B2k · · · B2 )−1 A2k B2k−1 · · · B1 and A∗2k+1 = (B2k+1 · · · B1 )−1 A2k+1 B2k · · · B2 , then the continued fractions A0 + K(Bn /An ) and A0 + K(I /A∗n ) are equivalent. It follows that the convergence of one of these continued fractions implies the convergence of the other continued fraction. 2.3.2. Convergence criteria of K(Bn /An ) In this section, we present a new convergence criteria on the continued fractions of the matrices. Theorem 4. Let (An ), (Bn ) be two sequences in Mm . If k(B2k−2 · · · B2 )−1 A−1 2k−1 B2k−1 · · · B1 k 6 α and k(B2k−1 · · · B1 )−1 A−1 2k B2k · · · B2 k 6 β for all k > 1, where 0 < α < 1, 0 < β < 1 and αβ 6 1/4, then the continued fraction K(Bn /An ) converges in Mm . Corollary 2. If we have for all k > 1, k−1 Y
−1 kB2i kkA−1 2k−1 k
i=1
k Y
kB2i−1 k 6 α
i=1
and k Y i=1
−1 kB2i−1 kkA−1 2k k
k Y
kB2i k 6 β,
i=1
where 0 < α < 1, 0 < β < 1 and αβ 6 1/4, then the continued fraction K(Bn /An ) converges in Mm . Proof of Theorem 4. The continued fractions K(Bn /An ) and K(I /A∗n ) are equivalent and an equivalence transformation does not change the classical approximants of continued fractions. So, from Theorem 3, we deduce the convergence of K(I /A∗n ). And the result of Theorem 4 is proved.
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Remark 2. We remark that we can replace the norm used in this paper by any matricial norm of matrices. The previous results can be generalized if the matrices An and Bn are not invert+ ible by replacing the inverses of An and Bn by their pseudo-inverses A+ n and Bn , respectively.
3. Examples 3.1. The matrix exponential We recall that Euler has proven that the continued fraction expansion of the exponential function ex exists for all real numbers x (see [4]), it was given by x2 2x x 2 x 2 , , ,..., ,... . ex = 1; 2 − x 6 10 4k − 2 Let A be an m × m matrix with real coefficients. We study the convergence of the continued fractions A0 + K(Bk /Ak ), where A0 = I, Ak = (4k − 2)I A1 = 2A and for all k > 2. B1 = 2I − A Bk = A2 We have for k > 2 k(B2k−1 · · · B3 )−1 A−1 2k (B2k · · · B4 )k =
1 8k − 2
and 1 kA2 k. 8k + 2 Let α = β = 1/2. We remark that the first inequality 1/(8k − 2) 6 1/2 is true for all k > 2 and for all matrices A, and when k is a large number, the second inequality k(B2k · · · B4 )−1 A−1 2k+1 (B2k+1 · · · B3 )k =
1 kA2 k 6 1/2 8k + 2 is already verified. According to Theorem 4, we affirm that the following continued fraction +∞ A2 2A , I; 2I − A (4k − 2)I k=2 converges for all m × m matrices A with real coefficients. The limit is the exponential matrix eA , and consequently we have 2A A2 A2 A2 A , , ,..., ,... . (20) e = I; 2I − A 6I 10I (4k − 2)I
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3.2. The function Log(I + A) It is well known that the continued fraction expansion of Log(1 + x), for all real x in ] − 1, 1[ is (see [5]) x x/2 x/6 2x/6 2x/10 , , , ,... , Log(1 + x) = 0; , 1 1 1 1 1 which is equivalent to i h x a x a x an x 2 3 , ,..., ,... , Log(1 + x) = 0; , 1 1 1 1 where k k a2k = and a2k+1 = for all k > 1. 2(2k − 1) 2(2k + 1) Let A be a matrix in Mm such that kAk < 1. We now study the convergence of the following continued fractions A0 + K(Bk /Ak ), where A0 = 0, Ak = I A1 = A and for all k > 2. B1 = I Bk = ak A For this example, we recall that the continued fraction K(I /An ) is equivalent −1 ∗ −1 to the continued fraction K(Bn∗ /I ), where B1∗ = A−1 1 and Bn = An−1 An for all n > 2. So in our proof we will use Theorem 3 and this equivalence. From the above-continued fraction expansion, we see that Bn∗ = an An . Since kAk < 1, then kBn∗ k = an kAk < 1/4 when n is a large number. According to Theorem 3, we affirm that the following continued fraction an A A A/2 A/6 2A/6 2A/10 , , , ,..., ,... 0; , I I I I I I converges for each m × m matrix A such that kAk < 1. Furthermore, the continued fraction expansion of Log(I + A) is A A/2 A/6 2A/6 2A/10 an A Log(I + A) = 0; , , , , ,..., ,... . I I I I I I Remark 3. As is well known (e.g., see [3]) every invertible matrix has a logarithm (not necessary real). Since, in this example kAk < 1, then the matrix I + A is invertible and its logarithm exists.
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