Continued Fraction Expansions of the quasi-arithmetic power means ...

Continued Fraction Expansions of the quasi-arithmetic power means of positive matrices with parameter (p,α) Ali Kacha Ibn Toufail University, Faculty of Sciences Kenitra Department of Mathematics Morocco

This work is in collaboration with B. Nejjar and S. Salhi.

(XXXème Journées Arithmétique 2017-CAEN)

3-7 juillet 2017

1 / 25

Plan 1

Introduction

2

Definitions and notations

3

Main result

4

References


3-7 juillet 2017

2 / 25

Plan 1

Introduction

2


3

Main result

4

References


3-7 juillet 2017

2 / 25

Plan 1

Introduction

2


3

Main result

4

References


3-7 juillet 2017

2 / 25

Plan 1

Introduction

2


3

Main result

4

References


3-7 juillet 2017

2 / 25

Plan 1

Introduction

2


3

Main result

4

References


3-7 juillet 2017

2 / 25

Introduction

The basic idea of this theory over real numbers is to give an approximation of various real numbers by the rational ones. Recently, the extension of continued fractions theory from real numbers to the matrix case has seen several developments and interesting applications. Further, in the convergent case, the continued fraction expansions have the advantage that they converge more rapidly than other numerical algorithms.


3-7 juillet 2017

3 / 25

Introduction



3-7 juillet 2017

3 / 25

Introduction



3-7 juillet 2017

3 / 25

Introduction The definition of the quasi-arithmetic power mean with parameter (p,α), defined for A > 0 and B > 0 is given as follows : fp,α (A, B) = A1/2 ((1 − α)I + α(A−1/2 BA−1/2 )p )1/p A1/2 . The class of quasi-arithmetic power means contain many kinds of means : the mean f1,α (A, B) is the α-weighed arithmetic mean. The case f0,α (A, B) is the α-weighed geometric mean (this case is understood that we take limit as p−→0). The case f−1,α (A, B) is the α-weighed harmonic mean. The mean fp,1/2 (A, B) is the power mean or binomial mean of order p. However, the rational exponents of matrices in fp,α (A, B) imposes many difficulties.


3-7 juillet 2017

4 / 25

Introduction The definition of the quasi-arithmetic power mean with parameter (p,α), defined for A > 0 and B > 0 is given as follows : fp,α (A, B) = A1/2 ((1 − α)I + α(A−1/2 BA−1/2 )p )1/p A1/2 . The class of quasi-arithmetic power means contain many kinds of means : the mean f1,α (A, B) is the α-weighed arithmetic mean. The case f0,α (A, B) is the α-weighed geometric mean (this case is understood that we take limit as p−→0). The case f−1,α (A, B) is the α-weighed harmonic mean. The mean fp,1/2 (A, B) is the power mean or binomial mean of order p. However, the rational exponents of matrices in fp,α (A, B) imposes many difficulties.


3-7 juillet 2017

4 / 25


Definition 1) Let A ∈ Mm . A is said to be positive semidefinite (resp. positive definite) if A is symmetric and ∀x ∈ Rm , < Ax, x >≥ 0 ( resp. ∀x ∈ Rm , x 6= 0 < Ax, x >> 0) where < ., . > denotes the standard scalar product of Rm .


Definition 2) Let {An }n≥0 and {Bn }n≥1 be two sequences of matrices in Mm . We denote the continued fraction expansion by B1 B1 Bn A0 + := A0 ; , ..., . B2 A1 An A1 + A2 + ... Bn +∞ Sometimes, we denote this continued fraction by A0 ; or An n=1 K (Bn /An ).

Definition 3) Let (An )n≥0 and (Bn )n≥1 be two sequences of matrices in Mm . We define the sequences (Pn )n≥−1 and (Qn )n≥−1 by P−1 = I, P0 = A0 Pn = An Pn−1 + Bn Pn−2 n ≥ 1. (1) and Qn = An Qn−1 + Bn Qn−2 Q−1 = 0, Q0 = I The matrix Pn /Qn is called the nth convergent of the continued fraction K (Bn /An ) .

Proposition 1) For any two matrices C and D with C invertible, we have n B1 D B2 C −1 Bk Bk n D = CA0 D; , . , C A0 ; Ak k =1 A2 Ak k =3 A1 C −1 Proposition Bk +∞ 2) Let A0 ; be a given continued fraction. Then Ak k =1 " #n Xk Bk Xk−1 Pn Bk n −2 = A0 ; = A0 ; Qn Ak k =1 Xk Ak Xk−1 −1

,

k =1

where X−1 = X0 = I and X1 , X2 , ..., Xn are arbitrary invertible matrices.

Proposition 1) For any two matrices C and D with C invertible, we have n B1 D B2 C −1 Bk Bk n D = CA0 D; , . , C A0 ; Ak k =1 A2 Ak k =3 A1 C −1 Proposition Bk +∞ 2) Let A0 ; be a given continued fraction. Then Ak k =1 " #n Xk Bk Xk−1 Pn Bk n −2 = A0 ; = A0 ; Qn Ak k =1 Xk Ak Xk−1 −1

,

k =1

where X−1 = X0 = I and X1 , X2 , ..., Xn are arbitrary invertible matrices.

Main result This section is devoted to give a continued fraction expansion of the quasi-arithmetic power mean with parameter (p, α) of two postive matrices A and B given by fp,α (A, B) = A1/2 ((1 − α)I + α(A−1/2 BA−1/2 )p )1/p A1/2 f1,α (A, B) = (1 − α)A + αB. Then, we restrict ourself to the case where p 6= 1 Lemma Let A, B ∈ Mm be a positive definite matrices, α a real number such that 0 ≤ α ≤ 1 and p 6= 1 be a strictly positive integer. Then ((1 − α)I + α(A     

−1/2

BA

−1/2 p 1/p

) )

Bn = I; An

+∞ , n=1

α L B1 = 2 A1/2 A−1/2 , p K

    A1 = −I − α A1/2 L A−1/2 , p K

(2)

and           

B2 =

where

L = A − B(A−1 B)p−1 ;

α2 ( p12

−

1)A1/2

L K

2

A−1/2 , ,

2 L 1 2 2 1/2  Bn = α ( p2 − (n − 1) )A A−1/2 , for all n ≥ 3,   K        An = −(2n − 1)I, for every n ≥ 2. K = (2 − α)A + αB(A−1 B)p−1 .

(3)

Demonstration

+∞ 2Φ(C) (1/p2 − k 2 )(Φ(C))2 ((1−α)I+α(A BA ) ) = I; , −pI − Φ(C) −(2k + 1)I k= (4) I−C . where C = (1 − α)I + α(A−1/2 BA−1/2 )p ) and Φ(C) = I+C But −1/2

−1/2 p 1/p

Φ(C) = = =

α(I−(A−1/2 BA−1/2 )p ) (2−α)I+(A−1/2 BA−1/2 )p αA−1/2 (A−B(A−1 B)p−1 )A−1/2 A−1/2 ((2−α)A+B(A−1 B)p−1 )A−1/2 αA1/2 KL A−1/2 ,

where L = A − B(A−1 B)p−1 and K = (2 − α)A + B(A−1 B)p−1 .

Theorem Let A, B ∈ Mm be two positive definite matrices, α a real number such that 0 ≤ α ≤ 1 and p be a strictly positive integer. Then a continued fraction expansion of fα,p (A, B) is given by : Bn +∞ , fp,α (A, B) = A; An n=1 where we set     

B1 = 2

αL , pK

    A1 = −A−1 − α L A−1 , pK

(5)

and             Bn =         

B2 =

α2 (1/p2

α2 (1/p2

− (n −

2

− 1)

L K

2

1)2 )

An = −(2n − 1)I,

L K

,

A−1 ,

for all n ≥ 3,

for every n ≥ 2.

where L = A − B(A−1 B)p−1 and K = (2 − α)A + αB(A−1 B)p−1 .

(6)

Remark : Let A, B ∈ Mm be two positive definite matrices, α a real number such that 0 ≤ α ≤ 1 and p 6= 1 be a strictly positive integer. 1) If p = 2 we have f2,α (A, B) = A1/2 (A−1/2 ((1 − α)A + αBA−1 B)A−1/2 )1/2 A1/2 . Accordingly f2,α (A, B) = A1/2 (A−1/2 C1 A−1/2 )1/2 A1/2 Where C1 = (1 − α)A + αBA−1 B. It follows that f2,α (A, B) = g2 (A, C1 ) is the geometric matrix mean of the matrices A and C1 . Hence f2,α (A, B) is the unique solution of the matrix equation XA−1 X = (1 − α)A + αBA−1 B.

2) If p = 3 we have f3,α (A, B) = A1/2 (A−1/2 ((1 − α)A + αB(A−1 B)2 )A−1/2 )1/2 A1/2 . Accordingly f3,α (A, B) = A1/2 (A−1/2 C2 A−1/2 )1/2 A1/2 = g3 (A, C2 ). Where C2 = (1 − α)A + αB(A−1 B)2 By virtue of Theorem, we conclude that f3,α (A, B) is the unique solution of XA−1 XA−1 X = (1 − α)A + αB(A−1 B)2 . (7) Such that X is a positive matrix. 3) If A and B are two positive matrices such that AB = BA, then 1

fp,α (A, B) = ((1 − α)Ap + αB p ) p . Therefore fp,α (A, A) = A

Example 1. Let us consider the function q

f (x) = f2,1/2 (x, 1) = 12 x 2 + 12 , Using the Theorem in real case, the first convergents of f (x) are given by :

f1 (x) =

f2 (x) =

f3 (x) =

(5x 2 + 3)x 7x 2 + 1

(29x 4 + 30x 2 + 5)x 41x 4 + 22x 2 + 1

(169x 6 + 245x 4 + 91x 2 + 7)x . 232x 6 + 227x 4 + 45x 2 + 1

Iteration 1 :

Iteration 2 :

Iteration 3 :

Example 2. Now, let us Ccnsider the matrices A and B such that 

   3 1 1 4 1 1 A =  1 3 1  and B =  1 4 1  1 1 3 1 1 4 The solution of the equation XA−1 X = 12 (A + BA−1 B) is X √ √  =1 √ √ 1 61 2 + 6√ 3 √13√2 √  1 61 2 − 1 13 2 6√ 6√ √ √ 1 1 61 2 − 6 6 13 2

√

√

1 6 √61√2 1 6 √61√2 1 6 61 2

√ √ − 16 √13√2 + 13 √13√2 − 16 13 2

√

√

1 6 √61√2 1 6 √61√2 1 6 61 2

√ √  − 61 √13√2 − 61 √13√2  , + 31 13 2

and thus

X = 3.54056667400999991 0.991056918009999954 0.991056918029  0.991056918009999954 3.54056667400999991 0.991056918029 0.991056917999999953 0.991056917999999953 3.540566673999

According to remark 3.4, it is clair that X = f2,1/2 (A, B) Using Theorem 3.2, we can obtain the following approximations of this solution

F 1 = 3.53413603176636304 0.993595491225823247 0.993595491225  0.993595491225822913 3.53413603176636393 0.993595491225 0.993595491225823024 0.993595491225823135 3.534136031766

F 2 = 3.54047817549171473 0.991099496616634212 0.991099496616  0.991099496616633657 3.54047817549171562 0.991099496616 0.991099496616633768 0.991099496616633879 3.540478175491

F 3 = 3.54056539643065093 0.991057551961279093 0.991057551961  0.991057551961278760 3.54056539643065138 0.991057551961 0.991057551961278871 0.991057551961278871 3.540565396430

F 4 = 3.54056665545622806 0.991056926558535056 0.991056926558  0.991056926558534501 3.54056665545622895 0.991056926558 0.991056926558534834 0.991056926558534945 3.540566655456

F 5 = 3.54056667379078149 0.991056917401399673 0.991056917401  0.991056917401399451 3.54056667379078149 0.991056917401 0.991056917401399562 0.991056917401399562 3.540566673790 We can clearly see that F5 is approximately the exact value of the solution X . This example poves and justify the impotance of our approach.

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