Dynamic Systems and Applications 12 (2003) THE EULER MINDING ANALOGON AND OTHER IDENTITIES FOR CONTINUED FRACTIONS IN BANACH SPACES

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1 Dynamic Systems Applications 12 (2003) THE EULER MINDING ANALOGON AND OTHER IDENTITIES FOR CONTINUED FRACTIONS IN BANACH SPACES ANDREAS SCHELLING Carinthia Tech Institute, Villacher Strasse 1, A 9800 Spittal/Drau, Austria a.schelling@cti.ac.at ABSTRACT. The Euler Minding formulae other identities for the numerators denominators of the approximants of noncommutative continued fractions in Banach spaces are proved. AMS (MOS) Subject Classification. 11J70, 39A INTRODUCTION We investigate expressions of the form (1.1) B 0 A 1 (B 1 A 2 (...) 1 ) 1, where A n, B n are elements of a complex Banach algebra M, called noncommutative continued fraction (ncf). Wynn [4] developed a number of identities for such expressions. One of his results he announced as Euler Minding analogon, in fact the scalar result is due to Stern [1]. In this paper we generalize the genuine Euler Minding theorem for ncf develop formulae for generating a ncf to given sequences of nth numerators denominators. Furtheron we compute transformations, which are effictive in convergence theory of ncf. 2. DEFINITIONS AND NOTATIONS Througout this paper M denotes a noncommutative Banach algebra with identity E M the set of its invertible elements. Definition 2.1. For k N let S k : N k M, N k M, S k (X) := A k (B k X) 1, where A k, B k M. If T n := S 1... S n (0) exists, we call R n := B 0 T n the nth approximant of (1.1). Received September 20, $03.5 c Dynamic Publishers, Inc.

2 230 ANDREAS SCHELLING If R n exists, it can be shown [4] that R n = P n Q 1 n P n Q n are computed by the recurrence relations (especially Q n M ), where (2.1) P 1 = E, P 0 = B 0, P n = P n 1 B n P n 2 A n for n 1, Q 1 = 0, Q 0 = E, Q n = Q n 1 B n Q n 2 A n for n 1. P n Q n are called the nth numerator denominator of (1.1), respectively. Remark 2.2. To ensure the existence of R n, Q n M is not sufficient, see [2]. Since this has no consequences on our investigations, we renounce to go into detail. Wynn stated in [4] the identity 3. THE MAIN RESULTS n 1 (3.1) P n Q 1 n = B 0 ( 1) k A 1 B1 1 Q 0 A 2 Q 1 2 Q 1 A 3 Q 1 3 Q k 1 A k1 Q 1 k1 k=0 announced it as the Euler Minding formulae for ncf. In fact this is the generalization of Stern s result for ordinary continued fractions [1]. In the scalar case the Euler Minding formulae enable us to compute both, the nth numerators denominator of a continued fraction, without the recurrence relations (2.1). An analogous result for ncf is the following: Theorem 3.1. For the nth numerator P n denominator Q n of (1.1) we have (3.2) (3.3) n P n = B 0 B 1 B n B 0 B k 2 A k B k1 B n 1 k<i n 1 1 k<i<j n 1 k=1 B 0 B k 2 A k B k1 B i 1 A i1 B i2 B n B 0 B k 2 A k B k1 B i 1 A i1 B i2 B j 1 A j1 B j2 B n... n Q n = B 1 B n B 1 B k 2 A k B k1 B n 2 k<i n 1 2 k<i<j n 1 k=2 B 1 B k 2 A k B k1 B i 1 A i1 B i2 B n B 1 B k 2 A k B k1 B i 1 A i1 B i2 B j 1 A j1 B j2 B n... where B m B d = E, whenever m > d. Remark 3.2. We have [ n1 ] sums in (3.2) (3.3). 2

3 EULER MINDING IDENTITIES 231 Proof of 3.1. For the moment let B n M, n N. Define a sequence (U n ) as follows: U 0 := B 0, U 1 := B 0 B 1 A 1 U n := U n 1 (B n B 1 n 1A n ) for n 2. Writing out U n as an explicite sum, we split U n up in two parts U n U n, where U n consists of all terms which contain at least one of the factors B 1 k, hence U n = U nu n. We obtain U 0 = U 0 = P 0 U 1 = U 1 = P 1 immediately by definition U 2 = U 1 (B 2 B 1 1 A 2 ) = U 1 B 2 B 0 A 2 A 1 B 1 1 A 2 = U 1B 2 U 0A 2 U 2, hence U 2 = U 1B 2 U 0A 2 = P 2. Further U n = U n 1 B n U n 1 B 1 n 1A n = U n 1 B n (U n 2 B n 1 U n 2 B 1 n 2A n 1 )B 1 n 1A n = (U n 1 U n 1)B n (U n 2 U n 2)A n U n 2 B 1 n 2A n 1 B 1 n 1A n = U n U n. For B n 1 does not appear in U n 2, the term U n 2 B 1 n 2A n 1 B 1 n 1 is part of U n, thus U n = U n 1B n U n 2A n, we obtain by (2.1) that P n = U n for all n 0. To compute U n we establish that U n = (B 0 B 1 A 1 )(B 2 B 1 1 A 2 ) (B n B 1 n 1A n ) yields a part of U n if (i) in two successive factors (B k 1 B 1 k 2 A k 1) (B k B 1 k 1 A k) we multiply B k 1 by B 1 k 1 A k; or (ii) it is B 0 B 1 B n or A 1 B 2 B 3 B n. Combining this we obtain (3.2). Then (3.3) follows immediately, for Q n may be developed from P n 1 if we add 1 to the indices of all its elements. Finally we do not need any B 1 k B k M. for computation of (3.2) (3.3), so the formulae are granted for For investigations of continued fractions, the continued fraction transformations play an important part. Equivalence transformation see [2, 3], contraction extension see [3, 4] for ncf has already been studied. Another important transformation is the following. Let (3.4) E A 1 (E A 2 (...) 1 ) 1 be a ncf with its nth approximants R n. We now want to create a ncf with the same approximants but in permuted succession, for example R 1, R 0, R 3, R 2,.... This transformations are very interesting in convergence theory of ncf, because the transformed ncf converges exactly if (3.4) converges. Theorem 3.3. Assume (3.4) with nth numerators P n denominators Q n, then the sequence of numerators respectively denominators of (3.5) B 0 A 1(B 1 A 2(...) 1 ) 1 is:

4 232 ANDREAS SCHELLING (a) P 1, P 0, P 3, P 2 (EA 3 ), P 5, P 4 (EA 5 ),..., respectively, Q 1, Q 0, Q 3, Q 2 (EA 3 ), Q 5, Q 4 (E A 5 ),... if (E A 2n1 ) M for all n N, we put B 0 = E A 1, B 2 = A 2, B 2n = (E A 2n 1 ) 1 A 2n for n 2, B 2n 1 = E for n 1 A 1 = A 1, A 3 = A 2 A 3, A 2n1 = (E A 2n 1 ) 1 A 2n A 2n1 for n 2, A 2n = E A 2n1 for n 1, (b) P 0, P 2, P 1, P 4, P 3 (E A 4 ), P 6, P 5 (E A 6 ),..., respectively Q 0, Q 2, Q 1, Q 4, Q 3 (E A 4 ), Q 6, Q 5 (E A 6 ),..., if (E A 2n ) M for all n N, we put B 1 = E A 2, B 3 = A 3, B 2n1 = (E A 2n ) 1 A 2n1 for n 2, B 2n = E for n 0 A 1 = A 1, A 2n1 = E A 2n2 for n 1, A 2 = A 2, A 4 = A 3 A 4, A 2n = (E A 2n 2 ) 1 A 2n 1 A 2n for n 3. Proof. (a) Denote P n Q n as the numerators denominators of (3.5). Then P 0 = P 1, P 1 = P 0, by the recurrence relations (2.1) we obtain immediately P 2 = P 3 P 3 = P 2 (E A 2 ). Let P 2n = P 2n1 P 2n1 = P 2n (E A 2n1 ). Then (2.1) implies P 2n2 = P 2n1B 2n2 P 2nA 2n2 = P 2n (E A 2n1 )(E A 2n1 ) 1 A 2n2 P 2n1 (E A 2n2 ) = P 2n1 P 2n A 2n2 P 2n1 A 2n2 = P 2n3 P 2n3 = P 2n2B 2n3 P 2n1A 2n3 = P 2n3 P 2n (E A 2n1 )(E A 2n1 ) 1 A 2n2 A 2n3 = P 2n2 P 2n1 A 2n3 P 2n A 2n2 A 2n3 = P 2n2 (E A 2n3 ). Induction also implies the assertion for Q n. (b) follows analogously as (a). Finally we generate a ncf, whose numerators denominators are given sequences. Theorem 3.4. Let (X n ) n=0 M, (Y n ) n=0 M Y 0 = E. If (X n Yn 1 X n 1 Yn 1) M for all n N, then there exists a unique ncf B 0 A 1 (B 1 A 2 (...) 1 ) 1 with nth numerators X n denominators Y n for all n 0. For its partial numerators respectively denominators we obtain A 1 = X 1 X 0 Y 1, A n = Y 1 n 2(X n 1 Y 1 n 1 X n 2 Y 1 n 2) 1 (X n 1 Y 1 n 1 X n Y 1 n )Y n, for all n 2, respectively B 0 = X 0, B 1 = Y 1, B n = Y 1 n 1(X n 2 Y 1 n 2 X n 1 Y 1 n 1) 1 (X n 2 Y 1 n 2 X n Y 1 n )Y n for all n 2.

5 EULER MINDING IDENTITIES 233 Proof. Obviously we have P 0 = X 0, P 1 = X 1 Q 0 = Y 0, Q 1 = Y 1. For n 2 we construct A n B n as follows. The recurrence relations (2.1) imply (3.6) X n = X n 1 B n X n 2 A n Y n = Y n 1 B n Y n 2 A n. Multiply the left equation in (3.6) by X 1 n 1 the right equation in (3.6) by Y 1 n 1 to obtain by subtraction hence Y 1 n 1Y n X 1 n 1X n = (Y 1 n 1Y n 2 X 1 n 1X n 2 )A n Y 1 n 2(X n 1 Y 1 n 1 X n 2 Y 1 n 2) 1 (X n 1 Y 1 n 1 X n Y 1 n )Y n = A n, where uniqueness follows from regularity of all factors on the left side of the equation. Analogously B n is generated by subtraction of the left equation in (3.6) the right equation in (3.6) multiplied by Xn 2 1 Yn 2, 1 respectively. Finally the above construction leads to P n = X n Q n = Y n for all n 0. Remark 3.5. From Theorem 3.4 computation of ncf, whose nth denominators are identically E nth numerators are the nth partial sum of the series C k, the nth partial product of an infinite product is possible without difficulties, under the hypothesis that the respective expressions are invertible elements. As a consequence of Theorem 3.4 we obtain the following result. Corollary 3.6. Let E A 1 (B 1 A 2 (...) 1 ) 1 be a ncf with A n M for all n N, nth numerators P n M denominators Q n M for all n 0. Then E A 1 (B 1 A 1 A 2 (B 2 A 3 (...) 1 ) 1 ) 1 has the nth numerators Q n denominators P n. REFERENCES [1] O. Perron. Die Lehre von den Kettenbrüchen. Dritte, verbesserte und erweiterte Aufl. Bd. II. Analytisch-funktionentheoretische Kettenbrüche. B. G. Teubner Verlagsgesellschaft, Stuttgart, [2] P. Pfluger. Matrizenkettenbrüche. PhD thesis, ETH Zürich, [3] A. Schelling. Matrizenkettenbrüche. PhD thesis, Universität Ulm, [4] P. Wynn. Continued fractions whose coefficients obey a noncommutative law of multiplication. Arch. Rational Mech. Anal., 12: , k=0