But X=0, while D' is not =0 (I); we have therefore for the development of the Function X,

By substituting r+h for x in the Function D', we should, in like manner, find the development of this Function to be

in which D' and D''' are the first and second derivatives of D'.

Now h may be taken so small, that the first term in each of the last two polynomials, shall be greater than the sum of all the other terms in each the effect of multiplying by a proper fraction being, to diminish the quantity multiplied.

The sign of each of the two polynomials will then be the same as that of its first term, D'h and D'.

They will therefore have the same sign when h is positive, and contrary signs when h is negative; that is, the signs of the two Functions, X and D', form a variation for x=r-h, and a permanence for x=r+h.

V. Let the number p be less than t, and let it increase continually, until it becomes equal to t; then, if p be substituted for x in the series of Functions

X, D', R', R' .....K,

the number of variations among the signs of the results cannot be affected by the vanishing, or change of signs, of any of the Functions between X and K (III); and the Function K can in no way be affected by the substitution, since this Function is entirely independent of x.

But when p increases from a number which is a little less, to one which is a little greater, than a root of the Equation X=0, the signs of X and D' will be changed from a variation to a permanence (IV).

One variation of signs will thus be lost whenever the increasing values of p, pass a root of the given Equation; and therefore the number of variations lost when the substituted numbers increase from p to t, is equal to the number of real roots of the given Equation comprised between the numbers Ρ and t.

Thus the Theorem is demonstrated.

EXAMPLE II.

To determine the number of the real roots of the Equation

x3-2x2-5x+6=0,

which is known to have no equal roots.

The series of Functions will be found to be

X = x3 2x2-5x+6

D'-3x2-4x —5

R' 19x-22

K=10611

All the real roots of the Equation are comprised between -( ∞ and +; and when infinity is substituted for x, it is evident that the result in each polynomial will have the same sign as the result in the first term of the polynomial.

negative; while the other two are between 0 and ∞, and are consequently positive.

Limits and Situation of the Real Roots of Equations.

(296.) The limits to a root of an Equation, are two numbers, one greater, and the other less, than the root; and the root is said to be situated between those limits. When the two numbers differ from each other by unity, they are the nearest integral limits of the root.

The limits of all the positive roots will be found by substituting for x in the series of Functions

X, D', R', R',...K,

the values, ∞, 0, 1, 2, 3, &c., until the row of signs becomes the same that results from substituting ; and by observing where one or more variations are lost, the situation of each of these roots will be determined by its nearest integral limits.

The limits and situation of the negative roots may be found, in like manner, by substituting the values ∞, 0, -1, -2, -3, &c.

Thus, to recur to the preceding Example, we shall find that

Here, it will be seen that ∞ and 4, when substituted for x, produce the same row of signs; all the positive roots are therefore comprised between 0 and 4.

Also

- ∞ and

-3 produce the same row of signs; which shows that all the negative roots are included between 0 and -3.

The substitution of 1, 3, or -2, reduces the Function X to 0; these numbers are the roots of the given Equation.

It will also be noticed, that in passing these roots, the Function X changes its sign, and we should know that the Equation has at least one real root between each of the two numbers which produce a change of sign in the Function X (290), even if the intermediate substitutions had not been made.

1. How many real roots has the Equation

x3-5x2+8x-1=0,

which is known to have no equal roots? How many imaginary roots has the same equation?

Ans. 1 real root; and two imaginary roots.

2. How many real roots has the Equation

x3-7x-6=0,

which is known to have no equal roots? How many of these roots are positive? and how many negative?

Ans. 3 real roots; 1 positive, and 2 negative.

3. How many real roots has the Equation

x3-7x+7=0,

which is known to have no equal roots? How many of these roots are positive? and how many negative ?

limits of each root?

What are the nearest integral

Ans. 3 real roots; 2 positive roots between 1 and 2, and 1 negative

root between -3 and -4.

HORNER'S METHOD

Of Solving the Higher Equations.

(297.) The general principles of this method, which bears the name of its discoverer, and which is much admired for its simplicity and elegance, may be elucidated as follows:

1. Let any quadratic Equation be represented by

Let the integral part, found by trial (291), of one of its positive roots be denoted by r, and the decimal part by y; so that x=r+y, or y=x—r.

If now r+y be substituted for x in the given Equation, that Equation will be transformed into another, of the same degree, in which the values of y will be less by r than the values of x in the given Equation. Making the substitution, we have

A(r+y)2+B(r+y)=N;

which, by developing, becomes

Ar2+2Ary+Ay2+Br+By=N;

or, by arranging the unknown, and transposing the known terms,

(2) Ay2+(B+2Ar)y=N—(Br+ Ar2).

Since y represents the decimal part of the root of the given Equation, y2 is of small value in comparison with y itself; and we should probably find the first figure in the value of y-which would be the first decimal figure in the root of the given Equation-by omitting the term in y2, and taking

Denoting the first figure in the value of y by d', and substituting d'+z for y, in Equation (2), that Equation will be transformed into another, in which the values of z will be less by d' than those of y in Equation (2).

The value of z, will commence in the second decimal place, and 22 will be very small in comparison with z; so that we should pretty certainly find the first figure in the value of z-which would be the second decimal figure in the root of Equation (1)—by dividing the second member of this last equation by the coefficient of z.

We might go on, in like manner, to find any number of decimal figures in the root of the given equation.

2. Let any cubic Equation be represented by

(1) Ax3+Bx2+ Cx=N.

Let the integral part found by trial (291), of one of its positive roots, be denoted by r, and the decimal part by y; so that x=r+y, or y=x—r.

By substituting r+y for x, we have

A(r+y)3+B(r+y)2+C(r+y)=N;

which, by developing, arranging the unknown, and transposing the known terms, becomes

(2) Ay3+(B+3 Ar)y2+(C+2Br+3Ar2)y=N−(Cr+Br2+ Ar3).

y3 and y2 are of small value in comparison with y; and we should probably find the first figure in the value of y-which would be the first decimal figure in the root of equation (1)-by omitting the terms in y3 and y2, and taking

N-(Cr+ Br2+ Ar3) y= C+2 Br+3 Ar2

By considering the following arrangement of quantities, we may discover a method by which equation (2), may be conveniently obtained from equation (1)—without the tedious process of substitution.

The trinomial under N, subtracted from N, gives the second member of Equation (2); the last trinomial under C, is the coefficient of y; the last binomial under B, is the coefficient of y2; and A is the coefficient of y3, the same as of x3, in Equation (1).

From these coefficients, and the second member, those in the next transformed or derived equation would be obtained, in the same manner in which these are obtained from Equation (1); and we should thence, in like manner, find the 2d decimal figure in the root of Equation (1); and so on.

The preceding method, applied to an equation of any of the higher degrees, results in the following