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3. Required the sum of

1 1

+ + &c. to infinity.

1.2 2.3 3.4'

1 1 1

Assume x=1+++++, &c. to infinity.

2 3 4

1

1 1

Then x

+ +=+ + &c.

Subtract the latter from the former, and

1 1 1 1

+ +· + &c. the sum required.

1.2 2.3 3.4 4.5

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The student who desires to pursue this subject, may consult Wood's Algebra, Article 411, &c.

DIFFERENTIAL METHOD.

97. In any series of quantities, a, b, c, d, e, &c., if each term be subtracted from the next following one, and each term of the series of differences be taken from the next, and so on, the following series will be obtained.

1st differences, ba, cb, dc, e-d, &c. 2d diff. c-2b+a, d—2c+b, e—2d+c, &c. Sd diff. d-Sc+3b-a, e-3d+3c-b, &c. 4th diff. e4d+6c-4b+a, &c.

Or these, 1st diff. a+b, b+c, -c+d, &c.

2d diff. a-2b+c, b-2c+d, &c.

3d diff. -a+3b-3c+d,—b+Sc-sd+e, &c.

4th diff. a-4b+6c-4d+e,

5th diff. ―a+5b-10c+10d-5e+f, &c.

By a little attention to these expressions, and a comparison of their formation with the powers of a binomial, we perceive that if A be taken to denote the first term of any (the nth) order of differences,

±A=a-nb+n.

N -1
2

c-n.

ท- -1 n-2
2 3

d, &c.

the sign+being used when n is an even number, and when n is odd.

If the differences of any order vanish, any one of the terms may be found by means of the others.

Suppose the 4th difference a-4b+6c—4d+e=0, and e was not known, we should find,

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98. Let D, D, D, D, &c. denote the first terms of the

1 2 3 4

By this method the computers of the Nautical Almanac verify their calculations of the moon's longitude, latitude, &c.

1st, 2d, 3d, 4th, &c. orders of differences; viz. D=—a+b;"

1

D-a-2b+c, D=-a+3b-3c+d, D=a-4b+6c-4d

2

3

4

+e, &c.; then we find b=a+D, c=a+2D+D, d=a+

1

3D+3D+D, e=a+4D+6D+4D+D.

- 1

2 3

1

2

3 4

Hence, the n+1th term of the series

n-1

n—1n-2

1 2

=a+nD+n. D+n.- n— 2D, &c.

2

1

2

Consequently the nth term

n

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=a+n—1.D+n— -1. ·D+n—i.

2 2

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1. Required the nth term of the series of odd num

bers, 1, 3, 5, 7, &c.

Here the 1st diff. are 2, 2, &c.; 2d diff. 0.

.. the nth term =1+2.n-1=2n−1.

99. Required the sum of n terms of the series a, b, c, d, &c.

This is manifestly the same as the n+1th term of the following:

0, 0+a, 0+a+b, 0+a+b+c, &c.

Hence by the last article, the required sum

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1. Required the sum of n terms of the square numbers, 1, 4, 9, 16, &c.

Here a=1, D=3, D=2,D=0.

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2. Required the sum of the cube numbers, 1, 8, 27, 64, &c. continued to n terms.

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3. Required the sum of 25 terms of the series 1, 3, 5, &c. Ans. 625.

4. Required the sum of 15 terms of the series 1, 16, 81, 256, 625, 1296, &c. Ans. 178312.

When the differences at length vanish, any term of the series, or the sum of any number, may be accurately determined by the methods used in this and the preceding articles: when the differences become small, but do not vanish, a near approximation can be made.

100. In article 98, the number n is supposed to be an integer, in which case, if the differences vanish, the rule is demonstrably correct; the same formula is, however, applied to the case where n is fractional.

Suppose a series p, q, r, s, &c. of equidifferent quantities and another series, a, b, c, d, &c., such that the terms of the latter shall be similar functions of the correlative terms of the former. It is proposed to find a term y, in the latter series, corresponding to v in the former; v-p being given.

Let q-p: v-p::1: x, and take as before, D, D, D,

&c. the first of the 1st, 2d, 3d, &c. differences.

Then assuming the series, (art. 98,)

1 2 3

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This gives, when x=0, y=a, x=1, y=a+D, x=2,

1

y=a+2D+D, &c.; but these expressions equal b, c, &c.

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as they ought to do. This formula is, therefore, presumed to be universally true, whether is integral or fractional.

Required to find the 64th term in the series 1, 4, 9, 16.

11

Here x= a=1, D=3, D=2, D=0.

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2. Let a, b, c, d, be arcs of a great circle intercepted between a fixed star and the moon's centre at noon and midnight of two successive days, it is required to find the distance at 15 hours from the first noon.

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101. From article 85, it is obvious that, a being any number, if a"=N, and am=M, an+m=NM, and an―m=

N

M

If, therefore, all the numbers used in calculation were expressed in powers of a, multiplication and division might be performed by the addition and subtraction of the exponents of a. These exponents, thus employed,

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