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n

then will n be the number of persons living, of that age, out of which

n-1 n-2 n-3 one dies every year ; and

&c. will be the

proba. bilities of his living 1, 2, 3, &c. years : hence, the present value of

7-1 n21-3 an annuity of £i to be paid during his life is

+

nrs

(n-1).x +&c. continued to n terms. The sum of the series

+

+

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n

n

n

(n-2).r? (n-3).rs
+
to n terms, was found to be

(n-1)=-nxl+2* to

n.(1-2) and the sum of the series

n-1 1-2

n-3 +

+ &c. to a nr 2

let I=

+

[ocr errors]

nrs

terms is

(n-1)-n+

7.(1-1)

; the present value of the annuity.

(3.) Cor. 1. This expression for the sum is the same with

n 17 n.(-1)

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an annuity of £i to continue certain for n years, then P=

T

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(4.) Cor. 2. The present value of the annuity to continue for ever,

q P

from the death of the proposed individual, is

n.(r-1) Por, the whole present value of the annuity to continue for ever,

is

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and if from this its value for the life of the individual be

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taken, the remainder

is the present value of the annuity to

n.(1-1) continue for ever, from the time of his death.

(5.) To find the present value of an annuity of £1, to be paid as long as two specified individuals are both living.

Find the probability that they will both be alive at the expiration of 1, 2, 3, &c. years, to the end of the tables ; call these probabilities

c

+

b a, 0, c, &c. and r the amount of 1 in one year ; then + + &c. is the present value of the annuity required.

(6.) To find the present value of an annuity of £1, to be paid as long as either of two specified individuals is living.

Find the probability that they will not both be extinct in 1, 2, 3, &c. years, to the end of the tables, and call these probabilities A, B, C,

А B then the present value of the annuity is + -+*+&c.

g2 (7.) Cor. If the annuity be M £, the present value is M times as

A B great as in the former case, or Mx +

+&c.

(

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(187.) These we the mathematical principles on which the values of annuities for lives are calculated, and the reasoning may easily be applied to every proposed case. But in practice, these calculations, as they require the combination of every year of each life with the corresponding years of every other life concerned in the question, will be found extremely laborious, and other methods must be adopted when expedition is required. Writers on this subject, are De Moivre, Mr. Baron Maseres, Mr, Morgan, in the Philosophical Transactions nd Dr. Waring.

METHOD OF DIFFERENCES.

Definitions. 1. A variable quantity is one that is continually changing its value, either by increasing or decreasing its magnitude.

2. A constant quan city is one that never changes its value.

3. The function of a variable quantity which changes by certain steps, is called an integral.

4. The quantity from which the function is raised is called the base of that function, or of that integral.

5. The difference between two similar functions of two quantities, of which the one to be taken away has any variable quantity, x for its base, and the other function from which it is taken, has the same base increased by a given quantity, is called the difference or increment.

Notation. 1. Invariable quantities are represented by the first letters, a, b, c. &c. of the Alphabet.

2. The base of the function of a variable quantity, is represented by r, and is supposed to be continually increased, at every step, by unity, which is, therefore, the difference or increment of x.

3. The differences of quantities which are supposed to be functions of 2, are represented by the same letter, which indicates that function with x not under a.

Thus, suppose x to be a function of x, then x is the difference or increment of %.

4. The operation of taking the differences of two variable quantities is denoted by A, and if the same operation be performed on the difference, supposing it variable; then the double operation is denoted by A; and if n successive operations are performed, then the symbol denoting these n operations is A.

Arioms. 1. A constant quantity has no difference or increment.

2. The difference of any whole quantity, is the sum of the differences of all the terms, or parts of that quantity.

Problem 1. (1.) To resolve an algebraic product, consisting of binomial factors,

1

2

of the form (x+a)(+)(+c) &c. into a series of factorials which shall have x alone, or x with any given additive number for the first factor of each, and of which the factors shall have any given common difference.

Let a, b, y, &c. be the given quantities in arithmetical progression, which are respectively to be added to x, in order to form the series of factorials, and let d be their common difference ; find n, the number of given factors, then the general form of the series will be (x+x)" + B(r+a)-1d+C(r+

ad+ &c. where we have only to find the co-efficients of the 2d, 3d, &c. terms, which will be found by the following

Rule. (2.) 1. From the given quantities a, b, c, a, &c. in the whole number of binomial factors, subtract a, b, y, &c. respectively, to the same number, and call the remainders the first order of differences.

2. From 1, c, d, &c. the whole number of factors, except the first, subtract a, b, &c. respectively, to the same number, and call these the second order of differences.

3. Proceed in the same manner, omitting one every time, until one only remain.

4. Then form a series of successive sums, by taking the first dif. ference of the first order, for the first term, the sum of the first and second for the second term, and so on, and the last term of this series, is the co-efficient B, of the second factorial.

5. Multiply the terms of the series of successive sums, respectively, by the second order of differences, and with the products thus found, form a second series of successive sums, in the same manner as before, and the last term or sum of this series will be the co-efficient of the third factorial.

6. Proceed in the same manner, until there be only a single product which will be the last co-efficient or term.

In these operations when negative quantities occur the sign is placed above, in order to obtain a more convenient arrangement, as is done in the indices of Logarithms.

The following tables exhibit the form of the operation, as now expressed by the rule.

The first table simply contains the additive part of the required binomial factors, and the other columns consist of the additive parts, which are to be found in the binomial factors of the given Algebraic product. In these last, we must observe that the first term of every new column, is the same as the second term of the preceding column, and the other terms follow in consecutive order, and end with the additive number to be found in the last factor of the given Algebraic product.

The operation of differences is carried from the first table to the second, and the first order of differences é, f', g', h', &c. are placed in the first column of the second table, and the successive sums are placed in a column on the right with a line between them. The factors d', g", g", &c. are the second differences; A, B, C, are the

meCli

real products of these factors in numbers : il", 2", 1", &c. are the successive sums of the products, A, B, C, &c. and so on.

b

с

e =l¿i B b

d &c. f' k"f"=Bk!"f"=B' &c. d &c.

I!

&c. d &c.

&c. &c. &c. &c.

&c.

| A = الى الان 1

A = الى الخ

с

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с

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Example 1. Resolve (x+1)(x+3)(x+5) into factorials, of which the first factor shall be x, and the factors shall increase by unity.

Here a=i, b=3 and c=5, also a=0, B=1 and y=2, for as the factors proposed to be resolved are limited in their number, so is each of the series a, b, c, &c. as also a, b, y. &c. Whence the operation, where for convenience the negations are marked above.

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gives (1 - 1)(x+3)(x+5)=2311+ 4x21 - 15.

Example 2. Resolve (2-1)(x+3)(x+4)(x+5) into factorials, so that the first factor of each may be », and the factors to increase by 2.

Here a=i, b=3, c=4, and d=5.

Also a=0, B=2, y=4, and d=6. whence 0 1 3 4 5

1 (3)=3

3 (4) = 12 12(5)=60
4
1 O (2)=0

3 (3) = 9
4
5

0 0(1)=0 6

2

3

5

21

5

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gives (2-1)(x+3)(x+4)(x+5)=242 — 2312—32212-21%-60.

If we wish to resolve an algebraic product of binomial factors into powers, we have only to make the first column a series of cyphers.

Problem 2. (3.) To resolve the reciprocal of an algebraic product, consisting of

С binomial factors, or

a quantity of the

(x+a)(x+1)(1+c) &c. A B

С into a series of the form Tefalt

(4)

(c+a)*

nt ild

+ &c.

* + 2id

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