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P nega.

46. Expansion of binomials with negative fractional exponents.

To obtain a formula for this case we have only to make n

tive in formula (18) obtained in the last article. This will be effected by making either p or n negative; therefore for n substitute —n, and we have

(1x)

x2+

Making n negative in (19), we have

(1±x) ̄*=1=—=—=x+23 (n+1)(2n+1)

=22 +

2.3.n3

n

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-x3+ &c. (23)

In these formulæ we may express the denominators as in (21).

2.

P p(p+n)

2.n3

For example, to expand (a—b)— or

3. (a+b)'

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(a—b)—3—a ̄1× (1—2) —*— 1 (1-2)+

b

[making x=-, n=2, in (23)]

α

1. (1—22)—3—1+3 2o + 35.6

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̧p(p+n)(p+2n),
2.3. n3

1+

=c(1+.

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1

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x3+ &c (22)

2.5 2.5.8
-204+· x6+ &c.
3.6.9

we have

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1 x:2 5 X4 5.9 X6
+:

·

4 y2 2.42 y* ' 2.3.48 ̊ y6
+2.3.43 28+ &c.)

b 6 b2 6.11 b3
+

2.52 a2 2.3.5 as

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47. Involution of polynomials.

For example, to expand (x+y+z)3. In formula (15) make

a=x+y, b=z, m=3,

(x+y+z)3=(x+y)3+3(x+y)2z+3(x+y)z2+z3.

Then expanding (x+y) &c.

(x+y)3=x3+3x2y+3xy2+y3,
3(x+y)3z=3x2z+6xyz+3y2z,
3(x+y)z2=3xz2+3yz2,

23=23;

The sum of these equations gives (x+y+z)3=x3+3x3y+3xy3+y3+3x2z+6xyz+3y3z+3xz2+

3yz2+23.

To expand (a+b+c-d), we have by our formula

(x+y)=x+4x3y+6x3y2+4xy3+y+

in which make x=a+b, y=c—d; we shall find

(a+b+c-d)=a* + 4a3b + 6a2b3 + 4ab3 + b + 4a3c +12a2bc +12ab2c+4b3c-4a3d-12a2bd-12ab2d-4b3d

+6a2c2 + 12abc2 + 6b3c2

-12b3cd+6a3d2+12abd2+6b3d2+4ac3—12ac'd

- 12a2cd

EXAMPLES.

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24abcd

+12acd2-4ad2+4bc3-12bc'd +12bcd-4bd3 +c2-4c3d+6c3d2-4cd3+d+.

1. (a+by+cy3)3=a3 + 3a2by+ (3a3c + 3ab2)y2+(6abc + b3)y3 +(3ac2+36°c)y*+3bc3y3+c3y®.

2. (1−x+x3)6—1—6x + 21x3-50x3+90x4—126x+141x6

-126x7+90x3-50x9+21x1o—6x11+x12.

12

3. (a3+a2b+ab3+b3)* = a1o+4a11b +10a1ob2 + 20ab3+31a3ba +40a7b5 + 44a6b® + 40a5b7 + 31aab3 +20a8b9+10ab10 + 4ab11+ 19.

3

EXTRACTION OF ANY ROOT OF A NUMBER.

48. Any root of a number may be found by the binomial theorem by dividing the number into two parts and considering it as a binomial. Thus, to find the 3d root of 28, divide it into 27 and 1; then

√28=(27+1)+=273(1+27)*=8(1+27)3.

1

Now to develop (1+27)3, in (20) Art. 45, make x=;

, 27

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n=3,

6641 The first three terms of this series are equal to or 1.0122 6561' nearly. The fourth term is less than .00001; so that if we wish to find the root to four places of decimals only, this term, as well as all the succeeding terms, may be neglected. Therefore we have

√28=3(1+2)+

=3x1.0122=3.0366 nearly.

27

We divided 28 into two such parts that one of them was an exact power of the third degree, and also the greatest contained in 28. The general method of proceeding is expressed in the following rule: To find the nth root of any number, first find by trial the nearest integral root (a); divide the given number into two parts, one of which is the nth power of (a); and consider these two parts as the terms of a binomial which may be developed by the binomial theorem.

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&c.

49. Sometimes it is expedient to use the power next greater than the given number. Thus, to find the fifth root of 30 we may put

√30=(32—2)3=323(1.

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Then by (20), Art. 45, making x=

1

16'

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2 4.9

3

2(1-1)-21-316-2.5 (10) -2.5 (10) -&c.]

=

2.3.53

=2(1-1

(taking three terms of the series,)

3159

3200

=3(1—.

(taking the first two terms,)

=

17 17 =3X= = 18 6

4

1

3200

=

=2x· =1.9744 to within .0001.

in 0.000001 or

2.83(1.

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2

-3(1-1) — &c.)
(+)2

2 9 2.22

=2.83

The greatest integral power of the 5th degree less than 30 is 1, which would lead to a divergent series.

1

106

;; n=5,

50. If the root is desired to a greater number of places, a greater number of terms of the series must be taken. By the following method, however, we may approximate to any degree of accuracy with the use of only the first two terms. Thus, to find the square root of 8, we have

√/8=(9—1)=3(1—1—1)+

=2.83 nearly.

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3 256000

✓8=[(2.83)3— 0.0089]3=2.83(1

1

Taking 2.83 as a first approximation, we now divide 8 into two parts, one of which is the square of 2.83. We then have

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89

2*(283)2

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89×2.83
2×283×283

=2.83-0.001572=2.828428, which is correct with

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Taking 2.828428 as a second approximation, we shall find

✓8=[(2.828428)3—0.000004951184]

2.828428(1–

=2.828428(1-
(1—

=2.828428

Then we have

or

4951184

(2828428),

1

1012

=2.828428-0.000000875254

4951184

2 (2828428),

4951184

2×2828428×1000000

=2.82842,71247,46, which is correct in the last

place; that is, within

It is readily shown that the third term will not affect the 12th place of decimals. This term will be

2.828428x

1 (4951184)9 2×22(2828428)+

which may be expressed approximately thus,

3 (5,000,000)*

X.

10 (8,000,000,000,000)

51. The approximation may be effected also in the following manner.*

Suppose the mth root of any number N is sought. Having found an approximate root a, let b be the difference between am and N, or

N=amb

b=mam-12

3×(5×106)9

75

10x(8x1012) 64×10131012 1019.

1

Then Nm will be equal to a, plus or minus some number which call z; that is, let

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1

Nm-a±z, or N=(az).

am±b=(a±z)m=am±mam—1z+

m(m-1)

2

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am-2z2+ &c.

* Francœur, Algèbre, no. 488.

(A)

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