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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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Revue Roumaine de Mathématiques Pures et Appliquées
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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[a,b^(˙),2a][a, \dot{b}, 2 a].
- Eithercca natural number> 1>1, different from a square,(r_(n))_(n=0)^(oo)\left(r_{n}\right)_{n=0}^{\infty}the continuation of the reductions of the development ofsqrtc\sqrt{c}in arithmetic continued fraction and(x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}the sequence of approximations ofsqrtc\sqrt{c}given by the recurrence relation
{:(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*}
J. Mikusinski proved [2] the following theorem
Theorem 1. So that we have
(2)
(2)
x_(n)=r_(2^(n)-1),n=0,1,dotsx_{n}=r_{2^{n}-1}, n=0,1, \ldots
it is necessary and sufficient that the number c be of the form
{:(3)c=a^(2)+(2a)/(b):}\begin{equation*}
c=a^{2}+\frac{2 a}{b} \tag{3}
\end{equation*}
where a andbbare 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 ofsqrtc\sqrt{c}is of the form[a,b^(˙),2^(˙)a][a, \dot{b}, \dot{2} a]. In the following we will first give a slightly different demonstration of the sufficiency of condition (3).
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 ofsqrtc\sqrt{c}is of the form[a,b^(˙),2^(˙)a][a, \dot{b}, \dot{2} a]. In the following we will first give a slightly different demonstration of the sufficiency of condition (3).
Natural numbersccof the form (3) are characterized by the property that ifa=[sqrtc]a=[\sqrt{c}], the number2a2 ais divisible byc-a^(2)c-a^{2}.
REV. ROUM. MATH. PURES ET APPL., VOLUME XIII, NO. 1, p. 79-83, BUCHAREST, 1968
2. Let us denote byr_(n)=(P_(n))/(Q_(n)),n=0,1,dotsr_{n}=\frac{P_{n}}{Q_{n}}, n=0,1, \ldots, the reductions of the periodic arithmetic continued fraction[a,b^(˙),2^(˙)a][a, \dot{b}, \dot{2} a]. The sequels(P_(n)),(Q_(n))\left(P_{n}\right),\left(Q_{n}\right)are solutions(u_(n))_(n=0)^(oo)\left(u_{n}\right)_{n=0}^{\infty}of the system of recurrent equations
2. Let us denote byr_(n)=(P_(n))/(Q_(n)),n=0,1,dotsr_{n}=\frac{P_{n}}{Q_{n}}, n=0,1, \ldots, the reductions of the periodic arithmetic continued fraction[a,b^(˙),2^(˙)a][a, \dot{b}, \dot{2} a]. The sequels(P_(n)),(Q_(n))\left(P_{n}\right),\left(Q_{n}\right)are solutions(u_(n))_(n=0)^(oo)\left(u_{n}\right)_{n=0}^{\infty}of the system of recurrent equations
{:(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.
with the initial values ​​respectivelyP_(0)=a,P_(1)=ab+1;Q_(0)=1P_{0}=a, P_{1}=a b+1 ; Q_{0}=1,Q_(1)=bQ_{1}=b.
But if(u_(n))_(n=0)^(oo)\left(u_{n}\right)_{n=0}^{\infty}is a solution of the system (4), the partial sequences(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}are solutions (v_(n)v_{n}) of the recurrence equation
{:(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*}
According to the well-known theory of equations of this form we havev_(n)=Ax^(n)+By^(n),n=0,1,dotsv_{n}=A x^{n}+B y^{n}, n=0,1, \ldots, Orx,yx, yare the roots of the characteristic equationz^(2)-(2ab+2)z+1=0z^{2}-(2 a b+2) z+1=0AndA,BA, Bare independent ofnn.
If we take into account (3) we have
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}
Orxi=a+sqrtc,eta=a-sqrtc\xi=a+\sqrt{c}, \eta=a-\sqrt{c}. So we have
{:(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*}
To get the suites(P_(n)),(Q_(n))\left(P_{n}\right),\left(Q_{n}\right)we determine the partial sequences(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}, using formulas (6), taking into account the initial values ​​respectivelyP_(0)=a,P_(2)=2a^(2)b+3aP_{0}=a, P_{2}=2 a^{2} b+3 a;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}=b,Q_(3)=2ab^(2)+2bQ_{3}=2 a b^{2}+2 b. In each of these cases these initial values ​​determine the coefficientsA,BA, Bcorrespondents.
By doing the calculations, we find thus
{:(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.
It is clear that these formulas are valid in general ifa,b,ca, b, care non-zero numbers such thatcb=a^(2)b+2ac b=a^{2} b+2 a.
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
3. Returning to our problem, consider the recurrence relation (1). We have
(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}=a
from which it follows that
{:(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*}
where well
{:(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*}
But, from (7) it follows that
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, \ldots
from which equality (2) follows immediately.
4. Noting that|eta| < xi|\eta|<\xi, formula (8) shows us that the sequence (x_(n)x_{n}) converges tosqrtc\sqrt{c}. But we can approximatesqrtc\sqrt{c}by other iterative sequences. Such a sequence(y_(n))_(n=0)^(oo)\left(y_{n}\right)_{n=0}^{\infty}is determined by the recurrence relation
4. Noting that|eta| < xi|\eta|<\xi, formula (8) shows us that the sequence (x_(n)x_{n}) converges tosqrtc\sqrt{c}. But we can approximatesqrtc\sqrt{c}by other iterative sequences. Such a sequence(y_(n))_(n=0)^(oo)\left(y_{n}\right)_{n=0}^{\infty}is determined by the recurrence relation
{:(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*}
where we assume thaty_(0) > 0y_{0}>0and thatalpha\alphais a positive rational number. By limiting ourselves only to rational approximations ofsqrtc\sqrt{c}, the case ofalpha\alphapositive and irrational is of no interest.
All terms of the sequence (y_(n)y_{n}) are positive and we have
(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, \ldots
from where
{:(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*}
and, since|(alpha-sqrtc)/(alpha+sqrtc)| < 1\left|\frac{\alpha-\sqrt{c}}{\alpha+\sqrt{c}}\right|<1, it follows that the sequence (y_(n)y_{n}) converges tosqrtc\sqrt{c}.
The convergence of the sequence (x_(n)x_{n}) towardssqrtc\sqrt{c}is of order 2 while that of the sequence(y_(n))\left(y_{n}\right)only of order 1.
The convergence of the sequence (x_(n)x_{n}) towardssqrtc\sqrt{c}is of order 2 while that of the sequence(y_(n))\left(y_{n}\right)only of order 1.
Let's suppose thatalpha=y_(0)=[sqrt() bar(o)]=a\alpha=y_{0}=[\sqrt{ } \bar{o}]=a, then from (11) it follows that
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, \ldots
continuing to posexi=a+sqrtc,eta=a-sqrtc\xi=a+\sqrt{c}, \eta=a-\sqrt{c}.
This formula shows us that if the numberccis of the form (3) and ifalpha=y_(0)=a\alpha=y_{0}=a, we have
(12)
This formula shows us that if the numberccis of the form (3) and ifalpha=y_(0)=a\alpha=y_{0}=a, we have
(12)
y_(n)=r_(n),n=0,1,dotsy_{n}=r_{n}, n=0,1, \ldots
so that the recurrence relation (10), with the initial conditiony_(0)=[sqrtc]y_{0}=[\sqrt{c}], gives us precisely the sequence of reductions of the arithmetic continued fraction ofsqrtc\sqrt{c}.
5. Let us now propose to determine the natural numbercc(non-square) and the rational number (positive)alpha\alphaso that the recurrence relation (10) gives precisely the sequence of reductions(r_(n))_(n=0)^(oo)\left(r_{n}\right)_{n=0}^{\infty}of the arithmetic continued fraction ofV bar(c)V \bar{c}.
5. Let us now propose to determine the natural numbercc(non-square) and the rational number (positive)alpha\alphaso that the recurrence relation (10) gives precisely the sequence of reductions(r_(n))_(n=0)^(oo)\left(r_{n}\right)_{n=0}^{\infty}of the arithmetic continued fraction ofV bar(c)V \bar{c}.
For this we designate bya=[sqrtc]a=[\sqrt{c}]and let's askc=a^(2)+kc=a^{2}+k. If[a,b,d,e,dots][a, b, d, e, \ldots]is the arithmetic continued fraction ofsqrtc\sqrt{c}Andr_(0),r_(1),r_(2),r_(3),dotsr_{0}, r_{1}, r_{2}, r_{3}, \ldots, are its successive reductions, of
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}
it results
(13)
(13)
b=(a+alpha)/(k)b=\frac{a+\alpha}{k}
Of
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}
it results
(14)
(14)
d=(alpha-a+kb)/(2ab+1-kb^(2))d=\frac{\alpha-a+k b}{2 a b+1-k b^{2}}
At the end of
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)}
taking into account (13) and (14), we deduce
{:(15)e=(2alpha)/(k):}\begin{equation*}
e=\frac{2 \alpha}{k} \tag{15}
\end{equation*}
But the numbersb,eb, eare integers (positive), it therefore follows that2b-e=(2a)/(k)2 b-e=\frac{2 a}{k}is also whole. It follows that2a2 ais divisible bykk, so that the numberccis of the form (3). In this casee=be=b, SOalpha=a\alpha=a.
So we have the property expressed by
Theorem 2. For they_(n),n=0,1,dotsy_{n}, n=0,1, \ldots, 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 ofsqrtc\sqrt{c}, it is necessary and sufficient that it be of the form (3) and thatalpha=[sqrtc]=a\alpha=[\sqrt{c}]=a.
6. Note that ifccis any positive number and if we have (1), withx_(0) > 0x_{0}>0, it results
Theorem 2. For they_(n),n=0,1,dotsy_{n}, n=0,1, \ldots, 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 ofsqrtc\sqrt{c}, it is necessary and sufficient that it be of the form (3) and thatalpha=[sqrtc]=a\alpha=[\sqrt{c}]=a.
6. Note that ifccis any positive number and if we have (1), withx_(0) > 0x_{0}>0, it results
(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, \ldots
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)^(oo)\left(x_{n}\right)_{n=0}^{\infty},(y_(n))_(n=0)^(oo)\left(y_{n}\right)_{n=0}^{\infty}are given by the recurrence relations (1), (10), with the initial valuesx_(0)=y_(0)=alphax_{0}=y_{0}=\alpha:
1^(@)1^{\circ}They converge towardssqrtc\sqrt{c}.
2^(@)2^{\circ}We havex_(n)=y_(2^(n)-1),n=0,1,dotsx_{n}=y_{2^{n}-1}, n=0,1, \ldots
Received on May 3, 1967
Institute of Calculation
Academy of the Socialist Republic of Romania, Cluj Branch
1^(@)1^{\circ}They converge towardssqrtc\sqrt{c}.
2^(@)2^{\circ}We havex_(n)=y_(2^(n)-1),n=0,1,dotsx_{n}=y_{2^{n}-1}, n=0,1, \ldots
Received on May 3, 1967
Institute of Calculation
Academy of the Socialist Republic of Romania, Cluj Branch
BIBLIOGRAPHY
- 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.
- Mikusinski, J. On Newton's method of approximation. Annales Polon. Math., I (1954), 184-194.
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