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like signs just before a passes through each real root, and the same sign just after the passage.

H

K

Because X and or have like signs just after the pas

h k

sage of a through a root of the equation X 0, and since they must have unlike signs just before a passes through the

H next successive root of the equation, it follows that and h

H

K k

must change their signs between every two successive real roots. of the equation X = 0; consequently, the equation = 0 h

K or = 0 must have a root, which lies between every two k

successive roots of the equation X = 0.

H

Since (by (2) at p. 228) we have = A + A1h + Al2 +

h

etc., in which A, A1, A2, etc., are independent of h and finite; it is clear when h is very small (or when x is increased by the successive additions of the very sinall numbers or quan

H

tities h, k, etc., as is here supposed), that the sign of must h

be the same as that of A, which is called the first derived function of X; consequently, since

H
h

changes its sign with

A, it follows that one root of the equation A = 0 must lie between every two successive roots of the equation X = 0.

If a and b are two particular values of x, and A and B the corresponding values of X, then if A and B have unlike signs, it is clear, from what has been shown, that an odd number of the roots of the equation X = 0 must lie between a and b; but if A and B have like signs, then an even number of the roots or no (real) roots of the equation X = 0 can lie between a and b.

Resuming the equation of the nth degree, x" + A1x2-1 + Agen-2 +.... + An_1x +. An = 0, from p. 234; then, since when n is an odd number, and An negative, the equation must have one positive root, it follows, if it has any more positive roots, that it must have an even number of them; and since the function " + A 10-1 + A xxxn−2 + . . . . + An has the same sign when x = = 0, that it has when we put

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x

for a, and make a unlimitedly great, it follows that the equation has either an even number of negative roots or no negative roots.

If n is an odd number and An positive, it may be shown in a similar way that the equation must have an odd number of negative roots, and that it has either an even number of positive roots or no such roots.

If n is an even number and An negative, since it has been shown that the equation must have a positive and negative root, it follows that the equation may have any odd number of positive or negative roots, according to the nature of the

case.

=

If n is an even number and An positive, then, since the sign of the function "A"-1 + Å1⁄2ïx2-2 + :... + An is the same when a 0 that it is when a is unlimitedly great, it follows that the equation has either an even number of positive roots or no such roots; and because the sign of the function, when is put for x and x is unlimitedly great, is the same as the sign of An, it follows, as before, that the equation has either an even number of negative roots or no negative roots.

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H K

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2d. If or is an infinitesimal, or reduced to 0, when x

h

k

passes through a root of the equation X = 0, then the equation must have equal roots, for otherwise it is clear, from

what has been done, that

H K

or

h k

H K

Because X and or

h

k

can not be reduced to 0.

are reduced to 0 by the same

value of x, it is clear that the equations X = 0 and A = 0 must exist at the same time, and that any equal roots must enter once oftener into the first of these equations than into the second.

Hence, if we find the greatest common divisor of X and A, and put it equal to 0, and find the roots of the resulting equation, they will clearly be the equal roots of the equation X=0; and by increasing the number of equal roots thus found by one, we shall clearly get the number of times that each equal root enters into the equation X = 0.

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Thus, if we take the equation a- 8x + 120, and change into x+h, we shall get (x + h)3 — (x + h)3 — 8

(x + h) + 12 = x3 — x2 8x + 12 + (3x2 — 2x — 8)h + (3x − h3; 1)h+hs; consequently, 3-2-8, the coefficient of the simple power of h, is clearly the value of A, the first derived function of the function -- 8x + 12. Since x3- x2 8x + 12 =203 2x2 + x2

-

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x(x − 2) — 6(x − 2) = (x2 + x 6) (x

8x+12=

x2(x − 2) +

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=3x(x-2)+4(x-2)=(3x+4) (x-2), it is plain that

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x3- x2 - 8x + 12 and, 3x2 2x 8 have x 2 for a common factor; consequently, they will both be reduced to 0 by putting a2 equal to 0 or x= 2, so that x = 2 is a common root of the equations a3- x2-8x+12=0 and 3x2- 2x 8=0.

Since the root 2 is contained once more in the equation 8x+12= 0 than in 3x2 2x- 8= 0, it is plain

that

x2

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8 + 12 must be divisible by (x — 2)2 = x2 4x+4; performing the division, we get x + 3 for the quotient, so that we shall have a x2 −8x + 12 = (x − 2) (x-2)(x+3)=0, which, being satisfied by putting x = 2, x = 2, or x = 3, it follows that the equation æ3 - x2 - 8x +120 has 2, 2, and - 3 for its roots.

The equal roots of any equation may be found in a similar way, and then the remaining roots may be found as in the example.

3d. Resuming the equation A' + H = 0, whose root is expressed by a +h, which is such that the value, a, of a reduces X to the corresponding very small quantity A', and that x= a + h reduces X to 0; then it is clear that h must be very small.

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clear that A the first derived function of X will be finite; consequently, since H= Ah + А1h2 + Аgh3 + etc., the equation A' + H0 will be reduced to A' + Ah + Â ̧h2 + etc. 0, (1), in which A' is supposed to be very small, while A is not supposed to be very small.

=

If A1, A2, etc., are not very great in comparison to A, then, on account of the supposed minuteness of h, the terms  ̧h2, Â1⁄2h3, etc., of (1) may be rejected, and (1) will be re

duced to A'+Ah=0 very nearly, or Ah - A'; and

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suppose that, in order to find one of the unequal roots a + h of X = 0, a must be so assumed that h, as found from (2), shall be very small; then, if we add the value of h thus found to a according to their signs, and represent the sum by a, we shall have a more correct value of a; then, using the corrected value of a, we may from (2) get a new value of h, and thence obtain a still more correct value of a; and so on, to any extent; where it may be noticed, that this process is generally called Newton's Method of Successive Substitutions.

If we change h in (1) into h+k, it will become A' + A(h + h) + A ̧(h + k)2 + A2(h + k) + etc. = 0, or, expanding the powers of h+k, and retaining only those terms which involve the simple power of k, on account of its supposed minuteness relatively to h, we shall get (A + 2A ̧h + 3a1⁄2h3 + etc.) × k (A' + Ah + A1h2 + Аgh3 + etc.), very nearly; consequently, dividing by the coefficient of k, we get A' + Ah + A2+ etc.

k =

=

nearly, (3).

A+ 2Ah + 3A2 + etc. Hence, having found h from (2) or in any other way, if we substitute its value for h in (3), and obtain a value of k, which is much less (numerically) than that of h, the approximation to the root (a + h) will clearly be correct; and if we take the sum of h and k (added according to their signs) for a new value of h, we may get a new value of k from (3), and so on, to any extent.

If the equation X = 0 has no equal roots, yet if A is very small at the same time that X = A' is very small, it is clear that the equations must have roots which are nearly equal to each other; and it is clear that in cases of this kind we can often find h more readily and correctly from (3) than from (2).

To illustrate what has been done, take the following

EXAMPLES.

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Ex. 1. To find the roots of the equation - 7x + 5 = 0. Putting xh for x in the equation, it becomes (x + h)2

− 7(x + h) + 5 = x2− 7x + 5+ (2x — 7)h + h2 = 0, or omitting h2 on account of its supposed minuteness, we shall have

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-

- (x2 — 7x + 5) nearly, which divided by 2x-7 x2- 7x+5

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which corresponds to (2). Putting

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= 0.2, and of course

1 for x in the equation, we get h

1 0.2 0.8 is a more correct value of x; consequently,

putting 0.8 for x in the equation, we get h=

0.4

0.04

- 5.4

=

= 0.0074, which added to 0.08 gives 0.8074 for one of the 54 roots of the equation; which is correct in all its figures.

=

36 42 +

12-7

5

To get the other root of the equation, we put 6 for x in the expression for h, and thence get h = 0.2; consequently, we have 6.2 for a more correct value of x. Substituting 6.2 for a in the value of h, we shall get h=— 0.04 = - 0.0074; consequently, we have 6.2 -0.0074 = 6.1926 for the remaining value of x, which is correct to four decimal places.

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Ex. 2.-To find the roots of x3 7x+7=0. Putting x = a + h, the equation becomes (a + h)3 — 7(a + h) + 7 = a3 — 7a + 7 + (3a2 — 7)h + 3ah2 + 73 = 0, and − changing hinto h+k, and retaining only the simple powers of k, the equation becomes a - 7a + 7 + (3a2 — 7)h + 3ah2 + h3 + (3a2 − 7 + 6ah + 3h2) × k=0 very nearly, or (3a2 -7+6ah+3h2) × k = — (a3 — 7a + 7 + (3a2-7)h + 3ah +) nearly, or dividing by the coefficient of , we shall a3 − 7a + 7 + (3a2 — 7)h + 3ah2 + h3 have k=

:

3a2 -7+6ah + 3h2

nearly, which corresponds to (3), A' being represented by a - 7a +7, A by 3a-7, A, by 3a, and A, by 1.

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We will now ascertain whether a3 7a+7 and 3a2 - 7 (which correspond to A' and A1) can simultaneously become very small, or whether the given equation has roots which are nearly equal to each other. To the end in view, we put 3a7 equal to 0, and thence get 3a-70 or a2 =

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2.333; and extracting the square roots of these equals, we

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