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and omit all but the first partial fraction, its value will be

come.

Again, omitting all but the first and second partial fractions, we find.

1

1

3

2+1

7

3

Again, including one more partial fraction, we obtain

2+1

13 30°

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Our successive values obtained in this way, are 1, 4, 3, and.

These values may be derived in the following manner: Take the first partial fraction for the first value, multiply both numerator and denominator by the denominator of the next partial fraction, and we get ; if we increase this denominator by 1, it will give the second value. Again, multiplying numerator and denominator by the denominator of the next partial fraction, we get ; if we increase this numerator by the numerator of the last value, also increase the denominator

by the denominator of the last value, we get 13, which is the third value. Again, multiplying both numerator and denominator of this value, by the denominator of the next partial fraction, and to the respective products add the numerator and denominator of the preceding value, we obtain the last value,

68 157

This last value is the true value of the continued fraction, whilst the other values are successive approximations.

From what has been said we derive the following rule, for finding the vulgar fraction equivalent to a continued fraction.

RULE.

Consider the symbol as a fraction; then write this symbol and the first partial fraction for the two first terms of the approximate values. Multiply the numerator and denominator of the second approximate value by the denominator of the next partial fraction, and to the respective products add the numerator and denominator of the next preceding approximate value, and the result will be the succeeding approximate value. Thus continue to multiply the last approximative value, by the denominator of the succeeding partial fraction, and to the products add the numerator and denominator of the preceding approximate value, the result will be the succeeding approximate value.

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In this example we find the successive approximate values to be f,,,,, and .

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57. We will now show the application of the foregoing principles of continued fractions, by the solution of several practical questions.

1. Express approximately the fractional part of 24 hours, by which the solar year, of 365 days 5 hours 48 minutes and 48 seconds, exceeds 365 days.

5 hours 48 minutes 48 seconds=20928 seconds.

24 hours 86400 seconds. Therefore the true value of the fraction required is 20238-188.

864

Now, converting 18 into a continued fraction, by rule un

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converted into its approximative values (by rule under Art.

39

56) gives f, 1, 75, 33, 1, 1, 18.

The fraction, agrees with the correction introduced into the Calendar by Julius Cæsar, by means of bissextile or leap

year.

The fraction is the correction used by the Persian Astronomers, who add 8 days in every 33 years, by having seven regular leap-years, and then deferring the eighth until five years.

2. The French metre is 39.371 inches. Required the approximative ratio of the English foot to the metre.

In this example, the true ratio is 1399. Operating with this fraction as in the last example, we find some of the first approximative values to be, o, 73, 73, th 1335.

Hence, the foot is to the metre, as 3 to 10, nearly; a more correct ratio is as 32 to 105.

3. What are some of the approximative values of the ratio of the diameter of a circle to its circumference?

If we take the value of the circumference of the circle whose diameter is 1, to ten decimals, we have the vulgar fraction 888888, given to find its approximative values.

0

314159265359

Proceeding with this, as in the former examples, we find some of the first approximative values to be,, }}}, &c.

58. Continued fractions have been the means of obtaining elegant approximations to the roots of surds.

The method of converting a surd into a continued fraction is too complicated for us to explain in this place.

The square root of 2, or, what is the same thing, the ratio of the diagonal of a square to its side, is found to cqual the fol

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