Continued Fractions and Matrices
Author(s): J. S. Frame
Source: The American Mathematical Monthly, Vol. 56, No. 2 (Feb., 1949), pp. 98-103
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2306169 .
Accessed: 28/02/2014 06:44
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .
http://www.jstor.org/page/info/about/policies/terms.jsp
.
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of
content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms
of scholarship. For more information about JSTOR, please contact [email protected].
.
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to
The American Mathematical Monthly.
http://www.jstor.org
This content downloaded from 158.110.11.166 on Fri, 28 Feb 2014 06:44:43 AM
All use subject to JSTOR Terms and Conditions
CLASSROOM NOTES
EDITED BY
C. B. ALLENDOERFER, HaverfordCollegeand InstituteforAdvancedStudy
forAdshouldbe sentto C. B. Allendoerfer,
forthisdepartment
Institute
All material
NewJersey.
vancedStudy,Princeton,
CONTINUED FRACTIONS AND MATRICES
MichiganState College
J. S. FRAME,
1. Introduction.The closest rational approximationP/Q (P, Q integers)to a
given positive real numberx, subject to the condition that the denominatorQ
shall not exceed a given positive integerN, is most easily found by means of
continued fractions.For example, successive approximationsto ir obtained by
this method are 3, 22/7, 333/106, 355/113, * - - . Continued fractionsmay also
be used to facilitatethe solution of diophantineequations.
It is our purpose to show how the principal theoremsin the theoryof continued fractions,and the actual computationof successive convergentsto a continued fractioncan be presentedquite simplyby the use of two-rowedmatrices
and their determinants.To a student familiarwith the multiplicationof tworowedmatricesa good introductionto continuedfractionscan be presentedin an
hour's lecture. For a student who has just been introduced to matrices,their
use in connectionwith continued fractionsprovides an easy application at the
elementarylevel.
2. Simple continued fractions.For a given positive real x let the integral
part be a, and the remainderri, with 0? r1<1. We definesuccessivelythe posi, an, (called partialdenominators)and the remaindersr2,
tive integersa2,a8,
r8p. . . rnso that
(1)
1
-=
rl
x = a,+ri,
1
a2+r2,**
rn-1
O< rk < 1.
=a+r,
and we write
(2)
x
a1
a2
+
a3 +
an-
+
an + rn
or
(2')
x=al+*a*,X
1
1
a2 +
a3 +
1
an-1 +
98
This content downloaded from 158.110.11.166 on Fri, 28 Feb 2014 06:44:43 AM
All use subject to JSTOR Terms and Conditions
1
an + rn
1949]
CLASSROOM
99
NOTES
Aftersimplifyingthe complex fraction(2) and collectingcoefficientsof r., we
have
P.
+
P +
x r=Sn
(3)
rnS.
Pn
Sh\
where Mn = (P
\Qn Tn
Q. + r.T.
is an integralmatrix,and where,forthe case n= 1, we have from(1)
a
(i)
(
)
Q, T,
to denote the nth denominatorin (3) by Dn;
It will be convenfient
(4)
Qn+ rnTn
Dn=
To obtain a recursionformulaforthe P's and Q's, we replace n by n -1 in equation (3), divide numeratorand denominatorby rn-1,and apply (1).
(S)
x =
Pn-1 + rn-lSn-1
=
+ rn) + Sn-1
Pn(an
+ rn)+
+ rn-,Tn-1 Qn-1(an
Qn-1
Tn71
=
(aizPn-1 + Sn-1) + rnPn-1
+ Tn-1)+ rnQn(anQn-1
On comparingcoefficientsof rnin (3) and (5) we have
(6)
Sn = Pn-1,
Pn = anPn-1+ Sn-1
Tn = Qn-ly
Qn = anQn-1+ Tn_l
Hence,
Pn = anPn-1+ Pn-2
Qn = anQn-1 + Qn-2.
and D1 = 1, so
Also from(4) and (5) the ratio Dn-l/Dn is seen to be rn-1,
(8)
1/Dn =
r1r2
. .
rn-l,
provided that none of the r's are 0. The set of equations (7) can most easily be
writtenin matrixform
(Pn
Pn-1
(P_1
Pn-2)
an I
Inll0an
By induction we then obtain the fundamentalrelation
(10)
M
=(0~
0
:) (1 Q0)(I
1
)
(1
0)(1
This content downloaded from 158.110.11.166 on Fri, 28 Feb 2014 06:44:43 AM
All use subject to JSTOR Terms and Conditions
1)
[February,
CLASSROOM NOTES
100
3. Theorems on continued fractions.Since the determinantof the matrix
Mn is the productof the determinantsof its factors,we obtain from(10) the importantrelations
Pn
(11)
Pn-1
(12)
(13)
x-
=
Qn-1
Qn
Qn_l QnQn-1
Qn
Pn
Pn
Pn + rnPn-1
Qn
Qn+ rnQn-1 Qn
=
=_
=-
-
rn(- 1)n1
QnDn
(-1)n-1
QnDn+l
of x, are
THEOREM I. The rationalfractionsPn/Qn,called the "convergents"
ofsuccessiveconvergents
is the
alternatelyless and greaterthanx, and thedifference
reciprocalof theproductof theirdenominators.Since thisproductbecomesinfinite
definedby (12) approachzero and alternatein sign as n inwithn, thedifferences
Its limitis x.
creases,so thesequencePn/Qnconverges.
The proofof these statementsfollowsdirectlyfrom(13) and (12).
Equation (3) may be replaced by the matrixequation
(14)
( 1)
Q
(~~~Qnn-1 ( l/Dn+l)
Solving for rn,which is the ratio Dn/Dn+l?,we have
(15)
(16)
rn =
-
X
n-Pn-1- Qn_l1X
an+1 + rn+i=
-Qnx + Pn
to
compute
Equation (16) makes it possible
an+, directlyif the matrix Mn is
from(10). For example, in the
obtained
known. The next matrix Mn+l is then
continued fractionexpansion for ir the partial denominator a5 is the largest
integerin (1067r-333)/(-1137r+355) =292.+
THEOREM II. A periodic continuedfractionrepresentsa root of a quadratic
equation.
Proof. If two differentremaindersare equal, equate them, using (15), and
solve forx.
from
THEOREM III. The rationalfractionPn/Qndefinedby (1) and (10) differs
x by not morethan 1/QnQn+l,and it approximatesthereal numberx moreclosely
thandoes any otherrationalfractionwitha denominatornot exceedingQn.
Proof.The difference(13) is equal in absolute value to 1/QnDn+l,whichby (4)
Suppose there were two integersA and B, with
is certainlyless than 1/QnQn+l.
0 <B <Qn, such that the fractionA/B were closer to x than Pn/Qn. Then we
should have
This content downloaded from 158.110.11.166 on Fri, 28 Feb 2014 06:44:43 AM
All use subject to JSTOR Terms and Conditions
1949]
(17)
1
1
n/
QnDn+l\
101
NOTES
CLASSROOM
~~
~~AX
1
1
A\
B
QxDn+l
Addingl/QnDn+lto each termand replacingx by its value in (13) gives
0<
( 1)n( n _-A) <
<
2
or
(18)
0 < (-l)n(BPn
- AQn) < 2B/Qn+1.
It is easilyshownthattheinequality(18) cannotbe satisfiedby integers
A, B
withB <Qn. For thenthe rightmemberis less than 2, and the integerin the
middlemembercould only be 1. By (11) the latterconditionis satisfiedby
takingA = P1-, B = Qn-i,and thisis the onlychoiceforwhich0< B <Qn, as
we shall see in ?4 below.However,withB =-Q., we are led from(18) to the
inequality1 <2Qn_1/Q.+,,
whichis impossibleby (7).
4. Continuedfractions
forrationalnumbers.If at somestagein (1) we have
every
rn=0, thenx reducesby (3) to the rationalnumberPn/Qn.Conversely,
rationalnumberis represented
by a terminating
continuedfraction.
The nextto
is important
thelastconvergent
insolvingthelineardiophantine
equation(19).
Giventwo integersPn and Qnwithoutcommonfactor,to findall pairsof
integersu and v whichsatisfytheequation
Pnu - Q.v = 1.
(19)
The solutionis givenby
(20)
v = (-l)p.,_
+ NP,
U = (-1)
+ NQn
Qn1
N anyinteger,
wherePn-l and Qn-, are obtainedfromthe continuedfractionexpansionof
Pn/Qn.
5. Generalcontinuedfractions.If the simplecontinuedfraction(2) is replaced by the moregeneralform
(21)
+ b2
a2 + b3
a3 + b4
an_1+
bn
an + rn
This content downloaded from 158.110.11.166 on Fri, 28 Feb 2014 06:44:43 AM
All use subject to JSTOR Terms and Conditions
102
[February,
CLASSROOM NOTES
in which r. is not necessarilyless than 1, and ak and bkare not necessarilypositive integers,then it is readily shown that equations (7) become
+
anPn-1
Pn=
(22)
bnPn_2
anQn-1+ bnQn-2
Qn(=
and the fundamentalrelation (10) may be written
(23)
Pn n-l
M =-
1\( a21\
10/\b2
QnQn-1
(anl\
0
\bn 0,
where we definebi=Qi=Po=1,
Qo=0.
By taking determinantsin (23), the differencebetween successive convergentsto x is seen to equal (- 1)%bb2. . . bn/QnQn-1.Since the sum of these
terms may not always converge, the question of convergencein this case is
more complicated than that for the simple continued fractions,and we shall
not discuss it at this time.
It may be of interestto include without proofthe expansion
(24)
x
tan-1
x
= 1+
X2
3 +
(2x)2
5 + (3x)2
.7 +
which is typical of many continued fractionexpansions for analytic functions
obtained fromhypergeometricseries.
Settingx = 1 in (24) we see that the value of r can be expressedas a continued
are given by a simple law
fractionin which the coefficients
(25)
4
-=1
7r
+
1
3+ 4
5 + 9
7+
This expansion is of interestbecause of its regularity,but it does not enjoy the
rapidity of convergence which characterizes the simple continued fractions.
Successive convergentsin (25) are given by
D
D
1 18 3 18 5 1 {7 1> 9 1
(6 Pn+1 PnA
1
9
-1
1
O 4
(26)Qn1
Qn
6
...(
{2n + 1 18
n2 11
We may derive from(26) an alternativeexpansion forthe convergentsto 4/w,
in whichthe partial numeratorsbi are all 1 but the partial denominatorsare not
all integers.This is as follows:
This content downloaded from 158.110.11.166 on Fri, 28 Feb 2014 06:44:43 AM
All use subject to JSTOR Terms and Conditions
1949]
(27)
103
CLASSROOM NOTES
:P+1 P"
Qnl1
Qn
C11)C31
I
0
1
(5/4 1) (28/9 1) (81/64 1)
0
1
1
0
0
1
0
(an+
1
1)
0
where
a2n =
=
(28)
lim
2.4 ... (2n-
(4n -1)
(2n
(4n + 1)(213
2))2
-1)2
a2n =
n= 00
lim
a2fl+1 =
4
-
Since each matrixin (27) has the determinant(-1), formulas(12) and (13) are
valid in this case. The limitsin (28) are easily obtained fromthe Wallis product
formulafor u-/2.A close estimate forthe errorof Pl/Qn is (V/2- 1)2n-1.
DERIVATION OF THE TANGENT HALF-ANGLE FORMULA
of Oregon
F. E. WOOD, University
The followingderivationof the formulafortan 0/2appears to be an improvement over standard derivations,forit gives the result directlywithout a complicated discussion of the appropriate algebraic sign.
From the equation
sin
2
= sin (
\
-=
,2
-
sin 0 cos -
-
cos 0 sin
one obtains
(1 + cos 0) sin -
0
2
0
= sin 0 cos -
2
.
Consequently:
tan-
0
2
=
sin 0/2
sin 0
cos0/2
1 + cos 0
-
This content downloaded from 158.110.11.166 on Fri, 28 Feb 2014 06:44:43 AM
All use subject to JSTOR Terms and Conditions
2