ional fractions which approximate to any
quantity, and we can also estimate the error of our approximation. Thus
a continued fraction equivalent to [pi] (the ratio of the circumference
to the diameter of a circle) is

1 1 1 1 1 1
3 + - -- -- --- -- --
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 d