approximation. Thus a continued fraction equivalent to π (the
ratio of the circumference to the diameter of a circle) is
| 3 + | 1 | 1 | 1 | 1 | 1 | 1 | . . . | ||||||
| 7 | + | 15 | + | 1 | + | 292 | + | 1 | + | 1 | + |
of which the successive convergents are
| 3 | , | 22 | , | 333 | , | 355 | , | 103993 | , &c., |
| 1 | 7 | 106 | 113 | 33102 |
the fourth of which is accurate to the sixth decimal place, since the error lies between 1/q4q5 or .0000002673 and a6/q4q6 or .0000002665.
Similarly the continued fraction given by Euler as equivalent to 1/2(e − 1) (e being the base of Napierian logarithms), viz.
| 1 | 1 | 1 | 1 | 1 | |||||
| 1 | + | 6 | + | 10 | + | 14 | + | 18 | + . . ., |
may be used to approximate very rapidly to the value of e.
For the application of continued fractions to the problem “To find the fraction, whose denominator does not exceed a given integer D, which shall most closely approximate (by excess or defect, as may be assigned) to a given number commensurable or incommensurable,” the reader is referred to G. Chrystal’s Algebra, where also may be found details of the application of continued fractions to such interesting and important problems as the recurrence of eclipses and the rectification of the calendar (q.v.).
Lagrange used simple continued fractions to approximate to the solutions of numerical equations; thus, if an equation has a root between two integers a and a + 1, put x = a + 1/y and form the equation in y; if the equation in y has a root between b and b + 1, put y = b + 1/z, and so on. Such a method is, however, too tedious, compared with such a method as Homer’s, to be of any practical value.
The solution in integers of the indeterminate equation ax + by = c may be effected by means of continued fractions. If we suppose a/b to be converted into a continued fraction and p/q to be the penultimate convergent, we have aq − bp = +1 or −1, according as the number of convergents is even or odd, which we can take them to be as we please. If we take aq−bp = +1 we have a general solution in integers of ax + by = c, viz. x = cq − bt, y = at − cp; if we take aq − bp = −1, we have x = bt − cq, y = cp − at.
An interesting application of continued fractions to establish a unique correspondence between the elements of an aggregate of m dimensions and an aggregate of n dimensions is given by G. Cantor in vol. 2 of the Acta Mathematica.
Applications of simple continued fractions to the theory of numbers, as, for example, to prove the theorem that a divisor of the sum of two squares is itself the sum of two squares, may be found in J. A. Serret’s Cours d’Algèbre Supérieure.
2. Recurring Simple Continued Fractions.—The infinite continued fraction
| a1 + | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | ||||||||||
| a2 | + | a3 | + . . . + | an | + | b1 | + | b2 | + . . . + | bn | + | b1 | + | b2 | + . . . + | bn | + | b1 | + . . ., |
where, after the nth partial quotient, the cycle of partial quotients b1, b2, . . ., bn recur in the same order, is the type of a recurring simple continued fraction.
The value of such a fraction is the positive root of a quadratic equation whose coefficients are real and of which one root is negative. Since the fraction is infinite it cannot be commensurable and therefore its value is a quadratic surd number. Conversely every positive quadratic surd number, when expressed as a simple continued fraction, will give rise to a recurring fraction. Thus
| 2 − √3 = | 1 | 1 | 1 | 1 | 1 | |||||
| 3 | + | 1 | + | 2 | + | 1 | + | 2 | + . . ., |
| √28 = 5 + | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | ||||||||
| 3 | + | 2 | + | 3 | + | 10 | + | 3 | + | 2 | + | 3 | + | 10 | + . . . |
The second case illustrates a feature of the recurring continued fraction which represents a complete quadratic surd. There is only one non-recurring partial quotient a1. If b1, b2, . . ., bn is the cycle of recurring quotients, then bn = 2a1, b1 = bn−1, b2 = bn−2, b3 = bn−3, &c.
In the case of a recurring continued fraction which represents √N, where N is an integer, if n is the number of partial quotients in the recurring cycle, and pnr/qnr the nrth convergent, then p2nr −Nq2nr = (−1)nr, whence, if n is odd, integral solutions of the indeterminate equation x2 − Ny2 = ±1 (the so-called Pellian equation) can be found. If n is even, solutions of the equation x2 −Ny2 = +1 can be found.
The theory and development of the simple recurring continued fraction is due to Lagrange. For proofs of the theorems here stated and for applications to the more general indeterminate equation x2 −Ny2 = H the reader may consult Chrystal’s Algebra or Serret’s Cours d’Algèbre Supérieure; he may also profitably consult a tract by T. Muir, The Expression of a Quadratic Surd as a Continued Fraction (Glasgow, 1874).
1. The Evaluation of Continued Fractions.—The numerators and denominators of the convergents to the general continued fraction both satisfy the difference equation un = anun−1 + bnun−2. When we can solve this equation we have an expression for the nth convergent to the fraction, generally in the form of the quotient of two series, each of n terms. As an example, take the fraction (known as Brouncker’s fraction, after Lord Brouncker)
| 1 | 12 | 32 | 52 | 72 | |||||
| 1 | + | 2 | + | 2 | + | 2 | + | 2 | + . . . |
Here we have
un+1 = 2un + (2n−1)2un−1,
whence
un+1 − (2n + 1)un = −(2n − 1){un − (2n − 1)un−1},
and we readily find that
| pn | = 1 − | 1 | + | 1 | – | 1 | + . . . ± | 1 | , |
| qn | 3 | 5 | 7 | 2n + 1 |
whence the value of the fraction taken to infinity is 1/4π.
It is always possible to find the value of the nth convergent to a recurring continued fraction. If r be the number of quotients in the recurring cycle, we can by writing down the relations connecting the successive p’s and q’s obtain a linear relation connecting
pnr+m, p(n−1)r+m, p(n−2)r+m
in which the coefficients are all constants. Or we may proceed as follows. (We need not consider a fraction with a non-recurring part). Let the fraction be
| a1 | a2 | ar | a1 | ||||
| b1 | + | b2 | + . . . + | br | + | b1 | + . . . |
| Let un ≡ | pnr+m | ; then un = | a1 | a2 | ar | , | ||
| qnr+m | b1 | + | b2 | + . . . + | br + un1 |
leading to an equation of the form Aunun−1 + Bun + Cun−1 + D = 0, where A, B, C, D are independent of n, which is readily solved.
2. The Convergence of Infinite Continued Fractions.—We have seen that the simple infinite continued fraction converges. The infinite general continued fraction of the first class cannot diverge for its value lies between that of its first two convergents. It may, however, oscillate. We have the relation pnqn−1 − pn−1qn = (−1)nb2b3. . .bn, from which
| pn | – | pn−1 | = (−1)n | b2b3 . . . bn | , |
| qn | qn−1 | qnqn−1 |
and the limit of the right-hand side is not necessarily zero.
The tests for convergency are as follows:
Let the continued fraction of the first class be reduced to the form
| d1 + | 1 | 1 | 1 | |||
| d2 | + | d3 | + | d4 | + . . ., |
then it is convergent if at least one of the series d3 + d5 + d7 + . . ., d2 + d4 + d6 + . . . diverges, and oscillates if both these series converge.
For the convergence of the continued fraction of the second class there is no complete criterion. The following theorem covers a large number of important cases.
“If in the infinite continued fraction of the second class an ≥ bn + 1 for all values of n, it converges to a finite limit not greater than unity.”
3. The Incommensurability of Infinite Continued Fractions.—There is no general test for the incommensurability of the general infinite continued fraction.
Two cases have been given by Legendre as follows:—
If a2, a3, . . ., an, b2, b3, . . ., bn are all positive integers, then
I. The infinite continued fraction
| b2 | b3 | bn | |||
| a2 | + | a3 | + . . . + | an | + . . . |
converges to an incommensurable limit if after some finite value of n the condition an ≮ bn is always satisfied.
II. The infinite continued fraction
| b2 | b3 | bn | |||
| a2 | − | a3 | − . . . − | an | − . . . |
converges to an incommensurable limit if after some finite value of n the condition is always satisfied, where the sign > need not always occur but must occur infinitely often.
The functions pn and qn, regarded as functions of a1, . . ., an, b2, . . ., bn determined by the relations
pn = anpn−1 + bnpn−2,
qn = anqn−1 + bnqn−2,
with the conditions p1 = a1, p0 = 1; q2 = a2, q1 = 1, q0 = 0, have been studied under the name of continuants. The notation adopted is
| pn = K( | b2, . . ., bn | ), | |
| a1, | a2, . . ., an |
and it is evident that we have
| qn = K( | b3, . . ., bn | ). | |
| a2, | a3, . . ., an |
The theory of continuants is due in the first place to Euler. The reader will find the theory completely treated in Chrystal’s Algebra, where will be found the exhibition of a prime number of the form 4p + 1 as the actual sum of two squares by means of continuants, a result given by H. J. S. Smith.
