Return to Article Details Convergence theorems on non-commutative continued fractions

L'ANALYSE NUMÉRIQUE ET LA THÉORIE DE L'APPROXIMATION Tome 5, N 0 2 , 1976 N 0 2 , 1976 N^(0)2,1976\mathrm{N}^{0} 2,1976N02,1976, pp. 165-180

CONVER GENCE THEOREMS ON NON-COMMUTATIVE, CONTINUED FRACTIONS

by

N. NEGOESCU(Iaşi)

Introduction

The three lemmas on real continued fractions of the first paragraph are essential in the proofs of the theorems of paragraph 5.
In second paragraph, we state the known concepts and propositions on the Banach algebras, necessary in the paragraphs 3-7.
The definitions of non-commutative continued fractions and their convergence of paragraph 3 are given according to WYMAN FAIR [1].
The new results of this note of paragraphs 1 , 4 − 7 1 , 4 − 7 1,4-71,4-71,4−7 suggest a general method which may be used to prove the theorems of convergence for non-commutative continued fractions. In paragraph 6 , we obtain a generalization of a classical worpitzKY's theorem on the convergence of some continued fractions of complex numbers. In paragraph 7 too, we give some consequences of the main theorem from paragraph 6.

1. Three lemmas on real continued fractions

Lemma 1.1. Let be
(1.1)
a 1 − a a 1 − a (a)/(1-a)\frac{a}{1-a}a1−a
a real continued fraction, where
0 < a ⩽ 1 4 . 0 < a ⩽ 1 4 . 0 < a <= (1)/(4).0<a \leqslant \frac{1}{4} .0<a⩽14.

Then

(i) the convergents of continued fraction (1.1) are positive and its sequence is strictly monoton increasing,
(ii) the convergents satisfy the inequalities
(1.2) p n q n ⩽ 1 2 n n + 1 , (1.2) p n q n ⩽ 1 2 n n + 1 , {:(1.2)(p_(n))/(q_(n)) <= (1)/(2)(n)/(n+1)",":}\begin{equation*} \frac{p_{n}}{q_{n}} \leqslant \frac{1}{2} \frac{n}{n+1}, \tag{1.2} \end{equation*}(1.2)pnqn⩽12nn+1,
(iii) the sequence of convergents of (1.1) is convergent to the number
(1.3) 1 2 [ 1 − ( 1 − 4 a ) 1 / 2 ] (1.3) 1 2 1 − ( 1 − 4 a ) 1 / 2 {:(1.3)(1)/(2)[1-(1-4a)^(1//2)]:}\begin{equation*} \frac{1}{2}\left[1-(1-4 a)^{1 / 2}\right] \tag{1.3} \end{equation*}(1.3)12[1−(1−4a)1/2]
which is less or equal 1 2 1 2 (1)/(2)\frac{1}{2}12.
Proof. (i) From the evident inequality
(1.4) 0 < a 1 < a 1 − a (1.4) 0 < a 1 < a 1 − a {:(1.4)0 < (a)/(1) < (a)/(1-a):}\begin{equation*} 0<\frac{a}{1}<\frac{a}{1-a} \tag{1.4} \end{equation*}(1.4)0<a1<a1−a
it follows, for the first two convergents of real continued fraction (1.1),
0 < p 1 q 1 < p 2 q 2 0 < p 1 q 1 < p 2 q 2 0 < (p_(1))/(q_(1)) < (p_(2))/(q_(2))0<\frac{p_{1}}{q_{1}}<\frac{p_{2}}{q_{2}}0<p1q1<p2q2
From inequality (1.4), it follows also
p 2 q 2 = a 1 − a < a 1 − a 1 − a = p 3 q 3 p 2 q 2 = a 1 − a < a 1 − a 1 − a = p 3 q 3 (p_(2))/(q_(2))=(a)/(1-a) < (a)/(1-(a)/(1-a))=(p_(3))/(q_(3))\frac{p_{2}}{q_{2}}=\frac{a}{1-a}<\frac{a}{1-\frac{a}{1-a}}=\frac{p_{3}}{q_{3}}p2q2=a1−a<a1−a1−a=p3q3
By induction, we now suppose that
(1.5) p n − 1 q n − 1 < p n q n . (1.5) p n − 1 q n − 1 < p n q n . {:(1.5)(p_(n-1))/(q_(n-1)) < (p_(n))/(q_(n)).:}\begin{equation*} \frac{p_{n-1}}{q_{n-1}}<\frac{p_{n}}{q_{n}} . \tag{1.5} \end{equation*}(1.5)pn−1qn−1<pnqn.
Since
(1.6) p n + 1 q n + 1 = p n + a p n − 1 q n ! + a q n − 1 , (1.6) p n + 1 q n + 1 = p n + a p n − 1 q n ! + a q n − 1 , {:(1.6)(p_(n+1))/(q_(n+1))=(p_(n)+ap_(n-1))/(q_(n)!+aq_(n-1))",":}\begin{equation*} \frac{p_{n+1}}{q_{n+1}}=\frac{p_{n}+a p_{n-1}}{q_{n}!+a q_{n-1}}, \tag{1.6} \end{equation*}(1.6)pn+1qn+1=pn+apn−1qn!+aqn−1,
then the function
x → p n + x p n − 1 q n + x q n − 1 x → p n + x p n − 1 q n + x q n − 1 x rarr(p_(n)+xp_(n-1))/(q_(n)+xq_(n-1))x \rightarrow \frac{p_{n}+x p_{n-1}}{q_{n}+x q_{n-1}}x→pn+xpn−1qn+xqn−1
has the derivative
q n p n − 1 − p n q n − 1 ( q n + x q n − 1 ) 2 q n p n − 1 − p n q n − 1 q n + x q n − 1 2 (q_(n)p_(n-1)-p_(n)q_(n-1))/((q_(n)+xq_(n-1))^(2))\frac{q_{n} p_{n-1}-p_{n} q_{n-1}}{\left(q_{n}+x q_{n-1}\right)^{2}}qnpn−1−pnqn−1(qn+xqn−1)2
which is strictly negative. From equality (1.6) , it follows
p n + 1 q n + 1 > p n q n p n + 1 q n + 1 > p n q n (p_(n+1))/(q_(n+1)) > (p_(n))/(q_(n))\frac{p_{n+1}}{q_{n+1}}>\frac{p_{n}}{q_{n}}pn+1qn+1>pnqn
This completes the induction and the proof of (i).
(ii) By the hypothesis of the lemma, we have
p 1 q 1 = a ⩽ 1 4 p 1 q 1 = a ⩽ 1 4 (p_(1))/(q_(1))=a <= (1)/(4)\frac{p_{1}}{q_{1}}=a \leqslant \frac{1}{4}p1q1=a⩽14
By induction, we now suppose that p n q n p n q n (p_(n))/(q_(n))\frac{p_{n}}{q_{n}}pnqn satisfies (1.2). Then
p n + 1 q n + 1 = a 1 − p n q n ⩽ 1 4 1 − 1 2 n n + 1 = 1 2 n + 1 n + 2 p n + 1 q n + 1 = a 1 − p n q n ⩽ 1 4 1 − 1 2 n n + 1 = 1 2 n + 1 n + 2 (p_(n+1))/(q_(n+1))=(a)/(1-(p_(n))/(q_(n))) <= ((1)/(4))/(1-(1)/(2)(n)/(n+1))=(1)/(2)(n+1)/(n+2)\frac{p_{n+1}}{q_{n+1}}=\frac{a}{1-\frac{p_{n}}{q_{n}}} \leqslant \frac{\frac{1}{4}}{1-\frac{1}{2} \frac{n}{n+1}}=\frac{1}{2} \frac{n+1}{n+2}pn+1qn+1=a1−pnqn⩽141−12nn+1=12n+1n+2
Therefore, (ii) is proved completely.
(iii) By (i) the sequence of the convergents is strictly monoton increa sing and by (ii) the same sequence is less than 1 2 1 2 (1)/(2)\frac{1}{2}12; therefore it is convergent.
The real continued fraction, being periodic, its value is given as throot, which is less than 1 2 1 2 (1)/(2)\frac{1}{2}12, of equation
x = a 1 − x x = a 1 − x x=(a)/(1-x)x=\frac{a}{1-x}x=a1−x, that is
x = 1 2 [ 1 − ( 1 − 4 a ) 1 / 2 ] ⩽ 1 2 x = 1 2 1 − ( 1 − 4 a ) 1 / 2 ⩽ 1 2 x=(1)/(2)[1-(1-4a)^(1//2)] <= (1)/(2)x=\frac{1}{2}\left[1-(1-4 a)^{1 / 2}\right] \leqslant \frac{1}{2}x=12[1−(1−4a)1/2]⩽12
Lemma 1.2. If
(1.7) a 1 1 − a 2 1 − a 3 1 − (1.7) a 1 1 − a 2 1 − a 3 1 − {:(1.7)(a_(1))/(1-(a_(2))/(1-(a_(3))/(1-))):}\begin{equation*} \frac{a_{1}}{1-\frac{a_{2}}{1-\frac{a_{3}}{1-}}} \tag{1.7} \end{equation*}(1.7)a11−a21−a31−
where 0 ⩽ a k ⩽ a ⩽ 1 4 0 ⩽ a k ⩽ a ⩽ 1 4 0 <= a_(k) <= a <= (1)/(4)0 \leqslant a_{k} \leqslant a \leqslant \frac{1}{4}0⩽ak⩽a⩽14, is a real continued fraction, then
(i) the convergents of this continued fraction are non-negative and
(ii) the convergents satisfy inequalities
(1.8)
p n q n ⩽ 1 2 n n + 1 . p n q n ⩽ 1 2 n n + 1 . (p_(n))/(q_(n)) <= (1)/(2)(n)/(n+1).\frac{p_{n}}{q_{n}} \leqslant \frac{1}{2} \frac{n}{n+1} .pnqn⩽12nn+1.
Proof. For the first convergent p 1 q 1 = a 1 p 1 q 1 = a 1 (p_(1))/(q_(1))=a_(1)\frac{p_{1}}{q_{1}}=a_{1}p1q1=a1, we have evident
0 ⩽ a 1 1 ⩽ a 1 0 ⩽ a 1 1 ⩽ a 1 0 <= (a_(1))/(1) <= (a)/(1)0 \leqslant \frac{a_{1}}{1} \leqslant \frac{a}{1}0⩽a11⩽a1
If we suppose
(1.9) a ′ = a 2 1 − a 3 1 − … a n + 1 1 ⩾ 0 (1.9) a ′ = a 2 1 − a 3 1 − … a n + 1 1 ⩾ 0 {:(1.9)a^(')=(a_(2))/(1-)(a_(3))/(1-dots)quad(a_(n+1))/(1) >= 0:}\begin{equation*} a^{\prime}=\frac{a_{2}}{1-} \frac{a_{3}}{1-\ldots} \quad \frac{a_{n+1}}{1} \geqslant 0 \tag{1.9} \end{equation*}(1.9)a′=a21−a31−…an+11⩾0
then, since a 1 ⩾ 0 a 1 ⩾ 0 a_(1) >= 0a_{1} \geqslant 0a1⩾0 by hypothesis, it follows that
(1.10) p n + 1 q n + 1 = a 1 1 − a ′ (1.10) p n + 1 q n + 1 = a 1 1 − a ′ {:(1.10)(p_(n+1))/(q_(n+1))=(a_(1))/(1-a^(')):}\begin{equation*} \frac{p_{n+1}}{q_{n+1}}=\frac{a_{1}}{1-a^{\prime}} \tag{1.10} \end{equation*}(1.10)pn+1qn+1=a11−a′
is non-negative.
(ii) If the real continued fraction (1.9) is less than n 2 ( n + 1 ) n 2 ( n + 1 ) (n)/(2(n+1))\frac{n}{2(n+1)}n2(n+1), then, from (1.10), we obtain
p n + 1 q n + 1 = a 1 1 − a ′ ⩽ 1 4 1 − n 2 ( n + 1 ) = 1 2 n + 1 n + 2 . p n + 1 q n + 1 = a 1 1 − a ′ ⩽ 1 4 1 − n 2 ( n + 1 ) = 1 2 n + 1 n + 2 . (p_(n+1))/(q_(n+1))=(a_(1))/(1-a^(')) <= ((1)/(4))/(1-(n)/(2(n+1)))=(1)/(2)(n+1)/(n+2).\frac{p_{n+1}}{q_{n+1}}=\frac{a_{1}}{1-a^{\prime}} \leqslant \frac{\frac{1}{4}}{1-\frac{n}{2(n+1)}}=\frac{1}{2} \frac{n+1}{n+2} .pn+1qn+1=a11−a′⩽141−n2(n+1)=12n+1n+2.
This completes the proof of lemma 2.2 .
Le m m a 1.3. For a > 1 4 a > 1 4 a > (1)/(4)a>\frac{1}{4}a>14, the real continued fraction
(1.11) a 1 − a 1 − a 1 − … (1.11) a 1 − a 1 − a 1 − … {:(1.11)(a)/(1-)(a)/(1-)(a)/(1-)dots:}\begin{equation*} \frac{a}{1-} \frac{a}{1-} \frac{a}{1-} \ldots \tag{1.11} \end{equation*}(1.11)a1−a1−a1−…
is divergent.
Proof. If we suppose that continued fraction (1.11) is convergent, then it converges to a real number.
On the other hand, real continued fraction (1.11), being periodic, its value is one of the complex numbers
1 2 [ 1 ± ( 1 − 4 a ) 1 2 ] 1 2 1 ± ( 1 − 4 a ) 1 2 (1)/(2)[1+-(1-4a)^((1)/(2))]\frac{1}{2}\left[1 \pm(1-4 a)^{\frac{1}{2}}\right]12[1±(1−4a)12]
which are the roots of the equation
x = a 1 − x x = a 1 − x x=(a)/(1-x)x=\frac{a}{1-x}x=a1−x
This contradiction proves the lemma.

2. Banach algebras

Definition 3.1. A Banach algebra with elements x , y , … x , y , … x,y,dotsx, y, \ldotsx,y,… is a Banach space, that is a complete normed space endowed with a multiplication, everywhere defined, associative, distributive with respect to every linear combination and such that
(2.1) ‖ x y ‖ ⩽ ‖ x ‖ ‖ y ‖ (2.1) ‖ x y ‖ ⩽ ‖ x ‖ ‖ y ‖ {:(2.1)||xy|| <= ||x||||y||:}\begin{equation*} \|x y\| \leqslant\|x\|\|y\| \tag{2.1} \end{equation*}(2.1)‖xy‖⩽‖x‖‖y‖
for any x , y x , y x,yx, yx,y of a a aaa.
Algebra & is said to be commutative if the multiplication is commutative.
A subspace B B B\mathfrak{B}B of C C C\mathfrak{C}C is called a subalgebra of a a a\mathfrak{a}a if it is also an algebra. It is easy to prove the following propositions:
Proposition 2.1. If ( x n ) x n (x_(n))\left(x_{n}\right)(xn) and ( y n ) y n (y_(n))\left(y_{n}\right)(yn) are convergent sequences, with respect the topology induced by the algebra norm, then ( x n y n ) x n y n (x_(n)y_(n))\left(x_{n} y_{n}\right)(xnyn) converges to x y x y xyx yxy, that is the multiplication is continuous.
Proposition 2.2. For every x x xxx of a Banach algebra and n ∈ N ∗ n ∈ N ∗ n inN^(**)n \in \mathrm{~N}^{*}n∈ N∗, w have
‖ x n ‖ ⩽ ‖ x ‖ n x n ⩽ ‖ x ‖ n ||x^(n)|| <= ||x||^(n)\left\|x^{n}\right\| \leqslant\|x\|^{n}‖xn‖⩽‖x‖n
Proposition 2.3. If there is in a an element, e e eee which satisfies the relation
(2.2) ex = xe = x (2.2) ex = xe = x {:(2.2)ex=xe=x:}\begin{equation*} \mathrm{ex}=\mathrm{xe}=\mathrm{x} \tag{2.2} \end{equation*}(2.2)ex=xe=x
then e e eee is unique and ‖ e ‖ ⩾ 1 ‖ e ‖ ⩾ 1 ||e|| >= 1\|e\| \geqslant 1‖e‖⩾1, if a ≠ { 0 } a ≠ { 0 } a!={0}\mathfrak{a} \neq\{0\}a≠{0}.
Proposition 2.4. If there is an element e e eee in e e eee which satisfies (2.2), then the expresion
‖ x ‖ ′ = sup { ‖ x u ‖ : ‖ u ‖ = 1 } ‖ x ‖ ′ = sup { ‖ x u ‖ : ‖ u ‖ = 1 } ||x||^(')=s u p{||xu||:||u||=1}\|x\|^{\prime}=\sup \{\|x u\|:\|u\|=1\}‖x‖′=sup{‖xu‖:‖u‖=1}
exists in a a a\mathfrak{a}a and it defines a norm equivalent with the initial norm such that
(2.3) ‖ e ‖ ′ = 1 (2.3) ‖ e ‖ ′ = 1 {:(2.3)||e||^(')=1:}\begin{equation*} \|e\|^{\prime}=1 \tag{2.3} \end{equation*}(2.3)‖e‖′=1
holds.
For the norm ‖ ⋅ ‖ ′ , Q ‖ ⋅ ‖ ′ , Q ||*||^('),Q\|\cdot\|^{\prime}, \mathcal{Q}‖⋅‖′,Q is also a Banach algebra with properties (2.2) and (2.3).
It is easy to verify that ‖ ⋅ ‖ ′ ‖ ⋅ ‖ ′ ||*||^(')\|\cdot\|^{\prime}‖⋅‖′ is a norm. Then
‖ e ‖ ′ = sup { ‖ e u ‖ : ‖ u ‖ = 1 } = sup { ‖ u ‖ : ‖ u ‖ = 1 } = 1 ‖ x y ‖ ′ = sup { ‖ x y u ‖ : ‖ u ‖ = 1 } = sup { ‖ x − y u ‖ y u ‖ ‖ : ‖ u ‖ = 1 } ‖ y u ‖ ⩽ ⩽ sup { ‖ x y u ‖ y u ‖ ‖ : ‖ u ‖ = 1 } sup { ‖ y u ‖ : ‖ u ‖ = 1 } = ‖ x ‖ ′ ‖ y ‖ ′ ‖ e ‖ ′ = sup { ‖ e u ‖ : ‖ u ‖ = 1 } = sup { ‖ u ‖ : ‖ u ‖ = 1 } = 1 ‖ x y ‖ ′ = sup { ‖ x y u ‖ : ‖ u ‖ = 1 } = sup x − y u ‖ y u ‖ : ‖ u ‖ = 1 ‖ y u ‖ ⩽ ⩽ sup x y u ‖ y u ‖ : ‖ u ‖ = 1 sup { ‖ y u ‖ : ‖ u ‖ = 1 } = ‖ x ‖ ′ ‖ y ‖ ′ {:[||e||^(')=s u p{||eu||:||u||=1}=s u p{||u||:||u||=1}=1],[||xy||^(')=s u p{||xyu||:||u||=1}=s u p{||x-(yu)/(||yu||)||:||u||=1}||yu|| <= ],[ <= s u p{||x(yu)/(||yu||)||:||u||=1}s u p{||yu||:||u||=1}=||x||^(')||y||^(')]:}\begin{gathered} \|e\|^{\prime}=\sup \{\|e u\|:\|u\|=1\}=\sup \{\|u\|:\|u\|=1\}=1 \\ \|x y\|^{\prime}=\sup \{\|x y u\|:\|u\|=1\}=\sup \left\{\left\|x-\frac{y u}{\|y u\|}\right\|:\|u\|=1\right\}\|y u\| \leqslant \\ \leqslant \sup \left\{\left\|x \frac{y u}{\|y u\|}\right\|:\|u\|=1\right\} \sup \{\|y u\|:\|u\|=1\}=\|x\|^{\prime}\|y\|^{\prime} \end{gathered}‖e‖′=sup{‖eu‖:‖u‖=1}=sup{‖u‖:‖u‖=1}=1‖xy‖′=sup{‖xyu‖:‖u‖=1}=sup{‖x−yu‖yu‖‖:‖u‖=1}‖yu‖⩽⩽sup{‖xyu‖yu‖‖:‖u‖=1}sup{‖yu‖:‖u‖=1}=‖x‖′‖y‖′
since y u / ‖ y u ‖ y u / ‖ y u ‖ yu//||yu||y u /\|y u\|yu/‖yu‖ has the norm 1 .
This norm is equivalent to the initial norm, because
‖ x ‖ ′ ⩽ ‖ x ‖ ⩽ sup { ‖ u ‖ : ‖ u ‖ = 1 } ‖ e ‖ ⩽ ‖ x ‖ ′ ‖ e ‖ , ‖ x ‖ ′ ⩽ ‖ x ‖ ⩽ sup { ‖ u ‖ : ‖ u ‖ = 1 } ‖ e ‖ ⩽ ‖ x ‖ ′ ‖ e ‖ , ||x||^(') <= ||x|| <= s u p{||u||:||u||=1}||e|| <= ||x||^(')||e||,\|x\|^{\prime} \leqslant\|x\| \leqslant \sup \{\|u\|:\|u\|=1\}\|e\| \leqslant\|x\|^{\prime}\|e\|,‖x‖′⩽‖x‖⩽sup{‖u‖:‖u‖=1}‖e‖⩽‖x‖′‖e‖,
where u = e / ‖ e ‖ u = e / ‖ e ‖ u=e//||e||u=e /\|e\|u=e/‖e‖.
Propositions (2.3) and (2.4) suggest:
Definition 2.2. A Banach algebra is said to have the unit e [2], if for any x ∈ Q x ∈ Q x inQx \in \mathfrak{Q}x∈Q we have the relation (2.2) and (2.3).
Definition 2.3. Let a be a Banach algebra with the element e e eee, wich satisfies (2.2). An element x ∈ A x ∈ A x inAx \in \mathscr{A}x∈A has a left inverse, if it exists an element x s − 1 ∈ A x s − 1 ∈ A x_(s)^(-1)inAx_{s}^{-1} \in \mathcal{A}xs−1∈A, such that x s − 1 x = e x s − 1 x = e x_(s)^(-1)x=ex_{s}^{-1} x=exs−1x=e.
An element x ∈ a x ∈ a x inax \in \mathfrak{a}x∈a has a right inverse, if it exists an element x d − 1 ∈ a x d − 1 ∈ a x_(d)^(-1)inax_{d}^{-1} \in \mathfrak{a}xd−1∈a, such that x d − 1 x = e x d − 1 x = e x_(d)^(-1)x=ex_{d}^{-1} x=exd−1x=e. The element x x xxx is called invertible, if it has a left inverse and a right inverse.
Proposition 2.5. The invertible elements in a Banach algebra a with an element e, wich satisfies relations (2.2) have following properties:
(i) If y z = x z = e y z = x z = e yz=xz=ey z=x z=eyz=xz=e, then x x xxx is invertible in a and y = z y = z y=zy=zy=z. Therefore, the inverse of an element, if it exists, is unique. It is noted x − 1 x − 1 x^(-1)x^{-1}x−1.
(ii) If x x xxx is invertible in C C C\mathbb{C}C and x y = e ( y x = e ) x y = e ( y x = e ) xy=e(yx=e)x y=e(y x=e)xy=e(yx=e), then y = x − 1 y = x − 1 y=x^(-1)y=x^{-1}y=x−1.
(iii) The element e e eee is invertible and it is the own inverse. The element 0 is not invertible.
(iv) If x x xxx is invertible in a a a\mathfrak{a}a, then c x c x cxc xcx is invertible in a for any scalar c ∈ C c ∈ C c inCc \in \mathcal{C}c∈C, with c ≠ 0 c ≠ 0 c!=0c \neq 0c≠0 and
( c x ) − 1 = 1 c x − 1 ( c x ) − 1 = 1 c x − 1 (cx)^(-1)=(1)/(c)x^(-1)(c x)^{-1}=\frac{1}{c} x^{-1}(cx)−1=1cx−1
(v) If x , y x , y x,yx, yx,y are invertible in a a a\mathcal{a}a, then the product is invertible and ( x y ) − 1 = y − 1 x − 1 ( x y ) − 1 = y − 1 x − 1 (xy)^(-1)=y^(-1)x^(-1)(x y)^{-1}=y^{-1} x^{-1}(xy)−1=y−1x−1.
It is also easy to prove
Proposition 2.6. If A A A\mathscr{A}A is a Banach algebra with unit e (definition 2.2) and x ∈ A x ∈ A x in Ax \in Ax∈A verifies inequality ‖ x ‖ < 1 ‖ x ‖ < 1 ||x|| < 1\|x\|<1‖x‖<1, then the element e − x e − x e-xe-xe−x is invertible in a and
(i)
( e − x ) − 1 = e + x + x 2 + … ( e − x ) − 1 = e + x + x 2 + … (e-x)^(-1)=e+x+x^(2)+dots(e-x)^{-1}=e+x+x^{2}+\ldots(e−x)−1=e+x+x2+…
(ii)
‖ ( e − x ) − 1 ‖ ⩽ 1 1 − ‖ x ‖ ( e − x ) − 1 ⩽ 1 1 − ‖ x ‖ ||(e-x)^(-1)|| <= (1)/(1-||x||)\left\|(e-x)^{-1}\right\| \leqslant \frac{1}{1-\|x\|}‖(e−x)−1‖⩽11−‖x‖
One also see that if a a a\mathbb{a}a is a Banach algebra with unit, e e eee, then
(i) { p ( x ) : ∀ { p ( x ) : ∀ {p(x):AA\{p(x): \forall{p(x):∀ polynomial p , x ∈ Q p , x ∈ Q p,x inQp, x \in \mathbb{Q}p,x∈Q is a commutative subalgebra of E E E\mathfrak{E}E, with unit e e eee.
(ii) { a e : a ∈ K } { a e : a ∈ K } {ae:a in K}\{a e: a \in K\}{ae:a∈K}, where H H H\mathscr{H}H is R R R\mathbf{R}R or C C C\mathcal{C}C, is a commutative subalgebra of a. isomorphic to II, with unit e.

3. Non-commutative continued fractions

Definition 3.1. The formal expression
a 1 ε + a 3 ε + a 3 e + a 1 ε + a 3 ε + a 3 e + (a_(1))/(epsi+(a_(3))/(epsi+(a_(3))/(e+)))\frac{a_{1}}{\varepsilon+\frac{a_{3}}{\varepsilon+\frac{a_{3}}{e+}}}a1ε+a3ε+a3e+
wich may be written also as
(3.1) a 1 c + a 2 e + a 3 e + … (3.1) a 1 c + a 2 e + a 3 e + … {:(3.1)(a_(1))/(c+)(a_(2))/(e+)(a_(3))/(e+dots):}\begin{equation*} \frac{a_{1}}{c+} \frac{a_{2}}{e+} \frac{a_{3}}{e+\ldots} \tag{3.1} \end{equation*}(3.1)a1c+a2e+a3e+…
where a ‖ a ‖ a_(||)a_{\|}a‖are elements of a non-commutative complex Banach algebra & with unit e e eee, is said to be a non-commutative continued fraction.
Definition 3.2. The element
(3.2) q n − 1 p n (3.2) q n − 1 p n {:(3.2)q_(n)^(-1)p_(n):}\begin{equation*} q_{n}^{-1} p_{n} \tag{3.2} \end{equation*}(3.2)qn−1pn
(supposing q n q n q_(n)q_{n}qn invertible that is q n − 1 p n ∈ A q n − 1 p n ∈ A q_(n)^(-1)p_(n)inAq_{n}^{-1} p_{n} \in \mathfrak{A}qn−1pn∈A ), where p n p n p_(n)p_{n}pn and q n q n q_(n)q_{n}qn are given by the formules
p n + 1 = p n + a n + 1 p n − 1 (3.3) q n + 1 = q n + q n + 1 q n − 1 p 1 = p 2 = a 1 , q 1 = e , q 2 = e + a 2 p n + 1 = p n + a n + 1 p n − 1 (3.3) q n + 1 = q n + q n + 1 q n − 1 p 1 = p 2 = a 1 , q 1 = e , q 2 = e + a 2 {:[p_(n+1)=p_(n)+a_(n+1)p_(n-1)],[(3.3)q_(n+1)=q_(n)+q_(n+1)q_(n-1)],[p_(1)=p_(2)=a_(1)","quadq_(1)=e","quadq_(2)=e+a_(2)]:}\begin{gather*} p_{n+1}=p_{n}+a_{n+1} p_{n-1} \\ q_{n+1}=q_{n}+q_{n+1} q_{n-1} \tag{3.3}\\ p_{1}=p_{2}=a_{1}, \quad q_{1}=e, \quad q_{2}=e+a_{2} \end{gather*}pn+1=pn+an+1pn−1(3.3)qn+1=qn+qn+1qn−1p1=p2=a1,q1=e,q2=e+a2
is called the n th n th  n^("th ")n^{\text {th }}nth  convergent of the continued fraction (3.1).
Definition 3.3. The non-commutative continued fracion (1.1) is said to be convergent, if q n q n q_(n)q_{n}qn are invertible for a sufficient large n n nnn and the sequence of elements (3.2) converges.

4. Relations for non-commutative continued fractions

Proposition 4.1. For the non-commutative continued fractions, we have the following identities:
(i) q 3 = e + a 2 + a 3 q 3 = e + a 2 + a 3 q_(3)=e+a_(2)+a_(3)q_{3}=e+a_{2}+a_{3}q3=e+a2+a3,
(ii) p n + 1 q n − p n q n + 1 = − a n + 1 ( p n q n − 1 − p n − 1 q n ) p n + 1 q n − p n q n + 1 = − a n + 1 p n q n − 1 − p n − 1 q n p_(n+1)q_(n)-p_(n)q_(n+1)=-a_(n+1)(p_(n)q_(n-1)-p_(n-1)q_(n))p_{n+1} q_{n}-p_{n} q_{n+1}=-a_{n+1}\left(p_{n} q_{n-1}-p_{n-1} q_{n}\right)pn+1qn−pnqn+1=−an+1(pnqn−1−pn−1qn),
(iii) p n + 1 q n − p n q n + 1 = ( − 1 ) n a n + 1 a n a n − 1 … a 3 a 1 a 2 p n + 1 q n − p n q n + 1 = ( − 1 ) n a n + 1 a n a n − 1 … a 3 a 1 a 2 p_(n+1)q_(n)-p_(n)q_(n+1)=(-1)^(n)a_(n+1)a_(n)a_(n-1)dotsa_(3)a_(1)a_(2)p_{n+1} q_{n}-p_{n} q_{n+1}=(-1)^{n} a_{n+1} a_{n} a_{n-1} \ldots a_{3} a_{1} a_{2}pn+1qn−pnqn+1=(−1)nan+1anan−1…a3a1a2,
(iv) q n + 2 = ( e + a n + 1 + a n + 2 ) q n − q n + 1 a n q n − 2 q n + 2 = e + a n + 1 + a n + 2 q n − q n + 1 a n q n − 2 q_(n+2)=(e+a_(n+1)+a_(n+2))q_(n)-q_(n+1)a_(n)q_(n-2)q_{n+2}=\left(e+a_{n+1}+a_{n+2}\right) q_{n}-q_{n+1} a_{n} q_{n-2}qn+2=(e+an+1+an+2)qn−qn+1anqn−2.
Proof.
(i) and (ii) follow directly from (3.3).
(iii) is a corollary of (ii).
(iv) From formulas (3.3) we obtain
q n + 1 = q n + a n + 1 q n − 1 = q n + a n − 1 ( q n − a n q n − 2 ) = = ( e + a n + 1 ) q n − a n + 1 a n q n − 2 q n + 1 = q n + a n + 1 q n − 1 = q n + a n − 1 q n − a n q n − 2 = = e + a n + 1 q n − a n + 1 a n q n − 2 {:[q_(n+1)=q_(n)+a_(n+1)q_(n-1)=q_(n)+a_(n-1)(q_(n)-a_(n)q_(n-2))=],[=(e+a_(n+1))q_(n)-a_(n+1)a_(n)q_(n-2)]:}\begin{aligned} q_{n+1} & =q_{n}+a_{n+1} q_{n-1}=q_{n}+a_{n-1}\left(q_{n}-a_{n} q_{n-2}\right)= \\ & =\left(e+a_{n+1}\right) q_{n}-a_{n+1} a_{n} q_{n-2} \end{aligned}qn+1=qn+an+1qn−1=qn+an−1(qn−anqn−2)==(e+an+1)qn−an+1anqn−2
and
q n + 2 = q n + 1 + a n + 2 q n = ( e + a n + 1 ) q n − a n + 1 a n q n − 2 + a n + 2 q n = = ( e + a n + 1 + a n + 2 ) q n − a n + 1 a n q n + 2 . q n + 2 = q n + 1 + a n + 2 q n = e + a n + 1 q n − a n + 1 a n q n − 2 + a n + 2 q n = = e + a n + 1 + a n + 2 q n − a n + 1 a n q n + 2 . {:[q_(n+2)=q_(n+1)+a_(n+2)q_(n)=(e+a_(n+1))q_(n)-a_(n+1)a_(n)q_(n-2)+a_(n+2)q_(n)=],[=(e+a_(n+1)+a_(n+2))q_(n)-a_(n+1)a_(n)q_(n+2).]:}\begin{aligned} q_{n+2} & =q_{n+1}+a_{n+2} q_{n}=\left(e+a_{n+1}\right) q_{n}-a_{n+1} a_{n} q_{n-2}+a_{n+2} q_{n}= \\ & =\left(e+a_{n+1}+a_{n+2}\right) q_{n}-a_{n+1} a_{n} q_{n+2} . \end{aligned}qn+2=qn+1+an+2qn=(e+an+1)qn−an+1anqn−2+an+2qn==(e+an+1+an+2)qn−an+1anqn+2.
Proposition 4.2. If q n − 1 , q n , q n + 1 q n − 1 , q n , q n + 1 q_(n-1),q_(n),q_(n+1)q_{n-1}, q_{n}, q_{n+1}qn−1,qn,qn+1 are invertible elements of Banach algebra A A A\mathscr{A}A with unit e, then
(4.1) q n + 1 − 1 p n + 1 − q n − 1 p n = − q n + 1 − 1 a n + 1 q n − 1 ( q n − 1 p n − q n − 1 − 1 p n − 1 ) . (4.1) q n + 1 − 1 p n + 1 − q n − 1 p n = − q n + 1 − 1 a n + 1 q n − 1 q n − 1 p n − q n − 1 − 1 p n − 1 . {:(4.1)q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n)=-q_(n+1)^(-1)a_(n+1)q_(n-1)(q_(n)^(-1)p_(n)-q_(n-1)^(-1)p_(n-1)).:}\begin{equation*} q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}=-q_{n+1}^{-1} a_{n+1} q_{n-1}\left(q_{n}^{-1} p_{n}-q_{n-1}^{-1} p_{n-1}\right) . \tag{4.1} \end{equation*}(4.1)qn+1−1pn+1−qn−1pn=−qn+1−1an+1qn−1(qn−1pn−qn−1−1pn−1).
Proof. From (3.3), we obtain
q ˙ n + 1 − 1 p n + 1 − q n − 1 p n = q n + 1 − 1 ( p n + 1 − q n + 1 q n − 1 p n ) = = q n + 1 − 1 a n + 1 ( p n − 1 − q n − 1 − 1 q n − 1 p n ) = − q n + 1 − 1 a n + 1 q n − 1 ( q n − 1 p n − q n − 1 − 1 p n − 1 ) q ˙ n + 1 − 1 p n + 1 − q n − 1 p n = q n + 1 − 1 p n + 1 − q n + 1 q n − 1 p n = = q n + 1 − 1 a n + 1 p n − 1 − q n − 1 − 1 q n − 1 p n = − q n + 1 − 1 a n + 1 q n − 1 q n − 1 p n − q n − 1 − 1 p n − 1 {:[q^(˙)_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n)=q_(n+1)^(-1)(p_(n+1)-q_(n+1)q_(n)^(-1)p_(n))=],[=q_(n+1)^(-1)a_(n+1)(p_(n-1)-q_(n-1)^(-1)q_(n)^(-1)p_(n))=-q_(n+1)^(-1)a_(n+1)q_(n-1)(q_(n)^(-1)p_(n)-q_(n-1)^(-1)p_(n-1))]:}\begin{gathered} \dot{q}_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}=q_{n+1}^{-1}\left(p_{n+1}-q_{n+1} q_{n}^{-1} p_{n}\right)= \\ =q_{n+1}^{-1} a_{n+1}\left(p_{n-1}-q_{n-1}^{-1} q_{n}^{-1} p_{n}\right)=-q_{n+1}^{-1} a_{n+1} q_{n-1}\left(q_{n}^{-1} p_{n}-q_{n-1}^{-1} p_{n-1}\right) \end{gathered}q˙n+1−1pn+1−qn−1pn=qn+1−1(pn+1−qn+1qn−1pn)==qn+1−1an+1(pn−1−qn−1−1qn−1pn)=−qn+1−1an+1qn−1(qn−1pn−qn−1−1pn−1)
Proposition 4.3. If q 2 , q 3 , … , q n + 1 q 2 , q 3 , … , q n + 1 q_(2),q_(3),dots,q_(n+1)q_{2}, q_{3}, \ldots, q_{n+1}q2,q3,…,qn+1 are invertible elements of Banach algebra & with unit e, then
(4.2) q n + 1 − 1 p n + 1 − q n − 1 p n = ( − 1 ) n q n + 1 − 1 a n + 1 q n − 1 ⋅ q n − 1 a n q n − 2 … q 3 − 1 a 3 q 1 ⋅ q 2 − 1 a 2 a 1 (4.2) q n + 1 − 1 p n + 1 − q n − 1 p n = ( − 1 ) n q n + 1 − 1 a n + 1 q n − 1 ⋅ q n − 1 a n q n − 2 … q 3 − 1 a 3 q 1 ⋅ q 2 − 1 a 2 a 1 {:(4.2)q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n)=(-1)^(n)q_(n+1)^(-1)a_(n+1)q_(n-1)*q_(n)^(-1)a_(n)q_(n-2)dotsq_(3)^(-1)a_(3)q_(1)*q_(2)^(-1)a_(2)a_(1):}\begin{equation*} q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}=(-1)^{n} q_{n+1}^{-1} a_{n+1} q_{n-1} \cdot q_{n}^{-1} a_{n} q_{n-2} \ldots q_{3}^{-1} a_{3} q_{1} \cdot q_{2}^{-1} a_{2} a_{1} \tag{4.2} \end{equation*}(4.2)qn+1−1pn+1−qn−1pn=(−1)nqn+1−1an+1qn−1⋅qn−1anqn−2…q3−1a3q1⋅q2−1a2a1
or
(4.3) q n + 1 − 1 p n + 1 − q n − 1 p n = ( − 1 ) n q n + 1 − 1 c n + 1 c n … c 3 a 2 a 1 (4.3) q n + 1 − 1 p n + 1 − q n − 1 p n = ( − 1 ) n q n + 1 − 1 c n + 1 c n … c 3 a 2 a 1 {:(4.3)q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n)=(-1)^(n)q_(n+1)^(-1)c_(n+1)c_(n)dotsc_(3)a_(2)a_(1):}\begin{equation*} q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}=(-1)^{n} q_{n+1}^{-1} c_{n+1} c_{n} \ldots c_{3} a_{2} a_{1} \tag{4.3} \end{equation*}(4.3)qn+1−1pn+1−qn−1pn=(−1)nqn+1−1cn+1cn…c3a2a1
where
(4.4) c n = a n q n − 2 q n − 1 − 1 . (4.4) c n = a n q n − 2 q n − 1 − 1 . {:(4.4)c_(n)=a_(n)q_(n-2)q_(n-1)^(-1).:}\begin{equation*} c_{n}=a_{n} q_{n-2} q_{n-1}^{-1} . \tag{4.4} \end{equation*}(4.4)cn=anqn−2qn−1−1.
Proof. By writing (4:1) for n n nnn; n − 1 , … , 1 n − 1 , … , 1 n-1,dots,1n-1, \ldots, 1n−1,…,1 and after necessary changes one obtains (4.2). Using notations (4.4), formula (4.2) becomes (4.3).
Proposition 4.4. If q n q n q_(n)q_{n}qn is invertible we have
(4.5) q n + 1 = ( e + c n + 1 ) q n (4.5) q n + 1 = e + c n + 1 q n {:(4.5)q_(n+1)=(e+c_(n+1))q_(n):}\begin{equation*} q_{n+1}=\left(e+c_{n+1}\right) q_{n} \tag{4.5} \end{equation*}(4.5)qn+1=(e+cn+1)qn
Proof. (1.3) and (4.4) imply
q n + 1 = q n + a n + 1 q n − 1 = e q n + a n + 1 q n − 1 q n − 1 q n = ( e + c n + 1 ) q n q n + 1 = q n + a n + 1 q n − 1 = e q n + a n + 1 q n − 1 q n − 1 q n = e + c n + 1 q n q_(n+1)=q_(n)+a_(n+1)q_(n-1)=eq_(n)+a_(n+1)q_(n-1)q_(n)^(-1)q_(n)=(e+c_(n+1))q_(n)q_{n+1}=q_{n}+a_{n+1} q_{n-1}=e q_{n}+a_{n+1} q_{n-1} q_{n}^{-1} q_{n}=\left(e+c_{n+1}\right) q_{n}qn+1=qn+an+1qn−1=eqn+an+1qn−1qn−1qn=(e+cn+1)qn
  1. Conditions for invertibility of elements q n q n q_(n)q_{n}qn and limitations for q n − 1 q n − 1 q_(n)^(-1)q_{n}^{-1}qn−1 and
q n + 1 − 1 p n + 1 − q n − 1 p n . q n + 1 − 1 p n + 1 − q n − 1 p n . q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n).q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n} .qn+1−1pn+1−qn−1pn.
theorem 5.1. If a n a n a_(n)a_{n}an are elements of a Banach algebra with unit e e eee, which verify the inequalities
‖ a n ‖ ⩽ 1 4 for n = 2 , 3 , 4 , … a n ⩽ 1 4  for  n = 2 , 3 , 4 , … ||a_(n)|| <= (1)/(4)" for "n=2,3,4,dots\left\|a_{n}\right\| \leqslant \frac{1}{4} \text { for } n=2,3,4, \ldots‖an‖⩽14 for n=2,3,4,…
then the norms of c n c n c_(n)c_{n}cn for the non-commutative continued fraction (1.1) satisfy to inequalities
‖ c n + 1 ‖ ⩽ 1 2 n n + 1 for n = 2 , 3 , 4 , … c n + 1 ⩽ 1 2 n n + 1  for  n = 2 , 3 , 4 , … ||c_(n+1)|| <= (1)/(2)(n)/(n+1)" for "n=2,3,4,dots\left\|c_{n+1}\right\| \leqslant \frac{1}{2} \frac{n}{n+1} \text { for } n=2,3,4, \ldots‖cn+1‖⩽12nn+1 for n=2,3,4,…
and the elements q n q n q_(n)q_{n}qn are invertible.
Proof. From proposition 2.5 (iii) q 1 = e q 1 = e q_(1)=eq_{1}=eq1=e is invertible. q 2 = e + a 2 q 2 = e + a 2 q_(2)=e+a_(2)q_{2}=e+a_{2}q2=e+a2 is also invertible because ‖ − a 2 ‖ = ‖ a 2 ‖ ⩽ 1 4 < 1 − a 2 = a 2 ⩽ 1 4 < 1 ||-a_(2)||=||a_(2)|| <= (1)/(4) < 1\left\|-a_{2}\right\|=\left\|a_{2}\right\| \leqslant \frac{1}{4}<1‖−a2‖=‖a2‖⩽14<1 (proposition 2.6).
From proposition 2.6 (ii), we have
‖ q 2 − 1 ‖ = ‖ ( e + a 2 ) − 1 ‖ ⩽ 1 1 − ‖ a 2 ‖ . q 2 − 1 = e + a 2 − 1 ⩽ 1 1 − a 2 . ||q_(2)^(-1)||=||(e+a_(2))^(-1)|| <= (1)/(1-||a_(2)||).\left\|q_{2}^{-1}\right\|=\left\|\left(e+a_{2}\right)^{-1}\right\| \leqslant \frac{1}{1-\left\|a_{2}\right\|} .‖q2−1‖=‖(e+a2)−1‖⩽11−‖a2‖.
Then
‖ c 3 ‖ = ‖ a 3 q 1 q 2 − 1 ‖ ⩽ ‖ a 3 ‖ ‖ ( e + a 2 ) − 1 ‖ ⩽ ‖ a 3 ‖ 1 − ‖ a 2 ‖ ⩽ 1 3 = 1 2 2 3 c 3 = a 3 q 1 q 2 − 1 ⩽ a 3 e + a 2 − 1 ⩽ a 3 1 − a 2 ⩽ 1 3 = 1 2 2 3 ||c_(3)||=||a_(3)q_(1)q_(2)^(-1)|| <= ||a_(3)||||(e+a_(2))^(-1)|| <= (||a_(3)||)/(1-||a_(2)||) <= (1)/(3)=(1)/(2)(2)/(3)\left\|c_{3}\right\|=\left\|a_{3} q_{1} q_{2}^{-1}\right\| \leqslant\left\|a_{3}\right\|\left\|\left(e+a_{2}\right)^{-1}\right\| \leqslant \frac{\left\|a_{3}\right\|}{1-\left\|a_{2}\right\|} \leqslant \frac{1}{3}=\frac{1}{2} \frac{2}{3}‖c3‖=‖a3q1q2−1‖⩽‖a3‖‖(e+a2)−1‖⩽‖a3‖1−‖a2‖⩽13=1223
and
q 3 = ( e + c 3 ) q 2 q 3 = e + c 3 q 2 q_(3)=(e+c_(3))q_(2)q_{\mathbf{3}}=\left(e+c_{\mathbf{3}}\right) q_{\mathbf{2}}q3=(e+c3)q2 is invertible (proposition 2.5 ( v ) 2.5 ( v ) 2.5(v)2.5(\mathrm{v})2.5(v) ) being the product of two invertible elements ( e + c 3 e + c 3 (e+c_(3):}\left(e+c_{3}\right.(e+c3 is invertible because ‖ − c 3 ‖ ⩽ 1 3 < 1 , q 2 − c 3 ⩽ 1 3 < 1 , q 2 ||-c_(3)|| <= (1)/(3) < 1,q_(2)\left\|-c_{3}\right\| \leqslant \frac{1}{3}<1, q_{2}‖−c3‖⩽13<1,q2 is also invertible by the preceding argument).
We suppose inductively that q k ( k = 1 , 2 , … , n + 1 ) q k ( k = 1 , 2 , … , n + 1 ) q_(k)(k=1,2,dots,n+1)q_{k}(k=1,2, \ldots, n+1)qk(k=1,2,…,n+1) are invertible and
‖ c k + 1 ‖ ⩽ ‖ a k + 1 ‖ 1 − ‖ a k ‖ 1 − ℓ ⋅ 1 − ‖ a a ‖ ⩽ 1 2 k k + 1 for k = 2 , 3 , … n + 1 c k + 1 ⩽ a k + 1 1 − a k 1 − ℓ ⋅ 1 − a a ⩽ 1 2 k k + 1  for  k = 2 , 3 , … n + 1 ||c_(k+1)|| <= (||a_(k+1)||)/(1-(||a_(k)||)/(1-ℓ*1-||a_(a)||)) <= (1)/(2)(k)/(k+1)" for "k=2,3,dots n+1\left\|c_{k+1}\right\| \leqslant \frac{\left\|a_{k+1}\right\|}{1-\frac{\left\|a_{k}\right\|}{1-\ell \cdot 1-\left\|a_{a}\right\|}} \leqslant \frac{1}{2} \frac{k}{k+1} \text { for } k=2,3, \ldots n+1‖ck+1‖⩽‖ak+1‖1−‖ak‖1−ℓ⋅1−‖aa‖⩽12kk+1 for k=2,3,…n+1
From notation (.4) and roposition 4.4, we obtain
‖ c n + 2 ‖ = ‖ a n + 2 q n q n + 1 − 1 ‖ ⩽ ‖ a n + 2 ‖ ‖ q n ( e + c n + 1 ) q n − 1 ‖ ⩽ ‖ a n + 2 ‖ 1 − ‖ a n + 1 ‖ 1 − ⋱ ⩽ ‖ a n + 2 ‖ ‖ ( e + c n + 1 ) − 1 ‖ ⩽ 1 − ‖ a 2 ‖ 1 − 1 c n + 2 = a n + 2 q n q n + 1 − 1 ⩽ a n + 2 q n e + c n + 1 q n − 1 ⩽ a n + 2 1 − a n + 1 1 − ⋱ ⩽ a n + 2 e + c n + 1 − 1 ⩽ 1 − a 2 1 − 1 {:||c_(n+2)||=||a_(n+2)q_(n)q_(n+1)^(-1)|| <= ||a_(n+2)||||q_(n)(e+c_(n+1))q_(n)^(-1)|| <= (||a_(n+2)||)/(1-(||a_(n+1)||)/(1-ddots)) <= ||a_(n+2)||||(e+c_(n+1))^(-1)|| <= (1-||a_(2)||)/(1-1):}\begin{gathered} \left\|c_{n+2}\right\|=\left\|a_{n+2} q_{n} q_{n+1}^{-1}\right\| \leqslant\left\|a_{n+2}\right\|\left\|q_{n}\left(e+c_{n+1}\right) q_{n}^{-1}\right\| \leqslant \frac{\left\|a_{n+2}\right\|}{1-\frac{\left\|a_{n+1}\right\|}{1-\ddots}} \leqslant\left\|a_{n+2}\right\|\left\|\left(e+c_{n+1}\right)^{-1}\right\| \leqslant \frac{1-\left\|a_{2}\right\|}{1-1} \end{gathered}‖cn+2‖=‖an+2qnqn+1−1‖⩽‖an+2‖‖qn(e+cn+1)qn−1‖⩽‖an+2‖1−‖an+1‖1−⋱⩽‖an+2‖‖(e+cn+1)−1‖⩽1−‖a2‖1−1
Then, from lemma 1.2, we hvae
‖ c n + 2 ‖ ⩽ 1 2 n + 1 n + 2 c n + 2 ⩽ 1 2 n + 1 n + 2 ||c_(n+2)|| <= (1)/(2)(n+1)/(n+2)\left\|c_{n+2}\right\| \leqslant \frac{1}{2} \frac{n+1}{n+2}‖cn+2‖⩽12n+1n+2
and q n + 2 = ( e + c n + 2 ) q n + 1 q n + 2 = e + c n + 2 q n + 1 q_(n+2)=(e+c_(n+2))q_(n+1)q_{n+2}=\left(e+c_{n+2}\right) q_{n+1}qn+2=(e+cn+2)qn+1 is invertible, because it is product of two invertible elements ( e + c n + 2 e + c n + 2 e+c_(n+2)e+c_{n+2}e+cn+2 is invertible since ‖ − c n + 2 ‖ = ‖ c n + 2 ‖ < 1 − c n + 2 = c n + 2 < 1 ||-c_(n+2)||=||c_(n+2)|| < 1\left\|-c_{n+2}\right\|=\left\|c_{n+2}\right\|<1‖−cn+2‖=‖cn+2‖<1 and q n + 1 q n + 1 q_(n+1)q_{n+1}qn+1 is invertible by the hypothesis).
theorem 5.2. If a a aaa are elements of a Banach algebra a a aaa with unit e e eee, vich verify inequalities
‖ a n ‖ ⩽ 1 4 for n = 2 , 3 , 4 , … a n ⩽ 1 4  for  n = 2 , 3 , 4 , … ||a_(n)|| <= (1)/(4)" for "n=2,3,4,dots\left\|a_{n}\right\| \leqslant \frac{1}{4} \text { for } n=2,3,4, \ldots‖an‖⩽14 for n=2,3,4,…
then
(5.1) ‖ q n − 1 ‖ ⩽ 2 n n + 1 for n = 1 , 2 , 3 , … (5.1) q n − 1 ⩽ 2 n n + 1  for  n = 1 , 2 , 3 , … {:(5.1)||q_(n)^(-1)|| <= (2^(n))/(n+1)" for "n=1","2","3","dots:}\begin{equation*} \left\|q_{n}^{-1}\right\| \leqslant \frac{2^{n}}{n+1} \text { for } n=1,2,3, \ldots \tag{5.1} \end{equation*}(5.1)‖qn−1‖⩽2nn+1 for n=1,2,3,…
Proof. The invertibility of the elements q n q n q_(n)q_{n}qn is a consequence of theorem 5.1.
From formula (4.5), we have
‖ q n − 1 ‖ = ‖ q n − 1 − 1 ( e + c n ) − 1 ‖ ⩽ ‖ q n − 1 − 1 ‖ 1 − ‖ c n ‖ ⩽ ‖ q n − 1 − 1 ‖ 1 − n − 1 2 n q n − 1 = q n − 1 − 1 e + c n − 1 ⩽ q n − 1 − 1 1 − c n ⩽ q n − 1 − 1 1 − n − 1 2 n ||q_(n)^(-1)||=||q_(n-1)^(-1)(e+c_(n))^(-1)|| <= (||q_(n-1)^(-1)||)/(1-||c_(n)||) <= (||q_(n-1)^(-1)||)/(1-(n-1)/(2n))\left\|q_{n}^{-1}\right\|=\left\|q_{n-1}^{-1}\left(e+c_{n}\right)^{-1}\right\| \leqslant \frac{\left\|q_{n-1}^{-1}\right\|}{1-\left\|c_{n}\right\|} \leqslant \frac{\left\|q_{n-1}^{-1}\right\|}{1-\frac{n-1}{2 n}}‖qn−1‖=‖qn−1−1(e+cn)−1‖⩽‖qn−1−1‖1−‖cn‖⩽‖qn−1−1‖1−n−12n
that is
‖ q n − 1 ‖ ⩽ 2 n n + 1 ‖ q n − 1 − 1 ‖ q n − 1 ⩽ 2 n n + 1 q n − 1 − 1 ||q_(n)^(-1)|| <= (2n)/(n+1)||q_(n-1)^(-1)||\left\|q_{n}^{-1}\right\| \leqslant \frac{2 n}{n+1}\left\|q_{n-1}^{-1}\right\|‖qn−1‖⩽2nn+1‖qn−1−1‖
By writing the inequalities for n − 1 , n − 2 , … , 2 n − 1 , n − 2 , … , 2 n-1,n-2,dots,2n-1, n-2, \ldots, 2n−1,n−2,…,2 and after their multiplication, we obtain
‖ q n − 1 ‖ ⩽ 2 n n + 1 ⋅ 2 ( n − 1 ) n ⋯ 4 3 q n − 1 ⩽ 2 n n + 1 ⋅ 2 ( n − 1 ) n ⋯ 4 3 ||q_(n)^(-1)|| <= (2n)/(n+1)*(2(n-1))/(n)cdots(4)/(3)\left\|q_{n}^{-1}\right\| \leqslant \frac{2 n}{n+1} \cdot \frac{2(n-1)}{n} \cdots \frac{4}{3}‖qn−1‖⩽2nn+1⋅2(n−1)n⋯43
that is
‖ q n − 1 ‖ ⩽ 2 n n + 1 q n − 1 ⩽ 2 n n + 1 ||q_(n)^(-1)|| <= (2^(n))/(n+1)\left\|q_{n}^{-1}\right\| \leqslant \frac{2^{n}}{n+1}‖qn−1‖⩽2nn+1
THEOREM 5.3. If a n a n a_(n)a_{n}an are elements of a Banach algebra o with unit e, wich verify the inequalities
‖ a n ‖ ⩽ 1 4 for n = 2 , 3 , 4 , … a n ⩽ 1 4  for  n = 2 , 3 , 4 , … ||a_(n)|| <= (1)/(4)" for "n=2,3,4,dots\left\|a_{n}\right\| \leqslant \frac{1}{4} \text { for } n=2,3,4, \ldots‖an‖⩽14 for n=2,3,4,…
then for non-commutative continued fraction (3.1), we have
‖ q n + 1 − 1 p n + 1 − q n − 1 p n ‖ ⩽ 2 ‖ a 1 ‖ ( n + 1 ) ( n + 2 ) for n = 1 , 2 , 3 , … q n + 1 − 1 p n + 1 − q n − 1 p n ⩽ 2 a 1 ( n + 1 ) ( n + 2 )  for  n = 1 , 2 , 3 , … ||q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n)|| <= (2||a_(1)||)/((n+1)(n+2))" for "n=1,2,3,dots\left\|q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}\right\| \leqslant \frac{2\left\|a_{1}\right\|}{(n+1)(n+2)} \text { for } n=1,2,3, \ldots‖qn+1−1pn+1−qn−1pn‖⩽2‖a1‖(n+1)(n+2) for n=1,2,3,…
Proof. The invertibility of q n q n q_(n)q_{n}qn follows from thorem 5.1. The relations (4.2) and (2.1) imply the inequality
‖ q n + 1 − 1 p n + 1 − q n − 1 p n ‖ ⩽ ‖ q n + 1 − 1 ‖ ‖ c n + 1 ‖ ‖ c n ‖ … ‖ c a ‖ ‖ a 2 ‖ ‖ a 1 ‖ q n + 1 − 1 p n + 1 − q n − 1 p n ⩽ q n + 1 − 1 c n + 1 c n … c a a 2 a 1 ||q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n)|| <= ||q_(n+1)^(-1)||||c_(n+1)||||c_(n)||dots||c_(a)||||a_(2)||||a_(1)||\left\|q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}\right\| \leqslant\left\|q_{n+1}^{-1}\right\|\left\|c_{n+1}\right\|\left\|c_{n}\right\| \ldots\left\|c_{a}\right\|\left\|a_{2}\right\|\left\|a_{1}\right\|‖qn+1−1pn+1−qn−1pn‖⩽‖qn+1−1‖‖cn+1‖‖cn‖…‖ca‖‖a2‖‖a1‖
Using theorems 5.2 and 5.1, we obtain
‖ q n + 1 − 1 p n + 1 − q n − 1 p n ‖ ⩽ 2 n + 1 n + 2 1 2 n n + 1 ⋅ 1 2 n − 1 n ⋯ 1 2 2 3 ⋅ 1 4 ‖ a 1 ‖ q n + 1 − 1 p n + 1 − q n − 1 p n ⩽ 2 n + 1 n + 2 1 2 n n + 1 ⋅ 1 2 n − 1 n ⋯ 1 2 2 3 ⋅ 1 4 a 1 ||q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n)|| <= (2^(n+1))/(n+2)(1)/(2)(n)/(n+1)*(1)/(2)(n-1)/(n)cdots(1)/(2)(2)/(3)*(1)/(4)||a_(1)||\left\|q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}\right\| \leqslant \frac{2^{n+1}}{n+2} \frac{1}{2} \frac{n}{n+1} \cdot \frac{1}{2} \frac{n-1}{n} \cdots \frac{1}{2} \frac{2}{3} \cdot \frac{1}{4}\left\|a_{1}\right\|‖qn+1−1pn+1−qn−1pn‖⩽2n+1n+212nn+1⋅12n−1n⋯1223⋅14‖a1‖
or
‖ q n + 1 − 1 p n − 1 − q n − 1 p n ‖ ⩽ 2 ‖ a 1 ‖ ( n + 1 ) ( n + 2 ) q n + 1 − 1 p n − 1 − q n − 1 p n ⩽ 2 a 1 ( n + 1 ) ( n + 2 ) ||q_(n+1)^(-1)p_(n-1)-q_(n)^(-1)p_(n)|| <= (2||a_(1)||)/((n+1)(n+2))\left\|q_{n+1}^{-1} p_{n-1}-q_{n}^{-1} p_{n}\right\| \leqslant \frac{2\left\|a_{1}\right\|}{(n+1)(n+2)}‖qn+1−1pn−1−qn−1pn‖⩽2‖a1‖(n+1)(n+2)

6. Generalized Worpitzky's theorem and some consequences

theorem 6.1. If a n a n a_(n)a_{n}an are elements of a Banach algebra a with unit e, wich satisfy the inequalities
‖ a n ‖ ⩽ 1 4 for n = 2 , 3 , 4 , … a n ⩽ 1 4  for  n = 2 , 3 , 4 , … ||a_(n)|| <= (1)/(4)" for "n=2,3,4,dots\left\|a_{n}\right\| \leqslant \frac{1}{4} \text { for } n=2,3,4, \ldots‖an‖⩽14 for n=2,3,4,…
then
(i) the non-commutative continued fraction (3.1) converges uniformly to an element x ∈ Q x ∈ Q x inQx \in \mathcal{Q}x∈Q,
(ii) the values of non-commutative continued fraction (3.1) and of its convergents are in the set defined by inequality
‖ a 1 − x ‖ ⩽ 1 2 ‖ x ‖ , a 1 − x ⩽ 1 2 ‖ x ‖ , ||a_(1)-x|| <= (1)/(2)||x||,\left\|a_{1}-x\right\| \leqslant \frac{1}{2}\|x\|,‖a1−x‖⩽12‖x‖,
where a 1 a 1 a_(1)a_{1}a1 is a fixed element from A A A\mathfrak{A}A,
(iii) 1 4 1 4 (1)/(4)\frac{1}{4}14 is the ",best" constant c > 0 c > 0 c > 0c>0c>0 such that the non-commutative continued fraction (3.1) converges for
‖ a n ‖ ⩽ c ( n = 2 , 3 , … ) a n ⩽ c ( n = 2 , 3 , … ) ||a_(n)|| <= c(n=2,3,dots)\left\|a_{n}\right\| \leqslant c(n=2,3, \ldots)‖an‖⩽c(n=2,3,…)
Proof. (i) In view of the identity
q n + 1 − 1 p n + k − q n − 1 p n = ( q n + k − 1 p n + k − q n + k − 1 − 1 p n + k − 1 ) + + ( q n + k − 1 − 1 p n + k − 1 − q n + k − 2 − 1 p n + k − 2 ) + … + ( q n + 1 − 1 p n + 1 − q n − 1 p n ) q n + 1 − 1 p n + k − q n − 1 p n = q n + k − 1 p n + k − q n + k − 1 − 1 p n + k − 1 + + q n + k − 1 − 1 p n + k − 1 − q n + k − 2 − 1 p n + k − 2 + … + q n + 1 − 1 p n + 1 − q n − 1 p n {:[q_(n+1)^(-1)p_(n+k)-q_(n)^(-1)p_(n)=(q_(n+k)^(-1)p_(n+k)-q_(n+k-1)^(-1)p_(n+k-1))+],[+(q_(n+k-1)^(-1)p_(n+k-1)-q_(n+k-2)^(-1)p_(n+k-2))+dots+(q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n))]:}\begin{gathered} q_{n+1}^{-1} p_{n+k}-q_{n}^{-1} p_{n}=\left(q_{n+k}^{-1} p_{n+k}-q_{n+k-1}^{-1} p_{n+k-1}\right)+ \\ +\left(q_{n+k-1}^{-1} p_{n+k-1}-q_{n+k-2}^{-1} p_{n+k-2}\right)+\ldots+\left(q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}\right) \end{gathered}qn+1−1pn+k−qn−1pn=(qn+k−1pn+k−qn+k−1−1pn+k−1)++(qn+k−1−1pn+k−1−qn+k−2−1pn+k−2)+…+(qn+1−1pn+1−qn−1pn)
the subadditivity of the norm and theorem 5.3, we can write
‖ q n + k − 1 p n + k − q n − 1 p n ‖ ⩽ 2 [ 1 ( n + 1 ) ( n + 2 ) + 1 ( n + 2 ) ( n + 3 ) + ⋯ + + 1 ( n + k ) ( n + k + 1 ) ] ‖ a 1 ‖ . q n + k − 1 p n + k − q n − 1 p n ⩽ 2 1 ( n + 1 ) ( n + 2 ) + 1 ( n + 2 ) ( n + 3 ) + ⋯ + + 1 ( n + k ) ( n + k + 1 ) a 1 . {:[||q_(n+k)^(-1)p_(n+k)-q_(n)^(-1)p_(n)|| <= 2[(1)/((n+1)(n+2))+(1)/((n+2)(n+3))+cdots+:}],[{:+(1)/((n+k)(n+k+1))]||a_(1)||.]:}\begin{aligned} \left\|q_{n+k}^{-1} p_{n+k}-q_{n}^{-1} p_{n}\right\| & \leqslant 2\left[\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\cdots+\right. \\ & \left.+\frac{1}{(n+k)(n+k+1)}\right]\left\|a_{1}\right\| . \end{aligned}‖qn+k−1pn+k−qn−1pn‖⩽2[1(n+1)(n+2)+1(n+2)(n+3)+⋯++1(n+k)(n+k+1)]‖a1‖.
One also see that we have the inequality
(6.1) ‖ q n + k − 1 p n + k − q n − 1 p n ‖ ⩽ 2 ‖ a 1 ‖ ( 1 n + 1 − 1 n + k + 1 ) = 2 k ‖ a 1 ‖ ( n + 1 ) ( n + k + 1 ) (6.1) q n + k − 1 p n + k − q n − 1 p n ⩽ 2 a 1 1 n + 1 − 1 n + k + 1 = 2 k a 1 ( n + 1 ) ( n + k + 1 ) {:(6.1)||q_(n+k)^(-1)p_(n+k)-q_(n)^(-1)p_(n)|| <= 2||a_(1)||((1)/(n+1)-(1)/(n+k+1))=(2k||a_(1)||)/((n+1)(n+k+1)):}\begin{equation*} \left\|q_{n+k}^{-1} p_{n+k}-q_{n}^{-1} p_{n}\right\| \leqslant 2\left\|a_{1}\right\|\left(\frac{1}{n+1}-\frac{1}{n+k+1}\right)=\frac{2 k\left\|a_{1}\right\|}{(n+1)(n+k+1)} \tag{6.1} \end{equation*}(6.1)‖qn+k−1pn+k−qn−1pn‖⩽2‖a1‖(1n+1−1n+k+1)=2k‖a1‖(n+1)(n+k+1)
Therefore q n + k − 1 p n + k − q n − 1 p n ∈ A q n + k − 1 p n + k − q n − 1 p n ∈ A q_(n+k)^(-1)p_(n+k)-q_(n)^(-1)p_(n)inAq_{n+k}^{-1} p_{n+k}-q_{n}^{-1} p_{n} \in \mathfrak{A}qn+k−1pn+k−qn−1pn∈A tends in norm to 0 , for every natural number k k kkk, when n n nnn tends to ∞ ∞ oo\infty∞, such that { q n − 1 p n } q n − 1 p n {q_(n)^(-1)p_(n)}\left\{q_{n}^{-1} p_{n}\right\}{qn−1pn} is a Cauchy sequence. Banach algebra θ θ theta\thetaθ with unit e being complete with respect the metric topology induced by the algebra norm, it follows that sequence { q n − 1 p n } q n − 1 p n {q_(n)^(-1)p_(n)}\left\{q_{n}^{-1} p_{n}\right\}{qn−1pn} converges to an element x ∈ A x ∈ A x inAx \in \mathcal{A}x∈A.
(ii) First we shall prove that
‖ x ‖ ⩽ 2 ‖ a 1 ‖ and ‖ q n − 1 p n ‖ ⩽ 2 ‖ a 1 ‖ . ‖ x ‖ ⩽ 2 a 1  and  q n − 1 p n ⩽ 2 a 1 . ||x|| <= 2||a_(1)||quad" and "quad||q_(n)^(-1)p_(n)|| <= 2||a_(1)||.\|x\| \leqslant 2\left\|a_{1}\right\| \quad \text { and } \quad\left\|q_{n}^{-1} p_{n}\right\| \leqslant 2\left\|a_{1}\right\| .‖x‖⩽2‖a1‖ and ‖qn−1pn‖⩽2‖a1‖.
Let us remark that
x = q 1 − 1 p 1 + ( q 2 − 2 p 2 − q 1 − 1 p 1 ) + ( q 3 − 1 p 3 − q 2 − 1 p 2 ) + … x = q 1 − 1 p 1 + q 2 − 2 p 2 − q 1 − 1 p 1 + q 3 − 1 p 3 − q 2 − 1 p 2 + … x=q_(1)^(-1)p_(1)+(q_(2)^(-2)p_(2)-q_(1)^(-1)p_(1))+(q_(3)^(-1)p_(3)-q_(2)^(-1)p_(2))+dotsx=q_{1}^{-1} p_{1}+\left(q_{2}^{-2} p_{2}-q_{1}^{-1} p_{1}\right)+\left(q_{3}^{-1} p_{3}-q_{2}^{-1} p_{2}\right)+\ldotsx=q1−1p1+(q2−2p2−q1−1p1)+(q3−1p3−q2−1p2)+…
and
q n − p n = q 1 − p 1 + ( q 2 − 1 p 2 − q 1 − 1 p 1 ) + ( q 3 − p 3 − q 2 − 1 p 2 ) + … + ( q n − 1 p n − q n − 1 − 1 p n − 1 ) . q n − p n = q 1 − p 1 + q 2 − 1 p 2 − q 1 − 1 p 1 + q 3 − p 3 − q 2 − 1 p 2 + … + q n − 1 p n − q n − 1 − 1 p n − 1 . {:[q_(n)^(-)p_(n)=q_(1)^(-)p_(1)+(q_(2)^(-1)p_(2)-q_(1)^(-1)p_(1))+(q_(3)^(-)p_(3)-q_(2)^(-1)p_(2))+dots],[+(q_(n)^(-1)p_(n)-q_(n-1)^(-1)p_(n-1)).]:}\begin{gathered} q_{n}^{-} p_{n}=q_{1}^{-} p_{1}+\left(q_{2}^{-1} p_{2}-q_{1}^{-1} p_{1}\right)+\left(q_{3}^{-} p_{3}-q_{2}^{-1} p_{2}\right)+\ldots \\ +\left(q_{n}^{-1} p_{n}-q_{n-1}^{-1} p_{n-1}\right) . \end{gathered}qn−pn=q1−p1+(q2−1p2−q1−1p1)+(q3−p3−q2−1p2)+…+(qn−1pn−qn−1−1pn−1).
By this remark and theorem 5.3, we obtain
‖ x ‖ ‖ x ‖ ||x||\|x\|‖x‖
or
‖ x ‖ q n − 1 p n ‖ ⩽ { 1 + 2 [ ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ⋯ ] } ‖ a 1 ‖ . ‖ x ‖ q n − 1 p n ⩽ 1 + 2 1 2 − 1 3 + 1 3 − 1 4 + ⋯ a 1 ‖ . ||x||q_(n)^(-1)p_(n)|| <= {1+2[((1)/(2)-(1)/(3))+((1)/(3)-(1)/(4))+cdots]}||a_(1)||.\|x\| q_{n}^{-1} p_{n}\left\|\leqslant\left\{1+2\left[\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots\right]\right\}\right\| a_{1} \| .‖x‖qn−1pn‖⩽{1+2[(12−13)+(13−14)+⋯]}‖a1‖.
Therefore, we proved
‖ x ‖ ⩽ 2 ‖ a 1 ‖ and also ‖ q n − 1 p n ‖ ⩽ 2 ‖ a 1 ‖ . ‖ x ‖ ⩽ 2 a 1  and also  q n − 1 p n ⩽ 2 a 1 . ||x|| <= 2||a_(1)||quad" and also "quad||q_(n)^(-1)p_(n)|| <= 2||a_(1)||.\|x\| \leqslant 2\left\|a_{1}\right\| \quad \text { and also } \quad\left\|q_{n}^{-1} p_{n}\right\| \leqslant 2\left\|a_{1}\right\| .‖x‖⩽2‖a1‖ and also ‖qn−1pn‖⩽2‖a1‖.
Now in order to prove (ii), we remark that we can write every convergent of the non-commutative continued fraction (3.1) and its value, denoted by x x xxx, as follows
(6.2) x = a 1 ( e + 1 4 y ) − 1 (6.2) x = a 1 e + 1 4 y − 1 {:(6.2)x=a_(1)(e+(1)/(4)y)^(-1):}\begin{equation*} x=a_{1}\left(e+\frac{1}{4} y\right)^{-1} \tag{6.2} \end{equation*}(6.2)x=a1(e+14y)−1
with
(6.3) y = y 1 e + a 2 e + ⋯ and ‖ y 1 ‖ ⩽ 1 . (6.3) y = y 1 e + a 2 e + ⋯  and  y 1 ⩽ 1 . {:(6.3)y=(y_(1))/(e+)(a_(2))/(e+)cdots" and "||y_(1)|| <= 1.:}\begin{equation*} y=\frac{y_{1}}{e+} \frac{a_{2}}{e+} \cdots \text { and }\left\|y_{1}\right\| \leqslant 1 . \tag{6.3} \end{equation*}(6.3)y=y1e+a2e+⋯ and ‖y1‖⩽1.
Hence
x ( e + 1 4 y ) = a 1 x e + 1 4 y = a 1 x(e+(1)/(4)y)=a_(1)x\left(e+\frac{1}{4} y\right)=a_{1}x(e+14y)=a1
or, according to (2.1),
‖ a 1 − x ‖ ⩽ 1 4 ‖ x ‖ ‖ y ‖ a 1 − x ⩽ 1 4 ‖ x ‖ ‖ y ‖ ||a_(1)-x|| <= (1)/(4)||x||||y||\left\|a_{1}-x\right\| \leqslant \frac{1}{4}\|x\|\|y\|‖a1−x‖⩽14‖x‖‖y‖
From the first part of the proof, it follows ‖ y ‖ ⩽ 2 ‖ y ‖ ⩽ 2 ||y|| <= 2\|y\| \leqslant 2‖y‖⩽2 and therefore, we have
(6.4) ‖ a 1 − x ‖ ⩽ 1 2 ‖ x ‖ (6.4) a 1 − x ⩽ 1 2 ‖ x ‖ {:(6.4)||a_(1)-x|| <= (1)/(2)||x||:}\begin{equation*} \left\|a_{1}-x\right\| \leqslant \frac{1}{2}\|x\| \tag{6.4} \end{equation*}(6.4)‖a1−x‖⩽12‖x‖
(iii) In order to prove that 1 4 1 4 (1)/(4)\frac{1}{4}14 is the „best" constant c > 0 c > 0 c > 0c>0c>0 such that the non-commutative continued fraction (3.1) converges for ‖ a n ‖ ⩽ c ( n = 2 , 3 , … ) a n ⩽ c ( n = 2 , 3 , … ) ||a_(n)|| <= c(n=2,3,dots)\left\|a_{n}\right\| \leqslant c (n=2,3, \ldots)‖an‖⩽c(n=2,3,…), it is sufficient to observe that in the Banach subalgebra A 1 ⊂ Q A 1 ⊂ Q A_(1)subQ\mathfrak{A}_{1} \subset \mathfrak{Q}A1⊂Q of the real numbers multiplied by e ∈ A e ∈ A e inAe \in \mathfrak{A}e∈A, the continued fractions with a n = a e a n = a e a_(n)=aea_{n}=a ean=ae and a > 1 / 4 a > 1 / 4 a > 1//4a>1 / 4a>1/4 diverge according to lemma 1.3.
Coro11ary 6.1. (Approximation theorem) In the conditions of theorem 6.1, we have
‖ x − q n − 1 p n ‖ ⩽ 2 ‖ a 1 ‖ n + 1 x − q n − 1 p n ⩽ 2 a 1 n + 1 ||x-q_(n)^(-1)p_(n)|| <= (2||a_(1)||)/(n+1)\left\|x-q_{n}^{-1} p_{n}\right\| \leqslant \frac{2\left\|a_{1}\right\|}{n+1}‖x−qn−1pn‖⩽2‖a1‖n+1
Proof. From
x − q n − 1 p n = ( q n + 1 − 1 p n + 1 − q n − 1 p n ) + ( q n + 2 − 1 p n + 2 − q n + 1 − 1 p n + 1 ) + ⋯ x − q n − 1 p n = q n + 1 − 1 p n + 1 − q n − 1 p n + q n + 2 − 1 p n + 2 − q n + 1 − 1 p n + 1 + ⋯ x-q_(n)^(-1)p_(n)=(q_(n+1)^(-1)p_(n+1)-q_(n)^(-1)p_(n))+(q_(n+2)^(-1)p_(n+2)-q_(n+1)^(-1)p_(n+1))+cdotsx-q_{n}^{-1} p_{n}=\left(q_{n+1}^{-1} p_{n+1}-q_{n}^{-1} p_{n}\right)+\left(q_{n+2}^{-1} p_{n+2}-q_{n+1}^{-1} p_{n+1}\right)+\cdotsx−qn−1pn=(qn+1−1pn+1−qn−1pn)+(qn+2−1pn+2−qn+1−1pn+1)+⋯
and theorem 5.3, we obtain
‖ x − q n − 1 p n ‖ ⩽ 2 ‖ a 1 ‖ [ ( 1 n + 1 − 1 n + 2 ) + ( 1 n + 2 − 1 n + 3 ) + … ] = 2 ‖ a 1 ‖ n + 1 . x − q n − 1 p n ⩽ 2 a 1 1 n + 1 − 1 n + 2 + 1 n + 2 − 1 n + 3 + … = 2 a 1 n + 1 . ||x-q_(n)^(-1)p_(n)|| <= 2||a_(1)||[((1)/(n+1)-(1)/(n+2))+((1)/(n+2)-(1)/(n+3))+dots]=(2||a_(1)||)/(n+1).\left\|x-q_{n}^{-1} p_{n}\right\| \leqslant 2\left\|a_{1}\right\|\left[\left(\frac{1}{n+1}-\frac{1}{n+2}\right)+\left(\frac{1}{n+2}-\frac{1}{n+3}\right)+\ldots\right]=\frac{2\left\|a_{1}\right\|}{n+1} .‖x−qn−1pn‖⩽2‖a1‖[(1n+1−1n+2)+(1n+2−1n+3)+…]=2‖a1‖n+1.
Corollary 6.2. If a n a n a_(n)a_{n}an are some elements of a Banach algebra with unit e, which satisfy the inequalities
‖ a n ‖ ⩽ M for n = 2 , 3 , … a n ⩽ M  for  n = 2 , 3 , … ||a_(n)|| <= M" for "n=2,3,dots\left\|a_{n}\right\| \leqslant M \text { for } n=2,3, \ldots‖an‖⩽M for n=2,3,…
and x x xxx is variable element of a a aaa, then the non-commutative continued fraction.
a 1 e + a 2 x e + a 3 x e + ⋯ ⋅ a n x e + a 1 e + a 2 x e + a 3 x e + ⋯ ⋅ a n x e + (a_(1))/(e+)(a_(2)x)/(e+)(a_(3)x)/(e+)cdots*(a_(n)x)/(e+)\frac{a_{1}}{e+} \frac{a_{2} x}{e+} \frac{a_{3} x}{e+} \cdots \cdot \frac{a_{n} x}{e+}a1e+a2xe+a3xe+⋯⋅anxe+
converges uniformly for ‖ x ‖ ⩽ 1 4 M ‖ x ‖ ⩽ 1 4 M ||x|| <= (1)/(4M)\|x\| \leqslant \frac{1}{4 M}‖x‖⩽14M.
Proof. We are in the conditions of theorem 6.1. and we have
‖ a n x ‖ ⩽ ‖ a n ‖ ‖ x ‖ ⩽ M ⋅ 1 4 M = 1 4 for n = 2 , 3 , 4 , … a n x ⩽ a n ‖ x ‖ ⩽ M ⋅ 1 4 M = 1 4  for  n = 2 , 3 , 4 , … ||a_(n)x|| <= ||a_(n)||||x|| <= M*(1)/(4M)=(1)/(4)" for "n=2,3,4,dots\left\|a_{n} x\right\| \leqslant\left\|a_{n}\right\|\|x\| \leqslant M \cdot \frac{1}{4 M}=\frac{1}{4} \text { for } n=2,3,4, \ldots‖anx‖⩽‖an‖‖x‖⩽M⋅14M=14 for n=2,3,4,…
Remark 6.1. Theorem 6.1. (i) is evidently true also in the case in which a 1 , a 2 , … , a n a 1 , a 2 , … , a n a_(1),a_(2),dots,a_(n)a_{1}, a_{2}, \ldots, a_{n}a1,a2,…,an of A A A\mathfrak{A}A are fixed, a n + j ∈ A a n + j ∈ A a_(n+j)inAa_{n+j} \in \mathfrak{A}an+j∈A with ‖ a n − j ‖ ⩽ 1 4 ( j = a n − j ⩽ 1 4 ( j = ||a_(n-j)|| <= (1)/(4)(j=\left\|a_{n-j}\right\| \leqslant \frac{1}{4}(j=‖an−j‖⩽14(j=
= 1 , 2 , … ) = 1 , 2 , … ) =1,2,dots)=1,2, \ldots)=1,2,…). This problem, in a particular case, is solved in proposition 7.2.
Remark 6.2. If in the formulas (3.3) we permute a n : − 1 a n : − 1 a_(n:-1)a_{n:-1}an:−1 with p n − 1 p n − 1 p_(n-1)p_{n-1}pn−1 in the first relation and a n + 1 a n + 1 a_(n+1)a_{n+1}an+1 with q n − 1 q n − 1 q_(n-1)q_{n-1}qn−1 in the second, then we have to permute q n − 1 q n − 1 q_(n)^(-1)q_{n}^{-1}qn−1 with p n p n p_(n)p_{n}pn in (3.2), in order that the methode be still applicable.

7. Particular cases of theorem 6.1

Theorem 6.1 is evidently true also in the case in which the Banach algebra A A A\mathscr{A}A with unit e e eee is commutative.
(i) a = R a = R a=Ra=\mathbf{R}a=R with e = 1 e = 1 e=1e=1e=1 and the norm ‖ ⋅ ‖ = | ⋅ | ‖ ⋅ ‖ = | ⋅ | ||*||=|*|\|\cdot\|=|\cdot|‖⋅‖=|⋅| is a commutative Banach algebra. In this case, according to theorem 6.1, the real continued fraction
(7.1) a 1 1 + a 2 1 + a 3 1 + ( a n ∈ R ) (7.1) a 1 1 + a 2 1 + a 3 1 + a n ∈ R {:(7.1)(a_(1))/(1+(a_(2))/(1+(a_(3))/(1+)))quad(a_(n)inR):}\begin{equation*} \frac{a_{1}}{1+\frac{a_{2}}{1+\frac{a_{3}}{1+}}} \quad\left(a_{n} \in \mathbf{R}\right) \tag{7.1} \end{equation*}(7.1)a11+a21+a31+(an∈R)
converges and the convergents and its value is in the interval
[ 2 | a 1 | 3 , 2 | a 1 | ] if | a n | ⩽ 1 4 ( n = 2 , 3 , … ) . 2 a 1 3 , 2 a 1  if  a n ⩽ 1 4 ( n = 2 , 3 , … ) . [(2|a_(1)|)/(3),2|a_(1)|]" if "|a_(n)| <= (1)/(4)quad(n=2,3,dots).\left[\frac{2\left|a_{1}\right|}{3}, 2\left|a_{1}\right|\right] \text { if }\left|a_{n}\right| \leqslant \frac{1}{4} \quad(n=2,3, \ldots) .[2|a1|3,2|a1|] if |an|⩽14(n=2,3,…).
(ii) a = C a = C a=C\mathfrak{a}=Ca=C, with e = 1 e = 1 e=1e=1e=1 and ‖ ⋅ ‖ = | ⋅ | ‖ ⋅ ‖ = | ⋅ | ||*||=|*|\|\cdot\|=|\cdot|‖⋅‖=|⋅|, is a commutative algebra. Then complex continued fraction (7.1) ( a n ∈ C ) a n ∈ C (a_(n)in C)\left(a_{n} \in C\right)(an∈C) converges to the value z z zzz from the disk
D 1 = { z ∈ C : | z − 4 3 a 1 | ⩽ 2 3 | a 1 | } D 1 = z ∈ C : z − 4 3 a 1 ⩽ 2 3 a 1 D_(1)={z in C:|z-(4)/(3)a_(1)| <= (2)/(3)|a_(1)|}D_{1}=\left\{z \in C:\left|z-\frac{4}{3} a_{1}\right| \leqslant \frac{2}{3}\left|a_{1}\right|\right\}D1={z∈C:|z−43a1|⩽23|a1|}
if | a n | ⩽ 1 / 4 ( n = 2 , 3 , … ) a n ⩽ 1 / 4 ( n = 2 , 3 , … ) |a_(n)| <= 1//4(n=2,3,dots)\left|a_{n}\right| \leqslant 1 / 4(n=2,3, \ldots)|an|⩽1/4(n=2,3,…).
In this case, continued fraction (7.1) can be considered as a complex function f f fff of complex variables a 2 , a 3 , … , ( a 1 a 2 , a 3 , … , a 1 a_(2),a_(3),dots,(a_(1):}a_{2}, a_{3}, \ldots,\left(a_{1}\right.a2,a3,…,(a1 is fixed) and
(7.2) z = f ( a 1 , a 2 , a 3 , … , a n , … ) (7.2) z = f a 1 , a 2 , a 3 , … , a n , … {:(7.2)z=f(a_(1),a_(2),a_(3),dots,a_(n),dots):}\begin{equation*} z=f\left(a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\right) \tag{7.2} \end{equation*}(7.2)z=f(a1,a2,a3,…,an,…)
Proposition 7.1. The values of the function f f fff fill the closed disk D 1 D 1 D_(1)\mathscr{D}_{1}D1 when every a n ( n = 2 , 3 , … ) a n ( n = 2 , 3 , … ) a_(n)(n=2,3,dots)a_{n}(n=2,3, \ldots)an(n=2,3,…) runs over closed disk D 0 = { z ∈ C : | z | ⩽ 1 4 } D 0 = z ∈ C : | z | ⩽ 1 4 D_(0)={z in C:|z| <= (1)/(4)}\mathscr{D}_{0}=\left\{z \in C:|z| \leqslant \frac{1}{4}\right\}D0={z∈C:|z|⩽14}.
Proof. Let be
(7.2) z = a 1 1 + 1 − a 2 1 − 1 − 1 1 4 1 − (7.2) z = a 1 1 + 1 − a 2 1 − 1 − 1 1 4 1 − {:(7.2)z=(a_(1))/(1+1-(a_(2))/(1-1-1))((1)/(4))/(1-):}\begin{equation*} z=\frac{a_{1}}{1+1-\frac{a_{2}}{1-1-1}} \frac{\frac{1}{4}}{1-} \tag{7.2} \end{equation*}(7.2)z=a11+1−a21−1−1141−
where a 1 ∈ C a 1 ∈ C a_(1)in Ca_{1} \in Ca1∈C is fixed and a 2 ∈ C a 2 ∈ C a_(2)in Ca_{2} \in Ca2∈C is a complex variable. It is easy to see that
z = a 1 2 a 2 + 1 z = a 1 2 a 2 + 1 z=(a_(1))/(2a_(2)+1)z=\frac{a_{1}}{2 a_{2}+1}z=a12a2+1
which is linear and this function transforms the circle | z | = 1 4 | z | = 1 4 |z|=(1)/(4)|z|=\frac{1}{4}|z|=14 in circle | z − 4 3 a 1 | = 2 3 | a 1 | z − 4 3 a 1 = 2 3 a 1 |z-(4)/(3)a_(1)|=(2)/(3)|a_(1)|\left|z-\frac{4}{3} a_{1}\right|=\frac{2}{3}\left|a_{1}\right||z−43a1|=23|a1|. Therefore, this function carries the disk D 0 D 0 D_(0)\mathscr{D}_{0}D0 on to the disk D 1 D 1 D_(1)\mathscr{D}_{1}D1.
Proposition 7.2. If A = C , e = 1 , ‖ ⋅ ‖ = | ⋅ | , a 1 , a 2 , … , a n A = C , e = 1 , ‖ ⋅ ‖ = | ⋅ | , a 1 , a 2 , … , a n A=C,e=1,||*||=|*|,a_(1),a_(2),dots,a_(n)\mathfrak{A}=C, e=1,\|\cdot\|=|\cdot|, a_{1}, a_{2}, \ldots, a_{n}A=C,e=1,‖⋅‖=|⋅|,a1,a2,…,an are fixed,
| a n − j | ⩽ 1 4 ( j = 1 , 2 , … ) a n − j ⩽ 1 4 ( j = 1 , 2 , … ) |a_(n-j)| <= (1)/(4)(j=1,2,dots)\left|a_{n-j}\right| \leqslant \frac{1}{4}(j=1,2, \ldots)|an−j|⩽14(j=1,2,…)
and
(7.3) | q n − 1 q n − 2 + 4 3 a n | > 2 3 | a n | (7.3) q n − 1 q n − 2 + 4 3 a n > 2 3 a n {:(7.3)|(q_(n-1))/(q_(n-2))+(4)/(3)a_(n)| > (2)/(3)|a_(n)|:}\begin{equation*} \left|\frac{q_{n-1}}{q_{n-2}}+\frac{4}{3} a_{n}\right|>\frac{2}{3}\left|a_{n}\right| \tag{7.3} \end{equation*}(7.3)|qn−1qn−2+43an|>23|an|
hen the complex continued fraction (7.1) converges and the values of function fill a closed disk D n ⊂ C D n ⊂ C D_(n)sub C\mathscr{D}_{n} \subset CDn⊂C.
Proof. If.
a = a n 1 + a n + 1 1 + a n + 2 1 + a = a n 1 + a n + 1 1 + a n + 2 1 + a=(a_(n))/(1+(a_(n+1))/(1+(a_(n+2))/(1+)))a=\frac{a_{n}}{1+\frac{a_{n+1}}{1+\frac{a_{n+2}}{1+}}}a=an1+an+11+an+21+
then, a fill the closed disk
D 1 n = { a ∈ C : | a − 4 3 a n | ⩽ 2 3 | a n | } . D 1 n = a ∈ C : a − 4 3 a n ⩽ 2 3 a n . D_(1n)={a in C:|a-(4)/(3)a_(n)| <= (2)/(3)|a_(n)|}.\mathscr{D}_{1 n}=\left\{a \in C:\left|a-\frac{4}{3} a_{n}\right| \leqslant \frac{2}{3}\left|a_{n}\right|\right\} .D1n={a∈C:|a−43an|⩽23|an|}.
Because z z zzz from (7.1) is given by linear function
z ↦ p n − 1 + z p n − 2 q n − 1 + z q n − 2 , z ↦ p n − 1 + z p n − 2 q n − 1 + z q n − 2 , z|->(p_(n-1)+zp_(n-2))/(q_(n-1)+zq_(n-2)),z \mapsto \frac{p_{n-1}+z p_{n-2}}{q_{n-1}+z q_{n-2}},z↦pn−1+zpn−2qn−1+zqn−2,
it follows, as in proposition 7.1, that the values of function f f fff fill a closed disck D n ⊂ C D n ⊂ C D_(n)sub C\mathscr{D}_{n} \subset CDn⊂C. Condition (7.3) implies that ∞ ≠ D n ∞ ≠ D n oo!=D_(n)\infty \neq \mathscr{D}_{n}∞≠Dn (condition ∞ ≠ D 1 ∞ ≠ D 1 oo!=D_(1)\infty \neq \mathscr{D}_{1}∞≠D1 is automatic satisfied in proposition 7.1.).
Remark 7.1. If a = C , e = 1 , ‖ ⋅ ‖ =∣ a = C , e = 1 , ‖ ⋅ ‖ =∣ a=C,e=1,||*||=∣\mathfrak{a}=C, e=1,\|\cdot\|=\mida=C,e=1,‖⋅‖=∣ and a 1 = a a 1 = a a_(1)=aa_{1}=aa1=a, then (i) of the theorem 6.1 gives classical worpitzky's theorem [3].
Remark 7.2. If Q = C , e = 1 , ‖ ⋅ ‖ = | ⋅ | Q = C , e = 1 , ‖ ⋅ ‖ = | ⋅ | Q=C,e=1,||*||=|*|Q=C, e=1,\|\cdot\|=|\cdot|Q=C,e=1,‖⋅‖=|⋅| and a 1 = 1 a 1 = 1 a_(1)=1a_{1}=1a1=1, then the propositions (ii) and (iii) of theorem 6.1 and propositions 7.1 are Wall and PAYDON's precisions [4] of Worpitzky's theorem.
(iii) A = C A = C A=C\mathfrak{A}=CA=C, with e = 1 , ‖ z ‖ = | x | + | y | e = 1 , ‖ z ‖ = | x | + | y | e=1,||z||=|x|+|y|e=1,\|z\|=|x|+|y|e=1,‖z‖=|x|+|y| and ‖ e ‖ = 1 ‖ e ‖ = 1 ||e||=1\|e\|=1‖e‖=1, is a commutative algebra. Theorem 6.1 shows that, if
‖ a n ‖ ⩽ 1 4 ( n = 2 , 3 , … ) a n ⩽ 1 4 ( n = 2 , 3 , … ) ||a_(n)|| <= (1)/(4)quad(n=2,3,dots)\left\|a_{n}\right\| \leqslant \frac{1}{4} \quad(n=2,3, \ldots)‖an‖⩽14(n=2,3,…)
(the complex numbers a n a n a_(n)a_{n}an belong of the square centred at 0 with the vertices at the points ± 1 4 ± 1 4 +-(1)/(4)\pm \frac{1}{4}±14 and ± i 4 ± i 4 +-(i)/(4)\pm \frac{i}{4}±i4 ), then the value of complex continued fraction (7.1) ( a n ∈ C a n ∈ C a_(n)in Ca_{n} \in Can∈C ) belongs to the set defined by the inequation
(7.4) | α 1 − x | + | β 1 − y | ⩽ 1 2 ( | x | + | y | ) , (7.4) α 1 − x + β 1 − y ⩽ 1 2 ( | x | + | y | ) , {:(7.4)|alpha_(1)-x|+|beta_(1)-y| <= (1)/(2)(|x|+|y|)",":}\begin{equation*} \left|\alpha_{1}-x\right|+\left|\beta_{1}-y\right| \leqslant \frac{1}{2}(|x|+|y|), \tag{7.4} \end{equation*}(7.4)|α1−x|+|β1−y|⩽12(|x|+|y|),
where z = x + i y z = x + i y z=x+iyz=x+i yz=x+iy and a 1 = α 1 + i β 1 a 1 = α 1 + i β 1 a_(1)=alpha_(1)+ibeta_(1)a_{1}=\alpha_{1}+i \beta_{1}a1=α1+iβ1.
For instance, if a 1 = 2 + 3 i a 1 = 2 + 3 i a_(1)=2+3ia_{1}=2+3 ia1=2+3i, then (7.4) shows that z z zzz belongs to convex hull of the points 1 3 + 3 i , 2 + 4 3 i , 7 + 3 i , 2 + 8 i 1 3 + 3 i , 2 + 4 3 i , 7 + 3 i , 2 + 8 i (1)/(3)+3i,2+(4)/(3)i,7+3i,2+8i\frac{1}{3}+3 i, 2+\frac{4}{3} i, 7+3 i, 2+8 i13+3i,2+43i,7+3i,2+8i.
(iv) A = C A = C A=C\mathcal{A}=CA=C, with e = 1 , ‖ z ‖ = max { | x | , | y | } e = 1 , ‖ z ‖ = max { | x | , | y | } e=1,||z||=max{|x|,|y|}e=1,\|z\|=\max \{|x|,|y|\}e=1,‖z‖=max{|x|,|y|} and ‖ 1 ‖ = 1 ‖ 1 ‖ = 1 ||1||=1\|1\|=1‖1‖=1, is a commutative algebra. Then, if ‖ a n ‖ ⩽ 1 4 ( n = 2 , 3 , … ) a n ⩽ 1 4 ( n = 2 , 3 , … ) ||a_(n)|| <= (1)/(4)(n=2,3,dots)\left\|a_{n}\right\| \leqslant \frac{1}{4}(n=2,3, \ldots)‖an‖⩽14(n=2,3,…) ( a n a n a_(n)a_{n}an runs a square with the sides parallel to the coordinate axes and centred at 0 ), the value of continued fraction (7.1) ( a n ∈ C a n ∈ C a_(n)in Ca_{n} \in Can∈C ) are in the set defined by the inequation
max { | α 1 − x | , | β 1 − y | } ⩽ 1 2 max { | x | , | y | } , max α 1 − x , β 1 − y ⩽ 1 2 max { | x | , | y | } , max{|alpha_(1)-x|,|beta_(1)-y|} <= (1)/(2)max{|x|,|y|},\max \left\{\left|\alpha_{1}-x\right|,\left|\beta_{1}-y\right|\right\} \leqslant \frac{1}{2} \max \{|x|,|y|\},max{|α1−x|,|β1−y|}⩽12max{|x|,|y|},
where z = x + i y z = x + i y z=x+iyz=x+\mathrm{i} yz=x+iy and a 1 = α 1 + i β 1 a 1 = α 1 + i β 1 a_(1)=alpha_(1)+ibeta_(1)a_{1}=\alpha_{1}+\mathrm{i} \beta_{1}a1=α1+iβ1.

REFEIGE NCES

[1] Ir a if W y ma 11, A theorem of convergence for the non-commutative fractions. Journal of Approximation theory, 5, 74-76 (1972).
[2] Garnir H. G., et Gobert, I, Algebre de Banach. Université de Liège, Séminaire d'Analyse Mathénatique et d'Algèbre, pp. 1-17 (1966-1967).
[3] Worpitzky, Untersuchungen über die Entwicketung der Monodronen und Monogenen Funktion durck Kettenbrüch-Friedrich-Gymnasim und Realschule. Jahresbericht, Berlin pp. 3-39 (1865).
[4] Paydon, J. I. and Wall, II.S., The continued fraction as a sequence of linear transformations. Duke Math Journal, 9, 360-372 (1942).
Facullatea de matematică și mecanică Universitatea din Iaşi
R. S. România

Copyright (c) 2015 Journal of Numerical Analysis and Approximation Theory

Creative Commons License

This work is licensed under a Creative Commons Attribution 4.0 International License.