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1

(b +1) d+ 6 1

ad (6c+1)+ ab +od+1 at 1

(6 ct-1)dto 1 =

; it

(ab + 1) cdtad + ab +1 ; and this method can easily be applied in any other case.

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318. EXAMPLES.

1. Find the value of the continued fraction

1

1 2+

1

3+2

Ans. 18

2. Find the value of the continued fraction

1
1

1

4+

3+2

Ans. 10

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An approximate value of this fraction is obviously obtained by omitting all its terms bcyond any assumed fraction, and obtaining the value of the resulting fraction, as in the previous article.

Thus we obtain, successively,

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and each of these values is easily shown to be more accurate than the preceding; for the second value is what the first becomes by substituting, for the denominator a, the

1 more accurate denominator a tý; the third is what the second becomes by substituting, for the denominator 6, the

1 more accurate denominator 6+

i

and so on.

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Approximate Values of Continued Fractions.

320. Theorem. The numerator of any approximate value, as the nth, is obtained from the numerators of the two preceding approximate values, the (n 1)st, and the (n 2)nd, by multiplying the (n 1)st numerator by the nth denominator contained in the given continued fraction, and adding to the result the numerator of the (n

2)nd approximate value.

The denominator of the nth approximate value is obtained in the same way from the two preceding denominators.

Demonstration. Let the (n-3)rd, (n - 2)nd, (n-1)st, and nth approximate values be, respectively,

K L M N

and K L M

N' and let the (n − 1)st and the nth denominators, contained in the given continued fraction, be p and q.

We shall suppose the proposition demonstrated for the (n − 1)st approximate value, and shall prove that it can thence be continued to the nth value; that is, we shall suppose it proved that

M

pL+K

M pL'+K Now it is plain, from the remarks at the end of the preceding article, that the nth value is deduced from the (n-1)st,

1 by changing p into pt ; which change, being made in

9 the preceding value, gives

N
N

p1
p+1)+K

(pL+K)a+L
(p L'+K')9+ L

Difference between successive Approximate Values.

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Hence we have, by substituting

M=pL+K,

M =pL'+K;
N Ma +L

N M'a + L
that is, the value required to satisfy the theorem.

If, therefore, it can be shown that the proposition is true for any approximate value, it follows that it must be true for every succeeding value. But the comparison of the values given in the preceding article shows that it is true for the third value, and therefore for every succeeding value.

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321. Theorem. If two succeeding approximate values are reduced to a common denominator equal to the product of their denominators, the difference of their numerators is unity.

Demonstration. Let the (n − 2)nd, (n — 1)st, and nth approximate values be

L M N M a+L

and L M

M'a+L the difference between the (n - 2)nd and (n − 1)st is

L M -L'M +

L'M' and that between the (n − 1)st and nth is M' NMN (MM-MM)a+ML-ML


M' N

M'N'
LM-L'M
=+

M'N
of both which differences the numerators are the same;
and, therefore, this is always the case.

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Approximate Value compared with True Value.

Now the first and second approximate values, as given in art. 319, are, when reduced to a common denominator, ab +1

ab

and
a (ab+1) a (a b + 1)

6: the difference of the numerators of which is 1; and, therefore, unity must always be the difference of two such numerators.

322. Theorem. - The approrimate values of a coninued fraction are alternately larger and smaller han its true value, the first being larger, the second smaller, and so on alternately.

Demonstration. Since, in the preceding demonstration, the subtraction of the (n − 1)st value from the (n − 2)nd, gave a fraction having the same numerator as that obtained, by its subtraction from the nth; we see that if the (n— 1 )st value is larger than the (n. 2)nd, it must also be larger than the nth ; and if the (n 1)st is smaller than the (n − 2)nd, it is also smaller than the nth.

But the true value is, by art. 319, nearer the (n-1)st value than the (n − 2)nd, and nearer the nth than the (n − 1)st; so that when the (n-1)st value is larger than the (n–2)nd, the true value must likewise be larger than the (n — 2)nd, and smaller than the (n — 1)st, and so on alternately; but when the (n − 1)st value is smaller than the (n − 2)nd, the true value must be smaller than the (n — 2)nd, and larger than the (n-1)st, and so on alternately.

Now the first value is, by the preceding article, larger than the second, and therefore the true value is smaller than the first, larger than the second, and so on alternately.

323. Theorem. Each approximate value of a continued fraction differs from the true value by a

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