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Let the continued fraction be denoted by
1 1 1
а2+ а2 + a +
then Pnqn-1-Pn-1 ¶n=(an Pn-1+Pn−2) In−1−Pn−1(An In−1 +In−2) =(−1)(Pn-1Jn—2—Pn−2 In−1) =(−1)2(Pn-2¶n—3—Pn−3Jn−2), similarly,
But P2-P12= (α, α2+ 1) −α2. α=1=(−1)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
Pn-1 Pn In-1~Pn-1 In 1
476. Each convergent is nearer to the continued fraction than any of the preceding convergents.
Let x denote the continued fraction, and Pn,
... x ~ Pn _ k ( Pn÷1 In~Pn In+1)
three consecutive convergents; then x differs from
taking the complete (n+2)th quotient in the place of an+2; denote this by k; thus
Pu÷l Pu÷2 In In÷1 In+2 Pn÷2 only in
Now k is greater than unity, and In is less than n+1; hence on both accounts the difference between Pn+1 and 2 is less than In+1 the difference between Pa and x; that is, every convergent is
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 Pn Pn+1, Pn+2 be three consecutive convergents, and let In In+1 In+2
k denote the complete (n+2)th quotient;
Now k is greater than 1, therefore the difference between the 1 continued fraction x, and any convergent, Pa is less than In In+1 and greater than
Again, since +1>In, the error in taking 1 1 less than and greater than
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
; that is, less than
; hence the larger In (an+19n+In-1) an+1 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.
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 1, we have only to calculate the successive convergents up to Pa, where 92 is greater
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
The successive convergents are
This last convergent which precedes the large quotient 25 is a very near approximation, the error being less than 1 therefore less than
25 × (100)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, P, Pa-1 two consecutive
In In-1 2 convergents, 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 P-1 [Art. 476]; and since x lies between Pn
it follows that must lie between
Pn-1 Yn-1' Hence
EXAMPLES XLV. a.
Calculate the successive convergents to
1 1 1 1 1 1 1+ 3+ 5+ 1+ 1+ 2
that is, an integer less than a fraction; which is impossible. Therefore P must be nearer to the continued fraction than 2.
1 1 1 1 1 1 1 2+ 2+ 3+ 1+ 4+ 2+ 6 1 1 1 1 1 1 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.
12. Find limits to the error in taking yards as equivalent 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+ which differs from the true value by less than 0001. 14. Show by the theory of continued fractions that 1 from 1.41421 by a quantity less than
729 2318 11. 4-316.
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.
-2, that is √19-2
after this the quotients 2, 1, 3, 1, 2, 8 recur; hence
1 1 1 1 1 1
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 19+4; this is 2, and the remain√19+4
We then multiply numerator and denominator by the surd conjugate to √19–2, so that after inverting the result we begin a new line with a rational