If the present value thus found be divided by the value of the life, the quotient will be the annuity payable yearly, By using the value of two or three lives found by the preceding rules, we may apply the same rule.

Exam. 9. How much must a person aged 32 years, pay, at 5 cent., to secure to his family at his death the sum of £1000? At 5 cent., the perpetuity of £1 is £20, and, by table IV. the value of a life of 32 years, is 12.854. The difference of these is 7·146; and this being divided by 21, the quotient is ·340286, the present value of £1 payable at the decease of the person. Then, 340286 × £1000 £340 5 8, the value required: and £340.286-12.854 £26.473 = £26 9 5, the annuity. Hence it appears, that at 5 cent. annum, if a person either at one payment deposit £340 5 8, or pay each year during life the sum of £26 9 5, his heirs at his death will be entitled to receive £1000. Ex. 10. At 6 cent. annum, how much year, must a person aged 66 years, pay during life, to entitle his heirs at his death to receive £1500? Answ. £128 13 10.

11. If a husband and his wife be each aged 45 years; how much annum must be paid during the longer of the two lives, to entitle the family after the decease of both, to receive £4000, compound interest being allowed at 6 cent. annum ?

Answ. £75 8 5.

Exam 10. How much must a man, aged 39, pay per year, at 6 cent., during his marriage, or during life, to entitle his wife aged 32 years, in case she survive him, to an annuity of £50 a year during the remainder of her life?

By rule II., the value of the two lives jointly is 8-774, and the value of a life of 32 years is, by table IV., 11.512. The difference of these, 2.738, is the present value of the reversion of an annuity of £1 to be paid to the wife after her husband's death, in case she survive him. The product of this by £50 is £136.9, or £136 18, the sum to be paid at a single payment. Let this be divided by 8774, the value of their lives jointly, the quotient £15603, or £15 12 0, is the annual payment. Hence, if there be paid annually, during the joint continuance of both lives, the sum of £15 12 0, or at present, in a single payment, £136 18, the wife, if she survive the husband, will be entitled to an annuity of £50 per annum during life.

This solution proceeded on the supposition in regard to the annual payments, that the first of these was not made till the end of the first year. If it be made at present, however, we must divide by the value of the joint continuance increased by a unit; in the present case by 9.774. In this case the annual payment would be found to be £140 13.

ว

Exam. 11. If a number of persons form themselves into an association for providing annuities of £40 annum for their surviving widows, and each person, at the age of 30, contribute £75 to the fund, and equal annual contributions during the rest of his life: how much must each of these annual contributions be, money being improvable at 6 † cent. Pannum, and the wife of each being supposed to be at an average three years younger than himself?

The value of a life of 27 years is 11.917, and the value of two lives of 27 and 30 jointly is 9-481. The difference of these, 2-436, multiplied by £40, gives for the entire sum to be paid at present, at a single payment, £97.44. But as only £75 are to be paid at present, there will still remain £22-44 to be paid by annual contributions, by dividing which by 9-481, we obtain £2 7 3 for each of these contributions.

Exam. 12. Suppose that, as in the last question, a man aged 30 years, pays £50 into a fund; what annual contribution must he pay during life to entitle his family to an annuity of £40 per annum for years after his death, interest being at 6 cent? The present value of an annuity of £40 for 8 years at 6 cent. iş £248-3918, (by rule II. of Annuities Certain.) Then, by rule VII. of this article, the present value of this sum is found to be £70.084 from which £50 being subtracted, the remainder, £20:084, is the sum still remaining to be paid at present; and this being divided by 11.682, the value of a life of 30, the quotient, £1 14 4, is the annual contribution required.

CONTINUED FRACTIONS.*

IT often happens that fractions, even when reduced to their simplest forms, are expressed in numbers inconveniently large; and hence it is often desirable to approximate their values in smaller numbers. Thus, if the fraction 4, which is in its lowest terms, be proposed, we may wish, even for the purpose of forming a more correct idea of its magnitude, to find other fractions, in smaller terms, which will be nearly of the same value. The method that most naturally presents itself for this purpose is to divide both terms by the smaller of them, as the smaller will by this means become 1, and we shall thus be enabled to compare the other with that number with which the

* For full information respecting the nature and applications of continued fractions, see Bonnycastle's larger Treatise on Algebra, Lacroix's Complément des Elémens d'Algèbre, Legendre's Essai sur la Théorie des Nombres, Barlow's Theory of Numbers, Lagrange's Additions to Euler's Algebra, and several other works. What is here given is one of the most interesting applications, and one which cannot fail to be useful to the inore advanced arithmetician.

M

mind can most easily compare it in respect of magnitude. By

this means the fraction becomes

1 2017'

the denominator of which,

being between 2 and 3, we conclude that the value of the given fraction is between and; and therefore is a first approxima tion to its value, being too great. Divide again both terms of the the fraction in the denominator by 217, and it will be ome

1

5217

fraction,

1

20

we shall have, for a second approximate value of the proposed or, which is too small, as instead of 27, we used in the denominator, which is greater than . To continue the ap proximation still farther, we divide both terms of the fraction by 30: the result is which is less than and greater than Hence, if instead of using in the preceding fraction, we use or its equal, we shall have for the next approximation to the value of the given fraction or which is too great, because

1

54

great; and therefore

1

236

2

1

51

or its equal, is too small, in consequence of being too must be too great, since the denomina tor is too small. By continuing the process in a similar manner, we find 1,0, and 11, for the succeeding fractions, the last of which is the given fraction,* and the first an approximate frac tion smaller, and the second another greater, than the given fraetion. These successive fractions have the remarkable property that each of them approaches more nearly than the one which precedes it, to the value of the given fraction. Thus, t, is nearly equal to ; more nearly equal to ; more nearly equal to ; still more nearly equal to 4;4; and, finally, more nearly still to That there is this continual approach to the true value of the frac tion, will appear evident, if it be considered that in each of the re sults in the preceding operations, a correction is made on the result which goes before it.

83

The several converging fractions above obtained, if the last or supplementary simple fraction be rejected from each, except when its numerator, like that of the rest, is a unit, may be written at full length, as follows:

* If the given fraction be finite, the last of the converging fractions will always be equal to it, as in this example.

In this form they are called continued fractions. It appears, therefore, that a CONTINUED FRACTION is that which has for its denominator a whole number with a fraction annexed, which fraction has also for its denominator a whole number with a fraction annexed, the denominator of which latter fraction is also a whole number with a fraction; and so on, however far the fraction may be continued; and each numerator is a unit.*

It will be seen by a review of the preceding processes, that the denominators of the continued fraction are the quotients which would be found in using the rule given in page 89, for finding the greatest common measure. Hence, we have the following rule:

RULE I. To convert a given simple fraction into a continued fraction: Divide the greater term by the less, the less by the remainder, &c. as in finding the greatest common measure: the quotients will be the denominators of the several fractions in the continued fraction, and the numerator of each will be a unit.'

The successive fractions which approach continually to the va lue of a given fraction expressed in large numbers, may be found by reducing it to a continued fraction, and perating in the way already employed in approximating the value of 111: the following inethod, however, will be found much preferable.

RULE II. "A fraction expressed by a great number of figures being given, to find all the fractions in less terms, which approach s near the truth, that it is impossible to approach nearer without employing greater terms:" Reduce the proposed fraction to a continued one, or at least find the quotients, by the preceding problem. Then write the several quotients in a line; and if the given fraction be greater than a unit, take the first quotient as the numerator, and a unit as the denominator, of the first fraction, which set below the second quotient; but if the given fraction be less than a unit, make the first quotient the denominator, and a unit the numerator of the first fraction. Then, for the second fraction, multiply both the terms of the first by the quotient which stands above it, and add a unit to the product of that term which was the first quo. tient: the result is the second fraction, which is to be set below the third quotient. To find the succeeding fractions, multiply the terms of each fraction, when found by the quotient, which stands above it, and to the products add separately the terms of the preceding fraction.

Exam. 1. The height of Slieve Donard, in county Down, is 2654

It is not essential to the nature of continued fractions that each numerator be a unit; but fractions of this kind only are used as instruments of calculation.

feet, and that of Ben Lomond 3262 feet: it is required to find a series of converging fractions, expressing as nearly as possible the ratio of the heights of these mountains.

Here the fraction expressing the height of Slieve Donard in relation to that of Ben Lomond, is 398, or, in its lowest terms, 1831, and the quotients found in the manner shown in the margin, are 1, 4, 2, 1, 2, 1, 4, 1, 4: consequently the continued fraction is

+

1327)1631(1

304)1327(4
1216

111)304(2 &c.

+

The successive converging fractions will be found thus: 1 4 2 1 2 1 4 1 4

1, 4, 11, 18, 45, 18, 227, 338, 1831,

Here, the quotients, or denominators, being arranged in succession, we take for the denominator of the first fraction, 1, the first quotient, and we make its numerator 1 also. Then, by multiplying both terms by 4, the quotient which stands above them, and adding 1 to the product resulting from the denominator, because it was the first quotient, we obtain , for the second fraction. We then multiply 4 by 2, the figure above it, and add to the product the preceding numerator; and we multiply 5 by 2, and add to the product the preceding denominator; and thus we obtain for the third fraction. We next multiply the terms of this fraction by 1, and add to the results 4 and 5 severally: we thus find the next fraction to be 8. In this way, rejecting the first fraction, t which is evidently far from the truth, we find that the height of Slieve Donard is nearly that of Ben Lomond; more nearly it still more nearly, ; more nearly still, 3, &c.; too small, too great, too small, & too great, &c.*

being

Exam. 2. The circumference of the circle whose diameter is 1, is found to be greater than 3.1415926, but less than 3.1415927:

* Thus, of 3262-2609; of 3262=26681; 1 of 3262-26503; of 3262—— 2655 of 3262-26554, which are evidently approaching 2654, the true value, be ing alternately greater and less.-It may be observed, that in all cases, the error of each fraction is a less part of the integer than a unit divided by the product of its own and the succeeding denominator; but a greater part than a unit divided by the product of its own denominator, and the sum of that and the succeeding denominator. Thus, in the 1 1 preceding example, the error of is less than-5X16 X3262.

5X11

X3262, but greater than