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ma

3%, etc.).

Substitute - m for m, and we shall have

m2 m3 log. (1–m)=M(-m

2 Subtracting Eq. (2) from Eq. (1), observing that log. (1+m)

1+m - log: (1-m)=log. we shall have

1-m'
1+m

ma mo
=2M (m+
1-m

.

5+, etc.) Now, since this is true for every value of m, put

1
m=; whence

P
and the preceding series, by substitution, becomes
log.
p+1p2M

it ). 2p)'+"

log.

+

3

1+m_p+1

2p+1'

1-m

1

1

=log. (p+1)–log. p=22(2+1+3(2}+13+5(2p+1)*+, etc.).

+

P

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.429. This series converges rapidly, and may be employed for the computation of logarithms in the Naperian or the common systems. It is only necessary to compute the logarithms of prime numbers directly, since the logarithm of any other number may be obtained by adding the logarithms of its several factors. Making p=1, 2, 4, 6, etc., successively, we obtain the following

Naperian or Hyperbolic Logarithms.
1 1 1
+

=0.693147 3.33 5.35

1 1 1 1 3

+

=1.098612 3.53'5.55 7.57+ log. 4=2 log. 2

=1.386294 1 1 + +

=1.609438 3.93 log. 6=log. 3+log. 2

=1.791759 1 1 + 5.135+7.187+....

)

=1.945910 log. 8=3 log. 2

=2.07912 log. 9=2 log. 3

=2.197225 log. 10=log. 5+log. 2

=2.302585 etc., etc.,

etc.

log. 2=2&+8.3+6.3+7+3+...

) log. 3=log.2 +2++ out that...

) log.5=log.4+2(+3.41+5.50 +7.94

7.gst ....

)

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+

log. 7=log.6+2 (13+3.13

15910

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430. To construct a Table of Common Logarithms.-In order to compute logarithms of the common system, we must first determine the value of the modulus. In Art. 427, we found

log. (1+m) M=

Nap. log. (1+m) If 1+m=a, the base of the system, then log. a=1, and we have

1
M=
Nap. log. a

ai that is, the modulus of any system is the reciprocal of the Naperian logarithm of the base of the system.

The base of the common system is 10, whose Naperian logarithm is 2.302585. Hence

1 M=

=.434294,

2.302585 which is the modulus of the common system.

We can now compute the common logarithms by multiply. ing the corresponding Naperian logarithms by .434294, Art. 427. In this manner was the table on pages 290–1 computed.

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431. Results.—The base of Briggs's system is 10.

Napier's

2.71828. The modulus of Briggs's system is 0.43429. Napier's

1. Since, in Briggs's system, all numbers are to be regarded as powers of 10, we have

100.301=2,
100.477 =3,

100.6024, etc. In Napier's system, all numbers are to be regarded as powers of 2.71828. Thus,

2.7180.693 =2, 2.7181.098=3, 2.7181.886 =4, etc.

CHAPTER XXI.

GENERAL THEORY OF EQUATIONS. 432. A cubic equation with one unknown quantity is an equation in which the highest power of this quantity is of the third degree, as, for example, 2c8 — 6x2 +8x— 1550. All equations of the third degree with one unknown quantity may be reduced to the form

X2 + ax? +bx+c=0. A biquadratic equation with one unknown quantity is an equation in which the highest power of this quantity is of the fourth degree, as, for example, x* — 6x3 +7x2 +5x—4=0. Every equation of the fourth degree with one unknown quantity may be reduced to the form

24 + ax: +6x2 +c+d=0. The general form of an equation of the fifth degree with one unknown quantity is

25 +axt +bx+cx? + date=0; and the general form of an equation of the nth degree with one unknown quantity is X" + Ax*-1+Barn-+Can-8+ ... + Tx+V=0. Cən3

(1.). This equation will be frequently referred to hereafter by the name of the general equation of the nth degree, or simply as Equation (1)

An equation not given in this form may be reduced to it by transposing all the terms to the first member, arranging them according to the descending powers of the unknown quantity, and dividing by the coefficient of the first term. In this equation n is a positive whole number, but the coefficients A, B, C, etc., may be either positive or negative, entire or fractional, rational or irrational, real or imaginary. The term V may be regarded as the coefficient of 200, and is called the absolute term of the equation.

It is obvious that if we could solve this equation we should

have the solution of every equation that could be proposed. Unfortunately, no general solution has ever been discovered; yet many important properties are known which enable us to solve any numerical equation.

433. Any expression, either numerical or algebraic, real or imaginary, which, being substituted for x in Equation (1), will satisfy it, that is, make the two members equal, is called a root of the equation.

It is assumed that Eq. (1) has at least one root; for, since the first member is equal to zero, it will be so for some value of 2, either real or imaginary, and this value of x is by definition a root.

Then we

434. If a is a root of the general equation of the nth degree, its first member can be exactly divided by x-a.

For we may divide the first member by x-a, according to the usual rule for division, and continue the operation until a remainder is found which does not contain x. Let Q denote the quotient, and R the remainder, if there be one. shall have

in + A.xn-1 + Bxn-3+.... +To+V=Q(x-a)+R. (2.) Now, if a is a root of the proposed equation, it will reduce the first member of (2) to 0; it will also reduce Q(xa) to 0; hence R is also equal to 0. But, by hypothesis, R does not contain it is therefore equal to 0, whatever value be attributed to x, and, consequently, the first member is exactly divisible by

1-2

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435. If the first member of the general equation of the nth degree is exactly divisible by xa, then a is a root of the equation.

For suppose the division performed, and let Q denote the quotient; then we shall have

ac" + Axn1+ Ban–2 + ... + Tx+V=Q(x-a). If, in this equation, we make x=a, the second member reduces to 0; consequently the first member reduces to 0; and, therefore, a is a root of the equation.

EXAMPLES.

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1. Prove that 1 is a root of the equation

X2 - 6x2 +11x-6=0. The first member is divisible by x-1, and gives 22 –5x+6=0. 2. Prove that 2 is a root of the equation

23 -X-6=0. The first member is divisible by #—2, and gives 22 +23+3=0. 3. Prove that 2 is a root of the equation

23-1122+36-36=0. 4. Prove that 4 is a root of the equation

23 + x2 - 34x+56=0. 5. Prove that -1 is a root of the equation

24 – 38X3 +210x2 +5383+289=0. 6. Prove that -5 is a root of the equation

25+ 6x4 - 10x3 - 112x2-2073-110=0. 7. Prove that 3 is a root of the equation

x? +36-14X – 14x4 +49x3 +49m2 - 36x-36=0.

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436. Every equation of the nth degree containing but one unknown quantity has n roots and no more.

Since the equation has at least one root, denote that root by a; then will the first member be divisible by x—a, and the

-Q tient will be of the form

21-1+ A'ach-+B'ach-3+.... +T'x+V', and the given equation may be written under the form (oc-a)(n-1 + A'an-3+.... + T'+V')=0.

(3.) Now equation (3) may be satisfied by supposing either of its factors equal to zero. If the second factor equals zero, we shall have

-1 + Ax4-2 +B'2-3+.... +To+V'=0.
A'an

(4.) Now equation (4) has at least one root; denote that root by b; then will the first member be divisible by a—b, and equation (4) can be written under the form

(cb)(xn-2+A" 22–8 +.... +T":+V')=0, which reduces Eq. (3) to the form of

(x-a)(x-)(x-+A"-8 +.... T'x+V'')=0.

- 2

-3

-2

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