Make x =

WP+1 =r+1, where r is put in the place of '/ľ

for the sake of simplicity. Remembering that P= ?”, we have

Summing the geometrical series found in the parenthesis, by [360, (B)] the equation becomes

Now since

+1

is less than unity, the expression which consti

tutes the first member of (4) is positive. Moreover, the negative

term,

n-1 r

must always be less than unity, whatever be the value of r; hence, no value of r, however great, will render the first member of (4) negative. Thus we have shown that if we substitute for x the quantity VP+1, or any greater value, the result will be positive in equation (3); the result will therefore be positive in eq. (2), and also in eq. (1). Hence, by (452, 2), VP+1 is a superior limit of the positive roots in any equation, which was to be proved.

In applying the principle just established, the absolute term must be regarded as the coefficient of x'; and if the equation is incomplete, the deficient terms must be counted, in finding n.

It should be observed also, that an equation having no negative term can have no positive roots. For, every positive number substituted for x will render the first member positive. That is, no positive value of x can reduce the first member to zero.

EXAMPLES.

1. Find the superior limit of the positive roots of the equation x*+-5x*+2x3—14x2-26x+10 = 0.

Here n = 3 and P = 26. Hence we have, in whole numbers, "/P+1 = 1/26+1 = 4, Aμ.

2. Find the superior limit of the positive roots of the equation x+5x-25x-12x+68-0. Ans. 6.

3. Find the superior limit of the positive roots of the equation x-5.-9x+12=0. Ans. 4.

4. Find the superior limit of the positive roots of the equation x2+x2+3x-8= 0. Ans. 3. 454. To determine the superior limit of the negative roots of an equation, numerically considered,

Change the signs of the alternate terms, counting the deficient terms when the equation is incomplete; then apply the preceding rule. For, according to (444), the positive roots in the new equation will be numerically the negative roots in the given equation.

EXAMPLES.

1. Find the superior limit of the negative roots of the equation x3-3x2+5x+7=0. Ans. /7+1=3, in whole numbers.

2. Find the superior limit of the negative roots of the equation x-15.-10x+24 = 0. Ans. 5.

3. Find the superior limit of the negative roots of the equation x-3x+2x+27x-4x2-1 = 0.

LIMITING EQUATION.

Ans. 4.

455. If there be one equation whose roots, taken in the order of their values, are intermediate between the roots of another, the former is said to be the limiting equation of the latter.

456. Any equation being given, its limiting equation may be formed by putting its first derived polynomial equal to zero.

1

If a, b, c,....k, l are the roots of the given equation X = 0, and a', b', c',....k' are the roots of the derived polynomial X1 = 0, each set being arranged in the order of their values, then we are to show that all these roots, taken together, and arranged in the order of their values, will be as follows:

a, a', b, b', c, c',....k, k', l.

In both equations, put x = x'+u, developing the terms, and ar

ranging the results according to the ascending powers of u. Observe that X is the first derived polynomial of X,; hence, adopt

2

ing the same notation as in (436), we have, from the two equations,

19

where, it will be observed, X', X'1, X'2, etc., represent what X, X1, X2, etc., become, when x' takes the place of x.

29

3)

Now suppose x' =r; that is, x = rr+u, r being any root of the given equation. Then X 0; and as X',, X2, X' X' = now receive definite values, the values of X and X1 may, or may not become zero by giving a particular value to u. Dropping X' from (1), and factoring the result, we have

where the different terms may be essentially positive or negative, according to the values or r and u, upon which they depend.

Now, it is evident that by causing u to diminish numerically, each term after the first, in the parenthesis, may be made as small as we please; and by making u sufficiently small, the sum of the terms containing u, in each parenthesis, may be made less than the first term X'; in which case the essential sign of the quantity in either parenthesis will depend upon the sign of X',. Thus, when u is indefinitely small, the signs of the functions, X and

upon the signs of u (X',) and X', respectively.

1

X,, will depend
Hence, when u

is negative, X and X, will have opposite signs; but when u is positive, X and X, will have the same signs.

=

457. Thus we have shown, that if we substitute in a given equation X 0, and its first derived polynomial X, 0, a quantity r—u, which is insensibly less than the root r, the results will have opposite signs; but if we substitute the quantity r+u, which is insensibly greater than the root r, the results will have the same sign.

458. Consider the quantity substituted in the two functions to be insensibly less than a, the least root of X = 0, and let it increase

till it is insensibly greater than ɑ. In passing the root a, the function X will change sign, (452,3); hence the signs of the func tions will be as follows:

Now let the substituted quantity increase from x = a+u to 2:= b-u, a value insensibly near to b, the next root of X= = 0. According to the principle already established (457), X and X, must now have opposite signs. And since X can not have changed its sign during the change of x from a+u to b-u, there must have been a change of sign in the function X,. Hence, by (452,3) one root of X10 is found between a+u and b-u, or between a and . In like manner, it can be shown that X, =0 has one root between and c, one between c and d, and so on. Hence the proposition is proved.

1

STURM'S THEOREM.

459. The object of Sturm's Theorem is to determine the number of the real roots of an equation, and likewise the places of these roots, or their initial figures when the roots are irrational.

NOTE.--This difficult problem, which for a long time baffled the skill of mathematicians, was first solved by M. Sturm, his solution being submitted to the French Academy in 1829.

460. We have seen, (435), that the equal roots of an equation may always be found and suppressed. Now let

m-1

X = x+4x2+Bx2+.... Tx+u= 0

1

represent any equation having no equal roots, and X1 = 0 its first derived polynomial, or its limiting equation.

We will now apply to the functions, X and X,, a process similar to that required for finding their greatest common divisor (105), but with this modification, namely; that we change the signs of the successive remainders, and neither introduce nor reject a negative factor, in preparing for division.

Denote the successive remainders, with their signs changed, by

R, R1, R2,....R-1, R. Since the given equation has no equal roots, there can be no common divisor between X and X1, (435); hence, if the process of division be continued sufficiently far, the last remainder, R, must be different from zero, and independent of x. Now in the several functions, X, X1, R, R1, R2,..........R2−1, Rn, let us substitute for x any number, as h, and having arranged the signs of the results in a row, note the number of variations of signs. Next substitute for x a number, h', greater than h, and again note the number of variations of signs. The difference in the number of variations of signs, resulting from the two substitutions, will be equal to the number of real roots comprised between h and h'.

This is Sturm's Theorem, which we will now demonstrate.

Q,

Let 2, 21, 22,... Q-1, Qn denote the quotients in the successive divisions. Now in every case, the dividend will be equal to the product of the divisor and quotient, plus the true remainder, or minus the remainder with its sign changed. Hence,

1-If any number be substituted for x in the functions X, X1, R, R1,.... Rn, no two of them can become zero at the same time. For, if possible, let such a value of h be substituted for x as will render X, and R zero at the same time. Then the second equation of (4) will give R, = 0; whence, the third equation will become R0; and tracing the series through, we shall have, finally, R0, which is impossible.

-

2. If any one of the functions become zero by substituting a particular value for x, the adjacent functions will have contrary signs for the same value.

For, suppose R, in the third equation to become zero; then this equation will reduce to R -R. That is, R and R have con

trary signs.

Having established these principles, suppose the quantity h, which