Non-commutative extrapolation algorithms 1 Introduction - CiteSeerX
Non-commutative extrapolation algorithms A. Salam Laboratoire d'Analyse Numerique et d'Optimisation, Universite des Sciences et Technologies de Lille, UFR IEEA, B^at M3, 59655 Villeneuve d'Ascq Cedex, France
Abstract This paper is made up of two general ideas. The rst is to extend the theory of general linear extrapolation methods to a non-commutative eld (or even a non-commutative unitary ring). The second one, by exploiting these new results, is to solve an old conjecture about Wynn's vector "-algorithm. Thus, thanks to the use of designants and Cli ord algebras, we show how the vectors "(kn) can be written as a ratio of two designants. This result allows us to nd, as a particular case, some well-known results and some other which are news.
Subject classi cations: AMS(MOS) 65B05 Keywords: Designant, Cli ord algebra, extrapolation, vector "-algorithm.
1 Introduction Until now, the extrapolation methods used and their applications (solution of systems of linear and nonliear equations, approximation of the limit or anti-limit of a sequence, acceleration of the convergence of a sequence,: : :, see [3]) have IR or Cl as the basic elds. The multiplicative law of IR or Cl is commutative. The study of many of these methods is based on the theory of determinants. These determinants are used for building up recursive algorithms corresponding to these extrapolation methods (E-algorithm [4], [18], "-algorithm [28],: : :). In this paper, we shall try to answer to the following question: Is it possible to extend these extrapolation methods when the eld is non-commutative ?
1
There are many diculties to overcome and, among them, the following : the determinants do not exist when the eld (or the unitary ring) is not commutative, i.e. : let IK denote a non-commutative eld (or ring), IMn (IK) the set of matrices whose coecients are in IK. By a determinant we mean an application, denoted by det
det : IMn (IK) ?! IK A ?! det(A) and satisfying fhe following conditions 1) det is multilinear with respect to the rows of A 2) det(A) = 0 () A singular 3) det(A B ) = det(A):det(B ): Dyson's theorem ([11],[21]) states that if there exists an application det satisfying 1),2),3) then the multiplicative law on IK is necessarily commutative. In [10], Dieudonne showed the important role played by the commutativity of the multiplicative law in the theory of determinants. Let IK be a eld , IK = IK ? f0g; C : the group of the commutators of IK: Dieudonne de ned an application det
det : IMn(IK) ?! (IK=C ) [ f0g satisfaying the conditions 1),2),3) and coinciding with the usual determinant when IK is commutative. In the non-commutative case, the value of det is not in IK , but in (IK =C ) [ f0g i.e. det is an equivalence class. For this reason, Dieudonne's de nition is not adapted to our purpose. In [23], Ore de ned some "determinant" of a system of linear equations whose coecients are in IK (non-commutative). But the disadvantage of this de nition is that there is no way of computing recursively these "determinants", which makes this de nition unuseful in practice. Many other de nitions were proposed by various authors (Dyson [11], Mehta [21],Artin [2]), but they always present some disadvantage for our need. However there is an old de nition which seems particularly adapted to our purpose. It is the notion of designants. They were proposed by Heyting in 1927 [19].
2
2 Designants We shall now give brie y the de nition and some properties of designants. For more informations, see [19], [26]. Let IK be a non-commutative eld. Consider the system of homogeneous linear equations in the two unknowns x1 ; x2 2 IK, with coecients on the right
8 > <xa > :xa
1 11
+ x2a12 = 0
1 21
+ x2a22 = 0
aij 2 IK i; j = 1; 2:
(1)
Suppose that a11 6= 0; then, by eliminiting the unknown x1 in the second equation of the system, we get : x2(a22 ? a12a?111a21) = 0: Set a a 11 12 r = a a = a22 ? a12 a?111 a21: 21
r
22
r is called the right designant of the system (1). The sux r indicates that the designant in question is a designant of the right system (1). It indicates also the direction of the computation. In the same way, consider the system of homogeneous linear equations in the two unknowns
x1; x2, with coecients on the left 8 > < a11x1 + a12x2 = 0
a 2 IK i; j = 1; 2: > : a x + a x = 0 ij 21 1
(2)
22 2
Suppose that a11 6= 0; then, by eliminiting the unknown x1 in the second equation of the system, we get : (a22 ? a21 a?111 a12)x2 = 0: Set a a 11 12 l = a a = (a22 ? a21 a?111 a12): 21
22
l
l is called the left designant of the system (2). The sux l indicates that the designant in question is a designant of the left system (2). It indicates also the direction of the computation. If r 6= 0 (resp. l 6= 0 ) the system (1) (resp. the system (2)) has only the trivial solution. If r = 0 (resp. l = 0 ) the system (1) (resp. the system (2)) has more than one solution.
3
In the general case, we proceed in the same way : consider the system of homogeneous linear equations in the n unknowns x1; x2; :::; xn, with coecients on the right :
8 > x a + x a + +xn a n = 0 > < x a + x a + +xn a n = 0 .. .. .. ... > . . . > : x an + x an + +xn ann =0 1 11
2 12
1
1 21
2 22
2
1
2
1
(3)
2
where aij 2 IK i; j = 1; ; n: By eliminating x1 from the (n ? 1) last equations, then x2 from the (n ? 2) last equations, and so on, we obtain xn r = 0. r is called the right designant of the system (3) and it is denoted by
a1n
a r = ... . . . an
.. : .
11
ann r
1
r has a meaning only if its principal minor
a a ;n? . . .. .. .. . an? ; an? ;n?
a a ;n? . . .. .. .. . an? ; an? ;n?
11
1
11
1
1
1
r
is di erent from 0. At its turn, this r.designant has a meaning only if its principal minor 11
1
21
2
1
2
r
is di erent from 0, and so on. Finally r has a meaning only if
a ; aa 11
11 21
a ; :::; a r
a11 .. .
12
22
an?1;1
a ;n? .. ... . an? ;n? r 1
1
1
1
are all di erent from 0: Similarly, consider the system of homogeneous linear equations in the n unknowns x1; x2; :::; xn, with coecients on the left : 8 > a11x1+ a12x2+ +a1nxn = 0 > < a21x1+ a22x2+ +a2nxn = 0 (4) .. .. .. ... > . . . > : an1x1+ an2x2+ +ann xn = 0
4
where aij 2 IK i; j = 1; ; n: By eliminating x1 from the (n ? 1) last equations, then x2 from the (n ? 2) last equations, and so on ,we obtain l xn = 0. l is called the left designant of the system (4) and it is denoted by
: l
a a . . . n l = .. . . .. an ann 11
1
1
l has a meaning only if
a ; aa 11
11 21
a ; :::; a l
a11 .. .
12
22
1
an?1;1
are all di erent from 0:
a ;n? .. ... . an? ;n? l 1
1
1
Now, let us enumerate brie y some fundamental properties of designants. For more details, see [19], [26] Let a a .11 . 1. n r = .. . . .. ;
an ann r 1
Apq : the r.designant of order (n ? 1) obtained from r by keeping the rows 1; 2; :::; n ? 2; p and the columns 1; 2; :::n ? 2; q; Ap: the r.designant of order p obtained from r by keeping the rows 1; 2; :::; p and the columns 1; 2; :::; p; Apqr the r.designant of orderp p + 1 obtained from r by keeping the rows 1; 2; :::; q and the columns 1; 2; :::; p; r. Thus Ap+1;p+1 = Ap+1 :
Property 1
a r = ... . . . an 11
1
p A = p .. ;p p . r An;p
a1n .. .
ann
App
+1 +1
...
;n
+1
.. .
Apn;n
+1
: r
Taking p = n ? 2, we obtain an identity analogous to Sylvester's identity [1] for designants
a a n r = ... . . . ... an ann 11
1
1
Ann?? ;n? Ann?? ;n = An? Ann;n? n;n? r 2 1
2
1
1
2 1 2
: r
Let D be the determinant of the system (3) where the eld IK is commutative. r the r.designant of the same system ( here r = l). We have
5
Property 2
a a . . . n D = .. . . .. an ann 11
1
1
= r :An? :::A :a : 1
2
11
We shall now propose an identity veri ed by the designants, which is the analogous of Schweins' identity for determinants
Property 2
a . .. ak
h1 a11
a k
11
...
ak
;
+1 1
.. .
1
.. .
hk+1
;k
+1
a a k . . . . . . . . . ak; ak;k 12
1
2
a . .. ak
a k
12
...
ak
;
+1 2
1
.. .
;k
+1
a1 .. .
r ak
.. .
a1 ?1
a k ...
ak
;
+1 1
.. .
1
.. .
ak+1
;k
+1
h1 a12 a1k a1 ?1 . . . . . .. ..
hk
r
..
..
ak;2 ak;k
h1 a11 .. .
ak+1 hk+1
ak r
Proof : see [26].
.. .
;
+1 1
..
ak r
a k ...
ak
1
.. .
;k
+1
r
?
=
a1 ?1 .. .
ak+1
:
r
To end this section, we give the
Property 3 a . . .. . . an 11
1
a1n
.. = 0 () the vectors column are linearly dependents on the left, .
ann r
which means that 91 ; ::; n 2 IK not all zero such that: 1c1 + ::: + ncn = 0 where ci is the ith column. In the same way, we have :
a a n . . . . . . . . . an ann 11
1
1
= 0 () the vectors rows are linearly dependents on the right l
6
which means that 91 ; ::; n 2 IK not all zero such that:
l11 + ::: + lnn = 0 where li is the ith row.
Proof: see [26]. Before going to the next section, let us just say that since designants do not satisfy the condition 1) above, they have properties completely di erent from those of determinants, but also some similarities with them (see [19], [26]).
3 Extrapolation in a non-commutative eld In this section, where IK is non-commutative, we will show that the designants play an important role for the non-commutative extension of linear extrapolation methods, for the noncommutative extension of the Shanks transformation, a role as important as the role of determinants when IK is commutative. These designants are the basis on which the non-commutative extension of E-algorithm and "-algorithm are built.
3.1 A general extrapolation method
In [4], Brezinski has proposed a general extrapolation method in IR orCl . The aim of this section is to see if it is possible to built an analogous method in a non-commutative eld (or even a non-commutative unitary ring). The main diculty arises from the fact that, in a non-commutative eld , the determinants do not exist. This diculty will be overcomed thanks to the use of designants. Let (Sn ) be a sequence of elements of IK. Assume that it satis es
Sn = a1g1(n) + ::: + ak gk (n) + S
(5)
where the sequences gi (n) are given sequences.
S is computed by solving the right linear system 8 > Sn = a1g1(n)+ +ak gk (n) + S > < Sn+1 = a1g1(n + 1)+ +ak gk (n + 1) + S
.. .. .. ... > . . . > : Sn k = a g (n + k)+ +ak gk (n + k) + S: +
1 1
7
(6)
We shall always assume in this paper that such a system is nonsingular, which is equivalent to the assumption that the r.designant of the system is di erent from 0. If the sequence (Sn ) has not the exact form (5), then the value of S obtained by solving the preceding system will depend on the indexes n and k: We shall denote it by r Ek (Sn ) and from (6), using the r.designants, it is easy to see that ?1 g1(n) gk (n) Sn g1(n) gk (n) 1 r E (S ) = g1 (n.+ 1) gk (n.+ 1) Sn.+1 g1(n.+ 1) gk (n.+ 1) 1. : (7) k n ... ... .. .. .. .. .. .. g (n + k) g (n + k) S g (n + k) g (n + k) 1 k 1 k n+k r 1 r In general S is an approximate value of the limit of the sequence (Sn ) if it converges or an approximate value of the antilimit when it diverges. Similarly, let (Sn ) be a sequence of elements of IK. Assume that Sn = g1(n)a1 + ::: + gk (n)ak + S where the sequences gi (n) are given sequences.
S is computed by solving the left linear system 8 > Sn = g1(n)a1+ +gk (n)ak + S > < Sn+1 = g1(n + 1)a1+ +gk (n + 1)ak + S
.. .. .. ... > . . . > : Sn k = g (n + k)a + +gk (n + k)ak + S: +
1
(8)
(9)
1
We shall always assume in this paper that such a system is nonsingular, which is equivalent to the assumption that the l.designant of the system is di erent from 0. If the sequence (Sn ) has not the exact form (8), then the value of S obtained by solving the preceding system will depend on the indexes n and k: We shall denote it by l Ek (Sn ) and from(9), using the r.designants, it is easy to see that ?1 g1(n) gk (n) 1 g1(n) gk (n) Sn l E (S ) = g1(n + 1) gk (n + 1) 1 g1 (n + 1) gk (n + 1) Sn+1 : (10) .. .. .. .. .. k n .. .. ... g (n + k) . g (n.+ k) 1. g (n.+ k) . g (n.+ k) S . 1 k n+k l 1 k l
3.2 A Recursive Algorithm
The main algorithmic problem is now to compute recursively the quantities r Ek (Sn ) without computing the r.designants appearing in (7).
8
In this section we shall present a non-commutative extension of the E-algorithm [4; 18]. This extension will allow us to compute recursively the r Ek (Sn )'s. This algorithm, which will be called r E -algorithm, is the following r E (n) = Sn 0 r g (n) = gi(n) 0;i
n = 0; 1; ::: i = 1; 2; ::: and n = 1; 2; :::
For k = 1; 2; ::: and n = 0; 1; ::: r E (n) = fr E (n+1) (r g (n+1))?1 ?r E (n) (r g (n) )?1 g f(r g (n+1) )?1 ? (r g (n) )?1 g?1 k?1;k k?1;k k?1;k k?1 k?1;k k?1 k r g (n) = fr g (n+1) (r g (n+1))?1 ?r g (n) (r g (n) )?1 g f(r g (n+1) )?1 ? (r g (n) )?1 g?1 k?1;k k?1;k k?1;k k?1;i k?1;k k?1;i k;i
i = k + 1; k + 2; :::
Theorem 1
r E (n) =r E (S ): k n k
Proof: we will simultaneously prove by induction that and
r E (n) =r E (S ) k n k
g (n) . r g n = .. k;i g (n + k) 1
( )
1
gi(n) g1(n)
gk (n)
.. .
.. .. . . gk (n + k) gi(n + k)
... r g (n + k)
.. .
1
gk (n)
1 .. .. . . gk (n + k) 1
? : r 1
It is easy to verify that the equality is true for k = 0. Assume that it is still true for k ? 1. Set
g (n) . .. A = g (n + k) 1
1
We have
g (n + 1) .. . A = g (n + k ? 1) g (n) g (n + k) 1
1
1
1
By applying Sylvester's identity, we obtain
.. .
gk (n)
Sn
.. .. . . gk (n + k) Sn+k
gk (n + 1)
: r
Sn+1
.. .. . . gk (n + k ? 1) Sn+k?1 : gk (n) Sn gk (n + k) Sn+k r .. .
AJ ?1 = B ? C
9
with
g (n + 1) . .. B = g (n + k) 1
1
g (n) .. C = . g (n + k ? 1) 1
1
and
.. .
gk? (n + 1) Sn g (n + 1) .. .. .. .. .. . . . . . gk? (n + k) Sn k r g (n + k) gk? (n) Sn g (n) .. .. .. . . . gk? (n + k ? 1) Sn k? r g (n + k ? 1) g (n + 1) g (n + 1) k . .. .. .. J = : . . g (n + k) gk (n + k) r 1
+1
1
1
+
1
1
1
1
+
1
1
gk (n + 1) ?1
.. . gk (n + k)
.. .
; r
? .. . gk (n + k ? 1) gk (n)
1
r
1
1
From the induction assumption, it is easy to see that
?1
?1
) and C =r Ekn?1 :(r gk(n?)1;k ) : B =r Ekn?+11 :(r gk(n?+1) 1;k Set
g (n) . .. A0 = g (n + k)
gk (n)
1 .. .. .. . . . g ( n + k ) 1 1 k In the same way, by applying Sylvester's identity, we obtain 1
?1
: r
?1
? (r gk(n?1;k ) : A0 = (r gk(n?+1 1;k ) Since r Ek (Sn ) = AA0 ?1 = AJ ?1 JA0 ?1 = AJ ?1 (A0J ?1 )?1 , the result follows. The same proof is (n) : valid for the r gk;i In the same way , the l E -algorithm allow us to compute the quantities l Ek (Sn ) . It is the following l E (n) = Sn 0 l g (n) = gi(n) 0;i
n = 0; 1; ::: i = 1; 2; ::: and n = 1; 2; :::
For k = 1; 2; ::: and n = 0; 1; ::: l E (n) = f(lg (n+1))?1 ? (lg (n) )?1 g?1 f(lg (n+1) )?1 l E (n+1) ? (lg (n) )?1 l E (n) g k?1 k?1;k k?1 k?1;k k?1;k k?1;k k r g (n) = f(lg (n+1) )?1 ? (l g (n) )?1 g?1 f(lg (n+1) )?1 l g (n+1) ? (l g (n) )?1 (l g (n) )?1 g k?1;k k?1;k k?1;i k?1;k k?1;k k?1;k k;i
i = k + 1; k + 2; :::
10
Theorem 2
l E (n) =l E (S ) k n k
Proof: similar to that of theorem 1.
Remarks 1) The r E and l E -algorithms coincide with the E -algorithm when IK is commutative. 2)It is easy to see that the r E -algorithm can be written as follows r E n = Sn
n = 0; 1; :::
r g (n) = gi (n) 0;i
i = 1; 2; ::: n = 0; 1; :::
0
For k = 1; 2; ::: n = 0; 1; :::
r E n =r E n ? (r E n ) (r g (n) )?1 r g (n) k k?1 k?1 k?1;k k?1;k r g n =r g n ? (r g n ) (r g (n) )?1 r g (n) k?1;k k?1;k k;i k?1;i k?1;i
i = k + 1; ::: where the operator acts on the upper indexes n. 3) Similarly l E n = Sn
n = 0; 1; :::
l g (n) = gi(n) 0;i
i = 1; 2; ::: n = 0; 1; :::
0
Fork = 1; 2; ::: n = 0; 1; :::
l E n =l E n ?l g (n) (lg (n) )?1 (l E n ) k?1 k k?1 k?1;k k?1;k l g n =l g n ?l g (n) (l g (n) )?1 (l g n ) k?1;i k;i k?1;i k?1;k k?1;k
i = k + 1; ::: In [26], we proved the non-commutative extension of the results given by Brezinski in [4] and Havie in [18]. The most important ones are
11
Theorem 3 If
Sn = S + a1g1(n) + a2g2(n) + ::: 8 n
then
r E n = S + ar g n r n k k+1 k;k+1 + ak+2 gk;k+2 + :::
Theorem 4 If
Sn = S + g1(n)a1 + g2(n)a2 + ::: 8 n
then
l E n = S +r g n a +r g n a + ::: k k;k+1 k+1 k;k+2 k+2
Theorem 5 There exist coecients Ajk;n such that k k k;n n X r n X (
Ek =
with
k X j =0
8n; k:
j =0
Sn+j A(j
8n; k:
)
)
( ) = and gk;i
j =0
gi(n + j )A(jk;n)
A(jk;n) = 1:
4 The Vector case
Let IK be a non-commutative eld. IKp can be considered as a vector space on the left by de ning the external law as follows IK IKp ?! IKp
0 x 1 0 x 1 0 0x 11 B@ ; B@ ... CA CA ?! : B@ ... CA = B@ ... CA : [13] 1
1
1
xp
xp
xp
Let (IKp ) be the dual of IKp . It is a vector space on the right (see [13]). Consider now the sequence (Sn ) of elements of IKp such that
Sn = S + a1g1(n) + ::: + ak gk (n)
(11)
where S; g1(n); :::; gk(n) 2 IKp ; a1 ; :::; ak 2 IK: For nding S in terms of gi(n); i = 1; :::; k, we must solve the system
8 > > < > > :
a1g1(n)+ a1g1(n)+
.. . a1g1(n + k ? 1)+
.. .
+ak gk (n)+ S = Sn +ak gk (n) = Sn .. .. . . +ak gk (n + k ? 1) = Sn+k?1 :
12
(12)
Let y be an element of (IKp ) and x an element of IKp . Denote by (x; y ) the value y (x) 2 IK. Applying y to all the equations of the system (12), except the rst one , we obtain the system 8 > a1(g1(n); y)+ +ak (gk (n); y ) = (Sn ; y ) > .. .. .. .. < . . . . (13) > a1(g1(n + k ? 1); y)+ +ak (gk (n + k ? 1); y) = (Sn+k?1 ; y ) > : a1g1(n)+ +ak gk (n)+ S = Sn :
De nition 1 Let uij ; i = 1; :::; k ? 1 and j = 1; :::; k be elements of IK and let vkj j = 1; :::; k be elements of IKp. u k u .. . . .. .. uk? ; uk?. ;k denotes the vector obtained by applying Sylvester's identity and it is v vk;k r k; called a vector r.designant. u k u ... . . . ... In the same way, denotes the vector obtained by applying Sylvester's uk? ; uk? ;k vk; vk;k l 11
1
11
1
1
11
1
11
1
1
identity and it is called a vector l.designant.
Remarks
1) This de nition is a generalization of the vector determinants [7] to the non-commutative case. 2)All the recursive scalar relations remain valid for the vector case. 3)This de nition has no meaning if the vectors vki ; i = 1; :::; k are in a row di erent from the last one. Now, if the vector sequence (Sn ) has not the exact form (11), then the value of S will depend on the indexes n and k, and we shall denote it by r Ek (Sn ):
Lemma 1
(g (n); y ) (S.n; y) .. .. r Ek (Sn ) = (Sn k? ; y) (g (n +.k ? 1); y) 1 0 (g (n); y ) (Sn; y) .. . . .. .. (Sn k? ; y) (g (n + k ? 1); y) . S g (n) 1
+
1
1
1
+
n
1
1
1
13
(gk (n); y ) .. .. . . (gk (n + k ? 1); y) 0 (gk (n); y ) .. . (gk (n + k ? 1); y ) r gk (n)
? 1
r
Proof: see [26] As in the scalar case, the question is now to built an algorithm which computes recursively the quantities r Ek (Sn ): (n) computed as follows Consider r Ek(n) and r gk;i
For k = 1; 2; ::: n = 0; 1; :::
rg n ;i
( ) 0
r E (n) = S
n = 0; 1; ::: = gi (n) i = 1; 2; ::: n = 0; 1; ::: 0
n
r E (n) =r E (n) ? (r E (n) ; y )(rg (n) ; y )?1(r g (n) ) k?1;k k?1;k k?1 k?1 k r g (n) =r g (n) ? (r g (n) ; y )(rg (n) ; y )?1(r g (n) ): k?1;k k?1;k k?1;i k?1;i k;i
For these quantities, we have the
Theorem 6
r E (S ) =r E (n) k n k
Proof: see [26], pp. 121: We also have the Theorem 7 If Sn = S + a1g1(n) + a2g2(n) + ::: then r E (n) = S + ak+1 r g (n) + ak+2 r g (n) + ::: 8k; n k;k+2 k;k+1 k Proof : see [26] For the left extension, we obtain similar results
Lemma 2
(g (n); y ) (Sn; y) . .. .. l E (S ) = k n (Sn k? ; y) (g (n +.k ? 1); y) Sn g (n) (g (n); y ) (Sn; y) .. .. .. (Sn .k? ; y) (g (n +.k ? 1); y) . 1 0 +
1
1
1
1
1
+
1
1
Theorem 8
l Ek (Sn ) =l E (n) k
Theorem 9 If Sn = S + g (n)a + g (n)a + ::: then l E n = S +l g n ak +l g n ak + ::: 8k; n k k;k k;k ( )
( ) +1
1
+1
1 ( ) +2
2
2
+2
14
.. .
(gk (n); y ) .. . (gk (n + k ? 1); y ) gk (n)
? (gk (n); y ) .. . (gk (n + k ? 1); y ) 0
1
l
l
5 Shanks's transformation We shall now give a non-commutative extension of Shanks transformation [27]. Let (Sn ) be a sequence of elements of a non-commutative eld IK satisfying
(
a0Sn+ a1Sn+1+ +ak Sn+k = S a0+ a1+ +ak = 1:
(14)
In order to express S in term of the elements of the sequence, we solve the system
8 a+ a+ > > > a S + a S n n + < a Sn+ a Sn + > .. .. > . > : a Sn k? + a S.n k + 1
+1
1
+
0
1
0
1
0
0
+
1
+1
.. .
+ak +ak Sn+k ?S +ak Sn+k .. . +ak Sn+2k?1
=1 =0 =0
(15)
= 0:
If the sequence (Sn ) has not the exact form (14), then the value of S will depend on the indexes n and k. We shall denote it by r "2k (Sn ). Assuming that the r.desighnant of the system is di erent of zero, we have
1 1 1 1 Sn Sn k 0 Sn r " (S ) = Sn Sn Sn k 0 k n . .. .. .. ... .. . . . Sn k? Sn k Sn k? 0 r 1 1 0 ? 1 Sn Sn k ?1 Sn Sn Sn Sn k 0 : .. .. ... ... ... . . S 0 r n k? Sn k Sn k? +1
+
+1
2
+
1
+
+
+2
1
1
+1
+
+1
+
The transformation
1
+
+
+2
1
rT
: (Sn ) ?! r "2k (Sn ) is called the right Shanks transformation. In the same way, we can built the left Shanks transformation l T . Let (Sn ) be a sequence of elements of a non-commutative eld IK satisfying
(
Sn a0+ Sn+1 a1+ +Sn+k ak = S a0+ a1+ +ak = 1:
15
(16)
In order to express S in term of the elements of the sequence, we solve the system
8 a+ a+ > > > a + S S n n a+ < Sn a + Sn a + > .. .. > . > : Sn k? a + Sn .k a + 0
+1 1
1 0
+
0
1
0
+
+1 1
.. .
1
+ak +Sn+k ak ?S +Sn+k ak .. . +Sn+2k?1 ak
=1 =0 =0
(17)
= 0:
If the sequence (Sn ) has not the exact form (16), then the value of S will depend on the indexes n and k. We shall denote it by l "2k (Sn ). Assuming that the l.desighnant of the system is di erent of zero, we have
1 1 S S n n l " (S ) = Sn S n k n . . . . . . Sn k? Sn 1 1 S S n n Sn Sn .. ... . Sn k? Sn k 2
+
+
1
The transformation
1
1
0 S ? 1 +1 n+k Sn+k 0 +1 .. .. ... . . Sn+2k?1 0 +k 1 1 Sn+k 0 +1 Sn+k 0 : +1 .. .. ... . . Sn+2k?1 0 l +
? l 1
lT
: (Sn ) ?! l "2k (Sn ) is called the left Shanks's transformation. Now the question is to built an algorithm wich computes recursively the quantities r "2k (Sn ), l "2k (Sn ) without computing explicitly the r.designants and the l.designants appearing in the expressions of these quantities.
6 The "?algorithm
In [28], Wynn gave a recursive algorithm, the "?algorithm, which allows the recursive implementation of Shanks transformation (with IK = IR ou Cl ). We will now show that this algorithm is still valid for the computation of r "2k (Sn ) and
l "2k (Sn ). We set r "2k+1 (Sn ) = (r "2k (Sn))?1
and l "2k+1 (Sn ) = (l "2k (Sn ))?1:
16
Let "(kn) be the quantities computed by
"(?n1) = 0 n = 0; 1; ::: "(0n) = Sn n = 0; 1; :::
for k = 0; 1; ::: n = 0; 1; :::
+ ("(kn+1) ? "(kn) )?1 : "(kn+1) = "(kn?+1) 1
The main result is
Theorem 10
"(kn) =r "k (Sn) =l "k (Sn) 8 k; n 2 IN:
Proof: for k = 0, we have ?1 r "0 (Sn ) = 1 1 1 0 = (0 ? 1 1 Sn )(?1 ? 0 1 Sn )?1 = Sn = "(n) . 0 Sn 0 r Sn ?1 r Similarly r "1 (Sn ) = (r "0 (Sn ))?1 = (Sn )?1 = "(1n) : Now, we will show that "2k+2 (Sn ) ? "2k (Sn+1 ) = ("2k+1 (Sn+1 ) ? "2k+1 (Sn ))?1 : In fact, "2k+2(Sn) ? "2k (Sn+1) = 1 1
1 1 0 ?1 1 1 Sn Sn+1 Sn+k+1 ?1 0 Sn Sn SSnn+1+1 SSnn++k+1 Sn+1 Sn+k+1 0 ? k+1 0 Sn . . .. .. .. .. . . . . .. .. .. .. .. .. . . . . Sn+k Sn+k+1 Sn+2k+1 0 r Sn+k Sn+k+1 Sn+2k+1 0 r 1 1 0 ?1 1 1 1 1 1 Sn+1 Sn+2 Sn+k+1 0 Sn+1 Sn+2 Sn+k+1 ?1 Sn+1 Sn+2 Sn+k+1 0 Sn+1 Sn+2 Sn+k+1 0 .. .. .. .. .. .. .. .. .. ... . . . . . . . . . S n+k Sn+k+1 Sn+2k 0 r Sn+k Sn+k+1 Sn+2k 0 r Applying the property 3 of section 2 (Schweins' identity) and taking the inverse, we have ("2k+2 (Sn ) ? "2k (Sn+1 ))?1 =
1 Sn Sn ... Sn k +
1
1 Sn k Sn k
+ +1
.. .
+ +1
Sn
.. .
k
+2 +1
0 ?1 0 .. . 0
1 Sn Sn ... r Sn k
+1 +1
+ +1
17
1 Sn k Sn k
+ +1
.. .
+ +1
Sn
.. .
k
+2 +1
0 ?1 0 .. . 0
1 0 0 .. . 0
? : r 1
(18)
Since, by de nition, "2k+1 (Sn ) = ("2k (Sn ))?1 , we get "2k+11(Sn+1)? "2k+1(1Sn) = 0 1 1 1 ?1 2Sn+1 2Sn+k+1 ?1 2Sn+1 2Sn+k+1 0 Sn+1 Sn+k+1 0 Sn+1 Sn+k+1 0 ? .. .. .. .. .. .. .. ... . . . . . . . 2S 2 2 2 r S S 0 S 0 n+2k r n+k n1+k n+21k 1 1 1 ?1 0 2Sn 2Sn+k ?1 2Sn 2Sn+k 0 Sn Sn+k 0 Sn Sn+k 0 : .. .. .. .. .. .. .. .. . . . . . . . 2S . 2 2 2 n+k?1 Sn+2k?1 0 r Sn+k?1 Sn+2k?1 0 r By factorizing the second factor of the rst term of the second member of the equality and then applying Sylvester's identity, we obtain "2k+1(Sn+1) ? "2k+1(Sn) =
Sn Sn k Sn Sn k .. .. ... . . S n k Sn k 2
2
2
+
2
+
+
+2
?1 S1 0 Sn .. . n . .. 0 r S n 2
2
1 Sn k Sn k
1 0 +1 + +1 2 0 +1 + +1 .. .. .. . . . 2 Sn+2k 0 +k
? : r 1
(19)
Carrying out some simple manipulations of designants and comparing (18) to (19) we obtain the result. In the same way, we show that
"2k+3 (Sn) ? "2k+1(Sn+1) = ("2k+2(Sn+1 ) ? "2k+2(Sn))?1 : In fact, "2k+3(Sn) ? "2k+1 (Sn+1) = 1 1 Sn+1 2Sn Sn 2Sn+1 .. .. ... . . 2S 2 n+k Sn+k+1 1 1 S S 2 n+1 n+2 2 Sn+1 Sn+2 .. .. ... . . 2 2 Sn+k Sn+k+1
0 1 Sn+k+1 ?1 Sn 2 Sn+k+1 0 2 Sn .. .. .. . . . 2 Sn+2k+1 0 r 2Sn+k 1 0 1 Sn+k+1 ?1 Sn+1 2 Sn+k+1 0 2Sn+1 .. .. .. . . . 2 Sn+2k 0 r 2 Sn+k 1
18
1
Sn+1 2 Sn+1 .. . 2 Sn+k+1 1 Sn+2 2Sn+2 .. . 2 Sn+k+1
1 Sn k Sn k 2
.. .
+ +1 + +1
.. .
1 0 0 .. . 0
? ? r 1
Sn k 1 1 ? Sn k 0 Sn k 0 : .. .. .. . . . Sn k 0 r 2
2
2
+2 +1
+ +1
+ +1
+2
1
Applying the property 3 of section 2 (Schweins' identity) and taking the inverse, we have ("2k+3 (Sn ) ? "2k+1 (Sn+1 ))?1 =
1 Sn Sn ... Sn k
1 Sn k Sn k
2
2
2
.. .
+ +1 + +1
.. .
Sn
+
2
k
+2 +1
1 0 0 .. . 0
1 Sn Sn .. . S n k r 2
2
1 Sn k Sn k
+1 +1
2
.. .
+ +1
.. .
Sn
+ +1
2
1 0 0 .. . 0
+ +1
k
+2 +1
0 ?1 0 .. . 0
? : r 1
(20)
Then
" k1 (Sn ) ? " k 1 (Sn) =1 1 Sn k 0 Sn Sn Sn k 0 Sn Sn .. .. .. .. ... . . . . S S S 0 n k n k r 1n k 1 1 1 S S 0 S n n k n Sn Sn k 0 Sn .. .. .. .. .. ... . . . . . Sn k Sn k 0 r Sn k 2 +2
+1
2 +2
+1
+1
+ +2
+1
+ +2
+1
+ +1
+2 +2
+ +1
1 Sn k Sn k
0 ?1 0 .. . 0
+ +2 + +2
.. .
Sn
.. .
k
+2 +2
? ? r 1
0 ?1 Sn+k+1 ?1 + +1 Sn+k+1 0 : + +1 .. .. . . Sn+2k+1 0 r + +2 +1 + By factorizing the second factor of the rst term of the second member of the equality and then applying Sylvester's identity, we obtain "2k+2(Sn+1) ? "2k+2(Sn) =
1 S n Sn ... S n k Sn Sn k +1
+1
+
+ +1
1 Sn k Sn k
+ +2
.. .
+ +2
.. .
Sn Sn Sn
k k +2k +2
+2 +1 + +1
0 ?1 0 .. . 0 0 0
1 0 0 .. . 0 0 0
1
1 Sn Sn ... Sn k
+1 +1
+ +1
1 Sn k Sn k
+ +2
.. .
+ +2
Sn
.. .
k
+2 +2
r
0 ?1 0 .. . 0
? : r 1
(21)
Carrying out some simple manipulations of designants and comparing (20) to (21) we obtain the result. Since "k+1 (Sn ) = "k?1 (Sn+1 ) + ("k (Sn+1 ) ? ("k (Sn ))?1 8k; n and "0 (Sn ) = "(0n) ; "1 (Sn ) = "(1n) , we get "(kn) = "k (Sn ) 8n; k (see [26], pp. 132-141 for a more detailed proof.)
Remarks
1) All these constructions remain valid if IK is a ring, but the assumption
19
H1 : element di erent from zero must be replaced everywhere by H2 : element invertible. 2) By simple manipulations of r.designants and l.designants, we can write
Sn Sn k? Sn Sn Sn k Sn Sn r " k (Sn ) = S.n .. .. .. .. .. . . . . S n k Sn k? Sn k r Sn k +
+1
+
+1
+1
+
+
2
+
+2
1
+
.. .
This expression is similar to the one given in [3] with determinants.
Sn Sn k? Sn k l " k (Sn ) = S.n .. .. .. . . S n k Sn k? +
+1
1
+
2
+
+2
1
1 1 .. . 1
1 ?1 1 + .. .. : . . Sn+2k?1 1 r In the same way, we get
Sn k? Sn k
1
1
? Sn Sn k? Sn Sn Sn k Sn : .. .. .. ... . . . S n k Sn k? Sn k l l 1
+
+1
+
1
+
+2
+1
1
+
3)From these expressions, we see that the quantities r "2k (Sn ) (resp l "2k (Sn )) can be computed by the r E ?algorithm (resp the l E ?algorithm) with the choice gi (n) = Sn+i?1 : In conclusion, designants appear to be a natural generalization of determinants and allow us 1)to extend extrapolation methods to more general sets (non-commutative eld, non-commutative unitary associative ring ). 2)in all the methods exposed above, taking IR or Cl or a commutative eld instead of IK, we obtain the same results that for the methods using determinants.
7 Applications In this section, we shall show how the use of the general theory of extrapolation, designants and Cli ord algebras allows us to answer an old open question on the vector "?algorithm. Let us rst give some properties of Cli ord algebras.
7.1 Real Cli ord Algebra
Let V be a real linear space, q its quatratic form. There exists ([9]) an algebra (C (V ); +; ; :) called Cli ord algebra, which satis es the following fundamental property: there exists a linear transformation ' : V ?! C (V ) such that 8x 2 V ('(x))2 = q (x):
20
From now on, we take V = IRd and q (x) = (x; x) where (.,.) denotes the usual scalar product. Let e1 ; :::; ed be an orthogonal basis of V and set Ei = '(ei ); i = 1; :::; d: The Cli ord algebra associated to V is generated by d elements Ei; i = 1; :::; d [2], [25] which satisfy the anti-commutation relations
Ei Ej + Ej Ei = 2ij E0;
(22)
E0 being the identity element, ij the Kronecker symbol and 1 i; j d. The real linear space spanned by the products Ej1 Ej2 ::: Ejr ; 0 j1 < j2 < ::: < jr d i.e. E0; E1; E2; :::; Ed; E1 E2; :::; Ed?1 Ed ; :::; E1 E2 ::: Ed forms the associative but noncommutative (d > 1) algebra C (V ). It is easy to see that C (V ) is of dimension 2d. However, it is not a division algebra (d 1) as is proved by the relation (E0 + E1 )(E0 ? E1) = 0: There are various matrix representations of Ei e.g. [20], [17]. However, all of them involve square matrices whose order rises rapidly with the dimension d typicaly this order is at least proportional to 2d=2: P P Furthermore, we may identify each x = di=1 xi ei 2 V , with X = di=1 xi Ei 2 C (V ) and each 2 IR with E0 2 C (V ) (see [2]). According to this identi cation, one can consider V and IR as subsets of C (V ). From the equation (22), we can easily establish
8X 2 V; 8Y 2 V X Y + Y X = 2(X:Y )
(23)
which admits the following two particular rules if X ? Y then
X Y = ?Y X;
(24)
X X = (X:X ): This last relation allows to conclude that, if X = 6 0, then X : X ?1 = (X:X )
(25)
if we take X = Y , then
(26)
Let us notice that this inverse is the same as the Moore-Penrose pseudo-inverse (see [24]).
7.2 The vector "-algorithm
The vector "-algorithm is a quite powerful method for accelerating the convergence of vector sequences. It has many other applications : solution of systems of linear equations [5], [12], systems of nonlinear equations [8] and the computation of eigenvalues of a matrix [6]. It occurs likewise in approximation theory (vector continued fractions [29], [30], vector interpolation and Pade approximants [14],[15], [16], [17]). Although it was the subject of intensive
21
research, it still lacks of theoretical results. In this section, we will show that the vector "-algorithm, like other extrapolation algorithms, realizes some extrapolation in C (V ), solves some system of linear equations in C (V ) and can be written as a "ratio" of two designants. We saw above that (C (V ); +; ) is an associative unitary ring. It is not commutative when d > 1. We also saw that V C (V ); IR C (V ). For each nonzero element X = Pdi=1 xiEi of P V, we saw that its inverse is X ?1 = jjXXjj2 , where jjX jj = ( di=1 x2i )1=2: The vector "-algorithm of Wynn can be written as follows
"(?n1) = 0 "(0n) = Sn 2 V
n = 0; 1; ::: n = 0; 1; ::: ?1
+ ("(kn+1) ? "(kn) ) "(kn+1) = "k(n?+1) 1
k = 0; 1; :::; n = 0; 1; ::::
The inverse considered here is the above-mentioned inverse. The "(kn) 's belong always to V, this is because the sum and the di erence are closed operations in V and the inverse of a nonzero element of V also belongs to V. Owing to the fact that (C (V ); +; ; :) is an algebra, the elements of C (V ) can be regarded as scalar elements of the ring (C (V ); +; ), and as vector elements of the vector space (C (V ); +; :): Let (Sn ) be a sequence of vectors of V (scalar elements of the ring (C (V ); +; ) such that
(
a0+ a1 + +ak =1 a0 Sn+ a1Sn+1+ +ak Sn+k = S
(27)
where S 2 V ; a0 ; a1; :::; ak 2 C (V ): By applying the theorem 10 for IK = (C (V ); +; ), we have the fundamental result
Theorem 11 The quantities "kn computed by the vector "-algorithm can be expressed as a ( )
"ratio" of two designants as follows
S S Sn n k? . n . n . .. . . . " k = . . . . Sn k Sn k? Sn k +
1
( ) 2
+
+2
1
+
"(2nk)+1 = ("2k (Sn))?1: We also have
22
S S n k? . n . .. . . . . . Sn k Sn k? r +
+
+2
1
1
1 .. . 1
?
1
r
Theorem 12 "nk = S () 9A ; :::; Ak 2 C (V ) such that 1
2
Sn = S + A1Sn + ::: + Ak Sn+k?1 :
Proof: It is a direct consequence of theorem 11. In fact, S S n+k?1 Sn . n . (n) .. .. .. "2k = S () .. . . Sn+k Sn+2k?1 Sn+k
S . n = S .. r Sn k +
Sn
1 .. .. ... . . Sn+2k?1 1 k?1
+
: r
Thanks to properties of designants showed in [26], [19], we have
S S n n k? .. .. .. = .. . ... .. . . . . Sn k? Sn k r Sn k Sn k? S S Sn ? S n n k? = 0: .. .. ... " nk = S () ... . . Sn k Sn k? Sn k ? S r
S . n n " k = S () .. Sn k ( ) 2
+
Sn
k?1
+
+2
Sn
1
+
+
+
+
+2
1
1
S
.. ; .
S r
1
( ) 2
+
+2
1
+
By applying the property 4 of section 2, the result follows.
Remark
The fundamental algebraic result of McLeod ([22]) is a particular case of the theorem 12 by imposing that the Ai 's belong to IR ( C (V )). From theorem 11, we can also deduce some other interesting results and obtain some other well known results as particular cases.
Theorem 13 Let "kn be the vectors obtained by applying the "-algorithm to the vector sequence ( )
(Sn ) of elements of V . Let A; B be two invertible elements of C (V ), C an element of V . Let "e(kn) be the vectors obtained by applying the "-algorithm to the vector sequence (A Sn B + C ) then
"e(2nk) = A "(2nk) B + C "e(2nk)+1 = B?1 "(2nk)+1 A?1: Proof: see [26], pp. 178.
Theorem 14 Let M be a square orthogonal matrix of dimension d d (M T M = I ). Let "ekn
( )
be the vector obtained by applying the vector "-algorithm to the vector sequence MSn + C where C is a vector of V and MSn the product of the matrix M by the vector Sn : then
23
"e(2nk) = M:"2(nk) + C "e(2nk)+1 = M:"(2nk)+1 : Proof: Let u be an endomorphism0 of V 1which matrix relatively to the basis e1 ; :::; ed is M .
x1
P Let x = di xi Ei ; X = B @ ... C A: =1
xd u(x) can be written in matrix form as M:X .
To show that the theorem 14 is a particular case of the theorem 13, let us use the following well known result (see [2]): for each orthogonal isomorphism u : V ?! V , there exists an invertible element B 2 C (V ) such that 8x 2 V; u(x) = "B x B ?1 where " = 1 or ? 1: So "2k (M:Sn + C ) can be written as "2k ("B Sn B ?1 + C ): According to theorem 13 we get
"2k ("B Sn B?1 + C ) = "B "2k (Sn) B?1 + C; which can be written in a matrix form as M:"2k (Sn ) + C: In the same way "2k+1 (M:Sn + C ) can be written as
"2k+1 ("B Sn B?1 + C ): Using theorem 13, it follows that
"2k+1("B Sn B?1 + C ) = (B?1 )?1 "2k+1(Sn) ("B)?1 : Now, since "?1 = "; (B ?1 )?1 = B ), we obtain
"2k+1("B Sn B?1 + C ) = ("B "2k+1(Sn) B?1 : In a matrix form, this last equatity can be expressed as M:"2k+1 (Sn ): We nally have "e(2nk) = M"(2nk) + C and "e(2nk)+1 = M"(2nk)+1 :
Acknowledgement
I am grateful to the Professor Claude Brezinski for guiding me during this work, for helpful discussions, valuable suggestions and encouragements.
References [1] A.C. Aitken, Determinants and Matrices, Oliver and Boyd, Edinburgh, 1965. [2] A.Artin, Geometric Algebra, Interscience, NewYork, 1966.
24
[3] C.Brezinski, M.Redivo Zaglia, Extrapolation Methods. Theory and Practice, North-Holland, Amsterdam, 1991. [4] C.Brezinski, A general extrapolation algorithm, Numer. Math., 35(1980)175-187. [5] C.Brezinski, Some results in the theory of the vector "-algorithm., Linear Alg. Appl., 8(1974)77-86. [6] C.Brezinski, Computation of the eigenelements of a matrix by the "-algorithm, Linear Alg. Appl., 11(1975)7-20. [7] C.Brezinski, Some determinantal identities in a vector space, with applications. In Pade Approximation and its Applications. H.Werner and H.J.B}unger,eds. LNM 1071 SpringerVerlag, Berlin,1984,pp.1-11. [8] C.Brezinski, Application de l'"-algorithme a la resolution des systemes non lineaires, C.R.Acad.Sci.Paris, 271A(1970)1174-1177. [9] R.Deheuvels, Formes Quadratiques et Groupes Classiques, Presses Universitaires de France, Paris, 1981. [10] J.Dieudonne, Les determinants sur un corps non commutatif, Bull.Soc.Math. France, 7(1943)27-45. [11] F.J.Dyson, Quaternion determinants, Helv.Phys.Acta, 45(1972)289-302. [12] E.Gekeler, On the solution of systems of equations by the epsilon algorithm of Wynn, Math.Comput., 26(1972)427-436. [13] R.Godement, Cours d'Algebre, Hermann, Paris,1966. [14] P.R.Graves-Morris, Vector-valued rational interpolants I, Numer. Math., 42(1983)331-348. [15] P.R.Graves-Morris, Vector-valued rational interpolants II, I.M.A.J.Num.Anal, 4(1984)209224. [16] P.R.Graves-Morris, C.D. Jenkins, Vector-valued rational interpolants III, Constr.Approx., 2(1986)263-289. [17] P.R.Graves-Morris, D.E. Roberts, From matrix to vector Pade approximants, J.Comp.Appl.Math., to appear. [18] T.Havie, Generalized Neville type extrapolation schemes, BIT, 19(1979)204-213. [19] A.Heyting, Die theorie der linearen Gleichungen in einer Zahlenspezies mit nichtkommutativer Multiplikation, Math.Ann., 98(1927)465-490. [20] G.N. Hile, P.Lounesto, Matrix representations of Cli ord algebras, Linear Alg. Appl.,128(1990)51-63.
25
[21] M.L.Mehta, Matrix Theory. Selected Topics and Useful Results, Les editions de Physique, Les Ulis, 1989. [22] J.B.McLeod, A note on the "-algorithm, Computing, 7(1971)17-24. [23] O.Ore, Linear equations in non-commutative elds, Ann.Math., 32(1931)463-477. [24] R.Penrose, A generalised inverse for matrices, Proc. Cambridge Phil. Soc., 51(1955),406413. [25] IR.Porteous, Topological Geometry, 2nd ed, Cambridge University Press, Cambridge, 1981. [26] A.Salam, Extrapolation: Extension et Nouveaux Resultats, Thesis, Universite des Sciences et Technologies de Lille,1993. [27] D.Shanks, Non linear transformations of divergent and slowly convergent sequences, J.Math.phys., 34(1955)1-42. [28] P.Wynn, On a device for computing the em (Sn ) transformation, MTAC, 10(1956)91-96. [29] P.Wynn, Vector continued fractions, Linear Alg. Appl., 1(1968)357-395. [30] P.Wynn, Continued fractions whose coecients obey a non-commutative law of multiplication., Arch.Rational Mech.Anal., 12(1963)273-312.
26
Abstract This paper is made up of two general ideas. The rst is to extend the theory of general linear extrapolation methods to a non-commutative eld (or even a non-commutative unitary ring). The second one, by exploiting these new results, is to solve an old conjecture about Wynn's vector "-algorithm. Thus, thanks to the use of designants and Cli ord algebras, we show how the vectors "(kn) can be written as a ratio of two designants. This result allows us to nd, as a particular case, some well-known results and some other which are news.
Subject classi cations: AMS(MOS) 65B05 Keywords: Designant, Cli ord algebra, extrapolation, vector "-algorithm.
1 Introduction Until now, the extrapolation methods used and their applications (solution of systems of linear and nonliear equations, approximation of the limit or anti-limit of a sequence, acceleration of the convergence of a sequence,: : :, see [3]) have IR or Cl as the basic elds. The multiplicative law of IR or Cl is commutative. The study of many of these methods is based on the theory of determinants. These determinants are used for building up recursive algorithms corresponding to these extrapolation methods (E-algorithm [4], [18], "-algorithm [28],: : :). In this paper, we shall try to answer to the following question: Is it possible to extend these extrapolation methods when the eld is non-commutative ?
1
There are many diculties to overcome and, among them, the following : the determinants do not exist when the eld (or the unitary ring) is not commutative, i.e. : let IK denote a non-commutative eld (or ring), IMn (IK) the set of matrices whose coecients are in IK. By a determinant we mean an application, denoted by det
det : IMn (IK) ?! IK A ?! det(A) and satisfying fhe following conditions 1) det is multilinear with respect to the rows of A 2) det(A) = 0 () A singular 3) det(A B ) = det(A):det(B ): Dyson's theorem ([11],[21]) states that if there exists an application det satisfying 1),2),3) then the multiplicative law on IK is necessarily commutative. In [10], Dieudonne showed the important role played by the commutativity of the multiplicative law in the theory of determinants. Let IK be a eld , IK = IK ? f0g; C : the group of the commutators of IK: Dieudonne de ned an application det
det : IMn(IK) ?! (IK=C ) [ f0g satisfaying the conditions 1),2),3) and coinciding with the usual determinant when IK is commutative. In the non-commutative case, the value of det is not in IK , but in (IK =C ) [ f0g i.e. det is an equivalence class. For this reason, Dieudonne's de nition is not adapted to our purpose. In [23], Ore de ned some "determinant" of a system of linear equations whose coecients are in IK (non-commutative). But the disadvantage of this de nition is that there is no way of computing recursively these "determinants", which makes this de nition unuseful in practice. Many other de nitions were proposed by various authors (Dyson [11], Mehta [21],Artin [2]), but they always present some disadvantage for our need. However there is an old de nition which seems particularly adapted to our purpose. It is the notion of designants. They were proposed by Heyting in 1927 [19].
2
2 Designants We shall now give brie y the de nition and some properties of designants. For more informations, see [19], [26]. Let IK be a non-commutative eld. Consider the system of homogeneous linear equations in the two unknowns x1 ; x2 2 IK, with coecients on the right
8 > <xa > :xa
1 11
+ x2a12 = 0
1 21
+ x2a22 = 0
aij 2 IK i; j = 1; 2:
(1)
Suppose that a11 6= 0; then, by eliminiting the unknown x1 in the second equation of the system, we get : x2(a22 ? a12a?111a21) = 0: Set a a 11 12 r = a a = a22 ? a12 a?111 a21: 21
r
22
r is called the right designant of the system (1). The sux r indicates that the designant in question is a designant of the right system (1). It indicates also the direction of the computation. In the same way, consider the system of homogeneous linear equations in the two unknowns
x1; x2, with coecients on the left 8 > < a11x1 + a12x2 = 0
a 2 IK i; j = 1; 2: > : a x + a x = 0 ij 21 1
(2)
22 2
Suppose that a11 6= 0; then, by eliminiting the unknown x1 in the second equation of the system, we get : (a22 ? a21 a?111 a12)x2 = 0: Set a a 11 12 l = a a = (a22 ? a21 a?111 a12): 21
22
l
l is called the left designant of the system (2). The sux l indicates that the designant in question is a designant of the left system (2). It indicates also the direction of the computation. If r 6= 0 (resp. l 6= 0 ) the system (1) (resp. the system (2)) has only the trivial solution. If r = 0 (resp. l = 0 ) the system (1) (resp. the system (2)) has more than one solution.
3
In the general case, we proceed in the same way : consider the system of homogeneous linear equations in the n unknowns x1; x2; :::; xn, with coecients on the right :
8 > x a + x a + +xn a n = 0 > < x a + x a + +xn a n = 0 .. .. .. ... > . . . > : x an + x an + +xn ann =0 1 11
2 12
1
1 21
2 22
2
1
2
1
(3)
2
where aij 2 IK i; j = 1; ; n: By eliminating x1 from the (n ? 1) last equations, then x2 from the (n ? 2) last equations, and so on, we obtain xn r = 0. r is called the right designant of the system (3) and it is denoted by
a1n
a r = ... . . . an
.. : .
11
ann r
1
r has a meaning only if its principal minor
a a ;n? . . .. .. .. . an? ; an? ;n?
a a ;n? . . .. .. .. . an? ; an? ;n?
11
1
11
1
1
1
r
is di erent from 0. At its turn, this r.designant has a meaning only if its principal minor 11
1
21
2
1
2
r
is di erent from 0, and so on. Finally r has a meaning only if
a ; aa 11
11 21
a ; :::; a r
a11 .. .
12
22
an?1;1
a ;n? .. ... . an? ;n? r 1
1
1
1
are all di erent from 0: Similarly, consider the system of homogeneous linear equations in the n unknowns x1; x2; :::; xn, with coecients on the left : 8 > a11x1+ a12x2+ +a1nxn = 0 > < a21x1+ a22x2+ +a2nxn = 0 (4) .. .. .. ... > . . . > : an1x1+ an2x2+ +ann xn = 0
4
where aij 2 IK i; j = 1; ; n: By eliminating x1 from the (n ? 1) last equations, then x2 from the (n ? 2) last equations, and so on ,we obtain l xn = 0. l is called the left designant of the system (4) and it is denoted by
: l
a a . . . n l = .. . . .. an ann 11
1
1
l has a meaning only if
a ; aa 11
11 21
a ; :::; a l
a11 .. .
12
22
1
an?1;1
are all di erent from 0:
a ;n? .. ... . an? ;n? l 1
1
1
Now, let us enumerate brie y some fundamental properties of designants. For more details, see [19], [26] Let a a .11 . 1. n r = .. . . .. ;
an ann r 1
Apq : the r.designant of order (n ? 1) obtained from r by keeping the rows 1; 2; :::; n ? 2; p and the columns 1; 2; :::n ? 2; q; Ap: the r.designant of order p obtained from r by keeping the rows 1; 2; :::; p and the columns 1; 2; :::; p; Apqr the r.designant of orderp p + 1 obtained from r by keeping the rows 1; 2; :::; q and the columns 1; 2; :::; p; r. Thus Ap+1;p+1 = Ap+1 :
Property 1
a r = ... . . . an 11
1
p A = p .. ;p p . r An;p
a1n .. .
ann
App
+1 +1
...
;n
+1
.. .
Apn;n
+1
: r
Taking p = n ? 2, we obtain an identity analogous to Sylvester's identity [1] for designants
a a n r = ... . . . ... an ann 11
1
1
Ann?? ;n? Ann?? ;n = An? Ann;n? n;n? r 2 1
2
1
1
2 1 2
: r
Let D be the determinant of the system (3) where the eld IK is commutative. r the r.designant of the same system ( here r = l). We have
5
Property 2
a a . . . n D = .. . . .. an ann 11
1
1
= r :An? :::A :a : 1
2
11
We shall now propose an identity veri ed by the designants, which is the analogous of Schweins' identity for determinants
Property 2
a . .. ak
h1 a11
a k
11
...
ak
;
+1 1
.. .
1
.. .
hk+1
;k
+1
a a k . . . . . . . . . ak; ak;k 12
1
2
a . .. ak
a k
12
...
ak
;
+1 2
1
.. .
;k
+1
a1 .. .
r ak
.. .
a1 ?1
a k ...
ak
;
+1 1
.. .
1
.. .
ak+1
;k
+1
h1 a12 a1k a1 ?1 . . . . . .. ..
hk
r
..
..
ak;2 ak;k
h1 a11 .. .
ak+1 hk+1
ak r
Proof : see [26].
.. .
;
+1 1
..
ak r
a k ...
ak
1
.. .
;k
+1
r
?
=
a1 ?1 .. .
ak+1
:
r
To end this section, we give the
Property 3 a . . .. . . an 11
1
a1n
.. = 0 () the vectors column are linearly dependents on the left, .
ann r
which means that 91 ; ::; n 2 IK not all zero such that: 1c1 + ::: + ncn = 0 where ci is the ith column. In the same way, we have :
a a n . . . . . . . . . an ann 11
1
1
= 0 () the vectors rows are linearly dependents on the right l
6
which means that 91 ; ::; n 2 IK not all zero such that:
l11 + ::: + lnn = 0 where li is the ith row.
Proof: see [26]. Before going to the next section, let us just say that since designants do not satisfy the condition 1) above, they have properties completely di erent from those of determinants, but also some similarities with them (see [19], [26]).
3 Extrapolation in a non-commutative eld In this section, where IK is non-commutative, we will show that the designants play an important role for the non-commutative extension of linear extrapolation methods, for the noncommutative extension of the Shanks transformation, a role as important as the role of determinants when IK is commutative. These designants are the basis on which the non-commutative extension of E-algorithm and "-algorithm are built.
3.1 A general extrapolation method
In [4], Brezinski has proposed a general extrapolation method in IR orCl . The aim of this section is to see if it is possible to built an analogous method in a non-commutative eld (or even a non-commutative unitary ring). The main diculty arises from the fact that, in a non-commutative eld , the determinants do not exist. This diculty will be overcomed thanks to the use of designants. Let (Sn ) be a sequence of elements of IK. Assume that it satis es
Sn = a1g1(n) + ::: + ak gk (n) + S
(5)
where the sequences gi (n) are given sequences.
S is computed by solving the right linear system 8 > Sn = a1g1(n)+ +ak gk (n) + S > < Sn+1 = a1g1(n + 1)+ +ak gk (n + 1) + S
.. .. .. ... > . . . > : Sn k = a g (n + k)+ +ak gk (n + k) + S: +
1 1
7
(6)
We shall always assume in this paper that such a system is nonsingular, which is equivalent to the assumption that the r.designant of the system is di erent from 0. If the sequence (Sn ) has not the exact form (5), then the value of S obtained by solving the preceding system will depend on the indexes n and k: We shall denote it by r Ek (Sn ) and from (6), using the r.designants, it is easy to see that ?1 g1(n) gk (n) Sn g1(n) gk (n) 1 r E (S ) = g1 (n.+ 1) gk (n.+ 1) Sn.+1 g1(n.+ 1) gk (n.+ 1) 1. : (7) k n ... ... .. .. .. .. .. .. g (n + k) g (n + k) S g (n + k) g (n + k) 1 k 1 k n+k r 1 r In general S is an approximate value of the limit of the sequence (Sn ) if it converges or an approximate value of the antilimit when it diverges. Similarly, let (Sn ) be a sequence of elements of IK. Assume that Sn = g1(n)a1 + ::: + gk (n)ak + S where the sequences gi (n) are given sequences.
S is computed by solving the left linear system 8 > Sn = g1(n)a1+ +gk (n)ak + S > < Sn+1 = g1(n + 1)a1+ +gk (n + 1)ak + S
.. .. .. ... > . . . > : Sn k = g (n + k)a + +gk (n + k)ak + S: +
1
(8)
(9)
1
We shall always assume in this paper that such a system is nonsingular, which is equivalent to the assumption that the l.designant of the system is di erent from 0. If the sequence (Sn ) has not the exact form (8), then the value of S obtained by solving the preceding system will depend on the indexes n and k: We shall denote it by l Ek (Sn ) and from(9), using the r.designants, it is easy to see that ?1 g1(n) gk (n) 1 g1(n) gk (n) Sn l E (S ) = g1(n + 1) gk (n + 1) 1 g1 (n + 1) gk (n + 1) Sn+1 : (10) .. .. .. .. .. k n .. .. ... g (n + k) . g (n.+ k) 1. g (n.+ k) . g (n.+ k) S . 1 k n+k l 1 k l
3.2 A Recursive Algorithm
The main algorithmic problem is now to compute recursively the quantities r Ek (Sn ) without computing the r.designants appearing in (7).
8
In this section we shall present a non-commutative extension of the E-algorithm [4; 18]. This extension will allow us to compute recursively the r Ek (Sn )'s. This algorithm, which will be called r E -algorithm, is the following r E (n) = Sn 0 r g (n) = gi(n) 0;i
n = 0; 1; ::: i = 1; 2; ::: and n = 1; 2; :::
For k = 1; 2; ::: and n = 0; 1; ::: r E (n) = fr E (n+1) (r g (n+1))?1 ?r E (n) (r g (n) )?1 g f(r g (n+1) )?1 ? (r g (n) )?1 g?1 k?1;k k?1;k k?1;k k?1 k?1;k k?1 k r g (n) = fr g (n+1) (r g (n+1))?1 ?r g (n) (r g (n) )?1 g f(r g (n+1) )?1 ? (r g (n) )?1 g?1 k?1;k k?1;k k?1;k k?1;i k?1;k k?1;i k;i
i = k + 1; k + 2; :::
Theorem 1
r E (n) =r E (S ): k n k
Proof: we will simultaneously prove by induction that and
r E (n) =r E (S ) k n k
g (n) . r g n = .. k;i g (n + k) 1
( )
1
gi(n) g1(n)
gk (n)
.. .
.. .. . . gk (n + k) gi(n + k)
... r g (n + k)
.. .
1
gk (n)
1 .. .. . . gk (n + k) 1
? : r 1
It is easy to verify that the equality is true for k = 0. Assume that it is still true for k ? 1. Set
g (n) . .. A = g (n + k) 1
1
We have
g (n + 1) .. . A = g (n + k ? 1) g (n) g (n + k) 1
1
1
1
By applying Sylvester's identity, we obtain
.. .
gk (n)
Sn
.. .. . . gk (n + k) Sn+k
gk (n + 1)
: r
Sn+1
.. .. . . gk (n + k ? 1) Sn+k?1 : gk (n) Sn gk (n + k) Sn+k r .. .
AJ ?1 = B ? C
9
with
g (n + 1) . .. B = g (n + k) 1
1
g (n) .. C = . g (n + k ? 1) 1
1
and
.. .
gk? (n + 1) Sn g (n + 1) .. .. .. .. .. . . . . . gk? (n + k) Sn k r g (n + k) gk? (n) Sn g (n) .. .. .. . . . gk? (n + k ? 1) Sn k? r g (n + k ? 1) g (n + 1) g (n + 1) k . .. .. .. J = : . . g (n + k) gk (n + k) r 1
+1
1
1
+
1
1
1
1
+
1
1
gk (n + 1) ?1
.. . gk (n + k)
.. .
; r
? .. . gk (n + k ? 1) gk (n)
1
r
1
1
From the induction assumption, it is easy to see that
?1
?1
) and C =r Ekn?1 :(r gk(n?)1;k ) : B =r Ekn?+11 :(r gk(n?+1) 1;k Set
g (n) . .. A0 = g (n + k)
gk (n)
1 .. .. .. . . . g ( n + k ) 1 1 k In the same way, by applying Sylvester's identity, we obtain 1
?1
: r
?1
? (r gk(n?1;k ) : A0 = (r gk(n?+1 1;k ) Since r Ek (Sn ) = AA0 ?1 = AJ ?1 JA0 ?1 = AJ ?1 (A0J ?1 )?1 , the result follows. The same proof is (n) : valid for the r gk;i In the same way , the l E -algorithm allow us to compute the quantities l Ek (Sn ) . It is the following l E (n) = Sn 0 l g (n) = gi(n) 0;i
n = 0; 1; ::: i = 1; 2; ::: and n = 1; 2; :::
For k = 1; 2; ::: and n = 0; 1; ::: l E (n) = f(lg (n+1))?1 ? (lg (n) )?1 g?1 f(lg (n+1) )?1 l E (n+1) ? (lg (n) )?1 l E (n) g k?1 k?1;k k?1 k?1;k k?1;k k?1;k k r g (n) = f(lg (n+1) )?1 ? (l g (n) )?1 g?1 f(lg (n+1) )?1 l g (n+1) ? (l g (n) )?1 (l g (n) )?1 g k?1;k k?1;k k?1;i k?1;k k?1;k k?1;k k;i
i = k + 1; k + 2; :::
10
Theorem 2
l E (n) =l E (S ) k n k
Proof: similar to that of theorem 1.
Remarks 1) The r E and l E -algorithms coincide with the E -algorithm when IK is commutative. 2)It is easy to see that the r E -algorithm can be written as follows r E n = Sn
n = 0; 1; :::
r g (n) = gi (n) 0;i
i = 1; 2; ::: n = 0; 1; :::
0
For k = 1; 2; ::: n = 0; 1; :::
r E n =r E n ? (r E n ) (r g (n) )?1 r g (n) k k?1 k?1 k?1;k k?1;k r g n =r g n ? (r g n ) (r g (n) )?1 r g (n) k?1;k k?1;k k;i k?1;i k?1;i
i = k + 1; ::: where the operator acts on the upper indexes n. 3) Similarly l E n = Sn
n = 0; 1; :::
l g (n) = gi(n) 0;i
i = 1; 2; ::: n = 0; 1; :::
0
Fork = 1; 2; ::: n = 0; 1; :::
l E n =l E n ?l g (n) (lg (n) )?1 (l E n ) k?1 k k?1 k?1;k k?1;k l g n =l g n ?l g (n) (l g (n) )?1 (l g n ) k?1;i k;i k?1;i k?1;k k?1;k
i = k + 1; ::: In [26], we proved the non-commutative extension of the results given by Brezinski in [4] and Havie in [18]. The most important ones are
11
Theorem 3 If
Sn = S + a1g1(n) + a2g2(n) + ::: 8 n
then
r E n = S + ar g n r n k k+1 k;k+1 + ak+2 gk;k+2 + :::
Theorem 4 If
Sn = S + g1(n)a1 + g2(n)a2 + ::: 8 n
then
l E n = S +r g n a +r g n a + ::: k k;k+1 k+1 k;k+2 k+2
Theorem 5 There exist coecients Ajk;n such that k k k;n n X r n X (
Ek =
with
k X j =0
8n; k:
j =0
Sn+j A(j
8n; k:
)
)
( ) = and gk;i
j =0
gi(n + j )A(jk;n)
A(jk;n) = 1:
4 The Vector case
Let IK be a non-commutative eld. IKp can be considered as a vector space on the left by de ning the external law as follows IK IKp ?! IKp
0 x 1 0 x 1 0 0x 11 B@ ; B@ ... CA CA ?! : B@ ... CA = B@ ... CA : [13] 1
1
1
xp
xp
xp
Let (IKp ) be the dual of IKp . It is a vector space on the right (see [13]). Consider now the sequence (Sn ) of elements of IKp such that
Sn = S + a1g1(n) + ::: + ak gk (n)
(11)
where S; g1(n); :::; gk(n) 2 IKp ; a1 ; :::; ak 2 IK: For nding S in terms of gi(n); i = 1; :::; k, we must solve the system
8 > > < > > :
a1g1(n)+ a1g1(n)+
.. . a1g1(n + k ? 1)+
.. .
+ak gk (n)+ S = Sn +ak gk (n) = Sn .. .. . . +ak gk (n + k ? 1) = Sn+k?1 :
12
(12)
Let y be an element of (IKp ) and x an element of IKp . Denote by (x; y ) the value y (x) 2 IK. Applying y to all the equations of the system (12), except the rst one , we obtain the system 8 > a1(g1(n); y)+ +ak (gk (n); y ) = (Sn ; y ) > .. .. .. .. < . . . . (13) > a1(g1(n + k ? 1); y)+ +ak (gk (n + k ? 1); y) = (Sn+k?1 ; y ) > : a1g1(n)+ +ak gk (n)+ S = Sn :
De nition 1 Let uij ; i = 1; :::; k ? 1 and j = 1; :::; k be elements of IK and let vkj j = 1; :::; k be elements of IKp. u k u .. . . .. .. uk? ; uk?. ;k denotes the vector obtained by applying Sylvester's identity and it is v vk;k r k; called a vector r.designant. u k u ... . . . ... In the same way, denotes the vector obtained by applying Sylvester's uk? ; uk? ;k vk; vk;k l 11
1
11
1
1
11
1
11
1
1
identity and it is called a vector l.designant.
Remarks
1) This de nition is a generalization of the vector determinants [7] to the non-commutative case. 2)All the recursive scalar relations remain valid for the vector case. 3)This de nition has no meaning if the vectors vki ; i = 1; :::; k are in a row di erent from the last one. Now, if the vector sequence (Sn ) has not the exact form (11), then the value of S will depend on the indexes n and k, and we shall denote it by r Ek (Sn ):
Lemma 1
(g (n); y ) (S.n; y) .. .. r Ek (Sn ) = (Sn k? ; y) (g (n +.k ? 1); y) 1 0 (g (n); y ) (Sn; y) .. . . .. .. (Sn k? ; y) (g (n + k ? 1); y) . S g (n) 1
+
1
1
1
+
n
1
1
1
13
(gk (n); y ) .. .. . . (gk (n + k ? 1); y) 0 (gk (n); y ) .. . (gk (n + k ? 1); y ) r gk (n)
? 1
r
Proof: see [26] As in the scalar case, the question is now to built an algorithm which computes recursively the quantities r Ek (Sn ): (n) computed as follows Consider r Ek(n) and r gk;i
For k = 1; 2; ::: n = 0; 1; :::
rg n ;i
( ) 0
r E (n) = S
n = 0; 1; ::: = gi (n) i = 1; 2; ::: n = 0; 1; ::: 0
n
r E (n) =r E (n) ? (r E (n) ; y )(rg (n) ; y )?1(r g (n) ) k?1;k k?1;k k?1 k?1 k r g (n) =r g (n) ? (r g (n) ; y )(rg (n) ; y )?1(r g (n) ): k?1;k k?1;k k?1;i k?1;i k;i
For these quantities, we have the
Theorem 6
r E (S ) =r E (n) k n k
Proof: see [26], pp. 121: We also have the Theorem 7 If Sn = S + a1g1(n) + a2g2(n) + ::: then r E (n) = S + ak+1 r g (n) + ak+2 r g (n) + ::: 8k; n k;k+2 k;k+1 k Proof : see [26] For the left extension, we obtain similar results
Lemma 2
(g (n); y ) (Sn; y) . .. .. l E (S ) = k n (Sn k? ; y) (g (n +.k ? 1); y) Sn g (n) (g (n); y ) (Sn; y) .. .. .. (Sn .k? ; y) (g (n +.k ? 1); y) . 1 0 +
1
1
1
1
1
+
1
1
Theorem 8
l Ek (Sn ) =l E (n) k
Theorem 9 If Sn = S + g (n)a + g (n)a + ::: then l E n = S +l g n ak +l g n ak + ::: 8k; n k k;k k;k ( )
( ) +1
1
+1
1 ( ) +2
2
2
+2
14
.. .
(gk (n); y ) .. . (gk (n + k ? 1); y ) gk (n)
? (gk (n); y ) .. . (gk (n + k ? 1); y ) 0
1
l
l
5 Shanks's transformation We shall now give a non-commutative extension of Shanks transformation [27]. Let (Sn ) be a sequence of elements of a non-commutative eld IK satisfying
(
a0Sn+ a1Sn+1+ +ak Sn+k = S a0+ a1+ +ak = 1:
(14)
In order to express S in term of the elements of the sequence, we solve the system
8 a+ a+ > > > a S + a S n n + < a Sn+ a Sn + > .. .. > . > : a Sn k? + a S.n k + 1
+1
1
+
0
1
0
1
0
0
+
1
+1
.. .
+ak +ak Sn+k ?S +ak Sn+k .. . +ak Sn+2k?1
=1 =0 =0
(15)
= 0:
If the sequence (Sn ) has not the exact form (14), then the value of S will depend on the indexes n and k. We shall denote it by r "2k (Sn ). Assuming that the r.desighnant of the system is di erent of zero, we have
1 1 1 1 Sn Sn k 0 Sn r " (S ) = Sn Sn Sn k 0 k n . .. .. .. ... .. . . . Sn k? Sn k Sn k? 0 r 1 1 0 ? 1 Sn Sn k ?1 Sn Sn Sn Sn k 0 : .. .. ... ... ... . . S 0 r n k? Sn k Sn k? +1
+
+1
2
+
1
+
+
+2
1
1
+1
+
+1
+
The transformation
1
+
+
+2
1
rT
: (Sn ) ?! r "2k (Sn ) is called the right Shanks transformation. In the same way, we can built the left Shanks transformation l T . Let (Sn ) be a sequence of elements of a non-commutative eld IK satisfying
(
Sn a0+ Sn+1 a1+ +Sn+k ak = S a0+ a1+ +ak = 1:
15
(16)
In order to express S in term of the elements of the sequence, we solve the system
8 a+ a+ > > > a + S S n n a+ < Sn a + Sn a + > .. .. > . > : Sn k? a + Sn .k a + 0
+1 1
1 0
+
0
1
0
+
+1 1
.. .
1
+ak +Sn+k ak ?S +Sn+k ak .. . +Sn+2k?1 ak
=1 =0 =0
(17)
= 0:
If the sequence (Sn ) has not the exact form (16), then the value of S will depend on the indexes n and k. We shall denote it by l "2k (Sn ). Assuming that the l.desighnant of the system is di erent of zero, we have
1 1 S S n n l " (S ) = Sn S n k n . . . . . . Sn k? Sn 1 1 S S n n Sn Sn .. ... . Sn k? Sn k 2
+
+
1
The transformation
1
1
0 S ? 1 +1 n+k Sn+k 0 +1 .. .. ... . . Sn+2k?1 0 +k 1 1 Sn+k 0 +1 Sn+k 0 : +1 .. .. ... . . Sn+2k?1 0 l +
? l 1
lT
: (Sn ) ?! l "2k (Sn ) is called the left Shanks's transformation. Now the question is to built an algorithm wich computes recursively the quantities r "2k (Sn ), l "2k (Sn ) without computing explicitly the r.designants and the l.designants appearing in the expressions of these quantities.
6 The "?algorithm
In [28], Wynn gave a recursive algorithm, the "?algorithm, which allows the recursive implementation of Shanks transformation (with IK = IR ou Cl ). We will now show that this algorithm is still valid for the computation of r "2k (Sn ) and
l "2k (Sn ). We set r "2k+1 (Sn ) = (r "2k (Sn))?1
and l "2k+1 (Sn ) = (l "2k (Sn ))?1:
16
Let "(kn) be the quantities computed by
"(?n1) = 0 n = 0; 1; ::: "(0n) = Sn n = 0; 1; :::
for k = 0; 1; ::: n = 0; 1; :::
+ ("(kn+1) ? "(kn) )?1 : "(kn+1) = "(kn?+1) 1
The main result is
Theorem 10
"(kn) =r "k (Sn) =l "k (Sn) 8 k; n 2 IN:
Proof: for k = 0, we have ?1 r "0 (Sn ) = 1 1 1 0 = (0 ? 1 1 Sn )(?1 ? 0 1 Sn )?1 = Sn = "(n) . 0 Sn 0 r Sn ?1 r Similarly r "1 (Sn ) = (r "0 (Sn ))?1 = (Sn )?1 = "(1n) : Now, we will show that "2k+2 (Sn ) ? "2k (Sn+1 ) = ("2k+1 (Sn+1 ) ? "2k+1 (Sn ))?1 : In fact, "2k+2(Sn) ? "2k (Sn+1) = 1 1
1 1 0 ?1 1 1 Sn Sn+1 Sn+k+1 ?1 0 Sn Sn SSnn+1+1 SSnn++k+1 Sn+1 Sn+k+1 0 ? k+1 0 Sn . . .. .. .. .. . . . . .. .. .. .. .. .. . . . . Sn+k Sn+k+1 Sn+2k+1 0 r Sn+k Sn+k+1 Sn+2k+1 0 r 1 1 0 ?1 1 1 1 1 1 Sn+1 Sn+2 Sn+k+1 0 Sn+1 Sn+2 Sn+k+1 ?1 Sn+1 Sn+2 Sn+k+1 0 Sn+1 Sn+2 Sn+k+1 0 .. .. .. .. .. .. .. .. .. ... . . . . . . . . . S n+k Sn+k+1 Sn+2k 0 r Sn+k Sn+k+1 Sn+2k 0 r Applying the property 3 of section 2 (Schweins' identity) and taking the inverse, we have ("2k+2 (Sn ) ? "2k (Sn+1 ))?1 =
1 Sn Sn ... Sn k +
1
1 Sn k Sn k
+ +1
.. .
+ +1
Sn
.. .
k
+2 +1
0 ?1 0 .. . 0
1 Sn Sn ... r Sn k
+1 +1
+ +1
17
1 Sn k Sn k
+ +1
.. .
+ +1
Sn
.. .
k
+2 +1
0 ?1 0 .. . 0
1 0 0 .. . 0
? : r 1
(18)
Since, by de nition, "2k+1 (Sn ) = ("2k (Sn ))?1 , we get "2k+11(Sn+1)? "2k+1(1Sn) = 0 1 1 1 ?1 2Sn+1 2Sn+k+1 ?1 2Sn+1 2Sn+k+1 0 Sn+1 Sn+k+1 0 Sn+1 Sn+k+1 0 ? .. .. .. .. .. .. .. ... . . . . . . . 2S 2 2 2 r S S 0 S 0 n+2k r n+k n1+k n+21k 1 1 1 ?1 0 2Sn 2Sn+k ?1 2Sn 2Sn+k 0 Sn Sn+k 0 Sn Sn+k 0 : .. .. .. .. .. .. .. .. . . . . . . . 2S . 2 2 2 n+k?1 Sn+2k?1 0 r Sn+k?1 Sn+2k?1 0 r By factorizing the second factor of the rst term of the second member of the equality and then applying Sylvester's identity, we obtain "2k+1(Sn+1) ? "2k+1(Sn) =
Sn Sn k Sn Sn k .. .. ... . . S n k Sn k 2
2
2
+
2
+
+
+2
?1 S1 0 Sn .. . n . .. 0 r S n 2
2
1 Sn k Sn k
1 0 +1 + +1 2 0 +1 + +1 .. .. .. . . . 2 Sn+2k 0 +k
? : r 1
(19)
Carrying out some simple manipulations of designants and comparing (18) to (19) we obtain the result. In the same way, we show that
"2k+3 (Sn) ? "2k+1(Sn+1) = ("2k+2(Sn+1 ) ? "2k+2(Sn))?1 : In fact, "2k+3(Sn) ? "2k+1 (Sn+1) = 1 1 Sn+1 2Sn Sn 2Sn+1 .. .. ... . . 2S 2 n+k Sn+k+1 1 1 S S 2 n+1 n+2 2 Sn+1 Sn+2 .. .. ... . . 2 2 Sn+k Sn+k+1
0 1 Sn+k+1 ?1 Sn 2 Sn+k+1 0 2 Sn .. .. .. . . . 2 Sn+2k+1 0 r 2Sn+k 1 0 1 Sn+k+1 ?1 Sn+1 2 Sn+k+1 0 2Sn+1 .. .. .. . . . 2 Sn+2k 0 r 2 Sn+k 1
18
1
Sn+1 2 Sn+1 .. . 2 Sn+k+1 1 Sn+2 2Sn+2 .. . 2 Sn+k+1
1 Sn k Sn k 2
.. .
+ +1 + +1
.. .
1 0 0 .. . 0
? ? r 1
Sn k 1 1 ? Sn k 0 Sn k 0 : .. .. .. . . . Sn k 0 r 2
2
2
+2 +1
+ +1
+ +1
+2
1
Applying the property 3 of section 2 (Schweins' identity) and taking the inverse, we have ("2k+3 (Sn ) ? "2k+1 (Sn+1 ))?1 =
1 Sn Sn ... Sn k
1 Sn k Sn k
2
2
2
.. .
+ +1 + +1
.. .
Sn
+
2
k
+2 +1
1 0 0 .. . 0
1 Sn Sn .. . S n k r 2
2
1 Sn k Sn k
+1 +1
2
.. .
+ +1
.. .
Sn
+ +1
2
1 0 0 .. . 0
+ +1
k
+2 +1
0 ?1 0 .. . 0
? : r 1
(20)
Then
" k1 (Sn ) ? " k 1 (Sn) =1 1 Sn k 0 Sn Sn Sn k 0 Sn Sn .. .. .. .. ... . . . . S S S 0 n k n k r 1n k 1 1 1 S S 0 S n n k n Sn Sn k 0 Sn .. .. .. .. .. ... . . . . . Sn k Sn k 0 r Sn k 2 +2
+1
2 +2
+1
+1
+ +2
+1
+ +2
+1
+ +1
+2 +2
+ +1
1 Sn k Sn k
0 ?1 0 .. . 0
+ +2 + +2
.. .
Sn
.. .
k
+2 +2
? ? r 1
0 ?1 Sn+k+1 ?1 + +1 Sn+k+1 0 : + +1 .. .. . . Sn+2k+1 0 r + +2 +1 + By factorizing the second factor of the rst term of the second member of the equality and then applying Sylvester's identity, we obtain "2k+2(Sn+1) ? "2k+2(Sn) =
1 S n Sn ... S n k Sn Sn k +1
+1
+
+ +1
1 Sn k Sn k
+ +2
.. .
+ +2
.. .
Sn Sn Sn
k k +2k +2
+2 +1 + +1
0 ?1 0 .. . 0 0 0
1 0 0 .. . 0 0 0
1
1 Sn Sn ... Sn k
+1 +1
+ +1
1 Sn k Sn k
+ +2
.. .
+ +2
Sn
.. .
k
+2 +2
r
0 ?1 0 .. . 0
? : r 1
(21)
Carrying out some simple manipulations of designants and comparing (20) to (21) we obtain the result. Since "k+1 (Sn ) = "k?1 (Sn+1 ) + ("k (Sn+1 ) ? ("k (Sn ))?1 8k; n and "0 (Sn ) = "(0n) ; "1 (Sn ) = "(1n) , we get "(kn) = "k (Sn ) 8n; k (see [26], pp. 132-141 for a more detailed proof.)
Remarks
1) All these constructions remain valid if IK is a ring, but the assumption
19
H1 : element di erent from zero must be replaced everywhere by H2 : element invertible. 2) By simple manipulations of r.designants and l.designants, we can write
Sn Sn k? Sn Sn Sn k Sn Sn r " k (Sn ) = S.n .. .. .. .. .. . . . . S n k Sn k? Sn k r Sn k +
+1
+
+1
+1
+
+
2
+
+2
1
+
.. .
This expression is similar to the one given in [3] with determinants.
Sn Sn k? Sn k l " k (Sn ) = S.n .. .. .. . . S n k Sn k? +
+1
1
+
2
+
+2
1
1 1 .. . 1
1 ?1 1 + .. .. : . . Sn+2k?1 1 r In the same way, we get
Sn k? Sn k
1
1
? Sn Sn k? Sn Sn Sn k Sn : .. .. .. ... . . . S n k Sn k? Sn k l l 1
+
+1
+
1
+
+2
+1
1
+
3)From these expressions, we see that the quantities r "2k (Sn ) (resp l "2k (Sn )) can be computed by the r E ?algorithm (resp the l E ?algorithm) with the choice gi (n) = Sn+i?1 : In conclusion, designants appear to be a natural generalization of determinants and allow us 1)to extend extrapolation methods to more general sets (non-commutative eld, non-commutative unitary associative ring ). 2)in all the methods exposed above, taking IR or Cl or a commutative eld instead of IK, we obtain the same results that for the methods using determinants.
7 Applications In this section, we shall show how the use of the general theory of extrapolation, designants and Cli ord algebras allows us to answer an old open question on the vector "?algorithm. Let us rst give some properties of Cli ord algebras.
7.1 Real Cli ord Algebra
Let V be a real linear space, q its quatratic form. There exists ([9]) an algebra (C (V ); +; ; :) called Cli ord algebra, which satis es the following fundamental property: there exists a linear transformation ' : V ?! C (V ) such that 8x 2 V ('(x))2 = q (x):
20
From now on, we take V = IRd and q (x) = (x; x) where (.,.) denotes the usual scalar product. Let e1 ; :::; ed be an orthogonal basis of V and set Ei = '(ei ); i = 1; :::; d: The Cli ord algebra associated to V is generated by d elements Ei; i = 1; :::; d [2], [25] which satisfy the anti-commutation relations
Ei Ej + Ej Ei = 2ij E0;
(22)
E0 being the identity element, ij the Kronecker symbol and 1 i; j d. The real linear space spanned by the products Ej1 Ej2 ::: Ejr ; 0 j1 < j2 < ::: < jr d i.e. E0; E1; E2; :::; Ed; E1 E2; :::; Ed?1 Ed ; :::; E1 E2 ::: Ed forms the associative but noncommutative (d > 1) algebra C (V ). It is easy to see that C (V ) is of dimension 2d. However, it is not a division algebra (d 1) as is proved by the relation (E0 + E1 )(E0 ? E1) = 0: There are various matrix representations of Ei e.g. [20], [17]. However, all of them involve square matrices whose order rises rapidly with the dimension d typicaly this order is at least proportional to 2d=2: P P Furthermore, we may identify each x = di=1 xi ei 2 V , with X = di=1 xi Ei 2 C (V ) and each 2 IR with E0 2 C (V ) (see [2]). According to this identi cation, one can consider V and IR as subsets of C (V ). From the equation (22), we can easily establish
8X 2 V; 8Y 2 V X Y + Y X = 2(X:Y )
(23)
which admits the following two particular rules if X ? Y then
X Y = ?Y X;
(24)
X X = (X:X ): This last relation allows to conclude that, if X = 6 0, then X : X ?1 = (X:X )
(25)
if we take X = Y , then
(26)
Let us notice that this inverse is the same as the Moore-Penrose pseudo-inverse (see [24]).
7.2 The vector "-algorithm
The vector "-algorithm is a quite powerful method for accelerating the convergence of vector sequences. It has many other applications : solution of systems of linear equations [5], [12], systems of nonlinear equations [8] and the computation of eigenvalues of a matrix [6]. It occurs likewise in approximation theory (vector continued fractions [29], [30], vector interpolation and Pade approximants [14],[15], [16], [17]). Although it was the subject of intensive
21
research, it still lacks of theoretical results. In this section, we will show that the vector "-algorithm, like other extrapolation algorithms, realizes some extrapolation in C (V ), solves some system of linear equations in C (V ) and can be written as a "ratio" of two designants. We saw above that (C (V ); +; ) is an associative unitary ring. It is not commutative when d > 1. We also saw that V C (V ); IR C (V ). For each nonzero element X = Pdi=1 xiEi of P V, we saw that its inverse is X ?1 = jjXXjj2 , where jjX jj = ( di=1 x2i )1=2: The vector "-algorithm of Wynn can be written as follows
"(?n1) = 0 "(0n) = Sn 2 V
n = 0; 1; ::: n = 0; 1; ::: ?1
+ ("(kn+1) ? "(kn) ) "(kn+1) = "k(n?+1) 1
k = 0; 1; :::; n = 0; 1; ::::
The inverse considered here is the above-mentioned inverse. The "(kn) 's belong always to V, this is because the sum and the di erence are closed operations in V and the inverse of a nonzero element of V also belongs to V. Owing to the fact that (C (V ); +; ; :) is an algebra, the elements of C (V ) can be regarded as scalar elements of the ring (C (V ); +; ), and as vector elements of the vector space (C (V ); +; :): Let (Sn ) be a sequence of vectors of V (scalar elements of the ring (C (V ); +; ) such that
(
a0+ a1 + +ak =1 a0 Sn+ a1Sn+1+ +ak Sn+k = S
(27)
where S 2 V ; a0 ; a1; :::; ak 2 C (V ): By applying the theorem 10 for IK = (C (V ); +; ), we have the fundamental result
Theorem 11 The quantities "kn computed by the vector "-algorithm can be expressed as a ( )
"ratio" of two designants as follows
S S Sn n k? . n . n . .. . . . " k = . . . . Sn k Sn k? Sn k +
1
( ) 2
+
+2
1
+
"(2nk)+1 = ("2k (Sn))?1: We also have
22
S S n k? . n . .. . . . . . Sn k Sn k? r +
+
+2
1
1
1 .. . 1
?
1
r
Theorem 12 "nk = S () 9A ; :::; Ak 2 C (V ) such that 1
2
Sn = S + A1Sn + ::: + Ak Sn+k?1 :
Proof: It is a direct consequence of theorem 11. In fact, S S n+k?1 Sn . n . (n) .. .. .. "2k = S () .. . . Sn+k Sn+2k?1 Sn+k
S . n = S .. r Sn k +
Sn
1 .. .. ... . . Sn+2k?1 1 k?1
+
: r
Thanks to properties of designants showed in [26], [19], we have
S S n n k? .. .. .. = .. . ... .. . . . . Sn k? Sn k r Sn k Sn k? S S Sn ? S n n k? = 0: .. .. ... " nk = S () ... . . Sn k Sn k? Sn k ? S r
S . n n " k = S () .. Sn k ( ) 2
+
Sn
k?1
+
+2
Sn
1
+
+
+
+
+2
1
1
S
.. ; .
S r
1
( ) 2
+
+2
1
+
By applying the property 4 of section 2, the result follows.
Remark
The fundamental algebraic result of McLeod ([22]) is a particular case of the theorem 12 by imposing that the Ai 's belong to IR ( C (V )). From theorem 11, we can also deduce some other interesting results and obtain some other well known results as particular cases.
Theorem 13 Let "kn be the vectors obtained by applying the "-algorithm to the vector sequence ( )
(Sn ) of elements of V . Let A; B be two invertible elements of C (V ), C an element of V . Let "e(kn) be the vectors obtained by applying the "-algorithm to the vector sequence (A Sn B + C ) then
"e(2nk) = A "(2nk) B + C "e(2nk)+1 = B?1 "(2nk)+1 A?1: Proof: see [26], pp. 178.
Theorem 14 Let M be a square orthogonal matrix of dimension d d (M T M = I ). Let "ekn
( )
be the vector obtained by applying the vector "-algorithm to the vector sequence MSn + C where C is a vector of V and MSn the product of the matrix M by the vector Sn : then
23
"e(2nk) = M:"2(nk) + C "e(2nk)+1 = M:"(2nk)+1 : Proof: Let u be an endomorphism0 of V 1which matrix relatively to the basis e1 ; :::; ed is M .
x1
P Let x = di xi Ei ; X = B @ ... C A: =1
xd u(x) can be written in matrix form as M:X .
To show that the theorem 14 is a particular case of the theorem 13, let us use the following well known result (see [2]): for each orthogonal isomorphism u : V ?! V , there exists an invertible element B 2 C (V ) such that 8x 2 V; u(x) = "B x B ?1 where " = 1 or ? 1: So "2k (M:Sn + C ) can be written as "2k ("B Sn B ?1 + C ): According to theorem 13 we get
"2k ("B Sn B?1 + C ) = "B "2k (Sn) B?1 + C; which can be written in a matrix form as M:"2k (Sn ) + C: In the same way "2k+1 (M:Sn + C ) can be written as
"2k+1 ("B Sn B?1 + C ): Using theorem 13, it follows that
"2k+1("B Sn B?1 + C ) = (B?1 )?1 "2k+1(Sn) ("B)?1 : Now, since "?1 = "; (B ?1 )?1 = B ), we obtain
"2k+1("B Sn B?1 + C ) = ("B "2k+1(Sn) B?1 : In a matrix form, this last equatity can be expressed as M:"2k+1 (Sn ): We nally have "e(2nk) = M"(2nk) + C and "e(2nk)+1 = M"(2nk)+1 :
Acknowledgement
I am grateful to the Professor Claude Brezinski for guiding me during this work, for helpful discussions, valuable suggestions and encouragements.
References [1] A.C. Aitken, Determinants and Matrices, Oliver and Boyd, Edinburgh, 1965. [2] A.Artin, Geometric Algebra, Interscience, NewYork, 1966.
24
[3] C.Brezinski, M.Redivo Zaglia, Extrapolation Methods. Theory and Practice, North-Holland, Amsterdam, 1991. [4] C.Brezinski, A general extrapolation algorithm, Numer. Math., 35(1980)175-187. [5] C.Brezinski, Some results in the theory of the vector "-algorithm., Linear Alg. Appl., 8(1974)77-86. [6] C.Brezinski, Computation of the eigenelements of a matrix by the "-algorithm, Linear Alg. Appl., 11(1975)7-20. [7] C.Brezinski, Some determinantal identities in a vector space, with applications. In Pade Approximation and its Applications. H.Werner and H.J.B}unger,eds. LNM 1071 SpringerVerlag, Berlin,1984,pp.1-11. [8] C.Brezinski, Application de l'"-algorithme a la resolution des systemes non lineaires, C.R.Acad.Sci.Paris, 271A(1970)1174-1177. [9] R.Deheuvels, Formes Quadratiques et Groupes Classiques, Presses Universitaires de France, Paris, 1981. [10] J.Dieudonne, Les determinants sur un corps non commutatif, Bull.Soc.Math. France, 7(1943)27-45. [11] F.J.Dyson, Quaternion determinants, Helv.Phys.Acta, 45(1972)289-302. [12] E.Gekeler, On the solution of systems of equations by the epsilon algorithm of Wynn, Math.Comput., 26(1972)427-436. [13] R.Godement, Cours d'Algebre, Hermann, Paris,1966. [14] P.R.Graves-Morris, Vector-valued rational interpolants I, Numer. Math., 42(1983)331-348. [15] P.R.Graves-Morris, Vector-valued rational interpolants II, I.M.A.J.Num.Anal, 4(1984)209224. [16] P.R.Graves-Morris, C.D. Jenkins, Vector-valued rational interpolants III, Constr.Approx., 2(1986)263-289. [17] P.R.Graves-Morris, D.E. Roberts, From matrix to vector Pade approximants, J.Comp.Appl.Math., to appear. [18] T.Havie, Generalized Neville type extrapolation schemes, BIT, 19(1979)204-213. [19] A.Heyting, Die theorie der linearen Gleichungen in einer Zahlenspezies mit nichtkommutativer Multiplikation, Math.Ann., 98(1927)465-490. [20] G.N. Hile, P.Lounesto, Matrix representations of Cli ord algebras, Linear Alg. Appl.,128(1990)51-63.
25
[21] M.L.Mehta, Matrix Theory. Selected Topics and Useful Results, Les editions de Physique, Les Ulis, 1989. [22] J.B.McLeod, A note on the "-algorithm, Computing, 7(1971)17-24. [23] O.Ore, Linear equations in non-commutative elds, Ann.Math., 32(1931)463-477. [24] R.Penrose, A generalised inverse for matrices, Proc. Cambridge Phil. Soc., 51(1955),406413. [25] IR.Porteous, Topological Geometry, 2nd ed, Cambridge University Press, Cambridge, 1981. [26] A.Salam, Extrapolation: Extension et Nouveaux Resultats, Thesis, Universite des Sciences et Technologies de Lille,1993. [27] D.Shanks, Non linear transformations of divergent and slowly convergent sequences, J.Math.phys., 34(1955)1-42. [28] P.Wynn, On a device for computing the em (Sn ) transformation, MTAC, 10(1956)91-96. [29] P.Wynn, Vector continued fractions, Linear Alg. Appl., 1(1968)357-395. [30] P.Wynn, Continued fractions whose coecients obey a non-commutative law of multiplication., Arch.Rational Mech.Anal., 12(1963)273-312.
26
