then Pan-1-Pn-19n=(anPn-1+P-2) In-1-Pn-1(An In-1+In-2)

When the continued fraction is less than unity, this result will still hold if we suppose that a1=0, and that the first convergent is zero.

NOTE. When we are calculating the numerical value of the successive convergents, the above theorem furnishes an easy test of the accuracy of the work.

COR. 1. Each convergent is in its lowest terms; for if pn and In had a common divisor it would divide Pn In-1-Pn-1Jn, or unity; which is impossible.

COR. 2. The difference between two successive convergents is a fraction whose numerator is unity, and whose denominator is the product of the denominators of these convergents; for

476. Each convergent is nearer to the continued fraction than any of the preceding convergents.

Let x denote the continued fraction, and Pn, Pu±1, Pu÷2

In In+1 In+2

In+2

three consecutive convergents; then x differs from Pa+2 only in taking the complete (n+2)th quotient in the place of an+2; denote this by k; thus kpn+1+Pn; X=; "kqn+1+qa

1

and

Pn+1

~x=

Pn+19n Pn In+1

In+1 In+1(kqn+1+qn) In+1(k¶n+1+qn)

Now k is greater than unity, and In is less than In+1; hence on both accounts the difference between P+1 and a is less than In+1 the difference between Pa and x; that is, every convergent is

In

nearer to the continued fraction than the next preceding convergent, and therefore nearer than any preceding convergent.

Combining the result of this article with that of Art. 471, it follows that

The convergents of an odd order continually increase, but are always less than the continued fraction;

The convergents of an even order continually decrease, but are always greater than the continued fraction.

477. To find limits to the error made in taking any convergent for the continued fraction.

Let P2, Pn+1, Pn+2 be three consecutive convergents, and let k denote the complete (n+2)th quotient;

In In+1 In+2

Now k is greater than 1, therefore the difference between the continued fraction x, and any convergent,

1

and greater than

In (qu+1+In)

Pn

1

is less than

In

In In+1

Again, since +1>In, the error in taking Pn instead of x is

478. From the last article it appears that the error in Pn instead of the continued fraction is less than

taking

In

1

In In+1

1

; that is, less than

1 ; hence the larger In (an+12n+2n-1) an+19n an+1 is, the nearer does Pn approximate to the continued frac

In

tion; therefore, any convergent_which_immediately precedes a large quotient is a near approximation to the continued fraction.

1

In

Again, since the error is less than it follows that in order to find a convergent which will differ from the continued fraction by less than a given quantity, we have only to calculate the successive convergents up to P, where q2 is greater

than a.

In

479. The properties of continued fractions enable us to find two small integers whose ratio closely approximates to that of two incommensurable quantities, or to that of two quantities whose exact ratio can only be expressed by large integers.

Example. Find a series of fractions approximating to 3.14159. In the process of finding the greatest common measure of 14159 and 100000, the successive quotients are 7, 15, 1, 25, 1, 7, 4. Thus 1 1 1 1 1 1 1 7+ 15+ 1+ 25+ 1+ 7+ 4

3.14159=3+

The successive convergents are

3 22 333 355

1' 7' 106' 113'

This last convergent which precedes the large quotient 25 is a very near approximation, the error being less than

1

and

25 × (113) 2'

480. Any convergent is nearer to the continued fraction than any other fraction whose denominator is less than that of the convergent.

Let x be the continued fraction,

convergents,

Pn Pn-1
In In-1

two consecutive a fraction whose denominator s is less than q•

If possible, let be nearer to x than P", then must be

nearer to x than Pn-1 [Art. 476]; and since z lies between Pr

In-1

In

that is, an integer less than a fraction; which is impossible. Therefore Pn must be nearer to the continued fraction than 2.

Express the following quantities as continued fractions and find the fourth convergent to each: also determine the limits to the error made by taking the third convergent for the fraction.

222

12. Find limits to the error in taking yards as equivalent

203

to a metre, given that a metre is equal to 1.0936 yards.

13. Find an approximation to

1 1 1 1 1 3+5+ 7+9+ 11+

1+

which differs from the true value by less than 0001.

RECURRING CONTINUED FRACTIONS.

481. We have seen that a terminating continued fraction with rational quotients can be reduced to an ordinary fraction with integral numerator and denominator, and therefore cannot be equal to a surd; but we shall prove that a quadratic surd can be expressed as an infinite continued fraction whose quotients recur. We shall first consider a numerical example.

Example. Express 19 as a continued fraction, and find a series of fractions approximating to its value.

after this the quotients 2, 1, 3, 1, 2, 8 recur; hence

It will be noticed that the quotients recur as soon as we come to a quotient which is double the first.

[Explanation. In each of the lines above we perform the same series of operations. For example, consider the second line: we first find the greatest integer in 19+4; this is 2, and the remain3 der is -2, that is 19–2 We then multiply numerator

19+4
3

and denominator by the surd conjugate to √19-2, so that after

inverting the result

3

5
19+2'

we begin a new line with a rational

denominator.]