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CHAPTER XLV.

CONTINUED FRACTIONS.

467. AN expression of the form a+▪

c+

m

n

quotient and p the remainder; thus

......

continued fraction; here the letters a, b, c, may denote any quantities whatever, but for the present we shall only con1 sider the simpler form a1+ where a1, a2, A3, are

m

b

1

d

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e+

a2+ azt... positive integers. This will be usually written in the more compact form 1 1 a2+ ag+

a1+

n

p

...

...

468. When the number of quotients a, a, a, is finite the continued fraction is said to be terminating; if the number of quotients is unlimited the fraction is called an infinite continued fraction.

It is possible to reduce every terminating continued fraction to an ordinary fraction by simplifying the fractions in succession beginning from the lowest.

469. To convert a given fraction into a continued fraction.

Let be the given fraction; divide m by n, let a1 be the

is called a

...

divide n by p, let a2 be the quotient and q the remainder; thus

n

q

1

P

= α2+ = = α2+=;
p

p

P

q

divide p by q, let Thus

on.

m

n

a3

be the quotient and r the remainder; and so

1

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=a,+

ag +

If m is less than n, the first quotient is zero, and we put

m

n

159)832(5
795

1

n

m

and proceed as before.

It will be observed that the above process is the same as that of finding the greatest common measure of m and n; hence if m and n are commensurable we shall at length arrive at a stage where the division is exact and the process terminates. Thus every fraction whose numerator and denominator are positive integers can be converted into a terminating continued fraction.

1 1

A2 + A3+

Example. Reduce to a continued fraction.

832
159

Finding the greatest common measure of 832 and 159 by the usual process, thus:

37)159(4
148

5+

11)37(3

33

4)11(2

8

3)4(1
3

1)3(3

We have the successive quotients 5, 4, 3, 2, 1, 3; hence

832

1 1 1 4+ 3+2+1+ 3

159

470. The fractions obtained by stopping at the first, second, third, quotients of a continued fraction are called the first, second, third, convergents, because, as will be shown in Art. 476, each successive convergent is a nearer approximation to the true value of the continued fraction than any of the preceding convergents.

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471. To show that the convergents are alternately less and greater than the continued fraction.

Let the continued fraction be a1+

...

1 1

a2+ a +

The first convergent is a,, and is too small because the part is omitted. The second convergent is a1+ and is

1

1

1

a2+ az +

a2

too great because the denominator a, is too small. The third 1 1 convergent is a1+ and is too small because a2+

1

2

a2+ az

ag

a1+

a2+ a + a 4+ then the first three convergents are

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great; and so on.

When the given fraction is a proper fraction a1=0; if in this case we agree to consider zero as the first convergent, we may enunciate the above results as follows:

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The convergents of an odd order are all less, and the convergents of an even order are all greater, than the continued fraction.

472. To establish the law of formation of the successive convergents.

Let the continued fraction be denoted by

1 1 1

;

is too

...

and we see that the numerator of the third convergent may be formed by multiplying the numerator of the second convergent by the third quotient, and adding the numerator of the first convergent; also that the denominator may be formed in a similar manner.

Suppose that the successive convergents are formed, in a similar way; let the numerators be denoted by P1, P2, P31 and the denominators by 91, 92, 93

Assume that the law of formation holds for the nth convergent; that is, suppose

Pn=anPn-1+Pn-2, ¶n=An¶n−1+¶n−2·

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The (n+1)th convergent differs from the nth only in having the quotient an+ in the place of an; hence the (n+1)th con

1

An+1

vergent

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Pn+1=an+1Pn+Pn−1, Jn+1=An+1&n + In−1,

we see that the numerator and denominator of the (n+1)th convergent follow the law which was supposed to hold in the case of the nth. But the law does hold in the case of the third convergent, hence it holds for the fourth, and so on; therefore it holds universally.

674

Example. Reduce to a continued fraction and calculate

313

the successive convergents. 674

1 1 1 1 1 6+ 1+ 1+ 11+ 2 The successive quotients are 2, 6, 1, 1, 11, 2.

By Art. 469, =2+ 313

The fifth convergent =

2 13 15 28 323 674

The successive convergents are 1' 6' 7' 13' 150' 313'

[Explanation. With the first and second quotients take the first and second convergents which are readily determined. Thus, in this example, 2 is the first convergent, and 2+1

6

13 or the second convergent. The numerator of the third con6 vergent, 15, equals the numerator of the preceding convergent, 13, multiplied by 1, the third quotient, plus 2, the numerator of the convergent next preceding but one. The denominator is formed in a similar manner: thus 7=1x6+1.

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473. If the fraction is a proper fraction, we may consider zero as the first convergent, and proceed as follows:

84

Reduce to a continued fraction, and calculate the suc

227

cessive convergents.

Proceeding as in Art. 469,

227)84(0
00

We obtain

1 1 1

1 1 1 1

2+ 1+ 2+ 2 + 1+ 3+ 2

The successive quotients are 0, 2, 1, 2, 2, 1, 3, 2.

Writing for the first convergent we have, [Art. 472],

1

84)227(2
168

0+

475. If

011 3
I' 2' 3' 8'

59)84(1

59

25)59 (...

7 10 37 84 19' 27' 100' 227

474. It will be convenient to call an the nth partial quotient; 1 the complete quotient at this stage being an+

1

An++ an+2+ We shall usually denote the complete quotient at any stage by k.

We have seen that

Pn _ AnPn−1+Pn-2
In anqn-1+9n-2

x=

Let the continued fraction be denoted by x; then x differs from Pa only in taking the complete quotient k instead of the partial

In

quotient an ; thus

kPn-1+Pn-2.
kqn-1+qn-2

Pn be the nth convergent to a continued fraction, then

qn

Pn ¶n-1 — Pn-19¶n = ( − 1)".

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