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Let the continued fraction be denoted by

1 1 1 a,+

. ;

+ 4z + 14+
then Prqn-1-Pn-19n=(an Pn-1+P. ) 9n-1-Pn-1(an In-1+ In-

=(-1)(Pr-19n-2-Pn-2n-1)
=(-1)(Pn-29n-3-Pn-3 In-w), similarly,

=(-1)-(P.11-P192). P20-2192=(q, ag+1) - 07.0=1=(-1); hence

Pran-1-Pn-191=(-1)". When the continued fraction is less than unity, this result will still hold if we suppose that ay=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 Pr In-1-Pn-19n, 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 dlenominators of these convergents; for Pn Pn-1 Pr (n-1~P~-191

1 In 91-1 In In-1 In 2n-1 476. Each convergent is nearer to the continued fraction than any of the preceding convergents. Let x denote the continued fraction, and Pn, Puul, Put?

In In+1 2n+? three consecutive convergents; then x differs from I'm+2 only in

In+2 taking the complete (n+2)th quotient in the place of an+2; denote this by k; thus

kan+i+an
...~

Pn _ ki Pn+19n ~Pn In+1) k
In (kW*+-+9n) qu(kqn+1+In)

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kpn+1+Pn;

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X=

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and Pn+1

1 Pn+19n ~PnIn+1 9n+1 In+1(kqn+1+In)(n+1(kqn+1+qn) Now k is greater than unity, and Պ»

is less than 9n+1; hence on both accounts the difference between Pn+1 and 2 is less than

912+1 the difference between Pn and x; that is, every convergent is

200 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 oddl order continually increase, but are always less than the continued fraction ;

. The convergents of an even oriler 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 Pr, Pn+1, Pn+2 be three consecutive convergents, and let

91 In+1 Yn+2
k denote the complete (n+2)th quotient;
then

kpn+1+Pn.
k9n+1+1
K:

1
an Ink4n+1+2)

In
In In+1+

k Now k is greater than 1, therefore the difference between the

1 continued fraction x, and any convergent, Pn is less than

1

In

InIn+1 and greater than

In(qr+1+9) Again, since qu+1>In, the error in taking Pn instead of x is 1

1

In less than and greater than

29**+1

Pn

.. 3

In?

478. From the last article it appears that the error in

1 taking

Pn instead of the continued fraction is less than
In

In9n+1

or

1

1 ; that is, less than hence the larger In(an+In+ 2n-1)

an+1912

Pn anti is, the nearer does

approximate to the continued fraction; therefore, any convergent which immediately precedes a large quotient is a near approximation to the continued fraction.

1 Again, since the error is less than it follows that in order to find a convergent which will differ from the continued

1

we have only to calculate the successive convergents up to Pn, where qno is greater

an

2

fraction by less than a given quantity 3

than a.

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 3:14159=3+

7+ 15+ 1+ 25+ 1+ 7+ 4 The successive convergents are

3 22 333 355
ī 7' 106' 113

i This last convergent which precedes the large quotient 25 is a

1 very near approximation, the error being less than

and 1

25 x (113) therefore less than

25 X (100)2

or .000004. 450. 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, Pn, Pn-1 two consecutive

In In-1 convergents, ? a fraction whose denominator s is less than que If possible, let ? be nearer to x than Pr, then

must be

S

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an

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

Pn Pn-1

In and it follows that - must lie between and 9n-1

In-1 Hence rPn-P.Pn-1, that is <

1

; In-1 9n 9n-1

9n9n-1 ... 79n-1~8p>-1<;

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

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S

S

EXAMPLES XLV. a.

Calculate the successive convergents to 1. 2+

1 1 1 1 1 1
1+ 3+ 5+ 1+ 1+ 2

1 1 1 1 1 1 1
2.
2+ 2+ 3+ 1+ 4+ 2+ 6

1 1 1 1 1 1
3. 3+

3+ 1+ 2+ 2+ 1+ 9 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. 253 251 1189

729
4.
5.
6.

7.
179
802
3927

2318
8. •37.
9. 1.139.

10. 3029.

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 1+

3+ 5+ 7 + 9+ 11+ which differs from the true value by less than .0001.

99 14. Show by the theory of continued fractions that differs

1

70 from 1.41421 by a quantity less than

11830

11. 4.316.

.....

RECURRING CONTINUED FRACTIONS.

=2+

=2+

V19—3=1+

19-3-3+

=1+

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.

3 19=4+ (19-4)=4+

719+4 19+4 19-2

5 3

3

19+2
V19+2

2
=1+
5

5

1973 19+3

5 =3+

=3 2

2

719+3 19+3

3 =1+

V19-2 5

5
19+2

1
V19—4
=2+
3

3

19+4 19+4=8+(v19--4)=8+. after this the quotients 2, 1, 3, 1, 2, 8 recur; hence

1 1 1 1 1 1 V19=4+

2 + 1+ 3+ 1+ 2+ 8+ 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 V19+4; this is 2, and the remain

3 der is 719+4

We then multiply numerator 3

3 and denominaior by the surd conjugate to V19-2, so that after

5 inverting the result

we begin a new line with a rational

v19+2' denominator.]

V19+23

=2+

-2, that is V19-2

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