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17. To find limits to the roots of "x-23 - 7x2+10x + 100."

It is easy to perceive that the coefficients of t and v for the positive and negative roots may be represented by 2x3-3x2-7x+5,-x3-7x+15x+20; and 2x3-3x2 +7x+5, x3-72-15x+20.

=

Putting a 0 and a1, separately, in the coefficients for the negative roots, they become 5 and 20, 7 and −1; consequently, since the signs of the coefficients do not both change between 0 and 1, and that 20÷54 is greater than 0, while 17 is less than 1, it follows that -1 and 0 are limits to one of the negative roots; and in a similar way the remaining negative root will be found to have 3 and 2 for limits.

In like manner, from the coefficients for the positive roots, we get 2, 2.3, and 2.3, 3 for their limits.

Concluding Remark.—Because we may find the imaginary roots of an equation in much the same way that we find the real roots, it follows that we can by our method find all the roots of any numerical algebraic equation with great facility.

SECTION XX.

SOLUTION OF EXPONENTIAL EQUATIONS.

(1.) AN exponential equation is one in which the unknown letter enters as an exponent.

Thus, 29, 350, a* = b, xc, x=d, are exponential equations.

If we take the exponential equation AB, (1); then, since A1 + A − 1, B = 1 + B − 1, if we put A-1 a B-1, = and B-1b, (1) will be reduced to (1 + a)* = 1 + b, (2), in which a and b may be positive or negative, according to the nature of the case.

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(2.) Putting ∞ = ? (3), (2) by substituting for æ, is easily

reduced to (1+ a)" = (1 + b), or (expanding the members

by the binomial theorem) we have 1+ya+Y (y−1) a2 +

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1. 2

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2(2-1,
1.2

b3 + etc., which, by arranging the terms according to the ascending powers of y and 2, is easily reduced to the form

ая +

a4

α

= (b–

+ etc.y + Ay2+ Ay3 + etc. =

2 3 4 5

b2 b3 b4 b5

+

+

3 4 5

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etc.)2+ B+ B,≈3 + etc., (4).

etc.) z

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etc., are called the hyperbolic logarithms of 1 + a

and 1+b, if we denote them by writing l' before (1 + a)

and (1 + b), we shall have l'(1 + a) = a

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(by substitution from (5) in (4)), we shall have l'(1+a) xy+ Ay+Ay+ etc. l'(1 + b) x 2 + B122 + B2+ etc.; and

=

if we divide the terms of this by z (since by (3) x =

= 2), it will become l'(1 + α) × x + Â ̧xy + A„xy3 + etc. = l'(1 + b) + B12 + B222 + etc., (6).

2,

Because (6) ought clearly to be satisfied without limiting the values of y and z, it is manifest (from the method of undetermined coefficients, see Sec. XV.) that we must have l'(1 + a) × x = l'(1 + b), or x = l'(1 + b) l'(1 + a)'

(7); consequently,

x equals the hyperbolic logarithm of 1 + b divided by that of 1+a, or (which is the same) x is found from (1) by dividing the hyperbolic logarithm of B by that of A.

Since (1a) x x = l'(1 + b)

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if we (for brevity) represent l'(1 + a) x x by v, we shall have

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+ + etc.; consequently, if we revert this 2 3 4

series (see (5) of the Reversion of Series, p. 436), we shall get

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24
2 2. 3 2.3 4

(2), gives (1 + a)* = 1 + v +
·

+ etc., which being put for b in

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+ + etc., or restoring the 2 2.3

value of v, we shall have (1 + a)* = 1 + l'′(1 + a)x +

(l'(1 + a)x)2, (l'(1 + a)x)3

1. 2

+

1.2.3

+ etc., (S), which is called the

Exponential Theorem; and it is clear that it is only another form of the Binomial Theorem.

Hence, if we represent the hyperbolic logarithm of A by

A', we shall have A1 + A'x +

(A'x)2

1.2
A'2002 A'3

quently, (1) is reduced to 1+ A'x + +

etc.; conse

1.2 1.2.3

+ etc. =

B, (9), which (after having found a near value of x by trial) will clearly enable us (as in the solution of Algebraic Equations) to find a to any required degree of exactness.

(3.) If we multiply 1 + a by 1 + b, we shall have (1 + a). (1 + b)=1+a+b+ab, or putting a+b+ abc we shall have (1+a). (1 + b) = 1 + c; consequently, if we raise the members of this equation to the power whose exponent is any arbitrary number, as t, we shall have (1 + a)'. (1 + b)' =(1+c)', which by (8) is reduced to (1 + l'(1+a)t + etc.). (1 + l'(1 + b)t + etc.) = (1 + l'(1 + c)t + etc.), or we shall have 1+ [(1 + a) + l'(1 + b)]t + etc. = 1 + l'(1 + c)t + etc., or l'(1 + a) + l' (1 + b) = l'(1 + c), (10), because t is arbitrary.

It follows, from (10), that the hyperbolic logarithm of a product equals the sum of the hyperbolic logarithms of its factors; and reciprocally, the hyperbolic logarithm of a quotient equals the hyperbolic logarithm of the dividend, minus that of the divisor.

Hence, too, the hyperbolic logarithm of a power equals that ́ of the root, multiplied by the index of the power; and reciprocally, if the hyperbolic logarithm of a power is divided by the index, the quotient equals the hyperbolic logarithm of the

root.

(4.) Supposing A in (1) to be invariable, while B and are variable, or receive different values, then clearly the values B may be regarded as being derived from A, considered as the root or base.

If we put m = and call m the modulus, then we shall

1 L'A'

(according to what has been shown) have x =

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(11); consequently, the exponent of A equals the hyperbolic logarithm of the corresponding power divided by the hyperbolic logarithm of A (the root), or (which is the same) multiplied by the modulus.

Because A and (of course) l'A are invariable, it is clear that the exponents of A may be taken to represent the hyperbolic logarithms of the corresponding powers B; consequently, if we call a the logarithm of B, taken to the base A, and represent the logarithm by writing before B, we 'B shall have x = IB, and (11) will be reduced to IB = = l'A ml'B, (12).

ΚΑ If for B in (12) we put A, we shall get A= =1; l'A

consequently, the logarithm of the root or base is always 1, or unity.

Hence, if e denotes the base of hyperbolic logarithms, since x2 ఇని l'e=1, we shall from (8) get e2 = 1 + x + + 1.2 1.2.3 X4 1.2.3.4

+

x =

+ etc., (13); consequently, if in this we put

1, we shall have e= 1+1+ +

: =

1 1 1 1.2 1.2.3 1.2.3.4

+

+ etc. = 2.7182818. the root or base of hyperbolic logarithms.

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If for A we put 1-a, and suppose a to be positive and less than 1, A will be positive and less than 1, and from (5)

-(a

we shall have l'A = l'(1 − a) = − (a +

a2 a3

2 3

2

+ + etc.), negative result; consequently, it is clear from (12), that if the base is less than 1, the logarithms of all positive numbers less than 1 will be positive, while the logarithms of all numbers greater than 1 will be negative; and reciprocally, if the

base is greater than 1, the logarithms of all numbers greater than 1 will be positive, while the logarithms of all positive numbers less than 1 will be negative.

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If we put B=1-6, we shall have l'Bl'(1- b) =

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6 + -+ + etc.); consequently, if for b we put 1c,

2 3

and suppose c to be an infinitesimal, so that for b we may put 1, we shall get 'B'[1 − (1 − e)] = l'c = l'o =

(1

1 1

1 + + + etc.),

2 3

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a negative result, which we will now

show to be the negative of an unlimitedly great number. Thus, supposing a to stand for any positive number, we

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(1 − a) (1 − 2) (1

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− a)

-

1 -α

(2− a)

= 1+

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+

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4

+ etc., (14); which is a series that

may be taken as a representative of 1. If a is an infinites

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= ∞ = an unlimitedly great number, it

follows that the series in the right member of (14) represents an unlimitedly great number; consequently, since the series in the right member of (14) (when a is an infinitesimal) can 1 1 1

not exceed the series 1 + + + + etc., it follows that the

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+ etc., represents an unlimitedly great num

the negative of an unlimitedly great posi

Hence, when the base is greater than 1 and finite, we shall

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