On the Continued Fractions of J. Mikusiński

Tiberiu Popoviciu
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Tiberiu Popoviciu
Institutul de Calcul

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T. Popoviciu, Sur les fractions continues de J. Mikusinski, Rev. Roumaine Math. Pures Appl., 13, 1968, pp. 79-83 (in French)

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1968 a -Popoviciu- Rev. Roum. Math. Pures Appl. – Sur les fractions continues de J. Mikusinski

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1968 a -Popoviciu- Rev. Roum. Math. Pures Appl. - On the continued fractions of J. Mikusinski
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ON THE CONTINUOUS FRACTIONS OF J. MIKUSINSKI

BY

TIBERIU POPOVICIU
(Cluj)

We give a new proof of the sufficiency of condition (3) of theorem 1 of J. Mikusinski [2]. We also give some other properties of arithmetic continued fractions of the form [ has , b ˙ , 2 has ] [ has , b ˙ , 2 has ] [a,b^(˙),2a][a, \dot{b}, 2 a][has,b˙,2has].
  1. Either c c ccca natural number > 1 > 1 > 1>1>1, different from a square, ( r n ) n = 0 ∞ r n n = 0 ∞ (r_(n))_(n=0)^(oo)\left(r_{n}\right)_{n=0}^{\infty}(rn)n=0∞the continuation of the reductions of the development of c c sqrtc\sqrt{c}cin arithmetic continued fraction and ( x n ) n = 0 ∞ x n n = 0 ∞ (x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}(xn)n=0∞the sequence of approximations of c c sqrtc\sqrt{c}cgiven by the recurrence relation
(1) x n + 1 = 1 2 ( x n + c x n ) , n = 0 , 1 , … , x 0 = [ V c ¯ ] (1) x n + 1 = 1 2 x n + c x n , n = 0 , 1 , … , x 0 = [ V c ¯ ] {:(1)x_(n+1)=(1)/(2)(x_(n)+(c)/(x_(n)))","n=0","1","dots","x_(0)=[V bar(c)]:}\begin{equation*} x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{c}{x_{n}}\right), n=0.1, \ldots, x_{0}=[V \bar{c}] \tag{1} \end{equation*}(1)xn+1=12(xn+cxn),n=0,1,…,x0=[Vc¯]
J. Mikusinski proved [2] the following theorem
Theorem 1. So that we have
(2)
x n = r 2 n − 1 , n = 0 , 1 , … x n = r 2 n − 1 , n = 0 , 1 , … x_(n)=r_(2^(n)-1),n=0,1,dotsx_{n}=r_{2^{n}-1}, n=0,1, \ldotsxn=r2n−1,n=0,1,…
it is necessary and sufficient that the number c be of the form
(3) c = a 2 + 2 a b (3) c = a 2 + 2 a b {:(3)c=a^(2)+(2a)/(b):}\begin{equation*} c=a^{2}+\frac{2 a}{b} \tag{3} \end{equation*}(3)c=has2+2hasb
where a and b b bbbare natural numbers.
J. Mikusinski demonstrated very simply the necessity of the condition and also demonstrated the sufficiency of the condition by noting that then the continued fraction of c c sqrtc\sqrt{c}cis of the form [ a , b ˙ , 2 ˙ a ] [ a , b ˙ , 2 ˙ a ] [a,b^(˙),2^(˙)a][a, \dot{b}, \dot{2} a][has,b˙,2˙has]. In the following we will first give a slightly different demonstration of the sufficiency of condition (3).
Natural numbers c c cccof the form (3) are characterized by the property that if a = [ c ] a = [ c ] a=[sqrtc]a=[\sqrt{c}]has=[c], the number 2 a 2 a 2a2 a2hasis divisible by c − a 2 c − a 2 c-a^(2)c-a^{2}c−has2.
REV. ROUM. MATH. PURES ET APPL., VOLUME XIII, NO. 1, p. 79-83, BUCHAREST, 1968
2. Let us denote by r n = P n Q n , n = 0 , 1 , … r n = P n Q n , n = 0 , 1 , … r_(n)=(P_(n))/(Q_(n)),n=0,1,dotsr_{n}=\frac{P_{n}}{Q_{n}}, n=0,1, \ldotsrn=PnQn,n=0,1,…, the reductions of the periodic arithmetic continued fraction [ a , b ˙ , 2 ˙ a ] [ a , b ˙ , 2 ˙ a ] [a,b^(˙),2^(˙)a][a, \dot{b}, \dot{2} a][has,b˙,2˙has]. The sequels ( P n ) , ( Q n ) P n , Q n (P_(n)),(Q_(n))\left(P_{n}\right),\left(Q_{n}\right)(Pn),(Qn)are solutions ( u n ) n = 0 ∞ u n n = 0 ∞ (u_(n))_(n=0)^(oo)\left(u_{n}\right)_{n=0}^{\infty}(un)n=0∞of the system of recurrent equations
(4) { u 2 n = 2 a u 2 n − 1 + u 2 n − 2 u 2 n + 1 = b u 2 n + u 2 n − 1 n = 1 , 2 , … (4) u 2 n = 2 a u 2 n − 1 + u 2 n − 2 u 2 n + 1 = b u 2 n + u 2 n − 1 n = 1 , 2 , … {:(4){[u_(2n)=2au_(2n-1)+u_(2n-2)],[u_(2n+1)=bu_(2n)+u_(2n-1)]quad n=1,2,dots:}:}\left\{\begin{array}{l} u_{2 n}=2 a u_{2 n-1}+u_{2 n-2} \tag{4}\\ u_{2 n+1}=b u_{2 n}+u_{2 n-1} \end{array} \quad n=1,2, \ldots\right.(4){u2n=2hasu2n−1+u2n−2u2n+1=bu2n+u2n−1n=1,2,…
with the initial values ​​respectively P 0 = a , P 1 = a b + 1 ; Q 0 = 1 P 0 = a , P 1 = a b + 1 ; Q 0 = 1 P_(0)=a,P_(1)=ab+1;Q_(0)=1P_{0}=a, P_{1}=a b+1 ; Q_{0}=1P0=has,P1=hasb+1;Q0=1, Q 1 = b Q 1 = b Q_(1)=bQ_{1}=bQ1=b.
But if ( u n ) n = 0 ∞ u n n = 0 ∞ (u_(n))_(n=0)^(oo)\left(u_{n}\right)_{n=0}^{\infty}(un)n=0∞is a solution of the system (4), the partial sequences ( u 2 n ) n = 0 ∞ , ( u 2 n + 1 ) n = 0 ∞ u 2 n n = 0 ∞ , u 2 n + 1 n = 0 ∞ (u_(2n))_(n=0)^(oo),(u_(2n+1))_(n=0)^(oo)\left(u_{2 n}\right)_{n=0}^{\infty},\left(u_{2 n+1}\right)_{n=0}^{\infty}(u2n)n=0∞,(u2n+1)n=0∞are solutions ( v n v n v_(n)v_{n}vn) of the recurrence equation
(5) v n = ( 2 a b + 2 ) v n − 1 − v n − 2 , n = 2 , 3 , … (5) v n = ( 2 a b + 2 ) v n − 1 − v n − 2 , n = 2 , 3 , … {:(5)v_(n)=(2ab+2)v_(n-1)-v_(n-2)","n=2","3","dots:}\begin{equation*} v_{n}=(2 a b+2) v_{n-1}-v_{n-2}, n=2,3, \ldots \tag{5} \end{equation*}(5)vn=(2hasb+2)vn−1−vn−2,n=2,3,…
According to the well-known theory of equations of this form we have v n = A x n + B y n , n = 0 , 1 , … v n = A x n + B y n , n = 0 , 1 , … v_(n)=Ax^(n)+By^(n),n=0,1,dotsv_{n}=A x^{n}+B y^{n}, n=0,1, \ldotsvn=HASxn+Byn,n=0,1,…, Or x , y x , y x,yx, yx,yare the roots of the characteristic equation z 2 − ( 2 a b + 2 ) z + 1 = 0 z 2 − ( 2 a b + 2 ) z + 1 = 0 z^(2)-(2ab+2)z+1=0z^{2}-(2 a b+2) z+1=0z2−(2hasb+2)z+1=0And A , B A , B A,BA, BHAS,Bare independent of n n nnn.
If we take into account (3) we have
x = a b + 1 + b c = b 2 a ξ 2 , y = a b + 1 − b c = b 2 a η 2 x = a b + 1 + b c = b 2 a ξ 2 , y = a b + 1 − b c = b 2 a η 2 x=ab+1+bsqrtc=(b)/(2a)xi^(2),y=ab+1-bsqrtc=(b)/(2a)eta^(2)x=a b+1+b \sqrt{c}=\frac{b}{2 a} \xi^{2}, y=a b+1-b \sqrt{c}=\frac{b}{2 a} \eta^{2}x=hasb+1+bc=b2hasξ2,y=hasb+1−bc=b2hasη2
Or ξ = a + c , η = a − c ξ = a + c , η = a − c xi=a+sqrtc,eta=a-sqrtc\xi=a+\sqrt{c}, \eta=a-\sqrt{c}ξ=has+c,η=has−c. So we have
(6) v n = ( b 2 a ) n ( A ξ 2 n + B η 2 n ) , n = 0 , 1 , … (6) v n = b 2 a n A ξ 2 n + B η 2 n , n = 0 , 1 , … {:(6)v_(n)=((b)/(2a))^(n)(Axi^(2n)+Beta^(2n))","n=0","1","dots:}\begin{equation*} v_{n}=\left(\frac{b}{2 a}\right)^{n}\left(A \xi^{2 n}+B \eta^{2 n}\right), n=0,1, \ldots \tag{6} \end{equation*}(6)vn=(b2has)n(HASξ2n+Bη2n),n=0,1,…
To get the suites ( P n ) , ( Q n ) P n , Q n (P_(n)),(Q_(n))\left(P_{n}\right),\left(Q_{n}\right)(Pn),(Qn)we determine the partial sequences ( P 2 n ) n = 0 ∞ , ( P 2 n + 1 ) n = 0 ∞ , ( Q 2 n ) n = 0 ∞ , ( Q 2 n + 1 ) n = 0 ∞ P 2 n n = 0 ∞ , P 2 n + 1 n = 0 ∞ , Q 2 n n = 0 ∞ , Q 2 n + 1 n = 0 ∞ (P_(2n))_(n=0)^(oo),(P_(2n+1))_(n=0)^(oo),(Q_(2n))_(n=0)^(oo),(Q_(2n+1))_(n=0)^(oo)\left(P_{2 n}\right)_{n=0}^{\infty},\left(P_{2 n+1}\right)_{n=0}^{\infty},\left(Q_{2 n}\right)_{n=0}^{\infty},\left(Q_{2 n+1}\right)_{n=0}^{\infty}(P2n)n=0∞,(P2n+1)n=0∞,(Q2n)n=0∞,(Q2n+1)n=0∞, using formulas (6), taking into account the initial values ​​respectively P 0 = a , P 2 = 2 a 2 b + 3 a P 0 = a , P 2 = 2 a 2 b + 3 a P_(0)=a,P_(2)=2a^(2)b+3aP_{0}=a, P_{2}=2 a^{2} b+3 aP0=has,P2=2has2b+3has; P 1 = a b + 1 , P 3 = 2 a 2 b 2 + 4 a b + 1 ; Q 0 = 1 , Q 2 = 2 a b + 1 ; Q 1 = b P 1 = a b + 1 , P 3 = 2 a 2 b 2 + 4 a b + 1 ; Q 0 = 1 , Q 2 = 2 a b + 1 ; Q 1 = b P_(1)=ab+1,P_(3)=2a^(2)b^(2)+4ab+1;Q_(0)=1,Q_(2)=2ab+1;Q_(1)=bP_{1}=a b+1, P_{3}=2 a^{2} b^{2}+4 a b+1 ; Q_{0}=1, Q_{2}=2 a b+1 ; Q_{1}=bP1=hasb+1,P3=2has2b2+4hasb+1;Q0=1,Q2=2hasb+1;Q1=b, Q 3 = 2 a b 2 + 2 b Q 3 = 2 a b 2 + 2 b Q_(3)=2ab^(2)+2bQ_{3}=2 a b^{2}+2 bQ3=2hasb2+2b. In each of these cases these initial values ​​determine the coefficients A , B A , B A,BA, BHAS,Bcorrespondents.
By doing the calculations, we find thus
(7) { P n = 1 2 ( b 2 a ) [ n + 1 2 ] ( ξ n + 1 + η n + 1 ) Q n = 1 2 c ( b 2 a ) [ n + 1 2 ] ( ξ n + 1 − η n + 1 ) n = 0 , 1 , … (7) P n = 1 2 b 2 a n + 1 2 ξ n + 1 + η n + 1 Q n = 1 2 c b 2 a n + 1 2 ξ n + 1 − η n + 1 n = 0 , 1 , … {:(7){[P_(n)=(1)/(2)((b)/(2a))^([(n+1)/(2)])(xi^(n+1)+eta^(n+1))],[Q_(n)=(1)/(2sqrtc)((b)/(2a))^([(n+1)/(2)])(xi^(n+1)-eta^(n+1))]quad n=0,1,dots:}:}\left\{\begin{array}{l} P_{n}=\frac{1}{2}\left(\frac{b}{2 a}\right)^{\left[\frac{n+1}{2}\right]}\left(\xi^{n+1}+\eta^{n+1}\right) \tag{7}\\ Q_{n}=\frac{1}{2 \sqrt{c}}\left(\frac{b}{2 a}\right)^{\left[\frac{n+1}{2}\right]}\left(\xi^{n+1}-\eta^{n+1}\right) \end{array} \quad n=0,1, \ldots\right.(7){Pn=12(b2has)[n+12](ξn+1+ηn+1)Qn=12c(b2has)[n+12](ξn+1−ηn+1)n=0,1,…
It is clear that these formulas are valid in general if a , b , c a , b , c a,b,ca, b, chas,b,care non-zero numbers such that c b = a 2 b + 2 a c b = a 2 b + 2 a cb=a^(2)b+2ac b=a^{2} b+2 acb=has2b+2has.
The reduction of a system of the form (4) to an equation of the type (5) was made by Th. Angheluță [1].
3. Returning to our problem, consider the recurrence relation (1). We have
x n − c x n + c = ( x n − 1 − c x n − 1 + c ) 2 , n = 1 , 2 , … , x 0 = a x n − c x n + c = x n − 1 − c x n − 1 + c 2 , n = 1 , 2 , … , x 0 = a (x_(n)-sqrtc)/(x_(n)+sqrtc)=((x_(n-1)-sqrtc)/(x_(n-1)+sqrtc))^(2),n=1,2,dots,x_(0)=a\frac{x_{n}-\sqrt{c}}{x_{n}+\sqrt{c}}=\left(\frac{x_{n-1}-\sqrt{c}}{x_{n-1}+\sqrt{c}}\right)^{2}, n=1,2, \ldots, x_{0}=axn−cxn+c=(xn−1−cxn−1+c)2,n=1,2,…,x0=has
from which it follows that
(8) x n − c x n + c = ( η ξ ) 2 n , n = 0 , 1 , … (8) x n − c x n + c = η ξ 2 n , n = 0 , 1 , … {:(8)(x_(n)-sqrtc)/(x_(n)+sqrtc)=((eta )/(xi))^(2^(n))","n=0","1","dots:}\begin{equation*} \frac{x_{n}-\sqrt{c}}{x_{n}+\sqrt{c}}=\left(\frac{\eta}{\xi}\right)^{2^{n}}, n=0,1, \ldots \tag{8} \end{equation*}(8)xn−cxn+c=(ηξ)2n,n=0,1,…
where well
(9) x n = c ξ 2 n + η 2 n ξ 2 n − η 2 n , n = 0 , 1 , … (9) x n = c ξ 2 n + η 2 n ξ 2 n − η 2 n , n = 0 , 1 , … {:(9)x_(n)=sqrtc(xi^(2^(n))+eta^(2^(n)))/(xi^(2^(n))-eta^(2^(n)))","n=0","1","dots:}\begin{equation*} x_{n}=\sqrt{c} \frac{\xi^{2^{n}}+\eta^{2^{n}}}{\xi^{2^{n}}-\eta^{2^{n}}}, n=0,1, \ldots \tag{9} \end{equation*}(9)xn=cξ2n+η2nξ2n−η2n,n=0,1,…
But, from (7) it follows that
r n = c ξ n + 1 + η n + 1 ξ n + 1 − η n + 1 , n = 0 , 1 , … r n = c ξ n + 1 + η n + 1 ξ n + 1 − η n + 1 , n = 0 , 1 , … r_(n)=sqrtc(xi^(n+1)+eta^(n+1))/(xi^(n+1)-eta^(n+1)),n=0,1,dotsr_{n}=\sqrt{c} \frac{\xi^{n+1}+\eta^{n+1}}{\xi^{n+1}-\eta^{n+1}}, n=0,1, \ldotsrn=cξn+1+ηn+1ξn+1−ηn+1,n=0,1,…
from which equality (2) follows immediately.
4. Noting that | η | < ξ | η | < ξ |eta| < xi|\eta|<\xi|η|<ξ, formula (8) shows us that the sequence ( x n x n x_(n)x_{n}xn) converges to c c sqrtc\sqrt{c}c. But we can approximate c c sqrtc\sqrt{c}cby other iterative sequences. Such a sequence ( y n ) n = 0 ∞ y n n = 0 ∞ (y_(n))_(n=0)^(oo)\left(y_{n}\right)_{n=0}^{\infty}(yn)n=0∞is determined by the recurrence relation
(10) y n + 1 = α y n + c y n + α , n = 0 , 1 , … (10) y n + 1 = α y n + c y n + α , n = 0 , 1 , … {:(10)y_(n+1)=(alphay_(n)+c)/(y_(n)+alpha)","n=0","1","dots:}\begin{equation*} y_{n+1}=\frac{\alpha y_{n}+c}{y_{n}+\alpha}, n=0,1, \ldots \tag{10} \end{equation*}(10)yn+1=αyn+cyn+α,n=0,1,…
where we assume that y 0 > 0 y 0 > 0 y_(0) > 0y_{0}>0y0>0and that α α alpha\alphaαis a positive rational number. By limiting ourselves only to rational approximations of c c sqrtc\sqrt{c}c, the case of α α alpha\alphaαpositive and irrational is of no interest.
All terms of the sequence ( y n y n y_(n)y_{n}yn) are positive and we have
y n + 1 − c y n + 1 + c = α − c α + c ⋅ y n − c y n + c , n = 0 , 1 , … y n + 1 − c y n + 1 + c = α − c α + c ⋅ y n − c y n + c , n = 0 , 1 , … (y_(n+1)-sqrtc)/(y_(n+1)+sqrtc)=(alpha-sqrtc)/(alpha+sqrtc)*(y_(n)-sqrtc)/(y_(n)+sqrtc),n=0,1,dots\frac{y_{n+1}-\sqrt{c}}{y_{n+1}+\sqrt{c}}=\frac{\alpha-\sqrt{c}}{\alpha+\sqrt{c}} \cdot \frac{y_{n}-\sqrt{c}}{y_{n}+\sqrt{c}}, n=0,1, \ldotsyn+1−cyn+1+c=α−cα+c⋅yn−cyn+c,n=0,1,…
from where
(11) y n − c y n + c = ( α − c α + c ) n ⋅ y 0 − c y 0 + c , n = 0 , 1 , … (11) y n − c y n + c = α − c α + c n ⋅ y 0 − c y 0 + c , n = 0 , 1 , … {:(11)(y_(n)-sqrtc)/(y_(n)+sqrtc)=((alpha-sqrtc)/(alpha+sqrtc))^(n)*(y_(0)-sqrtc)/(y_(0)+sqrtc)","n=0","1","dots:}\begin{equation*} \frac{y_{n}-\sqrt{c}}{y_{n}+\sqrt{c}}=\left(\frac{\alpha-\sqrt{c}}{\alpha+\sqrt{c}}\right)^{n} \cdot \frac{y_{0}-\sqrt{c}}{y_{0}+\sqrt{c}}, n=0,1, \ldots \tag{11} \end{equation*}(11)yn−cyn+c=(α−cα+c)n⋅y0−cy0+c,n=0,1,…
and, since | α − c α + c | < 1 α − c α + c < 1 |(alpha-sqrtc)/(alpha+sqrtc)| < 1\left|\frac{\alpha-\sqrt{c}}{\alpha+\sqrt{c}}\right|<1|α−cα+c|<1, it follows that the sequence ( y n y n y_(n)y_{n}yn) converges to c c sqrtc\sqrt{c}c.
The convergence of the sequence ( x n x n x_(n)x_{n}xn) towards c c sqrtc\sqrt{c}cis of order 2 while that of the sequence ( y n ) y n (y_(n))\left(y_{n}\right)(yn)only of order 1.
Let's suppose that α = y 0 = [ o ¯ ] = a α = y 0 = [ o ¯ ] = a alpha=y_(0)=[sqrt() bar(o)]=a\alpha=y_{0}=[\sqrt{ } \bar{o}]=aα=y0=[o¯]=has, then from (11) it follows that
y n = c ξ n + 1 + η n + 1 ξ n + 1 − η n + 1 , n = 0 , 1 , … y n = c ξ n + 1 + η n + 1 ξ n + 1 − η n + 1 , n = 0 , 1 , … y_(n)=sqrtc(xi^(n+1)+eta^(n+1))/(xi^(n+1)-eta^(n+1)),n=0,1,dotsy_{n}=\sqrt{c} \frac{\xi^{n+1}+\eta^{n+1}}{\xi^{n+1}-\eta^{n+1}}, n=0,1, \ldotsyn=cξn+1+ηn+1ξn+1−ηn+1,n=0,1,…
continuing to pose ξ = a + c , η = a − c ξ = a + c , η = a − c xi=a+sqrtc,eta=a-sqrtc\xi=a+\sqrt{c}, \eta=a-\sqrt{c}ξ=has+c,η=has−c.
This formula shows us that if the number c c cccis of the form (3) and if α = y 0 = a α = y 0 = a alpha=y_(0)=a\alpha=y_{0}=aα=y0=has, we have
(12)
y n = r n , n = 0 , 1 , … y n = r n , n = 0 , 1 , … y_(n)=r_(n),n=0,1,dotsy_{n}=r_{n}, n=0,1, \ldotsyn=rn,n=0,1,…
so that the recurrence relation (10), with the initial condition y 0 = [ c ] y 0 = [ c ] y_(0)=[sqrtc]y_{0}=[\sqrt{c}]y0=[c], gives us precisely the sequence of reductions of the arithmetic continued fraction of c c sqrtc\sqrt{c}c.
5. Let us now propose to determine the natural number c c ccc(non-square) and the rational number (positive) α α alpha\alphaαso that the recurrence relation (10) gives precisely the sequence of reductions ( r n ) n = 0 ∞ r n n = 0 ∞ (r_(n))_(n=0)^(oo)\left(r_{n}\right)_{n=0}^{\infty}(rn)n=0∞of the arithmetic continued fraction of V c ¯ V c ¯ V bar(c)V \bar{c}Vc¯.
For this we designate by a = [ c ] a = [ c ] a=[sqrtc]a=[\sqrt{c}]has=[c]and let's ask c = a 2 + k c = a 2 + k c=a^(2)+kc=a^{2}+kc=has2+k. If [ a , b , d , e , … ] [ a , b , d , e , … ] [a,b,d,e,dots][a, b, d, e, \ldots][has,b,d,e,…]is the arithmetic continued fraction of c c sqrtc\sqrt{c}cAnd r 0 , r 1 , r 2 , r 3 , … r 0 , r 1 , r 2 , r 3 , … r_(0),r_(1),r_(2),r_(3),dotsr_{0}, r_{1}, r_{2}, r_{3}, \ldotsr0,r1,r2,r3,…, are its successive reductions, of
r 1 = a b + 1 b = a + 1 b = α a + a 2 + k a + α r 1 = a b + 1 b = a + 1 b = α a + a 2 + k a + α r_(1)=(ab+1)/(b)=a+(1)/(b)=(alpha a+a^(2)+k)/(a+alpha)r_{1}=\frac{a b+1}{b}=a+\frac{1}{b}=\frac{\alpha a+a^{2}+k}{a+\alpha}r1=hasb+1b=has+1b=αhas+has2+khas+α
it results
(13)
b = a + α k b = a + α k b=(a+alpha)/(k)b=\frac{a+\alpha}{k}b=has+αk
Of
r 2 = a b d + a + d b d + 1 = a + 1 b + 1 d = α ( a b + 1 ) + b ( a 2 + k ) a b + 1 + α b r 2 = a b d + a + d b d + 1 = a + 1 b + 1 d = α ( a b + 1 ) + b a 2 + k a b + 1 + α b r_(2)=(abd+a+d)/(bd+1)=a+(1)/(b+(1)/(d))=(alpha(ab+1)+b(a^(2)+k))/(ab+1+alpha b)r_{2}=\frac{a b d+a+d}{b d+1}=a+\frac{1}{b+\frac{1}{d}}=\frac{\alpha(a b+1)+b\left(a^{2}+k\right)}{a b+1+\alpha b}r2=hasbd+has+dbd+1=has+1b+1d=α(hasb+1)+b(has2+k)hasb+1+αb
it results
(14)
d = α − a + k b 2 a b + 1 − k b 2 d = α − a + k b 2 a b + 1 − k b 2 d=(alpha-a+kb)/(2ab+1-kb^(2))d=\frac{\alpha-a+k b}{2 a b+1-k b^{2}}d=α−has+kb2hasb+1−kb2
At the end of
v 3 = a + 1 b + 1 d + 1 e = a ( a b d + a + d ) + ( b d + 1 ) ( a 2 + k ) a b d + a + d + α ( b d + 1 ) v 3 = a + 1 b + 1 d + 1 e = a ( a b d + a + d ) + ( b d + 1 ) a 2 + k a b d + a + d + α ( b d + 1 ) v_(3)=a+(1)/(b+(1)/(d+(1)/(e)))=(a(abd+a+d)+(bd+1)(a^(2)+k))/(abd+a+d+alpha(bd+1))v_{3}=a+\frac{1}{b+\frac{1}{d+\frac{1}{e}}}=\frac{a(a b d+a+d)+(b d+1)\left(a^{2}+k\right)}{a b d+a+d+\alpha(b d+1)}v3=has+1b+1d+1e=has(hasbd+has+d)+(bd+1)(has2+k)hasbd+has+d+α(bd+1)
taking into account (13) and (14), we deduce
(15) e = 2 α k (15) e = 2 α k {:(15)e=(2alpha)/(k):}\begin{equation*} e=\frac{2 \alpha}{k} \tag{15} \end{equation*}(15)e=2αk
But the numbers b , e b , e b,eb, eb,eare integers (positive), it therefore follows that 2 b − e = 2 a k 2 b − e = 2 a k 2b-e=(2a)/(k)2 b-e=\frac{2 a}{k}2b−e=2haskis also whole. It follows that 2 a 2 a 2a2 a2hasis divisible by k k kkk, so that the number c c cccis of the form (3). In this case e = b e = b e=be=be=b, SO α = a α = a alpha=a\alpha=aα=has.
So we have the property expressed by
Theorem 2. For the y n , n = 0 , 1 , … y n , n = 0 , 1 , … y_(n),n=0,1,dotsy_{n}, n=0,1, \ldotsyn,n=0,1,…, given by the recurrence relation (10), where c is a natural number (not square) and a is a positive rational number, are the successive reductions of the arithmetic continued fraction of c c sqrtc\sqrt{c}c, it is necessary and sufficient that it be of the form (3) and that α = [ c ] = a α = [ c ] = a alpha=[sqrtc]=a\alpha=[\sqrt{c}]=aα=[c]=has.
6. Note that if c c cccis any positive number and if we have (1), with x 0 > 0 x 0 > 0 x_(0) > 0x_{0}>0x0>0, it results
x n − c x n + c = ( x 0 − c x 0 + c ) 2 n , n = 0 , 1 , … x n − c x n + c = x 0 − c x 0 + c 2 n , n = 0 , 1 , … (x_(n)-sqrtc)/(x_(n)+sqrtc)=((x_(0)-sqrtc)/(x_(0)+sqrtc))^(2^(n)),n=0,1,dots\frac{x_{n}-\sqrt{c}}{x_{n}+\sqrt{c}}=\left(\frac{x_{0}-\sqrt{c}}{x_{0}+\sqrt{c}}\right)^{2^{n}}, n=0,1, \ldotsxn−cxn+c=(x0−cx0+c)2n,n=0,1,…
Taking into account also formula (11), we see that we can state the
Theorem 3. If e and a are positive numbers and if the sequences ( x n ) n = 0 ∞ x n n = 0 ∞ (x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}(xn)n=0∞, ( y n ) n = 0 ∞ y n n = 0 ∞ (y_(n))_(n=0)^(oo)\left(y_{n}\right)_{n=0}^{\infty}(yn)n=0∞are given by the recurrence relations (1), (10), with the initial values x 0 = y 0 = α x 0 = y 0 = α x_(0)=y_(0)=alphax_{0}=y_{0}=\alphax0=y0=α:
1 ∘ 1 ∘ 1^(@)1^{\circ}1∘They converge towards c c sqrtc\sqrt{c}c.
2 ∘ 2 ∘ 2^(@)2^{\circ}2∘We have x n = y 2 n − 1 , n = 0 , 1 , … x n = y 2 n − 1 , n = 0 , 1 , … x_(n)=y_(2^(n)-1),n=0,1,dotsx_{n}=y_{2^{n}-1}, n=0,1, \ldotsxn=y2n−1,n=0,1,…
Received on May 3, 1967
Institute of Calculation
Academy of the Socialist Republic of Romania, Cluj Branch

BIBLIOGRAPHY

  1. Angheluță, Th. Integration of a class of linear equations with finite differences. Bulletin of the Soc. of Sciences of Cluj, Vol. III (1926), 94-104.
  2. Mikusinski, J. On Newton's method of approximation. Annales Polon. Math., I (1954), 184-194.

google scholar link

[MR0226243, Zbl 0172.06701]

1968

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